# Mathematical equivalence between Liénard-Wiechert potential and 4-potential in Rindler coordinates

I'm studying the problem of the radiation of an uniformly accelerated point charge:

$$x^{\mu}(\lambda)\to(g^{-1}\sinh g\lambda,0,0,g^{-1}\cosh g\lambda)$$

I found that when a point charge is moving along the $z$ axis with a constant acceleration $g$, the components of the 4-potential can be found using Rindler coordinates:

$$z=Z\cosh g\tau, \qquad t=Z\sinh g\tau \qquad \mathrm{I}$$ $$z=Z\sinh g\tau, \qquad t=Z\cosh g\tau \qquad \mathrm{II}$$ $$z=-Z\cosh g\tau, \qquad t=-Z\sinh g\tau \qquad \mathrm{III}$$ $$z=-Z\sinh g\tau, \qquad t=-Z\cosh g\tau \qquad \mathrm{IV}$$

with the metric

$$ds^{2}=\epsilon(-g^{2}Z{}^{2}\, d\tau^{2}+dZ^{2})+dx^{2}+dy^{2}$$

where $\epsilon=+1$ in regions I and III, and $\epsilon=-1$ in regions II and IV. And the numbers indicate the space-time region:

Now, I understand that the potentials obtained with the Rindler coordinates must be equivalent to the Liénard-Wiechert potential, because the change of coordinates (Minkowski->Rindler) is equivalent to change from an inertial frame of reference to an accelerated one. The problem is, the article

Radiation from a uniformly accelerated charge. D G Boulware. Ann. Phys. 124 no.1 (1980), pp. 169–188. Available on CiteSeerX

finds that the components of the 4-potential (Rindler coordinates) are:

$$A_{\tau}=-\frac{eg}{4\pi}\frac{\epsilon Z^{2}+\rho^{2}+g^{-2}}{\left[\left(\epsilon Z^{2}+\rho^{2}+g^{-2}\right)^{2}-4 \epsilon Z^{2}g^{-2}\right]^{1/2}}$$

where $\rho^{2}:=x^{2}+y^{2}$. $A_x=A_y=0$, and

$$A_{Z}=-\frac{e}{4\pi Z}$$

In regions I and II.

My question is: How can I conclude that this components are actually equivalent to the components of the Liénard-Wiechert potential?

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Any GR textbook would give you a full set of rules to perform coordinate tramsforms on motions, fields, potentials, and Maxwell's equations. Then it is just a long simple calculation. And remember, finally you can get something that differs from the sought potential by a pure gauge summand. –  firtree May 13 '13 at 5:28
@firtree Oh, I think I get it. In this case I must transform the Liénard-Wiechert potential using Rindler change of coordinates and I should obtain the $A_\tau$ and $A_Z$, right? –  Anuar May 13 '13 at 15:26
Yes but first you should get the Liénard-Wiechert potential for this particular motion of charge. –  firtree May 13 '13 at 15:39
(Please don't forget @firtree or I'm not notified.) Yes it is possible to consider the charge in the accelerated frame as one uniformly moving (or just stationary) but that does not simplify the work. Now gravitation is present, and the Coulomb's law doesn't hold true! Field lines become "heavy" and "drooping", and finally you have to solve Maxwell's equations for the curved space-time (can be found in your GR textbook) - which is more intricate than the use of the known Liénard-Wiechert potential which is already a solution. Be warned :-) –  firtree May 13 '13 at 16:28
It is not the familiar Coulomb field, actually. It is some different field, though static as well. It is close to the Coulomb field in the close neighborhood of the charge, but the gravitation changes it in the farther regions. You can plot it by some software to get convinced. And, $\mathbf{E}$ is a gauge-independent vector, so no gauge choice could change it. –  firtree Jun 4 '13 at 22:48