# Electromagnetism and Relativity

1. Aug 22, 2011

### PineApple2

Hello. Here is the problem:
A charge q is moving at speed v0 to the right in the lab frame. At t=0 the particle passes at the origin (0,0,0). calculate the field at (-d,0,0) at the instant it passes the origin, both in the lab frame and in the charge frame. are they equal?

My solution:

In the lab frame, the field is
$$\frac{q}{\gamma^2 d^2}$$
in the negative x direction. By the transformation of parallel fields, $$E_{||}'=E_{||}$$ so the result in the charge frame should be the same.
However working in the charge frame,
$$E=-\frac{q}{d'^2}, \qquad d'=d/\gamma$$
because of length contraction. therefore the field in the charge frame is
$$E=-\frac{q}{(d/\gamma)^2}=\frac{\gamma^2 q}{d^2}$$
these are not equal, the gamma here is in the "wrong place".
what is going on here?
thanks.

Last edited: Aug 22, 2011
2. Aug 23, 2011

### vela

Staff Emeritus
The spacetime point $x^\mu = (ct, x, y, z) = (0, -d, 0, 0)$ doesn't correspond to the point $x'^\mu = (ct', x', y', z') = (0, -d/γ, 0, 0)$ as you've assumed. You need to use the Lorentz transformations to find the correct $x'^\mu$.