Notes for Part 1: Basic Formalism of Electrodynamics

Dr Leek Meng Lee

Version: 9th Jan 2015

Abstract

The logic flow for this part is:

1. We want to state the conservation laws in electrodynamics clearly.

2. We want to formulate electrodynamics (the time-dependent EM) using the potentials, hopingto take advantage that there are only 4 quantities, ( and ~A) to solve for.

3. From the solutions, which are the retarded potentials, we then discover that information takesspeed of light, c, to travel.

4. We then derive the electrodynamic E and B-fields (Jefimenkos equations) which has Coulombslaw and Biot-Savarts law as static special cases.

5. We apply the solutions to some standard applications and discover velocity corrections (rela-tivistic corrections) to the fields.

6. Finally, the new acceleration terms in the fields are shown to give rise to radiation.

Contents

1 Conservation Laws 21.1 Charge continuity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Energy continuity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Momentum continuity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Potential Formulation of Electrodynamics 52.1 Electrodynamics in terms of potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Gauge transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2.1 Coulomb gauge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2.2 Lorenz gauge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.3 Solutions are Retarded Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 The fields: Jefimenkos equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.5 Standard Application 1: Point Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.5.1 Retarded potentials: Lienard-Wiechert potentials . . . . . . . . . . . . . . . . . . . 142.5.2 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.5.3 Example: Point charge in constant velocity . . . . . . . . . . . . . . . . . . . . . . 18

2.6 Standard Application 2: Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.6.1 From Point Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.6.2 From Hertzian Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.6.3 From Arbitrary Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1

1 Conservation Laws

1.1 Charge continuity equation

The global conservation of charge means that the total charge in the universe is a constant. The strongerstatement is the local conservation of charge which states: if the total charge in some volume changes,then that amount of charge must have entered or exited through the closed surface bounding that volume.We want to obtain the mathematical expression for that statement.

dQ

dt=

~J d ~A (1)

| write Q(t) =(~r, t)dV

(~r, t)

tdV =

~J d ~A (2)

| use divergence theorem=

~ ~JdV (3)

| since it holds true for any volume(~r, t)

t= ~ ~J(~r, t) (4)

Note that we may obtain the same continuity equation from Maxwells equations but the displacementcurrent term was added to be compatible with the charge continuity equation so the argument may besomewhat circular.

1.2 Energy continuity equation

Conservation of energy is of course one of the most sacred conservation laws we believe in. Actually in thecontext of relativity, conservation of energy and conservation of momentum is one combined conservationlaw as we will see in part 2 1. Now we will handle conservation of energy and conservation of momentumseparately.

Consider work done by electric and magnetic forces on charge q,

dW = ~F d~l = q( ~E + ~v ~B) ~vdt (5)| as expected, magnetic forces do no work= q ~E ~vdt (6)= dV ~E ~vdt (7)| recall that ~v = ~J

dW

dt=

~E ~JdV (8)

Thus ~E ~J is the power (per unit volume) delivered to the charges.

dW

dt=

~E ~JdV =

~E (

1

0~ ~B 0

~E

t

)dV (9)

| where we replace ~J using Ampere-Maxwell law

=

(1

0~E (~ ~B) 0 ~E

~E

t

)dV (10)

| use identity ~ ( ~E ~B) = ~B (~ ~E) ~E (~ ~B)

| then use Faradays law, ~E (~ ~B) = ~ ( ~E ~B) + ~B (

~B

t

)

| then rewrite ~B ~B

t=

1

2

t~B2 and ~E

~E

t=

1

2

t~E2

1Which includes rest mass energy.

2

= ddt

1

2

(0 ~E

2 +1

0~B2)dV 1

0

( ~E ~B) d ~A (11)

| define the EM field energy density, uEM = 12

(0 ~E

2 +1

0~B2)

| and define the Poynting vector, ~S = 10

( ~E ~B)

= ddt

uEMdV

~S d ~A (12)

where the closed surface bounds volume V . This is Poyntings theorem or work-energy theorem ofelectrodynamics. We will derive the differential version or the energy continuity equation shortly.

We first need to interprete Poyntings theorem clearly:

dWdt is the work done (per unit time) on the charges in volume V by EM forces ddt

uEMdV is the change in energy (per unit time) stored in the fields

~S d ~A is the energy (per unit time) that flowed out through the surfaceTo get the differential version, suppose we can write a mechanical energy density umech, then

dW

dt=

d

dt

umechdV , then

d

dt

umechdV = d

dt

uEMdV

~S d ~A (13)

| use divergence theorem on the last term| since the equation holds for any volume

t(umech + uEM) = ~ ~S (14)

which is the continuity equation for energy and the Poynting vector is the energy current vector. If there

are no sources in volume V , we haveuEMt

= ~ ~S which simply means that the change in energystored in the field in volume V is equal to the amount of energy that flowed in or out of the surfacebounding volume V .

1.3 Momentum continuity equation

Let us look at a quick example to see why the EM field also carries momentum.Consider 2 charges moving with constant v as shown in the figure.

Figure 1: Two charges moving with speed v and exerting forces on each other.

The calculation of E-field and B-field due to moving charges is not from Coulombs law and Biot-Savarts law. This calculation will be done later in part 1 but the essential results are easily stated:the E-field is still radial and the B-field is still circular. Thus the electric force and magnetic force are

3

as drawn in the figure and we can see that the net force on each charge is equal in magnitude but notopposite to each other.

This appears to be a violation of Newtons third law which is essential in the conservation of momen-tum. Thus if we realise that the EM fields also carry momentum then the conservation of momentumwould be the conservation of field momentum + mechanical momentum.

We shall now derive the momentum continuity equation. Consider the total electromagnetic force oncharges in volume V ,

~F =

( ~E + ~v ~B)dV (15)

| recall that ~v = ~J=

( ~E + ~J ~B)dV (16)

We can denote ~f = ~E+ ~J ~B as the force per unit volume. We want to eliminate and ~J and expressthe RHS completely in terms of the fields. This will allow us to discover the field momentum (and theMaxwell Stress tensor).

