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I am interested in the behavior of the proper volume when we switch the frame of reference. For example, I know that the proper volume element for the Schwarzschild and Kerr black holes may be extracted easily from the formulae:

$V=\sqrt{g}dr d\theta d\phi = \frac{r^{5/2} \sin (\theta )}{\sqrt{r-2 m}}$

(Where we only picked the spatial components of the $g_{\mu \nu}$ for the determinant)

However, I am interested in seeing how this proper volume changes when according to different observers. For example, I know that the local frame of reference can be expressed with a tetrad:

$e^m_\mu=\left( \begin{array}{cccc} \sqrt{\frac{r-2 m}{r}} & 0 & 0 & 0 \\ 0 & \sqrt{\frac{r^2}{r^2-2 m r}} & 0 & 0 \\ 0 & 0 & \sqrt{r^2} & 0 \\ 0 & 0 & 0 & \sqrt{r^2} \sin (\theta ) \\ \end{array} \right)$

Which fulfills: $e_{c}{}^{b} e_{e}{}^{d} g_{bd} = \eta_{c e}$ (the tetrad transformation into the local frame of reference)

Now I am interested in finding out what the proper volume $V_{\rm local}$ is in this local frame of reference given by the tetrad. Is this possible?

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Well, to calculate volumes, you need a hypersurface.

Let $\Sigma$ be such a hypersurface, locally described as $$ x^\mu=\Phi^\mu(\xi), $$ where $\xi=(\xi^1,\xi^2,\xi^3)$ are local coordinates on the hypersurface. Let us assume $\Sigma$ is spacelike, since that's what you seem to be going for.

The induced metric is then positive definite and is given by $$\gamma_{ij}=g_{\mu\nu}\frac{\partial\Phi^\mu}{\partial\xi^i}\frac{\partial\Phi^\nu}{\partial\xi^j},$$

the volume element is then $$d\Sigma=\sqrt{\det\gamma}d^3\xi.$$

Let $e_{(a)}=\{e_{1},e_{2},e_{3}\}$ be an orthonormal frame on $\Sigma$, so we have $$ \delta_{ab}=\gamma_{ij}e^i_{(a)}e^j_{(b)}, $$taking the determinant of both sides gives $$ 1=\det\gamma\cdot(\det e)^2, \\ \sqrt{\det\gamma}=\det(e^{-1}), $$ is we denote the dual frame with $\theta^{(a)}=\theta^{(a)}_id\xi^i$ (we have $e^i_{(a)}\theta^{(b)}_i=\delta^b_a$), we have $$\sqrt{\det\gamma}=\det\theta,$$ so the volume element is $$ d\Sigma=\det\theta d^3\xi. $$

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  • $\begingroup$ Ah, I see, so in this case the proper volume would be: $d\Sigma = det (e_m^\mu)^-1 dx dy dz=r^2 \sin (\theta ) \sqrt{\frac{r}{r-2 m}} dx dy dz$? Just to confirm I understood it properly. $\endgroup$ – Otto Mar 7 '17 at 9:12
  • $\begingroup$ Nevermind, I got it now! For clarification to others it may be helpful to note how I repeated the above derivation: We have $d\xi^m=e^m_{\ \ \mu} dx^\mu$, and if we set $dx^0=0$, we get the above result for $d^3 \xi=\sqrt{det\gamma}^{-1} dx^1 dx^2dx^3$ $\endgroup$ – Otto May 7 '17 at 8:18

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