I have a follow-up question from my answer to a previous question.
The electric field of a point charge moving at a constant velocity, as derived from the Liénard-Wiechert equation, reads $$ E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|^2},$$ where $ n =\overline{r-r_s}$ and $ \beta = v/c $.
There is nothing controversial about this, it is repeated in various textbooks and the derivation gets taught. For example, these notes or these ones.
In contrast, here is the electric field for the same point charge moving at constant velocity derived from the relativistic field tensors, $$ E = -\frac{e}{4\pi\epsilon_0} \frac{\gamma}{(1+u_r^2\gamma^2/c^2)^{3/2}}\mathbf r,$$ from this link.
Here is the E field for constant velocity derived by Griffiths, $$ E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-\beta^2\sin^2\theta)^\frac{3}{2}}\frac{n-\beta}{|r-r_s|},$$ as reviewed in Wikipedia.
Griffith's version is symmetric about $y$, Liénard-Wiechert is not.
If $n \cdot \beta = |\beta| \sin\theta$, then
$ \sqrt{1-\beta^2\sin^2\theta} = \sqrt{(1-|\beta|\sin\theta)(1+|\beta|\sin\theta)} $ in one case, and
$ (1-n \cdot \beta) = (1-|\beta|\sin\theta) = \sqrt{(1-|\beta|\sin\theta)^2} $ in the other case.
Presumably someone has made a sign error. One or the other of these formulas is wrong.
Have I misunderstood this? Are they both right?