${}$Magnetism and Special Relativity

Question

I was reading a book about Special Relativity. There is a chapter where it explains why magnetism is just an electrostatic force because of lenght contraction.
The book, uses two reference systems $S$ and $S'$. In the first one, is "at rest", and sees the electrons drifting with velocity $v$ (for simplicity we will assume that our test-charge, $Q$, move also with velocity $v$).

Then, $S'$ is the reference frame of the test-charge $Q$, hence it "sees" itself and electrons with $v = 0$ and the protons drifting to the left with velocity $-v$. The book states that the density of positive charges increases: $$ V = 2\pi r \ell \\ V' = 2\pi r \ell' \\ \ell' = \frac{\ell}{\gamma} \\ V' = \frac{2\pi r \ell}{ \gamma} \\ V' = \frac{V}{\gamma} \\ \rho_0 = \frac{Q}{V} \\ \rho'_+ = \frac{Q}{V'} = \frac{\gamma Q}{V} = \gamma \rho_0 \\ $$ This makes perfect sense to me, because $Q$ is moving, hence its surroundings experience length contraction, concentrating more positive charges into the same amount of volume (assuming the wire is infinite). But, here comes the thing that does not make sense to me, the book also says that the electrons separation increases, because in the $S'$ frame they are no longer suffering from length contraction. But I don't quite see this, they shouldn't have more space between them, they have zero velocity (in the S' frame), hence $\gamma$ should be equal to one, hence no contraction or dilatation, no?

Answer 1

It's simply the converse of length contraction. In $S$, the spacing between the electrons was contracted compared to its rest frame $S'$, so when you switch to $S'$, you have an effective length dilation of value $\gamma$.

Mathematically, you obtain the following. Let $\lambda_+,\lambda_-$ be the respective linear charge densities of the nuclei and the electrons in $S$. From neutrality, you have $\lambda_+ = -\lambda_-$. In $S'$, they are transformed to: $$ \lambda_- \to \lambda_-' = \frac{\lambda_-}{\gamma} \\ \lambda_+ \to \lambda_+' = \gamma\lambda_+ $$ Note that this means that there still is a current. In $S$, it's $I = \lambda_-v$, and in $S'$, it's $I'=-\lambda_+'v = -\gamma I$. In particular for low velocities, you recover $I'=-I$ as expected.

Btw, the remaining current means that you still have some magnetism in $S'$. In fact, the previous formula proves that the switch actually enhances the current, and therefore increases magnetism as well. Not only is magnetism in $S$ is not entirely described by electrostatics in $S'$, but it's rather the electrostatics that is present to compensate for the increase of magnetism. You'll need to switch to a reference frame $S''$ going at a different speed $u$ with respect to $S$ to have no current, and you can anticipate that for low $v$, it should be $v/2$.

By some lucky algebra, it turns out that it is also the exact relativistic result. Let the corresponding factor be $\gamma_u=(1-u^2/c^2)^{-1/2}$. In $S''$, the electrons are going at velocity: $$ v' = \frac{v-u}{1-uv/c^2} $$ with corresponding $\gamma' = (1-v'^2/c^2)^{-1/2}$. You get: $$ \lambda_+''= \gamma_u\lambda_+ \\ \lambda_-'' = \frac{\gamma'}{\gamma}\lambda_- \\ \lambda_-'' = \gamma_u(1-uv/c^2)\lambda_- $$ with the simplification coming from $\gamma' = \gamma_u\gamma(1-uv/c^2)$. You then get: $$ I'' = -\lambda_+'' u+\lambda_-'' v'\\ =\gamma_u(-u +(1-uv/c^2)v')\lambda_+ $$

The equation $I''=0$ is equivalent to: $$ v-2u=0 $$ which you can solve for $u$ to obtain: $$ u = v/2 $$

@Brian Bi's answer is entirely satisfactory, it's just that I had started to write my answer, so might as well post it.

Hope this helps.

Answer 2

Lorentz transformations are reversible. You can reason about the "old" frame by treating the "new" frame as if it's the old one, and then doing a boost in the other direction.

If the separation between electrons in the "new" frame (where the electrons are stationary) is $d$, then in the "old" frame, which is moving with velocity $-v$ relative to the "new" one, that length appears to be contracted to $d/\gamma$.

Based on this, it's clear that in the "new" frame the electrons are $\gamma$ further apart than they were in the "old" frame.