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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.

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    $\begingroup$ There are some problems when dealing with forces in different reference frames. One thing to remember is that because the measurement of time is different the time derivative in the usual definition of force (dp/dt) causes a disagreement. I think its more intuitive to compare the proper acceleration observed in the two frames (second derivative of position with respect to proper time). The invariant quantities in SR are scalars. So you need to be comparing the magnitude of 4-vectors to check for consistency (either proper acceleration or a 4-vector force). $\endgroup$ – Paul T. May 17 '16 at 16:40
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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.

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  • $\begingroup$ Thanks for this. I had overlooked the fact that the charge density $\sigma$ in the unprimed reference frame was itself unprimed. But doesn't that mean that the solution to a problem asking for the total electromagnetic force on the particle given some particular rest charge density $\sigma'$ would be $F=q\sigma'/2\epsilon_0 \gamma$ (not $1/\gamma^2$), since the Lorentz force doesn't account for the length contraction? $\endgroup$ – JAustin May 27 '16 at 2:22
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    $\begingroup$ I think what you said is correct. I'm not sure why you're using primed and unprimed variables in one equation (although it can be done). The Lorentz force doesn't account for length contraction like you say, that part needs to come from Coulomb's Law. $\endgroup$ – Faraday7000 May 28 '16 at 11:04
  • $\begingroup$ Thanks again. I was just using unprimed constants because I had erroneously assumed that the Lorentz Force Law accounted for all relativistic effects. Of course, a parallel plate capacitor doesn't often move at relativistic speeds, but still worth noting. Thanks for your help. $\endgroup$ – JAustin May 28 '16 at 13:06
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    $\begingroup$ No problem, enjoy your relativity!...always use primed/unprimed notation, will save a life of pain ;) $\endgroup$ – Faraday7000 May 28 '16 at 15:08

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