Charge conservation and lenght contraction for a current carrying wire vs zero current wire I'm reading through the Feynman lectures (specifically: http://www.feynmanlectures.caltech.edu/II_13.html, scroll down to 
"13–6 The relativity of magnetic and electric fields") and I'm having the following issues with what Feynman is saying:
Consider a wire with a zero current; it is electrically neutral because the number of electrons equals the number of protons. Now, when we get a current going through the wire somehow (using a cell, or maybe it's a superconductor, doesn't matter), we expect the electron density to become greater because the lengths of moving objects contract (the moving object is the electron 'gas').
there are two issues with this, the first is that this doesn't happen under laboratory conditions, no net charge is observed in a conducting wires, but relativistic reasoning suggests that the electron density would rise by a factor of $\frac{1}{\sqrt{1-v^2/c^2}}$. So there should be a net negative charge.
The second issue concerns the conservation of charge, if the electron density does increase as Feynman describes it to, where do these new electrons come from? For this, it's easier to imagine a neutral isolated superconductor with zero current, in which we induce a current, a current that should increase the negative charge density, but there is no source for this new charge and charge conservation appears broken.
Relativity and Current in Wire that answer leaves it to the reader to 'choose' which case is electrically neutral and which isn't, and that doesn't really help when considering a wire where the current is 'switched on', because both cases would have to be true, and that's not possible because the charge densities change.
My initial guess was that electrons in a wire a going in circles (or at least a loop), so their velocity isn't constant and therefore the contraction laws don't apply. But even if the wire is infinitely long and straight, the same issues arise - there is no source of negative charge for the density to increase, and there should be a net charge.
Further, the electrons would accelerating, though for a short time, when the current is turned on and during this time basic relativity would not apply. If that is the answer, I would like to know why my arguments don't work for accelerating electrons.
My other guess would be that it's not the electron gas that contracts, instead the individual electrons (they can be modelled as balls) contract. This works in principle, because the overall current is individual moving electrons. In that case, the longitudinal diameter of the individual balls would shrink, but the overall density would remain the same. This view means that a wire with a current would remain neutral, which is good because that is what we observe. 
However, with this assumption the magnetic force on a test particle comes out twice as large as the electric force for low velocities, so the magnetic force is no longer a relativistic electric force. (For this, I did the same calculation as Feynman, but I assumed that the negative charge density does not change). (Note that the positive charge would still change in the ball view because in that case the whole 'world' is moving, not just the positive charge).
Anyway, I'm fairly certain the second guess can't be right, because it messes up a lot more than it fixes.
 A: The density of the electrons will not change when you turn the current on, they will keep the same density, because they are not attached to a rigid stick. If they were, the stick would break when it accelerates, or expand if it is elastic. 
Feynman's example is for a wire that has a current and is neutral. 
Then, in such a case, an observer moving at the same speed than the charges will see the wire charged. The density of the charges at rest with the wire will increase to the moving observer, but not the net charge, because the wire will look shorter. This density will also increase on the return part of the wire, the one that closes the circuit (assumed to be very far away). 
The density of the moving charges will decrease along the segment of the wire that you are looking, but will decrease on the far away return part of the wire that closes the circuit (the electrons there are moving on the opposite direction at twice the speed than the electrons in the rest frame of the wire).  
Thus, the wire close to you will look positively charged (assuming the negative charge is the one that moves in the circuit), but the other part of the wire will look negatively charged. So the net charge will still be zero.
Update: There are two different thing here, one is length contraction/expansion of an object when observed by observers moving at different speeds. Another,  is  something that changes its speed (accelerate).  It is not the case that the accelerated object will just look contracted. The problem is that perfect  rigidity in relativity is impossible. If a rigid belt (with electrons glued to it) starts to roll when you switch the current on, then either  the belt breaks or expands (or, if they are free, you can accelerate them individually in such a way that the distances between them stays constant). The electrons will not move as a rigid object, so after a transient  there is no way a priori to tell what distances the electrons will keep to each other unless you do the actual calculations considering the electric fields, etc. An easier way is to use the fact that you mentioned: the wire will not look charged. Notice that, for a circular wire you can conclude that this is the case due to the symmetry of the configuration. 
