There is an anology that might help explain the situation, which will allow me to recylcle an old post for a new purpose.
Suppose we have two parallel wires, we treat electrons as classical particles, and the two parallel wires contain a total of 2N electrons, N electrons per wire. Let the length of the wire be L in its rest frame. We than have a classical case of particles in a box. And I've analyzed the case of particles in a box. (But I'll have to note that this analysis hasn't been double checked - at least not AFAIK). The density of electrons per unit length is N/L in the rest frame of the wire.
If the electrons move at a velocity v, and we move along with the electrons, we should observer the following, per the analysis I did in
https://www.physicsforums.com/showpost.php?p=965449&postcount=1
In the post we have two velocities, u and v, v being the velocity of the electrons, and u being the velocity of the observer relative to the box. In your example u=v, so I will use v everywhere. I will also assume that c=1.
Double-checking the orignial post, I seem to have found than n1 and n2 are interchanged. You might want to triple-check it. I get
n1 = N(1-v^2) electrons moving right
n2 = N(1+v^2) electrons moving left
for a total of 2N electrons.This equation was arrived at by equating the current of electrons existing one wire is equal to the current of electrons entering the other wire
If we redo just the essentials of the analysis, we are arguing the following.
In one wire, the electrons are standing still. The Lorent contracted wire, of length L' = L/sqrt(1-v^2) is moving with a velocity v. If this wire has n2 electrons, all of the electrons will exit the wire in a time T of L'/v. The current will therefore be
I2 = n2 / T = n2 v / L'
In another wire, the electrons are moving at a velocity of 2v / (1-v^2), the relativistic sum of v+v. The wire is moving with a velocity of v. If this wire has n1 electrons, all of the electrons will exit the wire in a time T of
L' /(2v/(1-v^2) - v) = (L'/v) (1-v^2)/(1+v^2)
So the current exiting the wire will be
<br />
I1 = n1 \left( \frac{2v}{1-v^2} - v \right) \frac{1}{L'}<br />
The equation says that the two currents are equal, i.e<br />
n2 \frac{v}{L'} = n1 \left( \frac{2v}{1-v^2} - v \right) \frac{1}{L'}<br />
we also have n1 + n2 = 2N. Checking the solution, we find
n1 = 1-v^2
n2 = 1+v^2
(It appears something got reversed in the original post, interchanging n1 and n2 as I mentioned)
works, and that the current entering and exiting each wire in the moving frame is the same, namely
(1+v^2) (v/L')
the length of the two wires (or the box) are Lorentz contractred in the moving frame so L' = L/sqrt(1-v^2) as previously mentioned.
Thus we can say the the density of the electrons in the wires are, respectively
(N/L) (1+v^2)/sqrt(1-v^2)
(N/L) (1-v^2)/sqrt(1-v^2)
Neither wire has a density of (N/L) in the moving frame. One wire has a greater density (a surplus of electrons) the other has a lesser density (a deficit of electrons).
