Green’s Functions and Fourier Transforms
A general approach to solving inhomogeneous wave equations like
1 ∂2
2
∇ − 2 2 V (x, t) = −ρ(x, t)/ε0
c ∂t
(1)
is to use the technique of Green’s (or Green) functions. In general, if L(x) is a linear
differential operator and we have an equation of the form
L(x)f (x) = g(x)
(2)
then we find a function G(x, x′ ) (called the Green’s function for the problem)
satisfying
L(x)G(x, x′ ) = δ(x − x′ ) .
(3)
If f0 (x) is a solution to the homogeneous equation L(x)f0 (x) = 0, then it is then
easy to see that the general solution to (2) is
Z
f (x) = f0 (x) + G(x, x′ )g(x′ ) dx′
(4)
because acting on both sides of this with L(x) yields
Z
L(x)f (x) = L(x)f0 (x) + L(x)G(x, x′ )g(x′ ) dx′
=
Z
δ(x − x′ )g(x′ ) dx′ = g(x)
as desired. Any necessary boundary conditions are incorporated into f0 (x) and the
solution for G(x, x′ ).
Before solving (3), let us show that G(x, x′ ) is really a function of x − x′ (which
will allow us to write the Fourier transform of G(x, x′ ) as a function of x − x′ ).
This is a consequence of translational invariance, i.e., that for any constant a we
have G(x + a, x′ + a) = G(x, x′ ). If we take the derivative of both sides of this with
respect to a using the chain rule we have
∂G ∂(x + a)
∂G ∂(x′ + a)
dG(x + a, x′ + a)
=
+ ′
da
∂x
∂a
∂x
∂a
=
∂G(x, x′ ) ∂G(x, x′ )
= 0.
+
∂x
∂x′
Now change coordinates to Y = x + x′ and y = x − x′ and again use the chain rule.
1
This yields
0=
=
∂G ∂y
∂G ∂Y
∂G ∂y
∂G ∂Y
+
+
+
′
∂Y ∂x
∂Y ∂x
∂y ∂x
∂y ∂x′
∂G ∂G ∂G ∂G
+
+
−
∂Y
∂Y
∂y
∂y
=2
∂G
∂Y
and therefore ∂G/∂Y = 0 so that G(x, x′ ) = G(y) = G(x − x′ ) as claimed.
Next, to solve (3) for G(x, x′ ) we write both sides as Fourier transforms
Z
′
1
δ(x − x′ ) =
eik(x−x ) d4 k
(2π)4
and
G(x, x′ ) =
1
(2π)4
Z
where (with x0 = ct)
′
e
eik(x−x ) G(k)
d4 k
′
′
(5)
′
eik(x−x ) = eik0 c(t−t ) e−ik·(x−x ) .
For the particular case of the wave equation (1) we have
L(x) = ∇2 −
1 ∂2
c2 ∂t2
so that (3) becomes
Z
′
1 ∂2
1
e
∇2 − 2 2 G(x, x′ ) =
(−k2 + k02 )eik(x−x ) G(k)
d4 k
4
c ∂t
(2π)
Z
′
1
eik(x−x ) d4 k .
:=
4
(2π)
Equating the integrands we then see that
e
G(k)
=
Using this in (5) we have
1
.
+ k02
−k2
1
(2π)4
Z
d4 k
1
=
(2π)4
Z
eik0 c(t−t ) e−i|k||x−x | cos θ
d k dk0
k02 − k2
1
(2π)4
Z
d3 k dk0
G(x, x′ ) =
=
′
eik(x−x )
−k2 + k02
′
′
′
′
3
2
eik0 c(t−t ) e−i|k||x−x | cos θ
.
(k0 − |k|)(k0 + |k|)
(6)
Now recall the Cauchy-Goursat theorem
I
f (z) dz = 0
C
and the Cauchy integral formula
2πif (z0 ) =
I
C
f (z)
dz
z − z0
where the function f (z) must be analytic within and on the contour C. To evaluate
(6), we want to first perform the integration along the real axis in the k0 -plane
where the integrand has poles at k0 = ± |k|. In general there are several ways to
go around these poles, but the physical solution we are looking for is that the wave
propagates in the forward time direction, i.e., for t > t′ . This dictates that we
choose to displace the poles by adding a small term to the denominator in such a
way that we get a nonzero answer for t > t′ but 0 for t < t′ . This means that we
add a term −iε to k0 to obtain
Z
′
′
1
eik0 c(t−t ) e−i|k||x−x | cos θ
3
G(x, x′ ) =
.
(7)
d
k
dk
0
(2π)4
[k0 − (|k| + iε)][k0 + (|k| − iε)]
The part of the numerator that we care about for now may be written as
0
′
0
0
′
′
0
′
eik0 c(t−t ) = ei(kRe +ikIm )c(t−t ) = eikRe c(t−t ) e−kIm c(t−t ) .
