When studying the potential of an uniformly moving charge in vacuum, Feynman proposes to apply a Lorentz transformation on the Coulomb potential, which reads in the rest frame
ϕ′(r′,t′)=q4πϵ01r′,
where |r′|=r′. In a frame with constant velocity v along the x-axis, he obtains the following expression: ϕ(r,t)=γq4πϵ01√(γ(x−vt))2+y2+z2
by transforming ϕ=γ(ϕ′+A′xvc2), where γ=1√1−v2c2 and the vector potential A′ vanishes within the rest frame. Another Lorentz transformation of the time and space coordinates (r′,t′)→(r,t) yields (1). I suspect that (1) describes the potential at a given point for the instantaneous time t. What I am wondering is how this formula is connected to the expression of Liénard and Wiechert, namely ϕ(r,t)=q4πϵ01|r−x(tret)|−1cv(tret)⋅(r−x(tret)),
where x(tret) describes the position of the charge and v(tret)=ddtx(t)|t=tret its velocity at the retarded time tret(r,t)=t−|r−x(tret)|c, respectively.
In the case of uniform motion, we have x(t)=(vt,0,0)⊺.
How do I get now from (2) to (1)?
My idea is to actually calculate an explicit expression for the retarded time and plug it into (2), which should yield (1) if I understand it correctly. By asserting that c^2(t-t_{ret})^2 = (x-vt_{ret})^2+y^2+z^2 , t_{ret} can be found in terms of solving the quadratic equation, leading to the solutions
t_{ret}^\pm = \gamma\left(\gamma(t-\frac{vx}{c^2})\pm\sqrt{\gamma^2(t-\frac{vx}{c^2})^2-t^2+\frac{r^2}{c^2}}\right) = \gamma\left(\gamma t'\pm\sqrt{\gamma^2t'^2-\tau^2}\right) where t' is the Lorentz transformation of t and \tau = \frac{t}{\gamma} looks like some proper time. Plugging this into (2) looks nothing like (1), what am I missing?
Answer
If you look at Feynman Volume II Section 21-6, he walks through this calculation. Your idea and initial assertion look good; the trick is to manage the algebra to get to the final form you want.
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