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?
