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Variation of the action

  1. Nov 30, 2005 #1
    If I have the expression for a Poincaré invariant action, how do I find the variation of the action? Any help/hints much appreciated, thanks ;)
  2. jcsd
  3. Nov 30, 2005 #2

    Physics Monkey

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    If your action is of the form
    S = \int d\lambda\, L\left(X^\mu, \frac{d X^\mu}{d \lambda} \right),
    where [tex] \lambda [/tex] is any affine parameter, then the variation looks like
    \delta S = \int d \lambda \,\left( \frac{\partial L}{\partial X^\mu} - \frac{d}{d\lambda}\left(\frac{\partial L}{\partial (dX^\mu/d\lambda)}\right)\right) \delta X^\mu,
    where in the process I have integrated by parts and dropped the boundary terms.

    As usual, you obtain the equations of motion by requring that the variation of the action be zero for arbitrary variations of the path [tex] \delta X^\mu [/tex]. If you are doing your free particle action that we talked about before, you can get the free particle equations of motion by first performing the variation and then setting the affine parameter equal to the proper time.

    Consult a text on classical mechanics for further clarification about variational principles and Lagrangians.
    Last edited: Nov 30, 2005
  4. Nov 30, 2005 #3
    General Case
    The action functional in ordinary spacetime looks like the one from classical physics:
    \mathcal{S}[\phi] = \int d^4x \, \mathcal{L}[\phi, \partial_\mu \phi; x^\mu] \, \quad \mapsto
    \quad S = \int dt \, L(q(t), \dot{q(t)}; t) \, .
    The most general way to write [tex]\delta \mathcal{S}[/tex] -- without imposing any symmetries -- is to just vary the Lagrangian with respect to the field(s) [tex]\phi[/tex],
    its derivative [tex]\partial_\mu \phi[/tex], and the coordinates [tex]x^\mu[/tex]:
    \delta \mathcal{S} = \int \delta \phi \frac{\partial \mathcal{L}}{\partial \phi}
    + \delta(\partial_\mu \phi) \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}
    + \delta x^\mu \frac{\partial \mathcal{L}}{\partial x^\mu} \, .
    If you are dealing with global Poincare invariance, the variation with respect to the coordinates vanishes. Have you dealt with
    \mathcal{L} = \frac{1}{2} |\partial_\mu \phi|^2 + \frac{1}{2}m^2 \phi^2
    [/tex] yet?
    Last edited: Dec 1, 2005
  5. Dec 1, 2005 #4
    Thanks a lot guys, really great stuff! =) Physics Monkey, unfortunately I dont have access to a decent library at the mo, and no money to buy books =(

    I suppose I didnt really ask the right questions, sorry, but what is a variation? and what does the basic formula for the variation of an action look like, without integration by parts? Is it the second one given by bigplanet? Whats the difference between global and local invariance?

    Is there any website that has good notes on this kind of stuff?
    Last edited: Dec 1, 2005
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