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Why is a wire with a current not charged?

  1. May 12, 2006 #1
    I just read about how a magnetic field for a (neutral) wire with a current can be viewed as an electric field in a moving frame of reference because of lorentz contraction.

    I now wonder why starting with a neutral wire without current and stationary electrons a charge is not produced when a current is send through the wire and the electrons move.

    The argument should be that the charge density of the moving electrons is increased by the lorentz contraction of the "electronic fluid". Is it because the extra charge is removed by the battery when the current is turned on or is my argument simply wrong?
  2. jcsd
  3. May 12, 2006 #2
    the wire stays neutral because the same amount of electrons you pump into it is the amount of electron you pump out of it... inside the battery theres a current, you can imagine your wire as a ring with a potential gradient.
    the number of electrons in the wire is constant, it doesnt matter whether they move or not, and in what direction for that matter.
  4. May 12, 2006 #3
    Do you agree that the charge density of the electrons is increased when they move because of the lorentz contraction? Your answer does not answer why this increase in charge density does not produce a net charge density in the wire.

    I am well aware that the same amount of electrons is injected and removed from the wire in steady state. My talk about the battery was just one possible explanation and I dont know if it is true.
  5. May 12, 2006 #4


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    Basically, Gauss's law guarantees that there is a conserved charge for the wire, which does not depend on the frame of refrence. This is true because the wire, as a whole, is an isolated system. Thus if you consider a loop of wire carrying a current, it will have the same net charge in a boosted (i.e. moving) frame of reference as in a stationary frame of reference.

    However, Lorentz contraction does have an effect on the charge distribution.

    Code (Text):

    x      x
    x      x
    Consider the above loop of wire, assumed to be of zero resistance, carrying a current. In the frame of the wire, the electric field is zero everywhere.

    In a frame moving with respect to the wire, this is not true. If we boost the frame of reference so that the loop is moving up, Lorentz contraction changes the spacing of the electrons and protons.

    More specifically, if the electrons are moving up on the right side of the wire, if we boost "up" to match the velocity of the electrons, they move further apart, while the protons get closer together. Hence we get a positve charge on the right hand side of the loop.

    But on the left-hand side of the loop, the electrons and the protons both get packed more tightly. This effect is larger for the electrons, so we get a negative charge on the left side of the loop.

    Gauss's law says that the enclosed charge is equal to the intergal of the electric field over some enclosing surface. When we perform the intergal over a surface enclosing the entire loop, we find that the net charge remains zero - the contributions of the electric field on the right side of the loop to the intergal are always balnced out by the opposite electric field on the left side of the loop.

    Charge conservation also says that if you have a loop of wire that is intially uncharged, and a battery that has zero net charge, when close the circuit to start a current flowing, that it remains uncharged.

    You can create a charged loop of wire by charging it up with an external source (say an electrostatic generator) and keeping the wire isolated from ground. When you do that, connecting a battery to make a current flow will still not change the value of total charge, which is still set by Gauss's law, as long as you keep the system isolated from ground.
    Last edited: May 12, 2006
  6. May 12, 2006 #5
    Thanks for the answer. I agree with you about what happens when we change our velocity with respect to the current carrying wire: The charge distributions change.

    But my question is not about what happens when we change our velocity. I am wondering about what happens when we make the electrons move: If the charge density of moving electrons is higher that that of non-moving electrons (all other things being equal) then why do we not see a negative charge density on all four sides of the loop when the current is on?
  7. May 12, 2006 #6


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    Consider the system of the battery + loop of wire + switch. Let the switch be open.

    By Gauss's law, the above system has a total net charge. The charge can be computed by enclosing the system in a surface and integrating the electric field perpendicular to the surface.

    See for example

    When one closes the switch, charges move, but no charge crosses the enclosing boundary.

    Therfore the total charge of the system, the charge enclosed by the boundary, remains constant before and after the switch is closed.

    This is obviously incosistent with having all 4 sides of the loop having a negative charge density - Gauss's law forbids this from happening.
  8. May 13, 2006 #7
    Good argument about the total charge. I am still wondering what is wrong with the argument:

    If the charge density of the ions in the wire is ro_i and the charge density of the electrons is ro_e = - ro_i then assuming (and this may be wrong) that after current is turned on the charge density of the electrons IN THEIR OWN RESTFRAME is still ro_e, this gives a charge density of the electrons of ro_e_moving = ro_e/sqrt(1-v^2/c^2) in the rest frame of the wire (the lab system). This produces a net negative charge density in the lab system since ro_i + ro_e_moving < 0.
  9. May 13, 2006 #8


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    ro_e does not necessarily remain the same (and in fact it does not). The charges instead arrange themselves in such a way that in the rest frame of the conductor, the electric field in the interior of the conductor is zero (for a perfect conductor).
  10. May 13, 2006 #9
    Thanks. So the argument is right but my assumption was not. Is there any other way (besides using that the net charge density in the rest frame of the conductor is zero) to calculate the charge density of the electrons in their rest frame? So that the neutrality of the wire can be calculated directly?
  11. May 13, 2006 #10
    It all has to do with simultaneity. That's what's been missing from this thread so far. Let me explain.