Since the integration contour is to be closed at infinity in the k0 -plane, we see
0
that for t > t′ we must close it in the upper half-plane where kIm
> 0. Then the
integrand tends to 0 at infinity and only the portion along the real axis contributes
to the integral.
k0
C
×
×
For t < t′ we must close in the lower half-plane to get convergence, and there the
integral vanishes because no poles are enclosed.
The integral over k0 is
Z
′
eik0 c(t−t )
dk0
[k0 − (|k| + iε)] [k0 + (|k| − iε)]
C
3
where the poles enclosed in the upper half-plane by the contour C are at k0 = |k|+iε
and k0 = − |k| + iε. In our case we have two poles so we deform the contour as
shown below:
k0
e
C
C1
×
C2
×
e doesn’t enclose any poles, so the overall integral is zero.
The entire contour C
e is equal to the sum of our desired contour C that encloses the two
However, C
poles, plus the paths C1 and C2 . (The portions leading towards and away from
C1 and C2 are infinitesimally close and in opposite directions so they cancel each
other.) In other words, we have
I
I
I
I
.
+
+
=
0=
e
C
C2
C1
C
Note that here the paths C1 and C2 are in the clockwise direction, whereas by convention, the Cauchy formulas refer to paths taken in a counterclockwise direction,
and hence are of opposite signs. Therefore we have
I
I
I
+
=
C2
C1
C
where now the paths C1 and C2 are assumed to be counterclockwise. (This is
essentially the residue theorem in the case of simple poles.)
For the path C1 (i.e., the pole at k0 = − |k| + iε), the function f (z) in the
Cauchy integral formula corresponds to
0
′
eik c(t−t )
k0 − (|k| + iε)
and hence (after letting ε → 0)
I
′
= 2πi
C1
e−i|k|c(t−t )
.
−2 |k|
For C2 (the pole at k0 = |k| + iε), the function f (z) corresponds to
0
′
eik c(t−t )
k0 + (|k| − iε)
4
so we have
I
′
= 2πi
C2
ei|k|c(t−t )
.
2 |k|
Equation (7) now reads
2πi
G(x, x ) =
(2π)4
′
′ ′
e−i|k|c(t−t ) −i|k||x−x′ | cos θ
ei|k|c(t−t )
−
e
.
d k
2 |k|
2 |k|
Z
3
Choosing our z-axis to be along x − x′ (so the integration variable θ is the same as
the θ in cos θ) we obtain (now writing |k| = k for simplicity)
G(x, x′ ) =
=
=
i
2(2π)3
Z
i
2(2π)2
Z
∞
k dk
0
Z
1
d cos θ
−1
∞
Z
2π
′
′
′
dϕ [eikc(t−t ) − e−ikc(t−t ) ]e−ik|x−x | cos θ
0
′
′
′
k dk [eikc(t−t ) − e−ikc(t−t ) ]
0
−1
2
2(2π) |x − x′ |
Z
∞
′
′
e−ik|x−x | − eik|x−x |
−ik |x − x′ |
′
′
′
dk {eik(c(t−t )−|x−x |) − eik(c(t−t )+|x−x |)
0
′
′
′
′
− e−ik(c(t−t )+|x−x |) + e−ik(c(t−t )−|x−x |) } .
If we now let k → −k in the last two terms of the integrand this becomes
Z ∞
′
′
′
′ −1
G(x, x′ ) =
dk eik(c(t−t )−|x−x |) − eik(c(t−t )+|x−x |)
2
′
2(2π) |x − x | −∞
=
−1
δ(c(t − t′ ) − |x − x′ |) − δ(c(t − t′ ) + |x − x′ |) .
′
4π |x − x |
But t > t′ so the argument of the second delta function is always greater than zero,
and hence we are finally left with the retarded Green’s function
G(x, x′ ) =
−1
δ(c(t − t′ ) − |x − x′ |) .
4π |x − x′ |
Using this result, the solution to (1) becomes
Z
1
ρ(x′ , t′ )
V (x, t) =
δ(c(t − t′ ) − |x − x′ |) d4 x′
4πε0
|x − x′ |
Z
1
ρ(x′ , tret ) 3 ′
=
d x
4πε0
|x − x′ |
where
tret = t −
5
|x − x′ |
c
(8)
is called the retarded time. If we had wanted solutions for t < t′ , then we would
write k0 + iε and close the contour in the lower half-plane, and this would lead to
the solution
|x − x′ |
tadv = t +
c
which is called the advanced time. Note that we are also assuming that there is
no homogeneous solution to this problem because we want only that solution due
to the source charge density.
6