1 Arbitrarily-moving observer

Suppose that a charge q is moving along a world-line defined by x0(τ). The null vectorto a point at position x is given by

X2 = 0, where X ≡ (x − x0(τ)). (1)

The proper times τ1,2 connected to x define fields τ1,2(x)and the gradient is given by

∇τ =X

X·v (2)

where v = x0 is the velocity.

2 Black Hole Electrodynamics revisited

A point charge is stationary at radius b outside a black hole of mass M . Close to thecharge, the field is Coulomb, but the field strength varies around the charge like

√F2 =

q

4πε0

1

(ρ2 + bb−2M

z2), (3)

which shows that there is a gravitational ‘Lorentz contraction’ in the radial direction.The current J must be in the γ0 direction, so that the A field is ∝ g(γ0), and is

A =q

4πε0

√

1 − 2M/b

(ρ2 + 1

1−2M/bz2)

g(γ0). (4)

The G field calculated from this has no magnetic H component. There is, of course,no way to measure this effect locally, all atoms and measuring rods are shrunk by thisfactor, which corresponds exactly to the well-known gravitational redshift.

We can derive gauge-dependent Lorentz factors for the case where the charge is movingradially outward at map-velocity x:

A/R E-F γ2 =1

1 − x2 − 2Mb

(1 ± x)2; Newtonian γ2 =

1

1 − x2 − 2Mb

− 2x√

2M/b).

(5)I haven’t done it for the transverse case yet, but it can’t be difficult.

For the point charge in a k > 0 universe I get

√F2 =

q

4πε0δ2

√

α(1 + kb2/4) , (6)

where δ is the (small) map distance from the charge at r = b. I haven’t reconciled thiswith det h = α3(1 + kr2/4)3 yet.

1

3 Radiation from moving charges

Suppose that a charge q is moving along a world-line defined by x0(τ). An observer atspace-time position x will receive an electromagnetic influence from the charge when itlies on the observer’s past light-cone:

X2 = 0, where X ≡ (x − x0(τ)). (7)

In terms of the velocity v ≡ x0, the Lienard-Wiechert potential can be written:

A =q

4πε0

v

|X · v| . (8)

I had written that down for many years without the modulus sign and, indeed, it doesn’tneed it if you use the retarded potential and v is future-pointing. I like it there now,because of the alternative

∫

dτδ(−1

2X2) form, which makes it clear how it arises.

This leads to a general expression for the field F :

F =q

4πε0

X ∧ v + 1

2XΩvX

(X · v)3, (9)

where Ωv ≡ vv is the acceleration bivector projected in the direction of v. The quantityvv is a pure bivector because v2 = 1 and v · v = 0.

This formula displays a clean split into a Coulomb term proportional to 1/(distance)2

and a long-range radiation term proportional to 1/(distance). The radiation term isproportional to the rest-frame acceleration, projected down the null-vector X.

To illustrate this general formula we will look at some simple cases.

3.1 Uniformly moving charge

Assuming, without loss of generality, that the charge goes along the z axis, we have

x0(τ) = τ (γ0 cosh u + γ3 sinh u) ≡ v0τ , (10)

where the parameter u is the ‘rapidity’. Setting X2 = 0, we have a quadratic in τ , andwe must select the earliest root to find past light-cone:

τ 2 − 2τ(x · v0) + x2 = 0, (11)

τ = x · v0 − ((x · v0)2 − x2)

1

2 , (12)

hence finding

X · v = x · v0 − τ = ((x · v0)2 − x2)

1

2 =

√

x2 + y2 + cosh2 u(z − t tanhu)2 ≡ R . (13)

Because v = 0 we have:

F =q

4πε0

X ∧ v

(X · v)3, =

q

4πε0R3x ∧ v0, (14)

If you use the advanced potential, rather than the retarded one, you get the sameanswer, but only if you attend to the modulus sign in the potential mentioned above.

2

Figure 1: Electric field lines of an accelerated charge at t = 3a. The charge began itsacceleration at t1 = −10a and has thereafter accelerated uniformly.

