1
$\begingroup$

Determinant of Schwarzschild metric is $-r^4\sin^2\theta$ which is also the determinant of flat spacetime represented in the same coordinates. Is this just coincidence or is it always so that the determinant of a metric depends only on the coordinate system used and not the manifold itself?


EDIT: Determinant of interior Schwarzschild metric is not $-r^4\sin^2\theta$ even though that metric is represented in same coordinates as exterior Schwarzschild metric. So is it only in vacuum when determinant does not depend on geometry?

$\endgroup$
  • $\begingroup$ Related and possibly a duplicate: Calculating the determinant of a metric tensor $\endgroup$ – John Rennie Aug 28 at 11:22
  • $\begingroup$ @JohnRennie It does not give satisfying answer for my question. See my edit. $\endgroup$ – Kirby Aug 28 at 16:28
  • $\begingroup$ The interior Schwarzschild metric is a different metric to the exterior metric, so of course the determinant is different. $\endgroup$ – John Rennie Aug 28 at 17:03
  • $\begingroup$ Can you clarify your last question? Since "vacuum" equates to "vanishing Einstein tensor", it reads a bit like "Is it only in [specific geometric circumstance] that the determinant does not depend on geometry?" $\endgroup$ – J. Murray Aug 28 at 17:10
  • $\begingroup$ If the determinant of a metric depended only on coordinate system used then shouldn't interior and exterior (regular) Schwarzschild metrics have the same determinant because they are represented in the same coordinates? But they have not the same determinants... $\endgroup$ – Kirby Aug 29 at 5:56
3
$\begingroup$

is it always so that the determinant of a metric depends only on the coordinate system used and not the manifold itself?

Yes, this is always true. The equivalence principle says that locally, spacetime is always equivalent to flat spacetime. That means that locally, we can always choose Minkowski coordinates, and the metric will have the Minkowski form. Therefore we can always make the determinant of the metric be $-1$. By rescaling the coordinates, we can also make the determinant of the metric have any other desired (negative) value.

$\endgroup$
  • $\begingroup$ See my edit. Why interior Schwarzschild metric then does not have the same determinant as exterior (regular) Schwarzschild metric even though they are in same coordinates? $\endgroup$ – Kirby Aug 29 at 5:52
  • $\begingroup$ @ProfessorKirby: The determinant of the metric at a given point is what's arbitrary. The expression you gave for the determinant of the metric depends on the coordinates. It's not the same for different points, even outside the horizon. $\endgroup$ – Ben Crowell Aug 29 at 15:17
2
$\begingroup$

A somewhat higher-brow answer is that volume forms on a manifold have no local invariants.

Let the "model space" for a "volumed" geometry be $(\mathbb R^n,\mu)$, where $\mu=\mathrm dx^1\wedge...\wedge\mathrm dx^n$.

Let $M$ be an $n$ dimensional, smooth, orientable manifold with positively oriented volume form $\omega$. Then any $x\in M$ has some open neighborhood $U\subseteq M$ such that there is a smooth local diffeomorphism $\psi:U\rightarrow\mathbb R^n$ for which $\omega=\psi^\ast\mu$.

This can be restated in "lowbrow" coordinate language that one may always find a coordinate system where $\omega=\mathrm dx^1\wedge...\wedge\mathrm dx^n$.

So essentially all volume forms "look like one another" locally, and there is no analogue of the Riemann curvature tensor for volume forms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.