Dirac comment on tensor densities (Dirac GTR, p. 37)

[Kostik]

TL;DR Summary

Dirac comment on tensor densities (Dirac GTR, p. 37) -- why is $\int T^{\mu\nu}\sqrt{-g}d^4 x$ invariant?

Dirac (GTR, p. 37) shows simply that for a scalar function $S$
$$\int S\sqrt{-g}\,d^4 x = \int S'\sqrt{-g'}\,d^4 x'$$ and this works precisely because $S=S'$ for a scalar. But for a tensor $T^{\mu\nu}$ the same procedure gives
$$\int T^{\mu\nu}\sqrt{-g} \, d^4 x = \int x^\mu_{\,\, ,\alpha'}x^\nu_{\,\, ,\beta'}T^{\alpha'\beta'}\sqrt{-g'}\,d^4 x' .$$ Dirac defines a "density" (such as $S\sqrt{-g}$) as a quantity whose integral is invariant. But clearly $$\int T^{\mu\nu}\sqrt{-g}\,d^4 x \neq \int T^{\mu'\nu'}\sqrt{-g'}\,d^4 x' $$ because of the $x^\mu_{\,\, ,\alpha'}x^\nu_{\,\, ,\beta'}$ terms. What am I missing here? Thanks in advance.

Edit: I realize I am somewhat confused here. Is there some way to write this:
$$\int T^{\mu\nu}\sqrt{-g}\,d^4 x = \int T^{\alpha'\beta'}x^\mu_{\,\, ,\alpha'}x^\nu_{\,\, ,\beta'}\sqrt{-g'}\,d^4 x' = \int \left[ T^{\mu\nu} \right]^{'} \sqrt{-g'}\,d^4 x'\quad??$$ I hesitate to write $T^{\mu'\nu'}$ in the last term, because the $\mu$'s and $\nu$'s should balance.

 

[Orodruin]

So the first thing you should ask yourself is: "Does the expression make sense at all?"
What does
$$
\int T^{\mu\nu} \sqrt{-g}\, d^4x
$$
actually mean? In essence, you are adding together tensor components from different parts of your spacetime and this generally holds no particular meaning. The tensors belong to different tensor spaces and so the expression is generally meaningless.

The only way that you can provide the expression with meaning is if you have a flat spacetime, e.g., Minkowski space, where there is a unique identification between the tangent spaces of different events. Once we have concluded that, we should also note that what then makes sense is to integrate a tensor itself, not only its components, i.e.,
$$
\int T^{\mu\nu} e_{\mu\nu} \sqrt{-g}\, d^4 x
$$
where $e_{\mu\nu}$ is the tensor basis.

The only way you can be left with only the tensor components inside the integral is if the basis is constant and can be taken outside of the integral. If you are using coordinate bases, this only happens if your coordinate system is affine - effectively restricting you to different inertial frames on Minkowski space (there are some other affine systems of course, such as non-orthonormal ones, but that's a detail - if not for that detail, you would always have $-g = 1$ and could remove that from the expression).

Given the above, what you really expect is that the result of the integral should be the components of a rank 2 tensor under Lorentz transformations, which exactly follows the transformation properties that you found. You should not expect a tensor to have the same components in different inertial frames.

Edit: Note: This is a general reply. I do not have access to Dirac's text so I cannot comment directly to what he says.

 

[Kostik]

Dirac already addressed this: he said "...is a tensor if the domain of integration is very small. It is not a tensor if the domain of integration is not small, because then it consists of a sum of tensors located at different points..."

Apologies I omitted this excerpt, assuming a reader might open to the relevant page.

 

[Orodruin]

If the domain is small enough for curvature to be negligible, then - locally - as always, you have Minkowski space and the above applies.