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Diffeomorphism invariance of metric determinant

  1. Apr 27, 2010 #1
    Hi;
    I am pretty sure that
    sqrt(-g) is diffeomorphism-invariant.

    I am wondering if all powers of this are diffeo-invariant too. For example, are
    -g, g^2, etc. all invariants too?
     
  2. jcsd
  3. Apr 27, 2010 #2

    bcrowell

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    If you want to use less cumbersome terminology, you can just say "is a scalar" rather than "is diffeomorphism-invariant." That's what "scalar" means in GR.

    The answer to your question is yes, because any function of a scalar is also a scalar.

    [--D'oh -- that was incorrect -- sorry! --]
     
    Last edited: Apr 27, 2010
  4. Apr 27, 2010 #3
    Thank you for your reply. This is good news for me. To be sure, a valid action may take the form
    [tex]
    S = \int d^4x \sqrt{-g}\left(R+\sqrt{-g}\phi^2\right)
    [/tex]
    where
    [tex]
    \phi^2
    [/tex]
    is a scalar
    Is this correct?
     
  5. Apr 27, 2010 #4

    Stingray

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    The metric determinant is not a scalar. Think of the volume elements associated with (say) spherical and Cartesian coordinates in flat space.
     
  6. Apr 27, 2010 #5
    Yes. You are right. The second term in my action does not work.
     
  7. Apr 27, 2010 #6

    bcrowell

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    D'oh, thanks for the correctoin, Stingray. I was obviously not awake this morning when I posted #2 :-)
     
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