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Field of a moving charge inside charge's own frame?

  1. May 22, 2015 #1
    We all know that in the lab frame observing a charge moving along x-axis, we see an electric field ##E = \frac{Q}{4\pi \epsilon_0 x^2\gamma^2}##, picking up an extra factor of ##\gamma^2## at the bottom.

    Assuming no external fields are present, what would the charge see in its own rest frame? It should see zero electric field, right? Since in the charge's rest frame, it only sees the lab frame moving and nothing else.
     
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  3. May 22, 2015 #2

    PeterDonis

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    I don't think the ##\gamma^2## factor in the denominator is right; shouldn't it be ##\gamma## in the numerator?

    Also, there would be a magnetic field, since there is a nonzero current density due to the moving charge. And the electric field would not be static; at any given point in the lab frame, it would be time-dependent, since the charge is moving so the distance to it varies with time.

    No, certainly not. In the charge's rest frame, you would see the static electric field ##\frac{Q}{4 \pi \epsilon_0 r^2}##, without the extra factor of ##\gamma^2## and with no time dependence. But you would not see a magnetic field, since the charge is not moving in this frame.
     
  4. May 22, 2015 #3
    I should be more specific. Charge is moving in S frame but not in S'.

    Sorry I meant ##E = \frac{Q}{4\pi \epsilon_0 x^2\gamma^2}##. Yes there is a factor of ##\gamma## at the numerator, but there is also a factor of ##\gamma^3## at the denominator due to change of coordinates ##x'= \gamma x##.


    So in the charge's rest frame, we see a stationary charge and ##E'_{||} = \frac{Q}{4\pi \epsilon_0 x'^2}##. The maths turns out right.
     
    Last edited: May 22, 2015
  5. May 22, 2015 #4

    PeterDonis

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    I'm still not seeing this; what you are saying does not seem consistent with how electromagnetic fields Lorentz transform. See, for example, here:

    http://en.wikipedia.org/wiki/Classi...rmation_of_the_fields_between_inertial_frames

    We are in agreement on how the field looks in the charge's rest frame: a static E field with magnitude ##\frac{Q}{4 \pi \epsilon_0 r^2}##, and no magnetic field. (Actually, we may not be in complete agreement, since as you'll note, I put ##r^2##, not ##x^2##, in the denominator. The static E field in the charge's rest frame is spherically symmetric.) But I don't see how you are getting from that to your expression for the E field in the frame in which the charge is moving. (Also, as I noted before, there is a B field in that frame; you haven't given a formula for that at all.)
     
    Last edited: May 23, 2015
  6. May 23, 2015 #5

    pervect

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    I don't know that. For instance http://web.hep.uiuc.edu/home/serrede/P436/Lecture_Notes/P436_Lect_18p5.pdf (and also wiki) give the expression that the perpindicular component of E gets multiplied by ##\gamma##, (as opposed to ##\gamma^2## which is your claim), while the parallel component of E doesn't change magnitude at all. There is also a magnetic field generated by a moving charge, and an electric field generated by a moving magnetic field.

    In it's own rest frame in flat space-time, there is only an electric field, given by coulomb's law.
     
  7. May 23, 2015 #6
    I'm referring to the parallel component..
     
  8. May 23, 2015 #7

    PeterDonis

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    In the charge's rest frame, there's no such thing as a "parallel component"; the electric field is static and spherically symmetric.

    In the frame in which the charge is moving, as pervect says, the component of the electric field parallel to its motion is unchanged from the charge's rest frame; only the component perpendicular to the motion changes.
     
  9. May 24, 2015 #8

    pervect

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    None of my saved links with pictures of the E-field of a moving charge are still up:(. http://farside.ph.utexas.edu/teaching/em/lectures/node125.html has the equations though. The equations are there, it'd be too annoying to type them in latex. Most of the variables are self explanatory, ##u_r## is a bit mysterious until you read the previous page, "potential due to a moving charge", http://farside.ph.utexas.edu/teaching/em/lectures/node124.html where where it is defined as the radial component of the velocity, thus when r points in the direction of the velocity v, the value is v, when r points orthogonal to v, the value is zero.

    The treatment uses tensors, unfortunately, but hopefully you can still find the results useful. I think Griffith's textbook has a non-tensor treatment.

    As far as the question of what happens for a motionless charge goes, when you substitue ##\gamma =1## and ##u_r = 0##, you get the field of a motionless charge, which is just what you'd expect, a coulomb field pointing in the r direction whose magnitude is proportional to the inverse square of the radial distance. In the notation used, that is

    (the usual constants) * ##\frac{\vec{r}} {r^3}##, r being ##| \vec{r} | ##. Note the web-text used ##\mathbf{r}## where I use ##\vec{r}##.
     

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