Greenzs function for the wave equation
Non-relativistic case
1 The wave equations
In the Lorentz Gauge, the wave equations for the potentials in Lorentz Gauge and
Gaussian units are:
2 ~
~ 1 @ A = ¡ 4¼ ~j
r 2A¡
(1)
2
c @t2
c
and
1 @2 ©
r2 ©¡ 2
= ¡4¼½
(2)
c @t2
The Gauge condition is:
~ ¢A
~ + 1 @© = 0
r
(3)
c @t
2 The Greenzs function
For both potentials we have a wave equation of the form
1 @2©
= ¡4¼ (source)
c2 @t2
~ The corresponding
where © can be either the scalar potential or a cartesian component of A:
Greenzs function problem is:
r2 ©¡
1 @2G
= ¡4¼± (~x ¡ ~x 0) ± (t ¡ t0 )
c2 @t2
where the source is now a unit event located at ~x = ~x0 and happening at t = t0 : To solve this
equation we ®rst Fourier transform in time:
Z 1
1
G (~x ; t; ~x 0 ; t0 ) = p
G (~x ; !; ~x 0 ; t0 ) e ¡i!td!
2¼ ¡ 1
and the equation becomes
µ
¶
Z 1
!2
1
r2 + 2 G (~x; !; ~x0 ; t0 ) = ¡4¼ p
± (~x ¡ ~x0 ) ± (t ¡ t0 ) ei!t dt
c
2¼ ¡1
p
0
= ¡2 2¼± (~x ¡ ~x 0) ei!t
r2 G (~x; t; ~x 0; t0 ) ¡
1
0 p
So let G (~x ; !; ~x0 ; t0) = g (~x ; ~x 0) ei!t = 2¼ and then g satis®es the equation
¡ 2
¢
r + k2 g = ¡4¼± (~x ¡ ~x0 )
Now in free space without boundaries, g must be a function only of R = j~x ¡ ~x 0 j and must
posess spherical symmetry about the source point. Thus in spherical coordinates, we can
write:
³ ´
1 d2
2
~
(Rg)
+
k
g
=
¡4¼±
R
(4)
R dR2
~
For R 6= 0; the right hand side is zero. Then the function Rg satis®es the exponential
equation, and the solution is:
AeikR + Be ¡ikR
¢
1 ¡ ikR
g =
Ae
+ Be¡ ikR
(5)
R
Near the origin, where the delta-function contributes, the second term on the LHS is
negligible compared with the ®rst, and the eequation becomes:
Rg
=
r2 g = ¡4¼± (~x ¡ ~x 0)
and we know that this has solution
g=
This is consistent with equation 5 provided that
1
R
A+B=1
(You should convince yourself that this solution is correct by differentiating and stuf®ng
back into equation 4.)
Thus we have the solution
¢
1 ¡ ikR
0
G (~x ; !; ~x 0; t0 ) = p
Ae
+ Be¡ikR ei!t
2¼R
Now we do the inverse transform:
Z 1
¢
0
1
1 ¡ ikR
G (~x; t; ~x0 ; t0) = p
p
Ae
+ Be¡ikR ei!t e¡ i!t d!
2¼ ¡1 2¼R
Z 1
1
=
A exp (i! (R=c + t0 ¡ t)) + B exp (i! (¡R=c + t0 ¡ t)) d!
2¼ ¡1
=
A± (t0 ¡ (t ¡ R=c)) + B± (t0 ¡ (t + R=c))
(6)
The second term is usually rejected because it predicts a response to an event occurring in
the future. However, Feynman and Wheeler have proposed a theory in which both terms are
kept. They show that this theory can be consistent with observed causality provided that the
universe is perfectly absorbing in the in®nite future. The time t ¡ R=c that appears in the
®rst term is called the retarded time tret.
The symmetry of this Greenzs function is:
G (~x; t; ~x 0; t0 ) = G (~x 0 ; ¡t0 ; ~x ; ¡t)
(See Morse and Feshbach Ch 7 pg 834-835). Causality requires:
G (~x; ¡1; ~x0 ; t0 ) = 0
2
and
G (~x; t; ~x0 ; t0 ) = 0 for t < t0
3 The potentials
Now that we have the Greenzs function, we can solve our original equations.
Z
½ (~x0 ; t0 )
© (~x ; t) =
± (t0 ¡ tret ) dt0 d3 x0
(7)
R
Z
½ (~x0 ; tret ) 3 0
=
d x
(8)
R
in Lorentz Gauge, and similarly:
Z ~ 0
j (~x ; tret ) 3 0
~ (~x ; t) = 1
A
d x
(9)
c
R
Notice that these equations have the same form as the static potentials (equations 1.17 and
5.32 in Jackson).
4 Radiation from a moving point charge (non-relativistic
case)
4.1
The Lienard-Wiechert potentials
Our source is a point charge moving with velocity ~v (t) : Then the charge and current
densities are
½ (~x ; t) = q± (~x ¡ ~r (t))
and
~j (~x;t) = q~v ± (~x ¡ ~r (t))
Then from equation 7, we have:
Z
q± (~x0 ¡ ~r (t0))
© (~x ; t) =
± (t0 ¡ tret ) dt0 d3 x0
R
We do the integral over the spatial coordinates ®rst.
Z
± (t0 + R (t0 ) =c ¡ t) 0
© (~x ; t) = q
dt
R (t0 )
where R (t0 ) = j~x ¡ ~r (t0)j : Now to do the t0 integral, we must reexpress the delta-function.
