Green’s function for the wave equation
Non-relativistic case
January 2017
1 The wave equations
In the Lorentz Gauge, the wave equations for the potentials are (Notes 1 eqns 38 and
39):
1 ∂2A
− ∇2 A = µ0 j
(1)
c2 ∂t2
and
ρ
1 ∂ 2Φ
− ∇2 Φ =
(2)
c2 ∂t2
ε0
The Gauge condition is (Notes 1 eqn 37):
1 ∂Φ
∇·A+ 2
=0
(3)
c ∂t
In Coulomb Gauge we have the Gauge condition:
∇·A= 0
which leads to the wave equations (Notes 1, eqn on top of page 20, with ∇ · A = 0 )
ρ
∇2 Φ = −
ε0
(4)
(5)
and (Notes 1, eqn 36, with ∇ · A = 0)
1 ∂Φ
1 ∂ 2A
− ∇2 A = µ0 j − 2 ∇
2
2
c ∂t
c
∂t
(6)
2 Longitudinal and transverse currents
The Coulomb Gauge wave equation for A (6) is awkward because it contains the scalar
potential Φ. We can eliminate the potential Φ to obtain an equation for A alone. First we
separate the current into two pieces, called the longitudinal current J and the transverse
current Jt :
J = J + Jt
where
(7)
∇ · Jt = 0
and
(8)
∇×J =0
1
(Basically we are applying the Helmholtz theorem. We can always do this: See Lea
Appendix II). From charge conservation (Notes 1 eqn 2):
∂ρ
=0
∇·J +
∂t
and using equation (5) to eliminate ρ, this becomes:
∂ ∂ −ε0 ∇2 Φ = ∇ · ε0
∇Φ
∇·J =−
∂t
∂t
Thus, up to an irrelevant constant,
∂
J = ε0 ∇ Φ + ∇ × u
∂t
But from (8)
∇ × J = 0 + ∇ × ∇ × u = 0 = ∇ ∇ · u − ∇2 u
So we take u = 0, and then
∂
(9)
J = ε0 ∇ Φ
∂t
Using equation (9) in equation (6), we have:
1 ∂2A
∂Φ
∇2 A − 2 2 = −µ0 J − ε0 ∇
c ∂t
∂t
= −µ0 J − J = −µ0 Jt
(10)
In the Coulomb Gauge, the transverse current Jt is the source of A.
We can also use result (9) to express Jl in terms of J. Since equation (5) is the same as
in the static case, the solution is also the same (Notes 1 eqn 24). Thus
]
∂
1
ρ (x , t) 3
∇
d x
J (x, t) =
4π ∂t
|x − x |
] ∂
1
∂t ρ (x , t) 3
∇
d x
=
4π
|x − x |
]
1
∇ · J (x ) 3
d x
J (x, t) = − ∇
(11)
4π
|x − x |
where we used charge conservation in the last step. We can also express Jt in terms of J as
follows, using (11),
]
1
∇ ·J 3
∇
d x
Jt = J − J = J +
4π
|x − x |
2
Let’s work on the integral. We start with an "integration by parts":
#
$
]
]
]
∇ · J (x ) 3
J (x )
1
d x =
d3 x
∇
d3 x − J · ∇
|x − x |
|x − x |
|x − x |
]
]
1
J (x ) · n̂ 2
d x + J ·∇
d3 x (Notes 1 eqn 14)
=
|x − x |
S∞ |x − x |
]
J (x ) 3
d x
= 0+∇·
|x − x |
The surface integral is zero provided that J is localized. Then
#
$
]
1
J (x , t) 3
∇ ∇·
d x
Jt (x, t) = J (x, t) +
4π
|x − x |
$
&
%
#
]
]
1
J (x , t) 3
J (x , t) 3
2
d x +∇
d x
= J+
∇× ∇×
4π
|x − x |
|x − x |
$ ]
&
%
#
]
1
1
J (x , t) 3
2
3
d x + J∇
d x
= J+
∇× ∇×
4π
|x − x |
|x − x |
$ ]
&
%
#
]
1
J (x , t) 3
3
d x + J [−4πδ (x − x )] d x
= J+
∇× ∇×
4π
|x − x |
#
$
]
1
J (x , t) 3
∇× ∇×
d x
Jt (x, t) =
(12)
4π
|x − x |
We still have the rather unphysical result from equation (5) that Φ changes
instantaneously everywhere as ρ changes. In classical physics the potential is just a
mathematical construct that we use to find the fields, and it can be shown (e.g. J prob 6.20)
that E is causal even though Φ is not.
