Generation of EM waves
Susan Lea
Spring 2015
1
The Green’s function
In Lorentz gauge, we obtained the wave equation:
2
A =
4
J
c
(1)
The corresponding Green’s function for the problem satis…es the simpler
di¤erential equation
2
# #0
D x; x
=
(4)
#0
#
x
x
and the boundaries are at in…nity. Thus D can depend only on the vector
#
#0
z = x x : We …nd D by taking the 4-dimensional Fourier transform:
Z
##
1
#
e (k) e i k z d4 k
D
D z =
(2 )2
#
#
#
(Note that k z = !c c (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:
Z
1
(4) #
z =
d4 ke ik z
(2 )4
The transformed di¤erential equation reads:
#
#
e
k kD
#
k
1
=
1
(2 )2
and transforming back, we get:
#
D z
=
=
1
4
(2 )
1
(2 )4
Z
e
##
ik z
d4 k
#
#
3
i~k (~
x x
~0 )
k k
Z
d ke
Z
1
1
0
d! e i!(t t )
c ! 2 =c2 k 2
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
0
e i!(t t )
e
lim (! + ck) 2 2
= c2
2
!! ck
! =c
k
2
i(t t0 )( ck)
2ck
0
=
ei(t t )ck
c
2k
and
0
lim (!
!!ck
e i!(t t )
e
ck) 2 2
= c2
2
! =c
k
i(t t0 )(ck)
=c
2ck
e
i(t t0 )ck
2k
Then we have:
#
D z
=
1
(t
(2 )4
ic
=
(t
2 (2 )3
0
t)
t0 )
Z
Z
~
d3 keik (~x
~
d3 keik (~x
~
x0 )
( 2 i) c
~
x0 ) 1
k
e
e
0
i(t t0 )ck
c
2k
i(t t0 )ck
ei(t
ei(t t )ck
2k
t0 )ck
The …rst exponential represents an outgoing wave, while the second represents an incoming wave.
Now do the remaining integrals, we choose our polar axis for ~k along the
vector ~z = ~x ~x0 : Then:
Z 1 Z +1 Z 2
ic
1
0
0
#
0
D z
=
k 2 dkd d eikz
e i(t t )ck ei(t t )ck
(t t )
3
k
2 (2 )
0
1
0
Z 1 Z +1
ic
0
0
=
(t t0 )
kdkd eikz e i(t t )ck ei(t t )ck
2
2 (2 )
0
1
Z 1
+1
ic
eikz
0
0
0
(t t )
kdk
e i(t t )ck ei(t t )ck
=
2
ikz 1
2 (2 )
0
Z 1
c
0
0
=
(t t0 )
dk eikz e ikz e i(t t )ck ei(t t )ck
2
2 (2 ) z
0
Z 1
c
exp (ik (z c (t t0 ))) exp ik (z + c (t t0 ))
0
dk
(t
t
)
=
exp [ ik (z + c (t t0 ))] + exp ik (z c (t t0 ))
2 (2 )2 z
0
Z 1
c
0
=
(t t )
dk ( exp ik (z + c (t t0 )) + exp (ik (z c (t t0 ))))
2 (2 )2 z
1
c
0
=
(t t ) ( (z c (t t0 ))
(z + c (t t0 )))
4 z
The second term can never contribute since both z and t
Thus we …nally have:
D (z) =
c
4 z
(t
t0 ) (z
3
c (t
t0 ))
t0 are positive.
and thus the solution to equation (1) is:
Z
4
c
#
A x
=
(t t0 ) (z
c
4 z
Z
1
(t t0 ) (z c (t
=
z
t0 )) J
c (t
#0
t0 )) J
x
#0
x
d4 x0
d4 x0
(2)
Note: Jackson prefers to write this using the fully covariant expression
#
x
#0 2
x
=
n
(ct )
2
j~
x
0 2
x
~j
o
f(ct
R) (ct + R)g
1
=
f (ct
R) + (ct + R)g
2R
=
x
~0 j ; and used the result
t0 ; R = j~
x
where we wrote t = t
(f (x)) =
X
zeros
Thus
#
D z
=
1
2
(x xi )
jf 0 (x0 )j
#
t0 )
(t
x
#0 2
x
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:
Z
1
4
#
#0 2
#
#0
(t t0 )
x x
A x
=
J x d 4 x0
c
2
Z
2
#
#0 2
#0
(t t0 )
x x
J x d4 x0
(3)
=
c
2
Charge density and current for a point charge.
The charge and current densities are given by
(~x;t) = q (~x
4
~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)) =
We can write this covariantly as
Z
J (t0 ; ~x) = qc u
#
q
u
(~x
~r (t))
#
x
(4)
(5)
x0 ( ) d
where
x0 = (ct0 ; ~r (t0 ))
is the charge’s position vector at time t0 : To see why this works, note that
#
x
#
x 0 ( ) = (ct
and
(ct
ct0 ( )) =
Thus
=
q
~r ( ))
( (t)
(t0 ))
=
jcdt=d j
( (t)
c
( (t)
c
(~x
Z
u
u
(~x
J (t0 ; ~x) = qc
ct0 ( )) (~x
(t0 ))
(t0 ))
~r ( )) d
~r (t0 ))
which agrees with equation (4).
