electrodynamics california

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Princeton University Ph501 Electrodynamics Problem Set 1 Kirk T. McDonald (1998) [email protected] http://puhep1.princeton.edu/~mcdonald/examples/ References: R. Becker, Electromagnetic Fields and Interactions (Dover Publications, New York, 1982). D.J. Griffiths, Introductions to Electrodynamics, 3rd ed. (Prentice Hall, Upper Saddle River, NJ, 1999). J.D. Jackson, Classical Electrodynamics, 3rd ed. (Wiley, New York, 1999). The classic is, of course: J.C. Maxwell, A Treatise on Electricity and Magnetism (Dover, New York, 1954). For greater detail: L.D. Landau and E.M. Lifshitz, Classical Theory of Fields, 4th ed. (Butterworth- Heineman, Oxford, 1975); Electrodynamics of Continuous Media, 2nd ed. (Butterworth- Heineman, Oxford, 1984). N.N. Lebedev, I.P. Skalskaya and Y.S. Ulfand, Worked Problems in Applied Mathematics (Dover, New York, 1979). W.R. Smythe, Static and Dynamic Electricity, 3rd ed. (McGraw-Hill, New York, 1968). J.A. Stratton, Electromagnetic Theory (McGraw-Hill, New York, 1941). Excellent introductions: R.P. Feynman, R.B. Leighton and M. Sands, The Feynman Lectures on Physics, Vol. 2 (Addison-Wesely, Reading, MA, 1964). E.M. Purcell, Electricity and Magnetism, 2nd ed. (McGraw-Hill, New York, 1984). History: B.J. Hunt, The Maxwellians (Cornell U Press, Ithaca, 1991). E. Whittaker, A History of the Theories of Aether and Electricity (Dover, New York, 1989). Online E&M Courses: http://www.ece.rutgers.edu/~orfanidi/ewa/ http://farside.ph.utexas.edu/teaching/jk1/jk1.html

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Page 1: Electrodynamics California

Princeton University

Ph501

Electrodynamics

Problem Set 1

Kirk T. McDonald

(1998)

[email protected]

http://puhep1.princeton.edu/~mcdonald/examples/

References:

R. Becker, Electromagnetic Fields and Interactions (Dover Publications, New York, 1982).D.J. Griffiths, Introductions to Electrodynamics, 3rd ed. (Prentice Hall, Upper Saddle

River, NJ, 1999).J.D. Jackson, Classical Electrodynamics, 3rd ed. (Wiley, New York, 1999).

The classic is, of course:J.C. Maxwell, A Treatise on Electricity and Magnetism (Dover, New York, 1954).

For greater detail:L.D. Landau and E.M. Lifshitz, Classical Theory of Fields, 4th ed. (Butterworth-

Heineman, Oxford, 1975); Electrodynamics of Continuous Media, 2nd ed. (Butterworth-Heineman, Oxford, 1984).

N.N. Lebedev, I.P. Skalskaya and Y.S. Ulfand, Worked Problems in Applied Mathematics(Dover, New York, 1979).

W.R. Smythe, Static and Dynamic Electricity, 3rd ed. (McGraw-Hill, New York, 1968).J.A. Stratton, Electromagnetic Theory (McGraw-Hill, New York, 1941).

Excellent introductions:R.P. Feynman, R.B. Leighton and M. Sands, The Feynman Lectures on Physics, Vol. 2

(Addison-Wesely, Reading, MA, 1964).E.M. Purcell, Electricity and Magnetism, 2nd ed. (McGraw-Hill, New York, 1984).

History:B.J. Hunt, The Maxwellians (Cornell U Press, Ithaca, 1991).E. Whittaker, A History of the Theories of Aether and Electricity (Dover, New York,

1989).

Online E&M Courses:http://www.ece.rutgers.edu/~orfanidi/ewa/

http://farside.ph.utexas.edu/teaching/jk1/jk1.html

Page 2: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Problem 1 1

1. (a) Show that the mean value of the potential over a spherical surface is equal to thepotential at the center, provided that no charge is contained within the sphere.

(A related result is that the mean value of the electric field over the volume of acharge-free sphere is equal to the value of the field at its center.)

(b) Demonstrate Earnshaw’s theorem: A charge cannot be held at equilibrium solelyby an electrostatic field.1

(c) Demonstrate that an electrostatic field E cannot have a local maximum of E2,using the mean value theorem mentioned in part (a) – or any other technique.

Remark: An interesting example of nonelectrostatic equilibrium is laser trappingof atoms. Briefly, an atom of polarizability α takes on an induced dipole momentp = αE in an electric field. The force on this dipole is then (Notes, p. 26),F = ∇(p ·E) = α∇E2. Since an electrostatic field cannot have a local maximumof E2, it cannot trap a polarizable atom. But consider an oscillatory field, inparticular a focused light wave. The time-average force, 〈F〉 = α∇ 〈E2〉 drawsthe atom into the laser focus where the electric field is a maximum. See,http://puhep1.princeton.edu/~mcdonald/examples/tweezers.pdf

1http://puhep1.princeton.edu/~mcdonald/examples/EM/earnshaw_tcps_7_97_39.pdf

Page 3: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Problem 2 2

2. Calculate the potential φ(z) along the axis of a disk of radius R in two cases:

(a) The disk is a uniform layer of charge density σ, and

(b) The disk is a uniform dipole layer of dipole moment density p = pz per unit area.

Page 4: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Problem 3 3

3. Suppose the electric field of point charge q were E = qr/r2+δ where δ 1, rather thenE = qr/r2.

(a) Calculate ∇ ·E and ∇×E for r = 0. Find the electric potential for such a pointcharge.

(b) Two concentric spherical conducting shells of radii a and b are joined by a thinconducting wire. Show that if charge Qa resides on the outer shell, then the chargeon the inner shell is

Qb − Qaδ

2(a− b)[2b ln 2a− (a + b) ln(a + b) + (a− b) ln(a− b)] (1)

Page 5: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Problem 4 4

4. (a) Starting from the dipole potential φ = p · r/r3 explicitly show that

E =3(p · r)r − p

r3− 4πp

3δ3(r). (2)

Hint: to show the need for the δ3(r) term, consider the volume integral of E overa small sphere about the dipole. You may need a variation of Gauss’ theorem:

∫V

∇φ dVol =∮

Sφn dS, (3)

where n is the outward normal to the surface.

(b) The geometric definition of the “lines of force” is that this family of curves obeysthe differential equation:

dx

Ex

=dy

Ey

=dz

Ez

. (4)

For a dipole p = pz, find the equation of the lines of force in the x-z plane. It iseasiest to work in spherical coordinates. Compare with the figure on the cover ofthe book by Becker.

Page 6: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Problem 5 5

5. Find the two lowest-order nonvanishing terms in the multipole expansion of the po-tential due to a uniformly charged ring of radius a carrying total charge Q. Take theorigin at the center of the ring, and neglect the thickness of the ring.

Page 7: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Problem 6 6

6. (a) A long, very thin rod of dielectric constant ε is oriented parallel to a uniformelectric field Eext. What are E and D inside the rod?

(b) What are E and D inside a very thin disc of dielectric constant ε if the disc isperpendicular to Eext?

(c) Find E and D everywhere due to a sphere of fixed uniform polarization densityP. Then calculate

∫E ·D dVol for the two volumes inside and outside the sphere’s

surface.

Hint: this problem is equivalent to two oppositely charged spheres slightly dis-placed.

(d) Show that for any finite electret, a material with fixed polarization P,

∫all space

E · D dVol = 0. (5)

Page 8: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Problem 7 7

7. A spherical capacitor consists of two concentric conducting shells of radii a and b.The gap is half filled with a (non-conducting) dielectric liquid of constant ε. You mayassume the fields are radial. The inner shell carries charge +Q, the outer shell −Q.

Calculate E and D in the gap, and the charge distribution in the inner shell. Alsocalculate the capacitance, defined as C = Q/V , where V is the potential differencebetween the inner and outer shells.

Page 9: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Problem 8 8

8. (a) As a classical model for atomic polarization, consider an atom consisting of a fixednucleus of charge +e with an electron of charge −e in a circular orbit of radius aabout the nucleus. An electric field is applied at right angles to the plane of theorbit. Show that the polarizability α is approximately a3. (This happens to bethe result of Becker’s (26-6), but the model is quite different!)

Assuming that radius a is the Bohr radius, ∼ 5.3 × 10−9 cm, use the modelto estimate the dielectric constant ε of hydrogen gas at S.T.P. Empirically, ε ∼1 + 2.5 × 10−4.

(b) Another popular classical model of an atom is that the electron is bound to aneutral nucleus by a spring whose natural frequency of vibration is that of somecharacteristic spectral line. For hydrogen, a plausible choice is the Lyman line at1225 Angstroms. In this model, show that α = e2/mω2, and estimate ε. Recallthat e = 4.8 × 10−10 esu, and m = 9.1 × 10−28 g.

Page 10: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 1 9

Solutions

1. a) We offer two solutions: the first begins by showing the result holds for small spheres,and then shows the result is independent of the size of the (charge-free) sphere; thesecond applies immediately for spheres of any size, but is more abstract.

We consider a charge-free sphere of radius R centered on the origin.

In a charge-free region, the potential φ(r) satisfies Laplace’s equation:

∇2φ = 0. (6)

First, we simply expand the potential in a Taylor series about the origin:

φ(r) = φ(0) +∑

i

∂φ(0)

∂xixi +

1

2

∑i,j

∂2φ(0)

∂xi∂xjxixj + ... (7)

We integrate (7) over the surface of the sphere:

∮Sφ(r)dS = 4πR2φ(0) +

∑i

∂φ(0)

∂xi

∮SxidS +

1

2

∑i,j

∂2φ(0)

∂xi∂xj

∮SxixjdS + ... (8)

For a very small sphere, we can ignore all terms except the first, In this case, eq. (8)becomes

1

4πR2

∮Sφ(r)dS = φ(0), [R “small”], (9)

which was to be shown.

Does the result still hold at larger radii? One might expect that since “small” is notwell defined, (9) holds for arbitrary R, so long as the sphere is charge free.

Progress can be made staying with the Taylor expansion. By spherical symmetry, theintegral of the product of an odd number of xi vanishes. Hence, only the terms witheven derivatives of the potential survive. And of these, only some terms survive. Inparticular, for the 2nd derivative, only the integrals of x2

1, x22, and x2

3 survive, and these3 are all equal. Thus, the 2nd derivative term consists of

1

2

(∂2φ(0)

∂x21

+∂2φ(0)

∂x22

+∂2φ(0)

∂x23

) ∮Sx2

1dS, (10)

which vanishes, since ∇2φ(0) = 0.

It is less evident that the terms with 4rth and higher even derivatives vanish, althoughthis can be shown via a systematic multipole expansion in spherical coordinates, whichemphasizes the spherical harmonics Y m

l .

But by a different approach, we can show that the mean value of the potential over acharge-free sphere is independent of the radius of the sphere. That is, consider

M(r) =1

4πr2

∮SφdS =

1

∫d cos θ

∫dϕφ(r, θ, ϕ), (11)

Page 11: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 1 10

in spherical coordinates (r, θ, ϕ). Then

dM(r)

dr=

1

∫d cos θ

∫dϕ∂φ

∂r=

1

∫d cos θ

∫dϕ∇φ · r

=1

4πr2

∮S

∇φ · dS =1

4πr2

∫V∇2φdVol = 0, (12)

for a charge-free volume. Hence, the mean value of the potential over a charge-freesphere of finite radius is the same as that over a tiny sphere about the center of thelarger sphere. But, as shown in the argument leading up to (9), this is just the valueof the potential at the center of the sphere.

A second solution is based on one of Green’s theorems (sec. 1.8 Of Jackson). Namely,for two reasonable functions φ(r) and ψ(r),

∫V

(φ∇2ψ − ψ∇2φ

)dVol =

∮S

(φ∂ψ

∂n− ψ

∂φ

∂n

)dS, (13)

where n is coordinate normally outward from the closed surface S surrounding a volumeV .

With φ as the potential satisfying Laplace’s equation (6), the second term on the l.h.s.of (13) vanishes. We seek an auxiliary function ψ such that ∇2ψ = δ3(0), so the l.h.s.is just φ(0). Further, it will be helpful if ψ vanishes on the surface of the sphere ofradius R, so the second term on the r.h.s. vanishes also.

These conditions are arranged with the choice

ψ =1

(1

R− 1

r

), (14)

recalling pp. 8-9 of the Notes. On the surface of the sphere, coordinate n is just theradial coordinate r, so

∂ψ

∂n=

1

4πr2. (15)

Thus, we can evaluate the expression (13), and get:

ψ(0) =1

4πR2

∮SφdS, (16)

which means that the value of the potential φ at the center of a charge-free sphere ofany size is the average of the potential on the surface of the sphere.

b) The potential energy of a charge q at point r, due to interaction with an electrostaticfield derivable from a potential φ(r), is:

U = qφ(r). (17)

For the point r0 to be the equilibrium point for a particle, the potential φ should havea minimum there. But we can infer from part (a) that:

Page 12: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 1 11

Harmonic functions do not have minima,

harmonic functions being a name for solutions of Laplace’s equation (6).

Indeed, a minimum of U at point r0 would imply that there is a small sphere centeredon r0 such that φ(r0) is less than φ at any point on that sphere. This would contradictwhat we have shown in part (a): φ(r0) is the average of φ over the sphere.

We continue with an example of Earnshaw’s theorem. Consider 8 unit charges locatedat the corners of a cube of edge length 2, i.e., the charges are at the locations (xi, yi, zi)= (1,1,1), (1,1,-1), (1,-1,1), (1,-1, -1), (-1,1,1), (-1,1,-1), (-1,-1,1), (-1,-1,1). It is sug-gestive, but not true, that the electric field near the origin points inwards and couldtrap a positive charge.

The symmetry of the problem is such that a series expansion of the electric potentialnear the origin will have terms with only even powers, and we must go to 4rth order tosee that the potential does not have a maximum at the origin. To simplify the seriesexpansion, we consider the electric field, for which we need expand only to third order.

The electric potential is given by

φ =8∑

i=1

1√(xi − x)2 + (yi − y)2 + (zi − z)2

. (18)

The x component of the electric field is

Ex = −∂φ∂x

= −8∑

i=1

xi − x

[(xi − x)2 + (yi − y)2 + (zi − z)2]3/2

= − 1

33/2

8∑i=1

xi − x

[1 + (−2xix− 2yiy − 2ziz + x2 + y2 + z2)/3]3/2

≈ − 1

33/2

8∑i=1

xi

[1 + yiy + ziz +

−2x2 + y2 + z2 + 5yiyziz

3

]

− 1

33/2

8∑i=1

xi

[11yiy

3 + 11ziz3

54− x2yiy + x2ziz

6+

3y2ziz + 3z2yiy

2

]

− 1

33/2

8∑i=1

[2xyiy + 2xziz

3− 7x3

54+

7xy2 + 7xz2

6+

20xyiyziz

9

]

=28

81√

3(x3 − 9xy2 − 9xz2), (19)

noting that x2i = y2

i = z2i = 1 and that

∑xi = 0 =

∑xiyi, etc. Similarly,

Ey ≈ 28

81√

3(y3 − 9yx2 − 9yz2) and Ez ≈ 28

81√

3(z3 − 9zx2 − 9zy2). (20)

The radial component of the electric field is therefore

Er =E · rr

=xEx + yEy + zEz

r≈ 28

81√

3r[x4 + y4 + z4 − 18(x2y2 + y2z2 + z2x2)]. (21)

Page 13: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 1 12

Along the x axis, r = x and the radial field varies like

Er ≈ 28

81√

3r3 > 0, (22)

but along the diagonal x = y = z = r/√

3 it varies like

Er ≈ − 476

243√

3r3 < 0. (23)

It is perhaps not intuitive that the electric field is positive along the positive x axis,although Earnshaw assures us that the radial electric field must be positive in somedirection. A clue is to consider the point (1,0,0) on the face of the cube whose cornershold the charges. At this point the electric fields due to the 4 charges with x = 1 sumto zero, so the field here is due only to the 4 charges with x = −1, and now “obviously”the x component of the electric field is positive. The charges at the corners of the cubeforce a positive charge toward the origin along the diagonals, but cannot prevent thatcharge from escaping near the centers of the faces of the cube.

c) If E2 has a local maximum at some point P in a charge-free region, then there is anonzero r such that E2 < E2(P ) for all points (other than P ) within a sphere of radiusr about P . Consequently, E < E(P ) in that sphere.

Let z point along E(P ). Then the mean-value theorem can be written

∫EzdVol =

4πr3

3E(P ), (24)

for the sphere about P . In general, Ez ≤ E, and by assumption E < E(P ) for allpoints other than P within the sphere, so

∫EzdVol ≤

∫EdVol <

∫E(P )dVol =

4πr3

3E(P ), (25)

which contradicts eq. (24). Hence, E2 cannot be locally maximal at P .

However, E2 can take on a local minimum....

Page 14: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 2 13

2. a) The potential φ(z) along the axis of a disk of radius R of charge density σ (per unitarea) is given by the integral:

φ(z) =

R∫r=0

2πrdrσ√

r2 + z2= 2πσ

(√R2 + z2 − |z|

). (26)

Notice that φ(z) behaves as πσR2/|z| for |z| R, which is consistent with the ob-servation that in this limit the disk may be considered as a point charge q = πσR2.Notice also, that at z = 0 the potential is continuous, but it’s first derivative (−E)jumps from −2πσ at z = 0+ to 2πσ at z = 0−. This reflects the fact that at small |z|(|z| R) the potential may be calculated, in first approximation, as the potential forthe infinite plane with charge density σ.

b) A disk of dipole-moment density p = pz can be thought of as composed of a layerof charge density +σ separated in z from a layer of charge density −σ by a distanced = p/σ. Say, the + layer is at z = d/2, and the − layer is as z = −d/2. Then, thepotential φb at distance z along the axis could be written in terms of φa(z) found in(26) as

φb(z) = φa(z − d/2) − φa(z + d/2) → −d∂φa(z)

∂z= 2πp

(sign(z) − z√

R2 + z2

). (27)

In this, we have taken the limit as d→ 0 while σ → ∞, but p = σd is held constant.

At large z we get the potential of a dipole P = πpR2 on its axis. But near the plate(|z| R), the potential has a discontinuity:

φb(0+) − φb(0−) = 4πp. (28)

We may explain this in our model of the dipole layer as a system of two close plateswith charge density σ and distance d between them, where p = σd. The field betweenthe plates is Ez = −4πσ, and the potential difference potential between two plates is

Δφ = −Ezd = 4πp, (29)

as found in (28).

Page 15: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 3 14

3. a) For a charge q at the origin, the proposed electric field is

E = qr

r2+δ= q

r

r3+δ. (30)

This still has spherical symmetry, so we can easily evaluate the divergence and curl inspherical coordinates:

∇ · E = q∇ · rr3+δ

+ qr · ∇ 1

r3+δ=

3q

r3+δ− q

3 + δ

r3+δ= − qδ

r3+δ; (31)

∇ × E = q∇ × r

r3+δ+ ∇ 1

r3+δ× qr = 0 − (3 + δ)

r

r5+δ× qr = 0. (32)

Since ∇ × E = 0, the field can be derived from a potential. Indeed,

E = −∇φ, where φ = −r∫

∞Erdr = −q

r∫∞

dr

r2+δ=

1

δ + 1

q

rδ+1. (33)

b) Let us first compute the potential due to a the spherical shell of radius a that carriescharge Q, as observed at a distance r from the center of the sphere. We use (33) andintegrate in spherical coordinates (r, θ, φ) to find

φ(r) =Q

4πa2

1

1 + δ

∫ 1

−1

2πar2 d cos θ

(a2 + r2 − 2ar cos θ)1+δ2

=Q

2ar

(a + r)1−δ − |a− r|1−δ

1 − δ2 . (34)

Now consider the addition of a sphere of radius b < a that carries charge Qb. The totalpotential at r = a is

φa =Qa

2a2

(2a)1−δ

1 − δ2 +Qb

2ab

(a+ b)1−δ − (a− b)1−δ

1 − δ2 , (35)

while that at r = b is

φb =Qa

2ab

(a + b)1−δ − (a− b)1−δ

1 − δ2 +Qb

2b2(2b)1−δ

1 − δ2 . (36)

We require Qb such that φa = φb, (as guaranteed by the wire connecting the twospheres). However, we neglect terms of O(δ2), assuming δ to be small. Then, (35-36)lead to the relation

Qb = −Qa(b/a)(2a)1−δ − (a+ b)1−δ + (a− b)1−δ

−(a/b)(2b)1−δ + (a+ b)1−δ − (a− b)1−δ(37)

What about the factors of form xδ, which are approximately 1 for small δ? Let xδ ≈1 + ε. Taking logarithms, δ lnx ≈ ln(1 + ε) ≈ ε. Thus,

xδ ≈ 1 + δ lnx, so x1−δ ≈ x

1 + δ lnx≈ x(1 − δ lnx). (38)

Using (38) in (37), we find

Qb = − Qaδ

2(a− b)[2b ln 2a− (a+ b) ln(a + b)− (a− b) ln(a− b)] . (39)

Measurement of the ratio Qb/Qa provides a stringent test of the accuracy of the 1/r2

law for electrostatics.

Page 16: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 4 15

4. a) The first terms in E can be obtained by explicit differentiation, perhaps best doneusing vector components. For r > 0,

Ei = − ∂

∂xi

pjxj

r3=

3(pjxj)xi

r5− pi

r3. (40)

To justify the δ-term, consider the integral over a small sphere surrounding the (point)dipole: ∫

VEdVol = −

∫V

∇φdVol =∮

SφndS =

∮S

p · rr3

rdS, (41)

according to the form of Gauss’ law (3) given in the hint. Evaluating the last integral ina spherical coordinate system with z axis along p, we find that only the p componentis nonzero: ∫

EdVol = p

1∫−1

2πd cos θ cos2 θ =4πp

3. (42)

No matter how small the sphere, the integral (42) remains the same.

On the other hand, if we insert the field E from (40) in the volume integral, only thez component (along p) does not immediately vanish, but then

∫VEzdVol =

r∫0

2πrdr

1∫−1

d cos θp(3 cos2 θ − 1)

r3= 0, (43)

if we adopt the convention that the angular integral is performed first.

To reconcile the results (42) and (43), we write that the dipole field has a spike nearthe origin symbolized by −(4πp/3)δ3(r).

Another qualitative reason for the δ-term is as follows. Notice, that without this termwe would conclude that the electric field on the axis of the dipole (z-axis) would alwaysalong +z. This would imply that if we moved some distribution of charge from largenegative z to large positive z, then that charge would gain energy from the dipolefield. But this cannot be true: after all, the dipole may be thought as a system of twocharges, separated by a small distance, and for such a configuration the potential atlarge distances is certainly extremely small.

We can also say that the δ-term, which points in the −p direction, represents the largefield in the small region between two charges that make up the dipole.

b) In spherical coordinates with z along p, the dipole potential in the x-z plane is

φ(r, θ) =p cos θ

r2(44)

Then,

Er = −∂φ∂r

=2p cos θ

r3, and Eθ = − ∂φ

r∂θ=p sin θ

r3. (45)

Page 17: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 4 16

Thus, the differential equation of the field lines

dr

Er=rdθ

Eθ, implies

dr

r= 2

d sin θ

sin θ, (46)

which integrates to

r = C sin2 θ = Cx2

r2, (47)

etc.

Page 18: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 5 17

5. The multipole expansion of the potential φ about the origin due to a localized chargedistribution ρ(r) is

φ(r) =Q

r+

P · rr2

+1

2

Qij rirj

r3+ . . . (48)

where

Q =∫ρ(r)dVol, Pi =

∫ρridVol, Qij =

∫ρ(3rirj − δijr

2)dVol, . . . (49)

For the ring, Q is just the total charge, while the dipole moment P is zero because ofthe symmetry.

Let us find the quadrupole moment, Qij. Take the z axis to be along that of the ring.Then, the quadrupole tensor is diagonal, and due to the rotational invariance,

Qxx = Qyy = −1

2Qzz . (50)

We compute Qxx in spherical coordinates (r, θ, ϕ):

Qxx =∫ρ(3x2 − r2)dVol =

∫ 2π

0

Q

2πaa2(3 cos2 ϕ− 1)adϕ =

Qa2

2. (51)

Thus, the third term in the expansion (48) is:

1

2r3

(Qxx sin2 θ cos2 ϕ+Qyy sin2 θ sin2 ϕ+Qzz cos2 θ

)= −Qa

2

4r3(3 cos2 θ − 1). (52)

Page 19: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 6 18

6. a) Remember that E is defined as a mean electric field, due to both the external fieldand the microscopic charges.

In the presence of external field Eext, polarization P is induced in the rod. Since∇ × E = 0, the tangential electric field is continuous across the surface of the rod.This suggests that inside the rod, which is parallel to Eext, we have E = Eext.

To check for consistency, note that in this case, P is constant apart from the ends.There is no net polarization charge in the bulk of the rod, and no change in the electricfield from Eext. But at the ends there are charges ±Q, where Q = PA, and A isthe cross-sectional area of the rod. Since A is very small for the thin rod, and theends of a long rod are far away from most of the rod, these charges do not contributesignificantly to the electric field. Hence our hypothesis is satisfactory.

The electric displacement can then be deduced as D = εE = εEext inside the rod.

b) For a dielectric disk perpendicular to Eext, the normal component of the fields isnaturally emphasized. Recall that ∇ · D implies that the normal component of thedisplacement D is continuous across the dielectric boundary.

Outside the disk, where ε = 1, D = E = Eext is normal to the surface. (That E = Eext

may be justified by noting that the induced charges on two surfaces of the thin diskhave opposite signs and do not contribute to the field outside of the disk.) Thus, insidethe disk, D = Eext. Lastly, inside the disk E = D/ε = Eext/ε.

c) The problem of a dielectric sphere of uniform polarization density P is equivalentto two homogeneous spheres with charge densities ρ and −ρ, displaced by distance d,such that in the limit d→ 0, ρ→ ∞ but ρd = P.

Recall that for a sphere of uniform charge density ρ, the interior electric field is

E =4π

3ρr. (53)

Thus, inside the polarized sphere we have

E = lim

3ρ(r − d) − 4π

3ρr

= −4π

3limρd = −4πP

3. (54)

The displacement is

D = E + 4πP =8πP

3. (55)

Then, ∫inside

E · DdVol = −4πP

3

8πP

3

4πr3

3= −128π2r3P 2

27. (56)

Outside the sphere, the electric field is effectively due to two point charges q = 4πρr3/3separated by small distance d, where ρd = P. The external field is simply that of apoint dipole at the origin of strength

p =4πr3P

3. (57)

Page 20: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 6 19

Then,

E =3r(p · r) − p

r3, (58)

and D = E. Thus,

∫outside

E · DdVol =∫p2 + 3(p · r)2)

r6dVol (59)

= p2

∞∫r

4πr2dr

r6+ 3p2

∞∫r

2πr2dr

r6

1∫−1

cos2 θ d cos θ =8πp2

3r3=

128π3r3P 2

27,

and the sum of the inside and outside integrals vanishes.

d) Let us show, on general grounds, that for an electret the integral of E · D over thewhole space is zero. First, note that ∇ · D = 0 everywhere for an electret, and thatthe electric field can be derived from a potential, φ. Then

∫all space

E ·DdVol = −∫

V∇φ · DdVol = −

∫V

∇ · (φD)dVol = −∮

SφD · dS. (60)

But if electret occupies finite volume, then φ 1/r at large r, as seen from multipoleexpansion; in the same limit, D 1/r2, while dS r2. So the integral over the surfaceat infinity is zero.

What would be different in this argument if the material were not an electret? Ingeneral, we would then have ∇ · D = 4πρfree, and the 3rd step in (60) would have theadditional term

∫V φ∇ · D = 4π

∫V ρfreeφ = 8πU . Thus, the usual electrostatic energy

is contained in∫V E · D/8π, as expected.

This argument emphasizes that the work done in putting a field on an ordinary di-electric can be accounted for using only the free charges (which establish D). Oneneed not explicitly calculate the energy stored in the polarization charge distribution,which energy is accounted for via the modification to the potential φ in the presenceof the dielectric. But the whole argument fails for an electret, which remains polarized(energized) even in the absence of a free charge distribution.

Page 21: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 7 20

7. In the upper half of the capacitor (where there is no dielectric) we have via Gauss’ law:

Eup(r) = Dup(r) =4πQup

2πr2=

2Qup

r2, (61)

where Qup is the charge on the upper half of the inner sphere, assuming the fields areradial.

In the lower part we have:Edown(r) = Eup(r), (62)

which follows from the continuity of the tangential component of E across the boundarybetween dielectric and vacuum. Also,

Ddown(r) = εEdown(r), (63)

and

Ddown(r) =2Qdown

r2, (64)

as follows from ∇ · D = 4πρfree, where Qdown is the charge on the lower part of theinner sphere.

Combining (61-64),Qdown = εQup, (65)

holds for the “free” charge on the inner shell.

What about the total charge distribution, which include polarization charges in thedielectric? The polarization vector P obeys 4πP = (ε−1)E. Thus, the charge density,σ = 4πP · n, which appears microscopically on the boundary of the dielectric adjacentto the inner shell, equals 1 − ε times the charge density on the upper shell (sincen = −r). In other words, the total microscopic charge densities on the lower andupper parts of the shell are equal. This ensures that the electric field is the same inthe upper and lower part of the capacitor.

Of course,Qdown = Q−Qup. (66)

From (65) and (67),Q−Qup = εQup, (67)

and hence,

Qup =Q

1 + ε. (68)

This implies that the potential difference V between the shells is

V = −b∫

a

Edr = − 2Q

1 + ε

b∫a

dr

r2=

2Q

1 + ε

[1

b− 1

a

]. (69)

So, the capacitance is

C =Q

V=

1 + ε

2

ab

a− b. (70)

Page 22: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 8 21

8. a) In the presence of an electric field along the z direction, which is perpendicular tothe plane of the orbit of our model atom, the plane is displaced by a distance z. Wecalculate this displacement from the equilibrium condition: the axial component of theCoulomb force between the electron and the proton should be equal to the force eEon the electron of charge e due to the electric field. Namely,

Fz =e2

r2

z

r= eE, (71)

where r =√a2 + z2, and a is the radius of the orbit of the electron. The induced

atomic dipole moment p isp = ez = r3E ≈ a3E, (72)

where the approximation holds for small displacements, i.e., small electric fields. Sincep = αE in terms of the atomic polarizability α, we estimate that

α ≈ a3, (73)

where a is the radius of the atom.

The dielectric constant ε is related to the atomic polarizability via

ε− 1 = 4πNα, (74)

whereN is the number of atoms per cm3. For hydrogen, there are 2 atoms per molecule,and 6 × 1023 molecules in 22.4 liters, at S.T.P. Hence N = 2(6 × 1023)/(22.4 × 103) =5.4 × 1019 atoms/cm3. Estimating the radius a as the Bohr radius, 5.3 × 10−9 cm, wefind

ε− 1 ≈ 4π(5.4 × 1019)(5.3 × 10−9)3 ≈ 1.0 × 10−4. (75)

b) If an electric field E is applied to the springlike atom, then the displacement d ofthe electron relative to the fixed (neutral) nucleus is related by F = kd = eE, wherek = mω2 is the spring constant in terms of characteristic frequency ω. The induceddipole moment p is given by

p = ed =e2E

k=

e2E

mω2. (76)

Thus the polarizability α is given by

α =e2

mω2. (77)

The frequency ω that corresponds to the Lyman line at 1225

A is

ω =2πc

λ=

2π(3 × 1010)

1225 × 10−8= 1.54 × 1016 Hz. (78)

From (77),

α =(4.8 × 10−10)2

(9.1 × 10−28)(1.54 × 1016)2= 1.07 × 10−24 cm3. (79)

Page 23: Electrodynamics California

Princeton University 1998 Ph501 Set 1, Solution 8 22

Finally,ε− 1 = 4πNα = 4π(5.4 × 1019)(1.07 × 10−24) = 7.3 × 10−4 (80)

Thus, our two models span the low and high side of the empirical result. Of course,we have neglected the fact that the hydrogen atoms are actually paired in molecules.

Page 24: Electrodynamics California

Princeton University

Ph501

Electrodynamics

Problem Set 2

Kirk T. McDonald

(1998)

[email protected]

http://puhep1.princeton.edu/~mcdonald/examples/

Page 25: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Problem 1 1

1. Show that the electromagnetic energy of a dielectric subject to fields E and D = εE is

U =1

∫E · D dVol, (1)

by considering the model of atoms as springs (Problem 8b, set 1). The energy U thenhas two parts:

U1 =1

∫E2 dVol, (2)

stored in the electric field, and

U2 =∫nkx2

2dVol, (3)

stored in the spring-like atoms (n is the number of atoms per unit volume). Assume nis small so that the dielectric constant ε is nearly 1.

Page 26: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Problem 2 2

2. (a) Show that the energy of a quadrupole in an external electric field E,

Uquad = −1

6Qij

∂Ej

∂xi, (4)

in terms of its quadrupole tensor Qij, can be rewritten as

Uquad = −Qxx

4

∂Ex

∂x, (5)

if the quadrupole is rotationally symmetric about the x axis. Give an expressionfor the force F on the quadrupole.

(b) A rotationally symmetric quadrupole of strength Qxx (zero net charge, zero dipolemoment) is located at distance r from a point charge q. What is the force on thequadrupole if:

i. The x axis is along the line joining Qxx and q?

ii. The x axis is perpendicular to the line joining Qxx and q?

For your own edification, confirm your answer by considering the simple quadrupole:

−q1

2q1

−q1

Page 27: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Problem 3 3

3. The principle of an electrostatic accelerator is that when a charge e escapes from aconducting plane that supports a uniform electric field of strength E0, then the chargegains energy eE0d as it moves distance d from the plane. Where does this energy comefrom?

Show that the mechanical energy gain of the electron is balanced by the decrease inthe electrostatic field energy of the system.

Page 28: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Problem 4 4

4. (a) Two point dipoles of strength p are aligned along their line of centers, and distance2d apart. Calculate the force between the dipoles via F = (p ·∇)E, and by meansof the Maxwell stress tensor.

(b) A spherical conducting shell of radius a carries charge q. It is in a region of zeroexternal field. Calculate the force between two hemispheres in two different ways.

Page 29: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Problem 5 5

5. (a) Two coaxial pipes of radii a and b (a < b) are lowered vertically into an oil bath:

If a voltage V is applied between the pipes, show that the oil rises to height

h =(ε− 1)V 2

4πρg ln(

ba

)(b2 − a2)

, (6)

where g is the acceleration due to gravity.

(b) Recalling prob. 1(c) of set 1, discuss qualitatively how the force arises the pullsthe liquid up into the capacitor.

Page 30: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Problem 6 6

6. According to a theorem of Green, the potential φ(x) in the interior of a volume Vcan be deduced from a knowledge of the charge density ρ(x) inside that volume plusknowledge of the potential and the normal derivative ∂φ/∂n of the potential on thesurface S that bounds the volume,

φ(x) =∫

V

ρ(x′)R

dVol′ +1

∫S

[φ(x′)

∂n′

(1

R

)− 1

R

∂φ(x′)∂n′

]dS ′, (7)

where R = |x− x′| is the distance between the point of observation and the elementof the integrand. However, further insights of Green indicate that it suffices to specifyonly one of φ or ∂φ/∂n on the bounding surface to determine the potential within. Asa particular example, show that the potential within a charge-free sphere of radius a,centered on the origin, can be determined from knowledge of only the potential φ onits surface according to (Poisson, 1820)

φ(x) =a2 − x2

4πa

∫S

φ(x′)R3

dS ′. (8)

Green (1828) gave a derivation of Poisson’s integral (8) that can be generalized tomany other problems in electrostatics. Recall that a key step towards eq. (7) is theidentity

∫V

∇ · (ψ∇φ− φ∇ψ) dVol =∫

V(ψ∇2φ− φ∇2ψ) dVol

=∫

S(ψ∇φ− φ∇ψ) · dS =

∫S

(ψ∂φ

∂n− φ

∂ψ

∂n

)dS. (9)

For problems in which the interior of volume V is charge free the potential obeys∇2φ = 0 there. To have a nonzero potential φ inside V there must, of course, becharges on the surface of or exterior to volume V . If function ψ also obeys ∇2ψ = 0inside V (and so might be the potential for some other distribution of charges exteriorto V ), then the identity (9) reduces to

0 =∫

S

(ψ∂φ

∂n− φ

∂ψ

∂n

)dS. (10)

Hence, we could combine eqs. (7) and (10) to yield the relation

φ(x) =1

∫S

[φ(x′)

∂n′

(1

R+ ψ

)−(

1

R+ ψ

)∂φ(x′)∂n′

]dS ′

=1

∫S

[φ(x′)

∂G(x,x′)∂n′ −G(x,x′)

∂φ(x′)∂n′

]dS ′, (11)

where

G(x,x′) =1

R+ ψ. (12)

Page 31: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Problem 6 7

IF the Green’s function G(x,x′) vanishes on the surface S, then we have the desirablerelation between the potential φ in the interior of V and its value on the boundingsurface S,

φ(x) =1

∫Sφ(x′)

∂G(x,x′)∂n′ dS ′. (13)

Green noted that the auxiliary potential ψ can be thought of as due to exterior chargesthat bring the surface S to zero potential when there is unit charge at position x insidevolume V , and G as the total potential of that charge configuration. Further, we maythink of the bounding surface S as being a grounded conductor for the purposes ofdetermining the potentials ψ and G, in which case the “exterior” charges reside onthe surface S. Hence, it is plausible that these exist for interesting physical surfaces S(although it turns out that mathematicians have constructed examples of surfaces forwhich a Green’s function does not exist).

Since the function G is the potential for a specifiable charge configuration, the normalderivative −∂G/∂n corresponds to the electric field (whose only nonzero componentis En) at the surface S produced by those charges. If we consider surface S to bea grounded conductor when determining function G, then the charge density σG atposition x′ on that surface, caused by the hypothetical unit charge at x, would beσG(x,x′) = En/4π = −(1/4π)∂G/∂n. Green emphasized this phyisical interpretationin his original work, and wrote eq. (13) as

φ(x) = −∫

SσG(x,x′)φ(x′) dS ′. (14)

Turning at last to Poisson’s integral (8), we see that the needed Green’s function fora sphere corresponds to the potential at x′ due to unit charge at x in the presence ofa grounded conducting sphere of radius a. Use the method of images to construct theGreen’s function and its normal derivative, and thereby verify Poisson’s result.

Page 32: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Problem 7 8

7. A parallel-plate capacitor is connected to a battery which maintains the plates atconstant potential difference V0. A slab of dielectric constant ε is inserted between theplates, completely filling the space between them.

(a) Show that the battery does work Q0V0(ε− 1) during the insertion process, if Q0

is the charge on the plates before the slab is inserted.

(b) What is the change in the electrostatic energy of the capacitor?

(c) How much work is done by the mechanical forces on the slab when it is inserted?Is this work done by, or on, the agent inserting the slab?

Suppose the battery was disconnected before the dielectric was inserted.

(d) Repeat (b).

(e) Repeat (c).

Page 33: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Problem 8 9

8. (a) Find the “escape velocity” of an electron initially 1

A above a grounded conducting plate.

(b) Point electric dipoles p1 and p2 lie in the same plane at a fixed distance apart. Ifp1 makes angle θ1 to their line of centers, show that the equilibrium angle θ2 ofp2 is related to θ1 by

tan θ1 = −2 tan θ2. (15)

Page 34: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Problem 9 10

9. We may define the capacity of a single conductor with respect to infinity as C = Q/V ,where V is the potential (with respect to potential φ = 0 at ∞) when charge Q ispresent on the conductor.

Calculate the capacity of a conductor composed of two tangent spheres of radius a.

Page 35: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Problem 10 11

10. A grounded conducting sphere of radius a is placed in a uniform external field E = E0z.(This field changes after the sphere is added.)

This problem may be solved by the method of images if we suppose the field E0 is dueto two charges ±Q at positions z = ±R, with Q and R appropriately large.

(a) Show that the image of the source of E0 is then a dipole p = a3E0 located at thecenter of the sphere.

(b) Give an expression for the potential φ(r, θ) in spherical coordinates (r, θ, ϕ) cen-tered on the sphere. Sketch the electric field lines.

(c) Show that the induced charge distribution on the sphere is σ = 34πE0 cos θ.

(d) Show that the force between the two hemispheres with equator perpendicular toE0 is F = 9

16a2E0.

Page 36: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Problem 11 12

11. A hollow infinite rectangular conducting tube of sides a and b has two faces groundedand two faces at potentials V1 and V2 as shown:

a

b

φ=0

φ=0

φ=φ= V V21

x

y

Find the potential φ(x, y) inside the tube. Remember to use a sum of products of allsolutions to the separated equations which do not violate the boundary conditions.

Page 37: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Problem 12 13

12. A hollow rectangular conducting box has walls at x = 0 and a, at y = 0 and b, and atz = 0 and c. All faces are grounded except that at z = c, for which φ = V :

φ=0

x

y

z

φ=0φ=0

φ=0φ=0

φ=V

c

b

a

Find the potential φ(x, y, z) inside the box.

(Choose the signs of the separation constants carefully!)

Page 38: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 1 14

Solutions

1. The energy stored in a dielectric composed of spring like atoms can be written in twoparts,

U = U1 + U2 =1

∫E2 dVol +

∫nkx2

2dVol, (16)

where E is the applied electric field, and where n is the number of molecules per unitvolume.

The displacement x in the spring-like atom is related by kx = eEon atom, where e is thecharge of an electron. Then,

U2 =∫ne2E2

on atom

2kdVol =

1

2

∫ne2E2

on atom

mω2dVol =

1

2

∫nαE2

on atom dVol, (17)

where ω =√k/m is the frequency of oscillation of the electron of mass m, and α =

e2/mω2 is the atomic polarizability introduced in eq. (69) of set 1.

On p. 20 of the Notes, we argued that Eon atom = E + 4πP/3, in terms of the appliedfield E and the induced polarization P. But, P = nαEon atom, so

P =nα

1 − 4πnα/3E, and Eon atom = E

(1 +

1 − 4πnα/3

)≈ E, (18)

where the approximation holds for small n. In this case,

U2 ≈ 1

∫4πnαE2 dVol, (19)

and

U ≈ 1

∫(1 + 4πnα)E2 dVol ≈ 1

∫εE2 dVol =

1

∫E · D dVol, (20)

using the Lorenz-Lorentz approximation for the dielectric constant ε in terms of thepolarizability α, and supposing that D = εE.

Page 39: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 2 15

2. (a) As argued on p. 13 of the Notes, rotational symmetry of a charge distributionabout the x axis implies that its quadrupole tenson Qij can be written

Qij =

⎛⎜⎜⎜⎜⎜⎝Qxx 0 0

0 −Qxx/2 0

0 0 −Qxx/2

⎞⎟⎟⎟⎟⎟⎠ , (21)

and hence, from (4),

U = −Qxx

6

(∂Ex

∂x− 1

2

∂Ey

∂y− 1

2

∂Ez

∂z

)= −Qxx

6

(3

2

∂Ex

∂x− 1

2∇ · E

)= −Qxx

4

∂Ex

∂x,

(22)using ∇ · E = 0, assuming that the external field is produced by charges not atthe location of the quadrupole.

The force on the quadrupole is:

F = −∇U =Qxx

4

(∂2Ex

∂x2,∂2Ex

∂x∂y,∂2Ex

∂x∂z

). (23)

(b) i. Consider a point charge q at at the origin and the quadrupole at (x, y, z) =(R, 0, 0). The x-component of the electric field from q observed at (x, y, z) is

Ex =qx

r3, where r2 = x2 + y2 + z2. (24)

Then,∂Ex

∂x= q

r2 − 3x2

r5, (25)

and the force is evaluated from (23) at (R, 0, 0) as

F =(

3

2

qQxx

R4, 0, 0

). (26)

Let us check this for the simple quadrupole shown in the picture.

−q1

2q1

−q1Suppose the distance between −q1 and 2q1 is a. The force on the quadrupoledue to charge q at distance R from the center of the quadrupole, and alongthe latter’s axis, is

Fx = − q1q

(R− a)2+

2q1q

R2− q1q

(R + a)2= −6a2q1q

R4(1 + O(a/R)). (27)

This agrees with (26), since

Qij =∫ρ′(3r′ir

′j − r′2δij) dVol′ ⇒ Qxx =

∑2q′r

′2 = −4q1a2. (28)

Page 40: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 2 16

ii. If, instead, the quadrupole is at (0, R, 0) (but still oriented parallel to the xaxis), eqs. (23) and (25) combine to reveal that only the derivative ∂2Ex/∂x∂yis nonvanishing, and

F =(0,−3qQxx

4R4, 0). (29)

Again, we can directly compute the force on the simple quadrupole:

Fy = 2

(q1q

R2− q1qR

(R2 + a2)3/2

)≈ 3qq1a

2

R4= −3qQxx

4R4, (30)

using (28).

Page 41: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 3 17

3. Once the charge has reached distance d from the plane, the static electric field Ee atan arbitrary point r due to the charge can be calculated by summing the field of thecharge plus its image charge,

Ee(r, d) =er1

r31

− er2

r32

, (31)

where r1 (r2) points from the charge (image) to the observation point r, as illustratedbelow. The total electric field is then E0z + Ee.

The charge e and its image charge −e at positions (r, θ, z) = (0, 0,±d) withrespect to a conducting plane at z = 0. Vectors r1 and r2 are directed fromthe charges to the observation point (r, 0, z).

It turns out to be convenient to use a cylindrical coordinate system, where the obser-vation point is r = (r, θ, z) = (r, 0, z), and the charge is at (0, 0, d). Then,

r21,2 = r2 + (z ∓ d)2. (32)

The part of the electrostatic field energy that varies with the position of the charge isthe interaction term,

Uint =∫E0z · Ee

4πdVol

=eE0

∫ ∞

0dz∫ ∞

0πdr2

(z − d

[r2 + (z − d)2]3/2− z + d

[r2 + (z + d)2]3/2

)

=eE0

4

∫ ∞

0dz

⎛⎜⎝⎧⎪⎨⎪⎩

2 if z > d

−2 if z < d

⎫⎪⎬⎪⎭− 2

⎞⎟⎠

= −eE0

∫ d

0dz = −eE0d. (33)

When the particle has traversed a potential difference V = E0d, it has gained energyeV and the electromagnetic field has lost the same energy.

In a practical “electrostatic” accelerator, the particle is freed from an electrode atpotential −V and emerges with energy eV in a region of zero potential. However, theparticle could not be moved to the negative electrode from a region of zero potentialby purely electrostatic forces unless the particle lost energy eV in the process, leadingto zero overall energy change. An “electrostatic” accelerator must have an essentialcomponent (such as a battery) that provides a nonelectrostatic force that can absorbthe energy extracted from the electrostatic field while moving the charge from potentialzero, so as to put the charge at rest at potential −V prior to acceleration.

Page 42: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 4 18

4. (a) First, we calculate the force directly. The electric field from one of the dipoles,taken to be at the origin and with moment p = px, is

E =3(p · r)r − p

r3=

3pxr

r4− px

r3. (34)

For a second dipole at (x, y, z) = (2d, 0, 0), also with moment p = px, we have

F = (p · ∇)E = p∂E

∂x

∣∣∣∣∣(2d,0,0)

= −3p2

8d4x. (35)

The minus sign indicates that the dipole’s attract.

As an aside, we can also calculate F = −∇U , where U is the energy of interactionof the two dipoles. First, the energy of a charge q2 at position r2 in the field of adipole p1 at position r1 is

U = q2p1 · rr3

, (36)

where r = |r| = |r2 − r1|, as on p. 12 of the Notes. A point dipole p2 is the limitof a pair of charges ±q2 at positions r2 and r2 − s where s = sp2, and the productq2s is held constant at value p2. Thus, the interaction energy of two point dipolesis obtained from (36) as

U = lims→0, q2s=p

q2

(p1 · rr3

− p1 · r′r′3

)= (p2 ·∇2)

p1 · rr3

, (37)

where r′ = |r2 − s− r1|. For p1 = p2 = px separated by distance 2d along x, (37)reduces to

U = p2 ∂

∂x

∣∣∣∣∣x=2d

1

x2. (38)

Then, the force F = −∇U is along x with magnitude

F = −p2 ∂2

∂x2

1

x2

∣∣∣∣∣x=2d

= −3p2

8d4, (39)

as found in (35).

Now, let us calculate the force via the Maxwell stress tensor. The force on thecharges within a (closed) surface S is given by

Fi =∮

STijdSj , (40)

as on p. 33 of the Notes, where the Maxwell tensor in empty space is given by

Tij =1

(EiEj − 1

2δijE

2). (41)

In the problem with two dipoles, it is convenient to choose the surface as themidplane perpendicular to the line connecting two dipoles (the x axis), closing

Page 43: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 4 19

the surface at infinity around one of the dipoles. On this plane (x = d) the onlynonzero component of E is Ex, and this is twice Ex from the dipole at x = 0. Atradius r from (d, 0, 0) in the symmetry plane, the total field is then

Ex(r) = 2p

(3d2

[r2 + d2]5/2− 1

[r2 + d2]3/2

)= 2p

2d2 − r2

[r2 + d2]5/2. (42)

The Maxwell stress tensor is thus,

Tij =1

⎛⎜⎜⎜⎜⎜⎝E2

x 0 0

0 −E2x 0

0 0 −E2x

⎞⎟⎟⎟⎟⎟⎠ . (43)

We take our surface element to be dS = (2πrdr, 0, 0) in cylindrical coordinates,the sign of which implies that the surface S encloses the dipole at x = 0. Then,(40) and (43) indicate that only Fx is nonzero, and it is given by

Fx =1

4

∫ ∞

0r drE2

x = p2∫ ∞

0r dr

(r2 − 2d2)2

(r2 + d2)5

=p2

2

∫ ∞

0dt

(t− 2d2)2

(t+ d2)5=p2

2

∫ ∞

0dt

[(t+ d2) − 3d2]2

(t+ d2)5(44)

=p2

2

∫ ∞

0dt

[1

(t+ d2)3− 6d2

(t+ d2)4+

9d4

(t+ d2)5

]

=p2

2

[1

2d4− 2

d4+

9

4d4

]=

3p2

8d4.

This agrees with (35), noting that since the dipoles attract, the force on the dipoleat x = 0 is in the +x direction.

(b) The electric field outside the conducting sphere of radius a is E = qr/r2. Thepressure (= force per unit area) on the surface charges is P = σE/2, where σ isthe surface charge density; hence, P = q2r/8πa4. (The coefficient 1/2 is neededbecause E is the field outside the surface, while the field inside the sphere is zero,thus the average field inside the charge layer is E/2.) To find the force betweentwo hemispheres, we integrate the component of pressure normal to the equatorialplane (P cos θ) over one hemisphere:

F =∫ 1

02πa2 d cos θ

q2 cos θ

8πa4=

q2

8a2. (45)

Now, let us calculate the force using the Maxwell stress tensor. We integrateFz over the x-y plane separating our sphere into two hemispheres. Since dS =(0, 0, 2πr dr) there, and the only nonzero components of E on that surface areEx and Ey, only Tzz = −E2/8π = −q2/8πr4 contributes to the force. Integratingfrom r = a to ∞, we find

Fz =∫ ∞

a2πr drTzz =

q2

4

∫ ∞

a

dr

r3=

q2

8a2, (46)

in agreement with (45).

Page 44: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 5 20

5. (a) The electrical force F required to pull oil of density ρ into a cylindrical capacitorof inner and outer radii a and b, respectively, to height h above the bath is equalto the force of gravity:

F = ρgh(b2 − a2). (47)

A second relation for F can be computed from the balance of electrical energy,noting that the capacitor is held at constant voltage by a battery. Suppose weincrease the height of the oil by δh. Then, work Fδh is done on the oil, the energyU = CV 2/2 stored in the capacitor changes by δU , and the battery loses energyV δQ. Conservation of energy implies

0 = Fδh+ δU − V δQ. (48)

Since V = Q/C, we find for constant voltage,

V δQ = V 2δC = 2δ

(CV 2

2

)= 2δU. (49)

Together, (48) and (49) imply that

F = +∂U

∂h

∣∣∣∣∣V

. (50)

As the liquid is drawn into the capacitor, the energy for this must come fromelsewhere; yet, the energy of the capacitor increases because the battery losesenergy in twice the amount of work done on the liquid.

We now calculate the stored energy U by integrating the electric field energydensity. By cylindrical symmetry and Gauss’s law, the electric field between thepipes has form Er(r) = α/r, where α is fixed by

V =∫ b

aEr dr = α ln

b

a, or α =

V

ln ba

. (51)

Suppose the total height of the capacitor (above the bath) is H. Then, the energyof the electric field in the capacitor is:

U =1

8πεh∫ b

aE22πr dr +

1

8π(H − h)

∫ b

aE22πr dr, (52)

where the first term on the right is the contribution from the space filled with theoil whose dielectric constant is ε, while the second term is from the empty spaceabove. Evaluating the integrals:

2π∫ b

aE2r dr = 2π

V 2

ln2 ba

∫ b

a

dr

r= 2π

V 2

ln ba

, (53)

we then find:

U =1

4

V 2

ln ba

[(ε− 1)h+H] . (54)

Page 45: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 5 21

The force is obtained from (50) and (54):

Fel =∂U

∂h

∣∣∣∣∣V

=V 2(ε− 1)

4 ln ba

. (55)

Equating this to the force of gravity, (47) we obtain the height h of the oil column:

h =(ε− 1)V 2

4πρg ln(

ba

)(b2 − a2)

. (56)

(b) The force on the liquid arises from the effect of gradients of the electric field onthe molecular dipoles in the liquid. The spatially varying electric field E resultsin a bulk dielectric polarization given by

P = χE =ε− 1

4πE, (57)

where χ is the dielectric susceptibility and ε is the dielectric constant. The energydensity associated with the induced polarization is

u = −P · E = −ε− 1

4πE2, (58)

and so the force density on the liquid is given by

f = −∇u =ε− 1

4π∇E2. (59)

The gradient ∇E2 in the fringe field of the capacitor points from the outside tothe interior of the capacitor, with a generally vertical component for the liquidbelow the capacitor in the present problem.

It is interesting to consider a variant on this problem: a capacitor with horizontalplates completely immersed in a dielectric liquid. Here, the fringe fields of thecapacitor pull the liquid in from all sides, “trapping” it inside the capacitor. Thatis, work would be required to pull the liquid out of the capacitor in any direction.

Is this an example of electrostatic trapping – which is claimed not to exist? No!The “trapping” in the direction perpendicular to the capacitor plates is not pro-vided by purely electrostratic fields, but by the material of the capacitor plates(whose stability is not a result of purely electrostatic effects). See prob. 7 of set4 for further discussion.

We have concluded that the liquid is drawn into the interior of the capacitor andthat the liquid near the middle of the capacitor is forced up against the capacitorplates by electrostatic forces on the induced dipoles. If we drill a hole in thecenter of one capacitor plate, would liquid squirt out? (If yes, we would have aperpetual motion machine.) No, the fringe fields around the hole will pull liquidinto the interior of the capacitor creating a static equilibrium much as before.

Page 46: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 6 22

6. We work from eq. (13), for which we first need the potential φ at a point r inside agrounded conducting sphere of radius a when unit charge is located at x, also insidethe sphere. Then we need the normal derivative of this potential on the inner surfaceof the sphere, i.e.when |r| = r = a.

The image method for a grounded conducting sphere tells us that the potential insidethe sphere can be calculated as that due to unit charge at x together with charge −a/xat position x′ = a2x/x2. We denote the angle between vectors r and x as θ, so that

R = |r − x| =√r2 + 2rx cos θ + x2, (60)

and

R′ = |r − x′| =

√r2 + 2r

a2

xcos θ +

a4

x2. (61)

We see that when r = a, then

R′ =a

xR. (62)

The potential inside the sphere can now be written

φ(r) =1

R− a

R′x, (63)

The normal derivative of the potential on the inner surface of the sphere is the negativeof its radial derivative when r = a,

∂φ

∂n= −∂φ(r = a)

∂r=a + x cos θ

R3− a[a+ (a2/x) cos θ]

R′3x=a2 − x2

aR3, (64)

using eq. (62). Inserting this in eq. (13), we obtain Poisson’s integral,

φ(x) =a2 − x2

4πa

∫S

φ(x′)R3

dS ′. (65)

Page 47: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 7 23

7. (a) The capacitance C of a parallel-plate capacitor of area A, gap thickness d anddielectric constant ε is

C =εA

d≡ Q

V. (66)

Adding the dielectric increased the capacitance to

Cf = εC0, (67)

and hence the charge also increase, if the voltage is kept fixed. Thus, the workdone by the battery as the dielectric is inserted,

ΔWbatt = V0ΔQ = V 20 ΔC = V 2

0 (ε− 1)C0 = Q0V0(ε− 1), (68)

is positive.

(b) As the dielectric is inserted, the field energy U = CV 2/2 stored in the capacitorchanges by

ΔU =1

2ΔCV 2

0 =1

2C0V

20 (ε− 1) =

1

2Q0V0(ε− 1). (69)

(c) The work done by the battery, (68), is only partly accounted for in increase inthe field energy, (69). The rest of the work done by the battery is done on theexternal agent that held the dielectric during insertion (the external agent gainedenergy):

ΔWon agent =1

2Q0V0(ε− 1). (70)

(d) If the battery had been disconnected before the dielectric was inserted, then thecharge Q0 would be constant. From (66) we see that the final voltage would beonly V0/ε. Recalling (67), the change in the electrostatic field energy would thenbe

ΔU =1

2εC0

(V0

ε

)2

− 1

2C0V

20 =

1

2Q0V0

(1

ε− 1

)< 0. (71)

(e) By conservation of energy, the work done on the external agent that held thedielectric during insertion is equal and opposite to the change in stored energy.Hence the work done on the agent is again positive, but now with the value

ΔWon agent =1

2Q0V0

ε− 1

ε. (72)

That is, the dielectric is pulled into the capacitor whether or not the battery isstill connected.

Page 48: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 8 24

8. (a) The field energy associated with an electron at distance r from a grounded con-ducting plane is 1/2 that associated with the corresponding image charge, i.e.,with that electron plus a positron at distance −r, in the absence of the conductingplane. Hence,

U = −1

2

e2

2r= − e2

4r. (73)

The fields in the image solution have reality only outside the conducting plane;there is no energy associated with the “fictitious” image fields inside the conduc-tor.

Equation (73) indicates that an electron is “bound” to the conducting plane, andso to escape, must have a minimum velocity related by

vmin =

√2|U |m

=

√e2

2mr=

√e2c2

2mc2r= c

√re

2r, (74)

where re = e2/mc2 = 2.8 × 10−13 cm is the classical electron radius. Thus, forr = 1

A,

vmin

c=

√2.8 × 10−13

2 × 10−8= 0.0037. (75)

(Notice that the nonrelativistic approximation suffices.)

The “binding energy” can be estimated from (73) as

U = − e2

4mc2rmc2 = −re

r

mc2

4= −2.8 × 10−13

10−8

5.11 × 105 eV

4= −3.6 eV. (76)

(b) In equilibrium, the torque on dipole p2 must vanish, and so p2 will be directedalong the electric field created by dipole p1. The electric field of the latter is givenby

E =3(p1 · r)r − p1

r3. (77)

The projection of E onto the line connecting two dipoles is

E‖ = E · r = 2p1

r3cos θ1. (78)

The orthogonal projection isE⊥ = E− E‖, (79)

leading to

E⊥ = −p1

r3sin θ1, (80)

where the minus sign indicates that E⊥ is directed opposite to p1,⊥.

The angle of the field line, and hence of p2 is

tan θ2 =E⊥E‖

= −1

2tan θ1. (81)

Page 49: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 9 25

9. We solve the problem of the capacity of two tangent, conducting spheres of radii a bythe method of images.

We first find the image-charge distribution needed to bring one sphere to potential V ,but leaving the other at zero potential. Then, we complete the solution by superposingthe mirror distribution, obtained by reflection symmetry about the plane through thepoint of tangency of the two spheres.

(a) Place charge q = aV at the center of sphere 1, bringing its surface to otential V .

(b) To bring sphere 2 to zero potential, place charge −q(a/2a) = −q/2 at distancea2/2a = a/2 from the center of sphere 2, following the prescription on p. 41 ofthe Notes.

(c) The image charge (b) takes sphere 1 away from potential V . To bring it back, addan image charge (c) inside sphere 1 so that this sphere is at zero potential underthe effect of charges (b) and (c). That is, add charge −(−q/2)(a/(3a/2)) = +q/3at distance a2/(3a/2) = 2a/3 from the center of sphere 1, i.e., a/3 from the pointof contact.

(d) Add charge −q/4 at 3a/4 from the center of sphere 2 to bring it back to zeropotential.

(e) ....

q q-q/2 -q/2... ...

