# Why no electric field from a current in a wire? Length contraction.

1. Jan 27, 2014

### maajdl

I found a funny video on youtube, but I am not totally convinced by the argument.

It says that a pure magnetic field caused by a current in a wire can lead to a combined electric + magnetic field in a moving frame. It explains more or less convincingly that this can be understood as a consequence of length contraction observed in the moving frame. The contraction would break the neutrality of the wire.

I do not understand why the electronic current would not be contracted in the rest frame and why it would not also lead to an electric field. At time 1:35, the speaker says that positive and negative charges would be the same. Why?

Last edited by a moderator: Sep 25, 2014
2. Jan 27, 2014

### Staff: Mentor

That is the given initial condition, it is under experimental control. We are given that there is a current and no net charge on the wire in the lab frame. Then we transform to another frame to get the situation in that frame.

3. Jan 27, 2014

### maajdl

And how can that ubiquitous experimental fact be explained?
It seems also that in frame where the electrons are at rest, there would be then an electric field,
while this situation is similar to the original situation except for which charges are in motion.

4. Jan 27, 2014

### pervect

Staff Emeritus
Maxwell's equations are perfectly consistent with special relativity. In fact, the Lorentz transformations are just those transformation that leaves Maxwell's equations invariant.

And Maxwell's equations say that charge can't be created or destroyed, that the integral of the normal component of the electric field is zero.

So if you imagine a battery and a wire loop , it *could* be set up to have a charge, but it is presumed that the experimenter got rid of any such total charge (for example, by briefly touching a ground wire to the apparatus) before starting the experiment.

If you were asked to analyze the problem in terms of Maxwell's equation, I don't think you'd have any problem in saying that an uncharged wire was uncharged and didn't have any electric field initially, nor would you have any problem with the more subtle statment that in the lab frame, if the wire was uncharged before you connected it to the battery, it remained uncharged.

The surprising result is that in a moving frame you generate an electric dipole where one half of the wire becomes charged + and the other half becomes charged minus. But this doesn't violate maxwell's equations - total charge is conserved. In fact, the result is a result of Maxwell's equations.

5. Jan 27, 2014

### Staff: Mentor

Easy, experimenters ubiquitously choose not to charge the wires in their frame.

Yes. But the experimenters choose to have their wires uncharged in the protons frame.

There is simply no deep explanation for this, it is entirely a matter of experimental choice not to charge the wire in the lab frame. If you wanted to charge the wire in the lab frame then you could easily use the same process to find the resulting charge and current in other frames. Just specify the conditions in one frame and transform to any other frame. Here they are merely specified to be uncharged in the lab frame. No big deal.

Last edited: Jan 27, 2014
6. Jan 28, 2014

### maajdl

I feel I have no choice.
If I connect a battery to a wire, the charge on the wire will automatically be determined.
Why should it be zero?
I don't see how I could act separately on charge and current.
But I accept that as a convenient way to discuss the relation between relativity and the transformation of E and H fields.

Otherwise, I have no difficulty with SR and the Maxwell's equations.
I rather have difficulties explaining it to my grand mother.

Last edited: Jan 28, 2014
7. Jan 28, 2014

### Staff: Mentor

And the laws of physics offer you no other choice than to use a battery? No other devices are in existence which could offer a different choice regarding the charge on the wire?

Or are you simply stating your rejection of the concept of free will in general?

8. Jan 28, 2014

### maajdl

There is only one device: the wire.
An electric field drives the current.
Am I missing something?

Edit 1:
I got it!
If the battery + the wire are isolated, they could sustain a static charge.
In such a setup it is then indeed possible to chose both the current and the static charge.

Edit 2:
Now I ask myself what happens if the whole setup is isolated and neutral.
If I switch on a current in the wire, should this create a static charge due to length contraction?

Last edited: Jan 28, 2014
9. Jan 28, 2014

### A.T.

10. Jan 28, 2014

### A.T.

When the electrons start moving in the wire frame, their fields are contracted, but they still repulse each other, so their distances don't change. The same number of electrons still tries to spread out uniformly within the same length of the wire. Since the electron density doesn't change the wire stays neutral.

You cannot assume symmetry between electrons and cations here. The cations are fixed in a lattice, so their proper distances (distances measured in their frame) are fixed. The electrons are flowing freely and can change their proper distances, while keeping a fixed distance in the wire frame.

11. Jan 28, 2014

### Staff: Mentor

Excellent! You got it.

Basically in the lab you have two parameters to play with, specifically the voltage on each side of the wire. The average voltage determines whether or not the wire has a net charge (due to the self-capacitance of the wire), and the voltage difference determines whether or not the wire has a current.

You could think of attaching the negative terminal of the battery to a Van de Graaff generator. That would raise the negative terminal to a few hundred kV and the battery would raise the positive terminal a few V beyond that. The wire would then have both a current and a net charge in the lab frame.

If the whole setup is neutral then, by definition, it has no net charge. And if it is isolated then the amount of charge cannot change. So before and after the current is switched on the whole setup will remain neutral. Again, this is due to how the experimenter has chosen to set up the experiment.

12. Jan 28, 2014

### pervect

Staff Emeritus
The following picture of what a moving (relativistic term: boosted) current loop looks like might be of help.

The +++ and --- are charges, so a local excesss of electrons would be represented by ----. On the sides of the loop without any + or - symbols, the wires are locally neutral.

Note that when you consider a complete current loop, charge is conserved - it's just redistributed. You don't create or destroy any charge, but because you've changed the definition of simultaneity, the locations of the charges are no longer evenly distributed along the wire.

I'm not sure where you are with the math, the number-flux 4 vector or equivalently the charge density / current four-vector give you the mathematical tools you need to really analyze the problem - that's what I used to generate the diagram.