3.2 Accelerated charge

An accelerated charged particle is following the orbit

v(τ) = cosh gτγ0 + sinh gτγ3; x0(τ) = a (sinh gτγ0 + cosh gτγ3) , (15)

with a ≡ g−1. Define X ≡ x − x0 and seek the retarded solution of X2 = 0. We find

egτ =1

2a(z − t)

(

a2 + r2 − t2 −(

(a2 + r2 − t2)2 − 4a2(z2 − t2))

1

2

)

(16)

e−gτ =1

2a(z + t)

(

a2 + r2 − t2 +(

(a2 + r2 − t2)2 − 4a2(z2 − t2))

1

2

)

(17)

where x, y, z, t are the usual Cartesian coordinates and r ≡√

x2 + y2 + z2. The con-dition for the existence of a retarded solution is z + t > 0. The radiation from the chargecan now be calculated, and a crucial factor is the ‘effective distance’

X · v =((a2 + r2 − t2)2 − 4a2(z2 − t2))

1

2

2a. (18)

This factor vanishes if x = y = 0 and z2 − t2 = a2, i.e. on the path of the particle. Theother factor is

X ∧ v + 1

2XvvX =

1

2a(z2 − ρ2 − t2 − a2)σ3 +

zρ

aσρ +

tρ

aiσφ , (19)

3

Figure 2: Electric field lines of a charge at t = 5a. The charge accelerated for −0.2a <t < 0.2a, leaving an outgoing pulse of transverse radiation field.

where ρ ≡√

x2 + y2. The termination of the field lines at z = −t is a bit brutal, so,for the purposes of illustration, we suppose that the charge starts accelerating at t = t1,and stops again at t = t2. There are discontinuities in the electric field line direction onthe two appropriate light spheres. Figure(1) shows the field lines of a charge that beganaccelerating at t = −10a, so that the pattern is well developed, and shows clearly therefocusing of the field lines onto the ‘image charge’. Of course, the image is not actuallythere, and the field lines diverge after they cross the light sphere corresponding to thestart of the acceleration. In Figure(2) the acceleration is relatively brief, so that a pulseof radiation is sent outwards.

We now derive an approximate form for the pulse of radiation in the limit r a.The pulse is, to a good approximation, centred on the minimum of X · v, which occursat

t0 ≡√

r2 − a2 . (20)

The minimum value of X · v is ρ, except for a small region near z ≈ −r, which cannotsee the source because of the condition z + t > 0. Defining δt ≡ t − t0, we find

X · v ≈√

ρ2 +r2δt2

a2, (21)

X ∧ v + 1

2XvvX ≈ −rρ

a(σθ + iσφ) , (22)

4

Figure 3: Field lines of a rotating charge, with rapidity u = 0.1.

which is a pure, outgoing radiation field. The final formula is

F ≈ − q

4πε0ar sin2 θ

(

1 +δt2

a2 sin2 θ

)

−3

2

(σθ + iσφ) . (23)

3.3 Circular orbit

We now consider a charge moving around a circular orbit, with

v = cosh uγ0 + sinh u (− sin ωτγ1 + cos ωτγ2) , (24)

x0 = τ cosh uγ0 + a (cos ωτγ1 + sin ωτγ2) , (25)

with a ≡ ω−1 sinh u. We first locate the null vector X, noting that

t = τ cosh u +√

r2 + a2 − 2a(x cos ωτ + y sin ωτ) (26)

is connected by a null vector to the proper time τ . If we wish, for example, to calculatethe field at some fixed time t, then the retarded time τ cosh u is somewhere between

t −√

r2 + 2aρ + a2 < τ < t −√

r2 − 2aρ + a2 . (27)

At fixed x, y, z, the mapping bewteen t and τ is monotonic, so that a binary chop isthe above interval provides a reliable, if slow, method of locating the null vector. It iscertainly what I used...

5

Figure 4: Field lines of a rotating charge, with rapidity u = 0.4.

