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Variational technique

  1. May 11, 2008 #1
    Suppose I try to find variation of the term \phi^2 under variation of \phi i.e., (\delta \phi). Then I take derivative of \phi^2 with respect to \phi and multiply by \delta \phi. In case of more complicated objects containing derivative of \phi what is the procedure? for example:
    [tex]
    \delta(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})
    [/tex]
    what is the variation of this quantity under variation of \phi?
     
  2. jcsd
  3. May 11, 2008 #2
    Is the rule here
    [tex]
    \delta(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})=\frac{\partial(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})}{\partial \phi}\delta(\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi)
    [/tex]
    But then the derivative part creates problem. The part under square root depends on derivative of \phi and not on \phi itself, so the result is zero. I am confused.
     
    Last edited: May 11, 2008
  4. May 11, 2008 #3
    Always go back to the definition of the derivative:

    [tex]\delta F[\phi] = F[\phi + \delta \phi] - F[\phi][/tex].

    In any case, in your case the chain rule applies:
    [tex]

    \delta(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})=
    \frac{1}{ 2\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi} }
    \delta(\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi)

    [/tex]
     
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