Relativity and Current in Wire If an observer is stationary relative to a current-carrying wire in which electrons are moving, why does the observer measure the density of moving electrons to be the same as the density of electrons if there were no current in the wire?
I read the explanation of magnetic force as a consequence of special relativity. That is, when the observer moves with respect to a current carrying wire in the same direction as the flowing electrons, then s/he observes the density to decrease due to Lorentz expansion and observes the density of positive ions to increase due to Lorentz contraction. The imbalance of charges results in a force that can be explained by Coulomb's law.
That explanation mentions that when the observer is stationary, the density of positive ions and the density of electrons appears to be the same. That last point is what I am stuck on.
Why does the density of moving electrons appear to be the same as the density of stationary positive ions?
Clarification: I know that when the electrons are not moving, their charge density cancels out that of the stationary positive ions. When the electrons are moving, how could the charge density still cancel out that of the stationary positive ions?
Important: My assumption is that a current carrying wire will neither attract nor repel a charge that is stationary with respect to the wire. I am now not sure if that is right :-(
 A: 
Model: Let's simplify the model of a current in a wire, so we can be definite about what we are talking about. Take a wire (in the wire's frame) to have fixed positive charge density $\rho_{+}$ and assume the electrons at rest w.r.t the wire, with electron density $\rho_{-}$.
Introducing a current sets these electrons moving at some speed $v_\rm{drift}$ w.r.t wire, but leaves the positive charge fixed. We ask the following question:
What is the relationship between $\rho_{-}$ (the electron charge density at rest), and the electron density with current?

Answer: The density measured by the observer stationary w.r.t a current carrying wire is not the same as if the charges were stationary. They are related by a Lorentz transformation. Let's write the 4-current of the electrons when at rest, and when moving (with $c=1$):
$$J^\mu_\textrm{rest} = (\rho_{-},\vec{0})^\mu,\,\,\,\,\,J^\mu_\textrm{moving} = (\tilde{\rho},\vec{j})^\mu = {\Lambda(v)^\mu}_\nu J^\nu_\rm{rest}$$
where $\Lambda(v)$ is the Lorentz Boost between these two frames. Note in particular that $\boxed{\rho_{-} \neq \tilde{\rho}_{-}}$ because 
$$J^2_\textrm{rest} = J^2_\textrm{moving}~~\implies ~~ \rho_{-}^2 = \tilde{\rho}^2_{-}-\vec{j}.\vec{j},$$
and $~\vec{j}\neq \vec{0}$.
This means that when you set up your problem, we have two possible scenarios:
$(i)$ $\rho_{-}+\rho_{+} = 0$, that is we ask that the electron density in the electrons rest frame has the same magnitude as the positive charge density in the stationary wire.
$(ii)$ $\tilde{\rho_{-}}+\rho_{+} = 0$, that is we ask that the electron density in the wire's rest frame has the same magnitude as the positive charge density in the stationary wire. This is the situation of zero force on a stationary external charge you talked about in your edit.
So the question you have to ask yourself, is what situation do you want to deal with? It seems that for the "explanation of magnetic force as a consequence of special relativity" you are interested in, one should consider case $(ii)$ as this allows you to see how a test charge, moving parallel to the wire with velocity $v$, experiencing a force due to a pure magnetic force in one frame (wire rest frame) $F = q v\times B$, is the same force experienced by the charge in its rest frame, effected only by the electric force, $F = q E$, in that frame (as in this frame it isn't moving).

I hope this helps. If you need further explanation, don't hesitate to ask.

A: 
When the electrons are moving, how could the charge density still
  cancel out that of the stationary positive ions?

Recognize that the fixed charge is, well, fixed, while the mobile electrons are, well, mobile.  This is the crucial difference here.  The mobile electrons are free to "squeeze together" or "push apart" but the fixed charge is not.
Regardless of whether the electrons have an average drift velocity or not, with respect to the fixed charge of the wire, the mobile electron density may be such that the wire has net charge or not.
For example, connect a wire to one terminal of a battery.  The wire is unconnected at one end so, in steady state, there is no current through the wire.  But, it is certainly the case that the wire has net electric charge.
So, for this problem:
we choose the condition to be that, in the reference frame of the lab in which the constant current carrying wire is at rest, the wire is electrically neutral.
Then, the question is, in another reference frame relatively moving with respect to the lab (and thus, the wire), is the wire electrically neutral?
The answer is:  in general, no.
In the lab frame, the wire is neutral, we necessarily have that the fixed (positive) charge density equals the moving (negative) charge density.
Now, for example, in the frame of reference in which the negative charge moving in the lab frame is stationary, the density of the moving charge is decreased.  This is because, in the lab frame, lengths in the reference frame of the moving charge are contracted.  So the negative charge density decreases.
Also, in the same frame, the fixed charge density is increased since, in this frame, the wire is contracted.  So, the positive charge density increases.
It follows that, in this frame, the wire has a net positive charge density.

