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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?"

MartyMcFly
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    -1 Changing a question after it has been answered is impolite. – Ghoster Oct 06 '23 at 04:23
  • Sorry for that! However, I wrote in the title that it's meant to be constant in cartesian coordinates and I realized that I didn't make it clear that I want to disentangle coordinates and curvature of space-time. I know that it's dependent on the choice of coordinates. It's really tricky to disentagle physics-related curvature and a arbitrary choice of coordinates and I tried to make it clearer. – MartyMcFly Oct 06 '23 at 04:35
  • Cartesian spatial coordinates imply vanishing spatial curvature. You can’t have Cartesian coordinates in an arbitrarily curved spacetime. – Ghoster Oct 06 '23 at 04:48
  • Really??? But I could simply transform, for example, Schwarzschild metric, to cartesian coordinates. Then I would have cartesian coordinates - but curved space-time. – MartyMcFly Oct 06 '23 at 05:05
  • You seem to have a different definition of Cartesian coordinates than I do. What is it? – Ghoster Oct 06 '23 at 05:08
  • To me, cartesian coordinates are just x, y, z, t – MartyMcFly Oct 06 '23 at 05:29
  • That’s not a meaningful definition. Those are just names. – Ghoster Oct 06 '23 at 05:29
  • x, y, z, t and the jacobian is 1. Then the metric tensor is only dependent on the curvature... Isn't it? – MartyMcFly Oct 06 '23 at 06:15
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    @MartyMcFly A Jacobian matrix is defined for a particular coordinate transformation, not an individual coordinate system. In this context, Cartesian coordinates on a patch of spacetime are a set of coordinates such that the metric tensor takes the form $\mathrm{diag}(-1,1,1,1)$ in the induced coordinate basis. Obviously such a metric has vanishing Riemann curvature, which is a coordinate-independent statement. – J. Murray Oct 06 '23 at 13:21

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$\mathrm{det}(g)$ 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.

J. Murray
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  • Thank you for your answer! I added some edits to the question to clarify it. Regarding your answer, could you please explain what is "weight" of scalar density? And, please, how is a scalar density defined? I found here: https://physics.stackexchange.com/questions/721000/meaning-of-einsteins-condition-on-the-metric-tensor that Einstein started at the assumption that $\sqrt{-g} $ = 1. This is related to my question... Why did he do this? You say "it generally changes across space time". Could you please explain? – MartyMcFly Oct 06 '23 at 04:29
  • @MartyMcFly Sorry, weight $+2$, not $+1$. A scalar quantity remains invariant under changes of coordinate; a scalar density is scaled by $J^w$ under changes of coordinate, where $J$ is the determinant of the Jacobian matrix corresponding to the coordinate change and $w$ is the density's weight. In Cartesian coordinates, $\mathrm{det}(g)=-1$. If we switch to polar coordinates, the Jacobian determinant is $\mathrm r^2 \sin(\theta)$, and the metric determinant becomes $-\mathrm r^4 \sin^2(\theta)$, which is consistent with it being a scalar density of weight $+2$. – J. Murray Oct 06 '23 at 05:09
  • @MartyMcFly I've also edited my answer to address your follow-up questions. – J. Murray Oct 06 '23 at 05:10
  • Could you please explain what does "weight" and "scalar density" exactly mean? – MartyMcFly Oct 06 '23 at 05:42
  • @MartyMcFly I don't know how to explain it more exactly than I did in my previous comment. Can you explain which part of that you find unclear? – J. Murray Oct 06 '23 at 13:15
  • Thank you, now it's clear to me, your first comment was sufficient indeed. Another question: cartesian coordinates only work if the spacetime is flat. However, if I say that I approximate an arbitrarily curved spacetime with small pieces of a cartesian grid, "tangent spaces" - is that possible? – MartyMcFly Oct 23 '23 at 17:44
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    @MartyMcFly Given any point $p$, we can construct normal coordinates in a (possibly very small) neighborhood of $p$. In such a coordinate system, $g_{\mu\nu} = \delta_{\mu\nu} - \frac{1}{3}R_{\mu\nu\sigma\tau}x^\sigma x^\tau + O(x^3)$. – J. Murray Oct 24 '23 at 12:39