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Current through a wire and special relativity

  1. Jan 30, 2014 #1
    When I studied special relativity for the first time, I encountered the exemple
    of a infinite neutral wire in the laboratory frame of reference, R, through which a constant
    current is running. In this frame of reference, the electrons in the wire are moving
    with a velocity +v and the ions are stationary, the electric field is zero and there is
    a non zero magnetic field.
    Now, from the point of view of a frame of reference R' which is moving with a velocity
    +v relative to R, the electrons are stationary and the ions are moving with velocity -v,
    but the wire is not neutral because of Lorentz contraction : the ions are denser than
    in R and the electrons are less dense than in R. The wire is positively charged in R'
    and then the electric field is not equal to zero. This example illustrates how the magnetic
    field seen in R is a relativistic effect of the motions of charges.

    If this is coorect, I still have questions :
    1. The universe is neutral in R. If it had to be also neutral in R' and the wire is positively
    charged, where the negative charges needed to make the balance are located?
    2. Even though the electrons are moving in R, their density is considered equal to that
    of the ions which are not moving, in other words the Lorentz contraction has not been
    taken into account in R, why? Is there something special with this frame of reference? Is this related to the center of mass frame of reference?
  2. jcsd
  3. Jan 30, 2014 #2


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    Since the issue goes away if you consider a finite sized current loop, I've always assumed the problem was with the initial assumption that you had an infinitely long wire.
  4. Feb 1, 2014 #3
    1. Replace the infinite wire with a finite loop to solve that problem. The loop is a slightly harder but more realistic situation to deal with.
    2. The densities are equal in R because the problem states that the wire is neutral in this reference frame. That's part of the setup for the problem.
  5. Feb 1, 2014 #4


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    Lorentz contraction is taken into account. The E-fields of the electrons are contracted in R, but they are still repulsive, so the electrons still spread out evenly a keep a constant distance in R while they start moving.

    Here is a good diagram for closed loop:
  6. Feb 1, 2014 #5


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    It is a misconception from the very beginning that one assumes that there's no electric field outside a DC carrying wire, you find astonishingly often in the literature (even in the Feynman lectures!).

    For the complete calculation of the electric and magnetic field of DC conducting wires, see the very nice book

    http://www.ifi.unicamp.br/~assis/the-electric-force-of-a-current.pdf [Broken]

    There both the electric and magnetic field of stationary currents is calculated carefully. It's clear that any correct solution of the Maxwell equations can never lead to contradictions with the Poincare symmetry of SRT space-time, because they build a fully relativistic classical field theory.

    Additional violations of Lorentz symmetry that originate from the approximate non-relativistic treatment of the constitutive relations. E.g. usually, Ohm's Law is written as [itex]\vec{j}=\sigma \vec{E}[/itex] although the correct full equation is (in Heaviside-Lorentz units)
    [tex]\vec{j}=\sigma \left (\vec{E} +\frac{\vec{v}}{c} \times \vec{B} \right ).[/tex]
    The corresponding corrections are, however, totally negligible, because the electron drift velocities reached in usual household currents are tiny, even on everyday scales of velocities, let alone compared to the speed of light which is the relevant comparison here.
    Last edited by a moderator: May 6, 2017
  7. Feb 1, 2014 #6

    It should be possible to answer this question ... Let's see:

    If electrons are programmed to start their motion simultaneously in R, then in R' electrons will start their motion non-simultaneously.

    At one end of the wire, there are electrons shooting out of the wire at velocity +v. The electrons that have already left the wire form a cloud, it's an infinitely large cloud. The wire has been sprouting out electrons for an infinite time.

    At the other end the wire has lost all moving electrons, the last electron has moved a distance v*t, where t is a reasonable time, not infinite.

    That's how things are in frame R'.

    In frame R electrons start moving simultaneously at both ends of the infinitely long wire.
    Last edited: Feb 1, 2014
  8. Feb 1, 2014 #7


    Staff: Mentor

    In addition to jartsa's comments, the universe is not necessarily neutral in R'. You can prove that charge is both conserved and frame invariant, but the proof of its invariance has some important assumptions. One of those assumptions is that all charges and currents are 0 at spatial infinity. This assumption is clearly violated for an infinite current-carrying wire.

    See p. 94 here: http://www.phys.ufl.edu/~thorn/homepage/emlectures2.pdf
  9. Feb 2, 2014 #8
    Thank you all, your replies help a lot.
  10. Feb 18, 2014 #9

    I made a silly error. :eek: It's actually very simple:

    One end of the wire has been sprouting electrons for an infinitely long time, while at the other end the electrons will start their motion after an infinitely long time. (in the frame were the electrons do not start to move simultaneously)
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