# $E$-field of moving charge and transforming between frames

Let's say we are in the lab frame and have a charge $$e$$ moving at velocity $$\vec{v} = v_0 \hat{x}$$. In the charge's frame of reference (primed frame) it produces an E field of $$\vec{E'} = \frac{e}{4\pi \epsilon_0 {r'}^2}\hat{r'}$$, and has a parallel component $${E'_{\parallel }} = \frac{e}{4\pi \epsilon_0 {r'}^2}$$. If I want to find the E field at a position $$x' \hat{x}$$ away from the particle, I think it is $${E'_{\parallel }} = \frac{e}{4\pi \epsilon_0 {x'}^2}$$.

Now I want to know what the E field at this position is in the unprimed frame. We know the E fields transform with $${E'_{\parallel }} = E_{\parallel }$$. In the lab frame, lengths in the particles frame are contracted, so I would expect $${x' = \frac{1}{\gamma}x}$$. I would then think

$${E_{\parallel }} = {E'_{\parallel }}=\frac{e}{4\pi \epsilon_0 {x'}^2} = \frac{e}{4\pi \epsilon_0 {(\frac{1}{\gamma}x})^2} = \frac{e \gamma^2}{4\pi \epsilon_0 {x}^2} .$$

However, using the equation derived in Purcell and Morin (p. 238, 3rd ed.), $${E_{\parallel}} = \frac{e}{4\pi \epsilon_0 {x}^2} (1-\beta^2) = \frac{1}{\gamma^2} \frac{e}{4\pi \epsilon_0 {x}^2}.$$

I'm on board with Purcell's equation being correct, so I think I'm applying length contraction wrong. How should we transform between these two frames?

A rod of length $$L$$ in its rest frame will be contracted and measure $$L/\gamma$$ in the lab frame. Therefore your lab frame $$x$$ is equal to $$x'/\gamma$$. Substituting $$x'=\gamma x$$ gives the Purcell and Morin's result.