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Question about Lagrangian in electromagnetic interaction

  1. Aug 16, 2011 #1
    Sorry for a naive question.

    In EM textbook and QM path integral textbook, the action and Lagrangian in electromagnetic interaction are

    S = L dt = e(\phi – A v) dt ---equ.(1)

    But in QFT textbook, the action and Lagrangian density are

    S = L d^4x = A J d^4x ---equ.(2)

    As I understand, in equ.(2), J = \rho U = \rho \gamma V
    In which \rho is density, U is the 4-velocity=dx/d\tau, and V is the common velocity=dx/dt, \gamma is \sqrt (1-v^2/c^2).

    So equ.(2) will have a factor of \gamma, but equ.(1) does not have.

    So where is my mistake?
     
  2. jcsd
  3. Aug 16, 2011 #2

    bapowell

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    The 4-current is [itex]J^\alpha = \rho_0 u^\alpha[/itex], where [itex]\rho_0[/itex] is related to the charge density by [itex]\rho_0 = \rho/\gamma[/itex] and [itex]u^\alpha[/itex] is the 4-velocity, [itex]u^\alpha = \gamma(c,{\bf v})[/itex]. You'll find that the [itex]\gamma[/itex]'s cancel.
     
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