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Deriving the EOM for Proca Lagrangian

  1. May 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider the Proca Lagrangian

    [tex]L=-\frac{1}{16\pi}F^2-\frac{1}{c}J_{\mu}A^{\mu}+\frac{M^2}{8\pi}A_{\mu}A^{\mu}[/tex]​

    in the Lorentz guage [tex]\partial_{\mu}A^{\mu}=0[/tex]

    Find the equation of motion.


    2. Relevant equations

    [tex]F^2=F_{\mu\nu}F^{\mu\nu}[/tex]


    3. The attempt at a solution

    Well, first of all I'm not quite sure how E-L should look in this case. Clearly [tex]F_{\mu\nu}[/tex] is the part that is the derivatives of the dependent variables (A).
    What I have gotten to is
    [tex]\frac{\partial L}{\partial F_{\rho \sigma}} = -\frac{F^{\rho\sigma}}{8\pi}[/tex]
    [tex]\frac{\partial L}{\partial A_{\sigma}}=-\frac{1}{c} J^{\sigma} + \frac{M^2}{4 \pi}A^{\sigma}[/tex]

    So is this the E-L for 4D special relativity?
    [tex]\frac{\partial L}{\partial A_\sigma}-\partial_\sigma \left( \frac{\partial L}{\partial F_{\rho\sigma}} \right)=0[/tex]​

    Thanks for any help.
     
  2. jcsd
  3. May 23, 2009 #2

    George Jones

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    I would more comfortable not taking a short-cut, i.e., I would expand the F's in term of A's, but, after substituting for the L's in your final equation, you seem to get the correct equation of motion.
     
  4. May 23, 2009 #3
    Thanks so much for the quick reply.

    What I'm mostly unsure about is how to formulate E-L in terms of covariant or contravariant components. Do I take the derivative with respect to [tex]A^\sigma[/tex] or [tex]A_\sigma[/tex].

    I guess my question could be reformulated as:

    Should E-L look like
    [tex]\frac{\partial L}{\partial A_\sigma}-\partial^\sigma \left( \frac{\partial L}{\partial F_{\rho\sigma}} \right)=0[/tex]​
    or
    [tex]\frac{\partial L}{\partial A^\sigma}-\partial_\sigma \left( \frac{\partial L}{\partial F^{\rho\sigma}} \right)=0[/tex]​

    And based on what do I choose between the two?
     
    Last edited: May 23, 2009
  5. May 23, 2009 #4

    George Jones

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    Neither, :smile:. In each equation, [itex]\sigma[/itex] is a free index in one term and a dummy summed index in the other term, which is a no-no. In terms of index placement, an upstairs index in a denominator is like a downstairs index in a numerator, and a downstairs index in a denominator is like an upstairs index in a numerator.
     
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