3
Radiation from a moving charge- the potential
We insert the current vector (5) into the expression (3) for the potential:
Z
Z
2
#
#
#0 2
#0
#
0
A x
=
(t t )
x x
qc u
x
x 0 ( ) d d4 x0
c
Z
Z
#
#0 2
#0
#
0
= 2q
(t t )
x x
u
x
x 0 ( ) d d4 x0
5
#0
Using the delta function, the integration over x is easy:
Z
2
#
#
#
(t t0 ( ))
x x0 ( )
u d
A x = 2q
Now again we need to evaluate the delta function of a function- here the
#
#
x0
function f ( ) = x
f0 ( ) =
2
( )
2
dx0
(x
d
x0 ) =
and the zeros are those values of
panding this out, we get
c (t
t0 )2
x0 ) : The derivative
x0 ) (x
= (x
2v (x
for which (x
(~x
x0 )
x0 ) (x
x0 ) = 0: Ex-
~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
(6)
R=c = tret
the retarded time.
Then the denominator is
2v (x
x0 ) = 2 ( c; ~v )
c (t
~ =2
t0 ) ; R
c2 (t
or, using equation (6),
2
c2 R=c
~ = 2 Rc 1
~v R
6
~ R
^
t0 )
~
~v 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 covariantly as:
A
= q
u
u 4x
(7)
(c; ~v )
(c; ~v ) (c (t t0 ) ; ~x ~x0 )
1; ~
1; ~
q
= q
=
~
R 1 ~ R
^
c (t t0 ) ~ R
= q
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
dA
dx +
d
@x
d 0
7
0
(8)
Di¤erentiating the light cone constraint (x
r ( 0 )) (x
get:
2 (x
r ( 0 )) (dx
dr ( 0 )) = 0
r ( 0 )) = 0, we
(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
(x
r ( 0 )) dx
d 0 =
(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
d
qu
dx +
d
u x
d 0 u x
)
(
(
a
u @ x
dx + q
= qu
2
u x
[u x ]
dA
= @
qu
0
u
a
x +u ( u )
[u x ]2
Note that
@
x =
@
(x
@x
r ( 0 )) =
@
g
@x
(x
r ( 0 )) = g
Now we insert the expression (10) for d 0 ; and factor a bit:
dA
So
qdx
[u x ]2
qdx
=
[u x ]2
=
@A
q
=
@x
[u x ]2
u u g
+
u u +
x
x
u u +
a +u
x
8
a
u
a
c2
a +u
x
u u
"
u x"
a
u"
x
x"
c2
a
u"
x
x"
=g
)
d
0
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
c2
x a +
a
u " x"
x
(u
x
Now, using the notation in the diagram above,
#
x
#
u
and
#
= R (1; n
~)
= c 1; ~
#
d ~d
d~
;
+
dt
dt
dt
du
du
a =
=
=c
d
dt
#
and
d
=
dt
3~
9
d~
dt
!
u
x )
(11)
so
#
a =c
~
2~ d
;
dt
2
~
~ d
dt
2
!
~
~+d
dt
!
(12)
Then we can compute the dot products
u
~ n
^
x =c R 1
and
a
x
= c
!
~
d
2
~+
R (1; ~n)
dt
!
!
~
~
~
d
d
d
2~
2 ~
~ n
^
n
^
dt
dt
dt
!
~
~
d
d
2~
1 ~ n
^
n
^
dt
dt
~
2~ d
;
dt
2
= c 2R
= c 2R
~
~ d
dt
!
Then from (11), and writing
K=h
c R 1
we have
F
= K
= K
(
8
<
:
q
x a
(u
x u
x )
c R(1 ~ n
^)
x a
c2
c 2R
x a + (u
x
10
i2
~ n
^
x a +
2 ~ d~
dt
u
d~
n
^
1 ~ n
^
dt
0
~
c + 2 R ddt n
^
x )@
R 1 ~ n
^
(13)
)
19
=
~
3~ d A
(14)
dt ;
The top row is
8
<
0i
F
= K
x 0 ai
:
8
<
= K R ai
:
8
<
2
= KcR
:
19
=
~
n
^
i 0
i
0
0
i @ c+
3~ d A
xa + u x
u x
dt ;
R 1 ~ n
^
0
19
!
=
2 d~
~
~
c + R dt n
^
d
3~ d A
+ c R i ni @
ni c 4~
dt
dt ;
R 1 ~ n
^
0
!
!
i
2 d~
~
~
R dt n
^
d
i
i
2 ~ d
i 4~ d
i @c +
+
n
+
n
dt
dt
dt
R 1 ~ n
^
0
2
~
R ddt
where we used (12) for ~a: Then, inserting (13), the electric …eld is given by
Ei =
=
F 0i
2
KcR
8
<
:
i
ni
q
=
cR 1
~ n
^
2
8
<
:
0
~
c+
@ 2~ d +
dt
2R 1
ni
2
i
~
R ddt
~
2
^
R ddt n
2R
1
~ n
^
3
2~
~ n
^
c+
We may divide this result into two parts:
q
i
Ecoulomb
=
2 R2 1
~ n
^
n
^
ni
9
i=
d
dt ;
1
d~ A
dt
9
i=
d
dt ;
(15)
i
This …eld has the usual 1=R2 dependence 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.