The total charge needed to bring both spheres to potential V is double that describedin the sequence above. Hence,

Q = 2q(1 − 1

2+

1

3− 1

4+ ...

)= 2aV ln 2, (82)

and the capacitance isC = Q/V = 2a ln 2 = 1.386a. (83)

Note that since the dimensions of potential are [charge]/[length], capacitance has thedimension of [length] in Gaussian units. Thus, we expect that C ≈ a for this problem,since a is the only relevant length.

Page 50: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 10 26

10. (a) The uniform field E0 = E0z is approximated as being due to charges ±Q atz = ∓R, where Q → ∞ and R → ∞ in such a way as to keep Q/R2 constant.In the limit, the field in the region of the sphere is homogeneous and equal toE = (2Q/R2)z. According to the image method, we can make the potentialon the sphere vanish by adding charge q′ = −Qa/R at z = −a2/R and −q′ atz = a2/R. Thus, the perturbation to the field due to the sphere is effectively thatdue to a dipole with the moment

p = 2a2

RQa

Rz = a3E0. (84)

(b) The potential outside the sphere is thus,

φ = φ0 + φdipole = −E0r cos θ +E0a

3 cos θ

r2. (85)

The field lines bend in to be normal to the sphere at r = a:

p

(c) We find the surface charge density σ from the normal component of the electricfield at the surface of the sphere:

Er(a, θ) = − ∂φ

∂r

∣∣∣∣∣r=a

= 3E0 cos θ, (86)

and so

σ(θ) =Er(a, θ)

4π=

3E0 cos θ

4π. (87)

(d) The force acting on the surface charge density σ is F = σEr(a)r/2 = E2r (a)r/8π

(where the latter form follows immediately from the Maxwell stress tensor). Theforce on the right hemisphere is directed along z and is obtained by integratingthe z component of F:

Fz =1

∫ 1

02πa2 d cos θ(3E0 cos θ)2(cos θ) =

9

16a2E2

0 . (88)

Since the force on the hemisphere at z > 0 is positive, the hemispheres repel eachother.

Page 51: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 11 27

11. We seek solutions to Laplace’s equation in 2 dimensions, ∇2φ(x, y) = 0, of the formφ = X(x)Y (y). This leads to solutions of the form e±kxe±iky or e±ikxe±ky.

Since the boundary conditions include φ = 0 at y = 0 and b, it is advantageous toconsider functions Y of the type e±iky, which can be immediately restricted to theform:

Y (y) = sin ky, where k = nπ/b, n = 1, 2, . . . . (89)

This also fixes the separation constants k.

The general expression for the potential is now:

φ(x, y) =∑n

Xn(x)Yn(y) =∑n

(Ane

nπx/b +Bne−nπx/b

)sin

nπy

b. (90)

The boundary conditions at x = 0 and x = a are

φ(0, y) = V1 =∑n

(An +Bn) sinnπy

b, (91)

φ(a, y) = V2 =∑n

(Ane

nπa/b +Bne−nπa/b

)sin

nπy

b. (92)

A straigthforward approach to findAn andBn is to multiply (91) and (92) by sin(nπy/b)and integrate from y = 0 to b:

∫ b

0φ(0, y) sin

nπy

bdy = − bV1

nπcos

nπy

b

∣∣∣∣∣b

0

=bV1

⎧⎪⎨⎪⎩

2, n odd

0, n even=b

2(An +Bn) , (93)

and similarly,

bV2

⎧⎪⎨⎪⎩

2, n odd

0, n even=b

2

(Ane

nπa/b +Bne−nπa/b

). (94)

Thus, for n even, An = Bn = 0, while for n odd,

An =2

nπ sinh nπab

(V2 − V1e

−nπa/b), Bn =

2

nπ sinh nπab

(V1e

nπa/b − V2

). (95)

Finally, we get for the potential:

φ(x, y) =4

π

∑n odd

sin nπyb

nπ sinh nπab

[V2 sinh

nπx

b+ V1 sinh

nπ(a− x)

b

]. (96)

To verify that this solution satisfies the boundary conditions, note that (91) and (93)combine to yield the expansion:

1 =4

π

∑n odd

1

nsin

nπy

b. (97)

Page 52: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 11 28

We also note that the potential is symmetric about the midplanes, φ(x, y) =φ(a− x, y) = φ(x, b− y), which could have been invoked as far back as (90) to showthat only odd n contributes.

Remark: This problem could also usefully be solved as the superposition of two cases,each with three walls at potential zero and the fourth at a nonzero value. The form ofthe solution (96) displays this superposition.

Page 53: Electrodynamics California

Princeton University 1998 Ph501 Set 2, Solution 12 29

12. Since φ = 0 at x = 0, a and y = 0, b, solutions φ = X(x)Y (y)Z(z) must have the form

Xm(x) = sinmπx

a, and Yn(y) = sin

nπy

b, (98)

where n and m are positive integers (and odd, recalling the remark at the end ofproblem 9). The functions Z(z) then have the form e±kz . Since φ = 0 at z = 0, wecan make the further restriction:

Zmn(z) = sinh kmnz, (99)

where kmn is determined by inserting the trial solutions into Laplace’s equation, yield-ing

k2mn =

(mπ

a

)2

+(nπ

b

)2

. (100)

The general solution satisfying all the boundary conditions except for the one at theface z = c is:

φ(x, y, z) =∑m,n

Amn sinmπx

asin

nπy

bsinh

√(mπ

a

)2

+(nπ

b

)2

z. (101)

The remaining boundary condition tells us that

V =∑m,n

Amn sinmπx

asin

nπy

bsinh

√(mπ

a

)2

+(nπ

b

)2

c. (102)

To find Amn, multiply (102) by sin mπxa

sin nπyb

and integrate from 0 to a in x and from0 to b in y. Similarly to (93), we find

Amn =16V

mnπ2 sinh

√(mπa

)2+(

nπb

)2c

, (103)

for odd m and n, and 0 otherwise. Hence,

φ(x, y, z) =16V

π2

∑m,n odd

1

msin

mπx

a

1

nsin

nπy

b

sinh

√(mπa

)2+(

nπb

)2z

sinh

√(mπa

)2+(

nπb

)2c

. (104)

Note that we have demonstrated the expansion

1 =∑m

4

mπsin

mπx

a

∑n

4

nπsin

nπy

bfor 0 < x < a, 0 < y < b. (105)

Since this follows from (97), we could have used it to go from (102) to (103) withoutperforming the integrations.

Page 54: Electrodynamics California

Princeton University

Ph501

Electrodynamics

Problem Set 3

Kirk T. McDonald

(1999)

[email protected]

http://puhep1.princeton.edu/~mcdonald/examples/

Page 55: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Problem 1 1

1. A grid of infinitely long wires is located in the (x, y) plane at y = 0, x = ±na, n =0, 1, 2, . . .. Each line carries charge λ per unit length.

Obtain a series expansion for the potential φ(x, y). Show that for large y the field is

just that due to a plane of charge density λ/a. By noting that∞∑

n=1

zn

n= − ln(1− z) for

z complex or real, sum the series to show

φ(x, y) = −λ[2πy

a+ ln

(1 − 2e−2πy/a cos

2πx

a+ e−4πy/a

)]

= −λ ln[2(cosh

2πy

a− cos

2πx

a

)]. (1)

Show that the equipotentials are circles for small x and y, as if each wire were alone.

Page 56: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Problem 2 2

2. Two halves of a long, hollow conducting cylinder of inner radius b are separated bysmall lengthwise gaps, and kept at different potentials V1 and V2.

Give a series expansion for the potential φ(r, θ) inside, and sum the series to show

φ(r, θ) =V1 + V2

2+

V1 − V2

πtan−1

(2br sin θ

b2 − r2

)(2)

Note that∑

n odd

zn

n= 1

2ln 1+z

1−z, and Im ln z = phase(z).

Page 57: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Problem 3 3

3. The two dimensional region a < r < b, 0 ≤ θ ≤ α is bounded by conducting surfacesheld at ground potential, except for the surface at r = b.

Give an expression for φ(r, θ) satisfying these boundary conditions.

Give the lowest order terms for Er and Eθ on the surfaces r = a, and θ = 0.

As an application of the case α = 2π, consider a double gap capacitor designed for useat very high voltage (as in “streamer chamber” particle detectors):

The central electrode is extended a distance b beyond the ground planes, and is termi-nated by a cylinder of radius a b. Calculate the maximal electric field on the guardcylinder compared to the field E inside the capacitor, keeping only the first-order termderived above.

You may approximate the boundary condition at r = b as

φ(r = b) ⎧⎪⎨⎪⎩

Eh(1 − θ/θ0), |θ| < θ0,

0, θ0 < |θ|π,(3)

where θ0 ≈ h/b 1 and h is the gap height. Note that the surfaces r = a and θ = 0are not grounded, but are at potential Eh.

Answer: Emax 2Ehπ√

ab.

Page 58: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Problem 4 4

4. Find the potential distribution inside a spherical region of radius a bounded by twoconducting hemispheres at potential ±V/2 respectively. Do the integrals to evaluatethe two lowest-order nonvanishing terms.

Page 59: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Problem 5 5

5. Find the potential both inside and outside a spherical volume of charge of radius a inwhich the charge density varies linearly with the distance from some equatorial plane(Qtot = 0).

Page 60: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Problem 6 6

6. A uniform field E0 is set up in an infinite dielectric medium of dielectric constant ε.Show that if a spherical cavity is created, then the field inside the cavity is:

E =3ε

2ε + 1E0 . (4)

This problem differs from our discussion of the “actual” field on a spherical moleculein that the field inside the remaining dielectric can change when the cavity is created.The result could be rewritten as

Ecavity = E0 +4πP

2ε + 1, (5)

where P is the dielectric polarization. A Clausius-Mosotti relation based on this anal-ysis is, however, less accurate experimentally than the one discussed in the lectures.

Page 61: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Problem 7 7

7. A spherical capacitor consists of two conducting spherical shells of radii a and b, a < b,but with their centers displaced by a small amount c a. Take the center of thesphere a as the origin. Show that the equation of the surface of sphere b in sphericalcoordinates with z along the line of centers is

r = b + cP1(cos θ) + O(c2) . (6)

Suppose sphere a is grounded and sphere b is at potential V . Show that the electricpotential is

φ(r, θ) = V

[r − a

b − a

(b

r

)− abc

r2(b − a)

(r3 − a3

b3a3

)P1(cos θ) + O(c2)

]. (7)

What is the capacitance, to order c?

Page 62: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Problem 8 8

8. a) A charge Q is distributed uniformly along a line from z = −a to z = a at x = y = 0.Show that the electric potential for r > a is

φ(r, θ) =Q

r

∑n

(a

r

)2n P2n(cos θ)

2n + 1. (8)

b) A flat circular disk of radius a has charge Q distributed uniformly over its area.Show that the potential for r > a is

φ(r, θ) =Q

r

[1 − 1

4

(a

r

)2

P2(cos θ) +1

8

(a

r

)4

P4 − 5

64

(a

r

)6

P6 + . . .

]. (9)

For both examples, also calculate the potential for r < a.

Page 63: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Problem 9 9

9. A conducting disk of radius a carrying charge Q has surface charge density

ρ(r) =Q

2πa√

a2 − r2(10)

(both sides combined).

a) Show that the potential in cylindrical coordinates is

φ(r, z) =Q

a

∞∫0

e−k|z|J0(kr)sin ka

kdk . (11)

See section 5.302 of the notes on Bessel functions for a handy integral.

b) Show that the potential in spherical coordinates is (r > a):

φ(r, θ) =Q

r

∑n

(−1)n

2n + 1

(a

r

)2n

P2n(cos θ) . (12)

Note the relation for P2n(0) (see sec. 5.157) to obtain a miraculous cancelation.

Page 64: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Problem 10 10

10. A semi-infinite cylinder of radius a about the z axis (z > 0) has grounded conductingwalls. The disk at z = 0 is held at potential V . The “top” of the cylinder is open.Show that the electric potential inside the cylinder is

φ(r, z) =2V

a

∑l

e−klz

kl

J0(klr)

J1(kla). (13)

Refer to the notes on Bessel functions for the needed relations.

Page 65: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Problem 11 11

11. a) Calculate the electric potential φ everywhere outside a grounded conducting cylinderof radius a if a thin wire located at distance b > a from the center of the cylinder carriescharge q per unit length.

Use separation of variables. Interpret your answer as a prescription for the imagemethod in two dimensions.

b) Use the result of a) to calculate the capacitance per unit length between two con-ducting cylinders of radius a, whose centers are distance b apart.

Answer (in Gaussian units, the capacitance per unit length is dimensionless):

C =1

4 ln(

b+√

b2−4a2

2a

) . (14)

c) [A bonus.] You have also solved the problem: What is the resistance between twocircular contacts of radius a separated by distance b on a sheet of conductivity σ?

Apply voltage V between the contacts. Field E appears, and current density J = σEarises as well. The total current is

I =∫

J · dS = σ∫

E · dS = 4πσQin , (15)

by Gauss’ law, and Qin is the charge on one of the contacts needed to create the fieldE. But if t is the thickness of the sheet, then the disk is like a length t of the cylindricalconductor considered in part a). Therefore, Qin/t = V C with capacitance C as in partb), assuming that J and E are two-dimensional. Hence, I = 4πσtV C = V/R by Ohm’slaw, and (15) leads to

R =1

4πσtC=

R4πC

. (16)

where

R =1

σt=

L

σtL(17)

is the resistance of a square of any size on the sheet.

Page 66: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 1 12

Solutions

1. This problem is 2-dimensional, and well described in rectangular coordinates (x, y).We try separation of variables:

φ(x, y) =∑

X(x)Y (y) . (18)

Away from the surface y = 0, Laplace’s equation, ∇2φ = 0, holds, so one of X and Ycan be oscillatory and the other exponential. The X functions must be periodic withperiod a, and symmetric about x = 0. This suggests that we choose

Xn(x) = cos knx, with kn =2nπ

a. (19)

We first consider the regions y > 0 and y < 0 separately, and then match the solutionsat the boundary The Y functions are exponential, and should vanish far from the planey = 0. Hence we consider

Yn(y) =

⎧⎪⎨⎪⎩

e−kny, y > 0,

ekny, y < 0.(20)

However, we must remember that the case of index n = 0 is special in that the separatedequations are X

′′0 = 0 = Y

′′0 , so that we can have X0 = 1 or x, and Y0 = 1 or y. In the

present case, X0 = 1 is the natural extension of (19) for nonzero n, so we conclude thatY0 = ±y is the right choice; otherwise X0Y0 = 1, which is trivial. Then the potentialφ = ±y will be associated with a constant electric field in the y direction, which is tobe expected far from the grid of wires.

Combining X and Y , our series solution thus far is

φ(x, y) =

⎧⎪⎨⎪⎩

a0y +∑

n>0 an cos(2nπx/a) e−2nπy/a, y > 0,

−a0y +∑

n>0 an cos(2nπx/a) e2nπy/a, y > 0,(21)

where we have used continuity of the potential at y = 0 to use the same an for bothy > 0 and y < 0. Note, however, the sign change for a0, corresponding to the constantelectric field that points away from the wire plane.

At the boundary, y = 0, the surface charge density is

σ = λ∑n

δ(x − na) =1

4π(Ey(0

+) − Ey(0−)) =

1

(−∂φ(x, 0+)

∂y+

∂φ(x, 0−)

∂y

)

= − a0

2π+

1

a

∑n

nan cos(2nπx/a). (22)

We evaluate the an by considering the interval [−a/2 < x < a/2]. Multiplying bycos(2nπx/a) and integrating, we find

a0 = −2πλ

a, and an =

n. (23)

Page 67: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 1 13

The potential is then,

φ(x, y) = −2πλ|y|a

+ 2λ∑n>0

1

ncos(2nπx/a) e−2nπ|y|/a. (24)

To sum the series, we write it as

φ(x, y) = −2πλ|y|a

+ 2λRe∑n>0

1

ne2nπx/a e−2nπ|y|/a

= −2πλ|y|a

+ 2λRe∑n>0

1

n(e2πix/a e−2π|y|/a)n

= −2πλ|y|a

− 2λRe ln(1 − z), (25)

wherez = e2πix/a e−2π|y|/a. (26)

To take the real part, we note that if

ln(1 − z) ≡ u + iv, then 1 − z = eu eiv, |1 − z| = eu, (27)

and

Re ln(1 − z) = u = ln |1 − z|= ln |1 − e−2π|y|/a[cos(2πx/a) + i sin(2πx/a)]

= ln√

1 − 2 cos(2πx/a)e−2π|y|/a + e−4π|y|/a

=1

2ln[1 − 2 cos(2πx/a)e−2π|y|/a + e−4π|y|/a]. (28)

The potential is now

φ(x, y) = −λ ln e2π|y|/a − λ ln[1 − 2 cos(2πx/a)e−2π|y|/a + e−4π|y|/a]

= −λ ln[e2π|y|/a − 2 cos(2πx/a) + e−2π|y|/a]

= −λ ln[2 cosh(2π|y|/a)− 2 cos(2πx/a)]

= −λ ln[cosh(2π|y|/a)− cos(2πx/a)] − λ ln 2. (29)

For x, y small:

cosh2π|y|

a− cos

2πx

a≈ 1 +

1

2

(2πy

a

)2

+ . . . − 1 +1

2

(2πx

a

)2

+ . . . =1

2

(2πr

a

)2

(30)

where r2 = x2 + y2. Thus, at small x, y,

φ(x, y) → −2λ ln2πr

a, (31)

which is just the potential for an individual line charge λ. Close to each wire, theequipotentials are cylinders around this wire.

Page 68: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 2 14

2. For a 2-dimensional potential problem with cylindrical boundaries, it is appropriate touse polar coordinates (r, θ). In general, the potential could be expanded in a series ofterms in r±n cosnθ and r±n sinnθ. For a bounded potential in the region r < b, onlyfactors of rn can occur.

In the present problem, we measure θ from the plane that separates the two halfcylinders, and take 0 < θ < π on the half cylinder at potential V1. Since θ variesover the full range [0, 2π], n must be an integer. The average potential is (V1 + V2)/2,and the variable part of the potential has the symmetries φ(r,−φ) = −φ(r, φ), andφ(r, π − θ) − φ(r, θ). The first implies that only factors of sinnθ can occur, and thesecond tells us that n must be odd.

Thus, the potential has the form:

φ(r, θ) =V1 + V2

2+∑

n odd

anrn sinnθ. (32)

To fix the coefficients an, we use the boundary conditions at r = b, which can bewritten as ∑

n odd

anbn sinnθ =

V1 − V2

2sign(θ), (33)

where

sign(θ) ≡⎧⎪⎨⎪⎩

+1, 0 < θ < π,

−1, −pi < θ < 0.(34)

Thus, we have to learn how to decompose the function sign(θ) in Fourier series.

For a straightforward evaluation of the Fourier coefficients an, multiply (34) by sinnθand integrate from 0 to 2π:

πanbn = (V1 − V2)

∫ π

0sinnθ dθ =

2

n(V1 − V2). (35)

Thus,

φ(r, θ) =V1 + V2

2+

2

π(V1 − V2)

∑n odd

rn

nbnsinnθ

=V1 + V2

2+

2

π(V1 − V2)Im

∑n odd

(reiθ/b)n

n

=V1 + V2

2+

V1 − V2

πIm ln

1 + reiθ/b

1 − reiθ/b

=V1 + V2

2+

V1 − V2

πIm ln

1 − (r/b)2 + 2i(r/b) sin θ

1 + (r/b)2 − 2(r/b) cos θ

=V1 + V2

2+

V1 − V2

πtan−1

(2br sin θ

b2 − r2

). (36)

Page 69: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 2 15

In the above, we used the facts about logarithms stated in the problem; the second ofwhich follows from (27).

As a sidelight, we can compare (33) with (36) to learn that

sign(θ) =4

π

∑n odd

sinnθ

n. (37)

This is, of course, also the famous Fourier expansion of a square wave, since that is theresult of periodically extending the definition (33).

Page 70: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 3 16

3. The general possibilities for a series expansion for this problem are similar to those ofproblem 2. Since φ(r, 0) = 0 = φ(r, α), the angular factors can only be sin(nπθ/α).Then, since the radial extent includes neither the origin nor ∞, factors of both rnπ/α

and r−nπ/α can occur. Thus, the potential can be written

φ(r, θ) =∞∑

n=1

[an

(r

a

)πn/α

+ bn

(a

r

)πn/α]

sinπnθ

α. (38)

The use of factors r/a and a/r is convenient for enforcing the boundary conditionφ(a, θ) = 0, since this simply requires bn = −an.

For the electric field, we get:

Er = −∂φ

∂r= −∑

n

αan

[1

r

(r

a

)πn/α

+1

r

(a

r

)πn/α]

sinπnθ

α, (39)

Eθ = −1

r

∂φ

∂θ= −∑

n

αan

[1

r

(r

a

)πn/α

− 1

r

(a

r

)πn/α]cos

πnθ

α(40)

At r = a, Eθ = 0. At θ = 0, Er = 0, as expected. At r = a,

Er = −2π

∑n

nan sinπnθ

α. (41)

At θ = 0,

Eθ = − π

αr

∑n

nan

[(r

a

)πn/α

−(

a

r

)πn/α]. (42)

To obtain this, we ignored the small terms proportional to (a/b)n/2 compared to theterms proportional to (b/a)n/2 in φ(r = b).

We cannot determine the coefficients an until the boundary condition at r = b isspecified.

For the example of a “Streamer Chamber”, α = 2π. The surfaces r = a and θ = 0, 2πare at potential Eh rather than 0, but we can accommodate this by simply adding Ehto (38). To evaluate an, we use the boundary condition (3) at r = b. Since a b, wecan approximate the potential there as

φ(r = b) ≈ Eh +∑n

an

(b

a

)n/2

sinnθ

2=

⎧⎪⎨⎪⎩

Eh(1 − θ/θ0), |θ| < θ0,

0, θ0 < |θ|.(43)

Subtracting Eh from both sides, we find

∑n

an

(b

a

)n/2

sinnθ

2=

⎧⎪⎨⎪⎩

−Ehθ/θ0, |θ| < θ0,

−Eh, θ0 < |θ|.(44)

On multiplying by sin(nθ/2) and integrating from 0 to π, we find

nan =2Eh

π

(a

b

)n/2(

cosnπ

2− 2

nθ0sin

nθ0

2

). (45)

Page 71: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 3 17

From (41), the field on the surface of the guard cylinder is

Er(r = a) = −2Eh

πa

∑n

(a

b

)n/2(

cosnπ

2− 2

nθ0sin

nθ0

2

)sin

2. (46)

Since a/b is very small, it suffices to keep only the first term, which is maximal atθ = π:

Er,max(r = a) 2Eh

π√

ab. (47)

For reasonable values of a, b and h, we have Er,max<∼ E.

A sign that our approximations are somewhat delicate is obtained by evaluating Eθ(r =b) using (42). If we keep only the first term, we find that Eθ(r = b) ≈ Eh/πb, insteadof E. However, because of the form (45) of the an, the leading term at each order nin series (42) does not have any factors of a/b, and this series converges much moreslowly than does (41). The terms are of similar magnitude until nθ0 ≈ π, i.e., untiln ≈ πb/h, and Eθ(r = b) sums to E.

Page 72: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 4 18

4. This problem involves a spherical boundary, so we seek a solution in spherical coordi-nates (r, θ, ϕ). Since the problem has axial symmetry, the potential will be independentof ϕ, and of the general form

φ(r, θ) =∑n

[An

(r

a

)n

+ Bn

(a

r

)n+1]Pn(cos θ). (48)

The region of interest contains the origin, so we must have Bn = 0 for a finite potentialthere.

To find the An, we use the boundary condition at r = a: multiply (48) by Pn andintegrate over cos θ to find

An =2n + 1

2

1∫0

φ(a, θ)Pn(cos θ) d cos θ =2n + 1

4V[∫ 1

0(Pn(z) − Pn(−z)) dz

]. (49)

Since Pn(z) = (−1)nPn(−z), we get An = 0 for even n. For odd n, we get

An =2n + 1

2V∫ 1

0Pn(z) dz. (50)

Using the explicit expressions for the polynomials Pn, we find for the first 2 nonvan-ishing terms:

A1 =3

2V∫ 1

0zdz =

3

4V, (51)

A3 =7

2V∫ 1

0

1

2(5z2 − 3z)dz = − 7

16V. (52)

The potential is:

φ(r, θ) =3

4V

r

aP1(cos θ) − 7

16V(

r

a

)3

P3(cos θ) + . . . (53)

Page 73: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 5 19

5. This problem has an axially symmetric charge distribution ρ(r), so we can evaluatethe potential via the multipole expansion. This is the sum of two series, one forcontributions from charge at radius r′ < r and the other for charge at r′ > r:

φ(r, θ) =∑n

Pn(cos θ)

rn+1

∫ r

02πr

′2 dr′∫ 1

−1d cos θ′ρ(r′)r

′nPn(cos θ′)

+∑n

rnPn(cos θ)∫ ∞

r2πr

′2 dr′∫ 1

−1d cos θ′ρ(r′)

Pn(cos θ′)(r′)n+1

. (54)

In the present problem, the charge distribution is nonzero only for r < a, where it hasthe form ρ = ρ0z = ρ0r cos θ = ρ0rP1(cos θ).

We first evaluate the potential outside the sphere of radius a, for which we need onlythe first series of (54). The integral is

∫ a

02πr

′2 dr′∫ 1

−1d cos θ′ρ0r

′P1(cos θ′)r′nPn(cos θ′) =

⎧⎪⎨⎪⎩

2πa5

523ρ0, n = 1,

0, n = 1., (55)

using the orthogonality relation

∫ 1

−1d cos θ Pn(cos θ)Pm(cos θ) =

2δnm

2n + 1. (56)

Thus,

φ(r > a, θ) =4πa5ρ0

15

cos θ

r2, (57)

which the potential due to a dipole of strength p = 4πa4ρ0/15.

The total charge in the upper hemisphere is

Q0 =∫ a

02πr2 dr

∫ 1

0d cos θ ρ0rP1(cos θ) = 2π

a4

4

ρ0

2=

πa4ρ0

4, (58)

with −Q0 in the lower hemisphere. The effective height z0 of this charge is such thatthe dipole moment is p = 2Q0z0, so z0 = 8a/15.

For the potential inside the sphere, we must evaluate both series in (54), but we see ineach case that only the n = 1 term survives the angular integration. Therefore,

φ(r < a, θ) =P1(cos θ)

r2

∫ r

02πr

′2 dr′2

3ρ0r

′2 + rP1(cos θ)∫ a

r2πr

′2 dr′2

3ρ0

r′

r′2

= πρ0 cos θ

(2a2r

3− 2r3

5

). (59)

Expressions (57) and (59) give the same value at r = 1, as expected.

For possible instruction, we give a second solution for the potential inside the sphere,where Poisson’s equation applies:

∇2φ = −4πρ = −4πρ0rP1(cos θ). (60)

Page 74: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 5 20

Since this is a linear partial differential equation, a solution can be found in terms of aparticular solution φp to (60) plus the general solution to the homogeneous equation,

which is Laplace’s equation ∇2φh = 0 in the present case. Also, we must match oursolution for r < a to that found for r > a.

This problem has axial symmetry, so the general solution to the homogeneous equationfor r < a can be written

φh(r, θ) =∑n

Anr2Pn(cos θ). (61)

Since the solution for r > a involves only P1, we expect that the solution for r < a willalso. Then, φh = Ar cos θ.

Returning to Poisson’s equation, writing μ = cos θ, we get

1

r2

∂r

(r2 ∂φ

∂r

)+

1

r2

∂μ

[(1 − μ2)

∂φ

∂μ

]= −4πρ0rμ. (62)

We hope for a simple power law solution in r, and expect the angular function to bejust P1 = μ. That is, we try φ = Brnμ. Inserting this into (62), we learn that theparticular solution is φp = −2πρ0r

3μ/5. The complete solution then has the form

φ(r < a, θ) = φh + φp = Ar cos θ − 2πρ0r3

5cos θ. (63)

Matching this to (57) at r = a requires:

Aa − 2πρ0a3

5=

4πρ0a3

15, (64)

which gives A = 23πρ0a

2, and hence the solution (59).

Page 75: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 6 21

6. Let the potential be given by the function φ1(r, θ) inside the sphere (r < a) and φ2(r, θ)outside. The boundary conditions for the potential are

φ1(a, θ) = φ2(a, θ), (65)

∂rφ1(a, θ) = ε

∂rφ2(a, θ), (66)

where ε is the dielectric constant of the medium at r > a.

Since the asymptotic electric field at r → ∞ is E0 = E0z, the potential φ2 approaches

φ2(r → ∞) → −E0z = −E0rP1(cos θ), (67)

even after we have created the cavity at the origin. Hence, it is clear that the decompo-sition of φ in spherical harmonics should contain only terms proportional to P1(cos θ).Recalling the general form (48), we expect

φ1(r, θ) = Ar

aP1(cos θ), (68)

φ2(r, θ) =

[−E0r + B

(a

r

)2]P1(cos θ). (69)

From the boundary conditions at r = a, we get

A = −E0a + B = ε(−E0a − 2B). (70)

Solving for A and B, we get

φ1 = − 3ε

2ε + 1E0r cos θ = − 3ε

2ε + 1E0z, (71)

E(r < a) =3ε

2ε + 1E0z. (72)

The asymptotic value of polarization is related to E0 via

P =ε − 1

4πE0. (73)

Thus, we may rewrite E(r < a) in terms of E0 and asymptotic value of P as

E(r < a) = E0 +4πP

2ε + 1. (74)

Page 76: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 7 22

7. The equation of the surface of the sphere of radius b and center at z = c is

b2 = r2 − 2rc cos θ + c2 = r2(1 − 2

c

rcos θ

)+ c2, (75)

where r is measured from the origin in a spherical coordinate system. For small c, weapproximate this as

r = b + c cos θ + O(c2) = bP0(cos θ) + cP1(cos θ) + O(c2). (76)

Between the spherical shells of radii a and b, the potential φ has the general axiallysymmetric form (48). From the boundary condition φ(r = a) = 0, we conclude that

φ(r, θ) =∑n

An

[(r

a

)n

−(

a

r

)n+1]Pn(cos θ). (77)

The boundary condition on the outer shell can expressed via (76) in terms of P0(cos θ)and P1(cos θ). Hence, it is plausible that only A0 and A1 are nonzero in (77), whichthen reads

φ = A0

(1 − a

r

)+ A1

(r

a− a2

r2

)P1. (78)

[For a discussion that does not make this leap, see eqs. ]

The boundary condition on the outer shell now implies that

φ(r = b + cP1) = V

= A0

(1 − a

b + cP1

)+ A1

(b + cP1

a− a2

(b + cP1)2

)P1 (79)

≈ A0

(1 − a

b

)+

A0

ac

b2+ A1

(b

a− a2

b2

)P1

where we have dropped terms in P 21 in the last line of (79) as these lead to a correction

to A0 of order c2. Hence, the constant term on the last line of (79) equals V , whilethe coefficient of P1 must be zero. This determines the values of A0 and A1, and thepotential is

φ(r, θ) = V

[r − a

b − a

b

r− abc

r2(b − a)

(r3 − a3

b3 − a3

)P1(cos θ) + O(c2)

]. (80)

To find the capacitance C = Q/V , we have to find the charge ±Q on the sphericalshells. It is simpler to calculate this for the inner shell which is at r = q.

Q =∫

σ dS =∫

Er(r = a)

4πdS = −a2

2

∫ 1

−1d cos θ

∂φ(a, θ)

∂r. (81)

The terms in the integrand proportional to P1(cos θ) will integrate to zero, so thecapacitance is unchanged by a small offset c.

Page 77: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 7 23

For the record, Q = abV/(b − a) = CV so the capacitance is C = ab/(b − a) (which isa length, as always in Gaussian units).

As a footnote, we show how the boundary condition on the outer shell could be imple-mented without immediately assuming that only coefficients A0 and A1 are importantin (77). Neglecting terms of O(c2), we find

φ(r = b + c cos θ) = V

≈∑n

An

⎧⎨⎩(

b

a

)n (1 + n

c

bcos θ

)−(

b

a

)−n−1 (1 − (n + 1)

c

bcos θ

)⎫⎬⎭Pn(cos θ)

=∑n

An

⎡⎣(

b

a

)n

−(

b

a

)−n−1⎤⎦Pn(cos θ)

+c

b

∑n

An

⎡⎣n(

b

a

)n

+ (n + 1)

(b

a

)−n−1⎤⎦P1(cos θ)Pn(cos θ). (82)

At this point, we invoke the recurrence relation

P1(cos θ)Pn(cos θ) =n + 1

2n + 1Pn+1(cos θ) +

n

2n + 1Pn−1(cos θ). (83)

The constants An are determined from the requirement that the coefficients of Pn in(82) should be zero for n > 0, and the coefficient of P0 = 1 is V . This leads torecurrence relations for An of the following form (schematically):

(· · ·)An + c(· · ·)An+1 + c(· · ·)An−1 = 0, n > 0. (84)

By iteration, we find a solution with the property An ∝ O(cn). It is straightforwardto find the first two coefficients A0 and A1, which again gives (80).

Page 78: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 8 24

8. The problem concerns two examples of specified, axially symmetry charge distributions.Hence, the multipole expansion (54) can be used to calculate the electric potential.

a) The charge distribution is ρ = Q/2a along the line from z = −a to z = a. Thus,there is charge only for cos θ = 1 and −1, and the charge distribution is symmetric incos θ. Since Pn(−1) = (−1)nPn(1), the integrals in (54) will be nonzero only for evenn. For an observer at r > a, the multipole expansion simplifies to

φ(r > a, θ) =Q

2a

∑n

P2n(cos θ)

r2n+1· 2∫ a

0dz′(z′)2nP2n(1) =

Q

r

∑n

(a

r

)2n P2n(cos θ)

2n + 1. (85)

Similarly,

φ(r < a, θ) =Q

2a

∑n

P2n(cos θ)

r2n+1· 2∫ r

0dz′(z′)2nP2n(1)

+Q

2a

∑n

r2nP2n(cos θ) · 2∫ a

r

dz′

(z′)2n+1P2n(1)

=Q

a

∑n

P2n(cos θ)

1

2n + 1+

1

2n

[1 −(

r

a

)2n]

. (86)

As expected (85) and (86) agree at r = a.

b) As in example a), the charge distribution is symmetric in cos θ, so only even n willcontribute to the multipole expansion of the potential. The charge distribution on thedisc r < a, cos θ = 0 is ρ = Q/πa2. Hence, the potential for r > a is

φ(r > a, θ) =Q

πa2

∑n

P2n(cos θ)

r2n+1

∫ a

02πr′ dr′ (r′)2nP2n(0)

=Q

r

∑n

(−1)n(2n − 1)P2n(cos θ)

2n(n + 1)

(a

r

)2n

, (87)

which gives (9), noting that P2n(0) = (−1)2(2n − 1)/2n. Similarly,

φ(r < a, θ) =Q

πa2

∑n

P2n(cos θ)

r2n+1

∫ r

02πr′ dr′(r′)2nP2n(0)

+Q

πa2

∑n

r2nP2n(cos θ)∫ a

r2πr′ dr′

P2n(0)

(r′)2n+1.

=Qr

a2

∑n

(−1)nP2n(cos θ)

2n

[4n + 1

n + 1− 2(

r

a

)2n−1]. (88)

Again, these two expressions agree at r = a.

The charged surfaces in these examples are not conductors, so those surfaces are notequipotentials.

If you had forgotten the multipole expansion, you could have proceeded by first solvingthe simpler problem of the potential on the axis. For example, in the case of the charged

Page 79: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 8 25

disk,

φ(z > a) =Q

πa2

a∫0

2πr dr√r2 + z2

=2Q

a2

[√z2 + a2 − z

]

=Q

z

[1 − 1

4

(a

z

)2

+1

8

(a

z

)4

− 5

64

(a

z

)6

+ . . .

]. (89)

The potential φ(r, θ) is obtained from this simply by replacing z by r and multiplyingthe term in 1/rn by Pn(cos θ), etc.

Page 80: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 9 26

9. a) We expect any solution of Laplace’s equation having axial symmetry in cylindricalcoordinates (r, θ, z) to be a sum of expressions of the form

e±kzJ0(kr). (90)

In the example of a conducting disk, there are no boundary surfaces, so the separationconstant k will be continuous. The problem is symmetric about the plane z = 0, sowe look for a solution of the form

φ(r, z) =∫ ∞

0dk f(k)e−k|z|J0(kr). (91)

To find the Fourier coefficients f(k), we note that the electric field experiences a jumpacross the conducting disk at z = 0:

∂zφ(r < a, z = 0+) − ∂

∂zφ(r < a, z = 0−) = −4πσ. (92)

Hence,

− 2

∞∫0

kf(k)J0(kr) dk = −2Q

a

θ(a − r)√a2 − r2

, (93)

using expression (10) for the charge density σ. Given the integral relation

∫ ∞

0sin kaJ0(kr) dk =

⎧⎪⎨⎪⎩

0, r > a

1√a2−r2 , r < a

⎫⎪⎬⎪⎭ =

θ(a − r)√a2 − r2

. (94)

we find that f(k) = Q sin(ka)/ka, and the potential is given as in (11).

b) To give a solution in spherical coordinates for the potential due to a specified, axiallysymmetric charge distribution, we again use the multiple expansion (54). The presentproblem is quite similar to problem 8b, so we write

φ(r > a, θ) =Q

2πa

∑n

P2n(cos θ)

r2n+1

∫ a

02πr′ dr′

(r′)2n

√a2 − r′2

P2n(0)

=

√πQ

a

∑n

P2n(0)Γ(n + 1)

(2n + 1)Γ(n + 3/2)

(a

r

)2n+1

P2n(cos θ),

=Q

a

∑n

(−1)n

2n + 1

(a

r

)2n+1

P2n(cos θ), (95)

Page 81: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 10 27

10. This problem has boundary conditions on the potential that φ(r = a) = 0, and φ(z =0) = V , so a solution in cylindrical coordinates (r, θ, z) for r < a, z > 0 will have theform

φ =∑

l

Ale−klzJ0(klr). (96)

where J0(kla) = 0. At z = 0, we have

V =∑

l

AlJ0(klr). (97)

The J0(klr) are an orthogonal set of functions on the interval [0, a] upon integrationwith respect to dr2 rather than dr. Hence, we can evaluate the Fourier coefficients Al

by multiplying (97) by J0(kmr) and integrating:

∑l

Al

∫ a

0r dr J0(klr)J0(kmr) = Am

a2

2[J1(kma)]2

= V∫ a

0r dr J0(kmr) = V

a

kJ1(kma), (98)

using 5.297(3) and 5.294(7) of Smythe. With this, we obtain the expansion

φ(r, θ, z) =2V

a

∑l

e−klz

kl

J0(klr)

J1(kla). (99)

Page 82: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 11 28

11. This two dimensional problem involves cylinders about the z axis, so we use cylindricalcoordinates to discuss the potential φ(r, θ). The first conducting cylinder of radius ahas its axis along the z.

a) A wire at (r, θ) = (b, 0) carries charge density q per unit length.

We present three related solutions; the first two use Fourier series, where the firstdecomposes the potential into φ = φwire +φcylinder, while the second does not; the thirdsolution is more elementary.

In all cases, we have the symmetry φ(−θ) = φ(θ), so the Fourier expansion for thepotential contains terms in cosnθ, but not sin nθ.

The potential due to the wire has the general form

φwire(r, θ) =

⎧⎪⎨⎪⎩

a0 +∑

n=1 an

(rb

)ncosnθ, r < b,

a0 + b0 ln rb+∑

n=1 an

(br

)ncos nθ, r > b,

(100)

since this should not blow up at the origin, should be continuous at r = b, and canhave a logarithmic divergence at infinity.

The potential due to the conducting cylinder has the form

φcylinder(r > a, θ) = A0 + B0 ln r +∑n=1

An

rncosnθ. (101)

The cylinder is grounded, so the total potential (and not φcylinder) obeys φ(r = a) = 0.Hence,

a0 + A0 + B0 ln a = 0, An = −an

(a2

b

)n

, (102)

and so the potential due to the cylinder is

φcylinder(r > a, θ) = −a0 + B0 lnr

a−∑

n=1

an

(a2/b

r

)n

cosnθ, (103)

where coefficient B0 is not yet determined.

Comparing (103) to the form (100) of the potential due to the wire for r > b, we seethat these are the same for terms with n > 0, except for an overall − sign, and thesubstitution b → a2/b. Since we are still free to choose the value of B0, we set it to−b0 ln b/a, and the potential of the cylinder becomes

φcylinder(r > a, θ) = −a0 − b0 lnr

a2/b−∑

n=1

an

(a2/b

r

)n

cos nθ, (104)

and now the n = 0 terms are also related to those of eq. (100) in the same way as theterms with n > 0.

This suggests that the potential due to a grounded, conducting cylinder of radius a inthe presence of a line charge density q at r = b is the same that due to a line chargedensity −q at r = a2/b. This is the desired image method for cylindrical geometry.

Page 83: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 11 29

There was no need to evaluate the Fourier coefficients an to reach this conclusion!

In the second solution, we do not separate the potential into two parts, and we carryout the evaluation of the Fourier coefficients.

The cylinder r = a is at zero potential, so the most general form that satisfies theseconditions for a < r < b is

φ(r, θ) = b0 ln r − b0 ln a +∞∑

n=1

An

[(r

a

)n

−(

a

r

)n]cos nθ (a < r < b). (105)

Beyond the wire at r = b, we can only have the form

φ(r, θ) = c0 + d0 ln r +∞∑

n=1

Bn

(b

r

)n

cos nθ (r > b). (106)

As this problem is meant to represent a real 2-wire system, the energy per unit lengthmust be finite. Therefore, we must have d0 = 0, so that no field lines from the wireescape to infinity. This also means that the charge on the cylinder at r = a must be−q.

The potential is continuous at r = b, which leads to the conditions

c0 = b0 lnb

a, Bn = An

[(b

a

)n

−(

a

b

)n]. (107)

The remaining condition is obtained by considering a Gaussian surface (of unit lengthin z) that surrounds the cylindrical surface (b, θ):

4πqin =∫

E · dS =∫

b dθ (Er+ − Er−). (108)

For this we learn that

4πqδ(θ) = b(Er+ − Er−) = b

(−∂φ(b+)

∂r+

∂φ(b−)

∂r

)

= b0 +∑n

n cos nθ

Bn + An

[(b

a

)n

+(

a

b

)n]

. (109)

Multiply this by cos nθ and integrate over θ to find

b0 = 2q, Bn + An

[(b

a

)n

+(

a

b

)n]

=4q

n. (110)

Combining this with (107), we learn that

c0 = 2q lnb

a, An =

2q

n

(a

b

)n

, Bn =2q

n

[1 −(

a

b

)2n]. (111)

Page 84: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 11 30

For a < r < b we now have

φ(r, θ) = 2q lnr

a+ 2q

∑n

1

n

(a

b

)n [(r

a

)n

−(

a

r

)n]cosnθ

= 2q lnr

a+ 2qRe

∑n

1

n

[(r

b

)n

−(

a2

br

)n](eiθ)n

= 2q lnr

a− 2qRe

[ln

(1 − reiθ

b

)− ln

(1 − a2eiθ

br

)]

= 2q lnr

a− 2q ln

∣∣∣∣∣1 − reiθ

b

∣∣∣∣∣+ 2q ln

∣∣∣∣∣1 − a2eiθ

br

∣∣∣∣∣= 2q ln

b

a− 2q ln

∣∣∣b − reiθ∣∣∣+ 2q ln

∣∣∣∣∣r − a2eiθ

b

∣∣∣∣∣ (112)

= 2q lnb

a− 2q ln

√r2 − 2br cos θ + b2 + 2q ln

√√√√r2 − 2ra2

bcos θ +

(a2

b

)2

.

Recall that the potential at distance R from a line of charge q per unit length isφ = −2q lnR + const. Thus the second term of the last line of (112) is the potentialdue to the line of charge density q at (r, θ) = (b, 0). The second term is equal to thepotential due to a line of charge density −q at (r, θ) = (a2/b, 0).

Hence, we have demonstrated the cylindrical image method: the image in a conductingcylinder of radius a of a line of charge density q at radius b is a line of charge density −qat radius a2/b. In terms of distances r1 and r2 shown in the figure below, the potentialis then,

φ(r, θ) = 2q lnbr2

ar1. (113)

[We skip the demonstration that this prescription also works for r > b.]

For a third solution, we suppose that there exists an image wire carrying charge densityq′ at distance c from the center of the conducting cylinder, as shown in the figure.

Then the potential at an aribtrary point (r, θ) due to the two wires is

φ(r, θ) = φq + φq′ = K − 2q ln r1 − 2q′ ln r2

= K − q ln(r2 + b2 − 2br cos θ) − q′ ln(r2 + c2 − 2cr cos θ). (114)

Page 85: Electrodynamics California

Princeton University 1999 Ph501 Set 3, Solution 11 31

The cylinder is grounded, so

φ(r = a, θ) = 0 = K − q ln(a2 + b2 − 2ab cos θ) − q′ ln(a2 + c2 − 2ac cos θ). (115)

This can be arranged by putting q′ = −q, so that (115) simplifies to

K = q lna2 + b2 − 2ab cos θ

a2 + c2 − 2ac cos θ. (116)

If we take c = a2/b (inspired by our knowledge of the spherical image method), wefind that the argument of the logarithm becomes b2/a2, which is independent of θ, asdesired. Hence, K = 2q ln(b/a), and the potential is again (113).

b) Suppose we have two parallel conducting cylinders of radius a each, carrying charge+q and −q per unit length, whose axes are distance b apart. We want to find locationsfor two line charge densities +q and −q such that the fields from these lines chargesare the same as those due to the two conducting cylinders.

Clearly, these lines charges should be placed symmetrically in the plane containingthe axes of the cylinders, say distance c apart. Then the first line charge is distance(b+c)/2 from the center of the second cylinder. Its image charge would then be locateddistance 2a2/(b + c) from the center of the second cylinder. We want the image chargeto be the same as the second line charge, which is at distance (b− c)/2 from the centerof the second cylinder. Equating the two distances, we find that

c =√

b2 − 4a2. (117)

To find the capacitance C = q/ΔV , we need the potential difference ΔV betweenthe two cylinders. We can calculate the potential on a conducting cylinder at anyconvenient point. For example, consider the point on one cylinder closest to the other.This point is distance a − (b − c)/2 from one line charge, and distance (b + c)/2 − afrom the second line charge. The potential at this point is therefore

V = −2q ln[a − (b − c)/2] + 2q ln[(b + c)/2 − a] = 2q lnc + b − 2a

c − (b − 2a)

= 2q lnb + c

2a. (118)

The corresponding point on the second cylinder is at potential −V , so ΔV = 2V , andthe capacitance is

C =1

4 ln b+√

b2−4a2

2a

. (119)

Page 86: Electrodynamics California

Princeton University

Ph501 Midterm Examination

Electrodynamics

Kirk T. McDonald

(Oct. 23, 2000)

[email protected]

http://puhep1.princeton.edu/˜mcdonald/examples/

Page 87: Electrodynamics California

Princeton University Ph501 Midterm Exam Oct. 25, 2000 1

Please do all work in the exam booklets provided.

You may use either Gaussian or MKSA units on this exam.

1. (10 pts.) Show that the charge induced in a small area A on a grounded conductingplane by a point charge not in that plane is proportional to the solid angle subtendedat the point charge by area A.

2. (20 pts.) A hollow dielectric sphere of dielectric constant ε = 3 has inner radius onehalf its outer radius. When this sphere is placed in an initially uniform electric fieldE0, what is the resulting electric field strength at the center of the sphere?

3. (30 pts.) Two circular wires of radii a and b have a common center, and are free toturn on an insulating axis which is a diameter of both. Find the torque about thisdiameter required to hold the two wire loops at rest when their planes are at rightangles and they are carrying currents I and I ′, supposing that b ¿ a. Give both theleading term, and the first correction in a power of the small ratio b/a.

Hint: This requires evaluating the first correction to both the axial and transversemagnetic field components near the center of the larger loop. Recall that the torqueabout a point is ~τ = r× F where force F is applied at distance r.

Page 88: Electrodynamics California

Princeton University Ph501 Midterm Exam Oct. 25, 2000 2

Solutions

1. Let charge q be at perpendicular distance a from the grounded conducting plane. Thesmall area A has its center at distance r from the foot of the perpendicular to chargeq. The charge q′ induced in the area A is related by

q′ = σA =EA

4π, (1)

where E is the electric field strength at the surface of the conducting plane.

We calculate E using the image method, supposing that charge−q is located at distancea on the other side of the conducting plane from charge q. Then,

E = − 2q

R2

a

R= −2q

cos θ

R2, (2)

where R =√

a2 + r2 is the distance from charge q to area A, and θ is the angle betweenvector R and the perpendicular from q to the plane.

Combining eqs. (1) and (2), we have

q′ = −2qA cos θ

4πR2= −qΩ

2π, (3)

where Ω = A cos θ/R2 is the solid angle subtended by area A at charge q. For thewhole plane, Ω = 2π and q′ = −q.

2. This problem is closely related to that of a dielectric sphere in an otherwise uniformelectric field. We choose the z axis antiparallel to the initial field E0, with the originat the center of the dielectric sphere, where the potential is taken to be zero.

The potential of the initial field is then

φ0 = E0z = E0r cos θ = E0rP1(θ), (4)

where θ is the polar angle with respect to the z axis and P1 is the Legendre polynomialof order 1.

We recall from the case of a uniform dielectric sphere that the potential contains termsonly in P1, and we expect the same here.

Writing the inner radius of the sphere as a and the outer radius as b, we expect thatthe potential will have the form

φ1 = E0rP1 + Ar

aP1, (0 < r < a) (5)

φ2 = E0rP1 + Br

aP1 + C

b2

r2P1, (a < r < b) (6)

φ3 = E0rP1 + Db2

r2P1, (a < r < b) (7)

since the perturbation to field E0 must be finite at r = 0 and ∞.

Page 89: Electrodynamics California

Princeton University Ph501 Midterm Exam Oct. 25, 2000 3

The potential is continuous at r = a and b, so that

A = B + Cb2

a2, (8)

Bb

a+ C = D. (9)

Also, the normal component of the electric displacement D = εE is continuous at theboundaries, since ∇ ·D = 0. Hence,

∂φ1(a)

∂r= ε

∂φ2(a)

∂r, (10)

and

ε∂φ2(b)

∂r=

∂φ3(b)

∂r, (11)

which yields

E0 +A

a= εE0 + ε

B

a− 2ε

Cb2

a3, (12)

and

εE0 + εB

a− 2ε

C

b= E0 − 2

D

b. (13)

Inserting eq. (8) in (12), we get

ε− 1

aB − (2ε + 1)

b2

a3C = (1− ε)E0, (14)

while using eq. (9) in (13) gives

ε + 2

aB − 2(ε− 1)

bC = (1− ε)E0. (15)

These could be solved in general for A, B and C, but here we consider the particularcase that a = 1, b = 2 and ε = 3, for which eqs. (14) and (15) become

B − 14C = −E0, (16)

and5B − 2C = −2E0. (17)

We quickly find that

B = −13

34E0, C =

3

68E0, (18)

and from eq. (11),

A = B + 4C = − 7

34E0. (19)

The electric field strength at the center of the dielectric sphere is

E(0) = E0 + A =27

34E0. (20)

A dielectric sphere is not as effective as a conducting sphere in shielding its interiorfrom an external electric field.

Page 90: Electrodynamics California

Princeton University Ph501 Midterm Exam Oct. 25, 2000 4

3. (Problem 12, p. 448 of The Mathematical Theory of Electricity and Magnetism byJ. Jeans.)

The leading term of the torque is given by ~µ×B(0), where

µ =πI ′b2

c(21)

is the magnetic moment of the small loop of radius b that carries current I ′, and

B(0) =1

c

∫ I× dl

r2=

2πaI

ca2=

2πI

ca(22)

is the magnetic field at the center of the loops due to the current I in the loop of radiusa. When the two loops are at right angles, the vectors ~µ and B(0) are also at rightangles, so the magnitude of the leading term of the torque is

τ =2πII ′b2

c2a(23)

To evaluate the torque in greater detail, we consider the variation of the magnetic fieldover the small loop, and use the basic torque equation

~τ =∫

r× dF =1

c

∫r× [I ′dl′ ×B(due to I)]. (24)

We use a coordinate system in which the centers of the loops are at the origin, withthe axis of loop a is along the z axis. We take the sign of current I to be such thatthe resulting magnetic field at the origin is in the +z direction. The axis of loop b isdefined to be the y axis, and the sign of current I ′ is such that the magnetic moment~µ is along the +y axis. Then, we desire the x component of the torque ~τ about theorigin:

τx =1

c

∫br× [I ′bφdφ× (Bzz + Bρρ)]

∣∣∣∣x

=b2I ′

c

∫ 2π

0dφ cos φ(cos φBz + sin φBρ), (25)

where angle φ is measured in the x-z plane with respect to the z axis, such that for apoint on loop b, ρ = b sin φ and z = b cos φ.

If we don’t recall the results of problem 7, set 4, the magnitude of Bρ can be estimatedquickly using the Maxwell equation ∇ ·B = 0 and a “pillbox” surface of radius ρ andthickness dz whose axis is along the z axis:

0 =∫∇ ·BdVol =

∫B · dS

≈ πρ2(Bz(0, z + dz)−Bz(0, z)) + 2πρdzBρ(ρ, z).

≈ πρ2dz∂Bz(0, z)

∂z+ 2πρdzBρ(ρ, z). (26)

Page 91: Electrodynamics California

Ph501 Midterm Exam Oct. 25, 2000Princeton University 5

Hence,

Bρ(ρ, z) ≈ −ρ

2

∂Bz(0, z)

∂z. (27)

Then, near the center of loop a its magnetic field obeys ∇×B = 0, and in particular

∂Bz(ρ, z)

∂ρ=

∂Br(ρ, z)

∂z≈ −ρ

2

∂2Bz(0, z)

∂z2, (28)

using eq. (27). We can integrate this to find

Bz(ρ, z) ≈ Bz(0, z)− ρ2

4

∂2Bz(0, z)

∂z2, (29)

in agreement with the results of Problem 7, Set 4.

For points along the z axis the magnetic field due to loop a is

Bz(0, z) =1

c

∫ I× dl

r2

∣∣∣∣∣z

=2πa2I

c(a2 + z2)3/2≈ 2πI

ca

(1− 3z2

2a2

), (30)

where the approximation can be used when we evaluate the field on loop b for which|z| ≤ b ¿ a. Thus,

∂Bz(0, z)

∂z= − 6πa2zI

c(a2 + z2)5/2≈ −6πzI

ca3, (31)

and∂2Bz(0, z)

∂z2= −6πa2I(a2 − 4z2)

c(a2 + z2)7/2≈ −6πI

ca3, (32)

Using eqs. (27) and (31), the transverse magnetic field at a point on loop b is

Bρ(ρ, z) ≈ 3πIρz

ca3=

3πb2I cos φ sin φ

ca3, (33)

and eqs. (29), (30) and (32) give the axial field as

Bz(ρ, z) ≈ 2πI

ca

(1− 3z2

2a2

)+

3πIρ2

2ca3=

2πI

ca

(1− 3b2 cos2 φ

2a2

)+

3πb2I sin2 φ

2ca3. (34)

Combining eqs. (25), (33) and (34) we find

τx ≈ πb2II ′

c2a

∫ 2π

0dφ

(2 cos2 φ− 3b2 cos4 φ

a2+

3b2 cos2 φ sin2 φ

2a2+

3b2 cos2 φ sin2 φ

a2

)

=πb2II ′

c2a

∫ 2π

0dφ

(2 cos2 φ− 3b2 cos2 φ

a2+

15b2 sin2 2φ

8a2

)

=2π2b2II ′

c2a

(1− 9b2

16a2

). (35)

[The answer in MKSA units is obtained on setting c = 1 in the magnetic force equation,and replacing 1/c by µ0/4π in the Biot-Savart law, so 2π2/c2 → πµ0/2.]

Page 92: Electrodynamics California

Princeton University

Ph501

Electrodynamics

Problem Set 4

Kirk T. McDonald

(1999)

[email protected]

http://puhep1.princeton.edu/~mcdonald/examples/

Page 93: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 1 1

1. a) Child’s Law. Before the transistor era, a common device was a vacuum diode.This is a parallel plate capacitor with a potential difference V across a gap d, all ofwhich is inside a vacuum tube. The cathode (at φ = 0) is heated, so electrons canjump off and flow to the anode (at φ = V ). Positive charges have very low probabilityof leaving the anode and flowing to the cathode. The resulting one way flow of chargefrom cathode to anode is the diode action.

Consider a steady situation in which a constant current density j = ρ(x)v(x) flows, andwhere the electrons leave the cathode with velocity v(0) = 0. Here, ρ(x), 0 ≤ x ≤ d, isthe electron charge density.

Solve for the potential φ(x) via Poisson’s equation,

∇2φ = −4πρ. (1)

Show that

φ(x) = V(

x

d

)4/3

, and J = − 1

V 3/2

d2

√2e

m. (2)

where e and m are the magnitudes of the charge and mass of the electron, respectively.

Note that since the current density J = nev is constant, and v → 0 near the cathode,the charge density n → ∞ there. The field due to this large “space charge” distributionnear the cathode opposes the field due to the capacitor alone, and cancels it completelyvery close to the cathode. That is, E(x) ∝ xp with p > 0. Then, φ(x) = − ∫ E dx ∝x1+p rises more quickly than the simple linear relation for an ordinary capacitor.

b) Laser Driven Vacuum Photodiode. A vacuum photodiode is constructed in theform of a parallel-plate capacitor of area A, plate separation d. A battery maintainsconstant potential V between the plates. A short laser pulse illuminates that cathodeat time t = 0 with energy sufficient to liberate all of the surface-electron charge density.This charge moves across the capacitor gap as a sheet until it is collected at the anodeat time T . Then another laser pulse strikes the cathode, and the cycle repeats.

Estimate the average current density 〈j〉 that flows onto the anode from the battery,ignoring the recharging of the cathode as the charge sheet moves away. Then calculatethe current density and its time average when this effect is included.

You may suppose that the laser photon energy is equal to the work function of thecathode, so the electrons leave the cathode with zero velocity.

Page 94: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 2 2

2. Obtain a Legendre series expansion for the potential inside a conducting sphere ofradius a and conductivity σ when a current I enters at one pole through a fine wire,also of conductivity σ, and leaves through the other pole via a similar fine wire.

Define the potential as φ = 0 on the equator.

By noting that Pn(−μ) = (−1)nPn(μ), and referring to the expansion of 1/R given onp. 57 of the notes, show that

φ(r, θ) =I

2πσ

[1

R1− 1

R2+

1

2

∫ r

0

(1

R1− 1

R2

)d ln r

], (3)

where R1,2 is the distance from the “north” (“south”) pole to the point (r, θ, ϕ) inspherical coordinates. The integrals can be found in tables if desired.

Finally, suppose the wires have radius b a, and their surface of contact with thesphere is an equipotential. Show that the resistance of the sphere is that of a piece ofwire roughly b long.

Hint: Express the radial current density at r = a in terms of delta functions, δ(cos θ−1)and δ(cos θ + 1).

Page 95: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 3 3

3. Charge Distributions in a Wire that carries a Steady Current

a) A wire of circular cross-section carries a current I which is uniformly distributedacross the wire. We consider this current to be due to a number density ρ of freeelectrons moving with average drift velocity v. (In a typical situation, v 1cm sec−1!)Let ρ0 be the uniform number density of positive ions in the wire. For steady currentflow, there must be no radial force on the electrons. Use the Lorentz force law,

F = q(E +

v

c× B

)(4)

to find the relation between ρ0 and ρ such that the force vanishes. (As a check, youmay wish to do this problem via special relativity, but try it using Maxwell’s equationsand the Lorentz force.)

b) A resistor of resistance R, length l and cross-sectional area A carries a current I ,delivered by fine lead wires. Calculate the charge that accumulates on the end facesof the resistor in order to produce the field E which drives the current according toOhm’s law. Suppose that the current in the wire varies with time. Show that theconduction current inside the resistor is different from that in the lead wires, but thatMaxwell’s concept of displacement current restores the continuity of “total current”.You may assume that l √

A so that the current density J is uniform inside the wire.

Page 96: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 4 4

4. A straw tube chamber is a low cost version of a proportional counter. These devicesconsist of a pair of coaxial conducting cylinders with the region between the cylindersis filled with a gas such as argon. The inner cylinder of radius a is the anode, and isheld at potential V ; the outer cylinder of radius b is the cathode, and is grounded.

If a penetrating charged particle passes through the chamber, it will ionize about twogas molecules per mm of path length. The ionization electrons are pulled by theelectric field towards the anode. Close to the anode, the field is strong enough that theelectrons gain enough energy during one mean free path to ionize the molecule they hitnext, liberating one or more additional electrons. In a proportional chamber, the fieldis kept low enough that the resulting Townsend avalanche involves 104-106 molecules.

