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Maxwells equations from variational principle

  1. Sep 22, 2013 #1
    1. Hey,
    I have to find Maxwells equations using the variational principle and the electromagnetic action:

    [tex]S=-\intop d^{4}x\frac{1}{4}F_{\mu\nu}F^{\mu\nu}[/tex]
    by using

    [tex]\frac{\delta s}{\delta A_{\mu(x)}}=0

    therefore [tex]\partial_{\mu}F^{\mu\nu}=0

    3. I have had a go at the solution:

    [tex]S[\varphi]=-\intop d^{4}y\frac{1}{4}F_{\mu\nu}F^{\mu\nu}

    [tex]-\int d^{4}y\frac{1}{4}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})[/tex]

    [tex]\frac{\delta s}{\delta A_{\mu(x)}}=\frac{\delta s}{\delta A_{\mu(x)}}\int d^{4}y\frac{1}{4}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})

    [tex]=-\frac{1}{4}\frac{\delta s}{\delta A_{\mu(x)}}\int2(\partial_{\mu}A_{\nu}\partial^{\mu}A^{\nu}-\partial_{\mu}A_{\nu}\partial^{\nu}A^{\mu}

    [tex]=\frac{1}{2}\int d^{4}y\eta^{\mu\alpha}\eta^{\nu\beta}\frac{\delta s}{\delta A_{\mu(x)}}(\partial_{\mu}A_{\nu}\partial_{\beta}A_{\alpha}-\partial_{\mu}A_{\nu}\partial_{\alpha}A_{\beta}

    [tex]=\frac{1}{2}\int d^{4}y\eta^{\mu\alpha}\eta^{\nu\beta}A_{\nu}\frac{\delta s}{\delta A_{\mu(x)}}(-\partial_{\mu}\partial A_{\alpha}+\partial_{\mu}\partial_{\alpha}A_{\beta})

    [tex]=\frac{1}{2}\int d^{4}yA_{\nu}\frac{\delta s}{\delta A_{\mu(x)}}(-\eta^{\mu\alpha}\eta^{\nu\beta}\partial_{\mu}\partial_{\beta}A_{\alpha}+\eta^{\mu\alpha}\eta^{\nu\beta}\partial_{\mu}\partial_{\alpha}A_{\beta})

    [tex]=\frac{1}{2}\int d^{4}y\frac{\delta s}{\delta A_{\mu(x)}}(-A_{\nu}\partial_{\mu}\partial^{\nu}A^{\alpha}+A_{\nu}\partial_{\mu}\partial^{\mu}A^{\nu})

    [tex]=\frac{1}{2}\int d^{4}yA_{\nu}\frac{\delta s}{\delta A_{\mu(x)}}(\partial_{\mu}\partial^{\nu}A^{\alpha}-\partial_{\mu}\partial^{\mu}A^{\nu})

    [tex]=\frac{1}{2}\int d^{4}x\frac{\delta A_{\nu(y)}}{\delta A_{\mu(x)}}(\partial_{\mu}\partial^{\nu}\frac{\delta A^{\alpha(y)}}{\delta A_{\mu(x)}}-\partial_{\mu}\partial^{\mu}\frac{\delta A^{\nu(y)}}{\delta A_{\mu(x)}})

    I don't know if what I have done is right... or not.... I've continued with the problem but it leads to the wrong answer...so yes I'd like help in checking what I've done so far...

  2. jcsd
  3. Sep 23, 2013 #2
    Don't you think that variation will also be considered with respect to a term like ∂αAβ.This will finally give you lagrange's eqn from which you can get the eqn. you desire.
  4. Sep 23, 2013 #3


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    The trick is to realize that
    [tex]\delta [F_{\mu \nu} F^{\mu \nu}]=2 F^{\mu \nu} \delta F_{\mu \nu} = 4 F^{\mu \nu} \delta (\partial_\mu A_{\nu}).[/tex]
    Further you can use
    [tex]\delta \partial_{\mu} A_{\nu}=\partial_{\mu} \delta A_{\nu},[/tex]
    because in Hamilton's principle the space-time variables are not varied but only the fields (potential).

    The rest is partial integration to get the variation of the action functional.
  5. Sep 23, 2013 #4
    ImageUploadedByPhysics Forums1379942433.147187.jpg

    Done! Included the answer in case others are interested
  6. Sep 23, 2013 #5

    So modifying the equation so that it now reads:

    [tex]S=\intop d^{4}x[-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+(\partial_{\mu}-ieA_{\mu})\phi(\partial_{\mu}+ieA_{\mu})\phi^{*}-V\left[\phi\right]]

    How would I need to modify my approach? I figure I can just use the results from above for the first term in the integral...but not sure what to do for the next set of terms...
  7. Sep 25, 2013 #6
    Help please?
  8. Sep 25, 2013 #7


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    It's pretty simple. Just vary the addional interaction terms wrt. [itex]A^{\mu}[/itex]. Then you'll get the em. current of the KG field (at presence of the em. field, of course!) via
    [tex]\partial_{\mu} F^{\mu \nu} = j^{\nu}.[/tex]
  9. Sep 25, 2013 #8
    ImageUploadedByPhysics Forums1380120127.984183.jpg

    See picture: like this?
  10. Sep 26, 2013 #9

    So I have got this far.... ImageUploadedByPhysics Forums1380197837.615626.jpg
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