8
9
d~
<
i=
n
^
q
d
i
i
i
dt
(16)
Erad
=
2 : n
dt ;
1 ~ n
^
cR 1 ~ n
^
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:
#
!
"
~
~
d
d~
d
n
^
n
^ ~
= n
^ ~
n
^
1 ~ n
^
dt
dt
dt
11
19
=
~
4~ d A
dt ;
Thus
q
~ rad =
E
cR 1
^
3n
~ n
^
"
d~
dt
~
n
^
#
(17)
~ rad is perRemember: everything is evaluated at tretarded. Also notice that E
pendicular to n
^ ; and is proportional to the acceleration d~ =dt:
The magnetic …eld is given by:
~ =
B
F 23 ; F 13 ; F 12
For example:
Bx =
=
=
=
F 238
0
<
2 d~
R dt n
^
2 3
3 2
3
2
2
3 @ c+
3~
xa
xa + u x
u x
K
:
R 1 ~ n
^
8
19
0
<
2 d~
~ =
^
c + R dt n
3~ d A
K n2 a3 n3 a2 + u3 n2 u2 n3 @
:
dt ;
R 1 ~ n
^
8
3
2
3 ~ d~
>
n2 3 n3 2 +
n2 ddt
n3 ddt
<
dt
q
c R 1
~ n
^
2
>
:
3 2
+
n
2 3
n
The …rst and last terms cancel. Thus:
8
< c + 2 R d~ n
^
q
dt
~
B=
n
^
2 :
R 1 ~ n
^
c R 1 ~ n
^
c+
n
^
R(1 ~ n
^)
~+ n
^
Again we may divide the result into two parts:
q
~ B-S =
B
2 R2
1
~ n
^
3
~
2 R d~
dt
3 ~ d~
dt
19
d~ A=
dt ;
9
>
=
>
;
9
~
d =
dt ;
n
^
This term, which goes as 1=R2 ; is the usual Biot -Savart law result, with rel~ coulomb )
ativistic corrections. (Compare J eqn 11.152 and our expression for E
12
The radiation term is:
q
~ rad =
B
~ n
^
cR 1
=
2
q
n
^
d~
dt
: 1
2
~ n
^
cR 1
= n
^
8
<
~ rad
E
9
~
d =
dt ;
n
^
n
^ ~ +n
^
~ n
^
8
9
< d~ n
~
^ ~ d =
dt
+
: 1 ~ n
dt ;
^
where we used equation (16) for the electric …eld.
5
Radiated power
The Poynting vector for the radiation …eld is
c
4
c
=
4
c
=
4
~ =
S
=
~ rad
E
~ rad
B
~ rad
E
n
^
2
Erad
n
^
0
c B
@
4
4
q
~ n
^
cR 1
=
R2 c
~ rad
E
q2
~ n
^
1
6
^
3n
"
"
n
^
n
^
~
12
d~ C
^
A n
dt
d~
dt
~
n
^
#
#
2
n
^
Thus the power radiated per unit solid angle is:
dP
d
~ n
= R2 S
^
q2
=
4 c 1
~ n
^
6
13
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
=
n
^
4 c
n
^
d~
dt
#
2
q2 2 2
a sin
4 c3
=
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:
Z +1 Z 2
Z
q2 2
dP
2
d =
d
d
a 1
P =
3
d
4
c
1
0
q2 2
=
a
2c3
+1
3
=
3
1
2q 2 2
a
3c3
(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
R 0
From
problem
12.15
we
know
that
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 q2
a a
3 c3 0
2
=
2q @ 2
_
3 c
14
d~
_~ +
d
!2 1
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
=
But
d~
d
d~
d
3~
_ =
2 _~
So
() =
4
2
=
Giving
~
~ d
d
0
@_2 +
2
P =
!2
2q
3 c
2
2_2
0
@_2 +
4
2
~
~ d
d
2
~
~ d
d
!2 1
A
~
~ d
d
2
!2
!2 1
A
which is positive, as expected. Now let’s look at the two special cases:
1. ~ parallel to d~ =dt :
P =
2 q2
3 c
2_2
2
1+
2
1
=
2 q2
3 c
4_2
where _ = d =d = d =dt = a=c: Thus
2 q2
3 c3
P =
6 2
(20)
a
2. ~ perpendicular to d~ =dt :
2 q2
P =
3 c
2_2
15
2 q2
=
3 c3
4 2
a
(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
P =
2 q2
3 c3
3
F
m
and so
2
2. F~ perpendicular to ~ : In this case a = F=m and so
P =
2 q2
3 c3
2
F
m
2
So the power radiated per unit force is greater when F~ is perpendicular
~
to
(circular motion).
16