What is the time dependence, I(t), of the current that flows off the anode due to theavalanche of a single initial electron?

What is the spatial dependence, q(z) of the charge distribution induced on the anodeduring the time when the current is large, where the z axis is the chamber axis? Youmay restrict your attention to values of z far from the ends of the tube of length l.

Measurement of the charge distribution via a segmented cathode permits localizationin z of the ionization, and hence, of the initiating charged particle. [C. Leonidopoulos,C. Lu and A.J. Schwartz, Development of a Straw Tube Chamber with Pickup-PadReadout, Nucl. Instr. and Meth. A427, 465 (1999).]

You may ignore the tiny current that flows while the electron drifts towards the anode.The avalanche takes place so close to the anode, that the small remaining drift timefor the electrons to reach the anode may also be ignored. In this approximation, thesituation at t = 0 is that electrons of total charge −q0 reside on the anode in closeproximity to positive ions of total charge +q0. Current flows off the anode only whensome of the field lines from the positive ions detach from the electrons on the anode,and extend to the cathode where charge is induced to terminate these field lines. Thisoccurs only as the positive ions move away from the anode, with velocity related by

v = μE, (5)

where μ positive ion mobility.

I(t) via Reciprocity and Weighting Fields

This problem can be solved by an application of Green’s reciprocation theorem, whichstates that if a set of fixed conductors is at potentials Vi when carrying charges Qi,and at potentials V ′

i when carrying charges Q′i, then

∑i

ViQ′i =

∑i

V ′i Qi. (6)

To see this, we label the 3-dimensional potential distribution associated with charges Qi

by φ(r), and that associated with charges Q′i by φ′. The space outside the conductors

is charge free and with dielectric constant ε = 1. Then ∇2φ = 0 = ∇2φ′ outside theconductors.

Page 97: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 4 5

We invoke Green’s theorem (p. 37 of the Notes)

∫(φ∇2φ′ − φ′∇2φ)dvol =

∮(φ∇φ′ − φ′∇φ) · dS, (7)

where we take the bounding surface S to be that of the set of conductors. Hence,

0 =∑

i

∮(Vi∇φ′

i − V ′i ∇φi) · dSi = −4π

∑i

(ViQ′i − V ′

i Qi), (8)

using Gauss’ Law (in Gaussian units) that

4πQi =∮

Ei · dSi = −∮

∇φi · dSi. (9)

In the present problem, we have a small charge q0 at position r0(t) that moves underthe influence of the field due to conductors i = 1, ..., n that are held at potentialsVi. The charges Qi on the conductors obey Qi q0, so the motion of charge q0 isdetermined, to a very good approximation by the charges Qi on the conductors whenq0 = 0. Hence, the problem can be considered as the superposition of two situations:

A: charge q0 absent; conductors i = 1, ...n at potentials Vi.

B: charge q0 present; conductors i = 1, ...n grounded, with charges ΔQi on them.We are particularly interested in the charge on electrode 1, whose time rate of changeis the desired current I(t).

To use the reciprocation theorem, we suppose that in case B the charge resides on atiny conductor at position r0 that is at the potential V0 = φA(r0) obtained from caseA. Then, the charges and potentials in case B can be summarized as

B: q0, V0; ΔQi, Vi = 0, i = 1, ..., n.We solve the electrostatics problem for a third case,

C: q′0 = 0, V ′0(r0); Q1, V ′

1 = 1; ΔQi = 0, V ′i = 0, i = 2, ..., n.,

in which conductor 1 is held at unit potential, the charges on all other conductors atzero, and all other conductors are grounded except for the tiny conductor at positionr0. Again, we solve this problem as in case A, first ignoring the tiny conductor, thenevaluating V ′

0 as φC(r0).

The reciprocation theorem (6) applied to cases B and C implies that

0 = q0V′0 + ΔQ1 · 1. (10)

The current that moves off electrode 1 in case B is therefore,

I1 = −dΔQ1

dt= q0

dV ′0 (r0)

dt= q0∇V ′

0(r0) · dr0

dt= −q0Ew · v, (11)

where the velocity v of the charge is determined using the fields from case A, and

Ew = −∇V ′0(r0) = −∇φC(r0) (12)

Page 98: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 4 6

is called the weighting field. For the case of two conductors (plus charge q0) one ofwhich is grounded, the weighting field is the same as the field from case A, but ingeneral they are distinct.

As the present problem involves only two conductors, you may wish to find a solutionthat does not appear to use the initially cumbersome machinery of the reciprocationtheorem.

Page 99: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 5 7

5. Resistance of a Disk with Edge Contacts

Calculate the resistance between two contacts on the rim of a disk of radius a, thicknesst a, and conductivity σ, when each (perfectly conducting) contact extends for asmall distance δ around the circumference, and the distance along the chord betweenthe contacts is d δ.

Page 100: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 6 8

6. Some biological systems consist of two “phases” of nearly square fiber bundles of dif-fering thermal and electrical conductivities. Consider a circular region of radius a neara corner of such a system as shown below.

Phase 1, with electrical conductivity σ1, occupies the “bowtie” region of angle ±α,while phase 2, with conductivity σ2 σ1, occupies the remaining region.

Deduce the approximate form of lines of current density J when a background electricfield is applied along the symmetry axis of phase 1. What is the effective conductivityσ of the system, defined by the relation I = σΔφ between the total current I and thepotential difference Δφ across the system?

It suffices to consider the case that the boundary arc (r = a, |θ| < α) is held at electricpotential φ = 1, while the arc (r = a, π − α < |θ| < π) is held at electric potentialφ = −1, and no current flows across the remainder of the boundary.

Hint: When σ2 σ1, the electric potential is well described by the leading term of aseries expansion.

Page 101: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 7 9

7. A rectangular loop of size 2a by 2b carries a current I ′, and is free to rotate about anaxis that bisects the sides of length 2b. The axis is parallel to and distance d froma wire that carries current I . If the plane of the loop makes angle θ to the planecontaining the wire and the axis, and if the currents in the wire and in the side (oflength 2a) of the loop closest to the wire flow in the same direction, show that themagnitude of the torque on the loop is

N =8abdII ′

c

(b2 + d2) sin θ

b4 + d4 − 2b2d2 cos 2θ. (13)

What is its direction?

Page 102: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 8 10

8. Helmholtz Coils

a) Each of a pair of parallel, coaxial “Helmholtz” coils has radius a and carries a currentI in the same sense. Their centers are at z = ±b, where the z axis is the common axisof the coils. Calculate the magnetic field along the axis, and determine the separation2b such that the first, second and third derivatives of Bz with respect to z all vanishat the mid‘ axis. Thus, the field is very uniform at the center of the Helmholtz coils.

b) Suppose we desire an even more uniform field at the origin. Add a second pair ofHelmholtz coils of radius a′ = a/2. What current I ′ should flow in the second pair soas to cancel the 4rth derivative of Bz of the first pair? What fraction of the originalcentral field is lost in this configuration?

c) In some applications, it is more important that the field outside the coils be assmall as possible, rather than the field inside be highly uniform.

Give an expansion for the field along the axis of a set of Helmholtz coils as a functionof u = 1/z for z a, b. Identify the first two nonvanishing multipoles, and find thevalue of b for which the second of these can be made to vanish.

To cancel the leading multipole as well, add a second coil pair with a′ = 2a. Whatcurrent I ′ should flow in this pair? What fraction of the central field of the first pairis lost? What is the order of the first remaining nonzero multipole?

[See, E.M. Purcell, Am. J. Phys. 57, 18 (1989).1]

1http://puhep1.princeton.edu/~mcdonald/examples/EM/purcell_ajp_57_18_88.pdf

Page 103: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 9 11

9. Expansion of an Axially Symmetric Magnetic Field in Terms of the AxialField

Suppose a magnetic field in a current-free region is rotationally symmetric about thez-axis. Then,

B = Br(r, z)r + Bz(r, z)z (14)

in cylindrical coordinates. The axial field Bz(0, z) is often relatively easy to calculate.If we write

Bz(r, z) =∞∑

n=0

an(z)rn, and Br(r, z) =∞∑

n=0

bn(z)rn , (15)

then a0(z) = Bz(0, z). Use ∇ · B = 0 and ∇ × B = 0 to show that

Bz(r, z) =∑n

(−1)na(2n)0 (z)

(n!)2

(r

2

)2n

, (16)

and

Br(r, z) =∑n

(−1)n+1 a(2n+1)0 (z)

(n + 1)(n!)2

(r

2

)2n+1

, (17)

where

a(n)0 =

dna0

dzn. (18)

This magnetic field can also be deduced from the vector potential whose only nonzerocomponent is

Aφ(r, z) =∑n

(−1)n a(2n)0 (z)

(n + 1)(n!)2

(r

2

)2n+1

. (19)

For the example of Helmholtz coils, prob. 5, we know that

Bz(0, z) = B0 + B4z4 + . . . (20)

Give Bz and Br correct to fourth order in r and z.

Show also that, for small r, ∇ · B = 0 leads to the relation

Br(r, z) ≈ −r

2

∂Bz(0, z)

∂z. (21)

Remark. An electrostatic field with azimuthal symmetry about the z axis can also beexpanded according to eqs. (16)-(17). For example, consider a capacitor with circularplates centered about (r, θ, z) = (0, 0, 0). Then we can expand

Ez(0, 0, z) ≈ Ez(0, 0, 0) +z2

2

d2Ez(0, 0, 0)

dz2+ ... (22)

and

Ez(r, 0, 0) ≈ Ez(0, 0, 0) − r

2

d2Ez(0, 0, 0)

dz2+ ... (23)

Thus, if Ez has a maximum with respect to z at the origin, it is at a minimum withrespect to r, or vice versa. The field E cannot be at a maximum with respect to bothr and z, as shown in general in prob. 1(c) of set 1.

Page 104: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 10 12

10. Nonaxially Symmetric Magnetic Field in Terms of the Axial Field

In the previous problem, it was demonstrated how knowledge of a static, axial magneticfield leads to a complete characterization of the field if that field is axially symmetric.

A variant on the electro- or magnetostatic boundary value problem arises in acceleratorphysics, where a specified field, say B(0, 0, z), that is not axially symmetric is desiredalong the z axis. In general there exist static fields B(x, y, z) that reduce to the desiredfield on the axis, but the “boundary condition” B(0, 0, z) is not sufficient to insure aunique solution.

For example, find a field B(x, y, z) that reduces to

B(0, 0, z) = B0 cos kzx + B0 sin kzy (24)

on the z axis. In this, the magnetic field rotates around the z axis as z advances.

Show that the use of rectangular or cylindrical coordinates leads “naturally” to differentforms for B off the z axis.

One 3-dimensional field extension of (24) is the so-called helical wiggler, which obeysthe auxiliary requirement that the field at z + δ be the same as the field at z, butrotated by angle kδ. Show that this field pattern can be realized by a current-carryingwire that is wound in a helix of period λ = 2π/k.

See, B.M. Kincaid, A short-period helical wiggler as an improved source of synchrotronradiation, J. Appl. Phys. 48, 2684-2691 (1977);2 J.P. Blewett and R. Chasman, Orbitsand fields in the helical wiggler, J. Appl. Phys. 48, 2692-2698 (1977).3

2http://puhep1.princeton.edu/~mcdonald/examples/EM/kincaid_jap_48_2684_77.pdf3http://puhep1.princeton.edu/~mcdonald/examples/EM/blewett_jap_48_2692_77.pdf

Page 105: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 11 13

11. Axial Field of a Solenoid Magnet

A solenoidal coil of radius a and length l has n turns per unit length and carries acurrent I (in each turn). On the axis, show that

Bz(0, z) =2πnI

c(cos θ1 + cos θ2), (25)

where θ1 and θ2 are the angles between the axis and the ends of the solenoid at theobservation point. Near the midpoint of the solenoid (z = 0), show

Br(r, z) ≈ 288πnIa2rz

cl4. (26)

At the end of the coil (z = l/2), show that

Bz ≈ 2πnI

c≈ Bz(0, 0)

2, (27)

and

Br ≈ πnIr

ac. (28)

If one is interested in the fields near the end of a long solenoid (l a), it is oftensufficient to approximate the coil as semi-infinite, for which (25) leads to

Bz(0, z) =2πnI

c

(1 +

z√z2 + a2

), (29)

where z = 0 at the end of coil.

Page 106: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 12 14

12. a) A coil is wound on the surface of a sphere such that the magnetic field inside thesphere will be uniform. How should the turns be distributed?

b) What is the effective magnetic dipole moment of a sphere of uniform surface chargedensity σ which rotates with constant angular velocity ω about an axis of the sphere?

An electron has a permanent magnetic dipole moment of magnitude

μ =eh

2mc. (30)

Suppose the electron is a rotating spherical shell of charge with radius

a =e2

mc2, (31)

the “classical electron radius”. What is the velocity at the equator?

Page 107: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 13 15

13. Saturation and Hysteresis

a) A piece of iron saturates in a magnetic field of ≈ 20, 000 Gauss, when all availableelectron magnetic moments are aligned. The density of iron is 8 g/cm3. How manyelectrons per iron atom have been aligned to produce this field?

b) You can understand some aspects of the hysteresis curve of a ferromagnet via amodel consisting of two permanent dipoles separated by a fixed distance d, but free torotate. In the absence of any external field, what is the equilibrium orientation andenergy of the two dipoles?

Suppose a magnetic field B is applied at right angles to the line of centers of the dipoles.What is the minimum field strength needed to align the dipoles along B? Higher fieldsproduce no further change – saturation has occurred.

Suppose the dipoles were originally aligned parallel to their line of centers, and then afield B is applied antiparallel to the dipoles. What is the minimum value of B neededto flip the dipoles?

If the dipoles flip and later B is reduced to zero, the dipoles do not unflip – hysteresishas occurred.

Page 108: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 14 16

14. Magnetic Field Mapping. You may be familiar with the method of mapping theequipotentials in 2-dimensional electrostatic problems using conducting paper:

On the paper, J = σE, where J is the two dimensional current density (= current perunit length perpendicular to J), σ is the surface conductivity of the resistive paper,and E is the electric field in the paper. Outside of the sources and sink of current inthe patches of conducting paint, we have ∇ ·J = 0, so ∇ ·E = 0 also. The currents andfields are steady, so ∇× E = 0, and the electric field can be derived from a potential,E = −∇φ that obeys Laplace’s equation, ∇2φ = 0. The value of the potential φ atany point on the paper can be read directly with a voltmeter.

The boundary conditions are that

• J and E are perpendicular to the boundaries of the patches of conducting paint.

• J and E are parallel to the edges of the paper (where there is no conducting paint.Thus, the region outside the paper is like a dielectric with constant ε = 0. This isnot very physical, so make the paper much larger than the region used to modelthe problem of interest (if the boundaries are not entirely conducting).

The conducting paper technique can also be used to model 2-dimensional magnetostaticproblems due to current distributions that are normal to the paper.

Imagine that the regions of conducting paint represent the cross sections of infiniteconductors that are perpendicular to the paper. Let z label the unit vector normal tothe paper. Then, the vector potential due to our imagined currents would be

A =1

c

∫J

rdVol = Azz. (32)

(Here, J is due to the imagined current normal to the paper, not the surface currentsin the paper.)

Show that the observed potential φ on the paper, when a battery feeds current intoand out of the conductors on the paper, is proportional to the vector potential of theimagined situation:

Az = kφ, k = constant. (33)

Page 109: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Problem 14 17

Your analysis might include the following:

• Relate the magnetic field B of the imagined 2-dimensional current distribution toφ on the paper. When A = Azz, B is perpendicular to z. Show that lines of Bexactly follow equipotentials of φ.

• Relate B to E on the paper.

• Suppose current I from a battery enters a region of conducting paint on the paper,causing current density J to flow outwards:

Consider∮J×dl for a loop enclosing the region of conducting paint to determine

the constant k in (33).

• Show that the voltage difference between any two points on the paper is propor-tional to the magnetic flux passing between these points.

• Show that the boundary conditions at the edge of the paper are such that we mayconsider the region outside the paper as being iron of a very large permeability μ.

As an example, consider a long electromagnetic with an iron yoke:

Invoking symmetry, we could map this with an arrangement like:

Try it in the lab sometime!

Page 110: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 1 18

Solutions

1. a) The problem is 1-dimensional, so Poisson’s equation is

d2φ

dx2= −4πρ(x). (34)

Since an electron leaves the cathode at v = 0, when it reaches position x, it has energyeφ(x) = mv2/2, and velocity

v =

√2eφ

m. (35)

Since the current density J = ρv is constant, eq. (34) becomes

d2φ

dx2= −4π

J

v= −4πJ

√m

2eφ−1/2. (36)

We try (pray for) a power law solution, φ = axp, which quickly leads p = 4/3. Then,since φ(d) = V , the potential is

φ(x) = V(

x

d

)4/3

. (37)

Equation (36) can now be rearranged as

J = −φ′′

√2e

mφ1/2 = − 1

V 3/2

d2

√2e

m= − 1

6.36π

V 3/2

d2

√e

m. (38)

The electric space charge density ρ(x) follows from eqs. (34) and (37),

ρ(x) = −φ′′

4π=

V

9πd3√

dx2, (39)

which is very large close to the cathode at x = 0.

b) The initial electric field in the capacitor is E = V/d, so the initial surface chargedensity on the cathode is

σ = −E/4π = −V/4πd. (40)

The laser liberates this charge density at t = 0.

The average current density that flows onto the anode from the battery is

〈J〉 = −σ

T=

V

4πdT, (41)

where T is the transit time of the charge across the gap d. We first estimate T byignoring the effect of the recharging of the cathode as the charge sheet moves away fromit. In this approximation, the average field on the charge sheet is always E/2 = V/2d,

Page 111: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 1 19

so the acceleration of an electron is a = eV/2dm, and the time to travel distance d is

T =√

2d/a = 2d√

m/eV . Hence,

〈J〉 =1

V 3/2

d2

√e

m. (42)

This is close to Child’s Law (38).

[This sign difference between (38) and (42) is because the former is the current flowingoff the anode, while the latter is the current flowing onto it.]

We now make a detailed calculation, including the effect of the recharging of thecathode, which will reduce the average current density somewhat.

At some time t, the charge sheet is at distance x(t) from the cathode, and the anodeand cathode have charge densities σA and σC , respectively. All the field lines that leavethe anode terminate on either the charge sheet or on the anode, so

σ + σC = −σA. (43)

The magnitude of the electric field strength in the region I between the anode and thecharge sheet is

EI = 4πσA, (44)

and that in region II between the charge sheet and the cathode is

EII = −4πσC . (45)

The voltage between the capacitor plates is therefore,

V = EI(d − x) + EIIx = 4πσAd − Vx

d, (46)

using (40) and (43-45). Thus,

σA =V

4πd

(1 +

x

d

), σC = − V x

4πd2, (47)

and the time-dependent current density flowing onto the anode is

J(t) = σA =V x

4πd2. (48)

This differs from the average current density (41) in that x/d = T , since x varies withtime.

To find the velocity x of the charge sheet, we consider the force on it, which is due tothe average field set up by charge densities on the anode and cathode,

Eon σ = 2π(−σA + σC) = − V

2d

(1 +

2x

d

). (49)

Page 112: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 1 20

The equation of motion of an electron in the charge sheet is

mx = −eEon σ =eV

2d

(1 +

2x

d

), (50)

or

x − eV

md2x =

eV

2md. (51)

With the initial conditions that the electron starts from rest, x(0) = 0 = x(0), wereadily find that

x(t) =d

2(cosh kt − 1), (52)

where

k =

√eV

md2. (53)

The charge sheet reaches the anode at time

T =1

kcosh−1 3

2=

0.96

k, (54)

compared to T = 1/k as found above without the battery.

The average anodoe-current density is, using (41) and (54),

〈J〉 =V

4πdT=

V 3/2

4π cosh−1(3/2) d2

√e

m=

V 3/2

12.09 πd2

√e

m. (55)

The electron velocity is

x =dk

dsinh kt, (56)

so the anode-current density (48) is

J =1

V 3/2

d2

√e

msinh kt (0 < t < T ). (57)

Page 113: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 2 21

2. Although current is flowing inside the conducting sphere, it remains neutral. Hence,the potential satisfies Laplace’s equation, ∇2φ = 0.

We analyze the problem in spherical coordinates (r, θ, ϕ), with the origin at the centerof the sphere of radius a, and θ = 0 and π at the points of contact with the wires. Theproblem has axial symmetry, so φ will be independent of ϕ. We require the potentialto be well behaved at the origin, so it can be expressed in a Legendre series,

φ =∞∑

n=0

An

(r

a

)n

Pn(cos θ) . (58)

The convention that φ = 0 at the equator, θ = π/2, implies that An = 0 for n even.Therefore, we can write

φ =∑

n odd

An

(r

a

)n

Pn(cos θ) . (59)

To complete the solution, we need the boundary condition on φ at the surface of thesphere, r = a. We know that the radial component of the current density, jr is zeroat the surface, except for the contact points where the current enters and exits. SinceJ = σE = −σ∇φ, we obtain a condition on the derivative of the potential at theboundary,

∂φ

∂r

∣∣∣∣∣r=a

= −Er(r = a) = −Jr(r = a)

σ. (60)

In the limit of very fine wires, the current density Jr(r = a) is zero except at the poles,so we can express it in terms of Dirac δ functions. The current dI that crosses anannular region on the surface of the sphere of angular extent d cos θ centered on angleθ is given by

dI = 2πa2Jr(a, θ)d cos θ. (61)

Current I enters at cos θ = 1, and exits at cos θ = −1. Hence, the form

Jr(a, θ) =I

2πa2[−δ(cos θ − 1) + δ(cos θ + 1)] (62)

describes the entrance and exit currents upon integration of (61).

Combining (59-60) and (62), we have

∑n odd

nAn

aPn(cos θ) =

I

2πa2[δ(cos θ − 1) − δ(cos θ + 1)] . (63)

As usual, to evaluate the Fourier coefficients we multiply by Pn(cos θ) and integrateover d cos θ to find

2nAn

(2n + 1)a=

2I

2πa2σ. (64)

Thus, the Legendre series expansion for the potential is

φ(r, θ) =I

2πaσ

∑n odd

(2 +

1

n

)(r

a

)n

Pn(cos θ). (65)

Page 114: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 2 22

To express this series in closed form, we utilize the expansion for the distance R1

between the point (a, 0) and (r, θ) given on p. 57 of the notes:

1

R1=

1

a

∞∑n=0

(r

a

)n

Pn(cos θ) , (66)

Similarly, the distance R2 between the point (a, π) and (r, θ) is

1

R2=

1

a

∞∑n=0

(r

a

)n

Pn(cos(θ−π)) =1

a

∞∑n=0

(r

a

)n

Pn(− cos θ) =1

a

∞∑n=0

(−1)n(

r

a

)n

Pn(cos θ).

(67)Hence,

1

R1− 1

R2=

2

a

∑n odd

(r

a

)n

Pn(cos θ). (68)

It follows that ∫ r

0

(1

R1

− 1

R2

)dr

r=

2

a

∑n odd

1

n

(r

a

)n

Pn(cos θ). (69)

Then, (65) and (68-69) combine to give to the alternative form (3) for φ.

As we approach the “north” pole, R1 → 0, and (we claim; details given later) the firstterm in (3) dominates. That is, the potential diverges at the poles for the case of veryfine wires.

When considering actual wires of radius b, we suppose that our solution holds outsidethe region of contact between the wire and the sphere. Indeed, we expect that thepotential is constant in planes perpendicular to the axis of the wire, so that the interfacebetween the wire and the sphere is an equipotential. This cuts off the formal divergencein (3) near the poles.

In this way, the potential at the interface is obtained from (3) on putting R1 = b andneglecting all but the first term: φinterface = I/2πσb. The potential difference acrossthe sphere is twice this;

ΔV =I

πσb= I

b

σπb2≡ IR. (70)

Thus, the effective resistance of the sphere is R = b/(σπb2), which is also the resistanceof a piece of wire of radius b, length b, and conductivity σ.

To verify the claim that the first term of (3) dominates for small R1, we consider thepoint (r, θ) = (a − b, 0) for b a. Then, the first term of (3) is 1/b, and the secondterm is 1/(2a− b) which is negligible compared to the first. Inside the integral term of(3), we have R1 = a − r and R2 = a + r, so the integral is∫ a−b

0

(1

R1− 1

R2

)d ln r =

∫ a−b

0

2

a2 − r2dr =

1

aln

2a − b

b≈ 1

aln

2a

b. (71)

The ratio of the integral term to the first term of (3) is therefore

b

2aln

2a

b, (72)

which goes to zero as b becomes small.

Page 115: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 3 23

3. a) Take the axis of the wire to be the z axis of a cylindrical coordinate system, (r, θ, z).By rotational symmetry, there is no azimuthal component to the electric field, and bytranslation symmetry, the axial and radial components can only depend on r. Similarly,there is no radial or axial component to the magnetic field, and its azimuthal componentis only dependent on r.

Consider a cylindrical portion of the wire, of radius r and length l. The charge con-tained in this cylinder is then,

Q = e(ρ0 − ρ)πr2l, (73)

where e is the magnitude of the charge of an electron. From Gauss’ Law and (73), wehave

4πQ = 4π2er2l(ρ0 − ρ) =∮

E · dS = 2πrlEr(r), (74)

since the contributions from the flat end surfaces are be equal and opposite. Hence,

E = 2πe(ρ0 − ρ)rr + Ezz. (75)

The resulting radial force on the free electrons must be opposed by magnetic effectsassociated with the electron current, which we presume flows only in the z direction

The current density in the wire is

J = −ρevz, (76)

so from Ampere’s law,

2πrBθ = −4π

cρevπr2, and B = −2π

ceρvrθ. (77)

The radial component of the Lorentz force on an electron, which must vanish if thereis to be no current in radial direction, is then,

Fr = −e(Er − vz

cBθ

)= −e

[2πe(ρ0 − ρ)r + 2πeρ

v2

c2r

]= 0, (78)

using (75) and (77). Hence,

ρ0 = ρ

(1 − v2

c2

), (79)

and the positive charge density is less than the negative by one part in 1021 for v = 1cm/s.

b) In this problem, we treat the resistor as a kind of conductive capacitor. Since thecurrent density is uniform, the electric field E is also. To maintain this electric field,surface charge ±Q must reside at the ends of the resistor. From Gauss’ Law,

Q =EA

4π, (80)

where A is the cross-sectional area of the resistor.

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Princeton University 1999 Ph501 Set 4, Solution 3 24

If the current I into to resistor varies with time, then part of it goes to changing thecharge Q at the ends of the resistor, and part of it appears as the conduction currentIC across the resistor. Thus, IC is less than I according to

IC = I − Q. (81)

However, Maxwell advises us that inside the resistor we should also consider the dis-placement current,

ID =D

4π=

εE

4π= εQ = Q, (82)

using (80), in a medium whose dielectric constant ε is unity.

Combining (81) and (82), the “total” current inside the resistor is thus,

Itotal = IC + ID = I, (83)

which illustrates Maxwell’s notion that “total” currents are conserved.

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Princeton University 1999 Ph501 Set 4, Solution 4 25

4. The form of the current I(t) in a cylindrical “straw tube” can also be found withoutusing the reciprocation theorem, so we illustrate that first.

Elementary Solution for I(t)

The current that flows off the anode is equal to minus the rate of change of the chargeq(t) < 0 that remains on the anode as the positive ions of total charge q0 move outwardaccording to r(t).

The key to an elementary solution is that although the positive ions occupy a verysmall volume around the point (r, θ, z) = (r(t), 0, 0) in cylindrical coordinates, thecharge they induce on the cathode is exactly the same as if those ions were uniformlyspread out over a cylinder of radius r.

Because the superposition principle holds in electrostatics, the problem of the chamberwith voltage V on the anode plus ions at a fixed position between the anode andcathode can be separated into two parts. First, an empty chamber with voltage Von the anode, and second, a grounded chamber with positive ions inside. [That is,we decompose the problem into cases A and B of the discussion of the reciprocationtheorem, even though we won’t use that theorem here.]

For the second part, the radial electric field in the region a < r < r(t) can be calculatedfrom the charge q on the anode as

E(r) =2q(t)

rl, (84)

using Gauss’ Law, where l b is the length of the cylinder. Similarly, the electric fieldin the region r(t) < r < b is

E(r) =2(q0 + q(t))

rl. (85)

The potential difference between the inner and outer cylinder must be zero. Hence,

0 =2q(t)

l

∫ r(t)

a

dr

r+

2(q0 + q(t))

l

∫ b

r(t)

dr

r=

2q0

lln

b

r(t)+

2q(t)

lln

b

a, (86)

and so

q(t) = −q0ln(b/r(t))

ln(b/a). (87)

The current is

I(t) = −q(t) = − q0

ln(b/a)

v(t)

r(t). (88)

To calculate the dynamical quantities r(t) and v(t), we must return to the full problemof the ions in a chamber with voltage V . The electric field in the chamber is onlyslightly perturbed by the presence of the ions, and so is given by

E(r) =V

r ln(b/a). (89)

Page 118: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 4 26

According to (5), the positive ions have velocity

v(r) =μV

r ln(b/a), (90)

which integrates to give

r2(t) = a2 +2μV

ln(b/a)t. (91)

Inserting (90-91) in (88), we find

I(t) = − q0

2t0 ln(b/a)

1

1 + t/t0, (92)

where

t0 =a2 ln(b/a)

2μV. (93)

The idealized current pulse has a very sharp rise, and falls off rapidly over characteristictime t0, which is about 20 nsec in typical straw tube chambers.

I(t) via Reciprocity

Referring to the prescription in the statement of the problem, we first solve case C, inwhich the inner electrode is at unit potential and the outer electrode is grounded. Wequickly find that

VC(r) =ln(b/r)

ln(b/a). (94)

According to (11), the current off the inner electrode is therefore,

I(t) = −q0dVC

drv(r) = − q0

ln(b/a)

v(t)

r(t), (95)

as previously found in (88). We again solve for v and r(t) as in (89-91), which corre-sponds to the use of case A, to obtain the solution (92-93).

The Charge Distribution q(z) on the Cathode

The more detailed question as to the longitudinal charge distribution on the cathodecan be solved by the reciprocation method if we conceptually divide the cathode cylin-der into a ring of length dz at position z1 plus two cylinders that extends to z = ±l/2where l is the length of the cylinder. We label the ring as electrode 1, and calculatethe charge ΔQ1 = q(z)dz induced on this ring when the positive ion charge q0 is atposition (r0, 0, z0) in cylindrical coordinates (r, θ, z).

According to the prescription (10) given in the statement of the problem,

ΔQ1 = −q0VC(r0, 0, z0), (96)

where case C now consists of a cylinder of radius b grounded except for the ring atposition z1 at unit potential, and a grounded cylinder at radius a. For z not close to

Page 119: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 4 27

the ends of the cylinder, the end surfaces z = ±l/2 may be approximated as at groundpotential.

This problem is very similar to that discussed in sec. 5.36 of W.R. Smythe, Static andDynamic Electricity, 3rd ed. (Mcgraw-Hill, New York, 1968).

Laplace’s equation, ∇2φC(r) = 0 holds for the potential in the region a < r < b. Theproblem has azimuthal symmetry, so φC will be independent of θ. Since the planesz = ±l/2 are grounded, the longitudinal functions in the Fourier series expansion,

φC =∑n

Rn(r)Zn(z), (97)

must have the form Zn = sin 2nπz/l. The equation for the radial functions Rn(r)follows from Laplace’s equation as

d2Rn

dr2+

1

r

dRn

dr−(

2nπ

l

)2

Rn = 0. (98)

The solutions of this are the modified Bessel functions of order zero, I0(2nπr/l) andK0(2nπr/l). Both of these are finite on the interval a < r < b, so the expansion (97)will include them both.

The boundary condition that φC(a, θ, z) = 0 is satisfied by the expansion

φC =∑n

An

I0(2nπr/l)I0(2nπa/l)

− K0(2nπr/l)K0(2nπa/l)

I0(2nπb/l)I0(2nπa/l)

− K0(2nπb/l)K0(2nπa/l)

sin2nπz

l, (99)

where the form of the denominator is chosen to simplify the evaluation of the boundarycondition at r = b. Here, φC = 0, except of an interval dz long about z where it isunity. Hence, the Fourier coefficients are

An =2

lsin

2nπz1

ldz. (100)

In sum, the charge distribution q(z) on the cathode at radius b due to positive chargeq0 at (r0, 0, z0) follows from (96) and (98-99) as

q(z) = −2q0

l

∑n

I0(2nπr0/l)I0(2nπa/l)

− K0(2nπr0/l)K0(2nπa/l)

I0(2nπb/l)I0(2nπa/l)

− K0(2nπb/l)K0(2nπa/l)

sin2nπz

lsin

2nπz0

l. (101)

A numerical evaluation of (101) is illustrated in Fig. 1. As is to be expected, theinduced charge distribution on the cathode has characteristic width of order b−r0, thedistance of the positive charge from the cathode.

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Princeton University 1999 Ph501 Set 4, Solution 4 28

1.00.8

0.60.4

0.20.0

-1.0 -0.5 0.0 0.51.0

1.2

0.9

0.6

0.3

0.0

Z (cm)

R/R

cQ

(re

l.)

Figure 1: The induced charge distribution (101) on the cathode of a strawtube chamber of radius RC = 0.25 cm due to positive ion charge at radius R.

Page 121: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 5 29

5. This problem is posed on p. 363 of The Mathematical Theory of Electricity and Mag-netism, 5th ed., by James Jeans.

We will evaluate the resistance R via Ohm’s Law, R = V/I , by calculating the currentI that flows when a potential difference V is established between the two contacts.

For a thin disk, the current flow is 2-dimensional. Since J = σE, where J is the currentdensity and E is the electric field, the electric field is 2-dimensional also. And, sinceE = −∇φ, where φ is the electric potential, the potential is 2-dimensional as well.

The form of the 2-dimensional potential is well approximated (for distances more thanδ/2 from the centers of the contacts) by considering a cylinder of radius a, rather thanthe disk, with a line charge density λ that passes through the center of one contact,and line charge −λ that passes through the center of the other contact.

The electric field from the wire of charge density λ has magnitude

E1 =2λ

r1, (102)

according to Gauss’ law, where r1 is the distance from the wire to the observer. Thecorresponding electric potential is

φ1 = 2λ lnr1

r0, (103)

where r0 is a constant of integration. The potential due to the wire with charge density−λ is similarly

φ2 = −2λ lnr2

r0

, (104)

where r2 is the distance from the observer to wire 2. The potential at an arbitrarypoint is then given by

φ = φ1 + φ2 = 2λ lnr1

r2. (105)

The total potential difference between the two line charges is formally divergent. Tomake physical sense, we can suppose that expression (105) holds only for r1 and r2

greater than δ/2, the half width of the electrical contacts, and the potential is essentiallyconstant for smaller values of r1 and r2. That is, we approximate the contacts of widthδ by perfectly conducting wires of radii δ/2, as shown in the figure below. Then, thepotential of contact 2 is estimated from eq. (105) by setting r1 = d− δ/2 and r2 = δ/2

φ(contact 2) = 2λ lnd − δ/2

δ/2≈ 2λ ln

2d

δ. (106)

The potential at the surface of contact 1 is just the negative of eq. (106), so the potentialdifference is

V ≈ 4λ ln2d

δ. (107)

Page 122: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 5 30

We note that the current and the electric field must be tangential to the edge of the disk.We recall that the equipotentials of eq. (105) are circles, and that the correspondingelectric field lines are also circles which, of course, pass through the line charges. Hence,the boundary condition on the electric field at the edge of the disk is indeed satisfied.

To complete the solution, we must calculate the current I that is flowing. For this, wecan integrate the current density J across any surface between the two contacts. Forconvenience, consider a cylindrical surface of radius r centered on one of the contacts,such that δ/2 < r d. Since r d, this surface is essentially an equipotential, andthe electric field is essentially that due to the nearby charge density λ. Namely, theelectric field is normal to this surface, with magnitude

E =2λ

r. (108)

The current density across this surface is given by J = σE. Restricting the problemto a disk of thickness t, the relevant area of the surface is πrt, so the total current is

I = πrt · σ · 2λ

r= 2πσλt, (109)

which is independent of the choice of r.

Finally, the resistance is found by combining eqs. (107) and (109):

R =V

I≈ 4λ ln 2d/δ

2πσλt=

2

πσtln

2d

δ, (110)

independent of the radius a of the disk.

Page 123: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 6 31

6. The series expansion approach is unsuccessful in treating the full problem of a “checker-board” array of two phases if those phases meet in sharp corners as shown above. How-ever, an analytic form for the electric potential of a two-phase (and also a four-phase)checkerboard can be obtained using conformal mapping of certain elliptic functions; seeR.V. Craster and Yu.V. Obnosov, Checkerboard composites with separated phases, J.Math. Phys. 42, 5379 (2001).4 If the regions of one phase are completely surroundedby the other phase, rather lengthy series expansions for the potential can be given;see Bao Ke-Da, Jorger Axell and Goran Grimvall, Electrical conduction in checker-board geometries, Phys. Rev. B 41, 4330 (1990).5 The present problem is based onM. Soderberg and G. Grimvall, Current distribution for a two-phase material withchequer-board geometry, J. Phys. C: Solid State Phys. 16, 1085 (1983),6 and JosephB. Keller, Effective conductivity of periodic composites composed of two very unequalconductors, J. Math. Phys. 28, 2516 (1987).7

In the steady state, the electric field obeys ∇ × E = 0, so that E can be deducedfrom a scalar potential φ via E = −∇φ. The steady current density obeys ∇ · J = 0,and is related to the electric field by Ohm’s law, J = σE. Hence, within regions ofuniform conductivity, ∇ · E = 0 and ∇2φ = 0. Thus, we seek solutions to Laplace’sequations in the four regions of uniform conductivity, subject to the stated boundaryconditions at the outer radius, as well as the matching conditions that φ, E‖, and j⊥are continuous at the boundaries between the regions.

We analyze this two-dimensional problem in a cylindrical coordinate system (r, θ) withorigin at the corner between the phases and θ = 0 along the radius vector that bisectsthe region whose potential is unity at r = a. The four regions of uniform conductivityare labeled I , II , III and IV as shown below.

Since J⊥ = Jr = σEr = −σ∂φ/∂r at the outer boundary, the boundary conditions atr = a can be written

φI(r = a) = 1, (111)

4http://puhep1.princeton.edu/~mcdonald/examples/EM/craster_prsla_456_2741_00.pdf5http://puhep1.princeton.edu/~mcdonald/examples/EM/ke-da_prb_41_4330_90.pdf6http://puhep1.princeton.edu/~mcdonald/examples/EM/soderberg_jpc_16_1085_83.pdf7http://puhep1.princeton.edu/~mcdonald/examples/EM/keller_jmp_28_2516_87.pdf

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Princeton University 1999 Ph501 Set 4, Solution 6 32

∂φII(r = a)

∂r=

∂φIV (r = a)

∂r= 0, (112)

φIII(r = a) = −1. (113)

Likewise, the condition that J⊥ = Jθ = σEθ = −(σ/r)∂φ/∂θ is continuous at theboundaries between the regions can be written

σ1∂φI(θ = α)

∂θ= σ2

∂φII(θ = α)

∂θ, (114)

σ1∂φIII(θ = π − α)

∂θ= σ2

∂φII(θ = π − α)

∂θ, (115)

etc.

From the symmetry of the problem we see that

φ(−θ) = φ(θ), (116)

φ(π − θ) = −φ(θ), (117)

and in particular φ(r = 0) = 0 = φ(θ = ±π/2).

We recall that two-dimensional solutions to Laplace’s equations in cylindrical coordi-nates involve sums of products of r±k and e±ikθ, where k is the separation constantthat in general can take on a sequence of values. Since the potential is zero at theorigin, the radial function is only rk. The symmetry condition (116) suggests that theangular functions for region I be written as cos kθ, while the symmetry condition (117)suggests that we use sin k(π/2 − |θ|) in regions II and IV and cos k(π − θ) in regionIII . That is, we consider the series expansions

φI =∑

Akrk cos kθ, (118)

φII = φIV =∑

Bkrk sin k

2− |θ|

), (119)

φIII = −∑Akrk cos k(π − θ). (120)

The potential must be continuous at the boundaries between the regions, which requires

Ak cos kα = Bk sin k(

π

2− α

). (121)

The normal component of the current density is also continuous across these bound-aries, so eq. (114) tells us that

σ1Ak sin kα = σ2Bk cos k(

π

2− α

). (122)

On dividing eq. (122) by eq. (121) we find that

tan kα =σ2

σ1cot k

2− α

). (123)

Page 125: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 6 33

There is an infinite set of solutions to this transcendental equation. When σ2/σ1 1we expect that only the first term in the expansions (118)-(119) will be important, andin this case we expect that both kα and k(π/2 − α) are small. Then eq. (123) can beapproximated as

kα ≈ σ2/σ1

k(π2− α)

, (124)

and hence

k2 ≈ σ2/σ1

α(π2− α)

1. (125)

Equation (121) also tells us that for small kα,

Ak ≈ Bkk(

π

2− α

). (126)

Since we now approximate φI by the single term Akrk cos kθ ≈ Akr

k, the boundarycondition (111) at r = a implies that

Ak ≈ 1

ak, (127)

and eq. (126) then gives

Bk ≈ 1

kak(π2− α)

Ak. (128)

The boundary condition (112) now becomes

0 = kBkak−1 sin k

2− θ

)≈ k(π

2− θ)

a(π2− α)

, (129)

which is approximately satisfied for small k.

So we accept the first terms of eqs. (118)-(120) as our solution, with k, Ak and Bk

given by eqs. (125), (127) and (128).

In region I the electric field is given by

Er = −∂φI

∂r≈ −k

rk−1

akcos kθ ≈ −k

rk−1

ak, (130)

Eθ = −1

r

∂φI

∂θ≈ k

rk−1

aksin kθ ≈ k2θ

rk−1

ak. (131)

Thus, in region I , Eθ/Er ≈ kθ 1, so the electric field, and the current density, isnearly radial. In region II the electric field is given by

Er = −∂φII

∂r≈ −k

rk−1

kak(π2− α)

sin k(

π

2− θ

)≈ −k

rk−1

ak

π2− θ

π2− α

, (132)

Eθ = −1

r

∂φII

∂θ≈ k

rk−1

kak(π2− α)

cos k(

π

2− θ

)≈ rk−1

ak(π2− α)

. (133)

Page 126: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 6 34

Thus, in region II , Er/Eθ ≈ k(π/2 − θ) 1, so the electric field, and the currentdensity, is almost purely azimuthal.

The current density J follows the lines of the electric field E, and therefore behaves assketched below:

The total current can be evaluated by integrating the current density at r = a in regionI :

I = 2a∫ α

0Jrdθ = 2aσ1

∫ α

0Er(r = a)dθ ≈ −2kσ1

∫ α

0dθ = −2kσ1α = −2

√σ1σ2απ2− α

.

(134)In the present problem the total potential difference Δφ is -2, so the effective conduc-tivity is

σ =I

Δφ=

√σ1σ2απ2− α

. (135)

For a square checkerboard, α = π/4, and the effective conductivity is σ =√

σ1σ2. Itturns out that this result is independent of the ratio σ2/σ1, and holds not only forthe corner region studied here but for the entire checkerboard array; see Joseph B.Keller, A Theorem on the Conductivity of a Composite Medium, J. Math. Phys. 5,548 (1964).8

8http://puhep1.princeton.edu/~mcdonald/examples/EM/keller_jmp_5_548_64.pdf

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Princeton University 1999 Ph501 Set 4, Solution 7 35

7. Taking the wire that carries current I to be along the z axis, the magnetic field atdistance r from the wire is

B =2I

crθ. (136)

The force on an element dl of the loop that carries current I ′ is

dF =I ′

cdl × B. (137)

The torque about the axis of the loop due to that force element is

dN = sn× dF , (138)

where s is the distance from the axis to the element and n is the unit vector from theaxis to the element. Combining (136-138), the total torque on the loop is

N =2II ′

c

∮sn× (dl × θ)

r. (139)

The two sides of length 2b will have equal and opposite contributions which thereforecancel, leaving only the contributions from the two sides of length 2a. For the sidenearest the wire, s = b, and dl = zdl, since the currents flow in the same direction, sothat dl× θ = −rneardl (parallel currents attract),

sn× (dl × θ) = −bn× rneardl = −b sinαzdl = −bd sin θ

rnear

zdl, (140)

using the law of sines for the triangle shown.

The distance r from the wire to this side of the loop is

rnear =√

b2 + d2 − 2bd cos θ, (141)

so that the contribution of this side of the loop to the torque is

Nnear = −4abdII ′

c

sin θ

b2 + d2 − 2bd cos θz. (142)

For the side furthest the wire, again s = b, but dl = −zdl, so dl × θ = rfardl. Then,

sn × (dl × θ) = bn× rfardl = −b sinβzdl = −bd sin θ

rfarzdl. (143)

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Princeton University 1999 Ph501 Set 4, Solution 7 36

andrfar =

√b2 + d2 + 2bd cos θ, (144)

so that the contribution of the far side of the loop to the torque is

Nfar = −4abdII ′

c

sin θ

b2 + d2 + 2bd cos θz. (145)

The total torque on the loop is then,

N = Nnear + Nfar = −8abdII ′

c

(b2 + d2) sin θ

b4 + d4 − 2b2d2 cos 2θz, (146)

where the identity 2 cos2 θ = cos 2θ + 1 has been used. This torque is down the axis,that is, it acts to decrease θ and bring the loop into the plane of the wire and the axis.

Page 129: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 8 37

8. a) The magnetic field on the axis of the loop centered at z = b is obtained from theBiot-Savart law:

B1 =I

c

∮dl × r

r3=

2πIa2

c(a2 + (z − b)2)32

z, (147)

and that due to the other loop is

B2 =2πIa2

c(a2 + (z + b)2)32

z. (148)

The total field is then the sum of the (147) and (148). This is unchanged underz → −z, so all odd derivatives with respect to z automatically vanish at the origin.We can choose the separation b to cancel any desired even derivative at the origin.

We first accumulate a catalog of derivatives (some of which are needed in prob. 6):

cBz(z)

2πIa2=

1

(a2 + (z − b)2)32

+1

(a2 + (z + b)2)32

, (149)

cB ′z(z)

2πIa2= − 3(z − b)

(a2 + (z − b)2)52

− 3(z + b)

(a2 + (z + b)2)52

, (150)

cB′′z (z)

2πIa2=

12(z − b)2 − 3a2

(a2 + (z − b)2)72

+12(z + b)2 − 3a2

(a2 + (z + b)2)72

. (151)

cB′′′z (z)

2πIa2= −90(z − b)3 − 15a2(z − b)

(a2 + (z − b)2)92

− 90(z + b)3 − 15a2(z + b)

(a2 + (z + b)2)92

, (152)

cB′′′′z (z)

2πIa2=

540(z − b)4 − 450a2(z − b)2 + 15a4

(a2 + (z − b)2)112

+540(z + b)4 − 450a2(z + b)2 + 15a4

(a2 + (z + b)2)112

, (153)

The second derivative of Bz with respect to z at the origin is proportional to

4b2 − a2

(a2 + b2)72

, (154)

which vanishes when a = 2b. That is, the separation of a pair of Helmholtz coil isequal to their radius.

b) In a Helmholtz coil pair, the field at the origin is proportional to I/a according to(149), so the 4rth derivative at the origin is proportional to I/a5. If we add a secondHelmholtz pair with current I ′ and radius a′ = 2/a, the combined 4th derivative atthe origin is proportional to I/a5 + 32I ′/a5. Hence, the current I ′ = −I/32 willcancel the 4rth derivative at the origin. The field at the origin is then proportional toI/a− (I/32)(2/a) = 15I/16. That is, the central field has been reduced by 1/16.

c) Far outside the coils, the leading behavior of the magnetic field is due to the dipolemoment, 2πIa2/c, of the coils. If a second Helmholtz pair is placed at a′ = 2a to cancelthe dipole moment of the first pair, we need I ′ = −I/4. Since the central field of a

Page 130: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 8 38

Helmholtz pair varies as I/a, the combined central field will be I/a − (I/4)(1/2a) =7I/8a, i.e., 1/8 of the central field is lost to insure that the field far outside the coilsis extremely weak.

To characterize the field well outside a set of Helmholtz coils in more detail, we usethe variable u = 1/z, and expand about u = 0. From (149), and using a = 2b,

cB(0)z (u)

2πIa2=

u3

(1 − au + 5a2u2/4)32

+u3

(1 + au + 5a2u2/4)32

. (155)

Using the Taylor expansion,

1

(1 + ε)32

= 1 − 3

2ε +

15

8ε2 − 105

48ε3 + ...., (156)

we find that

B(0)z (u) =

4πIa2

c

(u3 +

75

16a3u6 + ...

). (157)

The leading term, u3 = 1/z3, is due to the dipole moment of the pair, which, of course,is proportional to Ia2. The next nonvanishing term, u6, is due to the hexadecupolemoment, which is proportional to Ia5. The brevity of this derivation hides that factthat the Helmholtz condition, a = 2b, served to cancel the octupole moment. Thequadrupole moment vanishes due to the symmetry of the coil pair.

Page 131: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 9 39

9. Since the divergence of the magnetic field vanishes, the proposed expansions (15) obey

∇ · B =1

r

∂Br

∂r+

∂Bz

∂z=∑n

[(n + 1)bnrn−1 + a(1)

n rn]

= 0, (158)

where a(m)(z) ≡ dma/dzm. For this to be true at all r, the coefficients of rn mustseparately vanish for all n. Hence,

b0 = 0, (159)

bn = − a(1)n−1

n + 1. (160)

Since the curl of the field vanishes,

(∇ × B)θ =∂Br

∂z− ∂Bz

∂r=∑n

(b(1)n rn − nanr

n−1)

= 0 , (161)

Again, the coefficient of rn must vanish for all n, so that

b(1)n = (n + 1)an+1. (162)

Using (162) in (160), we find

bn = − b(2)n−2

(n + 1)(n + 3). (163)

Since b0 vanishes, b2n vanishes for all n, and from (162), a2n+1 vanishes for all n. Then,using (163) in (162), we find

a2n = −a(2)2n−2

4n2. (164)

Repeatedly applying this to itself gives

a2n = (−1)n a(2n)0

22n(n!)2. (165)

Inserting this in (160), we get

b2n+1 = (−1)n+1 a(2n+1)0

22n+1(n + 1)(n!)2. (166)

Combining (165-166) with (15), we arrive at the desired forms (16-17) for the fields.

The axial field of a pair of Helmholtz coils has the form

a0(z) = B0 + B4z4 + ... (167)

The first four derivatives are

a(1)0 = 4B4z

3, a(2)0 = 12B4z

2, a(3)0 = 24B4z, a

(4)0 = 24B4. (168)

Page 132: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 9 40

From (165-166), the other non-vanishing functions through fourth order are

a2 = −3B4z2, a4 =

3B4

8, b1 = −2B4z

3, b3 =3B4

2z . (169)

The fields, correct to fourth order, are

Bz = B0 + B4z4 − 3B4r

2z2 +3B4

8r4 + . . . , (170)

Br = −2B4rz3 +

3B4

2r3z + . . . (171)

The constants B0 and B4 are obtained from the catalog of derivatives in prob. 5, using

Bz(0, z) = B0 + B4z4 = Bz(0, z) +

B′′′′z (0, z)z4

4!+ ... (172)

From (149),

B0 =4πIa2

c(a2 + b2)32

=32√

5πI

25ca, (173)

and from (153),

B4 =5πIa2(a4 − 30a2b2 + 36b4)

2c(a2 + b2)112

= −4864√

5πI

3125ca5= − 152

125a4B0 , (174)

using b = a/2. Since B4 < 0, the axial field Bz decreases as we move away from theorigin, as is to be expected.

These results are overly detailed for some purposes. If one is interested only in theleading behavior at small r, then (170-171) simplify to

Bz(r, z) ≈ Bz(0, z), Br(r, z) ≈ −r

2

∂Bz(0, z)

∂z. (175)

The result for Br also follows quickly from ∇ ·B = 0, according to eq. (158),

Br(r, z) = −∫ r

0r∂Bz(r, z)

∂zdr ≈ −

∫ r

0r∂Bz(0, z)

∂zdr = −r

2

∂Bz(0, z)

∂z. (176)

It is also instructive that the approximation (176) can be deduced quickly from theintegral form of Gauss’ law (without the need to recall the form of ∇ ·B in cylindricalcoordinates). Consider a Gaussian pillbox of radius r and thickness dz centered on(r = 0, z). Then,

0 =∫

B · dS ≈ πr2[Bz(0, z + dz) −Bz(0, z)] + 2πr dz Br(r, z)

≈ πr2 dz∂Bz(0, z)

∂z+ 2πr dz Br(r, z) , (177)

which again implies eqs. (175).

Page 133: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 10 41

10. We first seek a solution in rectangular coordinates, and expect that separation ofvariables will apply. Thus, we consider the form

Bx = f(x)g(y) cos kz, (178)

Bx = F (x)G(y) sinkz, (179)

Bz = A(x)B(y)C(z). (180)

Then∇ · B = 0 = f ′g cos kz + FG′ sin kz + ABC ′, (181)

where the ′ indicates differentiation of a function with respect to its argument. Equa-tion (181) can be integrated to give

ABC = −f ′gk

sin kz +FG′

kcos kz. (182)

The z component of ∇ × B = 0 tells us that

∂Bx

∂y= fg′ cos kz =

∂By

∂x= F ′G sin kz, (183)

which implies that g and F are constant, say 1. Likewise,

∂Bx

∂z= −fk sin kz =

∂Bz

∂x= A′BC = −f ′′

ksin kz, (184)

using (182-183). Thus, f′′ − k2f = 0, so

f = f1ekx + f2e

−kx. (185)

Finally,∂By

∂z= Gk cos kz =

∂Bz

∂y= AB ′C =

G′′

ksin kz, (186)

soG = G1e

ky + G2e−ky. (187)

The “boundary conditions” f(0) = B0 = G(0) are satisfied by

f = B0 cosh kx, G = B0 cosh ky, (188)

which together with (182) leads to the solution

Bx = B0 cosh kx cos kz, (189)

By = B0 cosh ky sin kz, (190)

Bz = −B0 sinh kx sin kz + B0 sinh ky cos kz, (191)

This satisfies the last “boundary condition” that Bz(0, 0, z) = 0.

However, this solution does not have helical symmetry.

Page 134: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 10 42

Suppose instead, we look for a solution in cylindrical coordinates (r, θ, z). We againexpect separation of variables, but we seek to enforce the helical symmetry that thefield at z + δ be the same as the field at z, but rotated by angle kδ. This symmetryimplies that the argument kz should be replaced by kz − θ, and that the field has noother θ dependence.

We begin constructing our solution with the hypothesis that

Br = F (r) cos(kz − θ), (192)

Bθ = G(r) sin(kz − θ). (193)

To satisfy the condition (24) on the z axis, we first transform this to rectangularcomponents,

Bz = F (r) cos(kz − θ) cos θ + G(r) sin(kz − θ) sin θ, (194)

By = −F (r) cos(kz − θ) sin θ + G(r) sin(kz − θ) cos θ, (195)

from which we learn that the “boundary conditions” on F and G are

F (0) = G(0) = B0. (196)

A suitable form for Bz can be obtained from (∇ ×B)r = 0:

1

r

∂Bz

∂θ=

∂Bθ

∂z= kG cos(kz − θ), (197)

soBz = −krG sin(kz − θ), (198)

which vanishes on the z axis as desired.

From either (∇ × B)θ = 0 or (∇ ×B)z = 0 we find that

F =d(rG)

dr. (199)

Then, ∇ ·B = 0 leads to

(kr)2 d2(krG)

d(kr)2+ kr

d(krG)

d(kr)− [1 + (kr)2](krG) = 0. (200)

This is the differential equation for the modified Bessel function of order 1. See,for example, M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions(National Bureau of Standards, Washington, D.C., 1964), sec. 9.6. Hence,

G = CI1(kr)

kr=

C

2

[1 +

(kr)2

8+ · · ·

], (201)

F = CdI1

d(kr)= C

(I0 − I1

kr

)=

C

2

[1 +

3(kr)2

8+ · · ·

]. (202)

Page 135: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 10 43

The “boundary conditions” (196) require that C = 2B0, so our second solution is

Br = 2B0

(I0(kr) − I1(kr)

kr

)cos(kz − θ), (203)

Bθ = 2B0I1

krsin(kz − θ), (204)

Bz = −2B0I1 sin(kz − θ), (205)

which is the form discussed by Blewett and Chasman.

For a realization of the axial field pattern (24), we consider a wire that carries currentI and is wound in the form of a helix of radius a and period λ = 2π/k. A suitableequation of this helix is

x1 = a sin kz, y1 = −a cos kz. (206)

The magnetic field due to this winding has a nonzero z component along the axis,which is not desired. Therefore, we also consider a second helical winding,

x2 = −a sin kz, y2 = a cos kz, (207)

which is offset from the first by half a period and which carries current −I . Thecombined magnetic field from the two helices has no component along their commonaxis.

The unit vector l1,2 that is tangent to helix 1(2) at a point

r′1,2 = (x′1,2, y

′1,2, z

′) = (±a sin kz′,∓a cos kz′, z′) (208)

has components

l1,2 =(±2πa cos kz′,±2πa sin kz′, λ)√

λ2 + (2πa)2, (209)

and the element dl′1,2 of arc length along the helix is related by

dl′1,2 = l′1,2dz′√

λ2 + (2πa)2

λ= dz′(±ka cos kz′,±ka sin kz′, 1). (210)

The magnetic field B at a point r = (0, 0, z) on the axis is given by

B(0, 0, z) =I

c

∫1

dl′1 × (r′1 − r)

|r′1 − r|3 − I

c

∫2

dl′2 × (r′2 − r)

|r′2 − r|3

=2Ia

c

∫ ∞

−∞dz′

[a2 + (z′ − z)2]3/2[x(k(z′ − z) sin kz′ + cos kz′)

+y(−k(z′ − z) cos kz′ + sin kz′)]

=2I

ca

∫ ∞

−∞dt

(1 + t2)3/2[x(kat sin(kat + kz) + cos(kat + kz))

+y(−kat cos(kat + kz) + sin(kat + kz))] (211)

=4Ik

c(x cos kz + y sin kz)

[1

ka

∫ ∞

0

cos kat

(1 + t2)3/2dt +

∫ ∞

0

t sin kat

(1 + t2)3/2dt

],

Page 136: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 10 44

where we made the substitution z′ − z = at in going from the second line to the third.Equation 9.6.25 of Abramowitz and Stegun tells us that

∫ ∞

0

cos kat

(1 + t2)3/2dt = kaK1(ka) , (212)

where K1 also satisfies eq. (200). We integrate the last integral by parts, using

u = sin kat, dv =t dt

(1 + t2)3/2, so du = ka cos kat dt, v = − 1√

1 + t2. (213)

Thus, ∫ ∞

0

t sin kat

(1 + t2)3/2dt = ka

∫ ∞

0

cos kat√1 + t2

dt = kaK0(ka) , (214)

using 9.6.21 of Abramowitz and Stegun. Hence

B(0, 0, z) =4Ik

c[kaK0(ka) + K1(ka)] (x cos kz + y sin kz). (215)

Both K0(ka) and K1(ka) have magnitudes ≈ 0.5e−ka for ka ≈ 1. That is, the fieldon the axis of the double helix is exponentially damped in the radius a for a fixedcurrent I .

Page 137: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 11 45

11. We analyze the magnetic field of the solenoid in cylindrical coordinates, (r, θ, z), withthe origin at the center of the solenoid and z axis along that of the solenoid.

First, the field at a point on the axis, (0, 0, z), to a current loop, with current dI ,centered on and perpendicular to the z-axis at z′ follows from the Biot-Savart law as

B(0, 0, z) =dI

c

∮dl × r

r3=

2πdIa2

c(a2 + (z − b)2)32

z, (216)

For the solenoid,dI = nI dz′ ,

where z′ runs from −l/2 to l/2, so the total field on the axis is

B(0, 0, z) =2πnIa2z

c

∫ l2

− l2

dz′

(a2 + (z − z′)2)32

=2πnIa2z

c

∫ l2−z

− l2−z

dz′

(a2 + z′2)32

=2πnI z

c

⎛⎝ l

2− z√

a2 + ( l2− z)2

+l2

+ z√a2 + ( l

2+ z)2

⎞⎠

=2πnI

c(cos θ1 + cos θ2)z , (217)

where θ1 is the angle between the z axis and the line joining the observation point,(0, 0, z) to the point (a, 0, l/2) on the end of the solenoid, etc.

