Just when you thought it was safe to go back into the classroom, along comes
Jaws himself. Green's functions are your *friends!*

The inhomogeneous Maxwell equations are now compactly written as &part#partial;_&alpha#alpha;F^&alpha#alpha;&beta#beta; = &mu#mu;_0 J^&beta#beta;. From the definition of the field strength tensor, this is A^&beta#beta;- &part#partial;^&beta#beta;(&part#partial;_&alpha#alpha;A^&alpha#alpha;) = &mu#mu;_0 J^&beta#beta; If the potentials satisfy the Lorenz condition, and therefore A^&beta#beta;= &mu#mu;_0 J^&beta#beta; Do you get the feeling that there is something mystical about space-time notations? Do you remember what a pain in the butt this was to derive the hard way?

To solve this inhomogeneous differential equation, we construct simultaneously a Green's function D(x,x') = &delta#delta;^(4)(x - x') and the associated integral equation over the source term: A^&alpha#alpha;(x) = A^&alpha#alpha;_I + &mu#mu;_0 &int#int;d^4x' D(x - x') J^&alpha#alpha;(x') (where the inhomogeneous term depends on the Green's function and is the ``boundary'' term or the free potential from inhomogeneous sources outside the region of integration).

Next week we will concentrate on the integral equation solutions themselves. Now let us see how to construct the appropriate (covariant) Green's function. As usual, the principle part of the Green's function can involve only the absolute distance between the points. Thus if we seek solutions to D(y) = &delta#delta;^(4)(y) .

There are several ways we could go about solving this equation. They are all equivalent at some level or another. For example, we have already solved this equation for a single fourier component in Chapter 9. We could transform this result and obtain a four dimensional result. However, a more general procedure is to construct the solution from scratch.

The four dimensional fourier transform of the desired Green's function is defined by D(y) = 1(2&pi#pi;)^4 &int#int;d^4k D(k) e^-ik ·y where . The four dimensional delta function is &delta#delta;^4(y) = 1(2&pi#pi;)^4 &int#int;d^4k d^-ik ·y so (taking the of under the integral and equating factors) D(k) = - 1k ·k . We therefore know that the Green's function has the form D(y) = -1(2&pi#pi;)^4 &int#int;d^4 k e^-ik ·yk ·k .

The integrand in this expression is singular when . Recall that the presence of singularities means that we have to decide how to treat them to get a well-defined result. There are several ways to do this, and each has a physical interpretation. If we integrate over the ``time'' component first, we get D(y) = - 1(2 &pi#pi;)^4 &int#int;d^3k e^i · &int#int; dk_0 e^-ik_0 y_0k_0^2 - &kappa#kappa;^2 where . Now the singularities live in a single 1-D integral that we can easily evaluate via contour integration and the method of residues provided that we select a suitable contour.

Let's do the integral carefully (in case your contour integration is bit
rusty). Note that the poles of this integral are both real. This means
that the integral is ambiguous - it can be assigned any of several
possible values depending on how we choose to evaluation it. It is
beyond the scope of these notes to evaluate the consequences of making
and physically interpreting each of these choices. Instead we will
choose to include *both* poles completely within a standard contour
closed in the upper or lower half plane respectively, and then take
limits such that the poles return to the real axis after the integral
because this particular choice leads us very simply to the advanced and
retarded forms of the Green's function that we already obtained when
discussing the fourier transform of the incoming or outgoing spherical
Green's functions for the Helmholtz equation.

First we have to decide which way to close the contour. Examining the
integrand, we note that if
the integrand vanishes
on a lower-half contour like
in the figure above. We displace the
poles down slightly so that they lie inside the contour
:
. Finally, let
be a
complex variable such that the real axis is
.
&conint#oint;_&Gamma#Gamma;dz e^-iy_0zz^2 - &kappa#kappa;^2 =
&int#int;_-&infin#infty;^&infin#infty;dk_0 e^-ik_0y_0k_0^2 - &kappa#kappa;^2
+
&int#int;_C dz e^-iy_0zz^2 - &kappa#kappa;^2
As noted, the integral over
clearly vanishes for
. Thus:
&int#int;_-&infin#infty;^&infin#infty;dk_0 e^-ik_0y_0k_0^2 - &kappa#kappa;^2 & = &
&conint#oint;_&Gamma#Gamma;dz e^-iy_0zz^2 - &kappa#kappa;^2

& = & _&epsi#epsilon;&rarr#to;0 (-2&pi#pi;i)Res
e^-i z y_0(z - (&kappa#kappa;- i&epsi#epsilon;))(z + (&kappa#kappa;+
i&epsi#epsilon;)

& = & -2&pi#pi;i { e^-i&kappa#kappa;y_02&kappa#kappa; +
e^i&kappa#kappa;y_0-2&kappa#kappa; }

& = & -2&pi#pi;sin(&kappa#kappa;y_0)&kappa#kappa;

We can then write the Green's function as
D(z) & = & &thetas#theta;(y_0)(2&pi#pi;)^3 &int#int;d^3k e^i**k** ·**z**
(&kappa#kappa;z_0)&kappa#kappa;

