Relativistic electromagnetism help: Current carrying wire

Question

I am looking at the first three pages of this file (https://www.mtholyoke.edu/courses/tdray/phys310/electromag.pdf).

In the lab frame, there is an infinitely long wire stretching from left to right, consisting of positive and negative charges with equal linear charge density $\lambda_+ = \lambda_-$ so that the wire is overall charge-neutral, $\lambda = \lambda_+ - \lambda_- = 0$. There is a test charge $q$ at some distance $r$ from the wire moving at speed $v = c\tanh{\beta}$ to the right. The positive charges in the wire are moving at a speed $u = c\tanh{\alpha}$ to the right and the negative charges are moving at the same speed $u= c\tanh{\alpha}$ to the left, so that the wire has a net positive current to the right of $I = 2\lambda_+ u$, leading to a (tangential $\hat{\phi}$) magnetic field at the moving test charge and thus a magnetic force (radially attractive in $\hat{r}$).

Now boost to the rest frame of the test charge. The wire's positive charges are now moving at $u_+ = c\tanh({\alpha-\beta})$ and the negative charges at $u_- = c\tanh({\alpha+\beta})$. The paper is trying to show that the force in the rest frame of the test charge now appears to be due to an electric force, because the linear charge densities have changed due to Lorentz-length contraction from the new velocities, and so $\lambda = \lambda_+ -\lambda_-$ is now nonzero.

My question is this: why is the wire not electrically neutral in the rest frame of the test charge also? I thought charge was a relativistic invariant. If we boost to the rest frame of the test charge than the spacing for both the positive and negative charges should contract by $\gamma_v$. I can agree that the relativistic velocity addition works as stated, but why are we using that to contract the charge densities, rather than the original boost factor?

Thanks for any help.

Answer 1

I can agree that the relativistic velocity addition works as stated, but why are we using that to contract the charge densities, rather than the original boost factor?

Because using the original boost factor would be a wrong way to contract a moving object.

For a moving object some boosts actually make the object longer.

If an object moves very fast, then even small boosts that decrease the speed increase the length quite a lot, and even small boosts that increase the speed decrease the length quite a lot.

As an other example think about how boosting effects the length of a light pulse. (Doppler effect)

Answer 2

As you say, charge is a relativistic invariant, but charge density is not. Charge density transforms as the timelike component of a four-vector called the four-current. The four-current is given by: $$J=(c\rho,j_x,j_y,j_z)$$

So in the lab frame the four-current is $J=(0,j,0,0)$ and in the charge frame it transforms as normal to $J’=(c\rho’,j’,0,0)=(\gamma j v, \gamma j, 0, 0)$

Now, the issue with this particular setup is the infinite length of the wire. The proof that charge is invariant subtly assumes that the four-current goes to zero at infinity (many EM proofs have that same assumption). Unfortunately, a long straight wire violates that assumption. So here it seems that charge is not invariant. If you used a more realistic current, e.g. a loop, then indeed you would see that the wire is neutral overall even though it contains some sections with positive charge density and others with negative charge density.

Answer 3

The best way to understand this problem is by setting a 1-D lattice of positive charges, with matching electrons moving to the right (or left), and the apply a Lorentz transformations. The are then 2 frames to consider: $S$ is the frame at rest w.r.t to the lattice, and $S'$ moves along with the electrons.

The lattice frame looks like this:

So here we see unit spaced ions (blue) moving forward in time, while electrons (red) move to the right. The key here is that the electron spacing in $S$ is still "one", so the overall wire is neutral.

In the $S'$ frame, those red-lines are parallel to the $t'$-axis, and the $x'$-axis is shown in magenta (with tic marks showing unit spacing in $S'$).

When you transform the above image to the $S'$ frame, the Minkowski diagram is:

The now-moving ion-lattice is contracted per length contraction ($\gamma=2$), while the electron spacing is diluted by $1/\gamma$. The electrons collectively are not a solid object, and do not undergo Lorentz contraction. They keep their coordinate spacing in $S$, and are thus spread out in $S'$. This effect is the basis of Bell's Spaceship Paradox: an object can not be uniformly accelerated from its rest frame without experiencing relativistic stresses.

Note also how relativity of simultaneity plays in: in the $S$ frame, there is a point where each ion is next to an electron, yielding manifest neutrality at a fixed time. In the $S'$ frame, the definition of "now" depends on position, so as you move to the right (left), those coincident positions events move further into $S'$'s past (future).