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Radiation from Point Charges

To summarize from the last chapter, two useful Green's functions for the inhomogeneous wave equation:

\Box A^\alpha = \mu_0 J^\alpha
\end{displaymath} (18.1)

D_r(x - x') = \frac{\theta(x^0 - x'^0)}{4\pi R} \delta(x^0 -
x'^0 - R)
\end{displaymath} (18.2)

(the retarded Green's function) and
D_a(x - x') = \frac{\theta[-(x^0 - x'^0)]}{4 \pi R} \delta(x^0 - x'^0 + R)
\end{displaymath} (18.3)

(the advanced Green's function). The integral equations associated with these Green's functions were:
A^\alpha(x) = A^\alpha_{\rm in}(x) + \mu_0 \int d^4x' D_r(x -
x') J^\alpha(x')
\end{displaymath} (18.4)

A^\alpha(x) = A^\alpha_{\rm out}(x) + \mu_0 \int d^4x' D_a(x -
x') J^\alpha(x') .
\end{displaymath} (18.5)

For the moment, let us ignore Dirac's observations and the radiation field and focus instead on only the ``normal'' causally connected retarded potential produced by a single charged particle as it moves in the absence of external potentials. This potential is ``causal'' in that the effect (the potential field) follows the cause (the motion of the charge) in time, where the advanced potential has the effect preceding the cause, so to speak. Let me emphasize that this is not a particularly consistent assumption (again, we the theory is manifestly time symmetric so ``past'' and ``future'' are pretty much arbitrary namings of two opposed directions), but it yields some very nice results, as well as some problems. In that case:

A^\alpha(x) = \mu_0 \int d^4x' D_r(x - x') J^\alpha(x')
\end{displaymath} (18.6)

where the four-current of a point charge $e$ is found from
J(\mbox{\boldmath$x$}',t) = \left\{\begin{array}{c}
c \rho(...
...{\boldmath$x$}' - \mbox{\boldmath$r$}(t)]
\end{displaymath} (18.7)

in the lab/rest frame $K$ or (in covariant form):
J^\alpha(x') = ec \int d\tau U^\alpha(\tau) \delta^{(4)}\left( [x' -
r(\tau)] \right)
\end{displaymath} (18.8)

U = \gamma\left(\begin{array}{c}
c \\
\end{array}\right) = \frac{dr}{d\tau}
\end{displaymath} (18.9)

Note that the $\delta$ function in these expressions simply forces the particle to be found at the correct location at each (proper) time. The $r(\tau)$ function is the trajectory of the particle. Its $\tau$ derivative is the four-velocity. This yields (when the $\gamma$'s have all been accounted for) the rest frame expression.

To do the integral, we need the ``manifestly covariant'' form of the retarded Green's function. Note that:

$\displaystyle \delta[(x - x')^2]$ $\textstyle =$ $\displaystyle \delta[(x^0 - x'^0)^2 - \vert\mbox{\boldmath$x$}- vx'\vert^2]$  
  $\textstyle =$ $\displaystyle \delta[(x^0 - x'^0 - R)(x^0 - x'^0 + R)]$  
  $\textstyle =$ $\displaystyle \frac{1}{2R} [\delta(x_0 - x_0' - R) + \delta(x_0 -
x_0' + R)]$ (18.10)

(where $R = \vert\mbox{\boldmath$x$}- \mbox{\boldmath$x$}'\vert$). In terms of this, $D_r$ is given by
D_r(x - x') = \frac{1}{2\pi} \theta(x_0 - x_0') \delta[(x - x')^2] .
\end{displaymath} (18.11)

Again, the second delta-function makes no contribution because of the opposing $\theta$-function. Thus
$\displaystyle A^\alpha(x)$ $\textstyle =$ $\displaystyle \frac{\mu_0 c}{2 \pi} \int d^4x' \theta(x_0 -
x_0') \delta([x - x']^2)$  
    $\displaystyle \quad \quad \times e \int d\tau U^\alpha(\tau)
\delta^{(4)}\left( [x' - r(\tau)] \right)$ (18.12)
  $\textstyle =$ $\displaystyle \frac{e\mu_0 c}{2\pi} \int d \tau U^\alpha(\tau) \theta[x_0 -
r_x(\tau)] \delta\left\{ [x - r(\tau)]^2 \right\} .$ (18.13)

The vector potential at a point gets a contribution only where-when that point lies on the light cone in the future (picked out by the $\theta$ function) of the world line of the charge (picked out be the $\delta$ function). The contribution is proportional to $eU^\alpha(\tau)$ at that (retarded) time. It dies off like $1/R$, although that is obscured by the form of the $\delta$ function.

