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Maxwells equations from variational principle

  • Thread starter smallgirl
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  • #1
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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
[/tex]

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





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]

[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]

[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]

[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]

[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]

[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]

[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]

[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]

[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)}})
[/tex]


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...




 

Answers and Replies

  • #2
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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.
 
  • #3
vanhees71
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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.
 
  • #4
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ImageUploadedByPhysics Forums1379942433.147187.jpg


Done! Included the answer in case others are interested
 
  • #5
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Hey,

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]]
[/tex]

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...
 
  • #6
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Help please?
 
  • #7
vanhees71
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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]
 
  • #8
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ImageUploadedByPhysics Forums1380120127.984183.jpg


See picture: like this?
 
  • #9
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Hey,

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

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