Lorentz force in different reference frames?

Question

I have been introducing myself to special relativity and relativistic electrodynamics, and became curious about the similarity of the electric and magnetic forces. I'm trying to show that the combined magnetic and electric Lorentz forces exerted by a moving charge sheet on a moving charge are equal to the electric force on the system in a comoving reference frame, but I've made a mistake somewhere. I'd appreciate some help.

An infinite (or very large) charged sheet will exert a purely electric force on an oppositely charged point charge some distance above it. The force will point in the negative z direction, and will be equal to

$$F = \frac{Q\sigma}{2\epsilon_0}$$

However, the force must presumably be the same in all reference frames, where the charged sheet and the point charge are moving to the right (in the positive x direction) with velocity $v$. In this frame, there will be both a magnetic and electric force.

By Ampere's Law/Lorentz Force Law, the magnetic force will point in the positive z direction, and will be equal to

$$F=\frac{Q\mu_0\sigma v^2}{2}$$

While the electric force will remain the same. The total force will be equal to

$$F=\frac{Q\sigma}{2\epsilon_0} - \frac{Q\mu_0\sigma v^2}{2}$$

Which simplifies to

$$F = \frac{Q\sigma}{2}\left(\frac{1}{\epsilon_0}-\mu_0 v^2\right)$$

And since $\mu_0=\frac{1}{\epsilon_0 c^2}$, this can be rewritten

$$F = \frac{Q\sigma}{2\epsilon_0}\left(1-\frac{v^2}{c^2}\right)$$

$$F = \frac{Q\sigma}{2\epsilon_0}\left(\frac{1}{\gamma^2}\right)$$

Now presumably this must be equal to the electrostatic force in the laboratory reference frame, so I tried transforming the electrostatic force using the special relativity identities for force and the length contraction of the charge density.

The moving force is equal to the rest force divided by the Lorentz factor, and the charge density should increase by the Lorentz factor.

So instead of the desired $$F = \frac{Q\sigma}{2\epsilon_0}\left(\frac{1}{\gamma^2}\right)$$

I get $$F = \frac{Q\sigma}{2\epsilon_0}\left(\frac{\gamma}{\gamma}\right)$$

Which is obviously wrong. I'm just trying to figure out where my mistake is. I am teaching myself the material, and don't yet have a solid understanding of the graphical or 4-vector theory of SR, so this may be a misunderstanding on my part.

Thanks.

Answer 1

The mistake you made was essentially transforming the force twice. In the frame in which the matter is stationary (I'll call this the primed frame), you correctly found:

$F'=q\sigma'/2\epsilon_0$

and in the frame in which the matter is moving (unprimed) you correctly found:

$F=q\sigma/2\epsilon_0 \gamma^2$

Since $\sigma'=\sigma/\gamma$ this is:

$F'=\gamma F$

and the momentum change is:

$dp'/dp=F'dt'/Fdt$

and since $dt=\gamma dt'$, we get:

$dp'=dp$

as desired.