# Length contraction in a current carrying wire?

1. Jul 30, 2009

### maartenrvd

Reading http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Length_Contraction I reminded something that always bothered me about this explanation: I would think that a current carrying wire would only appear electrically neutral when the observer is moving along the wire with half the speed of the electrons, and not when it is stationary relative to the protons in that wire.

Is this correct or am i missing something?

2. Jul 30, 2009

### Staff: Mentor

Hi maartenrvd, welcome to PF!

It is a measured fact that the wire is electrically neutral in the lab frame. As the link says "A current-carrying wire is observed to be electrically neutral in its rest frame" (emphasis added). The charge densities in other frames have to conform to that fact.

3. Jul 30, 2009

### maartenrvd

Then how is that possible?
I would assume that a stationary observer would see length contraction for the free electrons and not for the stationary positive charge in the wire. To me this should logically result in a net observed electric field.
However, since the velocities are very small this would only have a marginal effect.

4. Jul 30, 2009

### Staff: Mentor

Again, the fact that the wire is neutral and has no E field in the lab frame is observed. It is a fact. It is under experimental control. Consider it like an initial condition or a "given" in the problem. It has nothing to do with relativity.

What relativity explains is: Given the fact that a wire has a current in the lab frame and given the fact that a wire is neutrally charged in the lab frame, then what does it look like in other reference frames?

5. Jul 31, 2009

### maartenrvd

I appreciate the effort but your answer doesn't help me: I did not ask the question to just accept facts but to understand them.

My logic says otherwise so if anybody can show me the error I would appreciate that.

6. Jul 31, 2009

### vin300

Why do you say that?

7. Jul 31, 2009

### vin300

With relativistic considerations or not, the number of electrons on the conductor does not increase. They are only replaced and without a net charge there is no E field

8. Jul 31, 2009

### Staff: Mentor

Your logic is wrong because you think that you should be able to derive a given or a boundary condition.

Look, a scientist is sitting in a lab. They can drive a current through a wire (by connecting one end to the positive terminal and the other end to the negative terminal of a battery) and/or they can charge the wire (by connecting both ends to the positive terminal of a battery). It is something that they can choose arbitrarily at their whim (within the limits of their apparatus).

The question is: If the scientist chooses to have a current-carrying uncharged wire, then what does that wire look like in other frames? You cannot derive the fact that it is uncharged, that is simply part of the specification of the problem.

9. Jul 31, 2009

### Hans de Vries

You can find the math here:

section 2: Neutral wire condition from the relativistic charge field.
http://physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf

The electron density is equal to the ion density for a neutral current carrying wire.

Regards, Hans.

10. Jul 31, 2009

### Staff: Mentor

Hi Hans,

I was looking at the derivation. For equation 1 why did you set the magnetic force equal to the electric force? They are not equal in general.

11. Jul 31, 2009

### Hans de Vries

It's the derivation that the magnetostatic force is a relativistic effect of electrostatics,
that's why they have to be the same. It assumes that the wire is neutral in the rest frame.

Somebody did place a link to this paper at Wikipedia a long time ago
http://en.wikipedia.org/wiki/Relativistic_electromagnetism
and it seems that the derivation is published recently here
http://dx.doi.org/10.1119/1.3098206 by somebody who
came independently to the same conclusion.

Regards, Hans

It was first discussed here: https://www.physicsforums.com/showthread.php?t=133587

12. Jul 31, 2009

### DrGreg

I think you probably have a misunderstanding of what Lorentz contraction is. I think you have the idea that whenever you have two objects A and B that are initially stationary in some frame, and you then accelerate both objects, then the distance between A and B automatically shortens relative to the original frame. Well, the shortening isn't automatic. It all depends on what the relationship between A and B is. If there is a reason why the distance between A and B in the new frame must be equal to the original distance in the original frame, then, yes, their distance in the original frame will decrease.

But in this case there is no reason for that. In fact, we already know that the charge remains neutral, in the wire's rest frame, which means that the average separation of moving electrons must the the same as the average separation of (almost) stationary electrons. Lorentz contraction therefore tells us that in the frame in which the "moving" electrons are at rest, the electrons must be further apart.

13. Jul 31, 2009

### Per Oni

wikipedia:
Hans de Vries, end of part 2:
Both statements defeat the object of explaining magnetism as a result of relativity. According to relativity there should occur a length contraction in a moving train in the direction of movement. Charge per unit length should increase for moving ions according to relativity.