~f = ~E + ~J ~B (17)

| use Gauss law: = 0(~ ~E) and Ampere-Maxwell law: ~J = 10~ ~B 0

~E

t

= 0(~ ~E) ~E +(

1

0~ ~B 0

~E

t

) ~B (18)

| then use t

( ~E ~B) = ~E

t ~B + ~E

~B

t

| and Faradays law: ~B

t= ~ ~E so that

~E

t ~B =

t( ~E ~B) + ~E (~ ~E)

= 0

[(~ ~E) ~E ~E (~ ~E)

]+

1

0(~ ~B) ~B 0

t( ~E ~B) (19)

| rewrite (~ ~B) ~B = ~B (~ ~B)| introduce (~ ~B) ~B which is zero, to make the expression more symmetrical| recall the Poynting vector: ~S = 1

0( ~E ~B)

= 0

[(~ ~E) ~E ~E (~ ~E)

]+

1

0

[(~ ~B) ~B ~B (~ ~B)

] 00

~S

t(20)

| use identity: ~ ~E2 = 2( ~E ~) ~E + 2 ~E (~ ~E) = ~E (~ ~E) = 12~ ~E2 ( ~E ~) ~E

| a similar identity holds for ~B: ~B (~ ~B) = 12~ ~B2 ( ~B ~) ~B

= 0

[(~ ~E) ~E + ( ~E ~) ~E

]+

1

0

[(~ ~B) ~B + ( ~B ~) ~B

]1

2~(0 ~E

2 +1

0~B2) 00

~S

t(21)

| introduce the 3 3 matrix: Maxwell stress tensor| which is: Tij = 0

(EiEj 1

2ij ~E

2

)+

1

0

(BiBj 1

2ij ~B

2

)| the divergence of T is (~

T )j =

i=x,y,z

iTij

| and note that

i=x,y,z

iEi = ~ ~E and

i=x,y,z

EiiEj = ( ~E ~)Ej

| we have (~ T )j = 0(

(~ ~E)Ej + ( ~E ~)Ej 12j ~E2

)+

1

0

((~ ~B)Bj + ( ~B ~)Bj 1

2j ~B2

)

4

= ~ T 00 ~S

t(22)

The total force is obtained by integrating over volume V .

~F =

~ T dV 00

~S

tdV (23)

| use divergence theorem=

T d ~A 00 d

dt

~SdV (24)

Thus we can interpreteT as a generalized pressure. The diagonal elements, Txx, Tyy and Tzz are pressures

and the off-diagonal elements, Txy, Tyz, . . . are shears.We want to write it into the differential version which is the momentum continuity equation.

~F =

~ T dV 00 d

dt

~SdV (25)

| assume we can define a mechanical momentum density ~F = ddt

~pmechdV

| the field momentum density is ~pEM = 00~S = 1c2~S

d

dt

(~pmech + ~pEM)dV =

~ T dV (26)

t(~pmech + ~pEM) = ~

T (27)

which is the momentum continuity equation. We can thus also interprete T as the momentum fluxdensity.

1.4 Angular Momentum

We will only briefly mention angular momentum in this course. The angular momentum (density) isdefined in the usual way

lEM = ~r ~pEM = 0[~r ( ~E ~B)

](28)

2 Potential Formulation of Electrodynamics

2.1 Electrodynamics in terms of potentials

In electrostatics and magnetostatics, it is advantageous to solve the potential problem (by solving thePoisson or Laplace equation) then get the fields from the potentials. In electrodynamics, the advantageis smaller. So it is still worthwhile to express electrodynamics in terms of potentials. This means wewant the time dependent solutions of and ~A resulting from time dependent sources (~r, t) and ~J(~r, t).

Since ~B is still divergenceless in the time dependent problem, we still have

~B = ~ ~A (29)

Curl of ~E is Faradays law, so

~ ~E = ~B

t(30)

| insert ~B = ~ ~A and move it to LHS~

(~E +

~A

t

)= 0 (31)

| recall the identity ~ (~f) = 0

5

~E + ~A

t= ~ (32)

~E = ~ ~A

t(33)

When we insert this ~E into Gauss law,

~ ~E = 0

(34)

~ (~

~A

t

)=

0(35)

~2+ t

(~ ~A) = 0

(36)

which is the time dependent generalization of Poisson equation.Now using Ampere-Maxwell equation

~ ~B = 0 ~J + 00 ~E

t(37)

| insert ~B = ~ ~A and ~E = ~ ~A

t

~ (~ ~A) = 0 ~J 00~(

t

) 00

2 ~A

t2(38)

| use vector identity ~ (~ ~A) = ~(~ ~A) ~2 ~A

0 ~J =(~2 ~A 00

2 ~A

t2

) ~

(~ ~A+ 00

t

)(39)

The potential formulation has thus been achieved with those 2 equations. The 2nd equation is muchmuch more complicated than the magnetostatic version. We need to simplify the equations. This canbe achieved by using the gauge degree of freedom that these potentials possess.

2.2 Gauge transformations

Note that ~E and ~B are the physical fields and the and ~A potential fields are not. This gauge freedomexists because, according to the equations of electrodynamics, we can have different and ~A and yet thesame ~E and ~B fields are obtained. Lets work this out precisely.

suppose, ~A = ~A+ ~ and = + (40)

take curl, ~ ~A = ~ ~A+ ~ ~ (41)| the same ~B field is required

~B = ~B + ~ ~ (42)~ ~ = 0 (43)

= ~ = ~ (44)Now use,

~E = ~ ~A

t(45)

= ~( ) t

( ~A ~) (46)

= ~ ~A

t+ ~ + ~

t(47)

| we require ~ ~A

t= ~E, so

0 = ~ + ~t

(48)

6

0 = ~ + ~t

(49)

0 = ~( +

t

)(50)

= t

(51)

Thus,

~A = ~A+ ~ and = t

(52)

represent the allowed gauge transformations of and ~A.

2.2.1 Coulomb gauge

We will now choose the so-called Coulomb gauge and see what happens to the potential formulation ofelectrodynamics.

Take ~ ~A = ~ ~A+ ~ (~) and choose such that ~ ~A = 0 (Coulomb gauge).