3 The Green’s function
With either gauge we have a wave equation of the form
1 ∂ 2Φ
= (source)
c2 ∂t2
where Φ may be either the scalar potential (in Lorentz Gauge) or a Cartesian component
of A. (In Coulomb Gauge the scalar potential is found using the methods we have already
developed for the static case.) The corresponding Green’s function problem is:
∇2 Φ −
1 ∂ 2G
= −4πδ (x − x ) δ (t − t )
(13)
c2 ∂t2
where the source is now a unit event located at position x = x and happening at time t = t .
As usual, the primed coordinates x , t , are considered fixed for the moment. To solve this
∇2 G (x, t; x , t ) −
3
equation we first Fourier transform in time (see, eg, Lea pg 503):
] ∞
1
G (x, t; x , t ) = √
G (x, ω; x , t ) e−iωt dω
2π −∞
and the transformed equation (13) becomes
] ∞
ω2
1
2
δ (x − x ) δ (t − t ) eiωt dt
∇ + 2 G (x, ω; x , t ) = −4π √
c
2π −∞
4π
= − √ δ (x − x ) eiωt
2π
√
iωt
So let G (x, ω; x , t ) = g (x, x ) e / 2π and then g satisfies the equation
2
∇ + k2 g = −4πδ (x − x )
(14)
where k = ω/c. In free space without boundaries, g must be a function only of R = |x − x |
and must posess spherical symmetry about the source point1 . Thus in spherical coordinates
with origin at the point P with coordinates x , we can write:
1 d2
2
(Rg)
+
k
g
=
−4πδ
R
(15)
R dR2
For R = 0, the right hand side is zero. Then the function Rg satisfies the exponential
equation, and the solution is:
= AeikR + Be−ikR
1 ikR
Ae
R=0
(16)
g =
+ Be−ikR
R
Near the origin, where the delta-function contributes, the second term on the LHS of (14)
is negligible compared with the first, and equation (14) becomes:
Rg
∇2 g
−4πδ (x − x )
We recognize that this equation has the solution
1
R
This is consistent with equation (16) as R → 0, provided that
g=
A+B =1
Thus we have the solution
1 ikR
G (x, ω; x , t ) = √
+ (1 − A) e−ikR eiωt
Ae
2πR
You should convince yourself that this solution is correct by differentiating and stuffing back
into equation (15).
1
Note that the operator ∇2 + k2 is spherically symmetric.
4
Now we do the inverse transform:
] ∞
1
1 ikR
√
+ Be−ikR eiωt e−iωt dω
Ae
G (x, t; x , t ) = √
2π −∞ 2πR
] ∞
1
=
{A exp [iω (R/c + t − t)] + B exp [iω (−R/c + t − t)]} dω
2πR −∞
B
A
δ [t − (t − R/c)] + δ [t − (t + R/c)]
(17)
=
R
R
The second term is usually rejected (take B ≡ 0 and thus A = 1) because it predicts
a response to an event occurring in the future. However, Feynman and Wheeler2 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 infinite
future. (This now appears unlikely.) The time t − R/c that appears in the first term is called
the retarded time tret. Thus we take
1
1
δ [t − (t − R/c)] =
δ (t − tret )
(18)
G (x, t; x , t ) =
R
R
Causality (an event cannot precede its cause) requires that the symmetry of this Green’s
function is:
G (x, t; x , t ) = G (x , −t ; x, −t)
(See Morse and Feshbach Ch 7 pg 834-835). and also
and
G (x, −∞; x , t ) = 0
G (x, t; x , t ) = 0 for t < t
4 The potentials
Now that we have the Green’s function (18), we can solve our original equations.
