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All popular expositions (e.g. these ones) of relativistic electromagnetism claim univocally that electrons in motion become more dense due to the speed. They teach that Lorentz contraction of charges causes charge imbalance and wire with current charged. Thereby, no teacher says that the wire becomes negatively charged (because electrons move) in the lab frame, where we had it originally neutral, when no current was flowing. That is a question: is wire neutrality conserved, once we create a current of electrons in it?

It must be the case because, as Feynman points out, the lighter electrons better react on temperature changes and would charge the wire when heated. But, then answer how neutrality is persisted after you say that electron contraction takes place and it increases their density? Might be "some light" positive charges start moving in opposite direction, to compensate the growth of negative charge?.

I see that all discussions of "magnetism as electricity+relativism" avoid concerning this, most basic and most interesting case. Instead, they jump immediately to the case where you have an electric balance for a test charge in motion. Once finished with this, you are encouraged to answer how it is possible that density of charges is increased whereas volume of the loop is intact?

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"Positive ions must move to compensate the motion of free electrons!" Er...this is not required. Recall that a typical circuit is one or more loop(s), so everywhere there are electrons moving in to cover for the ones that just left. The net-positively charged substrate can just sit there. And it does when you're talking about current in, say, a copper wire. –  dmckee May 6 '13 at 19:24
    
What left-in, left-out? The whole loop becomes more negatively charged! You cannot cover electrons by electrons. Ok? –  Val May 6 '13 at 19:27
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I'm sorry if I wrote that poorly. At any point on the loop there are electrons that have present leaving the scene. Naively this would seem to leave a positive charge, but there are also electrons arriving. In a steady state there are just as many arriving as leaving, so there is no change. –  dmckee May 6 '13 at 19:30
    
Why arriving and leaving do not cancel each other? Why they add positive charge rather than negative. I bet they add even more negative. How does this affect the fact of relativistic charge density increase? Do you mean that it is always compensated in reality by arrivals? Then it is unobservable. Why do we talk about it? Can you answer in the Ehrenfest paradox. It is very interesting where electron density is coming from. –  Val May 6 '13 at 19:43
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If you have a closed circuit fountain of water, where does the water come from? That is the analogue of your question. Energy need is supplied by a small motor but the number of water molecules is the same. In a closed circuit the number of electrons and positive charges is equal and charge adds to zero. The electrons just move along continuously as @dmckee explains. The energy needed is supplied by an electric motor or an electric battery –  anna v May 7 '13 at 3:29

3 Answers 3

I don't hope too much to get the answer to the electric current density ernfest paradox.

That question didn't ask about the Ehrenfest paradox and isn't related in any way to the Ehrenfest paradox. The physicsforms link I gave in a comment provides a complete and correct analysis of your question.

Why discussions of relativistic origin of the magnetic field never discuss the simplest case: a neutral wire with current and test charge at rest?

They don't discuss this case because this case isn't interesting. The test charge experiences no electric or magnetic force, so it doesn't accelerate.

This is what should happen if only one polarity charges are moving in the wire. Right?

No, it happens in any frame that is in motion relative to the lab frame (where the lab frame is defined as the one in which the wire is electrically neutral), provided that the positive and negative charges are in different states of motion.

But, it forgets to mention that in the normal lab, the positive charges create a solid structure of the frame and, therefore, cannot move.

It's not a contradiction, because these treatments are not discussing a realistic model of a metal wire, in which the positive charges are atomic nuclei and the negative charges consist of some mixture of bound electrons and conduction electrons. To make the derivations more mathematically transparent, they make a model of a wire that doesn't have all these complicated physical characteristics.