~2+ t

(~ ~A) = 0

(53)

~2( +

t

)+

t

(~ ~A ~ (~)

)=

0(54)

| so the related terms cancel~2 +

t(~ ~A) =

0(55)

| recall that ~ ~A is chosen to be zero, then drop the prime~2 =

0(56)

which is just Poisson equation. The solution is already known in electrostatics. However in electrody-namics, we also need ~A to get ~E, so now we look at the other potential equation.(

~2 ~A 00 2 ~A

t2

) ~

(~ ~A+ 00

t

)= 0 ~J (57)(

~2( ~A ~) 00 2

t2( ~A ~)

) ~

(~ ( ~A ~) + 00

t

( +

t

))= 0 ~J (58)

note ~2(~) = ~(~ ~) ~ (~ (~)) and ~ (~)) = 0 |recall that ~ ~A is chosen to be zero |

~2 ~A 00 2 ~A

t2 00~

(

t

)= 0 ~J (59)

where we can drop the prime labels also. Thus solving for the ~A is very complicated, so Coulomb gaugedoes not look useful in getting solutions in electrodynamics.

2.2.2 Lorenz gauge

We again take ~ ~A = ~ ~A+ ~ (~) and choose such that ~ ~A = 00

t . This is the Lorenzgauge. Historically, this gauge was wrongly attributed to H. A. Lorentz. Thus the potential equationsbecome,

~2 + t

(~ ~A) = 0

(60)

~2 00 2

t2=

0(61)

7

and,

~2 ~A 00 2 ~A

t2= 0 ~J (62)

This gauge has the virtue that it decouples and ~A and treats them on an equal footing. In part 2, wewill also see that this gauge is Lorentz (not Lorenz!) invariant. In the next section, we will solve thepotential equations in this gauge.

2.3 Solutions are Retarded Potentials

We drop the primes from the potential equations,

~2 00 2

t2=

0and ~2 ~A 00

2 ~A

t2= 0 ~J (63)

This is the problem of solving an inhomogeneous partial differential equation and we will do it using themethod of Greens function.

The Greens function method is defined by,(~2 00

2

t2

)G(~r, t;~r, t) = 3(~r ~r)(t t) (64)

so the solution is

(~r, t) = 10

G(~r, t;~r, t)(~r, t)dV dt (65)

It is easy to see that when we apply

(~2 00

2

t2

), we get back the differential equation(

~2 00 2

t2

)(~r, t) = 1

0

(~2 00

2

t2

)G(~r, t;~r, t)(~r, t)dV dt (66)

= 10

3(~r ~r)(t t)(~r, t)dV dt (67)

= (~r, t)0

(68)

We shall solve G(~r, t;~r, t) in Fourier space.(~2 00

2

t2

)G(~r, t;~r, t) = 3(~r ~r)(t t) (69)

| use Fourier form: 3(~r ~r) =

d3~k

(2pi)3ei~k(~r~r)

| use Fourier form2: (t t) =

d

2piei(tt

)

| use Fourier form: G =

d3~k

dg(~k, )ei~k(~r~r)ei(tt

)

| note(~2 00

2

t2

)ei~k(~r~r)ei(tt

) =

(~k2 +

2

c2

)ei~k(~r~r)ei(tt

)

| where 00 = 1c2

, then comparing Fourier coefficients will give,

g(~k, ) = 1(2pi)4

1

k2 2c2(70)

where we wrote ~k2 = k2 for simplicity. Hence G(~r, t;~r, t) is solved.

G(~r, t;~r, t) = 1(2pi)4

d3~k

d1

k2 2c2ei~k(~r~r)ei(tt

) (71)

2The choice of different signs in the temporal and spatial Fourier forms is to cater for the Minkowski metric in SpecialRelativity.

8

Well, we still have to carry out the integrals. The integrand has 2 simple poles at = ck. We shalltreat as complex, so = R + iI and use Cauchys residue theorem:

f(z)dz = 2pii

Residues.

Therefore we need to choose a closed contour for the integration and making a choice of a closedcontour amounts to specifying a (physical) boundary condition for the Greens function, as we shall see.

Figure 2: Two simple poles of the integrand.

Choice 1 (Retarded): The choice to skip above the poles is equivalent to moving the poles infinites-imally into the lower half plane (LHP). This is equivalent to rewriting G as

Figure 3: Skipping above the poles is equivalent to moving the poles down.

G(~r, t;~r, t) = GR(~r, t;~r, t) (72)

= lim0

1(2pi)4

d3~k

d1

k2 (+i)2c2ei~k(~r~r)ei(tt

) (73)

| note (ck)2 ( + i)2 = (ck i)(ck + + i)

= lim0

1

(2pi)4

d3~k

dei~k(~r~r)ei(tt

)c2

( + ck + i)( ck + i) (74)

We still have not decided to close this straight path over UHP or LHP. If we close it in UHP, it doesnot enclose any poles so the integral is zero. Anyway to determine the domain in closing the integral inUHP where I > 0, we look at e

i(tt) = eiR(tt)eI(tt

). Thus we require t t < 0 so that theintegral is damped and therefore well defined.

Summarizing: GR(~r, t;~r, t) = 0 for t t < 0Now we choose to close this path over LHP. Closing over LHP, where I < 0, the damping of the integral(due to eI(tt

)) requires t t > 0. So now we can evaluate the contour integral,d

ei(tt)c2

( + ck + i)( ck + i) = 2pii

Residues (75)

9

Figure 4: Closing the contour over the lower half plane (LHP).

| the negative sign is due to the clockwise closed contour| break

d into

straight

d +

curve

d

| thencurve

d 0 due to damping and I

| andstraight

d =

d

dei(tt

)c2

( + ck + i)( ck + i) = 2piic2

[ei(cki)(tt

)

ck i ck + i +ei(cki)(tt

)

ck i+ ck + i

](76)

| take the limit 0=2piic2

2ck

(eick(tt) + eick(tt)

)(77)

=piick

(2i sin(ck(t t))) (78)

= 2pick

sin(ck(t t)) (79)

In summary, we have

GR(~r, t;~r, t) =

{0 t t < 0 c(2pi)3

d3~k e

i~k(~r~r)k sin(ck(t t)) t t > 0

(80)

or, GR(~r, t;~r, t) = (t t) c(2pi)3

d3~k

ei~k(~r~r)

ksin(ck(t t)) (81)

To finish the calculation, we need to carry out the k integral. We use spherical coordinates and align ~r~rto the z-axis so ~k (~r ~r) = k|~r ~r| cos = kR cos and the volume element is d3~k = k2 sin dkdd.