Modifying eqn 1.44 in Jackson ("Formal" Notes eqn 7) to include time dependence, and
with S → ∞, we get an integral over a volume in space-time rather than just space:
]
1
ρ (x , t ) G (x, t; x , t ) dt d3 x
Φ (x, t) =
4πε0
Thus, inserting (18), we have
]
1
ρ (x , t )
Φ (x, t) =
δ (t − tret ) dt d3 x
(19)
4πε0
R
]
1
ρ (x , tret ) 3
d x
(20)
=
4πε0
R (tret )
in Lorentz Gauge, and similarly:
]
µ
j (x , tret ) 3
A (x, t) = 0
d x
(21)
4π
R (tret )
Notice that these equations have the same form as the static potentials (equations 1.17 and
2
Reviews of Modern Physics, 1949, 21, 425
5
5.32 in Jackson, eqns 16 and 24 in Notes 1), but we must evaluate the source and the distance
R at the retarded time. This allows for the time for a signal to travel from the source to the
observer at speed c. Note that tret is a function of both x and x , as well as t.
Similar equations hold in Coulomb Gauge. The scalar potential (Notes 1 eqn. 24)
changes instantaneously as ρ changes, and the vector potential involves the transverse
current only in equation (21). In spite of this peculiarity, the fields E and B are causal.
(See problem 6.20 which demonstrates this in a relatively simple case. This problem may be
"relatively" simple, but it is not easy.)
When spatial boundaries are present the analysis is more complicated. We must use the
same kind of techniques that we used in the static case, expanding G in eigenfunctions.
However, the vector nature of A makes the problem much harder. (See Chapter 9 sections
6-12.)
5 Radiation from a moving point charge
5.1
The Lienard-Wiechert potentials
Here the source is a point charge q with position r (t) that is moving with velocity v (t) . The
charge and current densities are
and
ρ (x,t) = qδ [x − r (t)]
j (x,t) = qvδ [x − r (t)]
Because the source terms are delta-functions, it turns out to be easier to back up one step.
Then from equation (19), we have:
]
1
qδ [x − r (t )]
δ (t − tret ) dt d3 x
Φ (x, t) =
4πε0
R
We do the integral over the spatial coordinates first. Then
]
1
δ [t + R (t ) /c − t]
Φ (x,t) =
q
dt
4πε0
R (t )
where R (t ) = |x − r (t )| . To do the t integral, we must re-express the delta-function.
Recall (Lea eqn 6.10)
[
1
δ (x − xi )
δ [f (x)] =
(22)
|f
(x
i )|
i
where f (xi ) = 0. In this case:
R (t )
−t
f (t ) = t +
c
6
and, since v = dr/dt,
1 dR
1 d s
=1+
[x − r (t )] · [x − r (t )]
c dt
c dt
1 [x − r (t )] d
·
[x − r (t )]
= 1+
c |x − r (t )| dt
f (t ) = 1 +
v·R
v · R̂
v · [x − r (t )]
=1−
=1−
(23)
c |x − r (t )|
cR
c
The derivative f is always positive, since v < c. The function f is zero when t equals the
solution of the equation tret = t − R (tret ) /c. A space-time diagram shows this most easily:
tret is found from the intersection of the backward light cone from P with the charge’s world
line. There is only one root.
= 1−
Thus, evaluating the integral using (22) and (23), we get:
]
δ (t − tret )
1
1
q
dt =
q
Φ (x,t) =
v·R
4πε0
4πε
0
R (t ) 1 − cR
R 1−
and similarly
µ0
qv
A (x, t) =
4π R 1 − v·R cR
7
v·R cR
(24)
tret
(25)
tret
