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Equation of motion from the action

  1. Sep 18, 2015 #1
    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. jcsd
  3. Sep 19, 2015 #2

    Ben Niehoff

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    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##.
     
  4. Sep 19, 2015 #3
    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?
     
  5. Sep 19, 2015 #4

    Ben Niehoff

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    Very wrong. For one, no derivatives of F appear in the action, so you'll just get algebraic equations of motion that way.
     
  6. Sep 19, 2015 #5
    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.
     
  7. Sep 19, 2015 #6

    Ben Niehoff

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    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

    $$d \star F = 0$$
     
  8. Sep 19, 2015 #7
    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
     
  9. Sep 19, 2015 #8

    Ben Niehoff

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    At this point, you need to integrate by parts, so that ##\delta A_\mu## appears without any derivatives acting on it
     
  10. Sep 19, 2015 #9
    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
     
  11. Sep 19, 2015 #10

    Ben Niehoff

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    I don't suppose you considered writing out ##d \star F## in indices...
     
  12. Sep 19, 2015 #11
    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
     
  13. Sep 19, 2015 #12

    Ben Niehoff

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    Is this a homework problem? If you haven't done any GR, then how did you end up with an action for supergravity?
     
  14. Sep 19, 2015 #13
    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
     
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