GR(~r, t;~r, t) = c(2pi)3

(t t) 2pi0

d

pi0

d

0

dkeikR cos k sin sin(ck(t t)) (82)| the integral gives 2pi| the integral:

pi0

d sin eikR cos = pi0

d cos eikR cos =2

kRsin(kR)

= 2c(2pi)2R

(t t) 0

dk sin(kR) sin(ck(t t)) (83)

| write sin(kR) = 12i

(eikR eikR) and similarly for sin(ck(t t))

= (t t) 2c8piR

0

dk

2pi

(eik(R+c(tt

)) + eik(R+c(tt)) eik(Rc(tt)) eik(Rc(tt))

)10

| note 2nd term: 0

dk

2pieik(R+c(tt

)) k1=k 0

dk12pi

eik1(R+c(tt)) then

0

=

0

| so the 2nd term adds to 1st term. The 4th term adds similarly to the 3rd term= (t t) c

4piR

dk

2pi

[eik(R+c(tt

)) eik(Rc(tt))]

(84)

| recall the delta function Fourier representation:

dk

2pieik(R+c(tt

)) = (R+ c(t t))

| then use identity: c(R+ c(t t)) = (t t + R

c

)= (t t) 1

4piR

[

(t t + R

c

)

(t+ t + R

c

)](85)

| recall that the delta function is even| drop first delta function as we require t t > 0= (t t) 1

4piR

(t t R

c

)(86)

Choice 2 (Advanced): This will just be briefly mentioned as the steps are very similar and this isan unphysical solution. Skipping over the poles from below means,

Figure 5: Skipping below the poles is equivalent to moving the poles up.

Figure 6: Closing the contour over the upper half plane (UHP).

G(~r, t;~r, t) = GA(~r, t;~r, t) (87)

= lim0 1

(2pi)4

d3~k

d

1

k2 (i)2c2ei~k(~r~r)ei(tt

) (88)

For the same reasons of damping the integral, GA(~r, t;~r, t) = 0 for t t > 0. The path is to be closedover UHP (I > 0).

11

After evaluating the contour integral, we have in summary, 3

GA(~r, t;~r, t) =

{0 t t > 0c

(2pi)3

d3~k e

i~k(~r~r)k sin(ck(t t)) t t < 0

(89)

or, GA(~r, t;~r, t) = (t t) c(2pi)3

d3~k

ei~k(~r~r)

ksin(ck(t t)) (90)

Carry out the k integral in the same way to get,

GA(~r, t;~r, t) = (t t) 14piR

[

(t t + R

c

)

(t t R

c

)](91)

| drop the 2nd delta function s we require t t < 0= (t t) 1

4piR

(t t + R

c

)(92)

Thus the retarded solution of the scalar potential is

(~r, t) = 10

GR(~r, t;~r, t)(~r, t)dV dt (93)

| usually (t t) is left out and recall R = |~r ~r|=

1

4pi0

(~r, t)|~r ~r|

(t t |~r ~r

|c

)dV dt (94)

Similarly, we have

~A(~r, t) =04pi

~J(~r, t)|~r ~r|

(t t |~r ~r

|c

)dV dt (95)

where t = t |~r~r|c = tr is called the retarded time.

Figure 7: The information at the point of observation at time t is from the source at the retarded timetr.

To understand retarded time, we see that |~r~r|

c is the time for information to travel at speed cfrom the source to the point of observation. Thus the potential at time t is due to the informationfrom the source at tr. This is a clue that electrodynamics and relativity are related as relativity requirescausality where information takes a finite speed (maximum is c) to travel.

3The missing negative sign is due to the contour being closed in an anticlockwise manner.

12

2.4 The fields: Jefimenkos equations

Note that it appears that (~r, t) and ~A(~r, t) differs from their electrostatic and magnetostatic versionsjust by the retarded time. It is not that simple. If we take the electrostatic and magnetostatic versionsof the fields and put in the retarded time, the results are wrong.

~E(~r) =1

4pi0

(~r)|~r ~r|2 RdV

~E(~r, t) 6= 14pi0

(~r, t)|~r ~r|2 R

(t t |~r ~r

|c

)dV dt(96)

~B(~r) =04pi

~J(~r) R|~r ~r|2 dV

~B(~r, t) 6= 04pi

~J(~r, t) R|~r ~r|2

(t t |~r ~r

|c

)dV dt (97)

To get the fields, we need to use the retarded potentials and the equations: ~E = ~ ~At and~B = ~ ~A. We do the easier ~At term first,

~A

t=

t

04pi

~J(~r, tr)|~r ~r| dV

=04pi

~J(~r, tr)|~r ~r| dV

(98)

Then we calculate ~ but note that ~r is in the denominator as |~r ~r| and is hiding in tr = t |~r~r|

c .

~(~r, t) = 14pi0

( ~(~r, tr)|~r ~r| + (~r

, tr)~ 1|~r ~r|

)dV (99)

| write ~ = t

t

tr~tr =

t

(~|~r ~r|

c

)= 1

c~r ~r|~r ~r|

| and write ~ 1|~r ~r| = ~r ~r|~r ~r|3 =

R

|~r ~r|2

=1

4pi0

( c

R

|~r ~r| R

|~r ~r|2)dV (100)

~E(~r, t) = ~ ~A

t(101)

=1

4pi0

((~r, tr)|~r ~r|2 R+

(~r, tr)c|~r ~r| R

~J(~r, tr)c2|~r ~r|

)dV (102)

So there are 2 more terms extra as compared to our naive guess. This Jefimenkos equation for theE-field is essentially the time-dependent generalization of Coulombs law.

