Generation of EM waves
Susan Lea Spring 2015
1 The Green’s function
In Lorentz gauge, we obtained the wave equation:
2A = 4
c J (1)
The corresponding Green’s function for the problem satis…es the simpler di¤erential equation
2D #x;#x0 = (4) #x #x0
and the boundaries are at in…nity. Thus D can depend only on the vector
#z = #x #x0: We …nd D by taking the 4-dimensional Fourier transform:
D #z = 1 (2 )2
Z D (k) ee i
#k#zd4k
(Note that #k #z = !cc (t t0) ~k (~x ~x0) = ! (t t0) ~k (~x ~x0) : This explains the usual convention of transforming the time variable with the opposite sign from that for the space variable.)
Also recall the delta-function integral:
(4) #
z = 1
(2 )4 Z
d4ke ik z The transformed di¤erential equation reads:
#k #k eD #k = 1 (2 )2
and transforming back, we get:
D #z = 1
(2 )4
Z e i
#k #z
#k #k d4k
= 1
(2 )4 Z
d3kei~k (~x ~x0) Z 1
1
d!
c
e i!(t t0)
!2=c2 k2
We’ll do the frequency integral …rst. For t < t0;we close the contour with a big semi-circle in the upper half plane, while for t > t0 we close downward.
The integrand has two poles, at ! = ck; both on the real axis. Since the result must be zero for t < t0 (event precedes its source) we must evelaute the integral along the real axis by deforming the path to go above the two poles.
Contour for inverting the transform
For t > t0;we use the lower contour and we traverse it clockwise. Each of the poles is simple, and the residues are
!! cklim (! + ck) e i!(t t0)
!2=c2 k2 = c2e i(t t0)( ck)
2ck = cei(t t0)ck 2k
and
lim
!!ck(! ck) e i!(t t0)
!2=c2 k2 = c2e i(t t0)(ck)
2ck = ce i(t t0)ck 2k Then we have:
D #z = 1
(2 )4 (t t0) Z
d3kei~k (~x ~x0)( 2 i) ce i(t t0)ck
2k cei(t t0)ck 2k
= ic
2 (2 )3 (t t0) Z
d3kei~k (~x ~x0)1
k e i(t t0)ck ei(t t0)ck
The …rst exponential represents an outgoing wave, while the second repre- sents an incoming wave.
Now do the remaining integrals, we choose our polar axis for ~k along the vector ~z = ~x ~x0: Then:
D #z = ic
2 (2 )3 (t t0) Z 1
0
Z +1 1
Z 2 0
k2dkd d eikz 1
k e i(t t0)ck ei(t t0)ck
= ic
2 (2 )2 (t t0) Z 1
0
Z +1 1
kdkd eikz e i(t t0)ck ei(t t0)ck
= ic
2 (2 )2 (t t0) Z 1
0
kdk eikz ikz
+1
1
e i(t t0)ck ei(t t0)ck
= c
2 (2 )2z (t t0) Z 1
0
dk eikz e ikz e i(t t0)ck ei(t t0)ck
= c
2 (2 )2z (t t0) Z 1
0
dk exp (ik (z c (t t0))) exp ik (z + c (t t0)) exp [ ik (z + c (t t0))] + exp ik (z c (t t0))
= c
2 (2 )2z (t t0) Z 1
1
dk ( exp ik (z + c (t t0)) + exp (ik (z c (t t0))))
= c
4 z (t t0) ( (z c (t t0)) (z + c (t t0)))
The second term can never contribute since both z and t t0 are positive.
Thus we …nally have:
D (z) = c
4 z (t t0) (z c (t t0))
and thus the solution to equation (1) is:
A #x = 4 c
Z c
4 z (t t0) (z c (t t0)) J #x0 d4x0
= Z 1
z (t t0) (z c (t t0)) J #x0 d4x0 (2) Note: Jackson prefers to write this using the fully covariant expression
#x #x0 2 = n
(ct )2 j~x ~x0j2o
= f(ct R) (ct + R)g
= 1
2Rf (ct R) + (ct + R)g where we wrote t = t t0; R =j~x ~x0j ; and used the result
(f (x)) = X
zeros
(x xi) jf0(x0)j Thus
D #z = 1
2 (t t0) #x #x0 2
This is a covariant expression in spite of the explicit appearance of t and t0; since the step function serves to locate the result on the forward light cone from the source event, and this is an invariant statement. The expression for the potential becomes:
A #x = 4 c
Z 1
2 (t t0) #x #x0 2 J #x0 d4x0
= 2 c
Z
(t t0) #x #x0 2 J #x0 d4x0 (3)
