electromagnetism - From Liénard-Wiechert to Feynman potential expression

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

$ \phi'(\mathbf r',t') = \frac{q}{4\pi\epsilon_0} \frac{1}{r'} $,

where $ |\mathbf r'| = r' $. In a frame with constant velocity $ \mathbf v $ along the x-axis, he obtains the following expression: $$ \phi(\mathbf r, t) = \frac{\gamma q}{4\pi\epsilon_0} \dfrac{1}{\sqrt{(\gamma(x-vt))^2+y^2+z^2}} \tag 1 $$

by transforming $ \phi = \gamma\left(\phi'+\dfrac{A'_xv}{c^2}\right) $, where $ \gamma = \dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}} $ and the vector potential $ \mathbf A' $ vanishes within the rest frame. Another Lorentz transformation of the time and space coordinates $ (\mathbf r', t') \rightarrow (\mathbf 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 $$ \phi(\mathbf r, t)=\dfrac{q}{4\pi\epsilon_0}\dfrac{1}{|\mathbf r - \mathbf x(t_{ret})| - \frac{1}{c}\mathbf v(t_{ret})\cdot(\mathbf r - \mathbf x(t_{ret}))} \tag 2, $$

where $ \mathbf x(t_{ret}) $ describes the position of the charge and $ \mathbf v(t_{ret}) = \frac{d}{dt}\mathbf x(t)\bigg|_{t=t_{ret}} $ its velocity at the retarded time $ t_{ret}(\mathbf r,t) = t-\frac{|\mathbf r - \mathbf x(t_{ret})|}{c} $, respectively.

In the case of uniform motion, we have $ \mathbf x(t) = (vt,0,0)^\intercal $.

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.