For z a l, we use the next to last line of (217), and the Taylor expansion

1√1 + ε

= 1 − ε

2+

3ε2

8− 5ε3

16+ ..., (218)

to find that near the origin,

Bz(0, 0, z) ≈ 2πnI

c

(2 − 4a2

l2− 72a2z2

l4

). (219)

As noted as the end of prob. 6, the radial field near the axis can be obtained from theaxial field using (175). Hence,

Br(r, 0, z) ≈ 288πnIa2rz

cl4. (220)

Near the end of the solenoid at z = l/2,

cos θ1 = − sin(θ1 − π/2) ≈ −z − l/2

a, and cos θ2 ≈ l√

l2 + a2≈ 1 − a2

2l2.

(221)

Page 138: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 11 46

Then, from (217)

Bz(0, 0, z) ≈ 2πnI

c

(1 − a2

2l2− z − l/2

a

). (222)

Comparing with (219), we see that the axial field at the end of the solenoid is approx-imately 1/2 that at the center. The radial field at the end of the solenoid follows from(175) as

Br ≈ πnIr

ac. (223)

Page 139: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 12 47

12. a) As discussed on p. 98 of the Notes, a sphere of uniform magnetization has a uniformmagnetic field inside. As discussed on p. 93, the field associated with magnetizationdensity M can be thought of as arising from a magnetization current density, Jm =c∇×M, and a surface current density, Km = cM× n, where n is the outward normalfrom the bounding surface. For uniform magnetization, Jm = 0, while, if M = M z,then

Km = cM sin θ φ. (224)

Since this is to be produced by windings on the surface of the sphere, with the samecurrent flowing through each turn of the winding, the density of windings must beproportional to sin θ.

b) If the sphere has radius a and surface charge density σ and rotates with angularvelocity ω, its magnetic moment will be

μ =1

c

∫AreadI =

1

c

∫Area

dQ

T=

1

c

∫ π

0πa2 sin2 θ · σ2πa sin θa dθ

2π/ω

=πσωa4

c

∫ π

0sin3 θ dθ =

4πσωa4

3c=

Qωa2

3c, (225)

where Q = 4πσa2 is the total charge on the sphere.

For the classical model of the electron with magnetic moment μ = eh/2mc, the velocityat the equator is

v = ωa =3μc

Qa= 3

eh

2mcc1

e

mc2

e2=

3

2

hc

e2c =

3

2αc. (226)

Since the fine structure constant α = eh/c2, is ≈ 1/137, this velocity is more than twohundred times the speed of light!

Page 140: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 13 48

13. a) If n is the number density of alignable electrons, each of magnetic moment μ =eh/2mc, then the resulting bulk magnetic field strength is

B = 4πM = 4πnμ =2πneh

mc= 2πn

eh

m2c3mc2 =

2πnmc2

Bcrit

, (227)

where Bcrit = m2c3/eh = 4.4 × 1013 Gauss is the so-called QED critical field strength.Since mc2 = 0.511 MeV = 8.2 × 10−7 erg we have

n =2 × 104 · 4.4 × 1013

2π · 8.2 × 10−7= 1.7 × 1023/cm3. (228)

Iron has atomic weight A = 56 and mass density 8 g/cm3, so its number density is

natom =6 × 1023 · 8

56= 8.6 × 1022/cm2. (229)

Thus, two electrons per iron atom participate in its bulk magnetization.

b) The interaction energy U of two magnetic dipoles m1 and m2 of equal magnitudem separated by distance r2 − r1 = dz can be calculated by supposing that, say, dipole2 is held fixed while dipole 1 is brought into place from a large distance. The force ondipole 1 due to dipole 2 is then

F1 = ∇1(m1 · B2), (230)

which can be integrated to give the interaction energy

U = −∫

F1 · dr1 = −m1 · B2 = −m2 · B1, (231)

where the second form follows from a similar argument in which dipole 1 was held fixedwhile dipole 2 is moved into place. The field of a dipole is

B1 =3(m1 · r)r − m1

r3, (232)

so

U = −3(m1 · z)(m2 · z) − m1 · m2

d3. (233)

If dipole 1(2) makes angle θ1(2) to the z axis, and both lie in, say, the x-z plane, then

U(θ1, θ2) = −m2

d3[3 cos θ1 cos θ2 − cos(θ1 − θ2)]. (234)

This is a minimum for θ1 = θ2 = 0 or π, and

Umin = −2m2

d3. (235)

In the absence of an external field, the dipoles are aligned, and both are either parallelor antiparallel to the z axis.

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Princeton University 1999 Ph501 Set 4, Solution 13 49

We now add a uniform external magnetic field B that makes angle θ to the z axis inthe x-z plane. The interaction energy of the system is then,

U(θ1, θ2, θ) = −m2

d3[3 cos θ1 cos θ2−cos(θ1−θ2)]−mB[cos(θ−θ1)+cos(θ−θ2)]. (236)

First, consider when B is at right angles to the line of centers of the dipoles, θ = π/2.Then,

U(θ1, θ2) = −m2

d3[3 cos θ1 cos θ2 − cos(θ1 − θ2)] − mB[sin θ1 + sin θ2]. (237)

If the dipoles remain in their original orientation, say θ1 = θ2 = 0, then the interactionenergy U0 is still given by (235). Suppose the two dipoles rotate together towards B.Then,

U(θ1 = θ2) = −m2

d3[3 cos2 θ1 − 1] − 2mB sin θ1

=m2

d3

[3 sin2 θ1 − 2 − 2d3B

msin θ1

]. (238)

The dipoles will rotate from angle 0 to π/2 provided that U(θ1 = θ2) decreases mono-tonically along this path. Since (238) is a quadratic function of sin θ1, this requiresthat the minimum occur for sin θ1 ≥ 1. The critical condition is then

B =3m

d3, (239)

above which field strength the dipoles always align with the transverse B. Thus, (239)is the saturation magnetic field.

We note that for B < 3m/d3, the energy minimum occurs at sin θ1 = sin θ2 =d3B/3m < 1. From (237), it can be verified that this is the absolute minimum forall (θ1, θ2). So the equilibrium configuration falls short of full alignment with B untilthe field is larger than the saturation value (239).

Finally, we consider the case where the dipoles originally have θ1 = θ2 = 0, and externalfield B = −Bz is applied. We first suppose that the dipoles rotate together, θ1 = θ2,so from (236) the energy of an intermediate state is

U(θ1 = θ2) = −m2

d3

(3 cos2 θ1 − 1 − 2d3B

mcos θ1

). (240)

This form is concave downward in cos θ1, so the alignment can occur so long as themaximum occurs at cos θ1 ≥ 1. Thus, on this path the critical condition is again (239).

We might wonder whether the alternative path, θ2 = −θ1 leads to a lower critical field.From (236),

U(θ1 = −θ2) = −m2

d3

(cos2 θ1 + 1 − 2d3B

mcos θ1

). (241)

Page 142: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 13 50

This is also concave downward in cos θ1, so requiring the maximum to occur at cos θ1 =1 leads to the critical condition

B =m

d3. (242)

In our model of a permanent magnet as consisting of only 2 magnetic dipoles, wefind that it is favorable for the transition from one ferromagnetic (aligned dipoles)state to another at 180 due to application of an external field to pass through anantiferromagnetic state (anti-aligned dipoles). We leave it to statistical mechanics todecide whether this can occur in a system of a large number of magnetic dipoles.

Page 143: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 14 51

14. The technique of mapping two dimensional magnetostatic fields with conducting pa-per is NOT based on the analogy between electrostatics and magnetostatics that wasmentioned on p. 97 of the Notes. Rather, we start from (32-33). The 2-dimensionalmagnetic field derived from the vector potential Azz is

B = ∇ × A =

(∂Az

∂y,−∂Az

∂x, 0

). (243)

With the identification Az = kφ we can write

E = −∇φ = −(

∂φ

∂x,∂φ

∂y

)= −1

k

(∂Az

∂x,∂Az

∂y

)=

1

k(By, Bx) =

B × z

k. (244)

Hence, B is perpendicular to E and therefore parallel to equipotentials of φ.

When current I is feed into a region of conducting paint, it spreads out on the paperas described by current density J. Then, current conservation and Ohm’s law allow usto write

I z =∮

J × dl =∮

E

σ× dl =

1

∮(B × z) × dl =

z

∮B · dl =

4πIB

ckσz, (245)

where IB is the current needed to produce magnetic field B. Thus, if we set

k =4πIB

cσI, (246)

we can extract numerical values of B from the potential distribution φ on the paper.To be more precise, note that

Δφ = −∫

E · dl = −1

k

∫(B × z) · dl =

1

k

∫B · (dl × z) =

ΦB

k, (247)

where ΦB is the magnetic flux per unit length in z that passes between the end pointsof the integration.

As to the boundary conditions, first consider the patches of conducting paint. Theelectric field is perpendicular to the boundaries of these, and hence the model B isparallel to them. This is the magnetic boundary condition at a perfect conductor.

As mentioned in the statement of the problem, one can use a patch of conducting paintto simulate a surface on which the magnetic field is known to lie, thereby reducing theextent of the model.

The electric field is parallel to the edge of the conducting paper, and zero outside it,since E = J/σ. Hence, the model magnetic field is perpendicular to the edge of thepaper. Outside the paper, (244) and the vanishing of E would imply that B vanishesalso. This awkwardness can be avoided by supposing that we were actually modelingthe field H rather than B, and the paper corresponds to a region of permeabilityμ = 1, where B = H. Then, if we suppose that the region outside the paper has veryhigh permeability, the continuity of B⊥ at the boundary implies that H is extremely

Page 144: Electrodynamics California

Princeton University 1999 Ph501 Set 4, Solution 14 52

small outside the paper, which restores consistency of the actual electrical boundaryconditions with a class of magnetic boundary conditions.

Thus, we conclude that the conducting paper technique is particularly well suited formapping 2-dimensional magnetic fields in situations bounded by a high permeabilitymaterial, such as iron. Of course, the actual fields must not be so high that the ironsaturates and the effective permeability drops to near unity.

Page 145: Electrodynamics California

Princeton University

Ph501

Electrodynamics

Problem Set 5

Kirk T. McDonald

(1999)

[email protected]

http://puhep1.princeton.edu/~mcdonald/examples/

Page 146: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Problem 1 1

1. a) A charged particle moves in a plane perpendicular to a uniform magnetic field B.Show that if B changes slowly with time, the magnetic moment produced by the orbitalmotion of the charge remains constant. Show also that the magnetic flux through theorbit, Φ = πr2B is constant. These results are sometimes given the fancy name ofadiabatic invariants of the motion.

b) The Magnetic Mirror. Suppose instead, that the magnetic field is slightly non-uniform such that Bz increases with z. Then, if the charged particle has a smallvelocity in the z direction, it slowly moves into a stronger field. Again, we wouldexpect the flux through the orbit to remain constant, which means that the orbitalradius must decrease and the orbital velocity must increase. However, magnetic fieldswhich are constant in time cannot change the magnitude of the velocity, therefore vz

must decrease. If Bz increases enough, vz will go to zero, and the particle is “trapped”by the magnetic field. Write

v2 = v2z + v2

⊥ = v20 , (1)

where v⊥ is the orbital velocity and v0 is constant. Use the result of part a) to showthat

v2z(z) ≈ v2

0 − v2⊥(0)

Bz(z)

Bz(0). (2)

Page 147: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Problem 2 2

2. If one pitches a penny into a large magnet, eddy currents are induced in the penny,and their interaction with the magnetic field results in a repulsive force, according toLenz’ law. Estimate the minimum velocity needed for a penny to enter a long, 1-Tsolenoid magnet whose diameter is 10 cm.

You may suppose that the penny moves so that its axis always coincides with that ofthe magnet, and that gravity may be ignored. The speed of the penny is low enoughthat the magnetic field caused by the eddy currents may be neglected compared tothat of the solenoid. Equivalently, you may assume that the magnetic diffusion timeis small.

Page 148: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Problem 3 3

3. a) Diamagnetism. We consider a model of an atom in which the distance r from theelectron to the nucleus is somehow fixed, but the electron is free to orbit the nucleus.Then, if a field B is applied to the atom, an E.M.F. is induced around the orbit, whileB is changing, which generates a magnetic dipole moment m via the resulting motionof the electron. Show that

m = − e2r2

4mc2B , (3)

where e and m are the charge and mass of an electron, respectively. In bulk matter,with n atoms per unit volume, the magnetization M is then M = nm.

The magnetic susceptibility is defined by

M = χMH. (4)

Since B = μH, and also B = H + 4πM = (1 + 4πχM )H, we see that the diamagneticpermeability obeys μ < 1. Calculate χM = (μ − 1)/4π for hydrogen gas at S.T.P. andcompare with the measured value of −2.24 × 10−9.

b) In materials where B = μH, we claim that the magnetic energy is

Umag =1

∫B · H dVol =

1

∫B2 dVol − 1

2

∫B · M dVol . (5)

Use your analysis from part a) to show that the last term is just the kinetic energy ofthe electron’s motion induced by the field B.

Page 149: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Problem 4 4

4. a) A flip coil is a practical device for measuring magnetic fields. A coil whose axis isthe z-axis is flipped by 180 about the x-axis. The coil leads are connected to a chargeintegrator. Show that charge

Q =2Φ

R(6)

is collected in the flip, where Φ is the magnetic flux through the loop before (and after)flipping and R is the resistance of the integrator (plus coil).

b) A fancy flip coil is made by winding wire on the surface of a sphere such that theturns are distributed according to

dN ∝ sin θ dθ . (7)

(Recall prob. of set 4.) All turns are parallel to the x-y plane. For this coil, show that

Φ ∝∫

Bz dVol, (8)

the integration being over the interior of the sphere.

c) The field component Bz(r, θ, ϕ) obeys ∇2Bz = 0 inside the sphere, and so may beexpanded in a series of Legendre functions. However, Bz is not necessarily azimuthallysymmetric, so a slight generalization must be made:

Bz =∑m,n

Am,nrnPm

n (cos θ)e±imϕ , (9)

where n and m are integers, and the Pmn are the associated Legendre polynomials.

Note that P 0n(cos θ) = Pn(cos θ), the ordinary Legendre polynomials. Using this, show

that Φ ∝ Bz(0, 0, 0), so that the sin θ flip coil measures Bz at the center of the sphere,no matter how B varies over the sphere!

d) A sin θ coil is hard to build. Suppose we try to make do with a simple cylindricalcoil of radius a and height h. Show that if h =

√3a, all effects of the first, second

and third derivatives of the field vanish. With such a coil, accuracies of 1 in 104 maybe achieved. Hint: Expand Bz in rectangular coordinates and note that ∇ · B = 0,∇ ×B = 0 and hence ∇2B = 0.

Page 150: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Problem 5 5

5. A cylinder of dielectric constant ε rotates with constant angular velocity ω about itsaxis. A uniform magnetic field B is parallel to the axis, in the same sense as ω. Findthe resulting dielectric polarization in the cylinder and the surface and volume chargedensities, neglecting terms of order (ωa/c)2, where a is the radius of the cylinder.

Answer:

P =ε − 1

4πcεωBr (10)

where r is the radial vector out from the axis.

This problem can be conveniently analyzed by starting in the rotating frame. Consideralso the electric displacement D.

Page 151: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Problem 6 6

6. a) Show that the self- and mutual inductances of two circuits obey

L11L22 ≥ L212 (11)

by considering the magnetic energy

U =1

2L11I

21 +

1

2L22I

22 + L12I1I2. (12)

b) A toroidal coil of N turns has a circular cross-section of radius a; the central radiusof the coil is b > a. Show that the self-inductance is

L11 =8N2

c2(b−√

b2 − a2) sin−1 1√1 + 3b2

4a2

. (13)

c) A second circuit in the form of a single loop of radius > a links the toroid; the planeof the second circuit is the same as that of one of the turns of the toroid, and thatturn is entirely inside the new circuit. Calculate the mutual inductance L12 betweenthe toroid and the new circuit, and show that relation (11) is obeyed in this example.

Page 152: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Problem 7 7

7. a) A coaxial cable consists of a center wire of radius a surrounded by a thin conductingsheath of radius b > a. The region a < r < b is vacuum. Consider a circuit formed byjoining the two conductors at ±∞ to show that the self inductance per unit length is

L =2

c2

(1

4+ ln

b

a

). (14)

Assume the current is distributed uniformly within the center wire.

b) Suppose the axis of the sheath is a distance ε from the axis of the center wire.Calculate the self inductance accurate to terms in (ε/b)2.

Page 153: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Problem 8 8

8. a) A long cylinder of radius a has uniform magnetization M perpendicular to its axis.Find the magnetic fields B and H everywhere.

Let z be the axis of the cylinder and x the direction of the magnetization.

b) Suppose the cylinder is given a uniform velocity, v = vz, along its axis. Find theresulting charge density and electric field everywhere. You may ignore effects of order(v/c)2. You can check your result by noting that the Lorentz force on a charge at restwith respect to the cylinder should vanish.

Page 154: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Problem 9 9

9. An iron ring has a circular cross section of radius a, and average radius b a. However,the ring has a narrow gap from azimuth θ = 0 to h/b 1; the gap width is w. Atoroidal winding of N turns wraps around the ring.

Calculate the stored magnetic energy as a function of the current I in the windings andthe gap width w in a regime where the permeability of the iron is very large. Calculatethe force needed to keep the gap from closing.

Suppose the field in the gap were 15,000 Gauss, near the maximum that is readilyachieved in an iron core magnet. Express the force/area that tends to close the gap interms of atmospheric pressure.

Page 155: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Problem 10 10

10. Discuss the surface charges and flow of field energy in a cylindrical wire of radius a ofconductivity σ that carries current I distributed uniformly within the wire.

For definiteness, assume the current returns in a hollow conducting cylinder of innerradius b and very large outer radius. Then, the current density J and electric field Eare vanishingly small in the outer conductor, whose constant electrical potential maybe taken as zero.

Steps in the discussion: Find the magnetic field B everywhere. Find the electricpotential φ(r, θ, z) and electric field E first for r < a, and then for a < r < b. Defineφ(0, 0, 0) = 0 at the center of the wire.

Answer:

φ(a < r < b) = − Iz

πa2σ

ln(r/b)

ln(a/b). (15)

Find the surface charge density at r = a which is needed to shape the electric fieldinside the wire to be along z. When the current first begins to flow, the electric fieldis not yet uniform and free charge heads for the surface of the wire until the desiredstatic surface charge distribution is obtained.

A length l of the wire has resistance R = l/πa2σ and consumes power at the rate I2R.Show that the Poynting vector S = (c/4π)E × B at the surface of the wire providesthis power. Thus, according to Poynting, the power flows down the air gap and intothe side of the wire.

As Sommerfeld says, “Electromagnetic energy is transported without losses only innonconductors. ‘Conductors’ are nonconductors of energy, which is dissipated in Jouleheating.”

An alternative calculation of the surface charge density σ may be instructive. Considerfirst the question of how a tube of radius a of uniform axial electric field could be createdin the absence of the wire. A capacitor consisting of a pair of circular plates of radius ahas a very nonuniform field between the plates as their separation becomes large. Wewant the equipotentials to be perpendicular to the axis, and uniformly spaced, whichcould be approximately achieved by adding a set of conduting rings of radius a, spaceduniformly along the axis with potentials that vary linearly between the two end plates.The charge on a ring would be given by Q = CV , where C is he capacitance of a ring,and V is the desired potential of the ring.

The current-carrying wire is a kind of continuum limit of the above procedure. Thedesired potential inside the wire is φ(z) = −IRz. For the coaxial geometry of thepresent problem, calculate the capacitance per unit length between the wire of radiusa and the return conductor of radius b. Then calculate the charge per unit length, andthe surface charge density, on the wire via Q(z) = CV (z).

Page 156: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Problem11 11

11. Consider an air-core transformer in the form of two coaxial cylinders of length l andradii r1 < r2 l. Each cylinder is wrapped with Ni turns, and the total resistance ofcoil i is Ri.

a) Deduce the currents I1(t) and I2(t) in the coils when the primary coil 1 is driven byvoltage V1(t) = V0 cosωt. First, evaluate the self and mutual inductances, L1, L2 andM , and then solve the coupled circuit equations.

Calculate the time-average power dissipated in coil 2.

b) Evaluate the Poynting vector S to show that its time average is nonvanishing onlyfor r1 < r < r2, and that the total Poynting flux 2πrl 〈Sr〉 is just the power dissipatedin coil 2. What is the direction of S?

c) Consider coil 2 as the primary driven by voltage V2(t) = V0 cos ωt, and discuss therelation between the Poynting vector and the power dissipated in coil 1.

Page 157: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Problem 12 12

12. Feynman Disk Paradox. Consider a small coil centered on the origin that carries acurrent which sets up a magnetic dipole moment m = mz. A ring of radius a in theplane z = 0 has charge Q distributed uniformly on it. The ring is rigidly attached tothe coil, but the assembly is free to rotate about the z axis.

a) Calculate the initial angular momentum LEM in the electromagnetic field.

Use the multipole expansion for the potential of a ring of charge, pp. 58-59, to showthat

LEM,z =

2mQ/15ca, r < a,13mQ/15ca, r > a.

(16)

b) Now let the current in the coil decrease to zero. Calculate the field induced at thering, and the resulting torque to show that

Lmech,z =mQ

ca, (17)

once the moment m has vanished.

Hint: Since magnetic field lines always form loops, the flux through the ring is equaland opposite to that across the plane z = 0 outside the ring.

For yet another version of this problem, seehttp://puhep1.princeton.edu/~mcdonald/examples/feynman_cylinder.pdf

Page 158: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Problem 13 13

13. Consider particle with charge e and momentum P = Pz +P⊥ (P⊥ = 0) that is movingon average in the z direction inside a solenoid magnet whose symmetry axis is the z axisand whose magnetic field strength is Bz. Inside the solenoid, the particle’s trajectoryis a helix of radius R, whose center is at distance R0 from the magnet axis.

The longitudinal momentum Pz is so large that when the particle reaches the end ofthe solenoid coil, it exits the field with little change in its transverse coordinates. Thisbehavior is far from the adiabatic limit (c.f. Prob. 1) in which the trajectory spiralsaround a field line.

When the particle exits the solenoid, the radial component of the magnetic “fringe”field exerts azimuthal forces on the particle, and, in general, leaves it with a nonzeroazimuthal momentum, Pφ. Deduce a condition on the motion of the particle whenwithin the solenoid, i.e., on R, R0, Pz, P⊥, and Bz, such that the azimuthal momen-tum vanishes as the particle leaves the magnetic field region. Your result should beindependent of the azimuthal phase of the trajectory when it reaches the end of thesolenoid coil.

Hint: Consider the canonical momentum and/or angular momentum.

Page 159: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 1 14

Solutions

1. a) Since the particle moves in the plane perpendicular to the magnetic field , thevelocity v, the field B and the force F on the particle of mass m and charge q aremutually orthogonal. The orbit is a circle of radius r related by

F =mv2

r=

qvB

c, (18)

so long as B varies sufficiently slowly. Then,

r =mcv

qB(19)

and the magnetic moment due to this orbit is of magnitude

μ =πr2I

c=

πr2

c

qv

2πr=

qrv

2c=

mv2

2B. (20)

The vector μ is in the opposite direction to B, which can be considered as an exampleof Lenz’ Law.

If the field B varies with time, then an electric field is induced around the particle’sorbit as given by Faraday’s Law:

∮E · dl = 2πrE = −1

c

d

dt

∫B · dS = −πr2B

c. (21)

Thus,

E =rB

2c=

mvB

2qB, (22)

with E in the same direction as v. That is, if the magnetic field increases, the electricfield causes the particle to accelerate,

v =qE

m=

vB

2B. (23)

The solution to this isv ∝

√B , (24)

so that v2/B is constant, and hence the magnetic moment (20) is constant. The fluxlinked by the orbit,

Φ = πr2B =πm2c2v2

q2B, (25)

is also constant.

b) Suppose now that the magnetic field is constant in time, but varies in space. Forexample, consider a field that has azimuthal symmetry about the z axis, and Bz in-creasing with z. A charged particle with nonzero vz, moves along a kind of helix inthis field.

Page 160: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 1 15

The magnetic field at the z coordinate of the particle then varies as

dBz(z(t))

dt=

dBz

dzvz. (26)

If this change is slow, the analysis of part a) holds, and the particle’s motion varies soas to keep v2

⊥(z)/Bz(z) constant, so that

v2⊥(z) ≈ v2

⊥(0)Bz(z)

Bz(0). (27)

In writing this, we recall from prob. 6, set 4 that for magnetic fields with azimuthalsymmetry, Bz(r, z) ≈ Bz(0, z)−r2B

′′z (0, z)/4+ ..., and we ignore the radial dependence

for orbits with small r.

Of course, v2 = v2z + v2

⊥ remains constant as well, so we have

v2z(z) = v2

0 − v2⊥(0)

Bz(z)

Bz(0). (28)

The particle stops moving forward in z at the plane where

Bz(z) =v2

0

v2⊥(0)

Bz(0) > B(0) . (29)

Although the particle has vz = 0 at this plane, its v⊥ now equals v(0), so still there isa large Lorentz force in the −z direction, and the particle spirals its way back downthe z axis. Hence the term “magnetic mirror”.

A field configuration in which the axial field strength increases with |z| can trap chargedparticles near the origin. This is no contradiction to Earnshaw’s theorem, as a “mag-netic bottle” has no static equilibrium point, but relies on electrodynamics to trapparticles with constant, nonzero velocity.

Page 161: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 2 16

2. The penny has radius a and thickness Δz. For the motion as stated in the problem,the eddy current will flow in concentric rings about the center of the disk. Therefore,we first examine a ring of radius r and radial extent Δr.

The magnetic flux through the ring at position z is

Φ ≈ πr2Bz(0, z), (30)

whose time rate of change is

Φ = πr2Bz = πr2B ′zv, (31)

where ˙ indicates differentiation with respect to time, ′ is differentiation with respectto z, Bz stands for Bz(0, z), and v is the velocity of the center of mass of the ring.

The penny has electrical conductivity σ. Its resistance to currents around the ring is

R =2πr

σΔrΔz, (32)

so the (absolute value of the) induced current is

I =ER

cR=

σrB ′zvΔrΔz

2c, (33)

using Faraday’s law.

The azimuthal eddy current interacts with the radial component of the magnetic fieldto produce the axial retarding force. Close to the magnetic axis, we estimate the radialfield in term of the axial field according to

Br(r, z) ≈ r∂Br(0, z)

∂r= −r

2

∂Bz(0, z)

∂z≡ −rB ′

z

2, (34)

as can be deduced from the Maxwell equation ∇ ·B = 0, noting that on the magneticaxis ∂Br/∂r = ∂Bx/∂x = ∂By/∂y. Then, the retarding force on the ring is

ΔFz =2πrBrI

c= −πσr2BrB

′zvΔrΔz

c2≈ −πσr3(B ′

z)2vΔrΔz

2c2. (35)

Alternatively, we note that the kinetic energy lost by the penny appears as Jouleheating. Hence, for the ring analyzed above,

vΔFz =dU

dt= −I2R = −πσr3(B ′

z)2v2ΔrΔz

2c2, (36)

using eqs. (32) and (33), which agains leads to eq. (35).

The equation of motion of the ring is

dFz = −πσr3(B ′z)

2vΔrΔz

2c2= mv = 2πρrΔrΔz v′v, (37)

Page 162: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 2 17

where ρ is the mass density of the metal. We integrate this equation with respect toradius to find

− πσa4(B ′z)

2vΔz

8c2= πρa2Δz v′v, (38)

After dividing out the common factor πa2Δz v, we find

v′ = −σa2(B ′z)

2

8ρc2. (39)

For an estimate, we note that the peak gradient of the axial field of a solenoid ofdiameter D is about B0/D, and the gradient is significant over a region Δz ≈ D.Hence, on entering a solenoid the jet velocity is reduced by

Δv ≈ σa2B20

8c2ρD. (40)

The penny must have initial velocity v0 > Δv to enter the magnet.

A copper penny has a ≈ 1 cm, density ρ ≈ 10 g/cm3, electrical resistivity ≈ 10−6 Ω-cm, and therefore conductivity σ ≈ 9× 1017 Gaussian units. The minimum velocity toenter a 1-T = 104-G magnet with diameter D = 10 cm is then,

vmin ≈ 9 × 1017 · (1)2 · (104)2

8 · (3 × 1010)2 · 10 · 10 ≈ 125 cm/s. (41)

The case of a sphere rather than a disk has been presented in J. Walker and W.H. Wells,Drag Force on a Conducting Spherical Drop in a Nonuniform Magnetic Field, ORNL/TN-6976 (Sept. 1979).

Page 163: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 3 18

3. a) Assume that the electron of an atom is initially stationary and that the vector rbetween the electron and the nucleus is at a right angle to the direction of the increasingmagnetic field B = Bz. As the magnetic field is applied, an electric field is inducedaround a loop of radius r according to

E =rB

2c, (42)

which, as Lenz’ law decrees, will accelerate the electron to velocity v, given by

v =∫

v dt =∫ eE

mdt =

∫ erB

2mcdt =

erB

2mc, (43)

so that its magnetic dipole moment opposes the magnetic field. From the next to lastequality in (20), we have

m = −evr

2cz = −e2r2B

4mc2z = −r0r

2B

4z , (44)

where r0 = e2/mc2 = 2.8 × 10−13 cm is the classical electron radius.

With n atoms per unit volume, the magnetization is

M = nm = −nr0r2

4B. (45)

The magnetic susceptibility χM is related by

M ≡ χMH =χM

1 + 4πχM

B ≈ χMB , (46)

so

χM ≈ −nr0r2

4. (47)

Hence, the permeability, μ = 1 + 4πχM is less than one. For hydrogen, r is the Bohrradius, a0 = r0/α

2, and, at S.T.P., n ≈ 5.4 × 1019/cm3, so χM ≈ −1.1 × 10−10, whichis of the same order of magnitude as the stated value of −2.24 × 10−9.

b) In a volume V , there are N = nV electrons, and their kinetic energy is

T = N1

2mv2 =

nV e2r2B2

8mc2= −V

2M · B, (48)

using (43) and (45), which is just the second term in the expression (5) for the magneticenergy.

Page 164: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 4 19

4. a) As the coil flips, the total amount of flux cut by the coil is 2Φ. Then, since theE.M.F. E generated is proportional to the rate of change of flux, the charge integratedover time is

Q =∫

I dt =1

R

∫E dt =

1

cR

∫dΦ

dtdt =

1

cR

∫dΦ =

cR. (49)

b) The total magnetic flux through the turns, which are perpendicular to the z axis, is

Φ =∫ ∫ ∫

B · (z dx dy)dN. (50)

Since the density of turns obeys dN ∝ sin θ dθ, and dz = a sin θ dθ, where a the radiusof the sphere, we have dN ∝ dz. Hence, (50) becomes

Φ ∝∫

Bz dxdydz =∫

Bz dVol. (51)

c) Inserting the Legendre series (9) into (51), we have

Φ ∝∫ ∑

m,n

Am,nrnPm

n (cos θ)e±imϕr2 dr d cos θ dϕ. (52)

The integral over the azimuthal angle is∫ 2π

0e±imϕ dϕ = 2πδm0. (53)

The integral over the polar angle is then∫ 1

−1Pn(cos θ) d cos θ =

∫ 1

−1Pn(cos θ)P0(cos θ) d cos θ = 2δn0. (54)

The radial integral is just ∫ a

0r2 dr =

a3

3. (55)

Combining (52-55) then gives

Φ ∝ 4πa3

3A0,0 = Bz(0, 0, 0)Vol. (56)

d) The total flux linked by the coil is again given by (50), where now the density ofwindings is dN = ndz. Thus,

Φ = n∫

Bz dVol. (57)

In a Taylor expansion of Bz in rectangular coordinates about the center of the coil,the integral of odd-order terms will vanish because the cylinder is symmetrical underreflections. Hence, up to third order, the only terms which survive are the zeroth-orderterm, πa2hBZ(0, 0, 0), and the second-order term,

1

2

∑i

d2Bz

dx2i

∣∣∣∣∣0

∫x2

i dVol. (58)

Page 165: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 4 20

In current-free regions and static situations,

∇2Bz =∑

i

d2Bz

dx2i

= 0 , (59)

so (58) will vanish if the three integrals in the sum are equal. The x1 = x and x2 = yintegrals are automatically equal because of the symmetry of the cylinder, which meansthat for the term to vanish, we need

∫z2 dVol =

1

2

∫(x2 + y2) dVol =

1

2

∫r2 dVol (60)

⇒ πa2h3

12=

πa4h

4. (61)

Hence, we require that h =√

3a.

Page 166: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 5 21

5. The v×B force on an atom in the rotating cylinder is radially outwards, and increasinglinearly with radius, so we expect a positive radial polarization.

We begin our analysis in the rotating frame, in which any polarization charge densityis at rest and causes no additional magnetic field. Then, P′ = χE′, where E′ and P′

are the electric field and dielectric polarization in the rotating frame. If v = ωr c,then the electric field in the rotating frame is related to lab frame quantities by

E′ = E +v

c× B, (62)

where E is the electric field due to the polarization that we have yet to find. Sincepolarization is charge times distance, in the nonrelativistic limit the polarization is thesame in the lab frame and the rotating frame: P′ = P.

The velocity has magnitude v = ωr, and is in the azimuthal direction. Thus, v×B =ωBr, so that

P = χ(E +

ωB

cr)

. (63)

There are no free charges, so the electric displacement is zero:

D = 0 = E + 4πP. (64)

Thus, E = −4πP. Recalling that χ = (ε − 1)/4π, (63) leads to

P =ε − 1

4πcεωBr. (65)

The surface charge density is

σpol = P(a) · r =ε − 1

4πcεωBa, (66)

where a is the radius of the cylinder. As well as this surface charge density, there is avolume charge density,

ρpol = −∇ · P = −1

r

∂rPr

∂r= −ε− 1

2πcεωB, (67)

so that the cylinder remains neutral over all.

Both the surface and volume charge densities are proportional to v(r)/c, and are mov-ing at velocity v(r). Hence, the magnetic field created by these charges is of orderv2/c2, and we neglect it in this analysis.

This example is perhaps noteworthy in that a nonvanishing, static volume chargedensity arises in a charge-free, linear dielectric material. In pure electrostatics thiscannot happen, since P = χE together with ∇ ·D = 0 = ∇ · E + 4π∇ ·P imply thatρpol = −∇ · P = 0.

Page 167: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 5 22

We also offer an iterative solution. The axial magnetic field acts on the rotatingmolecules to cause a v × B force radially outwards. This can be described by aneffective electric field

E0 =ωB

cr. (68)

This field causes polarization

P0 = χE0 = χωB

cr. (69)

Associated with this is the uniform volume charge density

ρ0 = −∇ · P0 = −2χωB. (70)

According to Gauss’ Law, this charge density sets up a radial electric field

E1 = 2πρ0r = −4πχωBr. (71)

At the next iteration, the total polarization is

P1 = χ(E0 + E1) = χ(1 − 4πχ)ωB

cr. (72)

This causes additional charge density ρ2, which leads to additional electric field E2, ...

At the nth iteration, the polarization will have the form

Pn = knωB

cr. (73)

Then,ρn = −∇ · Pn = −2knωB, (74)

andEn+1 = 2πρnr = −4πknωBr. (75)

The effective electric field at iteration n + 1 is the sum of E0 due to the v × B forceand En+1 due to the polarization charge. Thus,

Pn+1 = χ(E0 + En+1) = χ(1 − πkn)ωB

cr. (76)

But by definition,

Pn+1 = kn+1ωB

cr. (77)

Hence,kn+1 = χ(1 − 4πkn). (78)

If this series converges to the value k, then we must have

k = χ(1 − 4πk), (79)

so that

k =χ

1 + 4πχ=

ε − 1

4πε, (80)

which again gives (10) for the polarization.

Page 168: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 6 23

6. a) In the expression (12) for the total magnetic energy in the circuits, let

I1 = −L12

L11I2. (81)

The total energy is then

U =1

2

(L22 − L2

12

L11

)I22 . (82)

Since the magnetic energy U is also given by∫

B2 dvol/8π, it must be non-negative.Hence, the factor in parentheses in (82) must be non-negative, and

L11L22 ≥ L212. (83)

b) The magnetic field due to current I in a toroid of N windings is azimuthal, and isconfined to the interior. Ampere’s Law gives the magnitude as

B(r) =2NI

cr, (84)

where r is the perpendicular distance from the axis. The self inductance L11 is relatedby NΦ1/cI , where Φ1 is the flux linked by one turn. Thus, for a toroid of central radiusb whose cross section is a circle of radius a,

L11 =2N2

c2

∫ a

−a

2dx√

a2 − x2

x + b. (85)

Substituting y = x + b,

L11 =4N2

c2

∫ b+a

b−a

dy√−y2 + 2by + a2 − b2

y.

=4N2

c2

[√−y2 + 2by + a2 − b2 − b sin−1 2(b − y)√

4a2 + 3b2

−√b2 − a2 sin−1 2by + 2a2 − 2b2

√4a2 + 3b2

]b+a

b−a

=8N2

c2(b−√

b2 − a2) sin−1 1√1 + 3b2

4a2

. (86)

c) To calculate the mutual inductance between the two circuits, we note that the secondloop links all the flux of the toroidal field, which we called Φ1 above. Hence,

L12 =Φ1

cI=

L11

N. (87)

If the second circuit has radius R, and is made of a wire of radius r0, then its selfinductance is

L22 =4πR

c2

(ln

8R

r0− 7

4

), (88)

Page 169: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 6 24

from p. 115b of the Notes. Hence, for the system of loop plus toroid,

L11L22

L212

=N2L22

L11

=πR

(ln 8R

r0− 7

4

)2(b −√

b2 − a2) sin−1 1√1+ 3b2

4a2

. (89)

The numerator is smallest when R = a, the minimum for which the second loop fullylinks the toroid. The denominator is largest when b = a and the toroid looks like adonut whose hole has shrunk to zero. Then,

L11L22

L212

∣∣∣∣∣min

=π(ln 8a

r0− 7

4

)2 sin−1

√47

. (90)

This expression equals unity when a = 1.06r0, i.e., when the second loop is alsoessentially a donut with no hole. However, the expression (88) for the self inductanceof a loop was deduced supposing that R r0. Since the general restriction (11) issatisfied using (88) for any R > 1.06r0, we infer that (88) is still reasonably accuratefor R only a few times r0.

Page 170: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 7 25

7. a) In cylindrical coordinates (r, θ, z), the magnetic field is azimuthal in a coaxial cablewhose axis is the z axis. When current I flows in the cable, whose solid inner conductorhas radius a and whose outer conductor is a cylindrical shell of radius b, the fieldstrength follows from Ampere’s law as

Bθ(r) =

⎧⎪⎨⎪⎩

2Ir/a2c, r ≤ a,2I/cr, a ≤ r ≤ b,0, r > b.

(91)

The energy per unit length along the cable of this magnetic field is

U =1

∫B2 dArea =

I2

2πc2

(∫ a

0

r2

a42πr dr +

∫ b

a

2πr dr

r2

)=

I2

c2

(1

4+ ln

b

a

)(92)

Since the energy can be expressed in terms of the self inductance L as U = 12LI2, we

obtain the result (14).

Alternatively, we can evaluate the self inductance as L = Φ/cI , where Φ is the magneticflux per unit length linked by the circuit. The flux linked for a < r < b is clearly

Φ(a < r < b) =∫ b

aBθ dr =

2I

c

∫ b

a

dr

r=

2I

cln

b

a. (93)

More care is required when discussing the region r < a. On p. 115a of the Notes wesaw that a consistent procedure for an extended current distribution is to average theflux linked by the various filamentary currents. In the present case, consider first afilament of area r′dr′dθ at (r′, θ). We can define the surface through which the flux isto be calculated as that portion of the shell of radius r′ that connects (r′, θ) with thepoint (r′, 0), plus the plane θ = 0 between r′ and a. Since the field is azimuthal, noflux is linked on the shell; all filaments on the same shell link the same flux. Thus,

Φ(r < a) =2I

c

1

πa2

∫ a

02πr′ dr′

∫ a

r′dr

r

a2=

2I

a2c

∫ a

0r′ dr′

(1 − r

′2

a2

)=

2I

4c(94)

Combining (93-94) and dividing by cI , we again arrive at (14).

b) It appears impossible to make an accurate estimate of the self inductance when theouter cylinder is off center by either of the methods used in part a). The reason is thatthe currents are no longer uniformly distributed over the surfaces of the cylinders, soit is hard to calculate the magnetic field properly.

A solution can be given for the closely related problem in which the inner conductor,as well as the outer conductor, is a cylindrical shell. Then, we know from transmissionline analysis (Lecture 13) that LC = 1/c2, where C is the capacitance per unit length.With some effort we then find that

L =2

c2

(ln

b

a− ε2

b2 − a2

), (95)

to O(ε2/b2). See my note An Off-Center “Coaxial” Cable (Nov. 21, 1999).http://puhep1.princeton.edu/~mcdonald/examples/coax.pdf

Page 171: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 7 26

Here we illustrate what happens if we follow the approachs of part a), assuming thecurrents are uniformly distributed over the two cylinders.

If the center of the outer cylinder is at (r, θ) = (ε, 0), then the surface of that cylinderfollows

b2 = r2 + ε2 − 2εr cos θ, (96)

or

r(θ) = ε cos θ +√

b2 − ε2 sin2 θ ≈ b + ε cos θ − ε2

2bsin2 θ. (97)

We first calculate the self inductance via the energy method. Inside the outer cylinderthe magnetic field is still given by the first two lines of eq. (91), but with r = b replacedby r(θ) from eq. (97). Outside the cylinder the field is not quite zero because themagnetic field vectors from the currents in the inner and outer cylinders have slightlydifferent magnitudes and directions. The vector from the center of the outer cylinder,(ε, 0) to a point (r, θ) has magnitude r′ ≈ r − ε cos θ, and makes angle ≈ (ε/r) sin θ tor. Hence, the magnetic field from the current in the outer cylinder is

B ≈ 2I

cr

(ε sin θ

r,−1 − ε cos θ

r

), (98)

and the total magnetic field outside the outer cylinder is

Boutside ≈ 2Iε

cr2(sin θ,− cos θ), (99)

so its magnitude is Boutside ≈ 2Iε/cr2.

The magnetic field energy per unit length along the axis is now

U =1

∫B2 dArea =

I2

2πc2

(∫ a

0

r2

a42πr dr +

∫ 2π

0dθ∫ r(θ)

a

r dr

r2+∫ 2π

0dθ∫ ∞

r(θ)

ε2r dr

r4

)

≈ I2

2πc2

2+∫ 2π

0dθ ln

[b

a

(1 +

ε

bcos θ − ε2

2b2sin2 θ

)]+

ε2

2

∫ 2π

0

b2

)

≈ I2

2πc2

2+∫ 2π

0dθ

(ln

b

a+

ε

bcos θ − ε2

2b2sin2 θ − ε2

2b2cos2 θ

)+

πε2

b2

)

=I2

c2

(1

4+ ln

b

a

)=

1

2LI2. (100)

Hence, we would conclude from the energy method that there is no change in theinductance to second order.

We contrast this with a calculation of the flux linked by the off-center coax. Thecontribution for r < a is again given by (94). For r > a but inside the off-centerouter cylinder, the magnetic field is still B = (0, 2I/cr). The flux through the region

Page 172: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 7 27

r(θ) > r > a varies with azimuth, so we average over filaments on the outer cylinder:

Φ(r(θ) > r > a) =2I

c

1

∫ 2π

0dθ∫ r(θ)

a

dr

r=

2I

2πc

∫ 2π

0dθ ln

[b

a

(1 +

ε

bcos θ − ε2

2b2sin2 θ

)]

≈ 2I

2πc

∫ 2π

0dθ

(ln

b

a+

ε

bcos θ − ε2

2b2sin2 θ − ε2

2b2cos2 θ

)

=2I

c

(ln

b

a− ε2

2b2

). (101)

Combining (94) and (101), we find that the self inductance is now

L =2

c2

(1

4+ ln

b

a− ε2

2b2

), (102)

to O(ε2/b2).

Comparing with the result (95), we infer that the calculation via the linked flux ismore accurate than that via the energy method when we use the incorrect assumptionof uniform current distributions.

Page 173: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 8 28

8. a) Since there are no free currents in the problem, ∇ × H = 0 and we can define amagnetic scalar potential such that H = −∇φ. As the cylinder is very long, we ap-proximate the problem as 2-dimensional: φ = φ(r, θ) in cylindrical coordinates (r, θ, z).

The source of the magnetic scalar potential is the imagined magnetic charges associatedwith the magnetization. Since M = M x, the volume charge density ρM = −∇ ·M = 0.However, at the surface of the cylinder at r = a, there is a density given by

σM = M · r = M cos θ. (103)

The potential is continuous at the boundary r = a, and Gauss’ law tells us that

4πσM = 4πM cos θ = Hr(r = a+)−Hr(r = a−) = −∂φ(r = a+)

∂r+

∂φ(r = a−)

∂r. (104)

The potential can be expanded as a harmonic series, but only the term in cos θ willcontribute in view of (104). Thus,

φ =

−Hr cos θ, r ≤ a,−H a2

rcos θ, r ≥ a,

(105)

satisfies continuity of the potential at r = a. Then, (104) also tells us that H = −2πM .

Inside the cylinder we have

φ(r < a) = 2πMx, (106)

H(r < a) = −2πM x = −2πM, (107)

B(r < a) = H + 4πM = 2πM. (108)

Outside the cylinder there is no magnetization, and

φ(r > a) = 2πMa2 cos θ

r, (109)

H(r > a) = B(r > a) =2πMa2

r2(cos θ r + sin θ θ). (110)

b) In case of a moving cylinder, the analysis of part a) holds in the rest frame of thecylinder. When the cylinder has velocity v = vz in the lab frame, there appears to bean electric field in the lab frame related by

E = −γv

c×B′ ≈ −v

c× B, (111)

where we ignore terms of order v2/c2, so the magnetic field B in the lab frame is thesame as the field B′ given by (108) and (110) in the rest frame. Regarding the sign in(111), we note that a charge which is at rest in the lab frame is moving with velocity−v in the rest frame of the magnetized cylinder, and so in the latter frame experiencesa Lorentz force −v/c× B.

Page 174: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 8 29

Thus,

E(r < a) = −2πMv

cy = −2πM

v

c(sin θ r + cos θ θ), (112)

E(r > a) =2πMva2

cr2(sin θ r − cos θ θ). (113)

There is an electric charge density on the surface of the cylinder given by

σ =1

[Er(r = a+) − Er(r = a−)

]=

Mv

csin θ . (114)

This can be thought of as arising from a polarization P related to the moving magne-tization by

P =v

c× M. (115)

See sec. 87 of Becker for a discussion of how M and P form a relativistic tensor.

Page 175: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 9 30

9. The magnetic induction B is related to the magnetic field H by B = μH, where μ isthe permeability. In the gap, μ = 1. The normal component of the magnetic inductionis continuous across the boundaries of the gap, since ∇ · B = 0. Thus,

Hgap = Bgap = Biron = μHiron. (116)

For a large permeability μiron, the magnetic field Hiron is negligible.

The magnetic field H at the center of the toroid is related by Ampere’s law as

∮H dl = Hgapw + Hiron(2πb − w) =

4πNI

c. (117)

With the neglect of the small quantity Hiron, we find

Hgap = Bgap = Biron ≈ 4πNI

cw. (118)

The magnetic energy is

U =1

∫B · H dVol ≈ πa2w

(4πNI

cw

)2

=2π2a2N2I2

c2w. (119)

The force tending to close the gap is

F = −dU

dw=

2π2a2N2I2

c2w2. (120)

The pressure can also be calculated via the Maxwell stress tensor as

Pgap =B2

gap

8π. (121)

If Bgap = 15, 000 Gauss, then

Pgap = 9 × 106 dyne/cm2 = 9 atmospheres. (122)

Page 176: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 10 31

10. The current density associated with a uniform current I in a wire of radius a whoseaxis is the z axis is

J =I

πa2z. (123)

Ohm’s law gives the electric field inside the wire as

E =J

σ=

I

πa2σz = IRz, (124)

where σ is the conductivity, and R = 1/πa2σ is the resistance per unit length of thewire.

The electric potential inside the wire is therefore,

φ(r < a) = −IRz, (125)

where we define φ(0, 0, 0) = 0.

For the region a < r < b, we suppose the potential satisfies separation of variables:

φ(a < r < b) = f(r)g(z). (126)

Continuity of the potential at r = a is satisfied by the form

φ(a < r < b) = −f(r)IRz. (127)

Substituting (127) into Laplace’s equation, ∇2φ = 0, we find that

1

r

d

drrdf

dr= 0, (128)

so f has the general solutionf = A + B ln r. (129)

The boundary conditions on the potential at r = a and b now require that f(a) = 1and f(b) = 0. Hence, f = ln(r/b)/ ln(a/b), and

φ(a < r < b) = −IRzln(r/b)

ln(a/b)= IRz

ln(r/b)

ln(b/a). (130)

The surface charge density σq at the surface of the wire is

σq =1

[Er(r = a+) − Er(r = a−)

]=

1

[−∂φ(r = a+)

∂r+

∂φ(r = a−)

∂r

]

= − IRz

4πa ln(b/a). (131)

The electric field is

E = −∇φ =

⎧⎨⎩

IRz, r < a,−IRzr/r ln(b/a) + IR ln(r/b)z/ ln(a/b), a < r < b,0, b < r.

(132)

Page 177: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 10 32

Figure 1: The solid curves show lines of Poynting flux S, and the dashed linesare the electric field E in the region between the wire and the outer conductor.Because the tangential component of E is continuous at the boundary r = a,and E = IRz for r < a, the field lines for r > a are bent towards positivez. For |z| < b the field lines leave positive surface charges at r = a and endon negative surface charges also at r = a; in loop circuits (b >∼ L) this is thegeneral behavior. From Electrodynamics by A. Sommerfeld (Academic Press,1952), p. 129.

The magnetic field follows from Ampere’s law:

Bθ(r) =

2Ir/a2c, r ≤ a,2I/cr, a ≤ r,

(133)

The Poynting vector is then,

S =c

4πE × B =

⎧⎪⎨⎪⎩−I2Rrr/2πa2, r < a,−I2R ln(b/r)r/2πr ln(a/b) − I2Rzz/2πr2 ln(b/a), a < r < b,0, b < r.

(134)The Poynting vector is radially inwards at the surface of the wire, and the energy fluxper unit length there is 2πaS(r = a) = I2R. That is, the Poynting flux energy thewire through its surface provides the I2R power loss to Joule heating.

The Poynting flux crossing a plane at constant z is

∫Sz dArea = − I2Rz

2π ln(b/a)

∫ b

a

2πr dr

r2= −I2Rz. (135)

Since the flux is zero at z = 0, we interpret (135) as indicating that the total Poyntingflux crossing a plane at constant z equals the power dissipated by the wire between 0and z. This flux exists in the region a < r < b, i.e., in the air (or vacuum) betweenthe conductors, rather than in the conductors themselves.

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Princeton University 1999 Ph501 Set 5, Solution 10 33

For the alternative calculation of the surface charge density, we note that the capaci-tance per unit length between the inner and outer conductors is

C =1

2 ln(b/a), (136)

so the charge per unit length needed to support the potential φ(a, z) = −IRz is

Q(z) = Cφ(z) = − IRz

2 ln(b/a), (137)

and the corresponding surface charge density is

σ =Q

2πa= − IRz

4πa ln(b/a), (138)

as previously found in eq. (131).

This argument helps us understand how the charge distribution and electric field inthe central region of the wire is insensitive to the physical details of the ends of thewire. The capacitance per unit length might be different from the expression (136)for a few wire diameters in z from the ends of the wire, but it is quite accurate overmost of the length of the wire. Hence, we are less surprised that the potential (130)was obtained without ever specifying the boundary conditions at the ends of the wire.Those boundary conditions only affect the potential very near the ends of the wire,and the potential over most of the wire must have the form (130) in any case.

The potential (130) can be thought of as a kind of zero-frequency mode of the cavitybetween the inner and outer conductors. This cavity more has a “natural” behaviorat the ends, found by inserting zend into eq. (130). We readily see that this radialpotential distribution would hold if the ends of the cable are terminated “naturally”in plates of uniform conductivity, so Er ∝ jr ∝ 1/r, and φ(r) ∝ ln r.

If the coaxial cable transmits energy from a source (battery) at one end to a load(resistor) at the other, there is a net momentum

∫dVol εμS/c2 stored in the fields,

which is very small due to the factor 1/c2. If the cable is an isolated system, then italso has an equal and opposite mechanical momentum. See,http://puhep1.princeton.edu/~mcdonald/examples/hidden.pdf

Page 179: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 11 34

11. a) If long coil 1 carries steady current I1, then the magnetic field inside that coil isaxial with magnitude

B1 =4πN1I1

cl, (139)

by an application of Ampere’s law, ignoring end effects. Outside the coil, the magneticfield is zero. The flux linked by coil 1 is therefore,

φ1 = N1πr21B1 =

4π2N21 r2

1I1

cl= cL1I1, (140)

so the self inductance of coil 1 is

L1 =4π2N2

1 r21

c2l. (141)

Similarly, the self inductance of coil 2 is

L2 =4π2N2

2 r22

c2l. (142)

The mutual inductance can be calculated via the flux linked in coil 2 when coil 1 carriescurrent I1. Since the magnetic field due to current I1 is zero outside coil 1, which isinside coil 2, we have

φ12 = N2πr21B1 =

4π2N1N2r21I1

cl= cMI1, (143)

so the mutual inductance is

M =4π2N1N2r

21

c2l. (144)

Since r2 > r1, we have L1L2 > M2.

In solving the coupled circuit equations in the presence of an oscillatory driving voltageat frequency ω, we use complex notation, and divide out the common factor eiωt. Thenthe symbols I1 and I2 are complex numbers such that the real current is Re I1e

iωt, etc.

The coupled equations are

V0 = I1R1 + I1L1 + I2M = I1R1 + iωI1L1 + iωI2M, (145)

0 = I2R2 + I2L2 + I1M = I2R2 + iωI2L2 + iωI1M. (146)

These are readily solved as

I1 =(R2 + iωL2)V0

R21 − ω2(L1L2 − M2) + iω(R1L2 + R2L1)

, (147)

I2 = − iωMV0

R21 − ω2(L1L2 − M2) + iω(R1L2 + R2L1)

, (148)

The time-average power dissipated in coil 2 is then,

〈P2〉 =|I2

2 |R2

2=

ω2M2R2V20

2[R2

1 − ω2(L1L2 − M2)]2+ ω2(R1L2 + R2L1)2

. (149)

Page 180: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 11 35

b) To calculate the Poynting vector S, we need the electric and magnetic fields. The(complex) magnetic field is

Bz(r) =

⎧⎪⎨⎪⎩

4π(N1I1 + N2I2)/cl, r < r1,4πN2I2/cl, r1 < r < r2,0, r < r2.

(150)

The electric field is azimuthal, as follows from Faraday’s law:

Eθ = − 1

2πrc

d

dt

∫ r

0Bz 2πr dr = − iω

rc

∫ r

0Bz r dr

=

⎧⎪⎨⎪⎩−2πiωr(N1I1 + N2I2)/c

2l, r < r1,−2πiω(r2

1N1I1 + r2N2I2)/c2lr, r1 < r < r2,

−2πiω(r21N1I1 + r2

2N2I2)/c2lr, r2 < r,

(151)

The Poynting vector is radial, and positive if both Eθ and Bz are positive. Its time-average value is 〈Sr〉 = (c/8π)ReE

θBz. For r < r1, EθBz is pure imaginary, so 〈Sr〉 = 0

here. Since Bz = 0 for r > r2, 〈Sr〉 = 0 here also. The remaining region gives

〈Sr(r1 < r < r2)〉 =c

8πRe

2πiω

c2lr(r2

1N1I1 + r2N2I

2)

4πN2I2

cl= −πr2

1ωN1N2

c2l2rIm(I

1I2)

=πr2

1ωN1N2

c2l2r

ωMR2V2

0

[R21 − ω2(L1L2 − M2)]

2+ ω2(R1L2 + R2L1)2

=ω2M2R2V

20

4πlr[R2

1 − ω2(L1L2 − M2)]2+ ω2(R1L2 + R2L1)2

=

1

2πrl〈P2〉 . (152)

Since 2πrl 〈Sr〉 is the power transported by the electromagnetic field across the cylinderof radius r and length l, we interpret the power consumed in the outer coil as flowingfrom the inner, driven coil.

c) If, instead, coil 2 is driven, then the solutions to the coupled equations are obtainedfrom (147-148) by swapping indices 1 and 2. Likewise, the power consumed in coil1 is obtained from (149) by the same swap of indices. The expressions (150-151) forthe electric and magnetic fields in terms of the currents remain the same, as does thefirst line of (152) for the Poynting vector. However, in the rest of (152) we must swapindices 1 and 2, and note the sign change that occurs in I

1 . Thus, we find

〈Sr(r1 < r < r2)〉 = − 1

2πrl〈P1〉 . (153)

Again, power flows from the driven coil to the load coil.

Page 181: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 12 36

12. a) The field angular momentum is given by

LEM =∫

r×Pfield dVol =1

4πc

∫r× (E×B) dVol =

1

4πc

∫[(r ·B)E− (r ·E)B] dVol.

(154)From the symmetry of the problem, we infer that the angular momentum will be alongthe z axis, and that the electric and magnetic field are independent of azimuth ϕ inspherical coordinates (r, θ, ϕ). Thus, we desire

LEM,z =1

2c

∫ ∞

0r3 dr

∫ 1

−1d cos θ(BrEz − ErBz). (155)

The magnetic field due to magnetic dipole mz is

B =3cos θr − z

r3m =

2cos θr + sin θθ

r3m. (156)

The components we need are

Br =2mP1(cos θ)

r3, and Bz = cos θBr − sin θBθ =

2mP2(cos θ)

r3. (157)

The electric field can be gotten from the electric potential φ of a charged ring, p. 59of the Notes with cos θ0 = 0:

φ =

⎧⎨⎩

Qa

∑n

(ra

)nPn(0)Pn(cos θ), r < a,

Qr

∑n

(ar

)nPn(0)Pn(cos θ), r > a.

(158)

Since Pn(0) = 0 for odd n, only even n terms contribute to the potential. The electricfield components are

Er = −∂φ

∂r=

⎧⎨⎩− Q

ar

∑n n

(ra

)nPn(0)Pn(cos θ), r < a,

Qr2

∑n(n + 1)

(ar

)nPn(0)Pn(cos θ), r > a.

(159)

Eθ = −1

r

∂φ

∂θ=

⎧⎨⎩− Q

ar

∑n

(ra

)nPn(0)P 1

n(cos θ), r < a,

−Qr2

∑n

(ar

)nPn(0)P 1

n(cos θ), r > a.(160)

Eϕ = 0, (161)

using the fact thatdPn(cos θ)

dθ= P 1

n(cos θ), (162)

where Pmn (cos θ) is an associated Legendre polynomial. See eq. (3.39) of Jackson.

We also need Ez = cos θEr − sin θEθ, for which it is useful to note two recurrencerelations (Gradshetyn and Ryzhik, 8.731.2 and 8.735.2):

cos θPn(cos θ) =(n + 1)Pn+1(cos θ) + nPn−1(cos θ)

2n + 1, (163)

sin θP 1n(cos θ) = n cos θPn(cos θ) − nPn−1(cos θ)

=n(n + 1)

2n + 1[Pn+1(cos θ) − Pn−1(cos θ)]. (164)

Page 182: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 12 37

Then

Ez =

⎧⎨⎩− Q

ar

∑n n

(ra

)nPn(0)Pn−1(cos θ), r < a,

Qr2

∑n(n + 1)

(ar

)nPn(0)Pn+1(cos θ), r > a.

(165)

so that

LEM,z =1

2c

∫ ∞

0r3 dr

∫ 1

−1d cos θ(BrEz −ErBz)

=mQ

c

⎧⎪⎨⎪⎩∫ a0 r3 dr 1

ar4

∑n n

(ar

)2Pn(0)

∫ 1−1 dμ [P2(μ)Pn(μ) − P1(μ)Pn−1(μ)]∫∞

a r3 dr 1r5

∑n(n + 1)

(ar

)2Pn(0)

∫ 1−1 dμ [P1(μ)Pn+1(μ) − P2(μ)Pn(μ)]

=mQ

c

⎧⎪⎨⎪⎩∫ a0 r3 dr 1

ar4 2(

ar

)2 (−12

) (25− 2

3

)∫∞a r3 dr 1

r5

[23− 3

(ar

)2 (−12

)25

]

=mQ

c

4

15a

∫ a0 r dr

23

∫∞a

drr2 + 3a2

5

∫∞a

drr4

=

2mQ15ac

, r < a,13mQ15ac

, r > a.(166)

where we have used the facts that P0(0) = 1, P2(0) = −1/2, and∫ 1−1 Pm(μ)Pn(μ) dμ =

2δmn/(2n + 1). Altogether,

LEM,z =mQ

ac, (167)

b) As the magnetic moment m goes to zero, the electromagnetic angular momentumvanishes. But, the consequent change in the flux through the charged ring results inan azimuthal electric field Eϕ around the ring, which causes a torque that increasesthe mechanical angular momentum:

dLmech,z

dt= aQEϕ = − aQ

2πac

dt. (168)

This integrates to

Lmech,z,final =Q

2πcΦinitial =

Q

2πc

∫ a

02πr dr Bz (169)

If we use (157) for Bz, the result diverges. However, this form does not correctlyaccount for the flux inside the small coil at the origin. We avoid this issue by notingthat, since ∇ · B = 0, the magnetic flux through the loop of radius a is the negativeof the flux across the plane z = 0 outside the loop. In that plane, cos θ = 0, and sinceP2(0) = −1/2, we have

Lmech,z,final = −Q

c

∫ ∞

ar dr Bz =

mQ

c

∫ ∞

a

dr

r2=

mQ

ac= LEM,z,initial. (170)

Remark: Equation (169) can be given another interpretation. The magnetic flux canbe expressed in terms of the vector potential:

Φinitial =∫

B · dS =∮

A · dl = 2πaAinitial,ϕ. (171)

Page 183: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 12 38

Thus,Lmech,z,final = (r ×Pfinal)z = aPfinal,ϕ = QAinitial,ϕ (172)

Since Pinitial,ϕ = 0 = Afinal,ϕ, we can write

[r ×

(P +

QA

c

)]z

= constant. (173)

This is the z component of the canonical angular momentum of a charged particle inan electromagnetic field. Hence, another view of the Feynman disk paradox is that itillustrates the conservation of canonical angular momentum.