Which takes me to a related subject. I read here, I think it was in this forum, that in particle accelerators just because of length contraction it is possible to push in more ions in the direction of travel the higher the ionic speed.
However I it could be I misread or the post was wrong.

14. Jul 31, 2009

### Staff: Mentor

But they do not have to be the same in general. In fact, as you say, the wire is neutral in the rest frame meaning that the electric force is 0 (and the magnetic force is not 0). Your equation certainly doesn't hold for the general case, I believe it only holds for one specific reference frame.

15. Jul 31, 2009

### Hans de Vries

First of all. The effect at low speed is almost all due to non-simultaneity
rather than to due to Lorentz contraction, The magnetic force is linear in
the velocity while Lorentz contraction is approximately quadratically
dependent on the velocity.

Secondly, The wire is neutral in the restframe. Which means that the
electron density is equal to the ion density. This taking into account
that the fields of moving electrons are transformed. The proof of this
is given in section 2. Remember that the electrons drift all at different
velocities, which are also constantly changing.

More over moving relativistic fields can be found here:
http://physics-quest.org/Book_Chapter_EM_LorentzContr.pdf

Regards, Hans

16. Jul 31, 2009

### Hans de Vries

That's of course correct. The electric force is in the rest frame of the test charge.

Regards, Hans

17. Jul 31, 2009

### Per Oni

,
Which effect are you pointing at?

True but what is the reason you mention this?

In section 2 you assume that the electrons have the same density as the ions at all speeds and then you go to great lengths to prove that therefore the electrical fields around the wire are the same.

We are talking about macroscopic effects so this statement serves no one.

26 pages? I've put forward a simple question.

Regards Willum.

18. Aug 1, 2009

### maartenrvd

Because only than the length contraction for the negative charge is the same as the length contraction for the positive charge.

@Hans

How do you come to the assumption that the electrons are evenly spaced (as seen from a moving observer). Would that also be the case for neutral particles? (and please do not assume that the speed of the electrons can be neglected just because it is very low).

I agree: that is the reference frame I am talking about. That is also why the force on the test-charge has a pure electrostatic and not a magnetic origin.

When I read this it exactly confirms my statement about the observed electric field around a current carrying wire: the electrons show contraction (at least the electrons closes to the test-charge) and the protons don't.
I don't see the difference of a test-charge being stationary relative to the 'moving' electrons or being stationary relative to the protons in the wire. Except of course that the polarity changes. As seen from the test-charge there is no magnetic force. So there has to be an e-field to create the force on the test-charge.

About experimental confirmation: This observed E-field would be quite small. By approximation derived from simple induction formula e=vxB * l, about 1 micro volt per cm for a local magnetic field strength of .1 tesla and an electron drift of 1 mm/s. You would have a hard time measuring this and distinguish this from induction effects.

19. Aug 1, 2009

### Staff: Mentor

Your calculations are way off. You didn't specify the configuration, but lets say we are measuring the E and B field 1 cm from a long straight wire.

The B-field is given by http://hyperphysics.phy-astr.gsu.edu/HBASE/magnetic/magcur.html#c2. For a B field of .1 T at a distance of 1 cm requires a current of 5 kA.

5 kA/(1 mm/s) gives a linear charge density of 5E6 C/m. Now, for the calculation of the E-field we only care about the excess charge. The Lorentz factor for v = 1 mm/s is γ-1 = 1.1E-16. That would give an excess charge of 5.6E-10 C/m (that is if there really were excess charge in the lab frame due to Lorentz contraction from the electron's rest frame).

The E-field is given by http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/elecyl.html. For a charge density of 5.6E-10 C/m and a distance of 1 cm we get an E-field of not just 1 uV/cm as you stated but about 1E7 uV/cm, or 1000 V/m. This would be easily detectable (if it were to happen).

Last edited: Aug 1, 2009
20. Aug 2, 2009

### maartenrvd

@Dalespam
This would indeed be the case when the whole wire would show the calculated length contraction, but that only counts for the part that is orthogonal to the observer. Then a small part of the wire shows length contraction adding to the observed field and the rest shows length "extension".

It would take me considerable time do the math but the overall observed E-field would be much less that what you mentioned and would indeed have to be the same as what I said it should be.
The reason that I used the that simple formula is just because of what I earlier mentioned: what is the difference of being stationary to the negative charge as being stationary to the positive charge in a wire (except that the polarity will change). For example: moving along the mentioned wire with 1 mm/sec and thus being stationary to the moving electrons, would produce a E-field of 1uV/cm, the other way around when being stationary to the protons would have to produce an E-field of the same magnitude but in opposite direction.
I still haven't heard conclusive reasons to assume otherwise. So again: what is wrong with this reasoning?

