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In spherical Schwarzschild coordinates, it's $A \cdot B$ = constant. Is there something similar in the Schwarzschild solution in $(x, y, z, t)$ coordinates? For example in Droste coordinates $g_{tt} \cdot $det$(g_{ij})$ = constant?

The determinant of the metric tensor is a scalar density of weight 2. If I got it correctly that means that $($det$(Jac))^2 \cdot $det $(g_{\mu\nu})$ = constant. Since the determinant of the Jacobian of $(x, y, z, t)$ is 1, det $g_{\mu\nu}$ should be constant then. Is this reasoning correct? However, the determinant of this:

$$g_{\mu \nu}^\rm D=\left( \begin{array}{cccc} \rm 1-\frac{r_s}{r} & 0 & 0 & 0 \\ 0 & \rm \frac{r_s \ x^2}{r^2 \ (r_s-r)}-1 & -\rm \frac{r_s \ x \ y}{r^2 \ (r-r_s)} & \rm -\frac{r_s \ x \ z}{r^2 \ (r-r_s)} \\ 0 & \rm -\frac{r_s \ x \ y}{r^2 \ (r-r_s)} & \rm \frac{r_s \ y^2}{r^2 \ (r_s-r)}-1 & \rm -\frac{r_s \ y \ z}{r^2 \ (r-r_s)} \\ 0 & \rm -\frac{r_s \ x \ z}{r^2 \ (r-r_s)} & \rm -\frac{r_s \ y \ z}{r^2 \ (r-r_s)} & \rm \frac{r_s \ z^2}{r^2 \ (r_s-r)}-1 \\ \end{array} \right)$$

doesn't end up in something constant.

Qmechanic
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MartyMcFly
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1 Answers1

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A Jacobian determinant is associated to a change of coordinates, not a single coordinate system, so talking about the Jacobian of a Cartesian coordinate chart doesn't make sense.

J. Murray
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  • Okay, thank you again, so if I change from spherical (where the determinant of the Schwarzschild metric is $r^2$sin$\theta$) to (x, y, z, t) - then the determinant of $g_{\mu\nu} $ should be $r^{-4} $sin$^{-2} \theta$? – MartyMcFly Nov 19 '23 at 14:50
  • It should be multiplied by that factor, yes. In spherical coordinates, $\mathrm{det}(g) = -r^4 \sin^2\theta$. To transform to Cartesian coordinates, you multiply it by $\big(1/(r^2 \sin\theta)\big)^2$ to get $\mathrm{det}(g) = -1$, as expected. – J. Murray Nov 19 '23 at 16:00
  • @MartyMcFly To be clear, your first comment is wrong - the determinant of the Schwarzschild metric in spherical coordinates is $\color{red}{-}r^\color{red}{4}\sin^\color{red}{2}(\theta)$, not $r^2\sin(\theta)$ – J. Murray Nov 19 '23 at 16:03
  • Okay, I see. cp-ed from axial. Sorry for that.... Seems that I have to go the way from spherical Schwarzschild to Droste Schwarzschild using that Jacobian, then calculating both determinants and having a look whether $(det(Jac)) ^2 \cdot $ det $g_{\mu\nu} $ is always the same. That's what my question is about. – MartyMcFly Nov 20 '23 at 06:43
  • @MartyMcFly I'm not sure how to communicate this better, but it simply does not make sense to ask whether $\mathrm{det}(J)^2 \mathrm{det(g)}$ is "the same" in two different charts, because $\mathrm{det}(J)$ is not defined in a chart. It is defined as part of the transformation from one chart to another. – J. Murray Nov 20 '23 at 06:56
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    The statement is that if $\mathrm{det}(g_{(x)})$ is the metric determinant in a coordinate chart $x$ and if the Jacobian corresponding to the transformation $x\rightarrow y$ is given by $J_{x\rightarrow y}$, then the metric determinant in the coordinate chart $y$ is given by $\mathrm{det}(g_{(y)}) = \mathrm{det}(J_{x\rightarrow y})^2 \mathrm{det}(g_{(x)})$. It doesn't make sense to talk about the Jacobian in one chart or another. – J. Murray Nov 20 '23 at 07:00