Is the volume in general relativity independent or dependent on the coordinates?

Question

The volume in curved space is calculated as:

$$V=4 \pi\int_{\Omega}r^2\sqrt{g_{rr}} d\Omega$$

Is this volume dependent or independent from the chosen coordinates? As I understand it should be independent from the chosen coordinates - however, I do not see how to get rid of the $g_{rr} $ as this is dependent on the chosen coordinates.

Edit: There seems to be on the one hand the measured volume as we measure it in certain coordinates and, on the other hand, the actual volume as a physical entity. The actual volume should be independent from the way we measure it. A length itself doesn't change whether I measure it in meter or in inch. It's only the number that changes. However, in the formula above we seem to calculate the volume in the chosen coordinate system, using the units of that coordinate system.

Therefore, the question would rather be: Is there a possibility to create something like the "metré des archives" (which is stored in Paris) for volumes? Probably a sphere with $r = 1m$, stored in Paris, therefore moving with a certain velocity in relation to all other objects (earth moving around the sun, which is moving around the center of gravity of the milky way and so on).

Answer 1

The actual volume should be independent from the way we measure it.

Untrue. As an example, one can mention the metré des archives, since that is easier to understand than dealing with volumes. The length of the metré des archives does depend on how you measure it. Namely, if you measure it at rest, you'll find it to be of roughly one meter ("roughly" being due to the fact that physical objects might deteriorate, dilate, and so on; and the definition of meter nowadays depends on the definition of second and on the prescribed value of the speed of light). However, if you measure it in a reference frame in which it moves relative to you at relativistic speeds, you'll notice it is considerably shorter, due to Lorentz contraction. This is not a property of the material, but rather of spacetime itself: space intervals are not coordinate invariant, nor need volumes be.

Answer 2

On any orientable pseudo-Riemannian manifold you have naturally defined volume n-form. You can integrate this to get a coordinate independent volume.

However, you are asking about spatial volume, not volume of spacetime. For this, you need to pick a spacelike hyperplane. Once you have this hyperplane, volume is again coordinate independent.

The dependence on coordinates comes in when you are making a choice about the hyperplane. You usually define the hyperplane as being natural to some particular observer and thus your choice depends on which observer you pick. This is basically what happens with length contraction in STR. Properties of the object are the same, but different observers are measuring its length along different spacetime axis, so no wonder they get different results.

In special cases though, like Schwarzschild solution, there is a symmetry that makes the choice possible without any reference to any observer.

Answer 3

The answer to this question will become more meaningful once we define the commonly used 3+1 decomposition of spacetime. Under the assumption that the spacetime manifold $\mathcal{M}$ is hyperbolic, it admits a split into a timelike coordinate - the coordinate time $t$, and a family of hypersurfaces $\{\Sigma_{t}, t \in \mathbb{R}\}$, each diffeomorphic to one another. The decomposition is, formally: $$ \mathcal{M} =\mathbb{R} \times \Sigma $$ which is by no means unique.

Now, with the full spacetime $\mathcal{M}$, you can associate the volume form (4-form) $\eta$, which in coordinates on $\mathcal{M}$ can be locally expressed as: $$ \eta_{g} = \sqrt{-\det g_{\mu\nu}}\; dx^{0}\wedge dx^{1} \wedge dx^{2} \wedge dx^{3},$$

The spacetime volumes (4-volumes) that you can measure with this form are coordinate independent - and the pseudo-Riemmanian volume form gives you a coordinate-invariant way of integrating scalar fields on the manifold.

$$ I[f] = \int_{\Omega}f\,\eta_{g}, \qquad \Omega \; \rm subset \,of \,\mathcal{M} $$.

The above coordinate invariance we can call full, 4-dimensional coordinate independence.

On the other hand, in the 3+1 split, once you choose the slicing, you can induce the spacetime metric $g$ on each hypersurface, obtaining a 3-metric $\gamma$. This $\gamma$ also has an associated volume form, namely:

$$ \eta_{\gamma} = \sqrt{\det \gamma_{ij}}\, dx^{1}\wedge dx^{2} \wedge dx^{3}$$.

Now, this volume form is also coordinate independent, but only with respect to the spatial coordinates - if we were to change the slicing - and equivalently, the notion of time - coordinate $t$, we would get a different 3-volume form altogether.

This is the GR way of stating that the measurement of spatial distance is dependent on the foliation - of which observers' foliations are a subset of.

To reiterate, the notion of spatial volumes is consistent within a choice of family of hypersufaces, but not invariant under the change of foliations (time coordinate). Once you pick your notion of time, the calculation of spatial volumes is independent of the spatial coordinates you choose. In turn, the 4-dimensional volumes are fully covariant. I hope this distinction is clear now.