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Many pointlessly argumentative comments deleted. –  dmckee May 7 '13 at 16:55
    
@Ben, this answer escapes the question. I have established pretty concrete conditions, asking if it is true that wire becomes charged. When you say "no, this is true for any frame of reference" you evade the question and, confirming my guess from one side, deny it from the other. When you say that it is not interesting you try to get out also. It is of the first importance to know if wire gets charged and test charge will also show electric force then. –  Val May 7 '13 at 18:32

Let us consider a lattice of static positive ions, through which electrons flow. In the laboratory reference frame, we will have equal charge densities and different velocities, thus their 4-currents would be:
$j_+^\mu=(\rho_+,0)$
$j_-^\mu=(\rho_-,\rho_-\mathbf{v}_-)=(-\rho_+,-\rho_+\mathbf{v}_-)$
and the total current, which creates an electromagnetic field, is their sum:
$j_\Sigma^\mu=j_+^\mu+j_-^\mu=(0,-\rho_+\mathbf{v}_-)$
So far, it is as we expected. Now, let us switch to the moving reference frame, having the velocity $\mathbf{v}_-$, so that electrons would appear static. That is done by the Lorentz transformations for 4-vector of current, and the result is ($\gamma=1/\sqrt{1-v_-^2}$ is the usual abbreviation):
$j_+^{\,\prime\mu}=(\gamma\rho_+,-\gamma\rho_+\mathbf{v}_-)$
$j_-^{\,\prime\mu}=(\gamma\rho_--\gamma\rho_-v_-^2,\gamma\rho_-\mathbf{v}_--\gamma\rho_-\mathbf{v}_-)=(-\rho_+/\gamma,0)$
(zero current of electrons, as we expected)
$j_\Sigma^{\,\prime\mu}=j_+^{\,\prime\mu}+j_-^{\,\prime\mu}=(\rho_+v_-^2\gamma,-\gamma\rho_+\mathbf{v}_-)$
This is exactly the same as we would get if applied Lorentz transformations immediately to $j_\Sigma^\mu$. No balance of the charge in a moving frame is left.

Now we can consider some other conducting system, say, a cell with electrolyte, adjusted in such a way that positive and negative ions in it have the same density and opposite velocity vectors. Now we repeat the same calculations and get:
$j_+^\mu=(\rho_+,\rho_+\mathbf{v}_+)=(\rho_+,-\rho_+\mathbf{v}_-)$
$j_-^\mu=(\rho_-,\rho_-\mathbf{v}_-)=(-\rho_+,-\rho_+\mathbf{v}_-)$
$j_\Sigma^\mu=j_+^\mu+j_-^\mu=(0,-2\rho_+\mathbf{v}_-)$
$j_+^{\,\prime\mu}=(\gamma\rho_++\gamma\rho_+v_-^2,-\gamma\rho_+\mathbf{v}_--\gamma\rho_+\mathbf{v}_-)=(\gamma\rho_+(1+v_-^2),-2\gamma\rho_+\mathbf{v}_-)$
$j_-^{\,\prime\mu}=(\gamma\rho_--\gamma\rho_-v_-^2,\gamma\rho_-\mathbf{v}_--\gamma\rho_-\mathbf{v}_-)=(-\rho_+/\gamma,0)$
$j_\Sigma^{\,\prime\mu}=j_+^{\,\prime\mu}+j_-^{\,\prime\mu}=(2\rho_+v_-^2\gamma,-2\gamma\rho_+\mathbf{v}_-)$

We got the same result! (With the factor of 2 for both frames.) That means, it does not depend on the way we construct the current. The fact holds true that if charges are balanced in some chosen reference frame, then they become unbalanced in other reference frames.

Now, why are charges balanced in the laboratory reference frame, what it the physical cause for this? The moving charges are ultimately supplied to the wire by some current source. That source produces the same amount of current carriers on one pole, as it absorbs on the other pole, and thus keeps the wire uncharged as a whole. Would it be different, that would take much energy to keep the wire in a charged state, because charged conductor has an electrostatic energy $q^2/2C$. That current source "calculates" the energy and balance of chanrge in its own reference frame. That's it: would the wire and the source move, they would keep some other balance, zero in a different reference frame.

Update: Let's consider contraction and stretch of individual charged particles. We follow the Wikipedia article. Figures are taken from it.

In the lab frame:

Negative charges are "the lattice" and they make a chain with positions $x_k=ka$, where $a$ is the lattice constant. Then their world-lines are given with equations
$x_{-,k}^\mu(t)=(t,ka\mathbf{e}_x)=(t,ka,0,0)$
Note that the lattice constant is not arbitrary here, and is fixed by the nature of the lattice (for example, if the lattice is made of Fe ions, $a$=0,286645 nm).

