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Elecromagnetic field of relativistic charge

  1. May 27, 2012 #1


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    Hi all,

    I'm trying to derive expressions for the EM-field of a charge moving at relativistic velocity.
    Let's make a rest frame for the charge in which it's velocity is equal to zero.
    The charge only produces an electric field with components:


    Where x' and y' are components of R'.

    Now to transform to another reference frame we have:

    [itex]E(y)=\gamma E'(y)[/itex]

    where [itex]\gamma[/itex] = gamma factor in relativity

    We will also have:

    [itex] x'= \gamma R cos(\theta)[/itex]
    [itex] y'=y=R sin(\theta)[/itex]
    [itex] R'=\gamma R {(1-\frac{v^2}{c^2}sin^2(\theta))^{(1/2)}}[/itex]

    Where [itex]\theta[/itex] is the angle between R and x.

    Putting this in above equations will give:

    [itex]E(x)= \frac{kq(1-\frac{v^2}{c^2})}{R^3{(1-\frac{v^2}{c^2}sin^2(\theta))^{(1,5)}}}R cos(\theta)[/itex]
    [itex]E(y)= \frac{kq(1-\frac{v^2}{c^2})}{R^3{(1-\frac{v^2}{c^2}sin^2(\theta))^{(1,5)}}}R sin(\theta)[/itex]
    And the resulting electric field:
    [itex]E= \frac{kq(1-\frac{v^2}{c^2})}{R^3{(1-\frac{v^2}{c^2}sin^2(\theta))^{(1,5)}}}R[/itex]

    And the magnetic field can be obtained by:
    [itex] B=v\times\frac{1}{c^2}E[/itex]

    I doubt this is right because when I set [itex]\theta[/itex]=0 the electric field in the direction of motion is reduced by [itex]\gamma^2[/itex] even though E'(x) should be equal to E(x)?

    I would gladly add a sketch of the situation if it isnt clear enough.
    Thank you very much!
    Last edited: May 27, 2012
  2. jcsd
  3. May 27, 2012 #2


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    nos, You can find the field of a moving charge derived in most books on E & M, such as Jackson.
  4. May 28, 2012 #3
    Your field eqn's are correct - see e.g. http://www.physicsinsights.org/moving_charge_1.html. The apparent discrepancy owes to confusing that E'x = Ex holds, but only for the case that the field (aka observation) point is immersed in the same longitudinal field in both frames. In the unprimed lab frame, where the charge is in motion along x, an observer δx distant from the charge in the primed frame is by SR length contraction a distance δx/γ from that charge in the lab frame. As obviously no relative displacement has occurred in the primed frame, the field strength is thus the same in both frames. Remove to a displacement δx in the lab frame, and by the inverse square law, the field strength reduction you worry about is readily explained. And for the transverse case, one has to include magnetic interaction when it comes to compare the net force on a test charge. And btw I see you made essentially the same query some time back in another thread.
    Last edited: May 28, 2012
  5. May 28, 2012 #4


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  6. May 28, 2012 #5


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    Hi everyone,

    Thank you all for replying:D

    Q-reeus,in the link you posted they derive expressions for the electric field in the x-axis(direction of motion) and y-axis which are:

    [itex] E(x) = q\frac{\gamma x}{(\gamma^2 x^2+y^2)^{3/2}}[/itex]
    [itex] E(y) = q\frac{\gamma y}{(\gamma^2 x^2+y^2)^{3/2}}[/itex]

    Are these equations equivalent to?:

    [itex] E(r) = q\frac{1-\frac{v^2}{c^2}}{(1-\frac{v^2 sin^2(\theta)}{c^2}}\frac{R}{R^3}[/itex]
  7. May 28, 2012 #6
    Apart from the matter of the right constant, which depends on the system of units used, the mathematical form looks correct to me, and can be checked against eq'n (45) here:
    www.physics-caluniv.in/anirban-kundu/reled.pdf [Broken] (note v there is normalized wrt c)
    [EDIT: On closer inspection, I see you have changed what was a correct form in #1 to something not actually right in #5!]
    Last edited by a moderator: May 6, 2017
  8. May 28, 2012 #7


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    oh I'm sorry. I see it. my mistake. It should have been the denominator tot the 3/2 power. And I have checked the equations are equivalent. Thank you so much!
  9. May 29, 2012 #8
    My pleasure nos. Figured it was probably just a typo as unclosed parentheses suggested. One minor quibble with the form for E in #5 is the typesetting used for R/R3 shown as R/R3 there. I know it's understood to mean numerator is a vector and denominator is a scalar, but strictly speaking it's showing both there as vectors according to normal convention. In #1 you used a larger R in the numerator which gave some sense of differentiation, but normal practice would have numerator R as plain bolded, and denominator R3 as unbolded italicized. Apart from that you are in better command of using LaTeX than myself!
  10. Jun 1, 2012 #9


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    oh thanks. you're right. one last question though. in my derivation I have excluded the z-component of the field. if i include this component too, will it give the same results?
  11. Jun 1, 2012 #10
    As the coordinate system is Cartesian, with x the axis of relative motion, the symmetry is such that interchange of y with z yields the same magnitude for results e.g. |Ey, By| = |Ez, Bz|. However, when doing a cross-product, there is a cyclic permutation sign convention to observe, so if say vx×Ey = Bz, then vx×Ez = -By etc.
    [Wiki article explains that thing here: http://en.wikipedia.org/wiki/Cross_product ]
    Last edited: Jun 1, 2012
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