# Equation of motion from the action

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1. Sep 18, 2015

### Emilie.Jung

Hello Physics Forums!

Supposing that we have an action that says:
$$L=\frac{1}{2}R-g_{C\bar{D}}\partial_{\mu}z^C\partial^{\mu}\bar{z}^D+\frac{1}{4} + \frac{1}{4}ImM_{IJ}F^I_{\mu\nu}\cdot F^{J\mu\nu} +\frac{1}{4}ReM_{IJ}F^I_{\mu\nu}\cdot \tilde{F}^{J\mu\nu}$$
where $$\tilde{F}^I_{\mu\nu}=\frac{1}{2}\epsilon_{\mu\nu}^{\mu'\nu'}F^I_{\mu'\nu'}$$
If I want to find the equation of motion of this, I know I can start from action and use the Euler Lagrange equation or directly use d*F=0 (Maxwell's equation).

This is one of the very rare times where I encounter general relativity as I am still a second year physics student. I have seen this sort of action (but way simpler) in my classical mechanics course) where we almost always use the Euler-Lagrange equation or vary the action with respect to the functional term and set it equal to zero and VOILA! you get the equation of motion you need. Here though I am finding a little of difficulty. The thing is that what was found for this action was the following answer:
$$d(ReM_{IJ}F^J+ImM_{IJ}\tilde{F}^J)=0.$$
I have tried so manyy ways to derive this but failed at it.

First, I tried to do it in the general way thinking I would avoid mistakes so I started by transforming the $\tilde{F}^{\mu\nu}$ to $\frac{1}{2}\epsilon^{\rho\sigma\mu\nu}F_{\rho\sigma}$ to get rid of the tilde and then deriving the action with respect to $F^{\kappa\lambda}$ and then set it equal to zero. BUT the procedure seems to take forever and I am not getting the answer above(It is too long I couldn't write it down, but if it matters I could type few of the details). I guess there is a better track to take, but I have no idea what it might be. On some websites, I found that instead of $\partial^{\mu}F_{\mu\nu}$ that we usually use in Classical Mechanics, there is now a G term, for example they write the maxwell equation as $\partial^{\mu}G_{\mu\nu}$ or something like that. You might relate to this better than I am. I am only trying to tell you what I have searched for before finally creating an account on this forum.

I am trying to practice those from the web in a spontaneous manner as my professor said she will bring us a bonus part in the coming exam that has to do with general relativity after she taught us a little of einstein indices operations. I know she might not bring this tough level but I am curious to know how things work in General Relativity and why if I followed the very classical way, I am not getting the answer above for the equation of motion? Any advice will be useful.

Yours,
Emilie.

2. Sep 19, 2015

### Ben Niehoff

The $F^I$ are not the fundamental fields. Presumably they are given by $F^I \equiv d A^I$. You need to vary the action with respect to the $A^I$, not the $F^I$.

3. Sep 19, 2015

### Emilie.Jung

But I know that if A is not explicitly shown in the action, I can just go ahead and vary the action with respect to the F. Is this wrong?

4. Sep 19, 2015

### Ben Niehoff

Very wrong. For one, no derivatives of F appear in the action, so you'll just get algebraic equations of motion that way.

5. Sep 19, 2015

### Emilie.Jung

What do you mean by "algebraic"equations of motion? Does this mean, I get this form $d(ReM_{IJ}F^J+ImM_{IJ}\tilde{F}^J)=0$? Is this the algebraic form? Because this is all I need.

6. Sep 19, 2015

### Ben Niehoff

I mean the EoM won't have any derivatives in them, and there are no dynamics.

No, because $d$ is a derivative.

Try something simpler first, like this:

$$S = -\frac14 \int F_{\mu\nu} F^{\mu\nu} \, d^4x, \qquad F \equiv d A.$$
You should be able to obtain

$$d \star F = 0$$

7. Sep 19, 2015

### Emilie.Jung

I will present what I did here:

$$S=-\frac{1}{4}\int{F_{\mu\nu}F^{\mu\nu}d^4x}$$
$$=-\frac{1}{4}\int{\delta F_{\mu\nu}F^{\mu\nu}d^4x+F_{\mu\nu}\delta F^{\mu\nu}d^x}$$
$$=-\frac{1}{4}\int{\delta(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})F^{\mu\nu}d^4x+F_{\mu\nu}\delta(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})d^4x}$$
I then expand terms,
$$=-\frac{1}{4}\int{(\partial_{\mu}\delta A_{\nu}F^{\mu\nu}-\partial_{\mu}\delta A_{\nu}F^{\nu\mu}+F_{\mu\nu}\partial^{\mu}g^{\nu\nu}\delta A_{\nu} -g^{\nu\nu}F_{\nu\mu}\partial^{\mu}\delta A_{\nu})d^4x}$$
$$=-\frac{1}{2}\int{(F^{\mu\nu}\partial_{\mu}+F^{\mu\nu}\partial_{\mu})\delta A_{\nu}}$$
setting this equal to zero, I get
$$F^{\mu\nu}\partial_{\mu}+F^{\mu\nu}\partial_{\mu}=0$$
where I do not get
$$d*F=0$$
Maybe I did a mistake somewhere.. @Ben Niehoff

8. Sep 19, 2015

### Ben Niehoff

At this point, you need to integrate by parts, so that $\delta A_\mu$ appears without any derivatives acting on it

9. Sep 19, 2015

### Emilie.Jung

Oh, then:
$$-\frac{1}{2}\int{2F^{\mu\nu}\partial_{\mu}\delta A_{nu}}$$
$$=-\int{F^{\mu\nu}\partial_{\mu}\delta A_{\nu}}$$
integration by parts
$$-F^{\mu\nu}\delta A_{\nu}+\int{\partial_{\mu}F^{\mu\nu}\delta A_{\nu}}$$
first terms goes away, and we set the second equal to zero to get
$$\partial_{\mu}F^{\mu\nu}=0$$
I am afraid I am not getting the dual of F here.. @Ben Niehoff

10. Sep 19, 2015

### Ben Niehoff

I don't suppose you considered writing out $d \star F$ in indices...

11. Sep 19, 2015

### Emilie.Jung

Sorry I don't know what you mean. As I have mentioned, I am a second year student and haven't really taken a course in General Relativity. If you may explain more by what you mean? @Ben Niehoff

12. Sep 19, 2015

### Ben Niehoff

Is this a homework problem? If you haven't done any GR, then how did you end up with an action for supergravity?

13. Sep 19, 2015

### Emilie.Jung

No this is not a homework. I am practicing different actions from the web in order to be prepared from my exam that will have a bonus question in my advanced classical mechanics course related to those (I mentioned this in my question). This I picked from an arxiv paper. In any case, I am solving a problem you set for me with an action different than the one I presented in my question, isn't this right? @Ben Niehoff