To evaluate this (and discover the embedded $R$), we use the rule (from way back at the beginning of the book, p. 30 in J1.2)

\delta[f(x)] = \sum_i \frac{\delta(x - x_i)}{\left\vert \left(
\frac{df}{dx}\right)_{x = x_i} \right\vert }
\end{displaymath} (18.14)

where the $x = x_i$ are the non-degenerate zeros of $f(x)$. $f(x)$ is assumed to be ``smooth''. Then if we let
f(\tau) = [x - r(\tau)]^2
\end{displaymath} (18.15)

(which is zero when $\tau = \tau_p$ in the past) then
\frac{d }{d\tau} [x - r(\tau)]^2 = -2[x - r(\tau)]_\beta U^\beta(\tau)
\end{displaymath} (18.16)

and therefore
\delta([x - r(\tau)]^2) = \frac{\delta(\tau - \tau_p)}{\lef...
...c{\delta(\tau - \tau_p)}{2[x -
r(\tau)]_\beta U^\beta(\tau)}
\end{displaymath} (18.17)

From this we see that

A^\alpha(x) = \frac{e\mu_0 c}{4\pi} \left.\frac{U^\alpha(\tau)}{U \cdot [x
- r(\tau)]} \right\vert _{\tau = \tau_p}
\end{displaymath} (18.18)

where $\tau_p$ is the proper time in the past of $x$ when the light cone of the charge contains the event $x$. This potential (and its other forms above) are called the Liénard-Wiechert potentials. In non-covariant form, they are obtained from the identity
$\displaystyle U \cdot (x - r)$ $\textstyle =$ $\displaystyle U_0[x_0 - r_0(\tau_p)] - {\bf U} \cdot [{\bf x} -
{\bf r}(\tau_p)]$  
  $\textstyle =$ $\displaystyle \gamma c R (1 - \mbox{$\boldmath\beta$} \cdot {\bf n})$ (18.19)

where n is a unit vector in the direction of ${\bf x} - {\bf r}(\tau)$ and where ${\boldmath\beta} = {\bf v}(\tau)/c$ as usual.

Recall that $A = (\phi/c,\mbox{\boldmath$A$})$. Thus:

A^0(x) = \frac{e\mu_0 c}{4\pi} \left.\frac{\gamma c}{\gamma ...
\cdot \hat{\mbox{\boldmath$n$}})}\right\vert _{\rm ret}
\end{displaymath} (18.20)

\phi(\mbox{\boldmath$x$},t) = cA^0 = \frac{e}{4\pi \epsilon_...
\cdot \hat{\mbox{\boldmath$n$}})}\right\vert _{\rm ret}
\end{displaymath} (18.21)

where all quantities (e.g. $\mbox{\boldmath$\beta$}, R$) must be evaluated at the retarded time where the event x is on the light cone of a point on the particle trajectory.


$\displaystyle \mbox{\boldmath$A$}(\mbox{\boldmath$x$},t)$ $\textstyle =$ $\displaystyle \frac{e\mu_0 c}{4\pi} \left.\frac{\gamma c\mbox{\boldmath$\beta$}...
\cdot \hat{\mbox{\boldmath$n$}})}\right\vert _{\rm ret}$  
  $\textstyle =$ $\displaystyle \frac{e}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}
\left. \frac{\mbox{...
... \mbox{\boldmath$\beta$}\cdot \hat{\mbox{\boldmath$n$}})}\right\vert _{\rm ret}$ (18.22)

where again things must be evaluated at retarded times on the particle trajectory. Note well that both of these manifestly have the correct non-relativistic form in the limit $\vert\mbox{\boldmath$\beta$}\vert « 1$.

We can get the fields from the 4-potential in any of these forms. However, the last few forms we have written are compact, beautiful, intuitive, and have virtually no handles with which to take vector derivatives. It is simpler to return to the integral form, where we can let $\partial^\alpha$ act on the $\delta$ and $\theta$ functions.