Now, we don't have to choose the wire to be electrically neutral in the lab frame.  But then the question becomes:  in another reference frame relatively moving with respect to the lab (and thus, the wire), is the wire charge density the same?
Again, the answer is, in general, no.

Imagine two identical, ideal wires with identical constant currents. In the lab frame in which both of these wires are at rest, one of these wires has zero net charge and the other wire has non-zero net charge. Do you accept or reject that this is possible?

I think this is not possible, but I will try to see why you might
  suggest this is possible.

Consider the following simple circuit at rest in the lab frame.

The "top" wire has a net charge density.  The "bottom" wire does not.  The current through each wire is identical.
A: Re your statement:
"That explanation mentions that when the observer is stationary, the density of positive ions and the density of electrons appears to be the same."
This part of the explanation looks to be incorrect (or at best misleading).  
The problem is most easily set up by assuming a neutral wire (canceling + and - charge densities) with current in a "lab" frame, and an "observer" charge moving parallel to it.  Once you boost to the observer's frame, the wire charge densities no longer cancel, and the observer feels a Coulomb force (contra that statement I quoted).
I think demonstrating that effect is the point of the other answers here:  if you start with a neutral current-carrying wire and then boost, the wire is no longer neutral.
Now you could start with a charged wire with current, such that the wire looks neutral in the observer's frame, (like the statement says).  But then there's no force on the observer charge: In the lab frame, the magnetic force cancels the coulomb force, and in the observer's frame, there's no net charge to attract it.  (i.e. Your important assumption is an important correct assumption.)
A: I think a good deal of confusion here comes from the fact that the wire is infinite, and so in some real sense the charge doesn't have to be conserved. You have an infinite number of negative charges and an infinite number of positive charges. The net charge on the wire is $\infty-\infty$, which is an indeterminate form that can equal any value.
So why then do we assume that a wire that is neutral before is neutral after? Because an infinite wire is not a physical configuration. Ultimately any real circuit is a finite loop. And since a finite loop has a finite number of charges, the net charge has to remain zero (in every reference frame, in fact, although the rest frame has the additional condition that the charge density is uniformly zero in the steady state).
So how is it possible that in the infinite case the charge density remains zero in one frane and not another? As with so many things in relativity, it comes down to simultaneity. 
To change the current from zero to $I$, we apply a uniform voltage gradient simultaneously across the whole wire and remove it when the current reaches the desired value. But of course in another frame this gradient is not simultaneous, so the current is greater in parts of the wire than others. This causes a build up of charge across the wire.
Note that conservation of charge was not violated! In a global sense, the charge was initially in the indeterminate form $\infty-\infty$, and it still is afterwards. In a local sense, the charge buildup was because of conservation of charge: the continuity equation gives $${d\lambda\over dt}={dI\over dx}$$
Or, in words, the gradient in the current causes a change in the charge density of the wire.
A: EDIT: I guess the reason I got down-voted was because the answer seemed to imply that any current carrying conductor would have net charge 0. I do not claim that. I presume that we start off with a neutral wire, because having it charged does not help illustrate anything relevant for this question. 
Also: Readers might get the impression from the answers that Charge Conservation does not apply here. Well, I think that it does. The Current density would have to be replaced by the 4-current version.
EDIT2: Does not apply to either wire in Alfred Cnaturi's diagram. This answer applies to a current carrying wire that is neutral in lab/conductor's frame of reference. I think that that is a fair assumption to answer this question.
If I understand your question correctly, then I would have to repeat Will's answer: There is not net charge in the wire.
Why? Because although the electrons (negative charges) are moving whilst the positive kernels are fixed, since there is a closed circuit in place, they are being replenished by new electrons, regardless of the section you are observing. (Just realised that bit is wrong if you ignore EDIT2 assumption, thanks to Alfred Centauri's excellent link)
Therefore the charge density cancels out that of the positive ions. This much does not require any knowledge of special relativity, and I am sure that this is not solely what you are after.
So, in the case of a moving frame (that of the electron, for instance), the charge density does change to that of rest frame (rest with respect to conductor). But then there has to be conservation of charge! Well, this change in charge density is due not due to accumulation of charge but rather the relativity of simultaneity affecting the measurement of charge density in the moving frame.
