# Electromagnetic Field Lagrangian - Field Equations

1. Aug 15, 2013

### ProfDawgstein

I was working on an exercise in Ohanian's book.

[Appendix A3, page 484, Exercise 5]

I guess he means charge conservation, but wrote $j^\nu = 0$.

The Lagrangian was given by $\mathcal{L}_{em} = -\frac{1}{16\pi} \left( A_{\mu ,\nu} - A_{\nu ,\mu} \right) \left( A^{\mu ,\nu} - A^{\nu ,\mu} \right)$.

Which should be the same as $\mathcal{L}_{em} = -\frac{1}{16\pi} F_{\mu\nu} F^{\mu\nu}$.

The Euler-Lagrange Equation is $\frac{\partial}{\partial x^\mu} \frac{\partial \mathcal{L}}{\partial A^{\nu}_{\ ,\mu}} - \frac{\partial \mathcal{L}}{\partial A^\nu} = 0$.

$\frac{\partial \mathcal{L}}{\partial A^\nu} = 0$ because there are no $A^\nu$ terms.

Now, because there are different components (covariant, contravariant), I insert some $\eta_{\mu\nu}$ to make them all look the same.
-> making the derivatives easier later

This leads to $\mathcal{L} = -\frac{1}{16\pi} (\eta_{\mu\alpha} A^{\alpha}_{\ ,\nu} - \eta_{\alpha\nu} A^{\alpha}_{\ ,\mu}) (\eta^{\beta\nu} A^{\mu}_{\ ,\beta} - \eta^{\beta\mu} A^{\nu}_{\ ,\beta})$.

Now just multiply...

Then I get this

$\mathcal{L} = -\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} A^{\alpha}_{\ ,\nu} A^{\mu}_{\ ,\beta} + ... + ... - ...$

Now I am supposed to do

$\frac{\partial \mathcal{L}}{\partial A^{\nu}_{\ ,\mu}}$

which means

$-\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} \frac{\partial}{\partial A^{\nu}_{\ ,\mu}} (A^{\alpha}_{\ ,\nu} A^{\mu}_{\ ,\beta})$

now product rule

...

then I get terms like this

$\frac{\partial A^{\alpha}_{\ ,\nu}}{\partial A^{\nu}_{\ ,\mu}}$

My problem:

-do I need to rename all of ${\partial A^{\nu}_{\ ,\mu}}$ or just the $,\mu$?

2) what do I get out of that?
-my thoughts were something like $\delta^a_b$

What I did was

$\frac{\partial A^{\alpha}_{\ ,\nu}}{\partial A^{\nu}_{\ ,\mu}} = \delta^\mu_\nu$

-ignoring the top (contravariant) indices
-just using the $,\mu$ and $,\nu$

I would be grateful if somebody could show the result of
$-\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} \frac{\partial}{\partial A^{\nu}_{\ ,\mu}} (A^{\alpha}_{\ ,\nu} A^{\mu}_{\ ,\beta})$

2. Aug 15, 2013

### WannabeNewton

Perhaps a simpler example will get you up to speed. Consider the Klein-Gordon Lagrangian density in flat space-time $\mathcal{L} = -\frac{1}{2}(\eta^{ab}\partial_{a}\varphi \partial_{b}\varphi + m^2 \varphi^2)$.

We have $\frac{\partial \mathcal{L}}{\partial \varphi} = -m^{2}\varphi$ and $\frac{\partial\mathcal{L}}{\partial(\partial_{a} \varphi)} = -\frac{1}{2}\{\eta^{bc}\partial_{b}\varphi \frac{\partial (\partial_{c}\varphi)}{\partial (\partial_{a}\varphi)} + \eta^{bc}\partial_{c}\varphi \frac{\partial (\partial_{b}\varphi)}{\partial (\partial_{a}\varphi)}\}\\ = -\frac{1}{2}\{\eta^{bc}\partial_{b}\varphi\delta^{a}_{c} + \eta^{bc}\partial_{c}\varphi \delta^{a}_{b}\}\\ = -\partial^{a}\varphi$
where $\frac{\partial (\partial_{a}\varphi)}{\partial (\partial_{b}\varphi)} = \delta^{b}_{a}$ was used. This gives the usual Klein-Gordon field equations $\partial_{a}\partial^{a}\varphi - m^{2}\varphi = 0$. The calculation for the electromagnetic field is extremely similar; you just have a 4-potential $A_{b}$ instead of a scalar potential $\varphi$ so that for example $\frac{\partial (\partial_{a}A_{b})}{\partial (\partial_{c}A_{d})} = \delta^{c}_{a}\delta^{d}_{b}$.