Now for the B-field,

~B(~r, t) = ~ ~A (103)| recall ~ (f ~A) = f(~ ~A) ~A (~f)

=04pi

( ~ ~J|~r ~r|

~J ~ 1|~r ~r|

)dV (104)

| we already know ~ 1|~r ~r| = ~r ~r|~r ~r|3 =

R

|~r ~r|2

| look at (~ ~J)x = Jzy Jy

z=Jzt

t

tr

try Jy

t

t

tr

trz

= Jztry Jy tr

z

| recall that ~tr = ~r ~r

|~r ~r|(1c

)= 1

cR

| hence, (~ ~J)x = Jz(1cRy

) Jy

(1cRz

)=

1

c( ~J R)x

=04pi

( ~J(~r, tr) Rc|~r ~r| +

~J(~r, tr) R|~r ~r|2

)dV (105)

13

This Jefimenkos equation for B-field is essentially the time-dependent generalization of Biot-Savartslaw.

These 2 Jefimenkos equations are essentially formal and have limited uses but they complete elec-trodynamics before we recast into a relativistic description.

2.5 Standard Application 1: Point Charge

2.5.1 Retarded potentials: Lienard-Wiechert potentials

Figure 8: The setup to calculate the retarded potentials of a moving charge.

We shall apply the retarded potentials (~r, t) and ~A(~r, t) to a moving charge of charge q.

Point charge trajectory: ~r0(t) Point charge velocity: ~v(t) = d~r0(t)dt Point charge density: (~r, t) = q3(~r ~r0(t)) Point charge current density: ~J(~r, t) = q~v(t)3(~r ~r0(t))

(~r, t) =1

4pi0

(~r, t)|~r ~r|

(t t |~r ~r

|c

)dV dt (106)

=q

4pi0

3(~r ~r0(t))|~r ~r|

(t t |~r ~r

|c

)dV dt (107)

| do the volume integration

=q

4pi0

(t t |~r~r0(t)|c )|~r ~r0(t)| dt

(108)

| because the dependence on t is implicit, we need an identity

| identity:ds(f(s))g(s) =

dsdf

ds

(f(s))dfds

g(s) =

df(f)

g(s)dfds

=g(s)dfds

f(s)=0

| so, f(t) = t t+ |~r ~r0(t)|

c= t t+ 1

c

(x x0(t))2 + (y y0(t))2 + (z z0(t))2

| so, df(t)

dt= 1 (x x0(t

))x0(t) + c

(x x0(t))2 + = 1 (~r ~r0(t

)) ~v(t)c|~r ~r0(t)| = 1

R(t) ~v(t)c

=qc

4pi0

1

R(t)c ~R(t) ~v(t)

f(t)=0 or t=tR(t)c =t

|~r~r0(t)|c

(109)

Then,

~A(~r, t) =q04pi

~v(t)3(~r ~r0(t))

|~r ~r| (t t |~r ~r

|c

)dV dt (110)

14

| so the only difference here is that g(s) = ~v(t)

|~r ~r0(t)| , so

=q04pi

cv(t)

R(t)c ~R(t) ~v(t)

t=tR(t)c =t

|~r~r0(t)|c

(111)

=~v(t)c2

(~r, t)t=tR(t)c =t

|~r~r0(t)|c

(112)

We can provide a nice physical (actually goemetrical) picture to the derivation. We write,

(~r, t) =1

4pi0

(~r, t)|~r ~r| (tr t

)dV dt (113)

| do the time integral formally=

1

4pi0

(~r, tr)|~r ~r| dV

(114)

The picture is that, information comes to the point of observation at speed c and the charge is movingsodV is not the total charge. This is a geometric effect and has nothing to do with relativity.

charge seen if charge distribution is not moving = shaded area = dAdr (115)

= dV (116)

Shifted area = ((~v R)dt)dA (117)| where dt = dr

c

15

= ~v RdrcdA (118)

=~v Rc

dV which is actually less

charge seen if charge distribution is moving with velocity ~v = shaded area

dq = dV ~v Rc

dV (119)dV =

1

1 ~vRcdq (120)

so roughly, 14pi0

1

|~r ~r0(tr)|dV (121)

=1

4pi0

1

|~r ~r0(tr)|

1

1 ~vRcdq (122)

=1

4pi0

1

|~r ~r0(tr)|1

1 ~vRc

dq (123)

anddq = q gives the same retarded potential expression. Again, this is to show that the origin of the

factor: 1/(1 ~vRc ) is a geometrical one.

2.5.2 Fields

The equations for the fields are ~E = ~ ~At and ~B = ~ ~A. The main difficulty in calculation isthat the differentiation needs to be done implicitly. Note the expressions

tr = t |~r ~r0(tr)|c

= t R(tr)c

= R(tr) = c(t tr) (124)

and, ~v(t) =d~r0(t)

dt(125)

We start by calculating ~(~r, t),

~(~r, t) = ~ qc4pi0

1

R(tr)c ~R(tr) ~v(tr)(126)

=qc

4pi0

1(Rc ~R ~v)2

~(Rc ~R ~v) (127)

| note ~R = ~|~r ~r0(tr)| = ~c(t tr) = c~tr| note ~(~R ~v) = (~R ~)~v + (~v ~)~R+ ~R (~ ~v) + ~v (~ ~R)| chain rule will be used to simplify all 4 terms| first term: (~R ~)~v =

(Rx

x+

)~v(tr) = Rx

d~v(tr)

dtr

trx

+ = ~a(~R ~tr)

| second term: (~v ~)~R = (~v ~)~r (~v ~)~r0(tr) = ~v ~v(~v ~tr)| third term: ~R (~ ~v) = ~R (~a ~tr)| fourth term: ~v (~ ~R) = ~v (~ ~r ~ ~r0(tr)) = ~v (~v ~tr)=

qc

4pi0

1(Rc ~R ~v)2

[c2~tr ~a(~R ~tr) ~v + ~v(~v ~tr) + ~R (~a ~tr) ~v (~v ~tr)