Page 184: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 13 39

13. The key to this problem is conservation of canonical momentum, P + eA/c, where Ais the vector potential (in Gaussian units).

It turns out to be even more effective to consider the canonical angular momentum,which is L = r × (P + eA/c).

We want Pφ = 0 outside the magnet. This implies Lz = rPφ = 0 also. Therefore, weneed r(Pφ + eAφ/c) = 0 inside the magnet.

A solenoid magnet with field Bz has vector potential Aφ = rBz/2. To see this, recallthat the integral of the vector potential around a loop is equal to the magnetic fluxthrough the loop: 2πrAφ = πr2Bz.

For a particle with average momentum in the z direction, its trajectory inside themagnet is a helix whose center is at some radius RG (called R0 in the statement of theproblem) from the magnetic axis. The radius RB (called R in the statement of theproblem) of the helix can be obtained from F = ma:

mv2⊥

RB= e

v⊥c

Bz, (174)

so

RB =eBz

cP⊥. (175)

The direction of rotation around the helix is in the −z direction (Lenz’ law).

Since the canonical angular momentum is a constant of the motion, we can evaluateit at any convenient point on the particle’s trajectory. In particular, we consider thepoint at which the trajectory is closest to the magnetic axis. As shown in Fig. 2, thispoint obeys r = RG −RB , and so

Lz = (RG − RB)P⊥ +eBz

2c(RG − RB)2 =

(R2

G − R2B

) eBz

2c. (176)

Note that R2G − R2

B is the product of the closest and farthest distances between thetrajectory and the magnetic axis.

Hence, the canonical angular momentum vanishes for motion in a solenoid field ifand only if RG = RB , i.e., if and only if the particle’s trajectory passes through themagnetic axis.

We also see that if the trajectory does not contain the magnetic axis, the canonicalangular momentum is positive; while if the trajectory contains the magnetic axis, thecanonical angular momentum is negative.

Page 185: Electrodynamics California

Princeton University 1999 Ph501 Set 5, Solution 13 40

R

R

R

R - R < 0B

G

R GMagnetic Axis

R - RG

G

B

B

B

R + RG B

R + RG B

a) L > 0 b) L < 0

P

P

zz

Figure 2: The projection onto a plane perpendicular to the magnetic axisof the helical trajectory a charge particle of transverse momentum P . Themagnetic field Bz is out of the paper, so the rotation of the helix is clockwise fora positively charged particle. a) The trajectory does not contain the magneticaxis, and Lz > 0. b) The trajectory contains the magnetic axis, and Lz < 0.

Page 186: Electrodynamics California

Princeton University

Ph501

Electrodynamics

Problem Set 6

Kirk T. McDonald

(2001)

[email protected]

http://puhep1.princeton.edu/~mcdonald/examples/

Page 187: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 1 1

1. a) The intensity of sunlight at the Earth’s orbit is ≈ 1.4 × 106 erg/s/cm2. What sizechunk of earth (ρ ≈ 5 g/cm3) could be levitated without orbiting, but at the radius ofthe Earth’s orbit?

b) Newton’s Rings. Explain briefly whether a dark or bright spot appears at thecenter, when viewing the reflected and transmitted fringe patterns in the apparatussketched on the left. Ignore multiple reflections inside the glass in both parts b) andc).

c) Lloyd’s Mirror. Explain whether a dark or bright spot appears at the base of thescreen in the apparatus sketched on the right.

Page 188: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 2 2

2. Carry out the derivation of Fresnel’s equations by matching the fields at the dielectricboundary, as discussed on p. 143 of the Notes. Deduce the four ratios:

E0r

E0i

,E0t

E0i

(1)

for Ei polarized parallel and perpendicular to the plane of incidence.

The derivation of Fresnel’s equations from Maxwell’s equation was first performed byHelmholtz.

Page 189: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 3 3

3. Fresnel’s Rhomb.

Linearly polarized light can be converted to circularly polarized light, and vice versa,with Fresnel’s rhomb: a piece of glass cut in the shape of a rhombic prism.

If the glass has index of refraction n = 1.5, show that the angle θ must be ≈ 50.2 or53.3.

The effect is based on the phase change of totally internally reflected light. Hint: write

E0r = E0ie−iφ, (2)

with φ⊥(φ‖) for E ⊥ (‖) to the plane of incidence, and show that

tan

(φ⊥ − φ‖

2

)= −

cos θi

√sin2 θi − 1/n2

sin2 θi

. (3)

Page 190: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 4 4

4. (a) More amplitude analysis. On pp. 141-142 of the Notes, we considered reflectionand transmission at a dielectric boundary, using the amplitudes i, r and t, which areproportional to the electric fields of the incident, reflected, and transmitted waves,respectively.

We found that if i = 1, then

|r|2 =sin2(θ1 − θ2)

sin2(θ1 + θ2), for E ⊥ to the plane of incidence, (4)

and

|r|2 =tan2(θ1 − θ2)

tan2(θ1 + θ2), for E ‖ to the plane of incidence. (5)

Can we deduce relations for the phases of r and t, and not merely their amplitudes?See sec. 33-6 of Vol. I of the Feynman Lectures on Physics.

Consider now an inverse situation:

If i′ = 1, then conservation of energy tells us that

|r|2 + |t|2 = |r′|2 + |t′|2 . (6)

We can also consider something even more peculiar:

Page 191: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 4 5

This can be regarded as the “time reversal” of the original situation.

But, we also recognize this as the superposition of two more ordinary configurations:

In particular, if a wave of amplitude r were incident from side 1, then the transmittedwave would have amplitude rt in terms of our original definitions.

Show that this implies that r is real, while t′ = t.

Hint: first deduce that r′ = −r and |t′| = |t|. Then, multiply the relation 1 = r2 + tt′

by its complex conjugate....

(b) Dielectric Slab. Consider a plate of thickness d of a dielectric with index ofrefraction n2, surrounded by a medium of index n1 = 1. A wave of unit amplitude isincident from below. Multiple reflections occur whose interference leads to the reflectedand transmitted waves, being the sums of the amplitudes at the dashed wavefrontsshown in the figure.

Show that the waves corresponding to a ray and its next higher-order neighbor have aphase difference 2Δ due to the different path lengths they have traveled, where

Δ =2πd cos θ2

λ2, (7)

while the phase lag for the first transmitted ray compared to the case of no plate isΔ −Δ′, where

Δ′ =2πd cos θ1

λ1. (8)

Define the total reflected and transmitted amplitudes to be R and T , respectively.

Page 192: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 4 6

Sum the partial amplitudes, and use the results of part (a) to show that

R =r(1 − e2iΔ)

1 − r2e2iΔ, and T =

(1 − r2)ei(Δ−Δ′)

1 − r2e2iΔ, (9)

which obey energy conservation: |R|2 + |T |2 = 1.

The ratio of the reflected to transmitted amplitudes is

R

T=

r(1 − e2iΔ)

(1 − r2)ei(Δ−Δ′) = −2ieiΔ′r sinΔ

1 − r2(10)

so there is a phase difference of Δ′−π/2 between amplitudes R and T . An experimentthat would be sensitive to this phase difference could involve a second beam, incidenton the beam splitter at angle θ1, but from the opposite side of the beam splitterfrom the original input beam. Then, the transmitted part of the first beam wouldinterfere with the reflected part of the second beam. If the two input beams are “inphase” at, say, the midplane of the beam splitter, they would have a phase differenceof Δ′ at the surface of the splitter onto which the second beam is incident. So, whenconsidering the interference of the two beams, the phase Δ′ found in eq. (10) woulddrop out, and we should say that there is an effective phase difference of 90

between the reflected and transmitted amplitudes in a beam splitter offinite thickness. This 90 phase difference plays an important role when comparingthe classical and quantum behavior of a beam splitter:http://puhep1.princeton.edu/~mcdonald/examples/bunching.pdf

An argument due to Feynman [Chaps. 30-31 of Vol. 1 of the Feynman Lectures onPhysics; pp. 282-285, Lecture 23 of the Notes] shows that R is i times a positivenumber, and hence that r is negative. Then, eqs. (4) and (5) lead to

r = −sin(θ1 − θ2)

sin(θ1 + θ2), for E ⊥ to the plane of incidence, (11)

and

r = −tan(θ1 − θ2)

tan(θ1 + θ2), for E ‖ to the plane of incidence. (12)

The expressions (9) for R and T are valid even if n1 > n2. Then, for large enough θ1,we expect total internal reflection.

In the Notes, we found for this case that

cos θ2 = i√

(n1/n2)2 sin2 θ1 − 1, (13)

so eq. (7) gives

Δ =2πid

λ2

√(n1/n2)2 sin2 θ1 − 1 ≡ i

d

δ. (14)

Thus,

T =(1 − r2)e−d/δe−iΔ′

1 − r2e−2d/δ→ 0 as d → ∞. (15)

Page 193: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 4 7

But for finite thickness d the transmitted amplitude is nonzero, even though we foundno wave motion in medium 2 which traveled normal to the boundaries. This phe-nomenon is called “tunneling”.

Page 194: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 5 8

5. (a) Antireflection Coatings

Suppose the slab of dielectric of prob. 4 separates media of indices n1 and n3.

Show that the reflected and transmitted amplitudes obey

R =r12 + r23e

2iΔ

1 + r12r23e2iΔ, T =

t12t23eiΔ

1 + r12r23e2iΔ, (16)

where r12 is the amplitude for a single reflection at boundary 1-2, etc.

Consider the special case of normal incidence. Show that if n2 =√

n1n3 and d = λ2/4,then R = 0, which is a prescription for an antireflection lens coating.

Show also that if d = λ2/2 the R is independent of n2, and if in addition n1 = n3 thenR vanishes.

(b) Dielectric Mirrors

Can we make a good mirror by applying an appropriate dielectric coating on a plateof glass?

Not with only two layers, but consider a multilayer mirror. For example, if a medium4 exists beyond medium 3, then the reflection at the 2-3 boundary could be describedby

R23 =r23 + r34e

2iΔ3

1 + r23r34e2iΔ3, T23 =

t23t34eiΔ3

1 + r23r34e2iΔ3, (17)

⇒ T =t12t23t34e

i(Δ2+Δ3)

(1 + r23r34e2iΔ3)(1 + r12R23e2iΔ2), etc. (18)

Then, for a stack of n 2-3 pairs, 1-2-3-2-3- · · · -3-2-1,

T =t12t23t32t23 · · · t32t21e

ni(Δ2+Δ3)

big mess=

|t12|2 |t23|2n eni(Δ2+Δ3)

big mess, (19)

where the big mess is not small if Δ2 = Δ3 = π/4, since in that case 1 + r23r32e2iΔ3 =

1 + r223, etc. Since |t23|2n → 0 for large n, T → 0 and R → 1.

The prescription given here for a multilayer dielectric mirror works well for only anarrow range of angles of incidence and a narrow range of wavelengths, since we requirethat Δ2 = Δ3 = π/4. Better prescriptions can be given that maintain very highreflectivity over a large range of parameters. See, for example, J.P. Dowling, Science282, 1841, (1998).1

1http://puhep1.princeton.edu/~mcdonald/examples/optics/dowling_science_282_1841_98.pdf

Page 195: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 6 9

6. In 1890, O. Wiener carried out an experiment that can be said to have photographedelectromagnetic waves.

a) A plane wave is normally incident of a perfectly reflecting mirror. A glass photo-graphic plate is placed on the mirror at a small angle α. The polarization of the waveis parallel to the line of intersection of the mirror and the plate.

The photographic emulsion is almost transparent – ignore attenuation and reflectionin it and in the glass.

When the plate is developed a striped pattern is observed.

Calculate the electromagnetic fields E and B for y > 0, where y = 0 is the surface ofthe mirror. Predict the position and spacing of the dark stripes that appear on thedeveloped “negative” plate.

b) Repeat the discussion for waves incident at 45. That is, calculate E and B fory > 0, and predict the pattern of blackening on the negative.

Distinguish the case of E ⊥ and ‖ to the plane of incidence.

Page 196: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 7 10

7. Two airplanes are flying at distance d apart at height h above the ocean whose dielectricconstant is ε. One plane sends signals to the other. Both airplanes have short verticalantennae. Ignore the curvature of the Earth.

Show that the ratio of the intensity of the signal reflected off the ocean to that of thedirect signal is

d6

(d2 + 4h2)3

⎛⎝√

(ε − 1)d2 + 4εh2 − 2εh√(ε − 1)d2 + 4εh2 + 2εh

⎞⎠

2

. (20)

In addition to facts about plane waves you need to know that

• The intensity of spherical broadcast waves falls off as 1/r2. Over small spatialregions (except close to the source) the spherical waves can be considered as planewaves.

• The amplitude (E field) of a broadcast wave varies linearly with the projectionperpendicular to the line of sight of the motion (acceleration) of the charges thatcause the wave (p. 141 of the Notes).

• Likewise, the current I excited in a receiving antenna varies as the projection ofE onto the antenna.

• The power of the received signal is, of course, I2R, where R is the resistance inthe receiving antenna.

Page 197: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 8 11

8. (a) Plasma with a dc magnetic field.

Consider the Earth’s ionosphere to be a plasma of uniform density with a static, uni-form magnetic field B0 (the Earth’s field) in the +z direction. Discuss the propagationof circularly polarized plane radio waves parallel (or antiparallel) to B0.

The response of an ionized electron of charge −e and mass m at position r to the waveof angular frequency ω is described by

mr + er

c× B0 = −eEei(kz−ωt), (21)

where for circularly polarized waves the electric field amplitude can be written

E± = E0(x ± iy). (22)

Show that

r± = − eE

mω(ω ∓ ωB), (23)

where

ωB =eB0

mc, (24)

and that this implies a dielectric constant for the plasma of

ε± = 1 −ω2

p

ω(ω ∓ ωB), (25)

where the plasma frequency ωp is given by

ω2p =

4πNe2

m(26)

for a plasma of number density N per cm3.

Show that for waves of circular polarization x+iy (called left-handed in optics althougha “photograph” of the electric field vector would show it to behave like a right-handedscrew),

vgroup = 2vphase = 2c

√ωωB

ωp. (27)

[See pp. 146a-d of the Notes for a discussion of group and phase velocity.]

It turns out that ωB ≈ ωp ≈ 107 Hz in the ionosphere. Estimate the difference inarrival times for signals of 105 and 2× 105 Hz originating simultaneously (in lightningflashes) at the opposite side of the Earth. This illustrates the “whistler” or “chirp”effect – as higher frequency waves arrive first.

What is the fate of waves of polarization x − iy?

Page 198: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 8 12

(b) Reflections and Mirages

In the ionosphere the density of ionized electrons actually increases with height: thelower atmosphere is screened from the Sun by the upper. Hence, the (frequency-

dependent) index of refraction decreases with height, since n(ω) ≈√

1 − (ωp/ω)2,where ωp is the plasma frequency.

Suppose at the bottom of the atmosphere, where n = 1, a radio wave propagatesupwards with angle θi to the vertical.

Use Snell’s law to show that the wave is reflected back downwards if the electron densityrises until ωp = ω cos θi at some height.

Mirages are a similar phenomenon in which higher temperatures in the air close to theEarth’s surface result in lower density (N = RT/P ) at lower height, and hence lower

index at lower height for optical frequencies (n ≈√

1 + ω2p/(ω

20 − ω2)), so downward

going light rays can be reflected upwards near the surface.

Page 199: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 9 13

9. A plane electromagnetic wave of angular frequency ω is normally incident on a goodconductor that occupies the region z > 0. Show that the Poynting vector 〈S〉 evaluatedat z just greater than zero is equal to the power (per unit area ⊥ to z) lost to Jouleheating in the conductor.

Page 200: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 10 14

10. A plane electromagnetic wave of angular frequency ω is normally incident on a thinconducting sheet of thickness a d, the skin depth. Ignoring reflection, show that therelative transmitted intensity is

T = 1 − 4π

cσa, (28)

where σ is the conductivity.

Use an energy argument as in prob. 8.

Extending the argument to show that the relative reflected intensity R is (2πσa/c)2.

A “trick” derivation is to note that a sheet of unit area and thickness a has resistance

R =l

σA=

1

σa · 1 =1

σa(29)

to the induced currents that flow in the plane of the sheet. Hence, power V 2/R ifabsorbed is a wave of voltage V per unit length crosses the sheet. But, the powercarried by a plane wave is V 2/Rvac where Rvac = 4π/c = 377 Ω. Thus, the fractionalpower absorbed is Rvac/R = 4πσa/c.

Page 201: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 11 15

11. Total Internal Reflection

A wave of frequency ω is incident at angle θi on the boundary between dielectrics ofindices n1 > n2. Find the time-averaged Poynting vector of the transmitted wave whensin θi > n2/n1, i.e., when total internal reflection occurs.

Verify that the transmitted surface wave satisfies the wave equation and Maxwell’sequations.

Page 202: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 12 16

12. The grating accelerator

In optics, a reflective grating is a conducting surface with a ripple. For example,consider the surface defined by

z = a sin2πx

d. (30)

The typical use of such a grating involves an incident electromagnetic wave with wavevector k in the x-z plane, and interference effects lead to a discrete set of reflectedwaves also with wave vectors in the x-z plane.

Consider, instead, an incident plane electromagnetic wave with wave vector in the y-zplane and polarization in the x direction:

Ein = E0xei(kyy−kzz−ωt), (31)

where ky > 0 and kz > 0. Show that for small ripples (a d), this leads to a reflectedwave as if a = 0, plus two surface waves that are attenuated exponentially with z.What is the relation between the grating wavelength d and the optical wavelength λsuch that the x component of the phase velocity of the surface waves is the speed oflight, c?

In this case, a charged particle moving with vx ≈ c could extract energy from thewave, which is the principle of the proposed “grating accelerator” [R.B. Palmer, ALaser-Driven Grating Linac, Part. Accel. 11, 81-90 (1980)].2

2http://puhep1.princeton.edu/~mcdonald/examples/accel/palmer_pa_11_81_80.pdf

Page 203: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 13 17

13. A radiofrequency quadrupole (RFQ) is a device for focussing beams of chargedparticles. The electric field in this device can be approximated as that derived fromthe quasistatic potential

φ(x, y, t) =E0

2d(y2 − x2) sin ωt, (32)

where d is a length and ω is the frequency of the field. The magnetic field is ignored inthis approximation. While the approximate fields do not satisfy Maxwell’s equations,there is little error for |x|, |y| λ, the wavelength of the radiofrequency waves.

Deduce the equations of motion for a particle of charge e and mass m in the radiofre-quency quadrupole. Consider solutions of the form

x(t) = f(t) + g(t) sinωt (33)

where g f and both f and g are slowly varying compared to sinωt. The parametersmay be assumed to satisfy the conditions that such solutions exist.

Complete the solution for the particular case that

x(0) = 0, x(0) = v0θ0, (34)

y(0) = 0, y(0) = 0, (35)

z(0) = 0, z(0) = v0, (36)

with θ0 1. At what distance along the z-axis is the first image of the beam ‘spot’,i.e., where the initially diverging beam is brought back to the z-axis?

Page 204: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Problem 14 18

14. Pulsar Timing

The distance from the Earth to a pulsar can be estimated by observing the dispersionof the radio-frequency pulses as they cross the interstellar medium.

a) Suppose the medium is a plasma of N electrons/cm3. What is the index of refractionn(ω) where ω is the angular frequency of a wave?

b) The pulsar emits a short pulse that contains a broad range of frequencies. Weobserve the pulse in a receiver that is tuned to a narrow band δω about an adjustablecentral frequency ω. We measure the time difference δt between the arrival of twocomponents of the pulse, centered at frequencies ω and ω + δω, where δω ω.

This can be done in a single receiver if the pulsar has a precise pulse rate – as is thecase. Pulsars are the most accurately periodic macroscopic phenomenon ever observed!

Which component, ω or ω + δω, arrives first, and by how much, as a function of theEarth-pulsar distance L?

Use the following representative values to calculate the distance L to pulsar “1913+16”:ω = 2000 MHz, δω/ω = 0.01, N = 0.04 electrons/cm3, and |δt| = 0.004 s.

A pulsar tidbit: Many pulsars occur in binary systems, including one such systemwhere the direction to the Earth lies very close to the plane of the orbit. By observingthe small general-relativistic pulse delays that occur when one of the binary partnersoccults the other, the eccentricity of the orbit can be determined to remarkable accu-racy. The data indicate that the orbit has radius ≈ REarth−Sun and a small eccentricitycorresponding to being out of round by less than 1 cm! The orbit is round to one partin 1013, which makes it the roundest object ever measured – and it’s at the other endof the galaxy (F. Camilo, Princeton Ph.D. thesis, ≈ 1994).

Page 205: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 1 19

Solutions

1. a) For a black chunk of earth of radius r levitated at distance RE = 1.5 × 1013 cm insunlight of intensity I = 1.4 × 106 erg/s/cm2,

FG =GMSm

R2E

=4πr3ρGMS

3R2E

= Frad = πr2 I

c. (37)

Hence

r =3IR2

E

4cρGMS=

3 · 1.4 × 106 · (1.5 × 1013)2

4 · 3 × 1010 · 5 · 6.7 × 10−8 · 2 × 1033≈ 1.2 × 10−5 cm, (38)

which is less than a wavelength of light!

b) The center of the reflected spot is dark, being the interference between the reflectionoff the bottom surface of the lens and the top surface of the glass plate. There is a180 phase difference between these two cases. (The reflections off the top of the lensand bottom of the glass plate give a “background” intensity that is independent ofposition.) The center of the transmitted spot is bright, as there is no phase shiftduring transmission across a dielectric boundary.

c) The fringe pattern is dark at the base of the screen, due to interference of the directand reflected rays. The latter undergo a 180 phase shift on reflection.

Page 206: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 2 20

2. We consider plane waves of the form

E = E0ei(k·r−ωt),

H =B

μ=

n

μk × E =

√ε

μk × E, (39)

where k =√

εμω/c = nω/c in media with dielectric constant ε, index n =√

εμ, andpermeability μ.

At a boundary between two dielectrics, the perpendicular components of D and B,and the parallel components of E and H are continuous. The incident and reflectedwaves are in medium 1, and the transmitted wave is in medium 2. The unit normalvector pointing into medium 2 is labeled n. Then the boundary conditions are

n21

μ1

(E0i + E0r) · n =n2

2

μ2

E0t · n, (40)

n1(ki × E0i + kr × E0r) · n = n2kt × E0t · n, (41)

(E0i + E0r) × n = E0t × n, (42)n1

μ1

(ki × E0i + kr × E0r) × n =n2

μ2

kt × E0t × n, (43)

Of course, Snell’s law tells us that

n1 sin θ1 = n2 sin θ2. (44)

(a) Polarization perpendicular to the plane of incidence (the plane containing ki, kr

and kt).

Relation (40) is satisfied identically. Both relations (41) and (42) yield

E0i + E0r = E0t. (45)

Relation (43) tells us that

n1

μ1

(E0i − E0r) cos θ1 =n2

μ2

E0t cos θ2 =n1

μ2

sin θ1 cos θ2

sin θ2E0t, (46)

Page 207: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 2 21

and hence,

E0i − E0r =μ1

μ2

sin θ1 cos θ2

cos θ1 sin θ2E0t. (47)

Adding (45) and (47), we find that

E0t

E0i=

2n1 cos θ1

n1 cos θ1 + μ1

μ2n2 cos θ2

→ 2 sin θ2 cos θ1

sin(θ1 + θ2)if μ1 = μ2 = 1. (48)

Then, (45) leads to

E0r

E0i=

E0t

E0i− 1 =

n1 cos θ1 − μ1

μ2n2 cos θ2

n1 cos θ1 + μ1

μ2n2 cos θ2

→ −sin(θ1 − θ2)

sin(θ1 + θ2)if μ1 = μ2 = 1. (49)

(b) Polarization parallel to the plane of incidence.

Relation (40) leads to

n21

μ1

(E0i − E0r) sin θ1 =n2

2

μ2

E0t sin θ2, (50)

which simplifies to

E0i − E0r =μ1n2

μ2n1E0t →

sin θ1

sin θ2E0t if μ1 = μ2 = 1. (51)

using Snell’s law. Similarly, relation (42) leads to

E0i + E0r =cos θ2

cos θ1E0t. (52)

Adding (51) and (52), we find that

E0t

E0i=

2n1 cos θ1

n2 cos θ1 + μ1

μ2n1 cos θ2

→ 2 sin θ2 cos θ1

sin θ1 cos θ1 + sin θ2 cos θ2=

4 sin θ2 cos θ1

sin 2θ1 + sin 2θ2

=2 sin θ2 cos θ1

sin(θ1 + θ2) cos(θ1 − θ2)if μ1 = μ2 = 1. (53)

Page 208: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 2 22

Combining this with (52), we have

E0r

E0i=

cos θ2

cos θ1

E0t

E0i− 1 =

n1 cos θ2 − μ1

μ2n2 cos θ1

n1 cos θ1 + μ1

μ2n2 cos θ2

→ 2 sin θ2 cos θ2

sin θ1 cos θ1 + sin θ2 cos θ2− 1

=sin θ2 cos θ2 − sin θ1 cos θ1

sin θ1 cos θ1 + sin θ2 cos θ2= −sin(θ1 − θ2) cos(θ1 + θ2)

sin(θ1 + θ2) cos(θ1 − θ2)

= −tan(θ1 − θ2)

tan(θ1 + θ2)if μ1 = μ2 = 1. (54)

(c) Normal Incidence.

Taking the limit of either polarization as θ1 → 0 and θ2 → 0, we find

E0t

E0i

=2n1

n2 + μ1

μ2n1

→ 2n1

n1 + n2

if μ1 = μ2 = 1, (55)

E0r

E0i=

n1 − μ1

μ2n2

n1 + μ1

μ2n2

→ n1 − n2

n1 + n2if μ1 = μ2 = 1. (56)

Page 209: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 3 23

3. For a linearly polarized wave, E⊥ and E‖ are in phase, while for a circularly polarizedwave their phase difference is Δφ = ±90.

In Fresnel’s rhomb, there are two internal reflections, each of which causes phasechanges Δφ⊥ and Δφ‖ for light polarized perpendicular and parallel to the plane ofincidence, respectively. Hence, if

Δφ⊥ − Δφ‖ = ±45 (57)

at each reflection, we will achieve the desired conversion of linearly into circularlypolarized light.

In case of total internal reflection where media 1 and 2 have indices n1 = n and n2 = 1,we use Snell’s law to write

sin θ2 = n sin θ1, and cos θ2 =√

1 − n2 sin2 θ1 = in√

sin2 θ1 − 1/n2. (58)

Then, eqs. (49) and (54) can be written as

E0r

E0i

∣∣∣∣⊥

= −sin(θ1 − θ2)

sin(θ1 + θ2)= −

sin θ1 · in√

sin2 θ1 − 1/n2 − cos θ1 · n sin θ1

sin θ1 · in√

sin2 θ1 − 1/n2 + cos θ1 · n sin θ1

=cos θ1 − i

√sin2 θ1 − 1/n2

cos θ1 + i√

sin2 θ1 − 1/n2, (59)

and

E0r

E0i

∣∣∣∣‖

= −tan(θ1 − θ2)

tan(θ1 + θ2)

=cos θ1 − i

√sin2 θ1 − 1/n2

cos θ1 + i√

sin2 θ1 − 1/n2·cos θ1 · in

√sin2 θ1 − 1/n2 − sin θ1 · n sin θ1

cos θ1 · in√

sin2 θ1 − 1/n2 + sin θ1 · n sin θ1

=cos θ1 − i

√sin2 θ1 − 1/n2

cos θ1 + i√

sin2 θ1 − 1/n2·

cos θ1

(cos θ1 + i

√sin2 θ1 − 1/n2

)− 1

− cos θ1

(cos θ1 − i

√sin2 θ1 − 1/n2

)+ 1

=cos θ1(1 − 1/n2 − 1) + i

√sin2 θ1 − 1/n2

− cos θ1(1 − 1/n2 − 1) + i√

sin2 θ1 − 1/n2

= −cos θ1 − in2

√sin2 θ1 − 1/n2

cos θ1 + in2√

sin2 θ1 − 1/n2. (60)

Both eqs. (59) and (60) have the form

a − ib

a + ib=

a2 − b2 − 2iab

a2 + b2≡ e−iΔφ, (61)

Page 210: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 3 24

so

tan Δφ =2ab

a2 − b2=

2(a/b)

1 − (a/b)2

= tan 2(Δφ/2) =2 tan(Δφ/2)

1 − tan2(Δφ/2), (62)

and hence,

tan(Δφ/2) =a

b. (63)

Thus,

tan(Δφ⊥/2) =

√sin2 θ1 − 1/n2

cos θ1, and tan(Δφ‖/2) = n2

√sin2 θ1 − 1/n2

cos θ1,

(64)and

tan(Δφ⊥/2 −Δφ‖/2) =tan(Δφ⊥/2) − tan(Δφ‖/2)

1 + tan(Δφ⊥/2) tan(Δφ‖/2)

=

√sin2 θ1 − 1/n2

cos θ1· 1 − n2

1 + n2 sin2 θ1−1/n2

cos2 θ1

= −cos θ1

√sin2 θ1 − 1/n2

sin2 θ1. (65)

The angle of incidence, θ1, is the same as angle θ shown in the figure for Fresnel’srhomb. Thus, condition (57) implies that

cos θ√

sin2 θ − 1/n2

sin2 θ= ± tan 22.5 = ±(

√2 − 1), (66)

or(3 − 2

√2) sin4 θ = (1 − sin2 θ)(sin2 θ − 1/n2), (67)

(4 − 2√

2) sin4 θ − (1 + 1/n2) sin2 θ + 1/n2 = 0, (68)

sin2 θ =1 + 1/n2 ±

√(1 + 1/n2)2 − 4(4 − 2

√2)/n2

2(4 − 2√

2). (69)

For n = 1.5, we find θ = 50.2 and 53.3.

Page 211: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 4 25

4. (a) Since the “time reversed” case is the superposition of the two cases shown belowthe former, we conclude that

i = 1 = r2 + tt′, (70)

and0 = rt + tr′. (71)

From eq. (71) we learn thatr′ = −r. (72)

Combining this with conservation of energy, eq. (6), we find that

|t′| = |t| . (73)

Given these relations, we can now write

r = r0eiα, t = t0e

iβ, and t′ = t0eiγ, (74)

where r0 and t0 are real and obey

r20 + t20 = 1. (75)

Inserting relations (74) into eq. (70), we find

1 = r20e

2iα + t20ei(β+γ). (76)

Multiplying eq.(76) by its complex conjugate yields

1 = r40 + t40 + 2r2

0t20 cos(2α − β − γ). (77)

In view of relation (75), we must have

2α = β + γ. (78)

Then, eq. (76) can be rewritten as

1 = (r20 + t20)e

2iα = e2iα. (79)

Hence,α = nπ, (80)

where n is an integer, from which we conclude that r (and r′) is real. Further,

β + γ = 2nπ, (81)

which implies thatt′ = t, (82)

according to the definitions (74).

(b) The phase difference 2Δ is that between points a and b in the figure

Page 212: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 4 26

Namely,

2Δ =2π · 2d

λ2 cos θ2

− 2πl sin θ1

λ1

, (83)

where l = 2d tan θ2. The wavelengths are related by n1λ1 = n2λ2, so eq. (83) becomes

2Δ =2π · 2d

λ2 cos θ2

(1 − n1 sin θ1 tan θ2 cos θ2

n2

)=

4πd cos θ2

λ2, (84)

using Snell’s law.

The phase lag for the first transmitted ray is that between points a and b in the figurebelow:

We note that l′ = d(tan θ1 − tan θ2), so that

φ =2πl′ sin θ1

λ1+

2πd

λ2 cos θ2− 2πd

λ1 cos θ1

Page 213: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 4 27

=2πd

λ2 cos θ2

− 2πd sin θ1 tan θ2

λ1

− 2πd

λ1 cos θ1

+2πd sin θ1 tan θ1

λ1

= Δ − Δ′, (85)

with

Δ′ =2πd cos θ1

λ1. (86)

The total reflected amplitude is

R = r + tr′t′e2iΔ + tr′3t′e4iΔ + ... = r − |t|2 re2iΔ

∑n=0

(r2e2iΔ)2

= r − (1 − r2)re2iΔ

1 − r2e2iΔ=

r(1 − e2iΔ)

1 − r2e2iΔ, (87)

using relations (72) and (82).

Similarly, the total transmitted amplitude is

T = tt′ei(Δ−Δ′) ∑n=0

(r2e2iΔ)2 =(1 − r2)ei(Δ−Δ′)

1 − r2e2iΔ. (88)

For a discussion of “tunneling” via the Poynting vector, see D. Mugani, Opt. Comm.175, 309 (2000).3

3http://puhep1.princeton.edu/~mcdonald/examples/EM/mugnai_oc_175_309_00.pdf

Page 214: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 5 28

5. (a) The reflected amplitude for the 3-layer medium is

R = r12 + t12r23t21e2iΔ + t12r23r21r23t21e

4iΔ + ...

= r12 + t12t21r23e2iΔ

∑j=0

(r21r23e

2iΔ)j

= r12 + t12t21r23e

2iΔ

1 − r21r23e2iΔ. (89)

From prob. 4, r21 = −r12, and t12t21 = |t12|2 = 1 − r212, so eq. (89) becomes

R = r12 + (1 − r212)

r23e2iΔ

1 + r12r23e2iΔ

=r12 + r23e

2iΔ

1 + r12r23e2iΔ. (90)

Similarly,

T = t12t23eiΔ + t12e

3iΔ + t12r23r21r23r21t23e5iΔ + ...

= t12t23eiΔ + t12r23r21t23e

3iΔ∑j=0

(r21r23e

2iΔ)j

= t12t23eiΔ + t12t23

r21r23e3iΔ

1 − r21r23e2iΔ

= t12t23eiΔ

(1 − r12r23e

2iΔ

1 + r12r23e2iΔ

)

=t12t23e

1 + r12r23e2iΔ. (91)

At normal incidence the reflected amplitude at the a-b boundary is

rab =na − nb

na + nb. (92)

If d = λ2/4, then Δ = 2πd/λ2 = π/2, and

R =r12 − r23

1 − r12r23, (93)

r12 − r23 =(n1 − n2)(n2 + n3) − (n2 − n3)(n1 + n2)

(n1 + n2)(n2 + n3)=

2(n1n3 − n22)

(n1 + n2)(n2 + n3), (94)

and R = 0 if n2 =√

n1n3.

If d = λ2/2, then Δ = π, and

R =r12 + r23

1 + r12r23

=(n1 − n2)(n2 + n3) + (n2 − n3)(n1 + n2)

(n1 + n2)(n2 + n3) + (n1 − n2)(n2 − n3)=

n1 − n3

n1 + n3

, (95)

which is independent of n2, and vanishes if media 1 and 3 are the same.

Page 215: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 6 29

6. a) The photographic plate intersects the mirror along the line x = y = 0. The incidentwave of frequency ω and polarization along the z axis has electric and magnetic fields

Ei = E0zei(−ky−ωt), Bi = −E0xei(−ky−ωt), (96)

where k = ωc = 2π/λ, c is the speed of light, and λ is the wavelength.

The reflected wave has electric field with a 180 phase change so as to satisfy theboundary condition that the tangential electric field vanish at the surface of the mirror.Hence,

Er = −E0zei(ky−ωt), Br = −E0xei(ky−ωt). (97)

The total fields are standing waves:

E = Ei + Er = Re(−2iE0z sin kye−iωt) = −2E0z sin ky sin ωt, (98)

B = Bi + Br = Re(−2E0x cos kye−iωt) = −2E0x cos ky cos ωt. (99)

The photographic plated is “exposed” by energy transfer between the electromagneticfields and the emulsion. Recall that a magnetic field cannot change the energy of acharged particle, while an electric field can. We conclude that it is the electric field (98)whose spatial dependence will determine the pattern of exposure of the photograph.Since the electric field energy depends on E2, the pattern of exposure will actuallyfollow the time average 〈E2〉 ∝ sin2 ky ∝ 1 − cos 2ky.

We also recall that the developed photographic “negative” would be black everywhereif it were unexposed. Exposure due to strong electric fields will result in transparentregions on the negative. The blackest stripes on the negative appear where the electricfield energy vanishes, i.e., at y = nλ/2.

For a plate making angle α to the x axis, y = s sinα, where distance s is measuredfrom the edge of the plate in contact with the mirror. Hence, the black stripes on thenegative appear at

s =nλ

2 sin α. (100)

If the incident wave had polarization along the x axis, the striping on the negativewould be parallel to the x axis with periodicity λ/2.

b) Now, the incident wave has a 45 angle of incidence, and the plane of incidence isthe x-y plane.

We first consider the case of polarization perpendicular to the plane of incidence. Then,the incident wave vector is

ki =k√2(x − y), (101)

and the electromagnetic fields are

Ei = E0zei(kx/

√2−ky/

√2−ωt), (102)

Bi = ki × Ei = −E0√2(x + y)ei(kx/

√2−ky/

√2−ωt). (103)

Page 216: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 6 30

The reflected wave has

kr =k√2(x + y), (104)

and the electromagnetic fields are

Er = −E0zei(kx/

√2+ky/

√2−ωt), (105)

Br = kr × Er = −E0√2(x− y)ei(kx/

√2+ky/

√2−ωt). (106)

The total fields are the waves:

E = Ei + Er = Re

(−2iE0z sin

ky√2ei(kx/

√2−ωt)

)

= 2E0z sinky√

2sin(kx/

√2 − ωt), (107)

B = Bi + Br = Re

(−√

2E0

(x cos

ky√2

+ iy sinky√

2

)ei(kx/

√2−ωt)

)

= −√

2E0

(x cos

ky√2

cos(kx/√

2 − ωt) + y sinky√

2sin(kx/

√2 − ωt)

). (108)

This is a travelling wave in the x direction, modulated in y by sin ky/√

2. Then,

⟨E2⟩∝ sin2 ky√

2∝ 1 − cos

√2ky, (109)

so the dark stripes appear on the plate at positions

s =

√2nλ

2 sin α. (110)

For polarization parallel to the plane of incidence,

Ei =E0√

2(x + y)ei(kx/

√2−ky/

√2−ωt), (111)

Bi = ki × Ei = E0zei(kx/

√2−ky/

√2−ωt), (112)

Er = −E0√2(x− y)ei(kx/

√2−ky/

√2−ωt), (113)

Br = kr × Er = E0zei(kx/

√2−ky/

√2−ωt), (114)

E = Ei + Er = Re

(−√

2E0

(ix sin

ky√2− y cos

ky√2

)ei(kx/

√2−ωt)

)

=√

2E0

(x sin

ky√2

sin(kx/√

2 − ωt) + y cosky√

2cos(kx/

√2 − ωt)

), (115)

B = Bi + Br = Re

(2E0z cos

ky√2ei(kx/

√2−ωt)

)

= 2E0z cosky√

2cos(kx/

√2 − ωt). (116)

Page 217: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 6 31

Thus, ⟨E2⟩∝ sin2 ky√

2+ cos2 ky√

2= 1, (117)

so the photographic plate would be uniformly exposed.

Page 218: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 7 32

7. Since the intensity I ∝ E2, and I ∝ 1/r2 for a wave that emanates from a localizedsource, we must have E ∝ 1/r for the electric field of the wave. Hence, if we label thefield strength of the source at airplane a as Ea, the field strength at airplane b is

Eb =Ea

d, (118)

where d is the distance between the two airplanes, taken to be at the same height h.

As shown in the figure, the reflected wave makes angle of incident θ1 with respect tothe surface of the ocean. The reflected angle is, of course, also θ1, while the transmittedangle θ2 obeys Snell’s law,

sin θ2 =n1

n2sin θ1 =

sin θ1√ε

=d/2l√

ε=

d√ε(d2 + 4h2)

, (119)

in the approximation that index n1 = 1 for air. For later use, we note that

tan θ1 =d

2h, (120)

and

tan θ2 =sin θ2√

1 − sin2 θ2

=d√

(ε − 1)d2 + 4εh2. (121)

The amplitude of the wave emitted at airplane a which is reflected by the ocean issmaller than that of the direct wave by the factor cosα = sin θ1.

Likewise, the amplitude of the currents excited in the antenna on airplane b by a wavethat makes angle α is smaller by the factor cosα than that due to the direct wave.

Since the antenna on airplane a is vertical, the polarization of the emitted wave is in avertical plane, which is also the plane of incidence of the wave with the ocean. Uponreflection, the wave suffers a loss of amplitude described by the ratio

Er

Ei

∣∣∣∣‖

= −tan(θ1 − θ2)

tan(θ1 + θ2)=

tan θ2 − tan θ1

tan θ2 + tan θ1· 1 − tan θ1 tan θ2

1 + tan θ1 tan θ2

=

√(ε − 1)d2 + 4εh2 − 2εh√(ε − 1)d2 + 4εh2 + 2εh

, (122)

Page 219: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 7 33

according to the Fresnel equation (54), and eqs. (120)-(121).

Also, the path length of the reflected wave between airplanes a and b is 2l, so theamplitude of the reflected wave has fallen off by factor 1/2l.

Altogether, the excitation in antenna b due to the reflected wave is proportional to

ER = Ea · cos α · 1

2l· Er

Ei

∣∣∣∣‖· cosα, (123)

while that due to the direct wave is proportional to

ED = Ea ·1

d. (124)

Therefore, the ratio of the intensity of the reflected to the direct signal is

IR

ID=

E2R

E2D

=d2

4l2sin4 θ1

tan2(θ1 − θ2)

tan2(θ1 + θ2)

=d6

(d2 + 4h2)3

⎛⎝√

(ε − 1)d2 + 4εh2 − 2εh√(ε − 1)d2 + 4εh2 + 2εh

⎞⎠

2

. (125)

Page 220: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 8 34

8. (a) The equation of motion of an ionized electron of chage −e and mass m in a circularlypolarized plane wave is

mr + er

c× B0z = −eE0(x ± iy)ei(kz−ωt). (126)

We seek solutions of a similar form:

r± = r0(x ± iy)ei(kz−ωt). (127)

Inserting eq. (127) into (126) we find

− mω2r0(x ± iy) − ieωB0

cr0(−y ± ix) = −eE0(x ± iy), (128)

r0

[−mω2 ± eωB0

c

](x ± iy) = −eE0(x ± iy), (129)

and hence,

r0 =eE0

mω(ω ∓ ωB), (130)

where

ωB =eB0

mc. (131)

Note that for propagation antiparallel to the direction of the magnetic field, ωB is anegative number.

Since r± measures the separation of electrons from positive ions, the resulting polar-ization density is

P± = −Ner± = − Ne2

mω(ω ∓ ωB)E ≡ χ±E, (132)

and the dielectric “constant” is

ε± = 1 + 4πχ± = 1 − 4πNe2

mω(ω ∓ ωB)= 1 −

ω2p

ω(ω ∓ ωB), (133)

where the square of the plasma frequency is given by

ω2p =

4πNe2

m. (134)

For radio waves with ω ωB ≈ ωp,

ε± ≈ ±ω2

p

ωωB. (135)

The phase velocity of the plane waves is related by

vphase =ω

k=

c

n=

c√ε. (136)

Page 221: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 8 35

Comparing eqs. (135) and (136) we see that the phase velocity is imaginary for waveswith polarization x− i y, which means that these waves are attenuated rapidly. Onlythe waves with polarization x+i y propagate in the ionosphere, and their phase velocityis

vphase = c

√ωωB

ωp. (137)

For propagation opposite to the direction of the magnetic field, the situation is reversed,and only wave with polarization x − i y survive. For the surviving waves, the wavevector is related to frequency by

k =ω

c

√ε+ =

ωp

c

√ω

ωB, (138)

and so the group velocity is given by

vgroup =dω

dk=

1

dk/dω= 2c

√ωωB

ωp= 2vphase . (139)

For waves with ω ≈ 105 Hz and ωB ≈ ωp ≈ 107 Hz, vphase ≈ c/10.

The difference in arrival times for pulses centered on frequencies ω1 = 105 and ω2 =2 × 105 Hz from the opposite side of the Earth (d = 2 × 109 cm, which used to be thedefinition of a centimeter) is

Δt =d

vg,1− d

vg,2=

d

2c

ωp√ω2ωB

(√ω2

ω1− 1

)

=2 × 109

2 · 3 × 1010

107

√2 × 105 · 107

(√2 − 1

)≈ 0.1 s. (140)

(b) For the radio wave to be reflected back downwards, there must be some height hsuch that θ(h) = 90. According to Snell,

ni sin θi = n(h) sin θ(h), (141)

so with ni = 1, sin θ(h) = 1, and n(h) =√

1 − (ωp(h)/ω)2, we need

ωp(h) = ω cos θi. (142)

Page 222: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 9 36

9. Inside the good conductor the plane wave has the form

E = E0e−βzei(βz−ωt), β =

√2πσμω

c, H =

√2πσ

μω(1 + i)z× E, (143)

where σ is the conductivity and μ is the permeability. Hence, the time-averaged Poynt-ing vector is

〈S〉 =c

8πRe(E × H) =

c

√2πσ

μω|E0|2 e−2βzz. (144)

The power lost per unit area to Joule heating is

〈P 〉 =∫ ∞

0〈J · E〉 dz =

1

2

∫ ∞

0Re(σE · E) dz =

σ |E0|2

2

∫ ∞

0e−2βz dz =

σ |E0|2

=cσ |E0|2

4√

2πσμω=

c

√2πσ

μω|E0|2 = 〈S(z = 0)〉 . (145)

Page 223: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 10 37

10. Ignoring reflections, the incident power is either absorbed or transmitted, so recallingprob. 8 we can write

〈Sout〉 = 〈Sin〉 − Joule heating = 〈Sin〉 −∫ a

0〈J · E〉 dz =

c |E0|2

8π− σ |E0|2

2

∫ a

0e−2z/d dz

≈ c |E0|2

8π− σa |E0|2

2, (146)

where the approximation holds since a d. The relative transmitted intensity is

T =〈Sout〉〈Sin〉

= 1 − 4πσa

c. (147)

To analyze the reflected intensity, we first consider the reflected and transmitted ampli-tudes. In particular, we focus on the magnetic field H, since its transverse componentis continuous across a metallic boundary. Furthermore, since the thickness of the sheetis much less than the skin depth, the magnitude H is essentially unchanged from oneside of the sheet to the other. That is,

Hi + Hr ≈ Ht, (148)

where i, r and t indicate incident, reflected and transmitted, respectively. We candeduce Ht from the transmitted intensity ratio T = |Ht|2 / |Hi|2,

|Ht| = |Hi|√T ≈ |Hi|

(1 − 2πσa

c

). (149)

Thus, from eq. (148)

Hr ≈ −2πσa

cHi, (150)

and

R =|Hr|2

|Hi|2≈(

2πσa

c

)2

. (151)

Page 224: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 11 38

11. According to Snell’s law, the angle θ2 of the transmitted wave obeys the formal relation

sin θ2 =n1

n2sin θ1. (152)

For the stated conditions, sin θ2 > 1. Then,

cos θ2 =√

1 − sin2 θ2 = i

√√√√n21

n22

sin2 θ1 − 1. (153)

We take the x axis to be normal to the 1-2 boundary, and the y axis along the boundaryin plane of incidence.

Formally, the transmitted wave vector kt has components

kx =n2ω

ccos θ2 = i

ω

c

√n2

1 sin2 θ1 − n22 ≡ iβ, ky =

n2ω

csin θ2 =

n1ω

csin θ1. (154)

The space-time dependence of the transmitted wave is therefore

ei(kt·r−ωt) = e−βxei(kyy−ωt), (155)

which describes a surface wave that propagates in the +y direction at phase velocityc/(n1 sin θ1) < c, and whose amplitude is significant only for x <∼ 1/β.

For incident electric field perpendicular to the plane of incidence, the Fresnel equation(48) and eq. (155) tell us that

Et = Etz =2 sin θ2 cos θ1

sin(θ1 + θ2)E0ize

−βxei(kyy−ωt). (156)

We can find the magnetic field via Faraday’s equation:

∇× Et = −1

c

∂Bt

∂t= i

ω

cBt, (157)

so that

Bt = −ic

ωx

∂Ez

∂y+ i

c

ωy

∂Ez

∂x=(n1 sin θ1x− iβ

c

ωy)

Et. (158)

The electric field (156) satisfies ∇ ·E = 0 since Ez does not depend on z. Similarly, themagnetic field (158) satisfies ∇ · B = 0, and hence Maxwell’s equations are satisfied.The wave equation is also obeyed by eq. (156) since

∇2Et = (β2 − k2y)Et =

n22ω

2

c2Et =

n22

c2

∂2Et

∂t2. (159)

Page 225: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 11 39

The time-averaged Poynting vector of the transmitted wave is

〈St〉 =c

8πRe(Et ×B

t ) =c

4 sin2 θ2 cos2 θ1

|sin(θ1 + θ2)|2|E0i|2 e−2βxn1 sin θ1 y. (160)

Now,

|sin(θ1 + θ2)|2 = |sin θ1 cos θ2 + cos θ1 sin θ2|2

= sin2 θ1 |cos θ2|2 + cos2 θ1 sin2 θ2

= sin2 θ1(sin2 θ2 − 1) + (1 − sin2 θ1) sin2 θ2

= sin2 θ2 − sin2 θ1

= sin2 θ2

(1 − n2

2

n21

), (161)

so that〈St〉 =

c

n1

1 − n22

n21

sin θ1 cos2 θ1 |E0i|2 e−2βx y. (162)

As expected, the Poynting vector is parallel to the y axis, so no energy is transmittedinto medium 2. However, the formal result (162) is that a non-negligible energy flowsin the thin layer near the bounding surface of medium 2.

For the case of the electric field parallel to the plane of incidence, the Fresnel equation(53) is

E0t(x = 0)

E0i=

2 sin θ2 cos θ1

sin θ1 cos θ1 + sin θ2 cos θ2(163)

However, the transmitted electric field cannot be only in the x direction, as this wouldnot satisfy ∇ · E = 0. There must be a y component as well, as we found for themagnetic field when the electric field was perpendicular to the plane of incidence.Hence, we expect that

Et =(n1 sin θ1x− iβ

c

ωy)

Ae−βxei(kyy−ωt), (164)

where A is chosen to satisfy eq. (163) at x = 0. That is,

(2n21 sin2 θ1 − n2

2) |A|2 =4 sin2 θ2 cos2 θ1

|sin θ1 cos θ1 + sin θ2 cos θ2|2|E0i|2

=4

n21

n22sin2 θ1 cos2 θ1

sin2 θ1(1 − sin2 θ1) +n2

1

n22sin2 θ1(

n21

n22sin2 θ1 − 1)

|E0i|2

=4cos2 θ1

(1 − n22

n21)((1 +

n21

n22) sin2 θ1 − 1)

|E0i|2 . (165)

Faraday’s law (157) gives us the magnetic field as

Bt = ic

ω

(∂Etx

∂y− ∂Ety

∂x

)= −n2

2Ae−βxei(kyy−ωt)z. (166)

Page 226: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 11 40

The time-averaged Poynting vector of the transmitted wave is

〈St〉 =c

8πRe(Et × B

t ) =c

8πn1n

22 sin θ1 |A|2 e−2βxy

=c

n1 sin θ1 cos2 θ1(1 − n2

2

n21

) ((1 +

n21

n22

)sin2 θ1 − 1

) |E0i|2 e−2βxy(2

n21

n22sin2 θ1 − 1

) . (167)

Perhaps the most interesting feature of the surface wave for the case of polarization inthe plane of incidence is that the electric field (164) includes a component along thedirection of propagation of the wave, and the wave velocity is less than c. A chargedparticle moving along with this wave would experience a continual force in the directionof motion, and would therefore be accelerated.

A related concept for particle acceleration by surface waves will be explored in prob. 12.

Page 227: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 12 41

12. The interaction between particle beams and diffraction gratings was first considered bySmith and Purcell, Phys. Rev. 62, 1069 (1953),4 who emphasized energy transfer fromthe particle to free electromagnetic waves. The excitation of surface waves by particlesnear conducting structures was first discussed by Pierce, J. Appl. Phys. 26, 627-638(1955),5 which led to the extensive topic of wakefields in particle accelerators. Thepresence of surface waves in the Smith-Purcell effect was noted by di Francia, NuovoCim. 16, 61-77 (1960).6 A detailed treatment of surface waves near a diffraction gratingwas given by van den Berg, Appl. Sci. Res. 24, 261-293 (1971).7 Here, we construct asolution containing surface waves by starting with only free waves, then adding surfacewaves to satisfy the boundary condition at the grating surface.

If the (perfectly) conducting surface were flat, the reflected wave would be

Er = −E0xei(kyy+kz z−ωt). (168)

However, the sum Ein + Er does not satisfy the boundary condition that Etotal mustbe perpendicular to the wavy surface (30). Indeed,

[Ein + Er]surface = 2iE0xei(kyy−ωt) sin kzz ≈ 2iakzE0xei(kyy−ωt) sin kxx, (169)

where the approximation holds for a d, and we have defined kx = 2π/d.

Hence, we require additional fields near the surface to cancel that given by (169). Forz ≈ 0, these fields therefore have the form

E = −akzE0xei(kyy−ωt)(eikxx − e−ikxx

). (170)

This can be decomposed into two waves E± given by

E± = ∓akzE0xei(±kxx+kyy−ωt). (171)

Away from the surface, we suppose that the z dependence of the additional waves canbe described by including a factor eik′

zz. Then, the full form of the additional waves is

E± = ∓akzE0xei(±kxx+kyy+k′zz−ωt). (172)

The constant k′z is determined on requiring that each of the additional waves satisfy

the wave equation,

∇2E± =1

c2

∂2E±∂t2

. (173)

This leads to the dispersion relation

k2x + k2

y + k′2z =

ω2

c2. (174)

4http://puhep1.princeton.edu/~mcdonald/examples/accel/smith_pr_92_1069_53.pdf5http://puhep1.princeton.edu/~mcdonald/examples/accel/pierce_jap_26_627_55.pdf6http://puhep1.princeton.edu/~mcdonald/examples/accel/toraldo_di_francia_nc_16_61_60.pdf7http://puhep1.princeton.edu/~mcdonald/examples/accel/vandenberg_asr_24_261_71.pdf

Page 228: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 12 42

The component ky of the incident wave vector can be written in terms of the angle ofincidence θin and the wavelength λ as

ky =2π

λsin θin. (175)

Combining (174) and (175), we have

k′z = 2πi

√√√√ 1

d2−(

cos θin

λ

)2

. (176)

For short wavelengths, k′z is real and positive, so the reflected wave (168) is accompanied

by two additional plane waves with direction cosines (kx, ky, k′z). But for long enough

wavelengths, k′z is imaginary, and the additional waves are exponentially attenuated in

z.

When surface waves are present, consider the fields along the line y = 0, z = π/2kz .Here, the incident plus reflected fields vanish (see the first form of (169)), and thesurface waves are

E± = ∓akze−π|k′

z |/2kzE0xei(±kxx−ωt). (177)

The phase velocity of these waves is

vp =ω

kx=

d

λc. (178)

When d = λ, the phase velocity is c, and k′z = iky according to (176). The surface

waves are then,

E± = ∓2πa cos θin

de−(π/2) tan θinE0xei(±kxx−ωt). (179)

A relativistic charged particle that moves in, say, the +x direction remains in phasewith the wave E+, and can extract energy from that wave for phases near π. Onaverage, the particle’s energy is not affected by the counterpropagating wave E−. Inprinciple, significant particle acceleration can be achieved via this technique. For asmall angle of incidence, and with a/d = 1/2π, the accelerating field strength is equalto that of the incident wave.

Page 229: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 13 43

13. This problem was abstracted from T.P. Wangler, Strong focusing and the radiofre-quency quadrupole accelerator, Am. J. Phys. 64, 177 (1996).8

The electric field in the RFQ can be obtained from the potential via E = −∇φ, so

Ex =x

dE0 sinωt, (180)

Ey = −y

dE0 sinωt. (181)

The equations of motion are

x =x

d

eE0

msinωt, (182)

y = −y

d

eE0

msinωt, (183)

z = 0. (184)

Then,z(t) = z0 + v0zt = v0t (185)

for the particular case specified.

For the x motion, we consider the form (33),

x = f + g sinωt + ωg cosωt, (186)

x = f + g sinωt + 2ωg cos ωt − ω2g sinωt. (187)

The x equation of motion now yields

f + 2ωg cosωt =

[−g + ω2g +

f + g sinωt

d

eE0

m

]sinωt. (188)

Since g is both small and slowly varying by hypothesis, we neglect the terms involvingg and g, leaving

f ≈[ω2g +

f

d

eE0

m

]sinωt +

g

d

eE0

msin2 ωt. (189)

In this, the coefficent of the rapidly varying term sinωt should vanish, and f shouldbe the average of the term in sin2 ωt. The first condition tells us that

g = − eE0

mω2df, (190)

which combines with the (averaged) second condtion to give a differential equation forf :

f = −1

2

(eE0

mωd

)2

f. (191)

8http://puhep1.princeton.edu/~mcdonald/examples/accel/wangler_ajp_64_177_96.pdf

Page 230: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 13 44

Thus,

f ≈ A cosΩt + B sinΩt, where Ω =eE0√2mωd

. (192)

Together we have

x(t) ≈ (A cosΩt + B sinΩt)(1 − eE0

mω2dsinωt

). (193)

The particular initial conditions (34)-(36) are satisifed by

x(t) ≈ v0θ0

ΩsinΩt

(1 − eE0

mω2dsinωt

). (194)

For this to be consistent we must have that

eE0

mω2d 1. (195)

Then, the beam returns to the z-axis at time t = π/Ω, corresponding to distancez = πv0/Ω.

The argument is similar for the y motion. The opposite sign of the electric field leadsto

g = +eE0

mω2df, (196)

and so

y(t) ≈ (C cos Ωt + D sinΩt)(1 +

eE0

mω2dsin ωt

). (197)

The particular initial conditions (34)-(36), however, require that both C and D vanish.

Experts will recognize that the dimensionless quantity

η ≡ eE0

mωc, (198)

where c is the speed of light, is a useful invariant of the field. In terms of this invariantthe condition of validity of the solution is

ηλ

2πd 1. (199)

If d is a characteristic aperture of the RFQ, we earlier required that λ d so the qua-sistatic approximation to the fields would be valid. Hence, the invariant field strengthη cannot be too large in the RFQ.

The physical meaning of the invariant η is that it is the ratio of the energy gain overdistance λ/2π to the electron rest energy mc2:

η =eE0

mωc=

eE0λ/2π

mc2. (200)

Thus, the RFQ should not impart relativistic transverse motion to the particles if it isto function as described above.

Page 231: Electrodynamics California

Princeton University 2001 Ph501 Set 6, Solution 14 45

14. a) We ignore the interstellar magnetic field (see prob. 7 for a discussion of the effect ofsuch fields), so the usual analysis of waves incident on free electrons applies:

mr = eEei(kz−ωt), (201)

r = − eE

mω2, (202)

p = Ner = −Ne2E

mω2= χE, (203)

ε = 1 + 4πχ = 1 − 4πNe2

mω2= 1 −

ω2p

ω2, (204)

where the plasma frequency ωp is given by

ω2p =

4πNe2

m=

4πNe2c2

mc2= 4πNr0c

2, (205)

where r0 = e2/mc2 = 2.8 × 10−13 cm is the classical electron radius. Then,

n(ω) =√

ε =

√1 −

ω2p

ω2. (206)

b) The propagation time for a wave of frequency ω over distance L is

t(ω) =n(ω)L

c, (207)

so the propagation-time difference for frequencies separated by δω is

δt =dt

dωδω =

L

cn

ω2p

ω2

δω

ω. (208)

The higher frequency takes longer to arrive.

