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Next: Radiation from Point Charges Up: Relativistic Dynamics Previous: The Symmetric Stress Tensor   Contents

Covariant Green's Functions

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

\begin{displaymath}
\partial_\alpha F^{\alpha \beta} = \mu_0 J^\beta .
\end{displaymath} (17.87)

From the definition of the field strength tensor, this is
\begin{displaymath}
\Box A^\beta - \partial^\beta (\partial_\alpha A^\alpha) = \mu_0 J^\beta
\end{displaymath} (17.88)

If the potentials satisfy the Lorentz condition, $\partial_\alpha A^\alpha =
0$ and therefore
\begin{displaymath}
\Box A^\beta = \mu_0 J^\beta
\end{displaymath} (17.89)

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

\begin{displaymath}
\Box D(x,x') = \delta^{(4)}(x - x')
\end{displaymath} (17.90)

and the associated integral equation over the source term:
\begin{displaymath}
A^\alpha(x) = A^\alpha_I + \mu_0 \int d^4x' D(x - x')
J^\alpha(x')
\end{displaymath} (17.91)

(where the inhomogeneous term $A^\alpha_I$ 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 $y^\alpha = x^\alpha -
x^{\prime \alpha}$ we seek solutions to

\begin{displaymath}
\Box D(y) = \delta^{(4)}(y) .
\end{displaymath} (17.92)

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

\begin{displaymath}
D(y) = \frac{1}{(2\pi)^4} \int d^4k \tilde{D}(k) e^{-ik \cdot y}
\end{displaymath} (17.93)

where $k \cdot y = k_0y_0 - \mbox{\boldmath$k$}\cdot \mbox{\boldmath$y$}$. The four dimensional delta function is
\begin{displaymath}
\delta^4(y) = \frac{1}{(2\pi)^4} \int d^4k d^{-ik \cdot y}
\end{displaymath} (17.94)

so (taking the $\Box$ of $D(y)$ under the integral and equating factors)
\begin{displaymath}
\tilde{D}(k) = - \frac{1}{k \cdot k} .
\end{displaymath} (17.95)

We therefore know that the Green's function has the form
\begin{displaymath}
D(y) = \frac{-1}{(2\pi)^4} \int d^4 k \frac{e^{-ik \cdot y}}{k \cdot k} .
\end{displaymath} (17.96)

The integrand in this expression is singular when $k \cdot k = 0$. 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 $k_0$ first, we get

\begin{displaymath}
D(y) = - \frac{1}{(2 \pi)^4} \int d^3k e^{i \mbox{\boldmath...
...ldmath$y$}} \int
dk_0 \frac{e^{-ik_0 y_0}}{k_0^2 - \kappa^2}
\end{displaymath} (17.97)

where $\left\vert \bf k \right\vert = \kappa$. 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.

Figure 17.1: Contours for evaluating the Green's function in 4-dimensions.
\begin{figure}
\centerline{\epsfbox{figures/contour_GF4.eps}}
\end{figure}

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 $y^0 = x^0 - x'^0 > 0$ the integrand vanishes on a lower-half contour like $C$ in the figure above. We displace the poles down slightly so that they lie inside the contour $\Gamma$: $\pm
\kappa \to \pm \kappa - i\epsilon$. Finally, let $z = k_0 + ik_i$ be a complex variable such that the real axis is $k_0$.

\begin{displaymath}
\oint_\Gamma dz \frac{e^{-iy_0z}}{z^2 - \kappa^2} =
\int_{-...
...- \kappa^2}
+
\int_{C} dz \frac{e^{-iy_0z}}{z^2 - \kappa^2}
\end{displaymath} (17.98)