]| use identity ~R (~a ~tr) ~a(~R ~tr) = (~R ~a)~tr| use identity ~v(~v ~tr) ~v (~v ~tr) = v2~tr=

qc

4pi0

1

(Rc ~R ~v)2[~v +

(c2 v2 + (~R ~a)

)~tr]

(128)

| lastly, we need c~tr = ~R = ~~R ~R =

~(~R ~R)2~R ~R

=(~R ~)~R+ ~R (~ ~R)

R

16

| then chain rule give (~R ~)~R = ~R ~v(~R ~tr) and ~ ~R = ~v ~tr| then, ~R (~v ~tr) = ~v(~R ~tr) (~R ~v)~tr and 2 terms cancel

| rearrange and make ~tr the subject to get ~tr = ~R

Rc ~R ~v

=qc

4pi0

1

(Rc ~R ~v)2

[~v

(c2 v2 + ~R ~a

) ~RRc ~R ~v

](129)

=qc

4pi0

1

(Rc ~R ~v)3[(Rc ~R ~v

)~v

(c2 v2 + ~R ~a

)~R]

(130)

Similarly, and we are not repeating the calculation, ~At is obtained as

~A

t=

qc

4pi0

1

(Rc ~R ~v)3[(Rc ~R ~v

)(R~ac ~v)

+(c2 v2 + ~R ~a

) R~vc

](131)

So the electric field is

~E(~r, t) = ~(~r, t) ~A(~r, t)

t(132)

=qc

4pi0

1

(Rc ~R ~v)3[(Rc ~R ~v

)(~v R~a

c+ ~v

)+(c2 v2 + ~R ~a

)(~R R~v

c

)]=

q

4pi0

R

(Rc ~R ~v)3[(Rc ~R ~v)~a+

(c2 v2 + ~R ~a

)(Rc ~v)

](133)

=q

4pi0

R

(Rc ~R ~v)3[(c2 v2)(Rc ~v) + (~R ~a)Rc (~R ~a)~v Rc~a+ (~R ~v)~a

](134)

| we can check backwards that the last 4 terms become ~R ((Rc ~v) ~a)=

q

4pi0

R

(Rc ~R ~v)3[(c2 v2)(Rc ~v) + ~R ((Rc ~v) ~a)

](135)

Term q4pi0 R(Rc~R~v)3 (c2 v2)(Rc ~v) is called the velocity field term and it falls as 1R2 .

Term q4pi0 R(Rc~R~v)3 ~R ((Rc ~v) ~a is called the acceleration field or the radiation field and itfalls as 1R .

For the magnetic field,

~B = ~ ~A (136)= ~

(~v

c2

)(137)

=1

c2

[(~ ~v) ~v (~)

](138)

| recall that ~ ~v = ~a ~tr = ~a~R

Rc ~R ~v| recall that ~ = qc

4pi0

1

(Rc ~R ~v)3[(Rc ~R ~v)~v (c2 v2 + ~R ~a)~R

]| then ~v ~v = 0 and ~v ~ = qc

4pi0

1

(Rc ~R ~v)3[(c2 v2 + ~R ~a)(~v ~R)

]=

1

c

q

4pi0

1

(Rc ~R ~v)3

[(Rc ~R ~v)2 ~a

~R

Rc ~R ~v + (c2 v2 + ~R ~a)(~v ~R)

](139)

= 1c

q

4pi0

1

(Rc ~R ~v)3~R

[~a(Rc ~R ~v) + ~v(c2 v2 + ~R ~a)

](140)

=1

c

q

4pi0

R

(Rc ~R ~v)3 R[(c2 v2)(~v) (~R ~a)~v Rc~a+ (~R ~v)~a

](141)

17

| insert Rc into first term since R Rc = 0 anyway| insert (~R ~a)Rc since R ~R = 0 anyway=

1

c

q

4pi0

R

(Rc ~R ~v)3 R[(c2 v2)(Rc ~v) + (~R ~a)Rc (~R ~a)~v Rc~a+ (~R ~v)~a

](142)

| we can check backwards that the last 4 terms become ~R ((Rc ~v) ~a)=

1

c

q

4pi0

R

(Rc ~R ~v)3 R[(c2 v2)(Rc ~v) + ~R ((Rc ~v) ~a)

](143)

Term 1c q4pi0 R(Rc~R~v)3 R (c2 v2)(Rc ~v) is called the velocity field term and it falls as 1R2 .

Term 1c q4pi0 R(Rc~R~v)3 R[~R ((Rc ~v) ~a)

]is called the acceleration field or the radiation field

and it falls as 1R .A quick comparison between the 2 fields give

~B(~r, t) =1

cR ~E(~r, t) (144)

Thus the magnetic field of a point charge is always perpendicular to the electric field and to the vectorR that points from the retarded position to the point of observation.

2.5.3 Example: Point charge in constant velocity

Before we start the example of a point charge with constant velocity, we quickly talk about a pointcharge that is stationary. This means we set ~a = 0 and ~v = 0. For the potentials,

(~r) =qc

4pi0

1

R(tr)c(145)

| where R(tr) is simply R=

q

4pi0R(as expected) (146)

~A(~r) =~v(tr)

c2

set ~v = 0= 0 (as expected) (147)

For the fields, after setting ~a = 0 and ~v = 0,

~E(~r) =q

4pi0

R

(Rc)3c2Rc =

q

4pi0R2R (which is Coulombs law as expected) (148)

~B(~r) =1

cR ~E = 1

cR q

4pi0R2R = 0 (as expected) (149)

For the point charge in constant velocity, the trajectory where the particle passes through the originat t = 0 is ~r0(t) = ~vt. So for this case, we can make the implicit dependence of tr to be explicit.

tr = t |~r ~r0(tr)|c

(150)

|~r ~vtr|2 = c2(t tr)2 (151)r2 2~r ~vtr + v2t2r = c2(t2 2ttr + t2r) (152)

tr =(c2t ~r ~v)(c2t ~r ~v)2 (c2 v2)(c2t2 r2)