Recall:
X
1
± (f (x)) =
± (x ¡ xi )
0
jf (xi )j
3
where f (xi ) = 0: In this case:
f (t0) = t0 +
R (t0 )
¡t
c
and, since ~v = d~r=dt;
1 dR
1 (~x ¡ ~r (t0 )) d
=1¡
¢
(~x ¡ ~r (t0 ))
0
c dt
c j~x ¡ ~r (t0 )j dt0
~
~v ¢ (~x ¡ ~r (t0 ))
~v ¢ R
1¡
= 1¡
0
c j~x ¡ ~r (t )j
cR
f 0 (t0 ) =
1+
=
The function f is zero when t0 = tret = t ¡ R=c: Thus, evaluating the integral, we get:
¯
¯
¯
q
´ ¯¯
© (~x; t) = ³
(10)
~
R 1 ¡ ~v¢cRR ¯
tret
And similarly
¯
¯
¯
~
³
´¯
A (~x; t) =
~ ¯
v ¢R
~
R 1 ¡ cR ¯
q~v
(11)
tret
These are the Lienhard-Wiechert potentials. It is convenient to use the shorthand
Ã
!
~
~
~v ¢ R
~v ¢ R
rv = R 1 ¡
=R¡
cR
c
4.2
(12)
Calculating the ®elds
In Lorentz Gauge, the ®elds are found using
~
~ = ¡ r©
~ ¡ 1 @A
E
c @t
and
~ =r
~ £A
~
B
But our expressions for the potentials are in terms of ~x and tret ; not ~x and t; so we have to
be very careful in taking the partial derivatives. We can put the origin at the instantaneous
position of the charge to simplify things. Then R = r: Our potential may be written:
© (~x ;t) ´ ª (~x ;tret )
Then
¯
~ ¯¯
d© = r©
const t
¢ d~x +
¯
@©
@ª
~ ¯¯
dt ´ rª
¢ d~x +
dtre t
@t
const t ret
@tret
But dtret = dt ¡ dr=c; so
¯
¯
@©
@ª dr
@ª
~ ¯¯
~ ¯¯
r©
¢ d~x +
dt ´ rª
¢ d~x ¡
+
dt
const t
@t
const tret
@tret c
@tret
4
~ must be modi®ed:
Thus the r¡component of r©
¯
¯
@© ¯¯
@ª ¯¯
1 @ª
=
¡
@r ¯ const t
@r ¯ const tret c @tret
(13)
Now we can calculate the ®elds:
~ =r
~ q = ¡ q rr
~ v
r©
rv
r2v
and
µ
¶
µ
¶
µ
¶
^
^
@
~r ¢ ~v
µ @
~r ¢ ~v
Á
@
~r ¢ ~v
~
rrv =
r¡
^r +
r¡
+
r¡
@r
c
r @µ
c
r sin µ @Á
c
We can choose our axes with polar axis along the instantaneous direction of ~v : Then
~r ¢ ~v = rv cos µ; and
³
´
´
^³
~ v = 1 ¡ v cos µ r^ + µ r v sin µ
rr
c
r
c
In the non-relativistic limit, v=c ¿ 1; to zeroth order in v=c; this is
~ v = r^
rr
We are also going to need
@rv
~r ¢ ~a
=¡
@t
c
Then we have
¯
~ ¯¯
¡ r©
~
1 @©
1 @A
r^ ¡
c @t
c @t
const tret
µ
¶
q
q
~r ¢ ~a
q @ ~v
r¡ 2 ¡
^
^r ¡
rv2
r
c
c @t rv
µ v
¶
µ
¶
q
~r ¢ ~a
q ~a
q~v
~r ¢ ~a
=
r 1+
^
¡
+ 2 r¡
rv2
c
c rv
crv
c
and again taking the non-relativistic limit, this becomes:
µ
¶
~ = q r^ 1 + ~r ¢ ~a ¡ q ~a
E
r2
c
cr
The ®rst term is the usual Coulomb ®eld. The other two terms depend on ~a : these are the
radiation ®eld.
~ rad = q [(^
E
r ¢ ~a) r^ ¡ ~a]
rc
q
=
(^
r £ (^
r £ ~a))
(14)
rc
~
E
=
+
5
4.3
Radiated power
The Poynting vector is
~
S
=
=
=
=
=
c ~
~ rad
E rad £ B
4¼
³
´
c ~
~ rad
E rad £ n
^£E
4¼
c 2
E ^n
4¼ rad
Ã
"
#!2
c
q
d~¯
n
^ £ ^n £
n
^
4¼ cR
dt
¯
"
#¯ 2
q 2 ¯¯
d~¯ ¯¯
^£ n
^£
^
¯n
¯ n
4¼R2 c ¯
dt ¯
Thus the power radiated per unit solid angle is:
dP
= R2 S~ ¢ n
^
d¯
Ã
! ¯2
q2 ¯¯
d~¯ ¯¯
=
¯ ^n£ ^n £
¯
4¼c ¯
dt ¯
q2 2 2
a sin µ
(15)
4¼c3
Thus is the Larmor formula, where a = dv=dt is the acceleration and µ is the angle between
~a and ^n: The total power radiated in the non-relativistic case is:
Z
Z +1
Z 2¼
¢
dP
q2 2 ¡
P =
d- =
d¹
dÁ
a 1 ¡ ¹2
3
d4¼c
¡1
0
µ
¶¯
2
3 ¯ +1
q 2
¹ ¯
2q2
=
a
¹¡
= 3 a2
(16)
3
¯
2c
3
3c
¡1
6