These are the Lienhard-Wiechert potentials. It is convenient to use the shorthand
$
#
v·R
v·R
=R−
rv = R 1 −
cR
c
so that
5.2
µ0 qv v
A (x, t) =
= 2Φ
4π rv tret
c
(26)
(27)
Calculating the fields
The fields are found using the usual relations (notes 1 eqns 35 and 15)
and
E = −∇Φ −
∂A
∂t
B =∇×A
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, using spherical coordinates, R = r. Our
potential may be written:
Φ (x, t) ≡ Ψ (x,tret )
A differential change in the potential due to a change in the coordinates is
∂Φ
∂Ψ
dt ≡ ∇Ψ
dΦ = ∇Φ
· dx +
· dx +
dtret = dΨ
∂t
∂tret
const t
const tret
But dtret = dt − dr/c, so
∂Ψ
∂Φ
∂Ψ dr
∇Φ
dt ≡ ∇Ψ
+
· dx +
· dx −
dt
∂t
∂tret c
∂tret
const t
const tret
This must be true for any dx and dt. Comparing the coefficient of dr on both sides, we see
that the r−component of ∇Φ must be modified:
∂Ψ 1 ∂Ψ
∂Φ =
−
(28)
∂r const t
∂r const tret c ∂tret
With this result, we can calculate the fields:
q
1 q
1
∇ =−
∇rv
(29)
∇Φ =
4πε0 rv
4πε0 rv2
where
∂
r·v
θ̂ ∂
r·v
φ̂ ∂
r·v
∇rv =
r−
r̂ +
r−
+
r−
∂r
c
r ∂θ
c
r sin θ ∂φ
c
We can choose our axes with polar axis along the instantaneous direction of v. Then
r · v = rv cos θ, and
θ̂ v
v
r sin θ
∇rv = 1 − cos θ r̂ +
c
r
c
In this coordinate system
v = vẑ = v r̂ cos θ − θ̂ sin θ
8
so
In the non-relativistic limit, v/c
also going to need
v
(30)
c
1, to zeroth order in v/c, this becomes ∇rv = r̂. We are
∇rv = r̂ −
r·a
∂rv
=−
∂t
c
Then, using (27), we have
∂A
1 ∂Φ
= − ∇Φ
r̂ −
+
c ∂t
∂t
const tret
1 ∂
1
∂Φ
= − ∇Φ
r̂ − 2 (vΦ)
+
c ∂t
c ∂t
const tret
1 ∂Φ
v
Φ ∂
= − ∇Φ
+
r̂ −
− 2 v
c ∂t
c
c ∂t
const tret
Inserting our results (29) , (30) and (31), we get
1
q
v
r·a
(r̂ − v/c) q
q a
E=
r̂
−
−
−
−
4πε0 rv2
c
c
rv2
c
rv c2
Combining the terms that involve the acceleration, we have
(r̂ − v/c) (r̂ · a) − a (1 − r̂ · v/c)
v
1 q
q
E =
r̂ −
+
2
2
4πε0 rv
c
4πε0 rv c
(1 − r̂ · v/c)
q
v
v
µ0 q
=
r̂ × r̂ −
r̂ −
+
×a
4πε0 rv2
c
4πrv (1 − r̂ · v/c)
c
(31)
E
(32)
(33)
Taking the non-relativistic limit3 v/c
1, (33) becomes:
q [r̂ × (r̂ × a)]
1
q
r̂
−
E=
4πε0 r2
c2
r
The first term is the usual Coulomb field which goes as 1/r2 . The second term depends on
the acceleration a: this is the radiation field.
µ q
(34)
Erad = 0 [r̂ × (r̂ × a)]
4π r
This term decreases as 1/r and dominates at large r. Note also that Erad is perpendicular to
r̂.
Eqn (33) is not quite correct if v/c is not small, as we have not done a correct
relativistic treatment of time. We are missing some factors of γ and 1 − β · r̂.
3
9
Next let’s calculate the magnetic field:
B = ∇ × A
const t
1 ∂
µ qv
=
∇
− r̂
× 0
c ∂t
4π rv
const tret
1
a
µ0
v ∂rv
1
q ∇
B =
− 2
× v − r̂ ×
4π
rv
c
rv
r ∂t
v
1
a
v r·a
v
µ0
r̂
+ 2
= − q 2 r̂ −
×v+ ×
4π rv
c
c
rv
rv c
µ0
µ0 q
q/c
r·a
v × r̂ −
=
(r̂ × a) (1 − v · r̂/c) + (r̂ × v)
4πrv2
4π rv (1 − v · r̂/c)
c
We can simpify this result using (32) for Erad and the fact that r̂ × r̂ ≡ 0.
µ0 q
µ
q
r̂ × [(r̂ − v/c) (r̂ · a) − a (1 − r̂ · v/c)]
B =
(35)
v × r̂ + 0
2
4πrv
4π rv (1 − v · r̂/c)
c
= BB-S + r̂ × Erad /c
In the limit v/c
1, the first term, which goes as 1/r2 , is the usual Biot-Savart law result .
