Maxwells equations from variational principle

  • Thread starter smallgirl
  • Start date
  • #1
80
0
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
1,024
32
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
Science Advisor
Insights Author
Gold Member
2021 Award
20,129
10,867
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
80
0
ImageUploadedByPhysics Forums1379942433.147187.jpg


Done! Included the answer in case others are interested
 
  • #5
80
0
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...
 
  • #7
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
20,129
10,867
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
80
0
Hey,

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

Related Threads on Maxwells equations from variational principle

Replies
9
Views
3K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
3K
Top