\partial^\alpha A^\beta = \frac{e\mu_0 c}{2\pi} \int d\tau
...0(\tau)] \partial^\alpha \delta\left( [x -
r(\tau)]^2 \right)
\end{displaymath} (18.23)

\partial^\alpha \delta[f] = \partial^\alpha f \cdot \left( ...
...a f \cdot \frac{d\tau}{df} \cdot
\frac{d}{d\tau} \delta[f] .
\end{displaymath} (18.24)

Again, we let $f = [x - r(\tau)]^2$. Then
\partial^\alpha \delta[f] = - \frac{(x - r)^\alpha}{U \cdot (x - r)}
\frac{d}{d\tau} \delta[f]
\end{displaymath} (18.25)

This is inserted into the expression above and integrated by parts:

$\displaystyle \partial^\alpha A^\beta$ $\textstyle =$ $\displaystyle - \frac{e\mu_0 c}{2\pi}\int d\tau
U^\beta(\tau) \theta[x_0 - r_0(\tau)] \frac{(x - r)^\alpha}{U \cdot (x -
r)} \frac{d}{d\tau} \delta[f]$  
  $\textstyle =$ $\displaystyle \frac{e\mu_0 c}{2\pi} \int d\tau \frac{d}{d\tau}\left\{
...pha}{U \cdot (x - r)} \right\}\theta[x_0
- r_0(\tau)] \delta([x - r(\tau)]^2) .$ (18.26)

There is no contribution from the $\theta$ function because the derivative of a theta function is a delta function with the same arguments
\frac{d}{d\tau} \theta(x_0 - r_0(\tau)) = \delta[x_0 - r_0(\tau)]
\end{displaymath} (18.27)

which constrains the other delta function to be $\delta(-R^2)$. This only gets a contribution at $R = 0$ (on the world line of the charge), but we already feel uncomfortable about the field there, which we suspect is infinite and meaningless, so we exclude this point from consideration. Anywhere else the result above is exact.

We can now do the integrals (which have the same form as the potential integrals above) and construct the field strength tensor:

F^{\alpha\beta} = \frac{e\mu_0 c}{4\pi} \left.\frac{e}{U \c...
...\beta U^\alpha}
{U \cdot (x-r)}\right\}\right\vert _{\rm ret}
\end{displaymath} (18.28)

This whole expression must be evaluated after the differentiation at the retarded proper time $\tau_p$.

This result is beautifully covariant, but not particularly transparent for all of that. Yet we will need to find explicit and useful forms for the fields for later use, even if they are not as pretty. Jackson gives a ``little'' list of ingredients (J14.12) to plug into this expression when taking the derivative to get the result, which is obviously quite a piece of algebra (which we will skip):

\mbox{\boldmath$E$}(\mbox{\boldmath$x$},t) = \frac{e\mu_0}{...
...ta$}\cdot \hat{\mbox{\boldmath$n$}})^3 R}
\right]_{\rm ret}
\end{displaymath} (18.29)

\mbox{\boldmath$B$}(\mbox{\boldmath$x$},t) = \frac{1}{c}(\hat{\mbox{\boldmath$n$}}\times \mbox{\boldmath$E$})
\end{displaymath} (18.30)

``Arrrgh, mateys! Shiver me timbers and avast!'', you cry out in dismay. ``This is easier? Nonsense!'' Actually, though, when you think about it (so think about it) the first term is clearly (in the low velocity, low acceleration limits) the usual static field:

\mbox{\boldmath$E$}\approx \frac{e}{4\pi\epsilon_0} \frac{\hat{\mbox{\boldmath$n$}}}{R^2}
\end{displaymath} (18.31)

Interestingly, it has a ``short'' range and is isotropic.

The second term is proportional to the acceleration of the charge; both E and B are transverse and the fields drop off like $R^{-1}$ and hence are ``long range'' but highly directional.

If you like, the first terms are the ``near'' and ``intermediate'' fields and the second is the complete ``far'' field; only the far field is produced by the acceleration of a charge. Only this field contributes to a net radiation of energy and momentum away from the charge.

With that (whew!) behind us we can proceed to discuss some important expressions. First of all, we need to obtain the power radiated by a moving charge.

next up previous contents
Next: Larmor's Formula Up: Relativistic Electrodynamics Previous: Covariant Green's Functions   Contents
Robert G. Brown 2007-12-28