For the example that ω = 2000 MHz, δω/ω = 0.01, N = 0.04 electrons/cm3, and|δt| = 0.004 s, we find that

ω2p = 4πNr0c

2 = 4π · 0.04 · 2.8 × 10−13 · (3 × 1010)2 ≈ 1.3 × 108, (209)

n =

√1 −

ω2p

ω2=

√1 − 1.3 × 108

4 × 1018≈ 1, (210)

and the distance to the pulsar is

L = cnδtω2

ω2p

ω

δω≈ 3× 1010 · 1 · 0.004 · 4 × 1018

1.3 × 108· 100 ≈ 3.7× 1020 cm ≈ 400 light years.

(211)

Page 232: Electrodynamics California

Princeton University

Ph501

Electrodynamics

Problem Set 7

Kirk T. McDonald

(2001)

[email protected]

http://puhep1.princeton.edu/~mcdonald/examples/

Page 233: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 1 1

1. Polarization Dependence of Emissivity

Deduce the emissive power Pν of radiation of frequency ν into vacuum at angle θ tothe normal to the surface of a good conductor at temperature T , for polarization bothparallel and perpendicular to the plane of emission.

Recall Kirchhoff’s law of heat radiation (as clarified by Planck, The Theory of HeatRadiation, chap. II, especially sec. 28) that

Aν= K(ν, T ) =

hν3/c2

ehν/kT − 1, (1)

where Pν is the emissive power per unit area per unit frequency interval (emissivity)and

Aν = 1 −R = 1 −∣∣∣∣E0r

E0i

∣∣∣∣2

(2)

is the absorption coefficient (0 ≤ Aν ≤ 1), c is the speed of light, h is Plank’s constantand k is Boltzmann’s constant.

Page 234: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 2 2

2. Rayleigh Resistance

A circular wire of conductivity σ and radius a d, where d(ω) λ is the skin depth,carries current that varies as I(t) = I0e

−iωt. As in prob. 9, set 6, consider the time-averaged Poynting vector at the surface of the wire. Relate this to the Joule loss 〈I2R〉to show that

R(ω) =a

2dR0, where R0 =

1

πa2σis the dc resistance per unit length. (3)

Page 235: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 3 3

3. Telegrapher’s Equation

Deduce the differential equation for current (or voltage) in a two-conductor trans-mission line that is characterized by resistance R (summed over both conductors),inductance L, capacitance C and leakage conductivity K, all defined per unit length.The leakage conductivity describes the undesirable current that flows directly from oneconductor to the other across the dielectric that separate them according to

Ileakage = KV, (4)

where V (x, t) is the voltage between the two conductors, taken to be along the x axis.

Deduce a relation among R, L, C and K that permits ‘distortionless’ waves of the form

e−γxf(x − vt) (5)

to propagate along the transmission line. Give expressions for v and γ in terms of R,L and C ; relate γ to the transmission line impedance defined by Z = V/I in the limitthat R and K vanish.

Page 236: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 4 4

4. Transmission Line Impedance

a) Consider a two-wire transmission line with zero resistance (and zero leakage current)that lies along the z axis in vacuum. We define the line impedance Z by V = ZI , whereI(z, t) is the current in each of the wires (+I in one, −I in the other), and V (z, t) isthe voltage difference between the two wires. Show that Z is real (⇒ V and I are in

phase), and that Z =√

L/C , where C and L are the capacitance and inductance perunit length.

b) Impedance Matching

A transmission line of impedance Z1 for z < 0 is connected to a line of impedance Z2

for z > 0. A wave Viei(k1z−ωt) is incident from z = −∞.

Show that the reflected wave Vre−i(k1z+ωt) (z < 0) and the transmitted wave Vte

i(k2z−ωt)

(z > 0) obeyVr

Vi=

Z2 − Z1

Z2 + Z1, and

Vt

Vi=

2Z2

Z2 + Z1. (6)

Note the boundary conditions on I and Z at z = 0.

Since Z1 and Z2 are real, the transmission line of impedance Z2 could be replaced bya pure resistance R = Z1 and no reflection would occur.

Even when line 2 is present, we can avoid a reflection by a kind of “antireflection”coating as discussed in prob. 5, set 6. Another way to deduce this is by considerationof the complex impedance Z(z) = V (z)/I(z) where V and I are the total voltage andtotal current. If the line for z < z0 < 0 were replaced by a source whose impedance isexactly Z(z0), then the waves for z > z0 would be unchanged.

Show that

Z(−l) = Z1Z2 − iZ1 tan k1l

Z1 − iZ2 tan k1l. (7)

Also show that when l = λ1/4 then Z(−l) = Z21/Z2, which is real. Hence, the source

at z = −l could be a transmission line of impedance Z0 = Z21/Z2.

That is, a quarter-wave section of impedance Z1 =√

Z0Z2 matches lines of impedancesZ0 and Z2 with no reflections.

c) Impedance Matching with Resistors

The quarter-wave matching of part b) works only at a single frequency. Impedancematching of transmission lines over a broad range of frequencies can be accomplishedwith appropriate resistors, as illustrated in the following examples. The key to ananalysis is that a transmission line of impedance Z acts on an input signal like a pureresistance of R = Z which is connected to ground potential.

Thus, a resistor of value R = Z1 − Z2 matches signals that moves from a line ofimpedance Z1 into one of impedance Z2 < Z1, as shown below

Page 237: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 4 5

However, if Z1 < Z2, then a matching circuit (for signals moving from line 1 into line2) can be based on a resistor R = Z1Z2/(Z2 −Z1) the connects the junction to ground,as shown below.

It might be desirable to have a circuit that matches lines 1 and 2 no matter whichdirection the signals are propagating. This could be accomplished with resistors R1 =√

Z1(Z1 − Z2) and R2 = Z2

√Z1/(Z1 − Z2) as shown below, assuming Z1 > Z2.

Another type of matching problem involves lines of a single impedance Z, where itis desired to split the signal into two parts with ratio A between the currents andvoltages. This could be accomplished with resistors R1 = Z/A and R2 = AZ as shownbelow.

In case of a 1:1 split, then R1 = R2 = Z. In this case, a reflectionless split could alsobe accomplished with three identical resistors of value R = Z/3 arranged as in thefigure below. If these resistors are mounted in a box with three terminals, the split isaccomplished properly no matter how the lines are connected.

Page 238: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 5 6

5. In the manufacture of printed circuit boards it is common to construct transmissionlines consisting of a “wire” separated from a “ground plane” by a layer of dielectric.Estimate the transmission line impedance for two typical configurations:

a) Wire over Ground

The wire has diameter d, centered at height h d above the conducting ground plane.The other space above the ground plane is filled with a dielectric of constant ε (= 4.7for G-10, a fiberglass-epoxy composite often used in circuit boards).

Estimate the capacitance C per unit length, and from this show the transmission lineimpedance is

Z ≈ 60 Ω√ε

ln4h

d. (8)

b) Stripline

A conducting strip of width w is at height h above the conductor.

It may be difficult to estimate the capacitance for w <∼ h, so consider the case thatw h to show that

Z ≈ 377 Ωh√εw

. (9)

Page 239: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 6 7

6. Off-Center Coaxial Cable

A “coaxial” transmission line has inner conductor of radius a and outer conductorof radius b, but the axes of these two cylinders are offset by a small distance δ b.Deduce the capacitance and inductance per unit length, and the impedance Z, accurateto order δ2/b2.

The (relative) dielectric constant and permeability of the medium between the twoconductors both unity. The relevant frequencies and conductivities are so large thatthe skin depth is small compared to δ.

Page 240: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 7 8

7. Cavity Q

a) A rectangular cavity with conducting walls of length Δx = Δy = a, Δz = l isexcited in the (1,1,0) mode:

Ez = E0 sinπx

asin

πy

ae−iωt, Ex = Ey + Hz = 0. (10)

Calculate the time-averaged force on each of the six faces.

b) The cavity Q (quality factor) is defined by

Q =〈stored energy〉

〈energy lost per cycle〉 . (11)

Show that

Q =al

2πd(a + 2l)≈ volume

d · surface area, (12)

where d is the skin depth. The approximate version of eq. (12) provides a reasonableestimate for the Q of the lowest mode of any cavity.

Page 241: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 8 9

8. Cavity Line Broadening

According to the definition of cavity Q given in prob. 4,

dU

dt= − ω0

2πQU, (13)

where U is the averaged energy stored in the cavity fields, whose angular frequency isω0. Thus, if the cavity were left to itself, the energy would die away:

U(t) = U0e−ω0t/2πQ. (14)

Since the field energy U is proportional to the square of the electric field E, we have

E(t) ∝ E0e−ω0t/4πQe−iω0t. (15)

This is not the behavior of a pure frequency ω0.

Perform a Fourier analysis of the electric field:

E(t) =∫

Eωe−iωt dω, (16)

supposing that the cavity is turned on at t = 0, which implies that

Eω =1

∫ ∞

0E(t)eiωt dt. (17)

Show that this leads to

Uω ∝ |Eω|2 ∝ 1

(ω − ω0)2 + (ω0/4πQ)2, (18)

which has the form of a resonance curve.

The damping due to resistive wall losses gives a finite width to the resonance: FWHMΔω = ω0/2πQ. Thus, the relation

Q =ω0

2πΔω(19)

gives additional meaning to the concept of cavity Q. That is, the Q is a measure ofthe sharpness of the cavity frequency spectrum.

Page 242: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 9 10

9. Rayleigh-Jeans Law

One of the most significant uses of a cavity was the measurement of the spectrumof the waves inside when the walls were “red hot”. Maxwell told us that if thermalequilibrium holds, each cavity mode carries energy kT (considering a mode as a kindof oscillator with two polarizations). Here, T is the temperature and k is Boltzmann’sconstant.

Show that the number of modes per unit interval of angular frequency in a cubicalcavity of edge a is

dN =a3ω2dω

π2c3, (20)

for frequencies such that the mode indices l, m, n are all large compared to one.

The famous hint is that each mode (l, m, n) corresponds to a point on a cubical lattice.

Then, the energy spectrum of the cavity radiation would be

dE =a3ω2kTdω

π2c3=

8πa3ν2kTdν

c3, (21)

where ν = ω/2π is the ordinary frequency.

The Rayleigh-Jeans expression (21) implies that the total energy of the cavity radiationgrows arbitrarily large as one include the contributions at high frequency.

Such behavior was, of course, not observed in the laboratory. Planck saw that thisrequires a rather fundamental change in our thinking...

Page 243: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 10 11

10. Right Circular Cavity

A simple mode of a right circular cavity is sketched below:

We expect that the electric field of angular frequency ω has the form

E = zE(r)e−iωt. (22)

Plugging in to the wave equation

∇2E =1

c2

∂2E

∂t, (23)

we see that1

r

∂r

r∂E

∂r+

ω2

c2E = 0, (24)

which we recognize as Bessel’s equation of order zero,

⇒ E ∼ J0

(ωr

c

). (25)

Alternatively, ignore the cylindrical walls initially, and simply consider the cavity tobe a parallel plate capacitor. This suggests that the time-dependent electric field ofangular frequency ω has the form

Ez = E0e−iωt. (26)

Show, however, that the time dependence of the displacement current induces an az-imuthal magnetic field

Hφ = − iωr

2cE0e

−iωt. (27)

Than, Faraday’s law tells us that a correction to E is induced by the time variation ofH...

Follow this logic enough to demonstrate the first and second corrections to Ez, whichform the first terms of the series

Ez = E0e−iωt

∞∑n=0

(−1)n

(n!)2

(ωr

2c

)2n

= E0J0

(ωr

c

)e−iωt. (28)

J0(x) oscillates, with zeros at x = 2.405, 5.520, 8.654, .... Hence, the boundary condi-tion that Ez = 0 at r = a is satisfied if

ωa

c= 2.405, 5.520, 8.654, ..., (29)

which describes an important class of modes of a right circular cavity.

Page 244: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 11 12

11. RF Cavity with Fields that vary Linearly with Radius

A simple rf cavity is a right circular cylinder of radius a and length d (see the precedingproblem), for which the TM0,1,0 mode has electromagnetic fields

Ez(r, θ, z, t) = E0J0(kr) cos ωt, (30)

Bθ(r, θ, z, t) = E0J1(kr) sin ωt, (31)

where ka = 2.405 is the first zero of the Bessel function J0.

Such a cavity is potentially interesting for particle acceleration in that the electricfield points only along the axis and is independent of z, so that a large fraction ofthe maximal energy eEd could be imparted to a particle of charge e as it traversesthe cavity. However, such cavities are not useful in practice for at least two reasons:the particles must pass through the cavity wall to enter or exit the cavity and therebysuffer undesirable scattering; the magnetic field does not vary linearly with radius, andso acts like a nonlinear lens for particles whose motion is not exactly parallel to theaxis.

Practical accelerating cavities have apertures (irises) of radius b in the entrance andexit surfaces, so that a beam of particles can pass through without encountering anymaterial. In this case, the electric field can no longer be purely axial. Deduce thesimplest electromagnetic mode of a cavity with apertures for which the transversecomponents of the electric and magnetic fields vary linearly with radius. Deduce alsothe shape of the wall of a perfectly conducting cavity that could support this mode.

Consider a cavity of extent −d < z < d, with azimuthal symmetry and symmetryabout the plane z = 0, that could be a unit cell of a repetitive structure. This impliesthat either Ez = 0 at (r, z) = (0, d) and (0,−d), or ∂Ez/∂z = 0 at these points.

Page 245: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 12 13

12. Reflex Klystron

The figure below shows a slice through a kind of cylindrical cavity used in generationof high strength radio frequency fields, the so-called reflex klystron.

This is something like a piece of (vacuum) coaxial cable of length h, inner radius aand outer radius b terminated with a conducting plate on the right, but with a smallgap d between the termination plate and the center conductor on the left. An electronbeam is made to pass along the axis of the cavity through small holes. The beam ismodulated at the cavity resonant frequency, and transforms its energy to the cavityfield if it crosses the gap ≈ 180 out of phase with the cavity field.

Estimate the lowest resonant frequency of the cavity.

Page 246: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 13 14

13. Guide Loss

Consider propagation of waves in the lowest TE mode of a rectangular waveguide withedges a < b, as shown in the figure below. The walls have conductivity σ and theinterior of the guide is at vacuum.

Due to Joule losses in the walls, the intensity of the propagating field dies out like e−βz

where

β =〈power loss per unit length along guide〉〈power transmitted down the guide〉 . (32)

For waves of wavelength λ show that

β =c

4π· 4π

a· 1√

σλc· 1 + 2a

b

(λ2b

)2

√1 −

(λ2b

)2, (33)

which could be minimized to find the best choice for λ.

Page 247: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Problem 14 15

14. Loop Coupling

A common way of feeding waves into a guide is shown in the figure below. The centerconductor of a coaxial cable is bent into a semicircle of radius r and “grounded” on theguide wall. Then, for waves with r λ it is a good approximation that the currentI0e

−iωt is constant over the loop.

a) Explain briefly why essentially no power is radiated into a TM mode by this coupler.

b) If r b < a, where a and b are the lengths of the edges of the guide, show that thepower radiated into the lowest TE mode is

〈P 〉 =4π

cI20

k

kg

a

b

(πr

2a

)4

, (34)

in each direction, independent of the position h of the coupling loop.

Page 248: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 1 16

Solutions

1. This solution is adapted from Born and Wolf, Principles of Optics, 7th ed., sec. 14.2.(see also Landau and Lifshitz, The Electrodynamics of Continuous Media, sec. 67),and finds application in the calibration of the polarization dependence of detectors forcosmic microwave background radiation (E.J. Wollack, Princeton Ph.D. dissertation,1994, Appendix C.1.1; C. Herzog, Princeton U. Generals Expt., 1999).

In eq. (2) we need the Fresnel equations of reflection that

E0r

E0i

∣∣∣∣⊥

=sin(θt − θi)

sin(θt + θi),

E0r

E0i

∣∣∣∣‖

=tan(θt − θi)

tan(θt + θi), (35)

where i, r, and t label the incident, reflected, and transmitted waves, respectively.

The solution is based on the fact that eq. (1) holds separately for each polarizationof the emitted radiation, and is also independent of the angle of the radiation. Thisresult is implicit in Planck’s derivation of Kirchhoff’s law of radiation, and is statedexplicitly in Reif, Fundamentals of statistical and thermal physics, sec. 9.14.

That law describes the thermodynamic equilibrium of radiation emitted and absorbedthroughout a volume. The emissivity Pv and the absorption coefficient Aν can dependon the polarization of the radiation and on the angle of the radiation, but the definitionsof polarization parallel and perpendicular to a plane of emission, and of angle relativeto the normal to a surface element, are local, while the energy conservation relationPν = AνK(ν, T ) is global. A “ray” of radiation whose polarization can be described asparallel to the plane of emission is, in general, a mixture of parallel and perpendicularpolarization from the point of view of the absorption process. Similarly, the angles ofemission and absorption of a ray are different in general. Thus, the concepts of paralleland perpendicular polarization and of the angle of the radiation are not well definedafter integrating over the entire volume. Thermodynamic equilibrium can exist only ifa single spectral intensity function K(ν, T ) holds independent of polarization and ofangle.

All that remains is to evaluate the reflection coefficients R⊥ and R‖ for the two polar-izations at a vacuum-metal interface. These are well known, but we derive them forcompleteness.

To use the Fresnel equations (35), we need expressions for sin θt and cos θt. Theboundary condition that the phase of the wave be continuous across the vacuum-metalinterface leads, as is well known, to the general form of Snell’s law:

ki sin θi = kt sin θt, (36)

where k = 2π/λ is the wave number. Then,

cos θt =

√√√√1 − k2i

k2t

sin2 θi. (37)

Page 249: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 1 17

To determine the relation between wave numbers ki and kt in vacuum and in theconductor, we consider a plane wave of angular frequency ω = 2πν and complex wavevector k,

E = E0ei(kt·r−ωt), (38)

which propagates in a conducting medium with dielectric constant ε, permeability μ,and conductivity σ. The wave equation for the electric field in such a medium is

∇2E − εμ

c2

∂2E

∂t2=

4πμσ

c2

∂E

∂t, (39)

where c is the speed of light. We find the dispersion relation for the wave vector kt oninserting eq. (38) in eq. (39):

k2t = εμ

ω2

c2+ i

4πσμω

c2. (40)

For a good conductor, the second term of eq. (40) is much larger than the first, so wewrite

kt ≈√

2πσμω

c(1 + i) =

1 + i

d=

2

d(1 − i), (41)

whered =

c√2πσμω

λ (42)

is the frequency-dependent skin depth. Of course, on setting ε = 1 = μ and σ = 0 weobtain expressions that hold in vacuum, where ki = ω/c.

We see that for a good conductor |kt| ki, so according to eq. (37) we may takecos θt ≈ 1 to first order of accuracy in the small ratio d/λ. Then the first of the Fresnelequations becomes

E0r

E0i

∣∣∣∣⊥

=cos θi sin θt/ sin θi − 1

cos θi sin θt/ sin θi + 1=

(ki/kt) cos θi − 1

(ki/kt) cos θi + 1≈ (πd/λ)(1 − i) cos θi − 1

(πd/λ)(1 − i) cos θi + 1, (43)

and the reflection coefficient is approximated by

R⊥ =∣∣∣∣E0r

E0i

∣∣∣∣2

⊥≈ 1 − 4πd

λcos θi = 1 − 2 cos θi

√ν

σ. (44)

For the other polarization, we see that

E0r

E0i

∣∣∣∣‖

=E0r

E0i

∣∣∣∣⊥

cos(θi + θt)

cos(θi − θt)≈ E0r

E0i

∣∣∣∣⊥

cos θi − (πd/λ)(1 − i) sin2 θi

cos θi + (πd/λ)(1 − i) sin2 θi

, (45)

so that

R‖ ≈ R⊥

(1 − 4πd

λ

sin2 θi

cos θi

)≈ 1 − 4πd

λ cos θi= 1 − 2

cos θi

√ν

σ. (46)

Page 250: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 1 18

An expression for R‖ valid to second order in d/λ has been given in Stratton, Electro-magnetic Theory, sec. 9.9. For θi near 90, R⊥ ≈ 1, but eq. (46) for R‖ is not accurate.Writing θi = π/2 − ϑi with ϑi 1, eq. (45) becomes

E0r

E0i

∣∣∣∣‖≈ ϑi − (πd/λ)(1 − i)

ϑi + (πd/λ)(1 − i), (47)

For θi = π/2, R‖ = 1, and R‖,min = (5 −√2)/(5 +

√2) = 0.58 for ϑi = 2

√2πd/λ.

Finally, combining eqs. (1), (2), (44) and (46) we have

Pν⊥ ≈ 4πd cos θ

λ3

ehν/kT − 1, Pν‖ ≈ 4πd

λ3 cos θ

ehν/kT − 1, (48)

andPν⊥Pν‖

= cos2 θ (49)

for the emissivities at angle θ such that cos θ d/λ.

The conductivity σ that appears in eq. (48) can be taken as the dc conductivity solong as the wavelength exceeds 10 μm. If in addition hν kT , then eq. (48) can bewritten

Pν⊥ ≈ 4πd kT cos θ

λ3 , Pν‖ ≈ 4πd kT

λ3 cos θ, (50)

in terms of the skin depth d.

Page 251: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 2 19

2. Inside a conductor the magnetic field at angular frequency ω is related to the electricfield by

H =c

μωd(1 + i)k × E, (51)

where c is the speed of light, μ is the permeability, and d = c/√

2πσμω is the skindepth. Hence, the time-averaged Poynting vector is

〈S〉 =c2

8πμωd|E|2 k, (52)

where for waves just inside the surface of the conductor k is very nearly normal to thesurface.

Since λ d, the electric field is very small inside the conductor, and Ampere’s lawapplied to a loop of radius a gives

2πaH0 ≈ 4π

cI, (53)

or2I

ac≈ H0 =

c

μωd(1 + i)k× E0. (54)

Thus,

|E0| ≈√

2μωd |I |ac2

, (55)

and eq. (52) gives

〈S0〉 ≈ μωd

4πa2c2|I |2 k. (56)

In terms of the effective resistance R per unit length of the wire, the average powerdissipated per unit length is

1

2|I |2 R = 2πa 〈S0〉 ≈ μωd

2ac2|I |2 . (57)

Hence, we identify the effective resistance at frequency ω as

R ≈ μωd

ac2=

μω

ac√

2πσμω·√

2πσμω√2πσμω

=1

2πadσ=

a

2d

1

πa2σ=

a

2dR0, (58)

where R0 = 1/πa2σ is the dc resistance (per unit length) of the wire to longitudinalcurrents.

Page 252: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 3 20

3. Referring to the sketch, Kirchoff’s rule for the circuit of length dz shown by dashedlines tells us

V (x) − I(Rdx) − V (x + dx) − (Ldx)∂I

∂t= 0, or − ∂V

∂x= L

∂I

∂t+ IR. (59)

Next, the charge dQ that accumulates on length dx of the upper wire during time dtis (Cdx)dV in terms of the change of voltage dV between the wires, which also can bewritten in terms of currents as

Q = (Cdx)dV = (I(x)− I(x + dx)− Ileakage)dt, so − ∂I

∂x= C

∂V

∂t+ KV. (60)

Together these imply the desired wave equation

∂2I

∂x2= LC

∂2I

∂t2+ (RC + KL)

∂I

∂t+ KRI. (61)

We seek solutions of the form

I = e−γxf(x − vt), (62)

for which∂I

∂t= −ve−γxf ′, and

∂2I

∂t2= v2e−γxf ′′, (63)

while

∂I

∂x= −γe−γxf + e−γxf ′, so

∂2I

∂x2= γ2e−γxf − 2γe−γxf ′ + e−γxf ′′. (64)

Inserting these into the wave equation we find

γ2f − 2γf ′ + f ′′ = v2LCf ′′ − v(RC + KL)f ′ + KRf. (65)

Page 253: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 3 21

This should be true for an arbitrary function f , so the coefficients of each derivativeof f must separately be equal:

v =

√1

LC, γ =

√KR, and 2

γ

v= RC + KL = 2

√RCKL, (66)

where we have used the first two relations in obtaining the second form of the third. Ingeneral, a + b = 2

√ab; this only holds when a = b. So we deduce the desired condition

RC = KL, (67)

for distortionless telegraphy.

With this condition, we can re-express γ as

γ = R

√C

L. (68)

Finally, we relate this to the impedance Z = V/I when R = 0 = K. For this, wesuppose that V = V0f(x − vt) and I = I0f , where V0 and I0 can be related by eitherof the first-order differential equations above. We quickly find that V0 = vLI0, so

Z =√

L/C . Then,

γ =R

Z, (69)

once we have arranged that RC = KL.

Remark: This problem was solved by O. Heaviside (1887) who argued that long-distance telegraph lines (including trans-Atlantic cables) should be designed to be‘distortionless’. Previous cables were fairly far from this ideal. However, long ca-bles are expensive so there was considerable hesitation to abandon the large existingcapital investment and implement the proposed improvements. Indeed, the editor ofthe journal that published Heaviside’s papers was fired for being too sympathetic toHeaviside’s views that were initially quite unpopular with industry. Heaviside, whowas unemployed for most of his life, could not be fired! Large-scale implementationof ‘distortionless” telegraphy occured only after 1900 following vigorous advocacy byM. Pupin of the U.S.A., for whom the physics building of Columbia U. is named.

A typical cable has RC KL. It costs a lot to reduce RC , although this was thedirection of industry prior to Heaviside. He noted that one shouldn’t even try toreduce leakage K, so long as the signal is not attenuated until it is undetectable – andthe distortion-free condition makes it much easier to detect small signals. Rather oneshould increase the inductance, or leakage, or both! This counterintuitive result didnot sit well with industry leaders, who, needless to say, were little guided by partialdifferential equations.

Page 254: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 4 22

4. a) For a two-wire transmission line with negligible resistance, the voltage V betweenthe wires is related to the current I in each of the wires by

∂V

∂z= L

∂I

∂t, (70)

which follows from Kirchhoff’s law applied to a short length of the line, and

∂V

∂t=

1

C

∂I

∂z, (71)

which follows from charge conservation, where C and L are the capacitance and in-ductance per unit length. For a wave of frequency ω moving in the +z direction, thewaveforms are

V = V0ei(kz−ωt), and I = I0e

i(kz−ωt). (72)

Substituting these forms into eqs. (70)-(71) we find

kV = −ωLI, and − ωV =kI

C, (73)

and hence,

V

I=

√L

C≡ Z. (74)

Since C and L are real numbers, the impedance Z is real in this case. This impliesthat V0/I0 is also real, and so the current and voltage are in phase.

b) When a transmission line of impedance Z1 that occupies z < 0 is connected to a lineof impedance Z2 for z > 0, the current and voltage will be continuous at the boundaryz = 0.

An incident wave of frequency ω from a source at z = −∞ has current Iiei(k1z−ωt). This

results in a reflected wave Ire−i(k1z+ωt) for z > 0 and a transmitted wave Ite

i(k2z−ωt) forz > 0. Of course, Vi = Z1Ii, Vr = Z1Ir, and Vt = Z2It.

Continuity of the current at z = 0 tells us that

Ii − Ir = It, and hence Vi − Vr =Z1

Z2Vt, (75)

since a positive value for Ir corresponds to current flowing in the −z direction.

Continuity of the voltage at z = 0 tells us that

Vi + Vr = Vt. (76)

Equations (75) and (76) yield

Vt =2Z2

Z2 + Z1Vi, and Vr =

Z2 − Z1

Z2 + Z1Vi. (77)

Page 255: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 4 23

At position z = −l the total voltage is

V (−l) = Viei(−k1l−ωt) + Vre

−i(−k1l+ωt) = Vie−iωt

(e−ik1l +

Z2 − Z1

Z2 + Z1eik1l

)

=Vie

−iωt

Z2 + Z1

((Z2 + Z1)e

−ik1l + (Z2 − Z1)eik1l)

=Vie

−iωt

Z2 + Z1

(Z2(e

ik1l + e−ik1l) − Z1(eik1l − e−ik1l)

)

=2Vie

−iωt

Z2 + Z1(Z2 cos k1l − iZ1 sin k1l) , (78)

and (again noting that positive Ir implies a negative current) the total current is

I(−l) = Iiei(−k1l−ωt) − Ire

−i(−k1l+ωt) =Vi

Z1e−iωt

(e−ik1l − Z2 − Z1

Z2 + Z1eik1l

)

=Vie

−iωt

Z1(Z2 + Z1)

((Z2 + Z1)e

−ik1l − (Z2 − Z1)eik1l)

=Vie

−iωt

Z1(Z2 + Z1)

(Z1(e

ik1l + e−ik1l) − Z2(eik1l − e−ik1l)

)

=2Vie

−iωt

Z1(Z2 + Z1)(Z1 cos k1l − iZ2 sin k1l) , (79)

Hence, we find the impedance

Z(−l) =V (−l)

I(−l)= Z1

Z2 cos k1l − iZ1 sin k1l

Z1 cos k1l − iZ2 sin k1l= Z1

Z2 − iZ1 tan k1l

Z1 − iZ2 tan k1l. (80)

When l = λ1/4 = π/2k1, then tan k1l → ∞ and Z(−l) = Z21/Z2 is a real number, which

permits the region π/2k1 < z < 0 of impedance Z1 to be an antireflection matchingsection between a line of impedance Z0 = Z2

1/Z2 and one of impedance Z2.

c) A signal propagates in line 1 without reflection is that line is terminated in (real)impedance Z1. Thus, for the circuit

We need Z1 = R + Z2, since the resistor is in series with line 2 (as viewed from line1). A proper match is possible so long as Z1 > Z2. The currents are the same inlines 1 and 2, so the transmitted voltage is V2 = V1Z2/Z1. However, this is smallerthan the transmitted voltage (77) for an unmatched line, because the matching resistordissipates more power than is “lost” to the reflection at an unmatched junction.

If Z1 < Z2, then a proper match to line 1 can be obtained with a resistor R in parallelwith line 2, as shown below,

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Princeton University 2001 Ph501 Set 7, Solution 4 24

provided the effective resistance of R and Z2 is again Z1, i.e.,

1

Z1=

1

R+

1

Z2⇒ R =

Z1Z2

Z2 − Z1. (81)

Here, the transmitted voltage is the same as that in line 1, but the transmitted currentis less: I2 = I1Z1/Z2.

To match lines 1 and 2 for signals moving in either direction, a combination of theabove two circuits can be used, as shown below for the case that Z1 > Z2:

Signals emanating from line 1 must be terminated in impedance Z1; hence,

Z1 = R1 +R2Z2

R2 + Z2. (82)

Likewise, signals from line 2 must be terminated in impedance Z2; hence,

Z2 =R2(R1 + Z1)

R1 + R2 + Z1. (83)

From eq. (82) we have

Z1Z2 = R1R2 + R1Z2 −R2(Z1 − Z2), (84)

while eq. (83) givesZ1Z2 = R1R2 −R1Z2 + R2(Z1 − Z2), (85)

Adding eqs. (84) and (84) we find

Z1Z2 = R1R2, (86)

while subtracting gives

R1 = R2Z1 − Z2

Z2. (87)

Solving these, we find

R1 =√

Z1(Z1 − Z2), and R2 = Z2

√Z1

Z1 − Z2. (88)

For signals emanating from line 1, the transmitted voltage is

V2 = V1

(1 −

√Z1 − Z2

Z1

), (89)

Page 257: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 4 25

while for signals emanating from line 2, the transmitted voltage is

V1 = V2Z1

Z2

(1 −

√Z1 − Z2

Z1

). (90)

In both cases, the transmitted voltage is less than the incident.

For a reflectionless split of the signal in a line of impedance Z we case use the circuit:

where the desired ratio of currents (and voltages) is

A =I1

I2=

R1 + Z

R2 + Z. (91)

Hence,R2 = AR1 + (A − 1)Z. (92)

Also, the two output lines must combine to terminate the input line in impedance Z,which tells us that

Z =(R1 + Z)(R2 + Z)

R1 + R2 + 2Z. (93)

Substituting eq. (92) in (93) we find

R1 = AZ, and R2 =Z

A. (94)

For a 50/50 split, A = 1 and R1 = R2 = Z.

If we perform this split with three identical resistors in the symmetric configuration

the matching condition is

Z = R +R + Z

2, (95)

and hence,

R =Z

3. (96)

Page 258: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 5 26

5. As shown on p. 156 of the Notes, the inductance and capacitance per unit length ofany two-conductor transmission line are related by

LC =ε

c2, (97)

where c is the speed of light in vacuum and the medium outside the conductors isfilled with a dielectric of constant ε. [The permeability μ is taken to be unity, and thefrequency of the waves of interest is high enough that the skin depth is small comparedto the transverse size of the conductors.] Thus, the impedance of the transmission linecan be expressed as

Z =

√L

C=

√ε

cC=

30√

ε Ω

C. (98)

a) Wire over Ground

The capacitance of the wire over ground is twice the capacitance of the wire plus itsimage wire (since C = Q/ΔV and ΔV in the present example is 1/2 that for the caseof two wires):

Recall prob. 11b, Set 3 to obtain the “exact” solution:

C =ε

2 ln(

2h+√

4h2−d2

d

) ≈ ε

2 ln 4hd

. (99)

Here, the presence of the dielectric medium leads to the factor ε in the numerator.Using the approximate form in eq. (99), which holds for d h, we find

Z ≈ 60 Ω√ε

ln4h

d. (100)

The approximate result can be obtained quickly as follows. When d h, the chargeon the wire is distributed nearly uniformly, so we may use the result that the potentialfor a uniformly charged wire embedded in a medium of dielectric constant ε variesas V (r) = 2(Q/ε) ln r. Thus, the potential difference between the wire of radius d/2and its image are distance 2h due to charge Q per unit length on the wire is ΔV ≈2(Q/ε) ln[2h/(d/2)]. In evaluating the capacitance, we suppose that the image wire

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Princeton University 2001 Ph501 Set 7, Solution 5 27

has charge −Q per unit length, which doubles the potential difference between the twowires. Finally, we recall that the voltage difference between the wire and the groundplane is 1/2 that between the wire and its image. Hence,

ΔV = 2(Q/ε) ln4h

d=

Q

C, (101)

which leads to the approximate result of eq. (99).

b) Stripline

If the strip width w is large compared to height h, then the capacitance per unit lengthis roughly

C ≈ εw

4πh, (102)

as follows from Gauss’ law, ∇ · D = 4πρfree, and the stripline impedance is

Z ≈ 120π Ω√ε

h

w=

377 Ω√ε

h

w. (103)

In practical circuit boards, w ≈ h, and we expect the impedance to be between theestimates (103) and (100). If we use the “exact” form of eq. (99) and take d = h = w,we find

C =ε

2 ln(2 +√

3)=

ε

2.6, (104)

and we estimate a lower bound on the impedance to be

Z ≈ 80 Ω√ε

. (105)

There does not appear to be a closed-form analytic solution to the present problem,but many numerical algorithms exist. See, for example,http://www.ideaconsulting.com/strip.htm

This program estimates the impedance of a stripline with h = w embedded in a thickdielectric medium to be

Z ≈ 110 Ω√ε

. (106)

Page 260: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 6 28

6. We use Gaussian units, and convert the impedance Z =√

L/C to MKSA units by

noting that 1/c = 30Ω, where c is the speed of light.

We don’t need to calculate both the capacitance C per unit length and the inductanceL per unit length, since in the case of a (perfectly conducting) transmission line theyare related by

LC =εμ

c2, (107)

where the dielectric constant ε and the permeability μ are unity in the present case. Theassumed smallness of the skin depth permits us to approximate the present transmissionline as perfectly conducting.

We first present two calculations of the capacitance (secs. a and b), and then a calcu-lation of the inductance (sec. c) as illustrations of various possible techniques.

a) The Capacitance Via the Image Method

It is expedient to use the image method for 2-dimensional cylindrical geometries. Recallthat in the case of a wire of charge q per unit length at distance b from a groundconducting cylinder of radius a, as shown in the figure, one can think of an image wireof charge −q at radius a2/b.

To apply this to the present problem, sketched in the figure below, note that the imagewires of charge ±q per unit length are both located to the left of the center of the innerconductor, say at distances ra and rb.

For the inner cylinder to be an equipotential, we must have

rb =a2

rb, (108)

and the outer cylinder is also an equipotential provided

rb + δ =b2

ra + δ, (109)

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Princeton University 2001 Ph501 Set 7, Solution 6 29

noting the offset by δ between the inner and outer cylinder. Combining eqs. (108) and(109), and noting that ra → 0 as δ → 0, we find

ra =b2 − a2 − δ2 −

√(b2 − a2 − δ2)2 − 4a2δ2

2δ. (110)

The capacitance is related by C = q/ΔV , where ΔV = Vb−Va is the potential differencebetween the two cylinders. Recall that the potential at distance r from a wire of chargeq per unit length is 2q ln r + constant. We evaluate the potentials at the points wherethe cylinders are closest to one another:

Va = 2q ln(a − ra) − 2q ln(rb − a) = 2q lna − ra

a2/ra − a= 2q ln

ra

a, (111)

using eq. (108), and

Vb = 2q ln(b − δ − ra) − 2q ln(rb − b + δ) = 2q lnb − ra − δ

b2/(ra + δ) − b= 2q ln

ra + δ

b, (112)

using eq. (109). Then,

ΔV = 2q ln

[a

b

(1 +

δ

ra

)]. (113)

When combined with eq. (110), this is an “exact” solution for any δ < b − a. Inparticular, as δ → b − a, then ra → a, and the cylinders touch with the result thatΔV = 0.

Here, we suppose that δ b − a, and expand δ/ra to second order:

δ

ra=

b2 − a2 − δ2 +√

(b2 − a2 − δ2)2 − 4a2δ2

2a2≈ b2 − a2

a2− b2δ2

a2(b2 − a2), (114)

so that

1 +δ

ra≈ b2

a2

(1 − δ2

b2 − a2

). (115)

The capacitance per unit length is therefore,

C =q

ΔV≈ 1

2(ln b

a− δ2

b2−a2

) , (116)

using eq. (113).

The inductance per unit length now follows from eq. (107):

L =2

c2

(ln

b

a− δ2

b2 − a2

), (117)

and the impedance is

Z =

√L

C≈ 2

c

(ln

b

a− δ2

b2 − a2

)= 60

(ln

b

a− δ2

b2 − a2

)Ω. (118)

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Princeton University 2001 Ph501 Set 7, Solution 6 30

Remark: The “exact” expression (113) is often written in a different fashion, which isconvenient for large δ, but perhaps less useful for small δ. The “exact” version of (114)leads to

1 +δ

ra=

b2 + a2 − δ2 +√

(b2 + a2 − δ2)2 − 4a2b2

2a2, (119)

which in turn leads to

C =q

ΔV=

1

2 lnb2+a2−δ2+

√(b2+a2−δ2)2−4a2b2

2ab

=1

2 cosh−1 a2+b2−δ2

2ab

. (120)

b) Capacitance Via Series Expansion of the Potential

The image method can be deduced by an application of series expansion techniques forthe electrostatic potential. In this section, we explore a direct use of such techniques.A full solution is long, and when we leave off some steps at the end, we get an answerthat is not quite correct.

We define the electrostatic potential φ to be zero on the inner conductor,

φ(r = a) = 0, (121)

and V on the outer conductor whose surface is approximately given by r = b + δ cos θ,

φ(r = b + δ cos θ) = V. (122)

The potential is symmetric about θ = 0:

φ(−θ) = φ(θ), (123)

so terms in sin nθ cannot appear in the series expansion of the potential:

φ(r, θ) = A0 ln r +∑n=1

(Anr

n +Bn

rn

)cos nθ. (124)

The capacitance C per unit length is, of course, given by C = Q/V , where the chargeQ per unit length on the inner conductor is given by

Q = 2πa∫ 2π

0σ(θ) dθ = 2πa

∫ 2π

0

Er(a, θ)

4πdθ =

a

2

∫ 2π

0

∂φ(a, θ)

∂rdθ =

A0

2. (125)

Thus,

C =A0

2V. (126)

Applying the boundary condition (121) to the general form (124), we have

0 = A0 ln a +∑n=1

(Ana

n +Bn

an

)cos nθ. (127)

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Princeton University 2001 Ph501 Set 7, Solution 6 31

Likewise, the boundary condition (122) yields

V = A0 ln(b + δ cos θ) +∑n=1

(An(b + δ cos θ)n +

Bn

(b + δ cos θ)n

)cos nθ. (128)

With considerable effort, the terms in eq. (128) of the form cosl θ cosmθ can be ex-pressed as sums of terms in the orthogonal set of functions cos nθ. Then, eqs. (127) and(128) can be combined to yield the Fourier coefficients An and Bn. Thus, subtractingeq. (127) from (128) and using the approximation (140), we have

V = A0

(ln

b

a+

δ cos θ

b− δ2 cos2 θ

2b2

)+ F (An, Bn, θ) (129)

IF the integral of F with respect to θ vanished, then integrating eq. (129) yields

V = A0

(ln

b

a− δ2

4b2

), (130)

and the capacitance would be

C =A0

2V≈ 1

2(ln b

a− δ2

4b2

) . (131)

However, we the presence of terms like A1 cos2 θ in F means that we cannot expect itsintegral to vanish, and eq. (131) is not quite correct.

c) Calculation of the Inductance

The calculation of the inductance is complicated by the fact that the currents in thisproblem are distributed over surfaces, rather than flowing in filamentary wires. Wewould like to use the relation,

Φ = cLI, (132)

where I is the total (steady) current flowing down the inner conductor (and back upthe outer conductor), and Φ is the magnetic flux per unit length linked by the circuit.From Ampere’s law, with the assumption that the currents are uniformly distributedon the inner and outer conductors, the azimuthal component Bθ of the magnetic fieldin the region between the two conductors is given by

Bθ(r) =2I

cr. (133)

If the cable were truly coaxial, the flux would be simply

Φ0 =∫ b

aBθ dr =

2I

cln

b

a, (134)

and the corresponding inductance would be

L0 =2

c2ln

b

a. (135)

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Princeton University 2001 Ph501 Set 7, Solution 6 32

Then, from eq. (107) the capacitance would be

C0 =1

2 ln(b/a), (136)

as is readily verified by an electrostatic analysis, and the transmission line impedancewould be

Z0 =

√L0

C0=

2

cln

b

a= 60 ln

b

aΩ. (137)

However, because the outer conductor is off center with respect to the inner, we cannotsimply use eq. (134). We can segment the currents on the conductors into filaments ofazimuthal extent dθ, and calculate the flux Φ(θ) linked the circuit element defined bythe segments centered on angle θ on the inner and outer conductors. Then, the effectiveinductance of the whole cable can be estimated from eq. (132) using the average ofΦ(θ):

L =1

2πcI

∫ 2π

0Φ(θ) dθ =

1

2πcI

∫ 2π

0dθ∫ rmax(θ)

aBθ(r) dr =

1

πc2

∫ 2π

0ln

rmax(θ)

adθ,

(138)using (133) and (134). The result holds only to the extent that the current distributionis independent of azimuth, as discussed in sec. d. However, there will be a smallazimuthal dependence to the current in this problem, so we will not obtain a completelycorrect result.

To complete the analysis, we need rmax(θ), the maximum radius about the centerof the inner conductor of magnetic field lines that are linked by the segment of theouter conductor at azimuthal angle θ. Assuming the currents is uniformly distributedover the inner and outer conductors, the magnetic field between the two conductors isentirely due to the current in the inner conductor, and the field is purely azimuthalabout the axis of the inner conductor as given by eq. (133). Then, the geometry shownin the figure tells us that

rmax(θ) = b + δ cos θ. (139)

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Princeton University 2001 Ph501 Set 7, Solution 6 33

This relation is “exact” to the extent that the currents are uniformly distributed;however, this is not actually the case in the present problem.

To use relation (139) in eq. (138), we approximate

lnrmax(θ)

a= ln

b + δ cos θ

a= ln

b

a+ln

(1 +

δ cos θ

b

)≈ ln

b

a+

δ cos θ

b− δ2 cos2 θ

2b2, (140)

which leads to

L ≈ 2

c2

(ln

b

a− δ2

4b2

). (141)

This result happens to agree with the result implied by sec. b, but differs somewhatfrom the more accurate result of sec. a.

d) The Magnetic Flux Linked by a Distributed Circuit

The magnetic flux through a filamentary circuit (one in which the conductors areidealized as wires) is well defined as

Φ =∫

B · dS, (142)

where the integral is taken over any surface bounded by the circuit. However, whenthe conductors of the circuit are distributed and have a finite cross sectional area A,then eq. (142) is not well defined.

We wish to show that a consistent definition of the flux through a distributed circuit isobtained by segmenting the conductors into a large number of circuits each with verysmall cross sectional area Ai, and defining

Φ =1

A

∑i

AiΦi, (143)

where the magnetic flux through subcircuit i is given by eq. (142).

We are interested in a definition of flux that gives consistency to the relation (132) inthe context of circuit analysis. In particular, if the circuit has total resistance R, andthe magnetic flux is changing, then we desire Faraday’s law to be written as

IR = E = −1

c

dt, (144)

which is the same form as holds for each of the filamentary subcircuits:

IiRi = Ei = −1

c

dΦi

dt. (145)

We suppose that the current flowing in subcircuit i is related to the total currentaccording to

Ii =Ai

AI, (146)

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Princeton University 2001 Ph501 Set 7, Solution 6 34

in which case the resistance of subcircuit i is given by

Ri =A

AiR. (147)

Then, we can combine eqs. (145)-(147) as

I =∑

i

Ii = −1

c

∑i

1

Ri

dΦi

dt= − 1

cRA

∑i

AidΦi

dt. (148)

Hence, the definition (143) leads to the desired relation (144) for the distributed circuit.

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Princeton University 2001 Ph501 Set 7, Solution 7 35

7. a) The force on the cavity walls can be evaluated via the Maxwell stress tensor. Recallthat for a good conductor the electric field is perpendicular to a conducting surface, andthe magnetic field is parallel. Also, the Maxwell stress associated with a perpendicularfield E is +E2/8π, while that with a parallel field H is −H2/8π. That is, the total,time-averaged force on a face of the cavity is given by

〈F 〉 =1

2

∫face

|E|2 − |H|28π

dArea. (149)

A positive value of F corresponds to an inward force.

The electromagnetic fields of the (1,1,0) mode are

Ex = 0, (150)

Ey = 0, (151)

Ez = E0 sinπx

asin

πy

ae−iωt, (152)

Hx = −iE0√

2sin

πx

acos

πy

ae−iωt, (153)

Hy = −iE0√

2cos

πx

asin

πy

ae−iωt, (154)

Hz = 0, (155)

using eq. (10), Faraday’s law

∇ ×E = −1

c

∂H

∂t, (156)

and the wave equation

∇2E =1

c2

∂2H

∂t2, (157)

which latter tells us that

2π2

a2=

ω2

c2. (158)

The force on each of the four faces perpendicular to the x or y axes is the same by thesymmetry of the problem, and can be calculated using the face at, say, x = 0 to be

〈F 〉 = − 1

16π

∫ a

0dy∫ l

0dz |Hy|2 = −alE2

0

64π. (159)

This force is outwards.

Likewise, the force on the two faces perpendicular to the z axis is the same, and is

〈F 〉 =1

16π

∫ a

0dx∫ a

0dy (|Ez|2 − |Hx|2 − |Hy|2) = 0. (160)

b) Turning to the cavity Q, we first calculate the time-averaged energy U in the cavity:

〈U〉 =1

2

∫dVol

|E|2 + |H|28π

=a2lE2

0

32π. (161)

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Princeton University 2001 Ph501 Set 7, Solution 7 36

The energy lost per cycle into the walls is the time-averaged power loss 〈P 〉 times theperiod T = 2π/ω.

The power 〈P 〉 lost in the cavity walls can be calculated by evaluating the componentof the (time-averaged) Poynting vector perpendicular to the walls:

〈P 〉 =∫

〈S〉⊥ to walls dArea =c

∫Re(E × H)⊥ to walls dArea

=c

∫Re(E‖ × H

‖) dArea. (162)

If the fields were actually those specified by eqs. (150)-(155), which assume perfectconductors, the power lost to the walls would be zero. For a good, but not perfect,conductor, it is an excellent approximation to suppose the cavity magnetic field is thatgiven by the perfect-conductor approximation, eqs. (153)-(155), but to take the electricfield near the conducting walls as having a small parallel component given by

E‖ at the walls = −ωd

2c(1 − i)n× H‖, (163)

whered =

c√2πσω

(164)

is the skin depth at frequency ω for the walls of conductivity σ, and n is the outwardnormal vector. This relation follows from the 4th Maxwell equation, evaluated justinside the surface of the conductor where J = σE,

∇ ×H =4π

cJ +

1

c

∂E

∂t=

4πσ

cE +

1

c

∂E

∂t≈ 4πσ

cE , (165)

the curl of which yields,

∇2H ≈ 4πσ

c2

∂H

∂t, (166)

where the approximations are valid for a good conductor. [This diffusion equation, dueto Lord Kelvin, was the basis of time-dependent electrodynamics in the era shortlybefore Maxwell clarified that if σ = 0 then waves can propagate with speed c withoutdiffusionlike distortion. It is amazing from a modern perspective that the first telegraphsystems were successfully constructed using eq. (166) as their theoretical model.]

Inside the conductor, and for waves of frequency ω and wave vector k of the formei(k·x−ωt), eq. (166) becomes

k2 ≈ 4πiσω

c2, (167)

so that

k ≈√

4πσω

c

1 + i√2

≡ 1 + i

d. (168)

Then, the Maxwell equation (165) becomes

ik × H ≡ ikn×H ≈ 4πσ

cE, (169)

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Princeton University 2001 Ph501 Set 7, Solution 7 37

which can also be written as eq. (163).

Inserting eq. (163) into (162), we find the general expression

〈P 〉 =ωd

16π

∫ ∣∣∣H‖∣∣∣2 dArea. (170)

Evaluating this for the present example, we find

〈P 〉into walls =ωdE2

0

32π(a2 + 2al). (171)

The energy lost per cycle is 〈P 〉 T = 2π 〈P 〉 /ω, so the cavity Q is

Q =〈U〉〈P 〉

ω

2π=

al

2πd(a + 2l)=

volume

πd · surface area. (172)

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Princeton University 2001 Ph501 Set 7, Solution 8 38

8. Given the electric field

E(t) = E0e−ω0t/4πQe−iω0t, t > 0, (173)

its Fourier components are given by

Eω =1

∫ ∞

0E(t)eiωt dt =

E0

∫ ∞

0ei(ω−ω0+iω0/4πQ)t dt

=iE0

1

ω − ω0 + iω0/4πQ,

(174)

where we ignore the oscillatory contribution associated with the limit t → ∞.

The Fourier analysis of the stored energy U therefore behaves as

Uω ∝ |Eω|2 ∝ 1

(ω − ω0)2 + (ω0/4πQ)2. (175)

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Princeton University 2001 Ph501 Set 7, Solution 9 39

9. The electric field of a standing wave mode of angular frequency ω inside a cubicalcavity of edge a has components of the form

Ex = E0 coslπx

asin

mπy

asin

nπz

ae−iωt, (176)

etc. The wave equation,

∇2E =1

c2

∂2E

∂t, (177)

yields the dispersion relation

k =π

a

√l2 + m2 + n2 =

ω

c. (178)

This leads to the interpretation that a mode (l, m, n) has a wave vector k whosecomponents are

π

a(l, m, n). (179)

The modes populate a cubical lattice in the first octant of k-space, with π/a as thelattice constant. For l, m, and n large, the number of modes in interval dω aboutfrequency ω = kc is equal to (a/π)3 times the volume of a shell of thickness dk = dω/cin the first octant of k-space – times two since there are two possible polarization ofthe electric field for each set of indices (l, m, n).

Thus,

dN = 2 · a3

π3· 1

8· 4π(ω/c)2(dω/c) =

a3ω2dω

π2c3. (180)

Jeans’ contribution to this result was to note that only indices in the first octantcorrespond to physical modes, and therefore Rayleigh’s original calculation was to bedivided by 8.

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Princeton University 2001 Ph501 Set 7, Solution 10 40

10. Given the zeroth order electric field of a right circular cavity,

Ez,(0) = E0e−iωt, (181)

we integrate the 4th Maxwell equation around a loop of radius r in the x-y plane tofind the first correction to the (initially zero) magnetic field,

∮H(1) · dl = 2πrHφ,(1) =

1

c

∫ ∂Ez,(0)

∂t· dS = − iωπr2

cE0e

−iωt, (182)

so that

Hφ,(1) = − iωr

2cE0e

−iωt. (183)

Next, we consider a loop in the x-z plane that includes the z axis and the line x = r,for which Faraday’s law tells us that

∮E(1) · dl = −hEz,(1) = −1

c

∫ ∂Hφ,(1)

∂t· dS =

ω2hr2

4c2E0e

−iωt, (184)

so that

Ez,(1) = −ω2r2

4c2E0e

−iωt. (185)

We now iterate, first using a loop of radius r in the x-y plane to find

∮H(2) · dl = 2πrHφ,(2) =

1

c

∫ ∂Ez,(1)

∂t· dS =

iω3πr4

4c3E0e

−iωt, (186)

so that

Hφ,(2) =iω3r3

8c3E0e

−iωt. (187)

Again, we consider a loop in the x-z plane that includes the z axis and the line x = r,for which Faraday’s law tells us that

∮E(2) · dl = hEz,(2) = −1

c

∫ ∂Hφ,(2)

∂t· dS =

ω4hr4

16c4E0e

−iωt, (188)

so that

Ez,(2) =ω4r4

16c4E0e

−iωt. (189)

Thus,

Ez = E0e−iωt

(1 − ω2r2

4c2+

ω4r4

16c4− ...

)= E0e

−iωt∞∑

n=0

(−1)n

(n!)2

(ωr

2c

)2n

= E0J0

(ωr

c

)e−iωt.

(190)

Page 273: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 11 41

11. We seek a standing wave solution where, say, the time dependence of Ez is cos ωt. Thecavity is symmetric about the plane z = 0, so we expect the z dependence of Ez tohave the form cos knz, where

kn =

⎧⎪⎨⎪⎩

(2n − 1)π/2d, if Ez(0,−d) = Ez(0, d) = 0,

nπ/d, if ∂Ez(0,−d)/∂z = ∂Ez(0, d)/∂z = 0.(191)

We can combine these two cases in the notation

kn = (2n − n0)π

2d, where

⎧⎪⎨⎪⎩

n0 = 1, if Ez(0,−d) = Ez(0, d) = 0,

n0 = 2, if ∂Ez(0,−d)/∂z = ∂Ez(0, d)/∂z = 0.(192)

where n = 1, 2, 3, ...

Our trial solution,Ez(r, z, t) = fn(r) cos knz cos ωt, (193)

must satisfy the wave equation

∇2Ez − 1

c2

∂2Ez

∂t2=

1

r

∂r

(r∂fn

r

)−(k2

n − ω2

c2

)fn = 0. (194)

This is the differential equation for the modified Bessel function of order zero, I0(Knr),where

K2n = k2

n − ω2

c2=[(2n − n0)

π

2d

]2−(

λ

)2

, (195)

the free-space wavelength at frequency ω is λ = 2πc/ω, and

I0(x) = 1 + (x/2)2 +(x/2)4

(2!)2+

(x/2)6

(3!)2+ · · · (196)

In the special case of kn = 0, eq. (194) reverts to that for the ordinary Bessel functionJ0, and the fields (30)-(31) are obtained. Since this form cannot exist in a cavity withapertures, we ignore it in further discussion.

A Fourier series for Ez with nonzero kn is then

Ez(r, z, t) =∞∑

n=1

anI0(Knr) cos knz cos ωt. (197)

The radial component of the electric field is obtained from

∇ · E =1

r

∂rEr

∂r+

∂Ez

∂z= 0, (198)

so that

Er(r, z, t) =1

r

∑n

ankn

∫rI0(Knr) dr sin knz cos ωt

=r

2

∑n

anknI1(Knr) sin knz cosωt, (199)

Page 274: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 11 42

using the fact that d(xI1)/dx = xI0, and where

I1(x) =2I1(x)

x= 1 +

(x/2)2

1!2!+

(x/2)4

2!3!+ · · · (200)

The azimuthal component of the magnetic field is obtained from

(∇× E)θ =∂Er

∂z− ∂Ez

∂r= −1

c

∂Bθ

∂t, (201)

so that

Bθ(r, z, t) =c

ω

∑n

an

(dI0(Knr)

dr− k2

nr

2I1(Knr)

)cos knz sinωt

=πr

λ

∑n

anI1(Knr) cos knz sin ωt, (202)

using the fact that I ′0(x) = I1(x).

We desire that the transverse fields Er and Bθ vary linearly with r. According toeqs. (199)-(200) and (202), this requires that Kn = 0. The simplest choice is n = 1,n0 = 1, so that kn = π/2d and d = λ/4. The fields are

Ez = E0 cosπz

2dcosωt, (203)

Er =πr

4dE0 sin

πz

2dcosωt, (204)

Bθ =πr

4dE0 cos

πz

2dsinωt. (205)

The cavity length is 2d = λ/2, and Ez vanishes on axis at the ends of the cavity. Thisconfiguration is called the π mode in accelerator physics. Since Er(z = ±d) = 0, thismode cannot exist in a structure with conducting walls at the planes z = ±d; aperturesare required.

The electric field is perpendicular to the walls of a perfectly conducting cavity. Ex-pressing the shape of the walls as r(z), we then have

dr

dz= −Ez

Er= −4d

πrcot

πz

wd, (206)

which integrates to the form

r2 = b2 −(

4d

π

)2

ln∣∣∣∣sin πz

2d

∣∣∣∣ , (207)

where b is the radius of the apertures at z = ±d. Near z = ±d, the profile is ahyperbola. Since r → ∞ as z → 0, no real cavity can support the idealized fields(203)-(205). However, it turns out that a cavity with maximum radius a = 0.4d has aFourier expansion (197) where a2 = 0.15a1, so the fields can be a good approximationto eqs. (203)-(205) in real devices.

Page 275: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 11 43

We can obtain additional formal solutions in which Kn = 0 for any value of n, andfor n0 either 1 or 2. However, these solutions are not really distinct from eqs. (203)-(205), but are simply the result of combining any number of λ/2 cells into a largerstructure. Such multicell π-mode structures are difficult to operate in practice, becausethe strong coupling of the fields from one cell to the next makes the useful range ofdrive frequencies extremely narrow. The main application of π-mode cavities is forso-called rf guns, in which a half cell has a surface at z ≈ 0 suitable for laser-inducedphotoemission of electrons, which are then accelerated further in one or a few moresubsequent cells. See K.T. McDonald, Design of the Laser-Driven RF Electron Gun forthe BNL Accelerator Test Facility, IEEE Trans. Electron Devices, 35, 2052 (1988).1

1http://puhep1.princeton.edu/~mcdonald/examples/EM/mcdonald_ieeeted_35_2052_88.pdf

Page 276: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 12 44

12. An estimate of the lowest rf frequency of the reflex klystron cavity can be made via anequivalent LC circuit:

ω ≈ 1√LC

=1√

L(C1 + C2), (208)

where C1 is the capacitance between the left termination plate and the disc of radiusa at the left end of the center conductor, C2 is the capacitance between the inner andouter conductor, and L is the self inductance between the inner and outer conduc-tor. The capacitances C1 and C2 are in parallel, and so are added to yield the totalcapacitance C .

C1 is estimated by the usual parallel-plate formula,

C1 ≈ Area

4π · height=

a2

4d. (209)

C2 is estimated by the capacitance of length h − d of a coaxial cable,

C2 ≈ h − d

2 ln b/a. (210)

The main interest in this type of cavity is for small gap d, so we write

C = C1 + C2 ≈ a2

4d

(1 +

2dh

a2 ln b/a

). (211)

L is estimated by the inductance of length h − d ≈ h of a coaxial cable,

L ≈ h − d

c2[C2/(h − d)]≈ 2h ln b/a

c2, (212)

recalling that the product of the capacitance per unit length and the inductance perunit length of a transmission line (in vacuum) is 1/c2.

The lowest cavity frequency is then estimated to be

ω ≈ 1√LC

≈ c

a

√2d

h(1 + 2dh/a2 ln b/a) ln b/a. (213)

For d small enough, we can neglect the capacitance C2, and eq. (213) simplifies to

ω ≈ 1√LC

≈ c

a

√2d

h ln b/a(d a2/h). (214)

Page 277: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 12 45

This result is small compared to ω = πc/h, the resonant frequency of a terminatedcoaxial cable of length h, which shows that it is possible to obtain low cavity frequencieswithout large cavity size.

The book Klystrons and Microwave Triodes by Princetonian D.R. Hamilton (Dover,1966) quotes (p. 75) a numerical analysis of the reflex klystron (dating from 1934) asclaiming that ω = 0.3c/a when b/a = 3, h/a = 3, and d/a = 0.32. Equation (213)yields ω ≈ 0.44c/a for these values, in reasonable agreement.