Once we have a satisfactory procedure for locating τ on the retarded light-cone, allis plain sailing for numerical simulation. Even so, I wouldn’t have found it at all easywithout a full range of STA routines. The plots show the fields lines in the equatorialplane of a rotating charge with ω = 1. For ‘low’ speeds, we get a gentle, wavy patternof field lines (Figure(3)). The case displayed in Figure(4) is intermediate (u = 0.4), anddisplays many interesting features. By u = 1 (Figure(5)) the field lines have concentratedinto synchrotron pulses, a pattern which continues therafter.

The first ingredient in the simulation is, of course, the effective distance X · v:

X · v = cosh u√

r2 + a2 − 2aρ cos(ωτ − φ) + ρ sinh u sin(ωτ − φ) . (28)

The other factor X ∧ v + 1

2XvvX is, frankly, a bit of a screenfull! The simulations were

produced simply by using the general formula numerically.

3.4 Synchrotron radiation

This is what all radioastronomers cut their teeth on! Synchrotron radiation from aradiogalaxy has, for example, a ≈ 108m and r ≈ 1025m. A power-series expansion ina/r is therefore fairly safe. Typical values of cosh u are 104 for electrons producing radioemission. In the limit r a, the relation between t and τ simplifies

t − r ≈ τ cosh u − a sin θ cos(ωτ − φ) , (29)

6

Figure 5: Field lines of a rotating charge, with rapidity u = 1.

where r, θ, φ are the polar coordinates of x, y, z. It doesn’t help that much, however,because the relation is still an implicit equation for τ . Equation (29) is the usual startingpoint for synchrotron theory.

The real simplification is the effective distance:

X · v ≈ cosh u (1 + tanh u sin θ sin(ωτ − φ)) , (30)

and the null vector X is just r(γ0 + γr). We can obviously ignore the X ∧ v term, andthe other easily simplifies to

XvvX

2r2≈ ω cosh u sinh u (cos θ cos(ωτ − φ))σθ + (sin θ tanh u + sin(ωτ − φ))σφ) (1+σr) .

(31)I think the STA derivation certainly beats what I remember of the traditional ‘bits andpieces’ approach!

Looking at these formulae, we can see all kinds of wonderful features, which I’ll tellyou about some other time...

To be continued.....

3.5 The current loop

This section is designed to practise elliptic integrals.... I follow the conventions of Grad-shteyn & Ryzhik, rather that Abramowitz & Stegun — there is a difference of a square

7

Figure 6: Electric field lines and equipotentials due to a ring of charge of radius a = 1.

root in the argument of the complete elliptic integral. The definition used is

K(k) ≡∫ π/2

0

dθ1

√

1 − k2 sin2 θ; E(k) ≡

∫ π/2

0

dθ√

1 − k2 sin2 θ . (32)

Consider the electrostatic potential of a loop with total charge q , having radius a:

A0 =q

8π2ε0

∫

2π

0

dφ1

√

r2 + a2 − 2aρ cos φ. (33)

We convert this into the canonical form for a complete elliptic integral of the first kind,defining parameters

D2 ≡ 1

2((a2 + r2) +

√

(a2 + r2)2 − 4a2ρ2) , p ≡ aρ

D2. (34)

The potential evaluates to

A0 =qK(p)

2π2ε0D. (35)

We employ the standard notation for the auxillary argument k′:

K ′(k) = K(k′) , k′ ≡√

1 − k2 , (36)

and will need the following relations for the derivative:

dK(k)

dk=

E(k)

kk′2− K(k)

k, (37)

8

Figure 7: Magnetic field lines of a current loop of radius a = 1.

d(K(k) − E(k))

dk=

kE(k)

k′2. (38)

Figure(6) shows the equipotentials and field lines for this case. The field lines were seededuniformly on the ring, which means they do not reflect the field strength properly by thetime that they get to the edge.

Chris reminds me about oblate spheroidal coordinates, which are

ρ = a cosh u cos v ; z = a sinh u sin v ; φ = φ . (39)

They satisfy the relations

r2 + a2 = a2(cosh2 u + cos2 v) ;√

(r2 + a2)2 − 4a2ρ2 = a2(cosh2 u − cos2 v) , (40)

so that the surfaces of constant D = a cosh u are oblate spheroids, and p = cos v. Thesecoordinates might tidy up the formulae for the components of the electric field.