2 Charge density and current for a point charge.
The charge and current densities are given by (~x;t) = q (~x ~r (t))
and
J (~~ x;t) = q~v (~x ~r (t)) and so the 4-vector is
J = (c ; ~v) = q (c; ~v) (~x ~r (t)) = q
u (~x ~r (t)) (4) We can write this covariantly as
J (t0; ~x) = qc Z
u #x #x0( ) d (5)
where
x0 = (ct0; ~r (t0))
is the charge’s position vector at time t0:To see why this works, note that
#x #x0( ) = (ct ct0( )) (~x ~r ( ))
and
(ct ct0( )) = ( (t) (t0))
jcdt=d j = ( (t) (t0)) c
Thus
J (t0; ~x) = qc Z
u ( (t) (t0))
c (~x ~r ( )) d
= q
u (~x ~r (t0)) which agrees with equation (4).
3 Radiation from a moving charge- the po- tential
We insert the current vector (5) into the expression (3) for the potential:
A #x = 2 c
Z
(t t0) #x #x0 2 qc Z
u #x0 #x0( ) d d4x0
= 2q Z
(t t0) #x #x0 2 Z
u #x0 #x0( ) d d4x0
Using the delta function, the integration over #x0 is easy:
A #x = 2q Z
(t t0( )) #x #x0( )
2
u d
Now again we need to evaluate the delta function of a function- here the function f ( ) = #x #x0( )
2
= (x x0) (x x0 ) : The derivative
f0( ) = 2dx0
d (x x0 ) = 2v (x x0 )
and the zeros are those values of for which (x x0) (x x0 ) = 0: Ex- panding this out, we get
c (t t0)2 (~x ~r (t0))2 = 0 t t0 = R=c where
R =j~x ~r (t0)j
Only the positive value contributes to the integral because of the factor (t t0( )) :(The event is on the positive light cone from the source.) Thus:
t = t0+ R=c or
t0 = t R=c = tret (6)
the retarded time.
Then the denominator is
2v (x x0 ) = 2 ( c; ~v) c (t t0) ; R = 2~ c2(t t0) ~v ~R or, using equation (6),
2 c2R=c ~v ~R = 2 Rc 1 ~ R^
Then
A #x = 2q Z
(t t0( )) #x #x0( ) 2 u ( ) d
= q (c; ~v) Rc 1 ~ R^
tre t
=
q 1; ~ R 1 ~ R^
tre t
These are the Lienard-Wiechert potentials. They may also be written co- variantly as:
A = q u
u 4x (7)
= q (c; ~v)
(c; ~v) (c (t t0) ; ~x ~x0)
= q
1; ~
c (t t0) ~ R~ = q R
1; ~ 1 ~ R^ using (6), and with
4x = x x0 all evaluated at the retarded time.
4 Fields due to a moving charge
To compute the …elds from the potentials (2 or 7), we need to take derivatives.
That is, we need to compare the potentials at two nearby events. The potential A changes as x changes both because of its explicit dependence on x but also because of the implicit dependence on 0(x ) that is because of the need to look back along the light cone to …nd the source event.
dA = @A
@x dx + dA
d 0d 0 (8)
Di¤erentiating the light cone constraint (x r ( 0)) (x r ( 0)) = 0, we get:
2 (x r ( 0)) (dx dr ( 0)) = 0 (9) and from the de…nition of the particle’s velocity
dr ( 0) = u d 0 So
(x r ( 0)) (dx u d 0) = 0
d 0 = (x r ( 0)) dx (x r ( 0)) u
= x dx
x u (10)
where I de…ned x (x r ( 0)) Thus from equation (8 and 7), we …nd
dA = @ qu
u x dx + d
d 0
qu
u x d 0
= qu
( u @ x
[u x ]2 )
dx + q
( a
u x u a x + u ( u ) [u x ]2
) d 0
Note that
@ x = @
@x (x r ( 0)) = @
@x g (x r ( 0)) = g = g Now we insert the expression (10) for d 0; and factor a bit:
dA = qdx
[u x ]2 u u g + x a u a x u u
u" x"
= qdx
[u x ]2 u u + x a + u c2 a x u" x"
So @A
@x = q
[u x ]2 u u + x a + u c2 a x u" x"
Now we compute the …eld tensor components F = @A
@x
@A
@x
The symmetric term u u will cancel in the subtraction, leaving
F = q
[u x ]2 x a x a + c2 a x
u" x" (u x u x ) (11)
Now, using the notation in the diagram above,
#x = R (1; ~n) u# = c 1; ~
and
#a = d#u
d = d#u
dt = c d dt; ~d
dt + d~
dt
!
