Are Electromagnetic Fields Equal in Different Frames?

[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
<br /> \frac{q}{\gamma^2 d^2}<br />
in the negative x direction. By the transformation of parallel fields, E_{||}&#039;=E_{||} so the result in the charge frame should be the same.
However working in the charge frame,
<br /> E=-\frac{q}{d&#039;^2}, \qquad d&#039;=d/\gamma<br />
because of length contraction. therefore the field in the charge frame is
<br /> E=-\frac{q}{(d/\gamma)^2}=\frac{\gamma^2 q}{d^2}<br />
these are not equal, the gamma here is in the "wrong place".
what is going on here?
thanks.

 

[vela]

The spacetime point x^\mu = (ct, x, y, z) = (0, -d, 0, 0) doesn't correspond to the point x&#039;^\mu = (ct&#039;, x&#039;, y&#039;, z&#039;) = (0, -d/γ, 0, 0) as you've assumed. You need to use the Lorentz transformations to find the correct x&#039;^\mu.