3. Aug 15, 2013

### dextercioby

You're making things more complicated than they are. And you should use indices consistently:

$$L= -\frac{1}{16\pi} F_{\mu\nu}F^{\mu\nu}= -\frac{1}{16\pi}\left(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\right)(...)$$

So

$$\frac{\partial L}{\partial A_{\alpha}} = 0$$

and

$$\frac{\partial L}{\partial \left(\partial_{\beta}A_{\alpha}\right)} = \left[\frac{1}{8\pi}\frac{\partial}{\partial \left(\partial_{\beta}A_{\alpha}\right)} \left(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\right)\right] (...)$$

Now use

$$\frac{\partial\left(\partial_{\mu}A_{\nu}\right)}{\partial \left( \partial_{\beta}A_{\alpha}\right)} = \delta^{\beta}_{\mu}\delta^{\alpha}_{\nu}$$

4. Aug 15, 2013

### ProfDawgstein

aah, thanks.

I was wondering if I get 2 $\delta$'s.

What should the derivative in the Lagrangian look like (to be compatible with the $\mathcal{L}$ terms)?

It shouldn't contain any indices, which are used inside the $\mathcal{L}$, right?

In my example $\frac{\partial A^{\alpha}_{\ ,\nu}}{\partial A^{\nu}_{\ ,\mu}}$

I should get $\delta^{\alpha}_{\nu} \delta^{\mu}_{\nu}$, right?

This does not look right, so I think I have to use something like $\frac{\partial \mathcal{L}}{\partial A^{\sigma}_{\ ,\tau}}$ ?

And in the 2nd part just the opposite?

$$\frac{\partial\left(\partial^{\mu}A^{\nu}\right)}{\partial \left(\partial^{\beta} A^{\alpha}\right)} = \delta^{\mu}_{\beta} \delta^{\nu}_{\alpha}$$

5. Aug 15, 2013

### dextercioby

The nu's must be on diagonal, if summed over.

6. Aug 15, 2013

### ProfDawgstein

Using a slightly different Euler-Lagrange Equation:

$\frac{\partial}{\partial x^\tau} \frac{\partial \mathcal{L}}{\partial A^{\sigma}_{\ ,\tau}} - \frac{\partial \mathcal{L}}{\partial A^\sigma} = 0$

Using this Lagrangian:

$\mathcal{L} = -\frac{1}{16\pi} (\eta_{\mu\alpha} A^{\alpha}_{\ ,\nu} - \eta_{\alpha\nu} A^{\alpha}_{\ ,\mu}) (\eta^{\beta\nu} A^{\mu}_{\ ,\beta} - \eta^{\beta\mu} A^{\nu}_{\ ,\beta})$

Then $\frac{\partial \mathcal{L}}{\partial A^\sigma} = 0$

Now for example:

$-\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} \frac{\partial}{\partial A^{\sigma}_{\ ,\tau}} (A^{\alpha}_{\ ,\nu} A^{\mu}_{\ ,\beta})$

$-\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} A^{\alpha}_{\ ,\nu} (\frac{\partial}{\partial A^{\sigma}_{\ ,\tau}} A^{\mu}_{\ ,\beta})$

and

$-\frac{1}{16\pi} \eta_{\mu\alpha} \eta^{\beta\nu} (\frac{\partial}{\partial A^{\sigma}_{\ ,\tau}} A^{\alpha}_{\ ,\nu}) A^{\mu}_{\ ,\beta}$

where

$(\frac{\partial}{\partial A^{\sigma}_{\ ,\tau}} A^{\mu}_{\ ,\beta}) = \delta^{\mu}_{\sigma} \delta^{\tau}_{\beta}$

and

$(\frac{\partial}{\partial A^{\sigma}_{\ ,\tau}} A^{\alpha}_{\ ,\nu}) = \delta^{\alpha}_{\sigma} \delta^{\tau}_{\nu}$

7. Aug 15, 2013

### dextercioby

Yes, using the partial notation ∂ for derivatives is better, it has more visibility.

8. Aug 15, 2013

### ProfDawgstein

Now I ended up with

$-4 k A_{\sigma}^{\ ,\tau} + 3 k A^{\tau}_{\ ,\sigma} + k A^{\mu}_{\ ,\mu}$

$k = \frac{1}{16\pi}$

where the last term looks like a divergence.

I wonder if this is the correct path...

9. Aug 15, 2013

### WannabeNewton

I must echo dexter here and say that you are making this much more complicated than it needs to be. I'm going to drop the $4\pi$ in the denominator and just write $\mathcal{L} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + A_{\mu}j^{\mu}$ (your exercise does it for the case of vanishing 4-current i.e. $j^{\mu} = 0$ but there's no real difference in the computation regardless). Hence $\frac{\partial \mathcal{L}}{\partial A_{\nu}} = j^{\nu}$. Now $\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}A_{\nu})} = -\frac{1}{4}\eta^{\alpha \rho}\eta^{\beta \sigma}\{\frac{\partial F_{\alpha \beta}}{\partial (\partial_{\mu}A_{\nu})}F_{\rho\sigma} +\frac{\partial F_{\rho\sigma}}{\partial (\partial_{\mu}A_{\nu})}F_{\alpha\beta} \}$ and $F_{\alpha\beta} = 2\partial_{[\alpha}A_{\beta]}$ so just evaluate.

