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Does J go to inf when v goes to c?

  1. Jul 10, 2009 #1
    In SR, does the current become infinite when the velocity of the charge approaches c?

    I'm thinking it doesn't because of the SR Lagrangian for a charged particle:
    [tex]L = -\frac{m c^{2}}{\gamma} - q \phi + q \vec{v} \cdot \vec{A}[/tex]
    doesn't have a gamma in the qv term.

    Another way of putting this is: Does the four current equal the charge times the four velocity?

    Thanks for your responses.
  2. jcsd
  3. Jul 11, 2009 #2


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    Hi GiftOfPlasma! :wink:
    Yes. :smile:
  4. Jul 11, 2009 #3
    So the Lagrangian should be:
    [tex]L = -\frac{m c^{2}}{\gamma} - q \phi + q {\gamma} \vec{v} \cdot \vec{A}[/tex]

  5. Jul 11, 2009 #4
    No, it shouldn't. The original one is ok, but you've probably lost [tex]c[/tex] somewhere. I do not know what you mean by 4-current... 4-current density? If so recall that the volume is also contracted is SR.
  6. Jul 11, 2009 #5
    Ok, I think I see what is going on, the original Lagrangian is for a charged particle.
    The classical particle is always a point, so it doesn't contract.
    Furthermore, the number of particles in a volume is the same in all reference frames.
    The volume is contracted with respect to an observer moving relative to the volume.
    So the relativistic observer sees a higher charge density and current density than a non-relativistic observer.

    I found a good explanation here:
    http://farside.ph.utexas.edu/teaching/jk1/lectures/node16.html" [Broken]

    Thanks for the effort.
    Last edited by a moderator: May 4, 2017
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