and
d
dt = 3~ d~
dt
so
#a = c 2 2~ d~
dt; 2 ~ d~ dt
!
~ + d~
dt
!
(12) Then we can compute the dot products
u x = c R 1 ~ ^n and
a x = c 2 2~ d~
dt; 2 ~ d~ dt
!
~ +d~
dt
!
R (1; ~n)
= c 2R 2~ d~ dt
2 ~ d~ dt
!
~ ^n d~
dt n^
!
= c 2R 2~ d~
dt 1 ~ ^n d~
dt n^
!
Then from (11), and writing
K = q
h
c R 1 ~ ^n i2 (13)
we have
F = K
( x a x a +
(u x u x )
c R(1 ~ ^n) c2 c 2R 2~ d~
dt 1 ~ ^n d~dt n^ )
= K 8<
: x a x a + (u x u x )
0
@ c + 2Rd~dt n^ R 1 ~ ^n
3~ d~ dt
1 A
9= (14);
The top row is
F0i = K 8<
: x0ai xia0+ ui x0 u0 xi 0
@ c + 2Rd~dt n^ R 1 ~ ^n
3~ d~ dt
1 A
9=
;
= K 8<
:R ai nic 4~ d~
dt
!
+ c R i ni 0
@ c + 2Rd~dt n^ R 1 ~ ^n
3~ d~ dt
1 A
9=
;
= KcR 8<
:
2 2 ~ d~
dt
!
i+d i dt
!
ni 4~ d~
dt + i ni 0
@c + 2Rd~dt n^ R 1 ~ ^n
4~ d~ dt
1 A
9=
; where we used (12) for ~a: Then, inserting (13), the electric …eld is given by
Ei = F0i
= 2KcR 8<
: ni i 0
@ 2~ d~
dt + c + 2Rd~dt n^
2R 1 ~ ^n
2~ d~ dt
1 A d i
dt 9=
;
= q
cR 1 ~ ^n 2 8<
: ni i c + 2Rd~dt n^
2R 1 ~ ^n
d i dt
9=
; (15)
We may divide this result into two parts:
Ecoulombi = q
2R2 1 ~ ^n 3
ni i
This …eld has the usual 1=R2dependence and is in fact the same as Jackson’s equation 11.154. (See Jackson page 664 for the details.) The other term is the radiation …eld: it depends on the particle’s acceleration.
Eradi = q
cR 1 ~ ^n 2 8<
: ni i
d~
dt n^ 1 ~ ^n
d i dt
9=
; (16)
It decreases as 1=R rather than 1=R2; and so dominates the Coulomb …eld at large distances. We can simplify the expression by noting that:
^ n
"
^
n ~ d~
dt
#
= n^ ~ n^ d~
dt
! d~
dt 1 ~ ^n
Thus
E~rad = q
cR 1 ~ ^n 3
^ n
"
^
n ~ d~
dt
#
(17)
Remember: everything is evaluated at tretarded. Also notice that ~Erad is per- pendicular to ^n; and is proportional to the acceleration d~ =dt:
The magnetic …eld is given by:
B =~ F23; F13; F12 For example:
Bx = F23
= K
8<
: x2a3 x3a2+ u3 x2 u2 x3 0
@ c + 2Rd~dt n^ R 1 ~ ^n
3~ d~
dt 1 A
9=
;
= K
8<
:n2a3 n3a2+ u3n2 u2n3 0
@ c + 2Rd~dt n^ R 1 ~ ^n
3~ d~ dt
1 A
9=
;
= q
c R 1 ~ ^n 2 8>
<
>:
3 ~ d~
dt n2 3 n3 2 + n2 ddt3 n3 ddt2 + 3n2 2n3 c+ 2R
d~
dt ^n R(1 ~ ^n)
3~ d~
dt
9>
=
>;
The …rst and last terms cancel. Thus:
B =~ q
c R 1 ~ ^n 2 8<
:
c + 2Rd~dt n^
R 1 ~ ^n ^n ~ + ^n d~
dt 9=
;
Again we may divide the result into two parts:
B~B-S= q
2R2 1 ~ ^n 3
~ n^
This term, which goes as 1=R2;is the usual Biot -Savart law result, with rel- ativistic corrections. (Compare J eqn 11.152 and our expression for ~Ecoulomb)
The radiation term is:
B~rad = q
cR 1 ~ ^n 2 8<
:
d~
dt n^
1 ~ ^n n^ ~ + ^n d~
dt 9=
;
= ^n q
cR 1 ~ ^n 2 8<
:
d~
dt n^
1 ~ ^n ~ + d~
dt 9=
;
= ^n E~rad
where we used equation (16) for the electric …eld.