Note that the simple coaxial cavity is not the limit of the reflex klystron cavity asd → 0, since C1 → ∞ in that case. Rather, the coaxial cavity obtains when C1 andd are both set to zero, in which case eq. (213) gives the estimate ω ≈ c/h, which is afactor of π smaller than the “exact” result.

Page 278: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 13 46

13. The electric field of a TE mode in a rectangular waveguide of edges a < b has the form

Ex = E0 cosmπx

asin

nπy

bei(kgz−ωt), (215)

Ey = E0 sinmπx

acos

nπy

bei(kgz−ωt), (216)

Ez = 0, (217)

where the guide wave number kg is found from the wave equation to obey

w = c

√k2

g +(

a

)2

+(

b

)2

. (218)

The lowest frequency mode has indices m = 0, n = 1, for which only Ex is nonzero.

The magnetic field of this mode follows from Faraday’s law,

∇× E = −1

c

∂H

∂t=

iωH

c, (219)

so that

Hx = 0, (220)

Hy =ckg

ωE0 sin

πy

bei(kgz−ωt), (221)

Hz =icπ

ωbE0 cos

πy

bei(kgz−ωt), (222)

and

kg =

√ω2

c2− π2

b2=

ω

c

√√√√1 −(

λ

2b

)2

. (223)

The time-averaged power transmitted down the guide follows from the Poynting vector,

〈P 〉transmitted =∫〈Sz〉 dx dy =

c

∫ExH

y dx dy =

c2

16πab

kg

ωE2

0

=c

16πabE2

0

√√√√1 −(

λ

2b

)2

. (224)

To find the time-averaged power loss in the walls, we recall the argument of prob. 4,which led to the general result (170),

〈P 〉lost in walls =ωd

16π

∫ ∣∣∣H‖∣∣∣2 dArea, (225)

where d = c/√

2πσω is the skin depth. For the present example, the power lost intothe walls per unit length along the guide is

〈P 〉lost = 2ωd

16π

(∫ a

0|Hz(x, 0, z)|2 dx +

∫ b

0[|Hy(0, y, z)|2 + |Hz(0, y, z)|2] dy

)

=ωd

8πE2

0

(c2π2

ω2b2a +

[c2k2

g

ω2+

c2π2

ω2b2

]b

2

)=

ωd

8πE2

0

(c2π2

ω2b2a +

b

2

)

=bc

16π

√c

σλE2

0

⎛⎝1 +

2a

b

2b

)2⎞⎠ . (226)

Page 279: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 13 47

Finally, the attenuation factor β is given by

β =〈P 〉lost

〈P 〉trans

=1

a

√c

σλ

1 + 2ab

(λ2b

)2

√1 −

(λ2b

)2=

c

4π· 4π

a·√

1

cσλ· 1 + 2a

b

(λ2b

)2

√1 −

(λ2b

)2. (227)

Page 280: Electrodynamics California

Princeton University 2001 Ph501 Set 7, Solution 14 48

14. a) The current in the semicircular loop creates a magnetic field that has a componentalong the axis of the guide. Since TM modes have no longitudinal magnetic field, thefield of the loop does not couple to these modes.

b) It is “derived” on pp. 170-171 of the Notes that an oscillatory, transverse currentdistribution J inside a waveguide excites a waveguide mode with normalized electricfield E0 to strength

E = −2π

cZE0

∫J · E0⊥dArea, (228)

where Z = k/kg for TE modes and Z = kg/k for TM modes.

Comparing with prob. 9, the lowest TE mode in the present problem (0 < x < a,0 < y < b < a) has

E0 =

√2

absin

πx

ay, (229)

where the normalization condition is∫

E20 dArea = 1.

Since the electric field of this mode is independent of y, the excitation of this mode bythe loop current will be independent of the position of the loop in y.

Taking θ to measure the angle around the semicircular loop, we have

∫J · E0⊥ dArea = −

∫ π

0r dθ I0E0 sin θ = −rI0

√2

ab

∫ π

0dθ sin

πr sin θ

asin θ

≈ −πr2

aI0

√2

ab

∫ π

0dθ sin2 θ = −π2r2

a

I0√2ab

, (230)

where the approximation holds for r a. The wave numbers are related by

kg =

√ω2

c2− π2

a2= k

√√√√1 −(

λ

2a

)2

(231)

According to eqs. (228)-(229), the strength of the field in the lowest TE mode is

E =2π

c

k

kg

π2r2

a

I0

absin

πx

ay, (232)

The (time-averaged) power in the lowest TE mode follows from the Poynting vector,

〈P 〉trans =∫

〈Sz〉 dx dy =c

8πZ

∫E2

⊥ dx dy =π

4c

k

kg

(π2r2

a

)2I20

ab

=4π

c

k

kg

(πr

2a

)4

I20

a

b=

c

(πr

2a

)4 I20√

1 −(

λ2a

)2

a

b. (233)

This calculation holds for TE waves excited in either direction down the guide.

As usual, to convert this result from Gaussian to SI units, simply replace 4π/c by377 Ω.

Page 281: Electrodynamics California

Princeton University

Ph501

Electrodynamics

Problem Set 8

Kirk T. McDonald

(2001)

[email protected]

http://puhep1.princeton.edu/~mcdonald/examples/

Page 282: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 1 1

1. Wire with a Linearly Rising Current

A neutral wire along the z-axis carries current I that varies with time t according to

I(t) =

⎧⎪⎨⎪⎩0 (t ≤ 0),

αt (t > 0), α is a constant.(1)

Deduce the time-dependence of the electric and magnetic fields, E and B, observed ata point (r, θ = 0, z = 0) in a cylindrical coordinate system about the wire. Use yourexpressions to discuss the fields in the two limiting cases that ct r and ct = r + ε,where c is the speed of light and ε r.

The related, but more intricate case of a solenoid with a linearly rising current isconsidered in http://puhep1.princeton.edu/~mcdonald/examples/solenoid.pdf

Page 283: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 2 2

2. Harmonic Multipole Expansion

A common alternative to the multipole expansion of electromagnetic radiation given inthe Notes assumes from the beginning that the motion of the charges is oscillatory withangular frequency ω. However, we still use the essence of the Hertz method whereinthe current density is related to the time derivative of a polarization:1

J = p. (2)

The radiation fields will be deduced from the retarded vector potential,

A =1

c

∫[J]

rdVol =

1

c

∫[p]

rdVol, (3)

which is a solution of the (Lorenz gauge) wave equation

∇2A− 1

c2

∂2A

∂t2= −4π

cJ. (4)

Suppose that the Hertz polarization vector p has oscillatory time dependence,

p(x, t) = pω(x)e−iωt. (5)

Using the expansionr = R − r′ · n + ... (6)

of the distance r from source to observer,

show that

A = −iωei(kR−ωt)

cR

∫pω(r′)

(1 + r′ · n

(1

R− ik

)+ ...

)dVol′, (7)

where no assumption is made that R source size or that R λ = 2π/k = 2πc/ω.

Consider now only the leading term in this expansion, which corresponds to electricdipole radiation. Introducing the total electric dipole moment,

P ≡∫

pω(r′) dVol′, (8)

1Some consideration of the related topics of Hertz vectors and scalars is given in the Appendix ofhttp://puhep1.princeton.edu/~mcdonald/examples/smallloop.pdf

Page 284: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 2 3

use

B = ∇ × A and ∇ × B =1

c

∂E

∂t(9)

to show that for an observer in vacuum the electric dipole radiation fields are

B = k2 ei(kR−ωt)

R

(1 +

i

kR

)n× P, (10)

E = k2ei(kR−ωt)

R

n× (P × n) + [3(n · P)n− P]

(1

k2R2− i

kR

). (11)

Alternatively, deduce the electric field from both the scalar and vector potentials via

E = −∇φ − 1

c

∂A

∂t, (12)

in both the Lorenz and Coulomb gauges.

For large R,

Bfar ≈ k2ei(kR−ωt)

Rn× P, Efar ≈ Bfar × n, (13)

while for small R,

Bnear ≈ ik

R2(n ×P)e−iωt, Enear ≈ 3(n ·P)n − P

R3e−iωt, (14)

Thus, Bnear Enear, and the electric field Enear has the shape of the static dipole fieldof moment P, modulated at frequency ω.

Calculate the Poynting vector of the fields of a Hertzian oscillating electric dipole (10)-(11) at all points in space. Show that the time-averaged Poynting vector has the sameform in the near zone as it does in the far zone, which confirms that (classical) radiationexists both close to and far from the source.

Extend your discussion to the case of an oscillating, point magnetic dipole by notingthat if E(r, t) and B(r, t) are solutions to Maxwell’s equations in free space (i.e., wherethe charge density ρ and current density J are zero), then the dual fields

E′(r, t) = −B(r, t), B′(r, t) = E(r, t), (15)

are solutions also.

Page 285: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 3 4

3. Rotating Electric Dipole

An electric dipole of moment p0 lies in the x-y plane and rotates about the x axis withangular velocity ω.

Calculate the radiation fields and the radiated power according to an observer at angleθ to the z axis in the x-z plane.

Define n towards the observer, so that n · z = cos θ, and let l = y × n.

Show that

Brad = p0k2 ei(kr−ωt)

r(cos θ y − i l), Erad = p0k

2 ei(kr−ωt)

r(cos θ l + i y), (16)

where r is the distance from the center of the dipole to the observer.

Note that for an observer in the x-y plane (n = x), the radiation is linearly polarized,while for an observer along the z axis it is circularly polarized.

Show that the (time-averaged) radiated power is given by

d 〈P 〉dΩ

=c

8πp2

0k4(1 + cos2 θ), 〈P 〉 =

2cp20k

4

3=

2p20ω

4

3c3. (17)

This example gives another simple picture of how radiation fields are generated. Thefield lines emanating from the dipole become twisted into spirals as the dipole rotates.At large distances, the field lines are transverse...

Page 286: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 4 5

4. Magnetars

The x-ray pulsar SGR1806-20 has recently been observed to have a period T of 7.5 sand a relatively large “spindown” rate

∣∣∣T ∣∣∣ = 8× 10−11. See, C. Kouveliotou et al., AnX-ray pulsar with a superstrong magnetic field in the soft γ-ray repeater SGR1806-20,Nature 393, 235-237 (1998).2

Calculate the maximum magnetic field at the surface of this pulsar, assuming it to bea standard neutron star of mass 1.4M = 2.8 × 1030 kg and radius 10 km, that themass density is uniform, that the spindown is due to electromagnetic radiation, andthat the angular velocity vector is perpendicular to the magnetic dipole moment of thepulsar.

Compare the surface magnetic field strength to the so-called QED critical field strengthm2c3/eh = 4.4× 1013 gauss, at which electron-positron pair creation processes becomehighly probable.

2http://puhep1.princeton.edu/~mcdonald/examples/EM/kouveliotou_nature_393_235_98.pdf

Page 287: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 5 6

5. Radiation of Angular Momentum

Recall that we identified a field momentum density,

Pfield =S

c2=

U

ck, (18)

and angular momentum density

Lfield = r × Pfield. (19)

Show that for oscillatory sources, the time-average angular momentum radiated in unitsolid angle per second is (the real part of)

d 〈L〉dt dΩ

=1

8πr3[E(n · B) − B(n · E)]. (20)

Thus, the radiated angular momentum is zero for purely transverse fields.

In prob. 2, eq. (11) we found that for electric dipole radiation there is a term in E withE · n ∝ 1/r2. Show that for radiation by an oscillating electric dipole p,

d 〈L〉dt dΩ

=ik3

4π(n · p)(n× p). (21)

If the dipole moment p is real, eq. (21) tells us that no angular momentum is radiated.However, when p is real, the radiation is linearly polarized and we expect it to carryno angular momentum.

Rather, we need circular (or elliptical) polarization to have radiated angular momen-tum.

The radiation fields (16) of prob. 2 are elliptically polarized. Show that in this casethe radiated angular momentum distribution is

d 〈L〉dt dΩ

= − k3

4πp2

0 sin θl, andd 〈L〉dt

=〈P 〉ω

z. (22)

[These relations carry over into the quantum realm where a single (left-hand) circularlypolarized photon has U = hω, p = hk, and L = h.

For a another view of waves that carry angular momentum, seehttp://puhep1.princeton.edu/~mcdonald/examples/oblate_wave.pdf

Page 288: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 6 7

6. Oscillating Electric Quadrupole

An oscillating linear quadrupole consists of charge 2e at the origin, and two charges−e each at z = ±a cos ωt.

Show that for an observer in the x-z plane at distance r from the origin,

Erad = −4k3a2esin(2kr − 2ωt)

rsin θ cos θ l, (23)

where l = y × n. This radiation is linearly polarized.

Show also that the time-averaged total power is

〈P 〉 =16

15ck6a4e2. (24)

Page 289: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 7 8

7. A single charge e rotates in a circle of radius a with angular velocity ω. The circle iscentered in the x-y plane.

The time-varying electric dipole moment of this charge distribution with respect to theorigin has magnitude p = ae, so from Larmor’s formula (prob. 2) we know that the(time-averaged) power in electric dipole radiation is

〈PE1〉 =2a2e2ω4

3c3. (25)

This charge distribution also has a magnetic dipole moment and an electric quadrupolemoment (plus higher moments as well!). Calculate the total radiation fields due to theE1, M1 and E2 moments, as well as the angular distribution of the radiated power andthe total radiated power from these three moments. In this pedagogic problem youmay ignore the interference between the various moments.

Show, for example, that the part of the radiation due only to the electric quadrupolemoment obeys

d 〈PE2〉dΩ

=a4e2ω6

2πc5(1 − cos4 θ), 〈PE2〉 =

8a4e2ω6

5c5. (26)

Thus,〈PE2〉〈PE1〉 =

12a2ω2

5c2∝ v2

c2, (27)

where v = aω is the speed of the charge.

Page 290: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 8 9

8. Radiation by a Classical Atom

a) Consider a classical atom consisting of charge +e fixed at the origin, and charge −ein a circular orbit of radius a. As in prob. 5, this atom emits electric dipole radiation⇒ loss of energy ⇒ the electron falls into the nucleus!

Calculate the time to fall to the origin supposing the electron’s motion is nearly circularat all times (i.e., it spirals into the origin with only a small change in radius per turn).You may ignore relativistic corrections. Show that

tfall =a3

4r20c

, (28)

where r0 = e2/mc2 is the classical electron radius. Evaluate tfall for a = 5 × 10−9 cm= the Bohr radius.

b) The energy loss of part a) can be written as

dU

dt= Pdipole ∝ e2aω4

c=

e2

a

(aω

c

)3

ω ∝ U(

v

c

)3

ω ∝ U

T

(v

c

)3

, (29)

ordipole energy loss per revolution

energy∝(

v

c

)3

. (30)

For quadrupole radiation, prob. 5 shows that

quadrupole energy loss per revolution

energy∝(

v

c

)5

. (31)

Consider the Earth-Sun system. The motion of the Earth around the Sun causes aquadrupole moment, so gravitational radiation is emitted (although, of course, thereis no dipole gravitational radiation since the dipole moment of any system of massesabout its center of mass is zero). Estimate the time for the Earth to fall into the Sundue to gravitational radiation loss.

What is the analog of the factor ea2 that appears in the electrical quadrupole moment(prob. 5) for masses m1 and m2 that are in circular motion about each other, separatedby distance a?

Also note that in Gaussian units the electrical coupling constant k in the force lawF = ke1e2/r

2 has been set to 1, but for gravity k = G, Newton’s constant.

The general relativity expression for quadrupole radiation in the present example is

PG2 =32

5

G

c5

m21m

22

(m1 + m2)2a4ω6 (32)

[Phys. Rev. 131, 435 (1963)]. The extra factor of 4 compared to E2 radiation arisesbecause the source term in the gravitational wave equation has a factor of 16π, ratherthan 4π as for E&M.

Page 291: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 9 10

9. Why Doesn’t a Steady Current Loop Radiate?

A steady current in a circular loop presumably involves a large number of electrons inuniform circular motion. Each electron undergoes accelerated motion, and accordingto prob. 5 emits radiation. Yet, the current density J is independent of time in thelimit of a continuous current distribution, and therefore does not radiate. How can wereconcile these two views?

The answer must be that the radiation is canceled by destructive interference betweenthe radiation fields of the large number N of electrons that make up the steady current.

Prob. 5 showed that a single electron in uniform circular motion emits electric dipole ra-diation, whose power is proportional to (v/c)4. But the electric dipole moment vanishesfor two electrons in uniform circular motion at opposite ends of a common diameter;quadrupole radiation is the highest multipole in this case, with power proportional to(v/c)6. It is suggestive that in case of 3 electrons 120 apart in uniform circular motionthe (time-dependent) quadrupole moment vanishes, and the highest multipole radia-tion is octupole. For N electrons evenly spaced around a ring, the highest multipolethat radiates in the Nth, and the power of this radiation is proportional to (v/c)2N+2.Then, for steady motion with v/c 1, the radiated power of a ring of N electrons isvery small.

Verify this argument with a detailed calculation.

Since we do not have on record a time-dependent multipole expansion to arbitraryorder, return to the basic expression for the vector potential of the radiation fields,

A(r, t) =1

c

∫[J]

rdVol′ ≈ 1

cR

∫[J] dVol′ =

1

cR

∫J(r′, t′ = t − r/c) dVol′, (33)

where R is the (large) distance from the observer to the center of the ring of radiusa. For uniform circular motion of N electrons with angular frequency ω, the currentdensity J is a periodic function with period T = 2π/ω, so a Fourier analysis can bemade where

J(r′, t′) =∞∑

m=−∞Jm(r′)e−imωt′, (34)

with

Jm(r′) =1

T

∫ T

0J(r′, t′)eimωt′ dt′. (35)

Then,A(r, t) =

∑m

Am(r)e−imωt, (36)

etc.

The radiated power follows from the Poynting vector,

dP

dΩ=

c

4πR2 |B|2 =

c

4πR2 |∇ × A|2 . (37)

However, as discussed on p. 181 of the Notes, one must be careful in going from aFourier analysis of an amplitude, such as B, to a Fourier analysis of an intensity that

Page 292: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 9 11

depends on the square of the amplitude. Transcribing the argument there to the presentcase, a Fourier analysis of the average power radiated during one period T can be givenas

d 〈P 〉dΩ

=1

T

∫ T

0

dP

dΩdt =

cR2

4πT

∫ T

0|B|2 dt =

cR2

4πT

∫ T

0B∑m

Bme−imωt dt

=cR2

∑m

Bm1

T

∫ T

0Be−imωt dt =

cR2

∞∑m=−∞

BmBm

=cR2

∞∑m=0

|Bm|2 ≡∞∑

m=0

dPm

dΩ. (38)

That is, the Fourier components of the time-averaged radiated power can be written

dPm

dΩ=

cR2

2π|Bm|2 =

cR2

2π|∇× Am|2 =

cR2

2π|imkn× Am|2 , (39)

where k = ω/c and n points from the center of the ring to the observer.

Evaluate the Fourier components of the vector potential and of the radiated powerfirst for a single electron, with geometry as in prob. 5, and then for N electrons evenlyspaced around the ring. It will come as no surprise that a 3-dimensional problem withcharges distributed on a ring leads to Bessel functions, and we must be aware of theintegral representation

Jm(z) =im

∫ 2π

0eimφ−iz cosφ dφ. (40)

Use the asymptotic expansion for large index and small argument,

Jm(mx) ≈ (ex/2)m

√2πm

(m 1, x 1), (41)

to verify the suppression of the radiation for large N .

This problem was first posed (and solved via series expansions without explicit mentionof Bessel functions) by J.J. Thomson, Phil. Mag. 45, 673 (1903).3 He knew that atoms(in what we now call their ground state) don’t radiate, and used this calculation tosupport his model that the electric charge in an atom must be smoothly distributed.This was a classical precursor to the view of a continuous probability distribution forthe electron’s position in an atom.

Thomson’s work was followed shortly by an extensive treatise by G.A. Schott, Electro-magnetic Radiation (Cambridge U.P., 1912), that included analyses in term of Besselfunctions correct for any value of v/c.

These pioneering works were largely forgotten during the following era of nonrelativis-tic quantum mechanics, and were reinvented around 1945 when interest emerged inrelativistic particle accelerators. See Arzimovitch and Pomeranchuk,4 and Schwinger.5

3http://puhep1.princeton.edu/~mcdonald/examples/EM/thomson_pm_45_673_03.pdf4http://puhep1.princeton.edu/~mcdonald/examples/EM/arzimovitch_jpussr_9_267_45.pdf5http://puhep1.princeton.edu/~mcdonald/accel/schwinger.pdf

Page 293: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 10 12

10. Spherical Cavity Radiation

Thus far we have only considered waves arising from the retarded potentials, and haveignored solutions via the advanced potentials. “Advanced” spherical waves convergeon the source rather than propagate away – so we usually ignore them.

Inside a cavity, an outward going wave can bounce off the walls and become an inwardgoing wave. Thus, a general description of cavity radiation should include both kindsof waves.

Reconsider your derivation in Prob. 1, this time emphasizing the advanced waves, forwhich t′ = t + r/c is the “advanced” time. It suffices to consider only the fields due tooscillation of an electric dipole moment.

If one superimposes outgoing waves due to oscillating dipole p = p0e−iωt at the origin

with incoming waves associated with dipole −p, then we can have standing waves –and zero total dipole moment.

Suppose all this occurs inside a spherical cavity with perfectly conducting walls atradius a. Show that the condition for standing waves associated with the virtualelectric dipole is

cot ka =1

ka− ka ⇒ ωmin = 2.74

c

a, (42)

where k = ω/c. [A quick estimate would be λmax = 2a ⇒ ωmin = πc/a.]

By a simple transformation, use your result to find the condition for standing magneticdipoles waves inside a spherical cavity:

tan ka = ka ⇒ ωmin = 4.49c

a. (43)

Page 294: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 11 13

11. Maximum Energy of a Betatron

A betatron is a circular device of radius R designed to accelerate electrons (charge e,mass m) via a changing magnetic flux Φ = πR2Bave through the circle.

Deduce the relation between the magnetic field B at radius R and the magnetic fieldBave averaged over the area of the circle needed for a betatron to function. Also deducethe maximum energy E to which an electron could be accelerated by a betatron in termsof B, Bave and R.

Hints: The electrons in this problem are relativistic, so it is useful to introduce thefactor γ = E/mc2 where c is the speed of light. Recall that Newton’s second law hasthe same form for nonrelativistic and relativistic electrons except that in the latter casethe effective mass is γm. Recall also that for circular motion the rest frame accelerationis γ2 times that in the lab frame.

Page 295: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 12 14

12. a) An oscillating electric dipole of angular frequency ω is located at distance d λaway from a perfectly conducting plane. The dipole is oriented parallel to the plane,as shown below.

Show that the power radiated in the direction (θ, φ) is

dP

dΩ= 4A sin2 θ sin2 Δ, (44)

where

Δ =2πλ

dsin θ cosφ, (45)

and the power radiated by the dipole alone is

dP

dΩ= A sin2 θ. (46)

Sketch the shape of the radiation pattern for d = λ/2 and d = λ/4.

b) Suppose instead that the dipole was oriented perpendicular to the conducting plane.

Show that the radiated power in this case is

dP

dΩ= 4A sin2 θ′ cos2 Δ, (47)

where

Δ =2πλ

dcos θ′. (48)

In parts a) and b), the polar angles θ and θ′ are measured with respect to the axes ofthe dipoles.

c) Repeat parts a) and b) for a magnetic dipole oscillator in the two orientations.

Page 296: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 13 15

13. In an array of antennas, their relative phases can be adjusted as well as their relativespacing, which leads to additional freedom to shape the radiation pattern.

Consider two short, center-fed linear antennas of length L λ, peak current I0 andfrequency ω, as discussed on p. 191 of the Notes. The axes of the antennas are collinear,their centers are λ/4 apart, and the currents have a 90 phase difference.

Show that the angular distribution of the radiated power is

dU

dtdΩ=

ω2

16πc3I20L

2 sin2 θ[1 + sin

2cos θ

)]. (49)

Unlike the radiation patterns of previous examples, this is not symmetric about theplane z = 0. Therefore, this antenna array emits nonzero momentum Prad. As aconsequence, there is a net reaction force F = −dPrad/dt. Show that

F = −1

c

dU

dt

(1 − π2

12

)z. (50)

A variant on this problem is athttp://puhep1.princeton.edu/~mcdonald/examples/endfire.pdf

Page 297: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 14 16

14. a) Consider a full-wave “end-fire” antenna whose current distribution (along the z axis)is

I(z) = I0 sin2πz

Le−iωt, (−L/2 < z < L/2), (51)

where L = λ = 2πc/ω.

Use the result of p. 182 of the Notes to calculate the radiated power “exactly”. Notethat the real part of the integral vanishes, so you must evaluate the imaginary part.Show that

dP

dΩ=

I20

2πc

sin2(π cos θ)

sin2 θ. (52)

Sketch the radiation pattern.

Use tricks like1

1 − u2=

1

1 + u+

1

1 − u(53)

to show that the total radiated power is

P =∫

dP

dΩdΩ =

I20

2c

∫ 4π

0

1 − cos v

vdv =

I20

2cCin(4π) = 3.11

I20

2c. (54)

(c.f. Abramowitz and Stegun, pp. 231, 244.)

b) Calculate the lowest order nonvanishing multipole radiation. You may need the factthat ∫

z2 cos z dz = (z2 − 2) sin z + 2z cos z. (55)

Show that to this order,

P =8π2

15

I20

2c= 5.26

I20

2c. (56)

which gives a sense of the accuracy of the multiple expansion.

Page 298: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 15 17

15. Scattering Off a Conducting Sphere

Calculate the scattering cross section for plane electromagnetic waves of angular fre-quency ω incident on a perfectly conducting sphere of radius a when the wavelengthobeys λ a (ka 1).

Note that both electric and magnetic dipole moments are induced. Inside the sphere,Btotal must vanish. Surface currents are generated such that Binduced = −Bincident, andbecause of the long wavelength, these fields are essentially uniform over the sphere.(c.f. Set 4, Prob. 8a).

Show that

Escat = a3k2 ei(kr−ωt)

r

[(E0 × n) × n− 1

2(B0 × n

], (57)

where n is along the vector r that points from the center of the sphere to the distantobserver.

Suppose that the incident wave propagates in the +z direction, and the electric fieldis linearly polarized along direction l, so E0 = E0 l and l · z = 0. Show that in this casethe scattering cross section can be written,

dΩ= a6k4

⎡⎣(1 − n · z2

)2

− 3

4(l · n)2

⎤⎦ . (58)

Consider an observer in the x-z plane to distinguish between the cases of electricpolarization parallel and perpendicular to the scattering plane to show that

dσ‖dΩ

= a6k4(

1

2− cos θ

)2

,dσ⊥dΩ

= a6k4

(1 − cos θ

2

)2

. (59)

Then, for an unpolarized incident wave, show

dΩ= a6k4

[5

8(1 + cos2 θ) − cos θ

], (60)

and

σ =∫

dΩdΩ =

10π

3a6k4. (61)

Sketch the angular distribution (60). Note that

dσ(180)dΩ

/dσ(0)

dΩ= 9, (62)

so the sphere reflects much more backwards than it radiates forwards.

Is there any angle θ for which the scattered radiation is linearly polarized for unpolar-ized incident waves?

Page 299: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Problem 16 18

16. Exotic radiation effects of charges that move at (essentially) constant velocity butcross boundaries between various media can be deduced from the radiation spectrumequation (14.70) from the textbook of Jackson (p. 182 of the Notes):

dU

dωdΩ=

ω2

4π2c3

[∫dt d3r n× J(r, t)eiω(t−(n·r)/c)

]2, (63)

where dU is the radiated energy in angular frequency interval dω emitting into solidangle dΩ, J is the source current density, and n is a unit vector towards the observer.

Consider the example of the sweeping electron beam in an (analog) oscilloscope. In thefastest of such devices (such as the Tektronix model 7104) the speed of the beam spotacross the face of an oscilloscope can exceed the velocity of light, although of coursethe velocity of the electrons does not. Associated with this possibility there should bea kind of Cerenkov radiation, as if the oscilloscope trace were due to a charge movingwith superluminal velocity.

As a simple model, suppose a line of charge moves in the −y direction with velocityu c, where c is the speed of light, but has a slope such that the intercept with thex axis moves with velocity v > c, as shown in the figure below. If the region y < 0 isoccupied by, say, a metal the charges will emit transition radiation as they disappearinto the metal’s surface. Interference among the radiation from the various chargesthen leads to a strong peak in the radiation pattern at angle cos θ = c/v, which is theCerenkov effect of the superluminal source – all of which can be deduced from eq. (63).

0.3a) A sloping line of charge moves in the −y direction with velocity vy = u csuch that its intercept with the x axis moves with velocity vx = v > c. As thecharge disappears into the conductor at y < 0 it emits transition radiation.The radiation appears to emanate from a spot moving at superluminal velocityand is concentrated on a cone of angle cos−1(c/v). b) The angular distributionof the radiation is discussed in a spherical coordinates system about the x axis.

Page 300: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 1 19

Solutions

1. The suggested approach is to calculate the retarded potentials and then take derivativesto find the fields. The retarded scalar and vector potentials φ and A are given by

φ(x, t) =∫

ρ(x′, t − R/c)

Rd3x′, and A(x, t) =

1

c

∫J(x′, t −R/c)

Rd3x′, (64)

where ρ and J are the charge and current densities, respectively, and R = |x− x′|.In the present case, we assume the wire remains neutral when the current flows (com-pare Prob. 3, Set 4). Then the scalar potential vanishes. For the vector potential, wesee that only the component Az will be nonzero. Also, J d3x′ can be rewritten as I dz forcurrent in a wire along the z-axis. For an observer at (r, 0, 0) and a current element at(0, 0, z), we have R =

√r2 + z2. Further, the condition that I is nonzero only for time

t > 0 implies that it contributes to the fields only for z such that (ct)2 > R2 = r2 + z2.That is, we need to evaluate the integral only for

|z| < z0 ≡√

(ct)2 − r2, (65)

which must be positive to have physical significance. Altogether,

Az(r, 0, 0, t) =α

c

∫ z0

−z0

(t√

r2 + z2− 1

c

)dz =

α

c

(t ln

ct + z0

ct − z0− 2z0

c

)

=2α

c

(t ln

z0 + ct

r− z0

c

). (66)

[The two forms tend to arise depending on whether or not one notices that the integrandis even in z.]

The magnetic field is obtained via B = ∇ × A. Since only Az is nonzero the onlynonzero component of B is

Bφ = −∂Az

∂r=

2αz0

c2r. (67)

[Some chance of algebraic error in this step!]

The only nonzero component of the electric field is

Ez = −1

c

∂Az

∂t= −2α

c2ln

z0 + ct

r. (68)

For long times, ct r, ⇒ z0 ≈ ct, and the fields become

Bφ ≈ 2αt

cr=

2I(t)

cr= B0(t), Ez ≈ −2α

c2ln

2ct

r= −B0

r

ctln

2ct

r B0, (69)

where B0(t) = 2I(t)/cr is the instantaneous magnetic field corresponding to currentI(t). That is, we recover the magnetostatic limit at large times.

For short times, ct = r + ε with ε r, after the fields first become nonzero we have

z0 =√

2rε + ε2 ≈ √2rε, (70)

Page 301: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 1 20

so

Bφ ≈ 2α

c2

√2ε

r, and Ez ≈ −2α

c2ln

r + ε +√

2rε

r≈ −2α

c2

√2ε

r= −Bφ. (71)

In this regime, the fields have the character of radiation, with E and B of equalmagnitude, mutually orthogonal, and both orthogonal to the line of sight to the closestpoint on the wire. (Because of the cylindrical geometry the radiation fields do not have1/r dependence – which holds instead for static fields.)

In sum, the fields build up from zero only after time ct = r. The initial fields propagateoutwards at the speed of light and have the character of cylindrical waves. But at afixed r, the electric field dies out with time, and the magnetic field approaches theinstantaneous magnetostatic field due to the current in the wire.

Of possible amusement is a direct calculation of the vector potential for the case of aconstant current I0.

First, from Ampere’s law we know that Bφ = 2I0/cr = −∂Az/∂r, so we have that

Az = −2I0

cln r + const. (72)

Whereas, if we use the integral form for the vector potential we have

Az(r, 0, 0) =1

c

∫ ∞

−∞I0 dz√r2 + z2

=2I0

c

∫ ∞

0

dz√r2 + z2

= −2I0

cln r + lim

z→∞ ln(z +√

z2 + r2). (73)

Only by ignoring the last term, which does not depend on r for a long wire, do werecover the “elementary” result.

Page 302: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 2 21

2. The expansionr = R − r′ · n + ... (74)

implies that the retarded time derivative of the polarization vector is

[p] = p(r′, t′ = t− r/c) ≈ −iωpω(r′)e−iω(t−R/c+r′·n/c) = −iωei(kR−ωt)pω(r′)e−ikr′·n

≈ −iωei(kR−ωt)pω(r′)(1 − ikr′ · n), (75)

where k = ω/c. Likewise,1

r≈ 1

R

(1 +

r′ · nR

). (76)

Then, the retarded vector potential can be written (in the Lorenz gauge)

A(L) =1

c

∫[p]

rdVol′ ≈ −iω

ei(kR−ωt)

cR

∫pω(r′)

[1 + r′ · n

(1

R− ik

)+ ...

]dVol′, (77)

The electric dipole (E1) approximation is to keep only the first term of eq. (77),

A(L)E1 = −iω

ei(kR−ωt)

cR

∫pω(r′) dVol′ ≡ −ik

ei(kR−ωt)

RP (Lorenz gauge). (78)

We obtain the magnetic field by taking the curl of eq. (78). The curl operation withrespect to the observer acts only on the distance R. In particular,

∇R =R

R= n. (79)

Hence,

BE1 = ∇× A(L)E1 = −ik∇ei(kR−ωt)

R× P = −ik

ei(kR−ωt)

R

(ikn− n

R

)× P

= k2ei(kR−ωt)

R

(1 +

i

kR

)n× P. (80)

The 4th Maxwell equation in vacuum tells us that

∇ ×BE1 =1

c

∂EE1

∂t= −ikEE1. (81)

Hence,

EE1 =i

k∇ × BE1 = ∇ ×

[ei(kR−ωt)

(ik

R2− 1

R3

)R × P

]

= ∇ei(kR−ωt)

(ik

R2− 1

R3

)× (R × P) + ei(kR−ωt)

(ik

R2− 1

R3

)∇ × (R × P)

= ei(kR−ωt)

[−k2

R− 3

(ik

R2− 1

R3

)]n× (n× P) − 2Pei(kR−ωt)

(ik

R2− 1

R3

)

= k2 ei(kR−ωt)

Rn× (P× n) + ei(kR−ωt)

(ik

R2− 1

R3

)[P− 3(P · n)n]. (82)

Page 303: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 2 22

We could also deduce the electric field from the general relation

E = −∇V − 1

c

∂A

∂t= −∇V + ikA. (83)

For this, we need to know the scalar potential V (L), which we can deduce from theLorenz gauge condition:

∇ · A(L) +1

c

∂V (L)

∂t= 0. (84)

For an oscillatory source this becomes

V (L) = − i

k∇ · A(L). (85)

In the electric dipole approximation (78) this yields6

V(L)

E1 = ei(kR−ωt)

(1

R2− ik

R

)(P · n) (Lorenz gauge). (86)

For small R the scalar potential is that of a time-varying dipole,

V(L)E1,near ≈

P · nR2

e−iωt. (87)

The electric field is given by

EE1 = −∇V(L)

E1 + ikA(L)E1

= (P · R)∇ei(kR−ωt)

(ik

R2− 1

R3

)+ ei(kR−ωt)

(ik

R2− 1

R3

)∇(P · R) + k2ei(kR−ωt)

RP

= ei(kR−ωt)

[−k2

R− 3

(ik

R2− 1

R3

)](P · n)n + ei(kR−ωt)

(ik

R2− 1

R3

)P

+k2 ei(kR−ωt)

RP

= k2 ei(kR−ωt)

Rn× (P × n) + ei(kR−ωt)

(ik

R2− 1

R3

)[P− 3(P · n)n], (88)

as before. The angular distribution in the far field (for which the radial dependence is1/R) is n×(P× n) = P−(P · n)n. The isotropic term P is due to the vector potential,while the variable term −(P · n)n is due to the scalar potential and is purely radial.Spherical waves associated with a scalar potential must be radial (longitudinal), butthe transverse character of electromagnetic waves in the far field does not imply theabsence of a contribution of the scalar potential; the latter is needed (in the Lorenzgauge) to cancel to radial component of the waves from the vector potential.

6Equation (86) is not simply the electrostatic dipole potential times a spherical wave because the retardedpositions at time t of the two charges of a point dipole correspond to two different retarded times t′.For a calculation of the retarded scalar potential via V (L) =

∫Vol

[ρ]/r, see sec. 11.1.2 of Introduction toElectrodynamics by D.J. Griffiths.

Page 304: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 2 23

We could also work in the Coulomb gauge, meaning that we set ∇ · A(C) = 0. Recall(Lecture 15, p. 174) that the “wave” equations for the potentials in the Coulomb gaugeare

∇2V (C) = −4πρ, (89)

∇2A(C) − 1

c2

∂2A(C)

∂t2= −4π

cJ +

1

c

∂∇V (C)

∂t. (90)

Equation (89) is the familiar Poisson equation of electrostatics, so the scalar potentialis just the “instantaneous” electric-dipole potential,

V(C)E1 =

P · nR2

e−iωt (Coulomb gauge). (91)

One way to deduce the Coulomb-gauge vector potential is via eq. (83),

A(C)E1 = − i

kEE1 − i

k∇V

(C)E1

= −ikei(kR−ωt)

Rn× (P× n) +

[ei(kR−ωt)

R2+

i(ei(kR−ωt) − e−iωt)

kR3

][P− 3(P · n)n]

≡ A(C)far + A(C)

near (Coulomb gauge). (92)

We learn that the far-zone, Coulomb gauge vector potential (i.e., the part of the vectorpotential that varies as 1/R) is purely transverse, and can be written as

A(C)far = −ik

ei(kR−ωt)

Rn× (P× n) (Coulomb gauge). (93)

Because the radiation part of the Coulomb-gauge vector potential is transverse, theCoulomb gauge is sometimes called the “transverse” gauge.

The Coulomb-gauge scalar potential is negligible in the far zone, and we can say thatthe radiation fields are entirely due to the far-zone, Coulomb-gauge vector potential.That is,

Efar = ikA(C)far = k2 ei(kR−ωt)

Rn× (P× n), (94)

Bfar = ∇ ×A(C)far = ik × A

(C)far = k2 ei(kR−ωt)

Rn× P. (95)

It is possible to choose gauges for the electromagnetic potentials such that some oftheir components appear to propagate at any velocity v, as discussed by J.D. Jackson,Am. J. Phys. 70, 917 (2002) and by K.-H. Yang, Am. J. Phys. 73, 742 (2005).7 The

7http://puhep1.princeton.edu/~mcdonald/examples/EM/jackson_ajp_70_917_02.pdfhttp://puhep1.princeton.edu/~mcdonald/examples/EM/yang_ajp_73_742_05.pdf

Page 305: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 2 24

potentials A(v) and V (v) in the so-called velocity gauge with the parameter v obey thegauge condition

∇ ·A(v) +c

v2

∂V (v)

∂t= 0. (96)

The scalar potential V (v) is obtained by replacing the speed of light c in the Lorenz-gauge scalar potential by v. Equivalently, we replace the wave number k = ω/c byk′ = ω/v. Thus, from eq. (86) we find

V(v)

E1 = −ei(k′R−ωt)

(ik′

R2− 1

R3

)(P ·R) (velocity gauge). (97)

Then, as in eq. (88) we obtain

− ∇V(v)

E1 = ei(k′R−ωt)

[−k′2

R− 3

(ik′

R2− 1

R3

)](P · n)n + ei(kR−ωt)

(ik′

R2− 1

R3

)P

= −k′2 ei(k′R−ωt)

R(P · n)n + ei(k′R−ωt)

(ik′

R2− 1

R3

)[P− 3(P · n)n]. (98)

The vector potential in the v-gauge can be obtained from eq. (83) as

A(v)E1 = − i

kEE1 − i

k∇V

(v)E1

= −ikei(kR−ωt)

Rn × (P × n) + ei(kR−ωt)

(1

R2+

i

kR3

)[P− 3(P · n)n]

−ik′2

k

ei(k′R−ωt)

R(P · n)n− ei(k′R−ωt)

(k′

kR2+

i

kR3

)[P− 3(P · n)n]. (99)

This vector potential includes terms that propagate with velocity v both in the nearand far zones. When v = c, then k′ = k and the velocity-gauge vector potential (99)reduces to the Lorenz-gauge potential (78); and when v → ∞, then k′ = 0 and thevelocity-gauge vector potential reduces to the Coulomb-gauge potential (92).8

Turning to the question of energy flow, we calculate the Poynting vector

S =c

4πE× B, (100)

where we use the real parts of the fields (80) and (82),

E = k2P n×(P× n)cos(kR − ωt)

R+P [3P · n)n−P]

[cos(kR − ωt)

R3+

k sin(kR − ωt)

R2

],

(101)

8Thanks to J.D Jackson and K.-H. Yang for discussions of the Coulomb gauge and the velocity gauge.See also, J.D. Jackson and L.B. Okun, Historical roots of gauge invariance, Rev. Mod. Phys. 73, 663 (2001),http://puhep1.princeton.edu/~mcdonald/examples/EM/jackson_rmp_73_663_01.pdf

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Princeton University 2001 Ph501 Set 8, Solution 2 25

B = k2P (n × P)

[cos(kR − ωt)

R− sin(kR − ωt)

kR2

]. (102)

The Poynting vector contains six terms, some of which do not point along the radialvector n:

S =c

k4P 2[n× (P× n)] × (n × P)

[cos2(kR − ωt)

R2− cos(kR − ωt) sin(kR − ωt)

kR3

]

+k2P 2[3(P · n)n− P] × (n× P)

[cos2(kR − ωt) − sin2(kR − ωt)

R4

+cos(kR − ωt) sin(kR − ωt)

(k

R3− 1

kR5

)]

=c

k4P 2 sin2 θn

[cos2(kR − ωt)

R2− cos(kR − ωt) sin(kR − ωt)

kR3

]

+k2P 2[4 cos θP + (3 cos2 θ − 1)n]

[cos2(kR − ωt) − sin2(kR − ωt)

R4

+cos(kR − ωt) sin(kR − ωt)

(k

R3− 1

kR5

)], (103)

where θ is the angle between vectors n and P. As well as the expected radial flowof energy, there is a flow in the direction of the dipole moment P. Since the productcos(kR − ωt) sin(kR − ωt) can be both positive and negative, part of the energy flowis inwards at times, rather than outwards as expected for pure radiation.

However, we obtain a simple result if we consider only the time-average Poyntingvector, 〈S〉. Noting that 〈cos2(kR − ωt)〉 =

⟨sin2(kR − ωt)

⟩= 1/2 and

〈cos(kR − ωt) sin(kR − ωt)〉 = (1/2) 〈sin 2(kR − ωt)〉 = 0, eq (103) leads to

〈S〉 =ck4P 2 sin2 θ

8πR2n. (104)

The time-average Poynting vector is purely radially outwards, and falls off as 1/R2

at all radii, as expected for a flow of energy that originates in the oscillating pointdipole (which must be driven by an external power source). The time-average angulardistribution d 〈P 〉 /dΩ of the radiated power is related to the Poynting vector by

d 〈P 〉dΩ

= R2n · 〈S〉 =ck4P 2 sin2 θ

8π=

P 2ω4 sin2 θ

8πc3, (105)

which is the expression often quoted for dipole radiation in the far zone. Here we seethat this expression holds in the near zone as well.

We conclude that radiation, as measured by the time-average Poynting vector, existsin the near zone as well as in the far zone.

Our considerations of an oscillating electric dipole can be extended to include an oscil-lating magnetic dipole by noting that if E(r, t) and B(r, t) are solutions to Maxwell’s

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Princeton University 2001 Ph501 Set 8, Solution 2 26

equations in free space (i.e., where the charge density ρ and current density J are zero),then the dual fields

E′(r, t) = −B(r, t), B′(r, t) = E(r, t), (106)

are solutions also. The Poynting vector is the same for the dual fields as for the originalfields,

S′ =c

4πE′ × B′ = − c

4πB× E = S. (107)

Taking the dual of fields (10)-(11), we find the fields

E′ = EM1 = −k2ei(kR−ωt)

R

(1 +

i

kR

)n× M, (108)

B′ = BM1 = k2 ei(kR−ωt)

R

n × (M× n) + [3(n · M)n− M]

(1

k2R2− i

kR

). (109)

which are also solutions to Maxwell’s equations. These are the fields of an oscillatingpoint magnetic dipole, whose peak magnetic moment is M. In the near zone, themagnetic field (109) looks like that of a (magnetic) dipole.

While the fields of eqs. (10)-(11) are not identical to those of eqs. (108)-(109), thePoynting vectors are the same in the two cases. Hence, the time-average Poyntingvector, and also the angular distribution of the time-averaged radiated power are thesame in the two cases. The radiation of a point electric dipole is the same as thatof a point magnetic dipole (assuming that M = P), both in the near and in the farzones. Measurements of only the intensity of the radiation could not distinguish thetwo cases.

However, if measurements were made of both the electric and magnetic fields, then thenear zone fields of an oscillating electric dipole, eqs. (10)-(11), would be found to bequite different from those of a magnetic dipole, eqs. (108)-(109). This is illustrated inthe figure on the previous page, which plots the ratio E/H = E/B of the magnitudesof the electric and magnetic fields as a function of the distance r from the center ofthe dipoles.

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Princeton University 2001 Ph501 Set 8, Solution 2 27

To distinguish between the cases of electric and magnetic dipole radiation, it sufficesto measure only the polarization (i.e., the direction, but not the magnitude) of eitherthe electric of the magnetic field vectors.

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Princeton University 2001 Ph501 Set 8, Solution 3 28

3. The rotating dipole p can be thought of as two oscillating linear dipoles oriented 90

apart in space, and phased 90 apart in time. This is conveniently summarized incomplex vector notation:

p = p0(x + iy)e−iωt, (110)

for a rotation from the +x axis towards the +y axis. Thus,

[p] = p(t′ = t − r/c) = −ω2p0(x + iy)ei(kr−ωt). (111)

The radiation fields of this oscillating dipole are given by

Brad =[p] × n

c2r= −k2p0e

−i(kr−ωt)

r(x + iy) × n

=k2p0e

−i(kr−ωt)

r(cos θ y − i l), (112)

Erad = Brad × n =k2p0e

−i(kr−ωt)

r(cos θ l + i y). (113)

The time-averaged angular distribution of the radiated power is given by

d 〈P 〉dΩ

=c

8πr2 |Brad|2 =

c

8πk4p2

0(1 + cos2 θ), (114)

since l and y are orthogonal. The total radiated power is therefore

〈P 〉 =∫

d 〈P 〉dΩ

dΩ =2c

3k4p2

0 =2

3c3ω4p2

0. (115)

The total power also follows from the Larmor formula,

〈P 〉 =1

2

2 |p|23c3

=2

3c3ω4p2

0, (116)

since |p| =√

2ω2p0 in the present example.

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Princeton University 2001 Ph501 Set 8, Solution 4 29

4. According to the Larmor formula, the rate of magnetic dipole radiation is

dU

dt=

2

3

m2

c3=

2

3

m2ω4

c3, (117)

where ω = 2π/T is the angular velocity, taken to be perpendicular to the magneticdipole moment m.

The radiated power (117) is derived from a decrease in the rotational kinetic energy,U = Iω2/2, of the pulsar:

dU

dt= −Iωω =

2

5MR2ω |ω| , (118)

where the moment of inertia I is taken to be that of a sphere of uniform mass density.Combining eqs. (117) and (118), we have

m2 =3

5

MR2 |ω| c3

ω3. (119)

Substituting ω = 2π/T , and |ω| = 2π∣∣∣T ∣∣∣ /T 2, we find

m2 =3

20π2MR2T

∣∣∣T ∣∣∣ c3. (120)

The static magnetic field B due to dipole m is

B =3(m · r)r− m

r3, (121)

so the peak field at radius R is

B =2m

R3. (122)

Inserting this in eq. (120), the peak surface magnetic field is related by

B2 =3

5π2

MT∣∣∣T ∣∣∣ c3

R4=

3

5π2

(2.8 × 1033)(7.5)(8 × 10−11)(3 × 1010)3

(106)4= 2.8×1030 gauss2.

(123)Thus, Bpeak = 1.7 × 1015 G = 38Bcrit, where Bcrit = 4.4 × 1013 G.

When electrons and photons of kinetic energies greater than 1 MeV exist in a magneticfield with B > Bcrit, they rapidly lose this energy via electron-positron pair creation.

Kouveliotou et al. report that Bpeak = 8 × 1014 G without discussing details of theircalculation.

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Princeton University 2001 Ph501 Set 8, Solution 5 30

5. The time-average field momentum density is given in terms of the Poynting vector as(the real part of)

〈P〉field =〈S〉c2

c

8πE × B. (124)

Hence, the time-averaged angular momentum density is

〈L〉field = r × Pfield =1

8πcr × (E × B) =

1

8πcr[E(n ·B) − B(n ·E)]. (125)

writing r = rn.

The time-average rate of radiation of angular momentum into solid angle dΩ is therefore

d 〈L〉dt dΩ

= cr2 〈L〉field =1

8πr3[E(n · B) − B(n · E)], (126)

since the angular momentum density L is moving with velocity c.

The radiation fields of an oscillating electric dipole moment p including both the 1/rand 1/r2 terms of eqs. (80) and (88) are,

E = kei(kr−ωt)

r

[(k +

i

r

)p−

(k +

3i

r

)(n · p)n

], (127)

B = k2 ei(kr−ωt)

r

(1 +

i

kr

)(n × p). (128)

Since n · B = 0 for this case, only the second term in eq. (126) contributes to theradiated angular momentum. We therefore find

d 〈L〉dt dΩ

= −k3r

(1 − i

kr

)(n× p)

(−2i

r

)(n · p) =

ik3

4π(n · p)(n× p), (129)

ignoring terms in the final expression that have positive powers of r in the denominator,as these grow small at large distances.

For the example of a rotating dipole moment (prob. 2),

p = p0(x + iy)e−iωt, (130)

we have

d 〈L〉dt dΩ

= Reik3p2

0

4π[n · (x + iy)][n× (x− iy)] = Re

ik3p20

4πsin θ[cos θ y + i l]

= − k3

4πp2

0 sin θ l. (131)

To find d 〈L〉 /dt we integrate eq. (130) over solid angle. When vector n is in the x-zplane, vector l can be expressed as

l = cos θ x− sin θ z. (132)

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Princeton University 2001 Ph501 Set 8, Solution 5 31

As we integrate over all directions of n, the contributions to d 〈L〉 /dt in the x-y planesum to zero, and only its z component survives. Hence,

d 〈L〉dt

= z∫

d 〈Lz〉dt dΩ

dΩ = 2πk3

4πp2

0 z∫ 1

−1sin2 θ d cos θ =

2k3

3p2

0 z

=2ck3

3ωp2

0 z =〈P 〉ω

z, (133)

recalling eq. (116) for the radiated power 〈P 〉. Of course, the motion described byeq. (130) has its angular momentum along the +z axis.

Page 313: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 6 32

6. The magnetic field radiated by a time-dependent, axially symmetric quadrupole isgiven by

B =[...

Q] × n

6c3r(134)

where the unit vector n has rectangular components

n = (sin θ, 0, cos θ), (135)

and the quadrupole vector Q is related to the quadrupole tensor Qij by

Qi = Qijnj. (136)

The charge distribution is symmetric about the z axis, so the quadrupole momenttensor Qij may be expressed entirely in terms of

Qzz =∫

ρ(3z2 − r2) dVol = −4a2e cos2 ωt = −2a2e(1 + cos 2ωt). (137)

Thus,

Qij =

⎛⎜⎜⎜⎜⎜⎝−Qzz/2 0 0

0 −Qzz/2 0

0 0 Qzz

⎞⎟⎟⎟⎟⎟⎠ , (138)

and the quadrupole vector can be written as

Q =

(−Qzz sin θ

2, 0, Qzz cos θ

)= −Qzz

2n +

3Qzz cos θ

2z

= a2e(1 + cos 2ωt)(n− 3z cos θ). (139)

Then,[...

Q] = 8ω3a2e sin 2ωt′(n− 3z cos θ), (140)

where the retarded time is t′ = t − r/c. Hence,

B =[...

Q] × n

6c3r= −4k3a2e

ry sin(2kr − 2ωt) sin θ cos θ, (141)

since z × n = y sin θ. The radiated electric field is given by

E = B× n = −4k3a2e

rl sin(2kr − 2ωt) sin θ cos θ, (142)

using y × n = l.

As we are not using complex notation, we revert to the basic definitions to find thatthe time-averaged angular distribution of radiated power is

d 〈P 〉dΩ

= r2 〈S〉 · n =cr2

4π〈E × B · n〉 =

2ck6a4e2

πsin2 θ cos2 θ, (143)

since l × y = n. This integrates to give

〈P 〉 = 2π∫ 1

−1

d 〈P 〉dΩ

d cos θ =16ck6a4e2

15. (144)

Page 314: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 7 33

7. Since the charge is assumed to rotate with constant angular velocity, the magneticmoment it generates is constant in time, and there is no magnetic dipole radiation.Hence, we consider only electric quadrupole radiation in addition to the electric dipoleradiation. The radiated fields are therefore

B =[p] × n

c2r+

[...

Q] × n

6c3r, E = B × n. (145)

The electric dipole radiation fields are given by eqs. (112) and 113) when we writep0 = ae.

The present charge distribution is not azimuthally symmetric about any fixed axis, sowe must evaluate the full quadrupole tensor,

Qij = e(3rirj − r2δij). (146)

to find the components of the quadrupole vector Q. The position vector of the chargehas components

ri = (a cos ωt, a sinωt, 0), (147)

so the nonzero components of Qij are

Qxx = e(3x2 − r2) = a2e(3 cos2 ωt − 1) =a2e

2(1 + 3 cos 2ωt), (148)

Qyy = e(3y2 − r2) = a2e(3 sin2 ωt − 1) =a2e

2(1 − 3 cos 2ωt), (149)

Qzz = −er2 = −a2e, (150)

Qxy = Qyx = 3exy = 3a2e sin ωt cosωt =3a2e

2sin 2ωt. (151)

Only the time-dependent part of Qij contributes to the radiation, so we write

Qij(time dependent) =3a2e

2

⎛⎜⎜⎜⎜⎜⎝cos 2ωt sin 2ωt 0

sin 2ωt − cos 2ωt 0

0 0 0

⎞⎟⎟⎟⎟⎟⎠ . (152)

The unit vector n towards the observer has components given in eq. (135), so thetime-dependent part of the quadrupole vector Q has components

Qi = Qijnj =3a2e

2(cos 2ωt sin θ, sin 2ωt sin θ, 0). (153)

Thus,

[...

Qi] =...

Qi(t′ = t − r/c) = −12a2eω3 sin θ(sin(2kr − 2ωt), cos(2kr − 2ωt), 0). (154)

It is preferable to express this vector in terms of the orthonormal triad n, y, andl = y × n, by noting that

x = n sin θ − l cos θ. (155)

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Princeton University 2001 Ph501 Set 8, Solution 7 34

Hence,

[...

Q] = −12a2eω3 sin θ(n sin θ sin(2kr − 2ωt)− l cos θ sin(2kr − 2ωt)+ y cos(2kr − 2ωt)).(156)

The field due to electric quadrupole radiation are therefore

BE2 =[...

Q] × n

6c3r= −2a2ek3

rsin θ(l cos(2kr − 2ωt) + y cos θ sin(2kr − 2ωt)),(157)

EE2 = BE2 × n =2a2ek3

rsin θ(y cos(2kr − 2ωt) − l cos θ sin(2kr − 2ωt)). (158)

The angular distribution of the radiated power can be calculated from the combinedelectric dipole and electric quadrupole fields, and will include a term ∝ k4 due only todipole radiation as found in prob. 2, a term ∝ k6 due only to quadrupole radiation,and a complicated cross term ∝ k5 due to both dipole and quadrupole field. Here, weonly display the term due to the quadrupole fields by themselves:

d 〈PE2〉dΩ

=cr2

4π〈EE2 × BE2 · n〉 =

ca4e2k6

2π(1 − cos4 θ), (159)

which integrates to give

〈PE2〉 = 2π∫ 1

−1

d 〈P 〉dΩ

d cos θ =8ca4e2k6

5. (160)

Page 316: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 8 35

8. a) The dominant energy loss is from electric dipole radiation, which obeys eq. (25),

dU

dt= −〈PE1〉 = −2a2e2ω4

3c3. (161)

For an electron of charge −e and mass m in an orbit of radius a about a fixed nucleusof charge +e, F = ma tells us that

e2

a2= m

v2

a= mω2a, (162)

so that

ω2 =e2

ma3, (163)

and also the total energy (kinetic plus potential) is

U = −e2

a+

1

2mv2 = − e2

2a(164)

Using eqs. (163) and (164) in (161), we have

dU

dt=

e2

2a2a = − 2e6

3a4m2c3, (165)

or

a2a =1

3

da3

dt= − 4e4

3m2c3= −4

3r20c, (166)

where r0 = e2/mc2 is the classical electron radius. Hence,

a3 = a30 − 4r2

0ct. (167)

The time to fall to the origin is

tfall =a3

0

4r20c

. (168)

With r0 = 2.8 × 10−13 cm and a0 = 5.3 × 10−9 cm, tfall = 1.6 × 10−11 s.

This is of the order of magnitude of the lifetime of an excited hydrogen atom, but theground state appears to have infinite lifetime.

This classical puzzle is pursued further in prob. 7.

b) The analog of the quadrupole factor ea2 in prob. 5 for masses m1 and m2 in circularorbits with distance a between them is m1r

21 + m2r

22, where r1 and r2 are measured

from the center of mass. That is,

m1r1 = m2r2, and r1 + r2 = a, (169)

so thatr1 =

m2

m1 + m2a, r2 =

m1

m1 + m2a, (170)

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Princeton University 2001 Ph501 Set 8, Solution 8 36

and the quadrupole factor is

m1r21 + m2r

22 =

m1m2

m1 + m2a2. (171)

We are then led by eq. (26) to say that the power in gravitational quadrupole radiationis

PG2 =8

5

G

c5

(m1m2

m1 + m2

)2

a4ω6. (172)

We insert a single factor of Newton’s constant G in this expression, since it has dimen-sions of mass2, and Gm2 is the gravitational analog of the square of the electric chargein eq. (26).

We note that a general relativity calculation yields a result a factor of 4 larger thaneq. (172):

PG2 =32

5

G

c5

(m1m2

m1 + m2

)2

a4ω6. (173)

To find tfall due to gravitational radiation, we follow the argument of part a):

Gm1m2

a2= m1

v21

r1= m1ω

2r1 = m2v2

2

r2, (174)

so that

ω2 =G(m1 + m2)

a3, (175)

and also the total energy (kinetic plus potential) is

U = −Gm1m2

a+

1

2m1v

21 +

1

2m2v

22 = −Gm1m2

2a(176)

Using eqs. (175) and (176) in (173), we have

dU

dt=

Gm1m2

2a2a = −32G4m2

1m22(m1 + m2)

5a5c5, (177)

or

a3a =1

4

da4

dt= −64G3m1m2(m1 + m2)

5c5. (178)

Hence,

a4 = a40 −

256G3m1m2(m1 + m2)

5c5t. (179)

The time to fall to the origin is

tfall =5a4

0c5

256G3m1m2(m1 + m2). (180)

For the Earth-Sun system, a0 = 1.5 × 1013 cm, m1 = 6 × 1027 gm, m2 = 2 × 1033 cm,and G = 6.7 × 10−10 cm2/(g-s2), so that tfall ≈ 1.5 × 1036 s ≈ 5 × 1028 years!

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Princeton University 2001 Ph501 Set 8, Solution 9 37

9. The solution given here follows the succinct treatment by Landau, Classical Theory ofFields, sec. 74.

For charges in steady motion at angular frequency ω in a ring of radius a, the currentdensity J is periodic with period 2π/ω, so the Fourier analysis (34) at the retardedtime t′ can be evaluated via the usual approximation that r ≈ R − r′ · n, where R isthe distance from the center of the ring to the observer, r′ points from the center ofthe ring to the electron, and n is the unit vector pointing from the center of the ringto the observer. Then,

[J] = J(r′, t′ = t − r/c) =∑m

Jm(r′)e−imω(t−R/c+r′·n/c

=∑m

eim(kR−ωt)Jm(r′)e−imωr′·n/c, (181)

where k = ω/c.