Now let the ring rotate, so that there is a current I. The vector potential is

Aφ =µ0I

4π

∫

2π

0

dφcos φ

√

r2 + a2 − 2aρ cos φ=

µ0I

4π

4D

aρ(K(p) − E(p)) , (41)

and we calculate the magnetic field using

B =1

ρ(−∂z(ρAφ)σρ + ∂ρ(ρAφ)σz) . (42)

9

At this point I consulted Lamb’s ‘Hydrodynamics’ (1924), which can always be reliedupon for a thorough treatment of difficult examples — in this case, the vortex ring.He uses the old-fashioned $ as radial coordinate, and (following Cayley (1896)) F1, E1

instead of K, E. He also points out the relations

D = r1 + r2; p =r1 − r2

r1 + r2

, (43)

where r1,2 =√

(ρ ± a)2 + z2 are the largest and smallest distances from the ring. Thefield lines of B are the surfaces of constant ρAφ, which are shown in Figure(7). Thecontours have been adjusted to make the seeding of field lines aesthetically pleasing(uniform in

√

ρAφ).There are interesting limiting forms for the fields in the disc z = 0, ρ < a:

Bz =µ0I

π

aE(ρ/a)

a2 − ρ2; Eρ =

q

2π2aε0

(

a2E(ρ/a)

ρ(a2 − ρ2)− K(ρ/a)

ρ

)

. (44)

In the limit ρ → a, we can evaluate an electromagnetic ‘velocity’, defined as the ratio ofthe Poynting vector to the total field energy. It evaluates to

2v0

1 + v2

0/c2

, (45)

where v0 ≡ I2πaq

is the effective velocity of the charged ring. There is no physical restric-tion on v0, which can exceed c if q → 0.

There is a weak divergence in the total field energy, of course...

3.6 The Bar Magnet

The current ring solution gives us enough power to solve for B and H inside and outsidea bar magnet, which we can model as a uniform cylinder of height l, radius a andmagnetisation M . The surface magnetisation current J s is given by the formula

J s = M×××××n , (46)

where n is the outward unit vector normal to the surface. I write down here the verysatisfying proof that the magnetisation current Jm is given by ∇×××××M . Any magnetisationcurrent that appears on the surface must, by definition have come from the interior, anyother current is considered to be ‘free’. So we can write

∮

|dS| J s×××××n +

∫

dτ Jm = 0 . (47)

Reassembling the vector area we find∮

dS×××××J s =

∫

dτ ∇×××××Jm =

∫

dτ Jm , (48)

which proves the result Jm = ∇×××××M .

10

For the magnetised cylinder, the surface current is Mσφ over the cylindrical face andzero on the ends. There are magnetisation poles ∇ · H = ∇ · M = ±M on the ends,which provides one way of finding H using the magnetic scalar potential of a disc. I’mpretty sure that I’ll find that in Lamb as well!

I didn’t do it that way, though. Using (??), we set up an array containing ρAρ, andthen convolve this array with a vertical line of length l. Hey presto! We have the fieldlines of B for either a short solenoid or a bar magnet. Contouring

√

ρAρ gives niceconstant seeding in the middle.

For H we have a choice. As mentioned above, we can model the magnetisation polesas discs and use the scalar potential, but there is another rather cunning method thatuses the vector potential. We have to form

H = B/µ0 − M , (49)

but we have used the trick of contouring ρAρ to get the field lines. Now, if we can expressM as the curl of something, then we can use the same trick for H. We need somethingthat has a constant curl, but is zero at ρ = a. The solution is

σφM(r − a2/r)/2. (50)

3.7 Oblate Spheroidal coordinates

Oblate spheroidal coordinates are

ρ = a cosh u cos v ; z = a sinh u sin v ; φ = φ . (51)

They satisfy the relations

r2 + a2 = a2(cosh2 u + cos2 v) ;√

(r2 + a2)2 − 4a2ρ2 = a2(cosh2 u − cos2 v) , (52)

so that the surfaces of constant D = a cosh u are oblate spheroids, and p = cos v. [Fig-ure to include here....] Separation of variables in Laplace’s equation, with cylindricalsymmetry for now yield

1

cosh u U(u)∂u(cosh u∂u U(u)) +

1

cos v V (v)∂v(cos v∂v V (v)) = 0. (53)

Comparing this with the separation that produces Legendre’s equation, we see that thesolutions are therefore Pl(j sinh u)Pl(sin v).