The second term, which goes as 1/r , is the radiation field. Notice that
Brad = r̂ × Erad /c
as expected for an EM wave. In the non-relativistic limit,
µ q
Brad = 0 a × r̂
4π rc
The Poynting flux for the radiation field is:
S
=
1
1
Erad × Brad =
Erad ×
µ0
µ0
=
2
Erad
r̂
µ0 c
(36)
#
r̂ × Erad
c
$
(37)
where from equation (34):
1 q
µ0 q
a |sin θ| =
a |sin θ|
4π r
4πε0 rc2
and θ is the angle between a and r̂. Thus, in the non-relativistic limit
2
1
q2
a2
1 q
S=
a
sin
θ
=
sin2 θ
2
2
2
3
µ0 c 4πε0 rc
r
(4π) ε0 c
Erad =
Notice that S ∝ 1/r2 , the usual inverse square law for light. S is the power radiated per unit
area of wavefront, S = dP/dA. Writing dA in terms of solid angle, dA = r2 dΩ, we find
the power radiated per unit solid angle is independent of distance:
q 2 a2
dP
= r2 S =
sin2 θ
dΩ
(4π)2 ε0 c3
10
(38)
Finally the total power radiated is
] 2π ] +1
]
q 2 a2
dP
dΩ =
1 − µ2 dφdµ
P =
2
dΩ
(4π) ε0 c3 0
−1
where as usual µ = cos θ. Thus
+1
q 2 a2
µ3 P =
µ−
(4πε0 ) 2c3
3 −1
2 q 2 a2
(39)
3 4πε0 c3
This result is called the Larmor formula.
In the relativistic case, we use (33) in (37) to get:
2
µ0 q
v
1 r̂ r̂ × r̂ −
S=
× a µ0 c 4πrv (1 − r̂ · v/c)
c
and
2
1
dP
µ0 q 2
r̂ × r̂ − v × a =
2
4
dΩ
c (4π) (1 − r̂ · v/c)
c
4
The denominator (1 − r̂ · v/c) indicates that the radiation is beamed into a small cone
around the velocity vector when v/c → 1. Note here that this derivation is not strictly correct
when v/c is not negligible, and it leads to the wrong power of (1 − r̂ · v/c) in the denominator. It does indicate qualitatively how the radiation is beamed. For the correct relativistic
derivation see Jackson Ch 14 or http://www.physics.sfsu.edu/~lea/courses/grad/radgen.PDF.
P
=
Example
A particle of charge q and mass m is moving in the presence of a uniform magnetic field
c. Find the power radiated.
B = B0 ẑ. Its speed v
First we compute the acceleration:
F = ma = qv × B
so
q
q
v×B =− B×v =ω×v
(40)
m
m
Only the component of v perpendicular to B contributes, and the motion is a circle with a
pointing toward the center. If the component of v parallel to B is not zero, the motion is a
c, so it will be ignored
helix. v remains unchanged and does not affect the radiation if v
from now on. (There are important beaming effects if v/c is NOT
1.)
a=
11
Choose coordinates as shown in the diagram above, and choose t = 0 when the particle is
on the x−axis. Then the angle χ between r̂ and a is found from
r̂ · â = cos χ = sin θ cos (ωt − φ)
where
qB ω=
m
is the cyclotron frequency (eqn 40). Thus from (38),
q
2 dP
q2
=
v⊥ B0
1 − sin2 θ cos2 (ωt − φ)
2
3
dΩ
m
(4π) ε0 c
Taking the time average, we get
<
q 4 (v⊥ B0 )2
dP
> =
dΩ
m2 (4π)2 ε0 c3
1−
sin2 θ
2
=
q 4 B02 K⊥ 1 + cos2 θ
2
3
3
16π ε0 m c
2
. Radiation is maximum along the direction
where K⊥ is the particle’s kinetic energy 12 mv⊥
of B, and minimum in the plane perpendicular to B. The total power radiated is (39):
2 2 q 4 (v B )2
q 4 B02 K⊥
2 q2 q
⊥ 0
v⊥ B0 =
=
P =
3
3
2
3 4πε0 c m
3 4πε0 c m
3πε0 c3 m3
The radiated power is proportional to the square of the magnetic field strength.
We can check the physical dimensions:
v 2
q 4 cuB
q 4 µ0 B02 v⊥ 2
⊥
=
P =
6πε0 µ0 m2 c c
3πε20 (mc2 )2 c
12
Since the energy of a pair of point charges is
U=
we have
2
P = (energy × length)
q2
4πε0 d
energy
energy length
1
=
2
volume time (energy)
time
as required.
Things get much more interesting as v → c. Synchrotron radiation is discussed in J Ch
14 and also in http://www.physics.sfsu.edu/~lea/courses/grad/radiation.PDF.
13