(c2 v2) (153)| to fix the sign, we set ~v = 0 and require tr < t| so the negative sign is the meaningful one

tr =(c2t ~r ~v)(c2t ~r ~v)2 (c2 v2)(c2t2 r2)

(c2 v2) (154)

18

In the Lienard- Wiechert potentials, we have the common denominator,

common denominator = R(tr)c ~R(tr) ~v(tr) (155)| recall that R(tr) = |~r ~r0(tr)| = |~r ~vtr| = c(t tr)= c2(t tr) (~r ~vtr) ~v (156)= c2(t tr) ~r ~v + v2tr (157)= (v2 c2)tr + c2t ~r ~v (158)| from above solution of tr write

((c2 v2)tr (c2t ~r ~v)

)=

=

(c2t ~r ~v)2 (c2 v2)(c2t2 r2) (159)| further rewrite: define ~rp = ~r ~vt where ~r0(t) = ~vt is the present position=

[c4t2 + (~r ~v)2 2c2t(~r ~v) + r2c2 c4t2 r2v2 + c2v2t2]1/2 (160)

=r2v2 cos2 r2v2 + c2r2p (161)

= crp

1 r

2v2

c2r2psin2 (162)

| use geometry as in figure 9

= crp

1 v

2

c2sin2 (163)

Figure 9: Somehow the field points from the present position but the information came from the retarded

position. Note the sine rule: sin rp =sin(1800)

r = sin2 r2p

= sin2 r2 .

The retarded potentials are

(~r, t) =q

4pi0rp

1 v2c2 sin2

and ~A(~r, t) =~v

c2(~r, t) =

q~v

4pi0c2rp

1 v2c2 sin2

(164)

We shall calculate the fields using the explicit formulae for ~E and ~B and start by setting ~a = 0 so onlythe velocity field term survives.

~E(~r, t) =q

4pi0

R

(Rc ~R ~v)3 (c2 v2)(Rc ~v) (165)

| recall that Rc ~R ~v = crp

1 v2

c2sin2

= note that R(Rc ~v) = ~RcR~v = (~r ~vtr)c c(t tr)~v = (~r ~vt)c = c~rp=

q

4pi0

c~rp(c2 v2)

c3r3p(1 v2c2 sin2

)3/2 (166)=

q

4pi0

1 v2c2(1 v2c2 sin2

)3/2 rpr2p (167)19

So ~E points along the line from the present position of the particle to the point of observation! The factthat the information came from the retarded position and yet the vector points from the present positionis due to an extraordinary coincidence which we will uncover in part 2.

The magnetic field is obtained by,

~B(~r, t) =1

cR ~E(~r, t) (168)

| rewrite R =~R

R=~r ~vtrR

=~rp + ~vt ~vtrc(t tr) =

~rpR

+~v

c

| since ~E is in the ~rp direction, ~rp ~E = 0=

1

c2

(~v ~E(~r, t)

)(169)

Thus the B-field is perpendicular to both ~v and ~E, so the B-field forms circles around the trajectory ofthe charge.

For a slow moving charge, where v

| use product rule: ~A ( ~B ~C) = ~B( ~A ~C) ~C( ~A ~B)=

1

0c

(~E2R (R ~E) ~E

)(179)

| then note that R (~R ((Rc ~v) ~a)) = 0| and R velocity field terms 1

R2does not cause radiation

| and so drop all velocity field terms 1

0c~E2R (180)

10c

(q

4pi0

R

(Rc ~R ~v)3

)2 [~R ((Rc ~v) ~a)

][~R ((Rc ~v) ~a)

]R (181)

To make the calculations easier, we carry on with the calculations in a frame where the charge isinstantaneously at rest at time tr, so set ~v = 0 but ~a 6= 0. We will generalise this in Part 2.

~S =1

0c

(q

4pi0

R

R3c3

)2 (~R (Rc ~a)

)(~R (Rc ~a)

)R (182)

| use product rule: ~A ( ~B ~C) = ~B( ~A ~C) ~C( ~A ~B)| so, ~R (R ~a) = R(~R ~a) ~a(~R R) = R(~R ~a)R~a

=1

0c

(q

4pi0

)21

R4c6c2(R(~R ~a)R~a

)(R(~R ~a)R~a

)R (183)

| recall 0 = 10c2

and carry out the dot product

=1

0c

( 0q4piR

)2 1R2

(R2a2 R2(R ~a)2

)R (184)

=1

0c

( 0q4piR

)2 (a2 (R ~a)2

)R (185)

| write R ~a = a cos =

1

0c

( 0q4piR

)2 (a2 a2 cos2 ) R (186)

=0q

2a2

16pi2c

sin2

R2R (187)

Thus power is emitted like a donut about the direction of instantaneous acceleration. No radiation isemitted in the forward or backward direction.

The power radiated into a patch of area R2 sin dd = R2d is

dP = ~S d ~A = |~S|R2d (188)dP

d=

0q2a2

16pi2csin2 (189)

The total power emitted is found by integrating over the entire solid angle.

P =

dP (190)

=

dP

dd or,

~S d ~A (191)

=0q

2a2

16pi2c

2pi0

d

pi0

d sin2 sin (192)

| note that 2pi0

d = 2pi and

pi0

sin3 d =4

3

=0q

2a2

6pic(193)

This is Larmors formula for slow moving, accelerating point charge. We shall see the Lienards gener-alised version of Larmors formula (for point charge) for any velocity in Part 2.

21

2.6.2 From Hertzian Dipole

Consider an electric dipole driven as shown in figure 10.

Figure 10: The setup for the Hertzian dipole radiation source.