5 Radiated power
The Poynting vector for the radiation …eld is S =~ c
4
E~rad B~rad
= c
4
E~rad n^ E~rad
= c
4 Erad2 ^n
= c
4 0
B@ q
cR 1 ~ ^n 3
^ n
"
^
n ~ d~
dt
#1 CA
2
^ n
= q2
4 R2c 1 ~ ^n 6
^ n
"
^
n ~ d~
dt
# 2
^ n
Thus the power radiated per unit solid angle is:
dP
d = R2S ^~ n
= q2
4 c 1 ~ ^n 6
^ n
"
^
n ~ d~
dt
# 2
(18)
For non-relativistic motion, 1; we retrieve the Larmor formula (704 wavemks notes eqn 35 with a units change):
dP
d = q2 4 c n^
"
^ n d~
dt
# 2
= q2
4 c3a2sin2
where a = dv=dt is the usual 3-acceleration and is the angle between ~a and
^
n: The total power radiated in the non-relativistic case is:
P =
Z dP d d =
Z +1 1
d Z 2
0
d q2
4 c3a2 1 2
= q2 2c3a2
3
3
+1
1
= 2q2
3c3a2 (19)
When is not small, there are important changes. The denominator of equation (18) gets small when ~ n^ is close to 1, that is for direction of propagation close to the particle’s velocity ~ : Thus the radiation is strongly beamed along the direction of the particle’s motion.
5.1 Covariant generalization
From problem 12.15 we know that R 0
dV transforms as a vector. Thus R 00
dV transforms in the same way as x0 = ct: Thus the ratio, which is the power, is a Lorentz invariant. But in the instantaneous rest frame of the particle,
a a = ~a ~a = a2 So we can write eqn (19) as
P = 2
3 q2 c3a a
= 2
3 q2
c 0
@ _2 _ ~ + d~
d
!21 A
where dot means d=d : The term is parentheses simpli…es as follows:
_2 _ ~ + d~
d
!2
= _2 _2 2+ 2_2+ 2 _ ~ d~
d
!
= 1
2_2 2_2 2 _ ~ d~
d But
_ = 3~ d~
d So
() = 4 ~ d~
d
!2
2_2 2 4 ~ d~
d
!2
= 2
0
@ _2 + 2 ~ d~
d
!21 A
Giving
P = 2 3
q2 c
2
0
@ _2+ 2 ~ d~
d
!21 A
which is positive, as expected. Now let’s look at the two special cases:
1. ~ parallel to d~ =dt : P = 2
3 q2
c
2_2 1 +
2
1 2 = 2
3 q2
c
4_2
where _ = d =d = d =dt = a=c: Thus P = 2
3 q2 c3
6a2 (20)
2. ~ perpendicular to d~ =dt : P = 2
3 q2
c
2_2 = 2 3
q2 c3
4a2 (21)
Thus we get more radiated power per unit acceleration if ~a is parallel to
~ (linear motion). However, if we relate the power to the force acting on the particle, (see exam problem),we get a di¤erent interpretation.
1. ~F parallel to ~ : In this case a = F=m 3 and so
P = 2 3
q2 c3
F m
2
2. ~F perpendicular to ~ : In this case a = F=m and so P = 2
3 q2 c3
2 F
m
2
So the power radiated per unit force is greater when ~F is perpendicular to ~ (circular motion).