Let’s try a conducting, earthed ellipsoid in a constant E field as an example. We seethat P1(j sinh u)P1(sin v) is indeed ∝ z as required, but, outside the ellipsoid, we needanother solution that approaches zero as r → ∞. We can construct one using the secondsolution

Q1(j sinh u) = 1

2j sinh u log

(

1 + j sinh u

1 − j sinh u

)

− 1 . (54)

The logarithm tends to jπ/2 for large u, so a suitable real solution is

U(u) ∝ (−Q1(j sinh u) + jπ/2P1(j sinh u)) = 1 + sinh u(tan−1 sinh u − π/2) . (55)

11

This function falls off smoothly from a value of unity at the disc u = 0, and has theasymptotic form 1/(3 sinh2 u). I have not used the simpler cot−1 sinh u construction,because neither Maple nor my C compiler have the arccot function — so you see it asI’ve programmed it. This combines to give the field outside as

V (u, v) = sin(v)(

−E0 sinh u + A(1 + sinh u(tan−1 sinh u − π/2))

. (56)

Naturally, this solution is also in Lamb’s monograph. At the surface u0, we now merelyrequire V (u, v) = 0, which implies

E0 sinh u0 = A(1 + sinh u0(tan−1 sinh u0 − π/2) . (57)

For field line plots, we need the Cartesian components of the field, so for completeness Iwrite the ‘dipole’ part down:

E =sin v cos v

(cosh2 u − cos2 v) cosh uσρ −

(

sinh u

cosh2 u − cos2 v+ tan−1 sinh u − π/2

)

σ3. (58)

All that works fine, and the plots will be included soon. There is a related problem(actually the one that Lamb addresses) of fluid flow aroud an ellipsoid, for which weneed the boundary condition U ′(u0) = 0. The form of solution is the same as (56), butwith A given by

E0 cosh u0 = A(tanh(u0) + cosh u0(tan−1 sinh u0 − π/2) . (59)

If the field is along one of the other principal axes, we must hack together some combi-nation of Q1

1and P 1

1that goes to zero at u → ∞. Sparing the details of the derivation,

we get

V (u, v, φ) = cos(v) cos(φ)(

−E0 cosh u + A(cosh u(tan−1 sinh u − π/2) + tanh u)

, (60)

with A given by

E0 sinh u0 = A(

1 + sinh u0(tan−1 sinh u0 − π/2) + sech2u0

)

(61)

for the conducting ellipsoid and by (59) for the fluid flow case. The field is a little morecomplicated and the dipole part is

E = −2 cos φsin v cos v

(cosh2 u − cos2 v) cosh uσ3

− sin φ(

sechu tanhu + tan−1 sinh u − π/2)

σφ

− cos φ

(

sinh u(cosh2 u + cos2 v)

cosh2 u(cosh2 u − cos2 v)+ tan−1 sinh u − π/2

)

σρ.

(62)

The equipotentials can be contoured and the streamlines followed in the usual numericalmanner, and look disturbingly plausible.

12

For completeness, we give the solution for a dielectric of relative permittivity ε. Foran applied field along the axis of symmetry, the values of the dipole strength A and theinternal field E1 are given by

A =E0(ε − 1) sinh u cosh2 u

1 + (ε − 1) cosh2(sinh u(tan−1 sinh u − π/2) + 1);

E1 =E0

1 + (ε − 1) cosh2(sinh u(tan−1 sinh u − π/2) + 1),

(63)

and for a field along the x-axis we have

A =E0(1 − ε) sinh u cosh2 u

1 + ε − (ε − 1) cosh2(sinh u(tan−1 sinh u − π/2) + 1);

E1 =2E0

1 + ε − (ε − 1) cosh2(sinh u(tan−1 sinh u − π/2) + 1),

(64)