We need to calculate the retarded potentials, then get the fields and finally get the power radiated.The retarded potentials are,

(~r, t) =1

4pi0

(~r, tr)|~r ~r| dV

(194)

| with (~r, tr) = q0 cos(t+r )3(~r d

2z) + (q0) cos(tr )3(~r +

d

2z)

| where t+r = tR+c

, R+ =

~r d2 z and tr = t Rc , R =

~r + d2 z

=1

4pi0

q0 cos((t R+c

))R+

q0 cos

((t Rc

))R

(195)| cosine rule: R2 = r2 +

(d

2

)2 2r

(d

2

)cos

(

1800 )

= r2 +

(d

2

)2 2r

(d

2

)cos

| Approximation 1: physical dipole perfect dipole d

[cos((t r

c

))+d

2ccos sin

((t r

c

))] 1r

(1 d

2rcos

)}(196)

| expand and 4 terms cancel pairwise, 4 terms add pairwise| denote p0 = q0d as the dipole moment=

p0 cos

4pi0r

{c

sin((t r

c

))+

1

rcos((t r

c

))}(197)

| Approximation 3: far field or radiation zone: r >> r >> c 1

r

rr

~A

t(210)

=p0 cos

4pi0c

[sin((t rc

))r2

r

cos((t r

c

))]r +

0p02

4pircos((t r

c

))z (211)

| recall that z = cos r sin and c2 = 100

| so cos cos((t r

c

))from

~A

tcancels that from

r

| drop cos sin((t rc

))r2

term as it is 1r2

0p02

4pi

sin

rcos((t r

c

)) (212)

which is a spherical wave expression.

~B = ~ ~A (213)=

1

r

[

r(rA) Ar

] (214)

| note that 1r

Ar

gives1

r2term which is not radiative, so drop it

=1

r

r

(r(0p0

4pirsin sin

((t r

c

)))) (215)

= 0p02

4pic

sin

rcos((t r

c

)) (216)

again a spherical wave expansion. The fields have the following features:

~E and ~B are in phase and they oscillate with (angular) frequency . The waves are travelling in the r direction while ~E and ~B are in the and directions respectively.

This means ~E and ~B are transverse and mutually perpendicular.

The amplitudes are in the ratio |~E|| ~B| = c which we already know from the vacuum solutions ofelectrodynamics.

Finally we can proceed to calculate the Poynting vector.

~S =1

0( ~E ~B) (217)

| note that = r

=0c

[p0

2

4pi

(sin

r

)cos((t r

c

))]2r (218)

We are usually more interested in the time-averaged power than the instantaneous power,

~Stime = 1T

T0

~Sdt (219)

=

2pi

2pi/0

0c

(p0

2

4pi

sin

r

)2cos2

((t r

c

))dtr (220)

| averaging cos2 over a period gives 12

=0p

20

4

32pi2c

sin2

r2r (221)

Power radiated into a patch of area r2 sin dd = r2d is

dP = ~Stime d ~A =~Stime r2d = 0p204

32pi2csin2 d (222)

24

Total Power = P =

dP (223)

=

dP

dd (224)

=0p

20

4

32pi2c

2pi0

d

pi0

sin3 d (225)

| recall that 2pi0

d = 2pi and

pi0

sin3 d =4

3

=0p

20

4

12pic(226)

This is Larmors formula for a Hertzian dipole.

2.6.3 From Arbitrary Distribution

We consider, in general, some configuration of charge and current localized in some finite volume nearthe origin.

Figure 11: The setup for an arbitrary distribution near the origin.

Again we start with the retarded potentials.

(~r, t) =1

4pi0

(~r, tr)|~r ~r| dV

(227)

| where cosine rule gives |~r ~r| = R =r2 + r2 2~r ~r

| Approximation 1: point of observation is much further away than source dimension: r

| so,(~r, t)dV = Q the total charge

| so,r ~rr

(~r, t)dV =r

r~r(~r, t)dV =

r ~p(t)r

| so,r ~rc

(~r, t)dV =d

dt

r

c~r(~r, t)dV =

r ~p(t)c

=1

4pi0

(Q

r+r ~p(t)r2

+r ~p(t)rc

)(230)

| Approximation 3: the first 2 terms, static monpole and dipole terms and are not radiative

14pi0

r ~p(t)rc

(231)

Now for ~A(~r, t),

~A(~r, t) =04pi

~J(~r, tr)R

dV (232)

| recall that we take the zeroth order approximation: R r| also recall the identity:

~J(~r, t r

c

)dV =

d~p

dt= ~p(t)

04pi

~p(t)

r(233)

We can proceed to calculate the fields,

~E(~r, t) = ~ ~A

t(234)

| in ~, after product rule, we drop the ~1r

term as it gives1

r2term

14pi0

~(r ~p(t)

)rc

04pi

~p(t)

r(235)

| so ~(r ~p(t)

)= ~pr = pr

rr =

pr

t

t

rr = pr

(1c

)= 1

c

(r ~p(t)

)r

=1

4pi0c2r ~p(t)r

r 04pi

~p(t)

r(236)

| recall that 1c2

= 00

=04pir

((r ~p(t)

)r ~p(t)

)(237)

| use product rule: ~A ( ~B ~C) = ~B( ~A ~C) ~C( ~A ~B)| so that r (r ~p(t)) = r(r ~p(t)) ~p(t)(r r)=

04pir

(r (r ~p(t))

)(238)

~B(~r, t) = ~ ~A (239)| use product rule: ~ (f ~A) = f(~ ~A) ~A (~f)| but ~1

rwill give

1

r2term and so we drop it

04pir

~ ~p(t) (240)

| use chain rule: ~ ~p(t) = (~t) ~p(t) and ~t = 1cr

= 04pirc

r ~p(t) (241)

26

To calculate power, we can choose to align ~p(t) along the z-axis,

~E(~r, t) = ~E(r, , t) =04pir

(r (r z)) p(t) (242)| recall z = cos r sin , r = and r =

=0p(t)

4pi

sin

r (243)

~B(~r, t) = ~B(r, , t) (244)

| recall that r z = sin

=0p(t)

4pic

sin

r (245)

The Poynting vector is,

~S =1

0( ~E ~B) = 0

c

(p(t)

4pi

sin

r

)2 = 0p(t)

2

16pi2c

sin2

r2r (246)

dP = ~S d ~A = |~S|r2 sin dd = 0p(t)2

16pi2csin3 d (247)

Total power, P =

dP

dd (248)

=0p(t)

2

16pi2c

2pi0

d

pi0

sin3 d (249)

=0p(t)

2

16pi2c 2pi 4

3(250)

=0p(t)

2

6pic(251)

27