As noted, the integral over $C$ clearly vanishes for $y_0 > 0$. Thus:
$\displaystyle \int_{-\infty}^\infty dk_0 \frac{e^{-ik_0y_0}}{k_0^2 - \kappa^2}$ $\textstyle =$ $\displaystyle \oint_\Gamma dz \frac{e^{-iy_0z}}{z^2 - \kappa^2}$  
  $\textstyle =$ $\displaystyle \lim_{\epsilon \to 0} (-2\pi i){\rm Res}
\frac{e^{-i z y_0}}{(z - (\kappa - i\epsilon))(z + (\kappa +
i\epsilon)}$  
  $\textstyle =$ $\displaystyle -2\pi i \left\{ \frac{e^{-i\kappa y_0}}{2\kappa} +
\frac{e^{i\kappa y_0}}{-2\kappa} \right\}$  
  $\textstyle =$ $\displaystyle -2\pi \frac{sin(\kappa y_0)}{\kappa}$ (17.99)

We can then write the Green's function as

$\displaystyle D(z)$ $\textstyle =$ $\displaystyle \frac{\theta(y_0)}{(2\pi)^3} \int d^3k e^{i{\bf k} \cdot {\bf z}}
\frac{\sin(\kappa z_0)}{\kappa}$  
  $\textstyle =$ $\displaystyle \frac{\theta(y_0)}{(2\pi)^3} \int_0^\infty \kappa^2 d\kappa
\int_...
... \int_0^{2\pi} d\phi e^{i\kappa R
\cos(\theta)} \frac{\sin(\kappa y_0)}{\kappa}$  
  $\textstyle =$ $\displaystyle \frac{\theta(y_0)}{(2\pi)^2} \int_0^\infty \kappa^2 d\kappa
\int_...
...) \frac{e^{i\kappa R \cos(\theta)}}{i\kappa R}
\frac{\sin(\kappa y_0)}{\kappa}$  
  $\textstyle =$ $\displaystyle \frac{\theta(y_0)}{2\pi^2 R} \int_0^\infty d\kappa \sin(\kappa
R) \sin(\kappa y_0)$  

where $R = \left\vert {\bf x} - {\bf x}' \right\vert$ is the spatial separation of the points $x$ and $x'$.

Using a trig identity (or if you prefer expanding the $\sin$'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 $-\infty$ we can write this as:

\begin{displaymath}
D(z) = \frac{\theta(y^0)}{4\pi R} \left\{\frac{1}{2\pi}\int...
...^{i(y_0 - R)\kappa} - e^{i(y_0 + R) \kappa} \right)
\right\}.
\end{displaymath} (17.100)

These remaining integrals are just one dimensional Dirac delta functions. Evaluating, we get:

\begin{displaymath}
D_r(x - x') = \frac{\theta(x^0 - x'^0)}{4\pi R} \left\{\delta(x^0 -
x'^0 - R) + \delta(x^0 - x'^0 + R)\right\}
\end{displaymath} (17.101)

where we have now labelled it with ``r'' for ``retarded''. The source event $x'$ is always at an earlier time than the observation event $x$. 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:
\begin{displaymath}
D_r(x - x') = \frac{\theta(x^0 - x'^0)}{4\pi R} \delta(x^0 -
x'^0 - R)
\end{displaymath} (17.102)

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:

\begin{displaymath}
D_a(x - x') = \frac{\theta[-(x_0 - x_0')]}{4 \pi R} \delta(x_0 - x_0' + R)
\end{displaymath} (17.103)

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):

\begin{displaymath}
A^\alpha(x) = A^\alpha_{\rm in}(x) + \mu_0 \int d^4x' D_r(x -
x') J^\alpha(x')
\end{displaymath} (17.104)

and
\begin{displaymath}
A^\alpha(x) = A^\alpha_{\rm out}(x) + \mu_0 \int d^4x' D_a(x -
x') J^\alpha(x') .
\end{displaymath} (17.105)

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::

\begin{displaymath}
A^\alpha_{\rm radiation} = A^\alpha_{\rm out} - A^\alpha_{\rm in} =
\frac{4 \pi}{c} \int d^4 x' D(x - x') J^\alpha(x')
\end{displaymath} (17.106)

where
\begin{displaymath}
D(z) = D_r(z) - D_a(z) .
\end{displaymath} (17.107)

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.
next up previous contents
Next: Radiation from Point Charges Up: Relativistic Dynamics Previous: The Symmetric Stress Tensor   Contents
Robert G. Brown 2013-01-04