Positive charges are "the flowing charges", and they make a chain with the same spacing, though moving. The positions $x_k=ka$ would be only their initial positions, and the world-lines are
$x_{+,k}^\mu(t)=(t,(ka+vt)\mathbf{e}_x)=(t,ka+vt,0,0)$
The requirement $x_{+,k}^x(0)=x_k$ is imposed in order for the wire to have zero charge as a whole.

In the moving frame:

To get this picture, we should take the world-lines of charges in the lab frame, and apply Lorentz transformations to them. For the negative ("lattice") charges, we have
$x_{-,k}'^\mu(t')=(t',\gamma ka-vt'-\gamma kav^2,0,0)=(t',ka/\gamma-vt',0,0)$
for the positive ("flow") charges, respectively,
$x_{+,k}'^\mu(t')=(t',\gamma ka,0,0)$
(please check these calculations yourself).

That shows that the negative ("lattice") charges are moving backwards with the speed $v$, and their distance is contracted ($a/\gamma<a$). At the same time, the positive ("flow") charges are static in the moving frame, and their distance is stretched ($\gamma a>a$). The factor $\gamma$ is a number greater than 1.

Finally, the negative ("lattice") charges are contracted with respect to the positive ("flow") charges ($a/\gamma<\gamma a$) in the frame of the positive ("flow") charges. And the positive ("flow") charges are not contracted with respect to the negative ("lattice") charges in the frame of the negative ("lattice") charges, they do have the same spacing ($a=a$).

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Your notation is very complex. Regarding the charge of electron current in wire, only lab frame with still wire, test charge at rest and moving electrons in the wire are interesting. What is your resolution? Will it be charged? Now, why cannot you just accelerate all electrons once and let them moving themselves, by inertia? Otherwise, resistence forces are missing in your picture, which complicate it even further. So, electrons are moving by inertia, without any current source. Ok? What are implications of that? –  Val May 8 '13 at 9:48
    
The notation should be familiar for you if you've read the Feynman lectures chapter 25 "Electrodynamics in Relativistic Notation". The only difference is that I put $\mu$ to the superscript instead of subscript - which is usual for other textbooks. (By rumour, Feynman personally disliked the superscript/subscript differences - and still did perfectly accurate calculations.) –  firtree May 8 '13 at 10:05
    
For the accelerated electrons, the answer would depend on the particular mode of acceleration, that is up to you to choose. Electrons are separate particles, and do not make up a single solid body, so they can be set in motion being denser or scarcer, and then move by inertia without any changes. So the wire can be either uncharged, or charged positive or negative, depending on your choice. But, as long as you fixed that choice for the lab frame, the charge would become well-defined in any other frame as well. –  firtree May 8 '13 at 10:17
    
Do you understand that my question was "why we see no charge increase despite density is increased?"? What our ability to complicate and compute has to do with this question? –  Val May 8 '13 at 10:34
    
Sorry, probably I didn't understand. Could you please make the question more clear? "Why we see..." - in what conditions? How do you define the mentioned values? Then how do you calculate them? Actually, in a closed loop the density of moving electrons is not increased. –  firtree May 8 '13 at 10:41
up vote 1 down vote accepted

I guess that when electrons start moving, their volume is condensed. So, though charge per volume is increased, the net charge conserves. If this is right, we can answer the next question: how it is possible to condense the electrons, without condensing the loop? Why people keep talking about source of current? Why space contraction is a bad compensation for the density increase?

I think that I have got the right and comprehensible answer is done in the comment, http://physics.stackexchange.com/questions/63534#comment128575_63774. Basically, inter-electron distances expand as they are accelerated. This expansion exactly compensates the Einstein contraction so that inter-electrons distances do not change in the lab frame and wire sustains neutrality. This is not easy realize after you are drilled the fact that objects are contracted when accelerate. This is true only for rigid bodies because they sustain their size in their proper frame and, thus, do not expand as the repelling electrons do.

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