    Consider a long straight neutral conducting wire (zero currentat first) lying on the x axis in frame S. Now let's get the electrons moving to form a current. As viewed in S let us accelerate all the electrons with the same value to a steady velocity. We start this acceleration of each electron at the same time as measured in S. If we get the electrons moving in this way then the electrons will be equally spaced in S and have their original spacing. Thus there is a current but no charge density. Now let us view this from S' which is moving in the +x direction. An observer at rest in S' determines that the electrons all have the same acceleration profile as each other (but not identical to those in S) but the electrons did not start moving at the same time as measured in S'. Therefore the charges as viewed in S' will bunch up on each other while the + charges will not bunch up since they remain unaccelerated. So the end result is a constant current in each frame and in S' a uniform charge density but zero charge density in S. This page may help you visualize some of this


    I made it for another reason but it should be of some help to you.

  12. May 13, 2006 #11
    Thank you very much. Thats exactly the argument I was missing for why the charge density of the electrons sholud be different in S' (after the current is turned on) from what it was in S.

    Just a minor thing: Since the charge density of the electrons is lower in their rest frame than in the lab frame I would think that the electrons will not bunch up but rather be spread out.
  13. May 13, 2006 #12


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    I would not expect all the electrons in the wire to move "at the same time" when the switch is closed.

    If we include a ground-plane, we can probably model the wire as a transmission line, and from that we can deduce the motion of the electrons from the known solution for currents and voltages in a transmission line.

    In a transmission line, the voltage and current should propagate through the line at the speed of light. (Slower if there are non-vacuum dielectrics involved).

    If we terminate the transmission line in its characteristic impedance, we won't have to deal with "bounces" (i.e. reflections at the end of the transmission line) and it will make for the cleanest thought experiment.

    For such a terminated transmission line, the end result is that an electron at a distance 'x' away from the switch should acquire, very rapidly, its drift velocity v after a time delay of x/c, as measured in the laboratory frame.

    In the crude model above, the acceleration is impulsive and the pulse is a square-wave. A more realistic model could include the rise-time of the pulse, this would give finite values for the rate of change of current which would be proprortional to acceleration.
    Last edited: May 13, 2006
  14. May 14, 2006 #13
    That was obviously not an exact description of a circuit by rather a thought experiment which takes one to the final charge/current configuration.

  15. May 18, 2006 #14
    That is indeed the case. Consider two of the electrons initially at rest in S, one at x = +L/2 and one at x = -L/2. At t = 0 they start to accelerate then come to a constant speed each with identical velocity profiles. Note: These are two simultaneous events seperated by a distance LThe charge density did not change in S. Consider the two events

    Event 1: spacetime coordinates (t = 0, x = L/2) = electron at x = +L/2 starts to accelerate

    Event 2: spacetime coordinates (t = 0, x = -L/2) = electron at x = -L/2 starts to accelerate

    Now view these events from the frame S' which is moving in the +x direction with speed v. Determine the sequence of the events as viewed from this frame by considering the time ordering of the events. From the transformation equation for time

    [itex] t' = \gamma(t - vx/c^2)[/itex]

    The event (t = 0, x = L/2) has spacetime coordinates in x' of

    [itex]t' = -\gamma vL/{2c^2})[/itex]

    The event (t = 0, x = -L/2) has spacetime coordinates in x' of

    [itex]t' = +\gamma vL/{2c^2})[/itex]

    So event 1 occurs before event 2 as viewed from an observer in S'. The electron on the + side the x' axis moves first while the electron at the - side of the x' axis sits there for a while and then starts to move with the same velocity profile as the first, just delayed in time. Thus the electrons will be more spread out as viewed from frame S.

    I hope that helps.


    ps - Sorry for not getting back sooner on this. Personal problems got in the way.
    Last edited: May 18, 2006
  16. May 18, 2006 #15
    Good explanation, Pete, how the lack of simultaneity changes the charge density viewed from the S' frame. Your site (given in your 1st post) also shows that concept in an alternative way clearly in the beginning (up to eqn. # 6).

    However, on that link you begin to try to explain the difference in charge density in a moving circuit (after eqn. # 8) in figure 5.

    Here you simply makes the figurative statement without proof that the top of the circuit becomes more negatively charged and the bottom more positive. That seems a bit ambiguous since you have now chosen a frame that is neither S or S' ? How do you justify that conclusion?

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