Last edited: Aug 15, 2013
10. Aug 15, 2013

### ProfDawgstein

I failed on one index...

Using this Euler-Lagrange:

$\frac{\partial}{\partial x^\tau} \frac{\partial \mathcal{L}}{\partial A^{\sigma}_{\ ,\tau}} - \frac{\partial \mathcal{L}}{\partial A^\sigma} = 0$

Now I have

$4 k A^{\tau}_{\ ,\sigma} - 4 k A_{\sigma}^{\ ,\tau}$

which looks a lot better.

Now I just raise the $\sigma$ index with $\eta^{\sigma\alpha}$

then I get $4 k ( A^{\tau ,\alpha} - A^{\alpha ,\tau} )$

which is $4 k F^{\alpha\tau}$

doing $\partial_\tau$ from Euler-Lagrange

leads to $4 k \partial_\tau F^{\alpha\tau} = 0$.

-----------------------------------------

I just wanted to try my way :)

Correction: He meant "current-free"...

Last edited: Aug 15, 2013
11. Aug 15, 2013

### WannabeNewton

Well as long as it worked out :)

12. Aug 15, 2013

### ProfDawgstein

Two pages and 400 indices later, yes.
It was a total mess, especially that little index mistake :D

NOW I will try your way.

Where

$\frac{\partial F_{\rho\sigma}}{\partial (\partial_{\mu}A_{\nu})} = \frac{\partial (\partial_\rho A_\sigma - \partial_\sigma A_\rho )}{\partial (\partial_{\mu}A_{\nu})} = \delta^{\mu}_{\rho} \delta^{\nu}_{\sigma} - \delta^{\mu}_{\sigma} \delta^{\nu}_{\rho}$

and

$\frac{\partial F_{\alpha\beta}}{\partial (\partial_{\mu}A_{\nu})} = \frac{\partial (\partial_\alpha A_\beta - \partial_\beta A_\alpha )}{\partial (\partial_{\mu}A_{\nu})} = \delta^{\mu}_{\alpha} \delta^{\nu}_{\beta} - \delta^{\mu}_{\beta} \delta^{\nu}_{\alpha}$

Somehow my lazyness prevents me from finishing this :D

This is indeed much nicer, thanks!

Thanks to you both!

[Another apple of knowledge thrown at me by Newton...]

Last edited: Aug 15, 2013
13. Aug 15, 2013

### WannabeNewton

Don't worry about doing it another way; if you got it once then it's fine. If you're interested: $\eta^{\alpha \rho}\eta^{\beta \sigma}\{\frac{\partial F_{\alpha \beta}}{\partial (\partial_{\mu}A_{\nu})}F_{\rho\sigma} +\frac{\partial F_{\rho\sigma}}{\partial (\partial_{\mu}A_{\nu})}F_{\alpha\beta} \}\\ = 2\eta^{\rho [\mu}\eta^{\nu]\sigma}F_{\rho\sigma} + 2\eta^{\alpha [\mu}\eta^{\nu]\beta}F_{\alpha\beta} = 4F^{\mu\nu}$
so $\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}A_{\nu})} = -F^{\mu\nu}$ hence $\partial_{\mu}F^{\nu\mu} = j^{\nu}$.

14. Aug 15, 2013

### ProfDawgstein

Actually I am working on it right now [hahah], but thanks :)

EDIT:

done

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

looks good

15. Aug 15, 2013

### ProfDawgstein

I was wondering about the whole thing using $\delta$ notation...

Using the Lagrangian

$\mathcal{L} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$

where

$\delta F_{\mu\nu} = \delta (\partial_\mu A_\nu - \partial_\nu A_\mu) = \delta (\partial_\mu A_\nu) - \delta (\partial_\nu A_\mu) = \partial_\mu (\delta A_\nu) - \partial_\nu (\delta A_\mu)$

Now the action

$I = \int \mathcal{L} \ d^4x$

The variation of the action

$\delta I = \int \delta (-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}) d^4x = 0$

where $F_{\mu\nu}$ goes to zero at the surface of the integration volume. [???]

Using the 'product rule'

$\delta I = \int -[(\frac{1}{4} (\delta F^{\mu\nu}) F_{\mu\nu}) + (\frac{1}{4} (\delta F_{\mu\nu}) F^{\mu\nu})] d^4x = 0$

where

$\delta F^{\mu\nu} = \delta (\partial^\mu A^\nu - \partial^\nu A^\mu) = \delta (\partial^\mu A^\nu) - \delta (\partial^\nu A^\mu) = \partial^\mu (\delta A^\nu) - \partial^\nu (\delta A^\mu)$

now plug in, do some multiplication and then partial integration, that's it?

The remaining term should be the field equation then?

16. Aug 15, 2013

### WannabeNewton

See section E.1 of Wald.