0
$\begingroup$

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?

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.