Determinant of metric tensor in Cartesian Coordinates constant in vacuum

Question

In 3D, the determinant of a metric tensor is related to the volume of the span of the basis vectors. In Cartesian Coordinates, it's just the volume of the cube.

I would guess that in 4D space-time, the determinant of the metric tensor in vacuum is related to the speed of light and, therefore, is constant. (Edit:in cartesian coordinates)

In the Schwarzschild equation, at least, this is true. (Edit: The determinant of the Schwarzschild solution is that of spherical coordinates, therefore, in cartesian coordinates, it would be constant.)

However, is that true for every arbitrary metric tensor in vacuum?

Is that probably related to the fact that in vacuum, there are "no sources", therefore $T_{\mu\nu} $ = 0, therefore $R_{\mu\nu} $ = 0?

Can one derive $|g_{\mu\nu}|$ = constant from $R_{\mu\nu} $ = 0?

EDIT: Of course, the determinant of the Schwarzschild metric in spherical coordinates is not constant, but that of spherical coordinates. That brought me to the guess that probably, in Cartesian Coordinates it would be constant. My question therefore better reads as

1. "is the determinant of the metric tensor only dependent on the choice of coordinates and not dependent on the underlying (physical) space-time?" Or

2. "Is the determinant of the metric tensor of an arbitrarily curved spacetime in vacuum in Cartesian coordinates always constant?"

Answer 1

$\mathrm{det}(g)$ is not coordinate-independent - it is a scalar density of weight $+2$ (or $-2$, depending on your convention) which generically changes across spacetime. In Minkowski space equipped with spherical polar coordinates, for example, $\mathrm{det}(g) = -r^4 \sin^2(\theta)$.

Is the determinant of the metric tensor only dependent on the choice of coordinates and not dependent on the underlying (physical) space-time?

I'm not quite sure how to answer this. At any particular point, $\mathrm{det}(g)$ can be made to take any (negative) value by appropriate change of coordinates. The precise way that $\mathrm{det}(g)$ varies from point to point depends on what coordinates you choose. The curvature and topology of the underlying manifold determine what coordinate systems are possible, so in that sense affect how $\mathrm{det}(g)$ can change, but that's rather indirect.

Is the determinant of the metric tensor of an arbitrarily curved spacetime in vacuum in Cartesian coordinates always constant?

Cartesian coordinates can exist only on manifolds (or at least, in neighborhoods) with vanishing Riemann curvature. In other words, if a Cartesian coordinate system exists in some patch of your spacetime, then that patch is flat.