This might be a very simple question. Is the following second derivative of the metric tensor a Lorentz scalar: $\partial_{\mu}\partial_{\nu}g^{\mu\nu}$ ? I know that for a vector field, $\partial_{\mu}A^{\mu}$ is a Lorentz scalar but wasn't sure whether the second derivative of the metric tensor as written above would still be a Lorentz scalar.
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$\begingroup$ Is the space Minkowskian or Riemannian ? In a Riemannian space neither $\partial_\mu\partial_\nu g^{\mu\nu}$ nor $\partial_\mu A^\mu$ would be a scalar. $\endgroup$– Frederic ThomasCommented Sep 8 at 19:24
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$\begingroup$ Hi @Frederic Thomas, why it wouldn't be a scalar for a Riemannian space? If that is the case, why would people write Lorentz scalars for example for graviton and stress-energy tensor coupling $h_{\mu\nu}T^{\mu\nu}$? $\endgroup$– physics_2015Commented Sep 8 at 21:04
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$\begingroup$ Hi physics_2015, How do you define Lorentz symmetry in curved space? $\endgroup$– Qmechanic ♦Commented Sep 9 at 0:06
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$\begingroup$ Hi @Qmechanic, so yes the Lorentz symmetry is valid for only Minkowski spacetime but I'm a bit confused about graviton stress-energy tensor coupling $h_{\mu\nu}T^{\mu\nu}$. It seems this coupling assumes a Minkowski spacetime while being used for a curved spacetime. I know that it assumes linear gravity in which $h_{\mu\nu}$ is a small perturbation but I'm still not very clear. Shall this be asked in a different question? $\endgroup$– physics_2015Commented Sep 9 at 23:30
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For the fluctuations $h^{\mu \nu}$ around a flat background, $\partial_\mu \partial_\nu h^{\mu \nu}$ is a Lorentz scalar. For highly curved backgrounds, it is not sensible to classify quantities by their Lorentz representations.
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1$\begingroup$ You could classify objects by representations of the internal Lorentz group (vielbein formalism) $\endgroup$– PraharCommented Sep 8 at 20:53
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$\begingroup$ Hi @Connor Behan , so does it mean that the entire Lagrangian formalism in quantum field theory is suited mostly for flat spacetime? $\endgroup$ Commented Sep 8 at 21:08
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$\begingroup$ Hi @Prahar , do you mean that regardless of the curvature of the background, it's always possible to create Lorentz scalars if one uses Vierbeins? $\endgroup$ Commented Sep 8 at 21:11
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1$\begingroup$ Curved space is harder of course but you can still consider Lagrangians. Normally, you demand that the terms are invariant under diffeomorphism (that act on matter fields as well as the metric) and then take the metric to be fixed. $\endgroup$ Commented Sep 8 at 22:19