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How does one construct the action which gives the equation of motion?

  1. Jul 5, 2007 #1
    I do not understand how people construct a suitable action which after variation will give the correct equation of motion. For example, the Einstein Hilbert action: S=integration[R d^4x] gives the equation of motion when varied with respect to [g_mu nu]. But no book I had read so far tells me how to construct this action. Can anybody help me with this?
    Second, if someone says that the action is constructed so that after variation it must give the equation of motion then why do I need to use this technique in the first place where first I have to know the equation of motion and then construct the action, then vary it and rederive the equation of motion which I already had?
    Third, how do I know that an action is the only action which will give the correct equation of motion?
     
  2. jcsd
  3. Jul 5, 2007 #2
    Actions are guessed. The possible choices are restricted by the symmetries you want your action to obey which automatically become symmetries of the equations of motion. It's much easier to impose symmetries (like Lorentz invariance) on the Lagrangian (a scalar function) than directly on the equations of motion. Only experiment is the final verification of your chosen action.

    Of course the usual assumption is that your system is describable by action which doesn't have to be the case 100% of the time. Another assumption is that if your equations obey certain symmetry, your action has to obey it too but that is not true in general. It's true in the reverse direction - a symmetry of the action automatically becomes a symmetry of the equations of motion.
     
    Last edited: Jul 5, 2007
  4. Jul 5, 2007 #3

    pervect

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    The fact that gravity acts like Newtonian gravity in the weak field limit also vastly limits the number of possible actions.

    I would think that Brans-Dicke gravity theories, for instance, could also be expressed in action form. (I haven't seen one written down, though, so I'm not positive). However, Brans-Dicke theory can be thought of as GR + an extra scalar field, and experimental evidence hasn't found any evidence of such a scalar field - i.e. Brans-Dicke has some adjustable parameters, and when the parameters of Brans-Dicke are adjusted to match experiment, it makes the same predictions that GR does. This doesn't rule out a scalar field, but basically means that it must be so small as to not have any known detectable effects.
     
  5. Jul 5, 2007 #4
    According to Wikipedia, Brans-Dicke theory does have an action:

    [tex]S=\frac{1}{16\pi}\int d^4x\sqrt{-g} \; \left(\phi R - \omega\frac{\partial_a\phi\partial^a\phi}{\phi} + \mathcal{L}_\mathrm{M}\right)[/tex]

    Although, I can't say I've seen it in a book. So I hope the appropriate pinch of salt is taken too.
     
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