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Length contraction in a current carrying wire?

  1. Jul 30, 2009 #1
    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. jcsd
  3. Jul 30, 2009 #2

    Dale

    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.
     
  4. Jul 30, 2009 #3
    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.

    I'm still puzzled about this.
     
  5. Jul 30, 2009 #4

    Dale

    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?
     
  6. Jul 31, 2009 #5
    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.
     
  7. Jul 31, 2009 #6
    Why do you say that?
     
  8. Jul 31, 2009 #7
    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
     
  9. Jul 31, 2009 #8

    Dale

    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.
     
  10. Jul 31, 2009 #9

    Hans de Vries

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    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.
     
  11. Jul 31, 2009 #10

    Dale

    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.
     
  12. Jul 31, 2009 #11

    Hans de Vries

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    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
     
  13. Jul 31, 2009 #12

    DrGreg

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    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.
     
  14. Jul 31, 2009 #13
    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.
     
  15. Jul 31, 2009 #14

    Dale

    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.
     
  16. Jul 31, 2009 #15

    Hans de Vries

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    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
     
  17. Jul 31, 2009 #16

    Hans de Vries

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

    Regards, Hans
     
  18. Jul 31, 2009 #17
    ,
    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.
     
  19. Aug 1, 2009 #18
    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.
     
  20. Aug 1, 2009 #19

    Dale

    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
  21. Aug 2, 2009 #20
    @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?
     
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