& = & &thetas#theta;(y_0)(2&pi#pi;)^3 &int#int;_0^&infin#infty;&kappa#kappa;^2 d&kappa#kappa;
&int#int;_-1^1 d((&thetas#theta;)) &int#int;_0^2&pi#pi; d&phis#phi;e^i&kappa#kappa;R
(&thetas#theta;) (&kappa#kappa;y_0)&kappa#kappa;

& = & &thetas#theta;(y_0)(2&pi#pi;)^2 &int#int;_0^&infin#infty;&kappa#kappa;^2 d&kappa#kappa;
&int#int;_-i&kappa#kappa;R^i &kappa#kappa;R d(i&kappa#kappa;R (&thetas#theta;)) e^i&kappa#kappa;R (&thetas#theta;)i&kappa#kappa;R
(&kappa#kappa;y_0)&kappa#kappa;

& = & &thetas#theta;(y_0)2&pi#pi;^2 R &int#int;_0^&infin#infty;d&kappa#kappa;(&kappa#kappa;
R) (&kappa#kappa;y_0)

where
is the spatial separation of the points
and
.

Using a trig identity (or if you prefer expanding the 's in terms of exponentials and multiplying out, then changing variables and exploiting the fact that only even terms survive) to extend the integral to we can write this as: D(z) = &thetas#theta;(y^0)4&pi#pi;R {12&pi#pi;&int#int;_-&infin#infty;^&infin#infty; d&kappa#kappa;( e^i(y_0 - R)&kappa#kappa; - e^i(y_0 + R) &kappa#kappa; ) }.

These remaining integrals are just one dimensional Dirac delta
functions. Evaluating, we get:
D_r(x - x') = &thetas#theta;(x^0 - x'^0)4&pi#pi;R {&delta#delta;(x^0 -
x'^0 - R) + &delta#delta;(x^0 - x'^0 + R)}
where we have now labelled it with ``r'' for ``retarded''. The source
event
is always at an *earlier time* than the observation event
. This means that the domain of the support of the Heaviside function
just happens to be disjoint from the support of the second delta
function. We can therefore simplify this to:
D_r(x - x') = &thetas#theta;(x^0 - x'^0)4&pi#pi;R &delta#delta;(x^0 -
x'^0 - R)
which is just what we got before from Fourier transforming the
outgoing stationary wave Green's function, as it should be.

If we had chosen the other contour, identical arguments would have led us to
the *advanced* Green's function:
D_a(x - x') = &thetas#theta;[-(x_0 - x_0')]4 &pi#pi;R &delta#delta;(x_0 - x_0' + R)
The other *possible* contours (enclosing only one or the other of
the two singularities, using a contour that avoids the singularities on
the real axis instead of displacing the singularities) would yield still
other possible Green's functions. Just as an arbitrary normalized sum
of outgoing and incoming Green's functions resulted in an acceptable
Green's function before, an arbitrary sum of advanced and retarded
Green's functions are acceptable here. However, the inhomogeneous term
of the integral equation is a functional of the Green's function
selected!

For what it is worth, the Green's functions can be put in covariant form. One
almost never uses them in that form, and it isn't pretty, so I won't bother
writing it down. We can now easily write down formal solutions to the wave
equation for *arbitrary* currents (*not* just harmonic ones):
A^&alpha#alpha;(x) = A^&alpha#alpha;_in(x) + &mu#mu;_0 &int#int;d^4x' D_r(x -
x') J^&alpha#alpha;(x')
and
A^&alpha#alpha;(x) = A^&alpha#alpha;_out(x) + &mu#mu;_0 &int#int;d^4x' D_a(x -
x') J^&alpha#alpha;(x') .
In these equations, the inhomogeneous terms are the radiation field incident
upon (radiated from) the *four*-volume of space-time containing the
four-current that are *not* connected to the four-current in that
four-volume by the retarded Green's function.

It is a worthwhile exercise to meditate upon what might be a suitable
form for the inhomogeneous terms if one considerst the integration
four-volume to be *infinite* (with no inhomogeneous term at all)
and then split the infinite volume up into the interior and exterior of
a finite four-volume, as we did with incoming and outgoing waves before,
especially when there are many charges and they are permitted to
interact.

Dirac noted that choosing a ``retarded'' Green's function, just as
choosing an ``outgoing wave'' Green's function before, results in a
somewhat misleading picture given that the actual physics is completely
time-reversal symmetric (indeed, independent of using a *mixed*
version of the Green's functions in either case). He therefore
introduced the ``radiation field'' as the *difference* between the
``outgoing'' and the ``incoming'' inhomogenous terms given the
contraint that the actual vector potential is the same regardless of
the choice of Green's function used::
A^&alpha#alpha;_radiation = A^&alpha#alpha;_out - A^&alpha#alpha;_in =
4 &pi#pi;c &int#int;d^4 x' D(x - x') J^&alpha#alpha;(x')
where
D(z) = D_r(z) - D_a(z) .
In some fundamental sense, only the radiation fields are ``physical'' -
they are the *change* in the vector potential at an event produced
symmetrically by any given four-current due to its past *and its
future* motion. This is a critical aspect of the interpretation of
radiation reaction as being produced by transfer of momentum both *to* a charge (event) from other charges in its past and *from a
charge* to those *same* charges in its future.