Diffeomorphism invariance of metric determinant

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Discussion Overview

The discussion revolves around the diffeomorphism invariance of the metric determinant, specifically the expression sqrt(-g) and its powers. Participants explore whether these quantities are invariant under diffeomorphisms, with implications for their use in action formulations in general relativity.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant asserts that sqrt(-g) is diffeomorphism-invariant and questions if all powers of this expression are also invariant.
  • Another participant suggests that using the term "scalar" is more straightforward than "diffeomorphism-invariant," stating that any function of a scalar is also a scalar.
  • A later reply expresses gratitude for the clarification and proposes a specific action involving sqrt(-g) and a scalar field, seeking confirmation of its validity.
  • Another participant challenges the notion that the metric determinant is a scalar, using the example of volume elements in different coordinate systems to illustrate their point.
  • One participant acknowledges the correction regarding the validity of the proposed action, indicating that the second term does not work as initially thought.
  • A participant admits to an error in their earlier post, indicating a lack of attention at the time of writing.

Areas of Agreement / Disagreement

The discussion contains multiple competing views regarding the invariance of the metric determinant and its powers, with no consensus reached on the validity of the proposed action involving these quantities.

Contextual Notes

Participants express uncertainty about the definitions and implications of diffeomorphism invariance and scalar quantities, highlighting the complexity of the topic.

Pacopag
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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?
 
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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:
Thank you for your reply. This is good news for me. To be sure, a valid action may take the form
<br /> S = \int d^4x \sqrt{-g}\left(R+\sqrt{-g}\phi^2\right)<br />
where
<br /> \phi^2 <br />
is a scalar
Is this correct?
 
The metric determinant is not a scalar. Think of the volume elements associated with (say) spherical and Cartesian coordinates in flat space.
 
Yes. You are right. The second term in my action does not work.
 
D'oh, thanks for the correctoin, Stingray. I was obviously not awake this morning when I posted #2 :-)
 

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