It is also important to note that the distance between the charge particles that expand or contract does not change, unlike the popular illustrations in books, instead the charge particles undergo the change in dimension.
A: The following is a relativistic take on a current in a wire. And it might not be necessarily an answer to your question but I'm adding it because everyone else seems to think it's relevant.
If you do an experiment where you measure the force on a stationary charge in the vicinity of a current-carrying wire you will measure that there is none. From this you can deduce that the net charge in the wire is zero.
In a wire carrying current the positive ions are bound and stationary so we'll take their frame as our stationary frame of reference. The linear charge density in the wire in the ion frame is then:
$$ \lambda=\lambda_-+\lambda_+=0$$
and the linear electron charge density is:
$$ \lambda_-=-\lambda_0$$
Now let's move to the electron frame of reference. In this frame the linear electron density must be:
$$ \lambda'_-=-\lambda_0/\gamma$$
This is simply follows from what we observe in the ion frame. If it were any different the linear electron charge density in the ion frame would be different:
$$ \lambda_-=\gamma\lambda'_-=\gamma(-\lambda_0/\gamma)=-\lambda_0$$
Additionally, when we move to the electron frame ions are no longer stationary and move. So their linear charge density changes:
$$ \lambda'_+=\gamma\lambda_0$$
Notice that this operation is inverse to that of linear electron charge density as here we are going from stationary to moving ions.
The total linear charge density in the electron frame is then:
$$\lambda'=\lambda'_++\lambda'_-−=\gamma\lambda_0−\lambda_0/\gamma$$
and this is not equal to zero.
If we have a stationary single electron in the ion frame beside the wire the sum of forces is experimentally shown to be zero. It follows that the same must hold in the electron frame.
In the electron frame the now "new" Coulomb force acting on the single electron has to be counteracted for the above to hold. The solution lies in the now moving ion charges. In this frame they cause the current which produces a magnetic field which now acts on our moving single electron (it was stationary in the ion frame). The forces cancel each other out exactly.
I've included a simplified graphic trying to demonstrate this.
This is the source(1) I was using to answer this question. It's where you will also find actual calculation of the forces in the electron frame and a bit more.
However, your question seems to be about electron density when there is and isn't current flowing.
Unfortunately I do not have a source for this but I will try to answer with what I think is going on. I think the crux of the situation lies in the first 4 equations above. We start with some linear electron charge density:
$$ \lambda^{v=0}_-=-\lambda_0$$
which we then speed up when we turn on the power and the current starts flowing. Now because of relativity the distance between electrons shortens resulting in a higher linear electron density from our perspective and the ions':
$$ \lambda^{v\neq0}_-=-\gamma\lambda_0.$$
Now the electrons that are stationary and want to join the current see it as too negative for what their potential is. Therefor they don't join in as fast and this results in the whole thing normalizing to:
$$\lambda^{v\neq0}_-=-\lambda_0.$$
So while we see the same linear electron density the actual number of electrons is smaller and it only appears not to be because of relativistic effects.
From the classical perspective, you need to look into what current is. Given that experiments show zero net charge, higher current means faster charge movement and not "more-charge" movement. At the moment I don't have the answer to microscopic reasons why this is the way it is. Most likely something to do with: more charge -> higher field -> higher force -> higher terminal velocity -> less charge.
(1) Current-Carrying Wires and Special Relativity by Paul van Kampen
Centre for the Advancement of Science and Mathematics Teaching and Learning &School of Physical Sciences
Dublin City University
A: Simplify the problem. Start with the effect of a moving negative charge on a stationary negative charge. We know the effect of a stationary positive charge on the stationary negative charge.
We have a correct formula for this, which is compatible with relativity.
In the frame where the target charge is stationary, and there is no acceleration, and the source velocity is perpendicular to the direction to the target, the formula simplifies considerably.
$E(r,t) = K\frac{(1-\beta^2)(n-\beta)}{R^2}  $
where $K$ is a few lumped fudge factors,
$n$ is the unit vector in the direction from the source when the force left it, and the target when the force arrives,
$R$ is the distance between the source when the force left it, and the target when the force arrives, and
$\beta$ is the source velocity when the force left it, divided by lightspeed.
The force is in a different direction! $n-\beta$ instead of $n$.
Also the vector sum $n-\beta$ has magnitude $\sqrt{1+\beta^2}$ so the total magnitude of the force is $\frac{(1-\beta^2)\sqrt{1+\beta^2}}{R^2}$ compared to $\frac{1}{R^2}$ when both are stationary.
So the total force is larger, but the force in the original direction is smaller.
Why do people not usually measure this? Because individual electrons may move very fast, but mostly they move in random directions and the effects cancel. The average drift velocity is very small, so the change in direction is very small and the change in magnitude also.