3.8 Prolate spheroidal coordinates

Yawn, yawn, yawn.... But it has to be done and written down for completeness. It’sactually simpler. Prolate spheroidal coordinates are

ρ = a sinh u sin v ; z = a cosh u cos v ; φ = φ , (65)

so that the surfaces of constant u are prolate spheroids. Separation of variables inLaplace’s equation, with cylindrical symmetry for ease yield

1

sinh u U(u)∂u(sinh u∂u U(u)) +

1

sin v V (v)∂v(sin v∂v V (v)) = 0. (66)

Comparing this with the separation above, we see that the solutions are this timePl(cosh u)Pl(cos v), so the beastly factor of j has disappeared. If you want to do prolatepotential problems, you can take the ones we did earlier and simply turn them through90 degrees — that would fool most people! If you think that’s cheating (and it certainlyis), then you can use the following l = 1 solutions that Maple has checked for me:

Φ0

1= cos v

(

1

2cosh u log

cosh u − 1

cosh u + 1+ 1

)

; (67)

Φ1

1= sin v cos φ

(

1

2sinh u log

cosh u − 1

cosh u + 1+ cothu

)

. (68)

The fields are

E10 =cos v sin v

sinh u(cosh2 u − cos2 v)σρ +

(

1

2log

cosh u − 1

cosh u + 1+

cosh u

cosh2 u − cos2 v

)

σ3 (69)

E11 = cos φ

(

1

2sinh u log

cosh u − 1

cosh u + 1+

cosh u(sinh2 u − sin2 v)

sinh2 u(cosh2 u − cos2 v)

)

σρ

− sin φ

(

1

2sinh u log

cosh u − 1

cosh u + 1+

cosh u

sinh2 u

)

σφ

− 2 cos φcos v sin v

sinh u(cosh2 u − cos2 v)σ3.

(70)

Here are the answers to the more obvious exercises that can be set.

13

• Conducting ellipsoid:

A =E0 cosh u

1

2cosh u log((cosh u − 1)/(cosh u + 1)) + 1

(along z axis),

E0 sinh u1

2sinh u log((cosh u − 1)/(cosh u + 1)) + cothu

(along x, y axes).

(71)

• Potential flow around ellipsoid:

A =E0 sinh u

1

2sinh u log((cosh u − 1)/(coshu + 1)) + cothu

(z axis),

E0 cosh u1

2cosh u log((cosh u − 1)/(cosh u + 1)) + 1 − cosech2u

(x, y axes).

(72)

• Dielectric ellipsoid (z axis):

E1

E0

=

(

1 − (ε − 1) sinh2 u

(

1

2cosh u log

cosh u − 1

cosh u + 1+ 1

))

−1

,

A

E1

= −(ε − 1) cosh u sinh2 u

(73)

• Dielectric ellipsoid (x, y axes):

E1

E0

=

(

1 + 1

2(ε − 1) coshu sinh2 u

(

1

2sinh u log

cosh u − 1

cosh u + 1+ cothu

))

−1

,

A

E1

= −1

2(ε − 1) cosh u sinh2 u

(74)

The solutions are indeed similar to the oblate case, but there are actually quite largedifferences of detail. For example, the field lines don’t notice the prolate spheroid untilthey’re almost upon it.

That’s enough orthogonal coordinates for now!!

3.9 The l = 0 solutions

There is an even simpler problem, having l = 0. The Newtonian potential

Φ = cot−1 sinh u (75)

satisfies ∇2Φ = 0 except on the disc u = 0 and has the contours of u for its equipotentials.Evaluating the gradient we find that the surface density is simply

σ = 1/ sin v =(

1 − ρ2)

−1/2

, (76)

14

which may be familiar to some readers... and integrates over the disc to give a total of4π. The field line are the lines of constant v, so they come into the disc at right angles— the disc is ‘self-supporting’ in the radial direction.

In prolate spheroidal coordinates the corresponding solution is

Φ = 1

2log

cosh u + 1

cosh u − 1, (77)

which is singular along the line u = 0, with the same density profile.

15