Maxwells equations from variational principle

[smallgirl]

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

$$S=-\intop d^{4}x\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$
by using

$$\frac{\delta s}{\delta A_{\mu(x)}}=0$$

therefore $$\partial_{\mu}F^{\mu\nu}=0$$





3. I have had a go at the solution:

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

$$-\int d^{4}y\frac{1}{4}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})$$

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

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

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

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

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

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

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

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


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...

 

[andrien]

Don't you think that variation will also be considered with respect to a term like ∂_αA_β.This will finally give you lagrange's eqn from which you can get the eqn. you desire.

 

[vanhees71]

The trick is to realize that
$$\delta [F_{\mu \nu} F^{\mu \nu}]=2 F^{\mu \nu} \delta F_{\mu \nu} = 4 F^{\mu \nu} \delta (\partial_\mu A_{\nu}).$$
Further you can use
$$\delta \partial_{\mu} A_{\nu}=\partial_{\mu} \delta A_{\nu},$$
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.

 

[smallgirl]

Done! Included the answer in case others are interested

 

[smallgirl]

Hey,

So modifying the equation so that it now reads:

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

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...

 

[smallgirl]

Help please?

 

[vanhees71]

It's pretty simple. Just vary the addional interaction terms wrt. $A^{\mu}$. Then you'll get the em. current of the KG field (at presence of the em. field, of course!) via
$$\partial_{\mu} F^{\mu \nu} = j^{\nu}.$$

 

[smallgirl]

See picture: like this?

 

[smallgirl]

Hey,

So I have got this far...