21. Aug 2, 2009

### Per Oni

Hi all.
Sometime ago I started a similar post so I’m quite interested in this one.
My understanding of the problem is this:

It is a fact that a stationary (dc) current carrying wire has no electric field around it (ignoring the one which causes the current to run in the first place).
The electric field we are interested in here is zero.

For me this is a paradox because according to relativity a length contraction takes place in a continuous line of moving charges. Then using Gauss’s law there should be an increase in the E-field. However, as stated E=0.

To solve this problem people have suggested that, when a current flows the line of charges spread out length ways and get contracted again because of relativity so that we end up with exactly the same charge density before and after.

I’ve got various reasons to doubt that explanation, for one: there’s no reason why these electrons should feel a force which spreads them out.

Now look at chapter 12.3.1 of Griffiths, adapted from Purcell. There’s a straight forward solution to this problem. First of all they accept that length contraction takes place in the normal relativistic way, both for stationary and moving wires. No spread out theory’s required.

OK, I know they cheated, they are using a +ve and a –ve current to balance things out. However I've got a feeling that they didn't like the spread out theory either. So here we are, for me the paradox goes on…….

22. Aug 2, 2009

### Staff: Mentor

No, it "counts" for the part which is parallel to the velocity. Since I specified a straight wire that implies that it "counts" for the whole wire.

I don't even know what you mean by "orthogonal to the observer". Usually an observer is considered to be a point, and there is no such thing as orthogonal to a point.

Then I suggest you take the time to do so, your present calculation is way off as I already demonstrated. Also, it appears that your simple formula is dimensionally inconsistent (the units on the left don't match the units on the right).

23. Aug 2, 2009

### Staff: Mentor

I don't understand your confusion here. You know the fact, that is it, end of story.

Do you understand what a boundary condition is and why boundary conditions are given instead of derived?

Here is an example that may help. Say you are doing a projectile problem. You can use Newton's laws of motion and a uniform force of gravity to derive a general form showing that projectiles travel in parabolic paths. But that, by itself, is insufficient to describe the path of any specific projectile. For that you need to provide two boundary conditions, typically an initial velocity and position. Say now that you are given that the projectile is initially launched horizontally at 100 mph from a height of 6 ft above the ground. You cannot go back and use Newton's laws of motion and a uniform force of gravity to explain why the projectile was launched horizontally. There is no explanation for it, it is simply a given boundary condition. It could have been some other value; the guy launching the projectile simply made a choice that he wanted to launch it horizontally. You cannot use Newton's laws to explain it, but given that condition you can say what will happen at other times.

Similarly here. Relativity describes the relationship in different inertial reference frames, but you have to start out with a complete description in one frame first. That complete description is the boundary condition. There is no relativistic reason for it, and you cannot use relativity to explain it. In this case, the current and the charge on the wire in the lab frame. The guy running the lab simply made a choice that he wanted an uncharged current-carrying wire; it could have been some other value but that was his choice. You cannot use relativity to explain it, but given that condition you can say what will happen in other reference frames.

Last edited: Aug 2, 2009
24. Aug 3, 2009

### Per Oni

No problem so far.

That was’nt his choice. He could have charged the wire if he wanted to by using the capacitance of this wire wrt another object or even wrt earth but we are not interested in any such charges. Ignoring those he cannot produce a charged wire simply by sending a current through. Nature therefore determines the neutrality of that wire and not some guy. Nature determines this boundary condition.

I'm just one of those guys who wants to know why nature behaves as it does.

25. Aug 3, 2009

### Staff: Mentor

This is still precisely analogous to the projectile problem. You could say:

"That was’nt his choice. The projectile could have been launched vertically if he wanted to by elevating the barrel of the launcher, but we are not interested in any such elevation. Ignoring those he cannot produce a vertically launched projectile simply by using a different speed of projectile. Nature therefore determines the horizontal initial velocity and not some guy. Nature determines this boundary condition."

In any case, I would disagree in both cases with the statements "That was’nt his choice" and "Nature determines this boundary condition" since in both cases there is something that the researcher could choose to do to obtain a different boundary condition. The fact that you are not interested in that change is also an arbitrary choice, determined by you and not nature.