# Constant determinant of metric tensor of Schwarzschild solution in $(x, y, z, t)$ coordinates?

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.

• 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$^{-2} \theta$? Nov 19, 2023 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. Nov 19, 2023 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 $r^2\sin(\theta)$ Nov 19, 2023 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. Nov 20, 2023 at 6:43
• 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. Nov 20, 2023 at 7:00