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Variational principles within GR

  1. Jan 28, 2013 #1
    Hi
    Can anyone explain why from varying a Lagrangian?
    [itex] \frac{\partial \mathcal{L}}{\partial (\partial_{\alpha }\phi)}(\frac{1}{2}\sqrt{-g}g^{\alpha\beta}\partial_{\alpha} \phi \partial_{\beta} \phi) = g^{\mu\nu}\partial_{\mu}\partial_{\nu}\phi + \frac{1}{\sqrt{-g}}(\partial_{\nu}\phi)\partial_{\mu}(g^{\alpha \beta}\sqrt{-g}) [/itex]

    I realise that the indices are dummies and so can be renamed accordingly but I don't understand why it isn't equal to:

    [itex] (\frac{1}{2}g^{\mu\nu} \partial_{\nu} \phi) + \frac{1}{\sqrt{-g}}(\partial_{\mu} \phi) \partial_{\mu}(\frac{1}{2}\sqrt{-g}g^{\mu\nu} \partial_{\nu} \phi)

    [/itex]
     
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  3. Jan 28, 2013 #2

    bcrowell

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    I don't understand your notation in the above expression. Should the [itex]\mathcal{L}[/itex] not be there? Isn't the expression in parens what [itex]\mathcal{L}[/itex] is? On the r.h.s. of the first equations, the beta doesn't make sense, because there are no other betas unbound anywhere else. Is [itex]\partial[/itex] a covariant derivative, or a plain partial derivative?

    Is this copied from messy lecture notes that you're having trouble reconstructing?
     
  4. Jan 28, 2013 #3
    yeah sorry the [itex]\mathcal{L}[/itex] was a mistake and should not be there, the action in full is:

    [itex] S= \int \sqrt{-g} [\frac{1}{2}g^{\alpha \beta}(\partial_{\alpha}\phi)(\partial_{\beta}\phi)-\frac{1}{2}m^{2}\phi^{2}] d^{4}X[/itex]

    The [itex] (\partial_{\alpha}\phi) [/itex] is a covariant derivative i think. It's just trying to find the:
    [itex] \frac{\partial \mathcal{l}}{(\partial_{\alpha}\phi)} [/itex] term in the E-L equations but i am having trouble with the GR notation.
     
  5. Jan 28, 2013 #4
    Try doing it in Minkowski spacetime first, so that you don't have to worry about the [itex]\sqrt{-g}[/itex] factor (it's just 1 in Special Relativity). After you've worked it out in flat spacetime, you'll probably find it easier to do it in the covariant case because you'll have experience with how to take derivatives of certain quantities.
     
  6. Jan 28, 2013 #5

    Bill_K

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    When you calculate (∂/∂φ)(½gαβφφ), the independent variable is φ. So it's like you're calculating (∂/∂x)(½Cxx). You apply the derivative to each φ in turn - it's like a quadratic. You'll get a factor of two from that, and what's left is just gαβφ.
     
  7. Jan 28, 2013 #6
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