We first consider a single electron, whose azimuth varies as φ = ωt + φ0, and whosevelocity is, of course, v = aω. The current density of a point electron of charge e canbe written in a cylindrical coordinate system (ρ, φ, z) (with volume element ρdρ dφ dz)using Dirac delta functions as

J = ρchargevφ = evδ(ρ − a)δ(z)δ(ρ(φ − ωt − φ0))φ. (182)

The Fourier components Jm are given by

Jm =1

T

∫ T

0J(r, t)eimt dt = evδ(ρ − a)δ(z)

eim(φ−φ0)

ρωTφ. (183)

Also,

r′ = ρ(cos φ x+sinφ y), n = sin θ x+cos θ z, and φ = − sinφ x+cosφ y. (184)

Using eqs. (183) and (184) in (181) and noting that ωT = 2π, we find

[J] =ev

2πρ

∑m

eim(kR−ωt)eim(φ−φ0−ωρ sin θ cosφ/c)δ(ρ − a)δ(z)φ. (185)

Inserting this in eq. (33), we have

A ≈ 1

cR

∫[J]ρ dρ dφ dz =

ev

2πcR

∑m

eim(kR−ωt−φ0)∫ 2π

0eim(φ−ωa sin θ cosφ/c)φ dφ

=∑m

Ame−imωt, (186)

so that the Fourier components of the vector potential are

Am =ev

2πcReim(kR−φ0)

∫ 2π

0eim(φ−v sin θ cosφ/c)(− sinφ x + cosφ y) dφ. (187)

Page 319: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 9 38

The integrals yield Bessel functions with the aid of the integral representation (40).The y part of eq. (187) can be found by taking the derivative of this relation withrespect to z:

J ′m(z) = − im+1

∫ 2π

0eimφ−iz cosφ cosφ dφ, (188)

For the x part of eq. (187) we play the trick

0 =∫ 2π

0ei(mφ−z cosφ)d(mφ − z cosφ)

= m∫ 2π

0eimφ−iz cosφ dφ + z

∫ 2π

0eimφ−iz cosφ sinφ dφ, (189)

so that

1

∫ 2π

0eimφ−iz cosφ sinφ dφ = −m

z

1

∫ 2π

0eimφ−iz cosφ dφ = − m

imzJm(z). (190)

Using eqs. (188) and (190) with z = mv sin θ/c in (187) we have

Am =ev

cReim(kR−φ0)

(1

imv sin θ/cJm(mv sin θ/c) x− 1

im+1J ′

m(mv sin θ/c) y

). (191)

We skip the calculation of the electric and magnetic fields from the vector potential,and proceed immediately to the angular distribution of the radiated power accordingto eq. (39),

dPm

dΩ=

cR2

2π|imkn× Am|2 =

ck2m2R2

2π|n× Am|2

=ck2m2R2

(cos2 θ |Am,x|2 + |Am,y|2

)=

ce2k2m2

(cot2 θJ2

m(mv sin θ/c) +v2

c2J

′2m(mv sin θ/c)

). (192)

The present interest in this result is for v/c 1, but in fact it holds for any value ofv/c. As such, it can be used for a detailed discussion of the radiation from a relativisticelectron that moves in a circle, which emits so-called synchrotron radiation. This topicis discussed further in Lecture 20 of the Notes.

We now turn to the case of N electrons uniformly spaced around the ring. The initialazimuth of the nth electron can be written

φn =2πn

N. (193)

The mth Fourier component of the total vector potential is simply the sum of compo-nents (191) inserting φn in place of φ0:

Am =N∑

n=1

ev

cReim(kR−φn)

(1

imv sin θ/cJm(mv sin θ/c)x − 1

im+1J ′

m(mv sin θ/c)y

)(194)

=eveimkR

cR

(1

imv sin θ/cJm(mv sin θ/c)x − 1

im+1J ′

m(mv sin θ/c)y

)N∑

n=1

e−i2πmn/N .

Page 320: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 9 39

This sum vanishes unless m is a multiple of N , in which case the sum is just N . Thelowest nonvanishing Fourier component has order N , and the radiation is at frequencyNω. We recognize this as Nth-order multipole radiation, whose radiated power followsfrom eq. (192) as

dPN

dΩ=

ce2k2N2

(cot2 θJ2

N (Nv sin θ/c) +v2

c2J

′2N (Nv sin θ/c)

). (195)

For large N but v/c 1 we can use the asymptotic expansion (41), and its derivative,

J ′m(mx) ≈ (ex/2)m

√2πm x

(m 1, x 1), (196)

to write eq. (195) as

dPN

dΩ≈ ce2k2N

4π2 sin2 θ

(e

2

v

csin θ

)2N

(1 + cos2 θ) NdPE1

dΩ(N 1, v/c 1). (197)

In eqs. (196) and (197) the symbol e inside the parentheses is not the charge but ratherthe base of natural logarithms, 2.718...

For currents in, say, a loop of copper wire, v ≈ 1 cm/s, so v/c ≈ 10−10, while N ≈ 1023.The radiated power predicted by eq. (197) is extraordinarily small!

Note, however, that this nearly complete destructive interference depends on the elec-trons being uniformly distributed around the ring. Suppose instead that they weredistributed with random azimuths φn. Then the square of the magnetic field at orderm has the form

|Bm|2 ∝∣∣∣∣∣

N∑n=1

e−imφn

∣∣∣∣∣2

= N +∑l =n

e−im(φl−φn) = N. (198)

Thus, for random azimuths the power radiated by N electrons (at any order) is justN times that radiated by one electron.

If the charge carriers in a wire were localized to distances much smaller than theirseparation, radiation of “steady” currents could occur. However, in the quantum viewof metallic conduction, such localization does not occur.

The random-phase approximation is relevant for electrons in a so-called storage ring,for which the radiated power is a major loss of energy – or source of desirable photonbeams of synchrotron radiation, depending on one’s point of view. We cannot expoundhere on the interesting topic of the “formation length” for radiation by relativisticelectrons, which length sets the scale for interference of multiple electrons. See, forexample, http://puhep1.princeton.edu/~mcdonald/accel/weizsacker.pdf

Page 321: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 10 40

10. We repeat the derivation of Prob. 1, this time emphasizing the advanced fields.

The advanced vector potential for the point electric dipole p = p0e−iωt located at the

origin is

AE1,adv =pcr

=p(t′ = t + r/c)

cr= −iω

e−i(kr+ωt)

crp0 = −ik

e−i(kr+ωt)

rp0, (199)

where k = ω/c.

We obtain the magnetic field by taking the curl of eq. (199),

BE1,adv = ∇ × AE1,adv = −ik∇e−i(kr+ωt)

r× p0 = −ik

e−i(kr+ωt)

r

(−ikn− n

r

)× p0

= k2 e−i(kr+ωt)

r

(−1 +

i

kr

)n × p0. (200)

Then,

EE1,adv =i

k∇× BE1,adv = −∇ ×

[e−i(kr+ωt)

(ik

r2+

1

r3

)r × p0

]

= −∇e−i(kr+ωt)

(ik

r2+

1

r3

)× (r × p0) − e−i(kr+ωt)

(ik

r2+

1

r3

)∇ × (r × p0)

= e−i(kr+ωt)

[−k2

r+ 3

(ik

r2+

1

r3

)]n × (n × p0) + 2p0e

−i(kr+ωt)

(ik

r2+

1

r3

)

= e−i(kr+ωt)

[−k2

r+ 3

(ik

r2+

1

r3

)](p0 · n)n +

[k2

r− ik

r2− 1

r3

]p0

. (201)

The retarded fields due to a point dipole −p are, from Prob. 1,

BE1,ret = −k2 ei(kr−ωt)

r

(1 +

i

kr

)n× p0, (202)

EE1,ret = ei(kr−ωt)

[k2

r+ 3

(ik

r2− 1

r3

)](p0 · n)n−

[k2

r+

ik

r2− 1

r3

]p0

.(203)

We now consider the superposition of the fields (200)-(203) inside a conducting sphereof radius a. The spatial part of the total electric field is then

EE1 =

[(k2

r− 3

r3

)(eikr − e−ikr) + 3

ik

r2(eikr + e−ikr)

](p0 · n)n

−[(

k2

r− 1

r3

)(eikr − e−ikr) +

ik

r2(eikr + e−ikr)

]p0

= 2i

[(k2

r− 3

r3

)sin kr + 3

k

r2cos kr

](p0 · n)n

−2i

[(k2

r− 1

r3

)sin kr +

k

r2cos kr

]p0. (204)

Page 322: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 10 41

Remarkably, this electric field is finite at the origin, although each of the fields (201)and (203) diverges there. We also recognize that this electric field could be expressedin terms of the so-called spherical Bessel functions,

j0(x) =sinx

x, j1(x) =

sinx

x2− cosx

x, j2(x) =

(3

x3− 1

x

)sinx− 3 cos x

x2, ... (205)

An expansion of the spherical cavity field in terms of spherical Bessel functions occurs“naturally” when we use the more standard approach to this problem, seeking solutionsto the Helmholtz wave equation via separation of variables in spherical coordinates. SeeElectromagnetic Theory by J.A. Stratton (McGraw-Hill, New York, 1941) for detailsof this method.

Because the sum of the magnetic fields (200) and (202) is purely transverse, this cavitymode is called a TM mode.

The boundary conditions at the surface of the sphere are that the radial componentof the magnetic field and the transverse component of the electric field must vanish.Since the magnetic fields (200) and (202) are transverse at any radius, we examine theelectric field at r = a. Of the terms in eq. (204), only those in p0 have transversecomponents, so the boundary condition is

0 =k

a2cos ka + sin ka

(k2

a− 1

a3

), (206)

or

cot ka =1

ka− ka ⇒ ka = 2.744. (207)

In case of a point magnetic dipole m = m0e−iωt at the origin, the fields have the same

form as for an electric dipole, but with E and B interchanged. That is, the advancedfields would be

EM1,adv = k2e−i(kr+ωt)

r

(−1 +

i

kr

)n× m0, (208)

BM1,adv = e−i(kr+ωt)

[−k2

r+ 3

(ik

r2+

1

r3

)](m0 · n)n +

[k2

r− ik

r2− 1

r3

]m0

,(209)

and the retarded field due to magnetic dipole −m would be

EM1,ret = −k2ei(kr−ωt)

r

(1 +

i

kr

)n ×m0, (210)

BM1,ret = ei(kr−ωt)

[k2

r+ 3

(ik

r2− 1

r3

)](m0 · n)n−

[k2

r+

ik

r2− 1

r3

]m0

.(211)

If the advanced and retarded magnetic dipole fields are superposed inside a spheri-cal cavity of radius a, the condition that the transverse electric field vanish at theconducting surface is

0 = eika[1 +

i

ka

]+ e−ika

[1 − i

ka

]= 2

[cos ka − sin ka

ka

], (212)

Page 323: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 10 42

ortan ka = ka ⇒ ka = 4.493. (213)

The electric field of this mode is purely transverse, so it is called a TE mode.

Clearly, other modes of a spherical cavity can be found by superposing the advancedand retarded fields due to higher multipoles at the origin.

Page 324: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 11 43

11. This problem is due to D. Iwanenko and I. Pomeranchuk, On the Maximal EnergyAttainable in a Betatron, Phys. Rev. 65, 343 (1944).9

The electron is held in its circular orbit by the Lorentz force due to the field B.Newton’s law, F = ma, for this circular motion can be written

F = γma =γmv2

R= e

v

cB. (214)

For a relativistic electron, v ≈ c, so we have

γ ≈ eRB

mc2. (215)

The electron is being accelerated by the electric field that is induced by the changingmagnetic flux. Applying the integral form of Faraday’s law to the circle of radius R,we have (ignoring the sign)

2πREφ =Φ

c=

πR2Bave

c, (216)

and hence,

Eφ =RBave

2c, (217)

The rate of change of the electron’s energy E due to Eφ is

dEdt

= F · v ≈ ecEφ =eRBave

2, (218)

Since E = γmc2, we can write

γmc2 =eRBave

2, (219)

which integrates to

γ =eRBave

2mc2. (220)

Comparing with eq. (215), we find the required condition on the magnetic field:

B =Bave

2. (221)

As the electron accelerates it radiates energy at rate given by the Larmor formula inthe rest frame of the electron,

dE

dt= −2e2p2

3c3= −2e2a2

3c3(222)

Because E and t are both the time components of 4-vectors their transforms fromthe rest frame to the lab frame have the same form, and the rate dE/dt is invariant.

9http://puhep1.princeton.edu/~mcdonald/examples/EM/iwanenko_pr_65_343_44.pdf

Page 325: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 11 44

However, acceleration at right angles to velocity transforms according to a = γ2a.Hence, the rate of radiation in the lab frame is

dEdt

= −2e2γ4a2

3c3= −2e4γ2B2

3m2c3, (223)

using eq. (214) for the acceleration a.

The maximal energy of the electrons in the betatron obtains when the energy loss (223)cancels the energy gain (218), i.e., when

eRBave

2=

2e4γ2maxB

2

3m2c3, (224)

and

γmax =

√3m2c3RBave

4e3B2=

√3R

4αc

Bave

B

Bcrit

B≈√

3R

4αcτ

Bcrit

B, (225)

where α = e2/hc = 1/137 is the fine structure constant, Bcrit = m2c3/eh = 4.4×1013 Gis the so-called QED critical field strength, and τ is the characteristic cycle time of thebetatron such that Bave = B/τ . For example, with R = 1 m, τ = 0.03 sec (30 Hz),and B = 104 G, we find that γmax ≈ 200, or Emax ≈ 100 MeV.

We have ignored the radiation due to the longitudinal acceleration of the electron,since in the limiting case this acceleration ceases.

Page 326: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 12 45

12. Since the dipole is much less than a wavelength away from the conducting plane, thefields between the dipole and the plane are essentially the instantaneous static fields.Thus, charges arranges themselves on the plane as if there were an image dipole atdistance d on the other side of the plane. The radiation from the moving charges onthe plan is effectively that due to the oscillating image dipole. A distant observer seesthe sum of the radiation fields from the dipole and its image.

The image dipole is inverted with respect to the original, i.e., the two dipoles are 180

out of phase.

Furthermore, there is a difference s in path length between the two dipole and thedistant observer at angles (θ, φ). We first calculate in a spherical coordinate systemwith z axis along the first dipole, and x axis pointing from the plane to that dipole.Then the path difference is

s = 2dx · n = 2d sin θ cos φ. (226)

This path difference results in an additional phase difference δ between the fields fromthe two dipoles at the observer, in the amount

δ = 2πs

λ=

4πd

λsin θ cos φ. (227)

If we label the electric fields due to the original and image dipoles as E1 and E2,respectively, then the total field is

E = E1 + E2 = E1(1 − eiδ), (228)

and, recalling eq. (46), the power radiated is

dP

dΩ=

|E|2|E1|2

dP1

dΩ=∣∣∣1 − eiδ

∣∣∣2 A sin2 θ = 2A sin2 θ(1 − cos δ) = 4A sin2 θ sin2 δ/2

= 4A sin2 θ sin2 Δ, (229)

where

Δ =δ

2=

2πd

λsin θ cos φ. (230)

Suppose we had chosen to use a spherical coordinate system (r, θ′, φ′) with the z′ axispointing from the plane to dipole 1, and the x′ axis parallel to dipole 1. Then the

Page 327: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 12 46

phase difference would have the simple form

Δ =δ

2=

π

λ2d z′ · n =

2πd

λcos θ′, (231)

but the factor sin2 θ would now become

sin2 θ = n2x + n2

y = n2z′ + n2

y′ = cos2 θ′ + sin2 θ′ cos2 φ′ = 1 − sin2 θ′ sin2 φ′. (232)

If d = λ/4, thendP

dΩ= 4A sin2 θ sin2

2sin θ cos φ

). (233)

In the “side” view, φ = 0, so the pattern has shape

sin2 θ sin2(

π

2sin θ

)(side view), (234)

while in the “top” view, θ = π/2 and the shape is

sin2(

π

2cos φ

)(top view). (235)

This pattern has a single lobe in the forward hemisphere, as illustrated below:

If instead, d = λ/2, then

dP

dΩ= 4A sin2 θ sin2 (π sin θ cos φ) . (236)

In the “side” view, φ = 0, so the pattern has shape

sin2 θ sin2 (π sin θ) (side view), (237)

Page 328: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 12 47

while in the “top” view, θ = π/2 and the shape is

sin2 (π cos φ) (top view). (238)

This pattern does not radiate along the line from the plane to the dipole, as illustratedbelow:

b) If the electric dipole is aligned with the line from the plane to the dipole, its imagehas the same orientation.

The only phase difference between the radiation fields of the dipole and its image isthat due to the path difference δ, whose value has been given in eqs. (230) and (231).It is simpler to use the angles (θ′, φ′) in this case, since the radiation pattern of a singledipole varies as sin2 θ′. Then,

E = E1 + E2 = E1(1 + eiδ), (239)

and, recalling eq. (46), the power radiated is

dP

dΩ=

|E|2|E1|2

dP1

dΩ=∣∣∣1 + eiδ

∣∣∣2 A sin2 θ′ = 2A sin2 θ′(1 + cos δ) = 4A sin2 θ′ cos2 δ/2

= 4A sin2 θ′ cos2 Δ, (240)

with

Δ =δ

2=

2πd

λcos θ′. (241)

This radiation pattern is axially symmetric about the line from the plane to the dipole.

Page 329: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 12 48

If d = λ/4, thendP

dΩ= 4A sin2 θ′ cos2

2cos θ′

). (242)

This pattern is a flattened version of the “donut” pattern sin2 θ′, as illustrated below:

If instead, d = λ/2, then

dP

dΩ= 4A sin2 θ′ cos2 (π cos θ′) . (243)

This pattern has a forward lobe for θ′ < π/6 and a “donut” for π/6 < θ′ < π/2, asillustrated below:

c) For a magnetic dipole with axis parallel to the conducting plane, the image dipolehas the same orientation, the image consists of the opposite charge rotating in theopposite direction, as shown below:

We use angles (θ, φ) and modify the argument of part a) to find

E = E1 + E2 = E1(1 + eiδ), (244)

Page 330: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 12 49

and, recalling eq. (46), the power radiated is

dP

dΩ=

|E|2|E1|2

dP1

dΩ=∣∣∣1 + eiδ

∣∣∣2 A sin2 θ = 2A sin2 θ(1 + cos δ) = 4A cos2 θ sin2 δ/2

= 4A sin2 θ cos2 Δ, (245)

where

Δ =δ

2=

2πd

λsin θ cos φ. (246)

If d = λ/4, thendP

dΩ= 4A sin2 θ cos2

2sin θ cosφ

). (247)

In the “side” view, φ = 0, so the pattern has shape

sin2 θ cos2(

π

2sin θ

)(side view), (248)

while in the “top” view, θ = π/2 and the shape is

cos2(

π

2cosφ

)(top view). (249)

This pattern, shown below, is somewhat similar to that of part a) for d = λ/2.

If instead, d = λ/2, then

dP

dΩ= 4A sin2 θ cos2 (π sin θ cosφ) . (250)

In the “side” view, φ = 0, so the pattern has shape

sin2 θ cos2 (π sin θ) (side view), (251)

while in the “top” view, θ = π/2 and the shape is

cos2 (π cosφ) (top view). (252)

This pattern, shown below, is somewhat similar to that of part b) for d = λ/2.

Page 331: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 12 50

Finally, we consider the case of a magnetic dipole aligned with the line from the planeto the dipole, in which case its image has the opposite orientation.

As in part b), the only phase difference between the radiation fields of the dipole andits image is that due to the path difference δ, whose value has been given in eqs. (230)and (231). We use the angles (θ′, φ′) in this case, since the radiation pattern of a singledipole varies as sin2 θ′. Then,

E = E1 + E2 = E1(1 − eiδ), (253)

and, recalling eq. (46), the power radiated is

dP

dΩ=

|E|2|E1|2

dP1

dΩ=∣∣∣1 − eiδ

∣∣∣2 A sin2 θ′ = 2A sin2 θ′(1 − cos δ) = 4A sin2 θ′ sin2 δ/2

= 4A sin2 θ′ sin2 Δ, (254)

with

Δ =δ

2=

2πd

λcos θ′. (255)

This radiation pattern is axially symmetric about the line from the plane to the dipole.

If d = λ/4, thendP

dΩ= 4A sin2 θ′ sin2

2cos θ′

). (256)

This pattern, shown below, is somewhat similar to that of part a) for d = λ/2.

Page 332: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 12 51

If instead, d = λ/2, then

dP

dΩ= 4A sin2 θ′ sin2 (π cos θ′) . (257)

This pattern is qualitatively similar to that for d = λ/4, shown just above, but themaximum occurs at a larger value of θ′.

Page 333: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 13 52

13. From p. 191 of the Notes we recall that a single, short, center-fed, linear antenna ofdipole moment

p(t) = iI0Le−iωt

2ω(258)

radiates time-averaged power (according to the Larmor formula)

dU1

dtdΩ=

〈p2〉 sin2 θ

4πc3=

ω2I20L

2 sin2 θ

32πc3. (259)

For the record, the current distribution in this short antenna is the triangular waveform

I(z, t) = I0e−iωt

(1 − 2 |z|

L

). (260)

The associated charge distribution ρ(z, t) is related by current conservation, ∇·J = −ρ,which for a 1-D distribution is simply

ρ = −∂I

∂z= −I0e

−iωt(∓ 2

L

), (261)

so that

ρ = ±2iI0e−iωt

ωL, (262)

and the dipole moment is given by

p =∫ L/2

−L/2ρz dz = i

I0Le−iωt

2ω, (263)

as claimed above.

Turning to the case to two antennas, we proceed as in the previous problem and writetheir combined electric field as

E = E1 + E2 = E1(1 − eiδ), (264)

where now the phase difference δ has contributions due to the path difference forradiation from the two antennas (whose separation is d = λ/4), as well as from theirintrinsic phase difference of 90. That is,

δ =2π

λ

λ

4cos θ +

π

2=

π

2(1 + cos θ). (265)

From eqs. (259), (264) and (265) we find

dU

dtdΩ=

dU1

dtdΩ

∣∣∣1 − eiδ∣∣∣2 =

ω2I20L

2 sin2 θ

16πc3(1 − cos δ)

=ω2I2

0L2 sin2 θ

16πc3

[1 + sin

2cos θ

)]. (266)

This angular distribution favors the forward hemisphere, as shown in the sketch:

Page 334: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 13 53

The total radiated power isdU

dt=

ω2I20L2

6c3. (267)

Associated with energy U radiated in direction n is momentum P = U n/c. Thus, theangular distribution of radiated momentum is

dP

dtdΩ=

ω2I20L

2 sin2 θ

16πc4

[1 + sin

2cos θ

)]n. (268)

On integrating this over solid angle to find the total momentum radiated, only the zcomponent is nonzero,

dPz

dt= 2π

ω2I20L2

16πc4

∫ 1

−1sin2 θ

[1 + sin

2cos θ

)]cos θ d cos θ

=2ω2I2

0L2

π2c4

(12

π2− 1

)=

12

π2c

dU

dt

(12

π2− 1

)≈ 0.26

c

dU

dt. (269)

The radiation reaction force on the antenna is Fz = −dPz/dt. For a broadcast antennaradiating 105 Watts, the reaction force would be only ≈ 10−4 N.

The radiation reaction force (269) cannot be deduced as the sum over charges of theradiation reaction force of Lorentz, Frad = 2e2v/3c3. Lorentz’ result is obtained byan integration by parts of the integral of the radiated power over a period. Thisprocedure can be carried out if the power is a sum/integral of a square, as holds for anisolated radiating charge. But it cannot be carried out when the power is the squareof a sum/integral as holds for (coherent) radiation by an extended charge/currentdistribution. Rather, the radiation reaction force on a current distribution must bededuced from the rate of radiation of momentum, as done here.

Page 335: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 14 54

14. According to p. 181 of the Notes, the power radiated from a known current distributionthat oscillates at angular frequency ω is given by

dPω

dΩ=

1

8πc

∣∣∣∣∫ Jω(r′) × k e−ik·r′ dVol′∣∣∣∣2 , (270)

where k = nω/c.

We take the z axis along the antenna, so the radiated power due to the current distri-bution

I(z) = I0 sin2πz

Le−iωt (−L/2 < z < L/2), (271)

is

dPω

dΩ=

1

8πc

∣∣∣∣∣∫ L/2

−L/2I0 sin

(2πz

L

)k sin θe−ikz cos θ dz

∣∣∣∣∣2

=k2I2

0

8πcsin2 θ

∣∣∣∣∫ ∣∣∣∣2 , (272)

where ∫=∫ L/2

−L/2sin(

2πz

L

)[cos(kz cos θ) − i sin(kz cos θ)] dz. (273)

For L = λ, we have k = 2π/L and∫=∫ π/k

−π/ksin kz[cos(kz cos θ) − i sin(kz cos θ)] dz. (274)

The real part of this integral vanishes, while

Im∫

= −2∫ π/k

0sin kz sin(kz sin θ) dz = −2

k

∫ π

0sin x sin(x sin θ) dx. (275)

Using the identity

2 sin A sinB = cos(A − B) − cos(A + B), (276)

we have

Im∫

=1

k

∫ π

0[cos[x(1 + cos θ)] − cos[x(1 − cos θ)] dx.

=1

k

sin[π(1 + cos θ)]

1 + cos θ− sin[π(1 − cos θ)]

1 − cos θ

=1

k

−sin(π cos θ)

1 + cos θ− sin(π cos θ)

1 − cos θ

=2 sin(π cos θ)

k sin2 θ. (277)

Inserting this in eq. (272), we find

dPω

dΩ=

I20

2πc

sin2(π cos θ)

sin2 θ. (278)

Page 336: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 14 55

The radiation pattern is sketched below:

The total radiated power is, setting cos θ = u,

P =∫

dPω

dΩdΩ =

I20

c

∫ 1

−1

sin2(π cos θ)

sin2 θd cos θ =

I20

2c

∫ 1

−1

1 − cos(2πu)

1 − u2du

=I20

4c

∫ 1

−1

(1 − cos(2πu)

1 + u+

1 − cos(2πu)

1 − u

)du =

I20

2c

∫ 1

−1

1 − cos(2πu)

1 + udu.(279)

To cast this in the form of a known special function, we let 1 + u = v/2π, so that

P =I20

2c

∫ 4π

0

1 − cos v

vdv =

I20

2cCin(4π) = 3.11

I20

2c, (280)

where Cin is the so-called cosine integral.

b) To calculate the radiation in the multipole approximation, we need to convert thecurrent distribution I(z)e−iωt to a charge distribution ρ(z, t). This is accomplished viathe continuity equation,

∂I

∂z= −ρ = iωρ. (281)

For the current distribution (271) we find

ρ = −2πi

ωLI0 cos

2πz

Le−iωt. (282)

The dipole moment of this distribution is

p =∫ L/2

−L/2ρz dz = 0, (283)

so there is no electric dipole radiation. As the current flows along a line, and not in aloop, there is no magnetic dipole radiation either.

The charge distribution is symmetric about the z axis, so its tensor quadrupole momentcan be characterized in terms of the single quantity

Qzz =∫ L/2

−L/22ρz2 dz = 4

∫ L/2

0ρz2 dz = −8πi

ωLI0

∫ L/2

0z2 cos

2πz

Ldz

= − L2i

π2ωI0

∫ π

0x2 cos x dx =

2L2i

πωI0, (284)

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Princeton University 2001 Ph501 Set 8, Solution 14 56

using the integral (55).

The total power radiated by the symmetric quadrupole moment is, according to p. 190of the Notes,

P =|Qzz|2 ω6

240c5=

L4ω4

30π2c4

I20

2c. (285)

When L = λ = 2πc/ω, this becomes

P =8π2

15

I20

2c= 5.26

I20

2c. (286)

In this example, higher multipoles must contribute significantly to the total power,which is given by the “exact” result (280).

Page 338: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 15 57

15. The scattering cross section is given by

dΩ=

power scattered into dΩ

incident power per unit area= r2 〈Sscat(θ, φ)〉

〈Sincident〉 = r2 |Escat|2E2

0

, (287)

where in the dipole approximation, the far-zone scattered electric field is

Escat = k2ei(kr−ωt)

r[(n× p0) × n − n ×m0] , (288)

and p0eiωt and m0e

−iωt are the electric and magnetic dipole moments induced in theconducting sphere by the incident wave.

Because the incident wavelength is large compared to the radius of the sphere, theincident fields are essentially uniform over the sphere, and the induced fields near thesphere are the same as the static fields of a conducting sphere in an otherwise uniformelectric and magnetic field. Then, from p. 57 of the Notes, the induced electric dipolemoment is given by

p0 = a3E0. (289)

For the induced magnetic dipole, we recall p. 98 of the Notes, remembering that aconducting sphere can be thought of a permeable sphere with zero permeability and adielectric sphere of infinite dielectric constant. Hence, the magnetic dipole moment is

m0 = −a3

2B0. (290)

Then

Escat = −k2a3ei(kr−ωt)

r

[n× (E0 × n) +

1

2(n× B0

], (291)

where n is along the vector r that points from the center of the sphere to the distantobserver.

For a wave propagating in the +z direction with electric field linearly polarized alongdirection l, E0 = E0l, and the magnetic field obeys B0 = z × E0. Then,

Escat = −k2a3E0ei(kr−ωt)

r

[n× (l × n) +

1

2n× (z × l)

]= −k2a3E0

ei(kr−ωt)

r

[l

(1 − (n · z)

2

)+

(n− z

2

)(n · l)

]. (292)

Inserting this in (287) we find

dΩ= k4a6

⎡⎣(1 − n · z2

)2

− 3

4(n · l)2

⎤⎦ . (293)

For an observer in the x-z plane, n · z = cos θ. Then for electric polarization parallel tothe scattering plane n · l = sin θ, while for polarization perpendicular to the scatteringplane n · l = 0.

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Princeton University 2001 Ph501 Set 8, Solution 15 58

Thus, eq. (293) yields

dσ‖dΩ

= k4a6(

1

2− cos θ

)2

,dσ⊥dΩ

= a6k4

(1 − cos θ

2

)2

. (294)

For an unpolarized incident wave,

dΩ=

1

2

(dσ‖dΩ

+dσ⊥dΩ

)= k4a6

[5

8(1 + cos2 θ) − cos θ

], (295)

and so

σ =∫

dΩdΩ =

10π

8k4a6

∫ 1

−1(1 + cos2 θ) d cos θ =

10πa2

3k4a4. (296)

From eqs. (294) we see that dσ⊥/dΩ is always nonzero, but dσ‖/dΩ = 0 for θ = π/3,so for this angle, the scattered radiation is linearly polarized parallel to the scatteringplane for arbitrary incident polarization.

Addendum: The Fields and Poynting Vector Close to the Sphere

Using the results of prob. 2 we can also discuss the fields close to the sphere. Inparticular, from eqs. (10) and (108) the scattered electric field at any position r outsidethe sphere is

Escat(r, t) = k2 ei(kr−ωt)

r

(n× p0) × n + [3(n · p0)n − p0]

(1

k2r2− i

kr

)−(1 +

i

kr

)n ×m0

= k2a3ei(kr−ωt)

r

(n× E0) × n + [3(n · E0)n− E0]

(1

k2r2− i

kr

)

Page 340: Electrodynamics California

Princeton University 2001 Ph501 Set 8, Solution 15 59

+1

2

(1 +

i

kr

)n× B0

, (297)

also using eqs. (289) and (290). In this Addendum, we suppose that the electric field ofthe incident plane wave is along the x-axis, so that E0 = E0x and B0 = E0y, while thepoint of observation is at r = (r, θ, φ). We express the electric field vector in sphericalcoordinates, noting that

n = r, (298)

x = sin θ cos φ r + cos θ cos φ θ − sinφ φ, (299)

y = sin θ sin φ r + cos θ sinφ θ + cosφ φ, (300)

z = cos θ r − sin θ θ. (301)

Thus,

Escat(r, t) = k2a3E0ei(kr−ωt)

r

cos θ cos φ θ − sin φ φ

+(2 sin θ cosφ r − cos θ cos φ θ + sin φ φ)(

1

k2r2− i

kr

)−1

2

(1 +

i

kr

)(cosφ θ − cos θ sinφ φ)

= k2a3E0

ei(kr−ωt)

r

2 sin θ cos φ

(1

k2r2− i

kr

)r

+cos φ[cos θ

(1 − 1

k2r2+

i

kr

)− 1

2

(1 +

i

kr

)]θ

− sin φ

[1 − 1

k2r2+

i

kr− cos θ

2

(1 +

i

kr

)]φ

. (302)

Similarly, using eqs. (8) and (109) the scattered magnetic field can be written

Bscat(r, t) = k2ei(kr−ωt)

r

(n× m0) × n + [3(n · m0)n− m0]

(1

k2r2− i

kr

)+(1 +

i

kr

)n× p0

= −k2a3ei(kr−ωt)

2r

(n ×B0) × n + [3(n · B0)n− B0]

(1

k2r2− i

kr

)−2(1 +

i

kr

)n× E0

= −k2a3E0

ei(kr−ωt)

2r

cos θ sinφ θ + cosφ φ

+(2 sin θ sinφ r − cos θ sin φ θ − cos φ φ)(

1

k2r2− i

kr

)−2(1 +

i

kr

)(sinφ θ + cos θ cos φ φ)

= −k2a3E0

ei(kr−ωt)

2r

2 sin θ sin φ

(1

k2r2− i

kr

)r

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Princeton University 2001 Ph501 Set 8, Solution 15 60

+ sin φ[cos θ

(1 − 1

k2r2+

i

kr

)− 2

(1 +

i

kr

)]θ

+cos φ[1 − 1

k2r2+

i

kr− 2 cos θ

(1 +

i

kr

)]φ

. (303)

On the surface of the sphere, r = a, the scattered electromagnetic fields are, to theleading approximation when ka 1,

Escat(r = a) ≈ E0e−iωt(2 sin θ cos φ r − cos θ cosφ θ + sinφ φ), (304)

Bscat(r = a) ≈ −E0

2e−iωt(2 sin θ sinφ r − cos θ sinφ θ − cosφ φ). (305)

In the same approximation, the incident electromagnetic fields at the surface of thesphere are

Ein(r = a) ≈ E0e−iωtx = E0e

−iωt(sin θ cosφ r + cos θ cos φ θ − sin φ φ), (306)

Bin(r = a) ≈ E0e−iωty = E0e

−iωt(sin θ sin φ r + cos θ sinφ θ + cosφ φ). (307)

Thus, the total electric field,

Etot(r = a) = Ein(r = a) + Escat(r = a) = 3E0e−iωt sin θ cos φ r, (308)

on the surface of the sphere is purely radial, and the total magnetic field,

Btot(r = a) = Bin(r = a) + Bscat(r = a) =3

2E0e

−iωt(cos θ sinφ θ + cos φ φ), (309)

is purely tangential, as expected for a perfect conductor.

The total charge density σtot on the surface of the conducting sphere follows fromGauss’ law as

σtot =Etot(r = a) · r

4π=

3E0

4πe−iωt sin θ cosφ =

3

2σscat, (310)

where σscat is the surface charge density corresponding to the scattered field (304).Similarly, the total current density Ktot on the surface of the sphere follows fromAmpere’s law as

Ktot =c

4πr × Btot(r = a) =

3cE0

8πe−iωt(− cos φ θ + cos θ sinφ φ) = 3Kscat, (311)

where Kscat is the surface charge density corresponding to the scattered field (305).

We can now discuss the energy flow in the vicinity of the conductor sphere from twoperspectives. These two views have the same implications for energy flow in the farzone, but differ in their description of the near zone.

First, we can consider the Poynting vector constructed from the total electromagneticfields,

Stot =c

4πEtot × Btot. (312)

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Because the tangential component of the total electric field vanishes at the surface ofthe sphere, lines of the total Poynting vector do not begin or end on the sphere, butrather they pass by it tangentially. In this view, the sphere does not absorb or emitenergy, but simply redirects (scatters) the flow of energy from the incident wave.

However, this view does not correspond closely to the “microscopic” interpretation thatatoms in the sphere are excited by the incident wave and emit radiation as a result,thereby creating the scattered wave. We obtain a second view of the energy flow thatbetter matches the “microscopic” interpretation if we write

Stot =c

4πEtot × Btot

=c

4π(Ein + Escat) × (Bin + Bscat)

=c

4πEin × Bin +

c

4π(Ein × Bscat + Escat × Bin) +

c

4πEscat ×Bscat

= Sin + Sinteraction + Sscat. (313)

Since the scattered fields (304)-(305) at the surface of the sphere include tangentialcomponents for both the electric and the magnetic field, the scattered Poynting vector,Sscat, has a radial component, whose time average we wish to interpret as the flow ofenergy radiated by the sphere. The scattered Poynting vector at any r is given by

〈Sscat〉 =c

8πRe(E

scat × Bscat)

=c

8πRe[(E

θ,scatBφ,scat − Eφ,scatBθ,scat)r + (E

φ,scatBr,scat − Er,scatBφ,scat)θ

+(Er,scatBθ,scat − E

θ,scatBr,scat)φ]

=c

k4a6E20

r2

⎧⎨⎩⎡⎣cos2 φ

(1

2− cos θ

)2

+ sin2 φ

(1 − cos θ

2

)2⎤⎦ r

− 1

k4r4

(cos θ

2r + sin θ θ

). (314)

The radial term of eq. (314) in square brackets is identical to the far-zone Poyntingvector. However, close to the sphere we find additional terms in 〈Sscat〉, so that in thenear zone 〈Srad〉 = 〈Sscat〉. Indeed, at the surface of the sphere we find

〈Sscat(r = a)〉 =c

8πE2

0

⎧⎨⎩k4a4

⎡⎣cos2 φ(

1

2− cos θ

)2

+ sin2 φ

(1 − cos θ

2

)2⎤⎦ r

−(

cos θ

2r + sin θ θ

)

≈ − c

8πE2

0

(cos θ

2r + sin θ θ

). (315)

Of course, the conducting sphere is not an energy source by itself, and the radiatedenergy is equal to the energy absorbed from the incident wave. For a description of

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Princeton University 2001 Ph501 Set 8, Solution 15 62

the flow of energy that is absorbed, we look to the time-average of the incident andinteraction terms of eq. (313). Lines of the incident Poynting vector,

〈Sin〉 =c

8πE2

0 z =c

8πE2

0(cos θ r − sin θ θ), (316)

enter and leave the sphere with equal strength, and are therefore not to be associatedwith energy transfer to the radiation fields. So, we look to the interaction term,

〈Sinteraction〉 =c

8πRe[(E

θ,scatBφ,in + Eθ,inBφ,scat − E

φ,scatBθ,in − Eφ,inBθ,scat)r

+(Eφ,scatBr,in + E

φ,inBr,scat − Er,scatBφ,in − E

r,inBφ,scat)θ

+(Er,scatBθ,in + E

r,inBθ,scat − Eθ,scatBr,in − E

θ,inBr,scat)φ]

=c

k2a3E20

r

[− cos[kr(1 − cos θ)]

cos θ

2k2r2

+(1 + cos θ)

[cos2 φ

(1

2− cos θ

)+ sin2 φ

(cos θ

2− 1

)]×(

sin[kr(1 − cos θ)]

kr− cos[kr(1 − cos θ)]

)]r

+

[cos[kr(1 − cos θ)]

sin θ

k2r2

(2 − 9

2cos2 φ

)+ . . .

+

[9

8cos[kr(1 − cos θ)]

sin 2θ sin 2φ

k2r2+ . . .

, (317)

where the omitted terms are small close to the sphere. Note that in the far zonethe time-average interaction Poynting vector contains terms that vary as 1/r timescos[kr(1 − cos θ)]. These large terms oscillate with radius r with period λ, and mightbe said to describe a radial “sloshing” of energy in the far zone, rather than a radialflow. It appears in practice that one cannot detect this “sloshing” by means of asmall antenna placed in the far zone, so we consider these terms to be unphysical.Nonetheless, it is interesting that they appear in the formalism.

At the surface of the sphere we have, again for ka 1,

〈Sinteraction(r = a)〉 =c

8πE2

0

[−cos θ

2r + sin θ

(2 − 9

2cos2 φ

)θ +

9

8sin 2θ sin 2φ φ

].

(318)The total Poynting vector on the surface of the sphere is the sum of eqs. (315), (316)and (318),

〈Stot(r = a)〉 =c

8πE2

0

(−9

2sin θ cos2 φ θ +

9

8sin 2θ sin 2φ φ

). (319)

The radial component of the total Poynting vector vanishes on the surface of the sphere,as expected for a perfect conductor.

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Princeton University 2001 Ph501 Set 8, Solution 15 63

This exercise permits an additional perspective, or possible relevance to thinking aboutradiation from antennas. Suppose that instead of knowing that a plane wave was inci-dent on the conducting sphere, we were simply given the surface current distributionKscat of eq. (311). Then, by use of retarded potentials, or the “antenna formula” (63),we could calculate the radiated power in the far zone, and would arrive at the usual ex-pression (314) (ignoring the terms that fall off as 1/r6). However, this procedure wouldlead to an incomplete understanding of the near zone. In particular, the excitation ofthe conducting sphere by an external plane wave leads to a total surface current thatis three times larger than the current Kscat. For a good, but not perfectly conductingsphere, an analysis based on Kscat alone would lead to only 1/9 the actual amount ofJoule heating of the sphere. And, if we attempted to assign some kind of impedanceor radiation resistance to the sphere via the form Prad = RradI

20/2 where I0 is meant

to be a measure of the peak total current, an analysis based only on knowledge thecurrent Kscat would lead to a value of Rrad that is 9 times larger than desired.

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Princeton University 2001 Ph501 Set 8, Solution 16 64

16. The possibility of radiation from superluminal sources was first considered by Heavisidein 1888. He considered this topic many times over the next 20 years, deriving mostof the formalism of what is now called Cerenkov radiation. However, despite being anearly proponent of the concept of a velocity-dependent electromagnetic mass, Heavisidenever acknowledged the limitation that massive particles must have velocities less thanthat of light. Consequently many of his pioneering efforts (and those of his immediatefollowers, Des Coudres and Sommerfeld), were largely ignored, and the realizable caseof radiation from a charge with velocity greater than the speed of light in a dielectricmedium was discovered independently in an experiment by Cerenkov in 1934.10

An insightful discussion of the theory of Cerenkov radiation by Tamm (J. Phys. U.S.S.R.1, 439 (1939), in English!)11 revealed its close connection with what is now called tran-sition radiation, i.e., radiation emitted by a charge in uniform motion that crosses aboundary between metallic or dielectric media. The present problem was inspired bya work of Bolotovskii and Ginzburg, Sov. Phys. Uspekhi 15, 184 (1972),12 on how ag-gregates of particles can act to produce motion that has superluminal aspects and thatthere should be corresponding Cerenkov-like radiation in the case of charged particles.The classic example of aggregate superluminal motion is the velocity of the point ofintersection of a pair of scissors whose tips approach one another at a velocity close tothat of light.

Here we consider the example of a “sweeping” electron beam in a high-speed analogoscilloscope such as the Tektronix 7104. In this device the “writing speed”, the velocityof the beam spot across the faceplate of the oscilloscope, can exceed the speed of light.The transition radiation emitted by the beam electrons just before they disappear intothe faceplate has the character of Cerenkov radiation from the superluminal beam spot,according to the inverse of the argument of Tamm.

Referring to the figure in the statement of the problem, the line of charge has equation

y =u

vx − ut, z = 0, (320)

so the current density is

J = −yNeδ(z)δ(t − x

v+

y

u

), (321)

where N is the number of electrons per unit length intercepting the x axis, and e < 0is the electron’s charge.

We also consider the effect of the image current,

Jimage = +y(−Ne)δ(z)δ(t − x

v− y

u

). (322)

We will find that to a good approximation the image current just doubles the amplitudeof the radiation. For u ∼ c the image current would be related to the retarded fields

10http://puhep1.princeton.edu/~mcdonald/examples/EM/cerenkov_pr_52_378_37.pdf11http://puhep1.princeton.edu/~mcdonald/examples/EM/tamm_jpussr_1_439_39.pdf12http://puhep1.princeton.edu/~mcdonald/examples/EM/bolotovskii_spu_15_184_72.pdf

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Princeton University 2001 Ph501 Set 8, Solution 16 65

of the electron beam, but we avoid this complication when u c. Note that the truecurrent exists only for y > 0, while the image current applies only for y < 0.

We insert the current densities (321) and (322) into eq. (63) and integrate using rect-angular coordinates, with components of the unit vector n given by

nx = cos θ, ny = sin θ cosφ, and nz = sin θ sinφ, (323)

as indicated in part b) of the figure. The current impinges only on a length L alongthe x axis. The integrals are elementary and we find, noting ω/c = 2π/λ,

dU

dωdΩ=

e2N2L2

π2c

u2

c2

cos2 θ + sin2 θ sin2 φ

(1 − u2

c2sin2 θ cos2 φ)2

⎛⎝sin[

πLλ

( cv− cos θ)

]πLλ

( cv− cos θ)

⎞⎠2

. (324)

The factor of form sin2 χ/χ2 appears from the x integration, and indicates that thisleads to a single-slit interference pattern.

We will only consider the case that u c, so from now on we approximate the factor1 − u2

c2sin2 θ cos2 φ by 1.

Upon integration over the azimuthal angle φ from −π/2 to π/2 the factor cos2 θ +sin2 θ sin2 φ becomes π

2(1 + cos2 θ).

It is instructive to replace the radiated energy by the number of radiated photons:dU = hωdNω. Thus

dNω

d cos θ=

α

ωN2L2 u2

c2(1 + cos2 θ)

⎛⎝sin[

πLλ

( cv− cos θ)

]πLλ

( cv− cos θ)

⎞⎠2

, (325)

where α = e2/hc ≈ 1/137. This result applies whether v < c or v > c. But for v < c,the argument χ = πL

λ( c

v− cos θ) can never become zero, and the diffraction pattern

never achieves a principal maximum. The radiation pattern remains a slightly skewedtype of transition radiation. However, for v > c we can have χ = 0, and the radiationpattern has a large spike at angle θC such that

cos θC =c

v,

which we identify with Cerenkov radiation. Of course the side lobes are still present,but not very prominent.

Discussion

The present analysis suggests that Cerenkov radiation is not really distinct from tran-sition radiation, but is rather a special feature of the transition radiation pattern whichemerges under certain circumstances. This viewpoint actually is relevant to Cerenkovradiation in any real device which has a finite path length for the radiating charge.The walls which define the path length are sources of transition radiation which isalways present even when the Cerenkov condition is not satisfied. When the Cerenkovcondition is satisfied, the so-called formation length for transition radiation becomes

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Princeton University 2001 Ph501 Set 8, Solution 16 66

longer than the device, and the Cerenkov radiation can be thought of as an interferenceeffect.

If L/λ 1, then the radiation pattern is very sharply peaked about the Cerenkovangle, and we may integrate over θ noting

dχ =πL

λd cos θ and

∫ ∞

−∞dχ

sin2 χ

χ2= π (326)

to find

dNω ∼ α

2π(Nλ)2dω

ω

L

λ

u2

c2

(1 +

c2

v2

). (327)

In this we have replaced cos2 θ by c2/v2 in the vicinity of the Cerenkov angle. Wehave also extended the limits of integration on χ to [−∞,∞]. This is not a goodapproximation for v < c, in which case χ > 0 always and dNω is much less than stated.For v = c the radiation rate is still about one half of the above estimate.

For comparison, the expression for the number of photons radiated in the ordinaryCerenkov effect is

dNω ∼ 2παdω

ω

L

λsin2 θC. (328)

The ordinary Cerenkov effect vanishes as θ2C near the threshold, but the superluminal

effect does not. This is related to the fact that at threshold ordinary Cerenkov radiationis emitted at small angles to the electron’s direction, while in the superluminal casethe radiation is at right angles to the electron’s motion. In this respect the movingspot on an oscilloscope is not fully equivalent to a single charge as the source of theCerenkov radiation.

In the discussion thus far we have assumed that the electron beam is well described by auniform line of charge. In practice the beam is discrete, with fluctuations in the spacingand energy of the electrons. If these fluctuations are too large we cannot expect thetransition radiation from the various electrons to superimpose coherently to producethe Cerenkov radiation. Roughly, there will be almost no coherence for wavelengthssmaller than the actual spot size of the electron beam at the metal surface, Thus therewill be a cutoff at high frequencies which serves to limit the total radiated energy toa finite amount, whereas the expression derived above is formally divergent. Similarlythe effect will be quite weak unless the beam current is large enough that Nλ 1.

We close with a numerical example inspired by possible experiment. A realistic spotsize for the beam is 0.3 mm, so we must detect radiation at longer wavelengths. Aconvenient choice is λ = 3 mm, for which commercial microwave receivers exist. Thebandwidth of a candidate receiver is dω/ω = 0.02 centered at 88 GHz. We takeL = 3 cm, so L/λ = 10 and the Cerenkov ‘cone’ will actually be about 5 wide, whichhappens to match the angular resolution of the microwave receiver. Supposing theelectron beam energy to be 2.5 keV, we would have u2/c2 = 0.01. The velocity of themoving spot is taken as v = 1.33c = 4 × 1010 cm/sec, so the observation angle is 41.If the electron beam current is 1 μA then the number of electrons deposited per cmalong the metal surface is N ∼ 150, and Nλ ∼ 45.

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Princeton University 2001 Ph501 Set 8, Solution 16 67

Inserting these parameters into the rate formula we expect about 7 × 10−3 detectedphotons from a single sweep of the electron beam. This supposes we can collect over allazimuth φ which would require some suitable optics. The electron beam will actuallybe swept at about 1 GHz, so we can collect about 7 × 106 photons per second. Thecorresponding signal power is 2.6 × 10−25 Watts/Hz, whose equivalent noise temper-ature is about 20 mK. This must be distinguished from the background of thermalradiation, the main source of which is in the receiver itself, whose noise temperatureis about 100K. A lock-in amplifier could be used to extract the weak periodic signal;an integration time of a few minutes of the 1-GHz-repetition-rate signal would sufficeassuming 100% collection efficiency.

Realization of such an experiment with a Tektronix 7104 oscilloscope would requirea custom cathode ray tube that permits collection of microwave radiation through aportion of the wall not coated with the usual metallic shielding layer.

Bremsstrahlung

Early reports of observation of transition radiation were considered by skeptics to bedue to bremsstrahlung instead. The distinction in principle is that transition radiationis due to acceleration of charges in a medium in response to the far field of a uniformlymoving charge, while bremsstrahlung is due to the acceleration of the moving chargein the near field of atomic nuclei. In practice both effects exist and can be separatedby careful experiment.

Is bremsstrahlung stronger than transition radiation in the example considered here?As shown below the answer is no, but even if it were we would then expect a Cerenkov-like effect arising from the coherent bremsstrahlung of the electron beam as it hits theoscilloscope faceplate.

The angular distribution of bremsstrahlung from a nonrelativistic electron will be sin2 θwith θ defined with respect to the direction of motion. The range of a 2.5-keV electronin, say, copper is about 5×10−6 (as extrapolated from the table on p. 240 of Studies inPenetration of Charged Particles in Matter, National Academy of Sciences – NationalResearch Council, PB-212 907 (Washington, D.C., 1964)), while the skin depth at88 GHz is about 2.5 × 10−5 cm. Hence the copper is essentially transparent to thebackward hemisphere of bremsstrahlung radiation, which will emerge into the samehalf space as the transition radiation.

The amount of bremsstrahlung energy dUB emitted into energy interval dU is justY dU where Y is the so-called bremsstrahlung yield factor. For 2.5-keV electrons incopper, Y = 3 × 10−4. The number dN of bremsstrahlung photons of energy hω ina bandwidth dω/ω is then dN = dUB/hω = Y dω/ω. For the 2% bandwidth of ourexample, dN = 6×10−6 per beam electron. For a 3-cm-long target region there will be500 beam electrons per sweep of the oscilloscope, for a total of 3×10−4 bremsstrahlungphotons into a 2% bandwidth about 88 GHz. Half of these emerge from the faceplateas a background to 7 × 10−3 transition-radiation photons per sweep. Altogether, thebremsstrahlung contribution would be about 1/50 of the transition-radiation signal inthe proposed experiment.

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Princeton University 2003 Ph304 Problem Set 10 1

Reading: Griffiths secs. 9.4-9.5.

1. Griffiths’ prob. 9.19.

2. Griffiths’ prob. 9.22. In part a), use dimensional analysis to determine the functionalform of the velocity of deep-water waves, which might depend on the wavelength ¸ , thedensity ½of water, and the acceleration g of gravity (which provides the restoring forceagainst wavy deformations), but which does not depend on the depth of the water.

3. Variant of Griffiths’ prob. 9.25. Consider a medium that is sufficiently well describedby a single resonance of angular frequency ! 0, and with a nonzero absorption coefficient° . The oscillator strength may be taken as unity. Show that at the central frequency,! = ! 0, the group velocity is given by

vg =1

dk=d!=

c1 ¡ ! 2

p=°2;

where ! p is the so-called plasma frequency,

! 2p =

Nq2

m²0:

If it happens that ° < ! p, then the group velocity is negative! However, since thisbizarre effect occurs for frequencies where absorption is very strong, it was thought to belargely irrelevant. But recent experiments involving a pair of closely spaced absorptionlines that are pumped into inverted populations has been able to demonstrate negativegroup velocity without attenuation.

The concept of negative group velocity requires careful thought, as one can be misledinto thinking it implies faster-than-light effects. See,http://puhep1.princeton.edu/˜mcdonald/examples/negativegroupvelocity.pdf

A related, but less controversial, phenomenon attainable with pumped pairs of absorp-tion lines is “slow light”:http://puhep1.princeton.edu/˜mcdonald/examples/slowlight.pdf

4. Griffiths’ prob. 9.29. You are asked to compute the so-called energy velocity,

vE =

RhSi ¢da

Rhui da

;

and compare this with the group velocity vg.

5. Griffiths’ prob. 9.30.

6. Griffiths’ prob. 9.38.

Princeton University

Ph304 Problem Set 10

Electrodynamics(Due in class, Wednesday Apr. 23, 2003)

Instructor: Kirk T. McDonald, Jadwin 309/361, x6608/[email protected]

http://puhep1.princeton.edu/˜mcdonald/examples/

AI: Matthew Sullivan, 303 Bowen Hall, [email protected]

Problem sessions: Sundays, 7 pm, Jadwin 303

Text: Introduction to Electrodynamics, 3rd ed.by D.J. Griffiths (Prentice Hall, ISBN 0-13-805326-X, now in 6th printing)

Errata at http://academic.reed.edu/physics/faculty/griffiths.html

Page 356: Electrodynamics California

Princeton University 2003 Ph304 Problem Set 10 1

Reading: Griffiths secs. 9.4-9.5.

1. Griffiths’ prob. 9.19.

2. Griffiths’ prob. 9.22. In part a), use dimensional analysis to determine the functionalform of the velocity of deep-water waves, which might depend on the wavelength ¸ , thedensity ½of water, and the acceleration g of gravity (which provides the restoring forceagainst wavy deformations), but which does not depend on the depth of the water.

3. Variant of Griffiths’ prob. 9.25. Consider a medium that is sufficiently well describedby a single resonance of angular frequency ! 0, and with a nonzero absorption coefficient° . The oscillator strength may be taken as unity. Show that at the central frequency,! = ! 0, the group velocity is given by

vg =1

dk=d!=

c1 ¡ ! 2

p=°2;

where ! p is the so-called plasma frequency,

! 2p =

Nq2

m²0:

If it happens that ° < ! p, then the group velocity is negative! However, since thisbizarre effect occurs for frequencies where absorption is very strong, it was thought to belargely irrelevant. But recent experiments involving a pair of closely spaced absorptionlines that are pumped into inverted populations has been able to demonstrate negativegroup velocity without attenuation.

The concept of negative group velocity requires careful thought, as one can be misledinto thinking it implies faster-than-light effects. See,http://puhep1.princeton.edu/˜mcdonald/examples/negativegroupvelocity.pdf

A related, but less controversial, phenomenon attainable with pumped pairs of absorp-tion lines is “slow light”:http://puhep1.princeton.edu/˜mcdonald/examples/slowlight.pdf

4. Griffiths’ prob. 9.29. You are asked to compute the so-called energy velocity,

vE =

RhSi ¢da

Rhui da

;

and compare this with the group velocity vg.

5. Griffiths’ prob. 9.30.

6. Griffiths’ prob. 9.38.

Princeton University

Ph304 Problem Set 10

Electrodynamics(Due in class, Wednesday Apr. 23, 2003)

Instructor: Kirk T. McDonald, Jadwin 309/361, x6608/[email protected]

http://puhep1.princeton.edu/˜mcdonald/examples/

AI: Matthew Sullivan, 303 Bowen Hall, [email protected]

Problem sessions: Sundays, 7 pm, Jadwin 303

Text: Introduction to Electrodynamics, 3rd ed.by D.J. Griffiths (Prentice Hall, ISBN 0-13-805326-X, now in 6th printing)

Errata at http://academic.reed.edu/physics/faculty/griffiths.html

Page 357: Electrodynamics California

Princeton University 2003 Ph304 Problem Set 11 1

Reading: Griffiths chap. 10.

1. Griffiths’ prob. 10.9, part a) only. Verify that the electric and magnetic fields first seenby an observer at distance r from the wire obey the radiation condition E = cB £ r.That is, consider the fields at time r=c + ². Show also that for t À r=c the magneticfield is that given by the instantaneous magnetostatic value: B ¼ ¹ 0kt=2¼r. However,the electric field never drops to its instantaneous static value (zero), because the everincreasing magnetic field continually induces more electric field.

2. Griffiths’ prob. 10.14. It may be simpler to proceed from eq. (10.39) than from (10.42).

3. Griffiths’ prob. 10.24. Hint: At what time does the observer at the origin first becomeaware of the fields of the moving charge?

4. Griffiths’ prob. 10.26. For another example of how electromagnetic field momentum isneeded for Newton’s 3rd law to be satisfied in electrodynamics, seehttp://puhep1.princeton.edu/˜mcdonald/examples/transmom2.pdf

5. Griffiths’ prob. 11.3. The “dipole” is that of sec. 11.1.2. Verify thatq

¹ 0=²0 = 377 Ω.

6. Griffiths’ prob. 11.4. Besides following Griffiths’ hint, another way to work the problemis to write the dipole moment as p = p0(x + i y)e¡ i!t , and use Griffiths’ eqs. (11.56)and (11.57).

Princeton University

Ph304 Problem Set 11

Electrodynamics(Due in class Wednesday Apr. 30, 2003)

Instructor: Kirk T. McDonald, Jadwin 309/361, x6608/[email protected]

http://puhep1.princeton.edu/˜mcdonald/examples/

AI: Matthew Sullivan, 303 Bowen Hall, [email protected]

Problem sessions: Sundays, 7 pm, Jadwin 303

Text: Introduction to Electrodynamics, 3rd ed.by D.J. Griffiths (Prentice Hall, ISBN 0-13-805326-X, now in 6th printing)

Errata at http://academic.reed.edu/physics/faculty/griffiths.html

Page 358: Electrodynamics California

Princeton University 2003 Ph304 Problem Set 11 1

Reading: Griffiths chap. 10.

1. Griffiths’ prob. 10.9, part a) only. Verify that the electric and magnetic fields first seenby an observer at distance r from the wire obey the radiation condition E = cB £ r.That is, consider the fields at time r=c + ². Show also that for t À r=c the magneticfield is that given by the instantaneous magnetostatic value: B ¼ ¹ 0kt=2¼r. However,the electric field never drops to its instantaneous static value (zero), because the everincreasing magnetic field continually induces more electric field.

2. Griffiths’ prob. 10.14. It may be simpler to proceed from eq. (10.39) than from (10.42).

3. Griffiths’ prob. 10.24. Hint: At what time does the observer at the origin first becomeaware of the fields of the moving charge?

4. Griffiths’ prob. 10.26. For another example of how electromagnetic field momentum isneeded for Newton’s 3rd law to be satisfied in electrodynamics, seehttp://puhep1.princeton.edu/˜mcdonald/examples/transmom2.pdf

5. Griffiths’ prob. 11.3. The “dipole” is that of sec. 11.1.2. Verify thatq

¹ 0=²0 = 377 Ω.

6. Griffiths’ prob. 11.4. Besides following Griffiths’ hint, another way to work the problemis to write the dipole moment as p = p0(x + i y)e¡ i!t , and use Griffiths’ eqs. (11.56)and (11.57).

Princeton University

Ph304 Problem Set 11

Electrodynamics(Due in class Wednesday Apr. 30, 2003)

Instructor: Kirk T. McDonald, Jadwin 309/361, x6608/[email protected]

http://puhep1.princeton.edu/˜mcdonald/examples/

AI: Matthew Sullivan, 303 Bowen Hall, [email protected]

Problem sessions: Sundays, 7 pm, Jadwin 303

Text: Introduction to Electrodynamics, 3rd ed.by D.J. Griffiths (Prentice Hall, ISBN 0-13-805326-X, now in 6th printing)

Errata at http://academic.reed.edu/physics/faculty/griffiths.html

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