- #1

TheMan112

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I'm trying to find the coordinates where the determinant of the Kerr metric goes towards infinity. This should give the ring singularity of a Kerr (rotating) black hole. So, I'm starting out with the standard form Kerr metric in Boyer-Lindquist coordinates:

[tex]ds^2=\frac{\Delta}{\rho^2}(dt-a sin^2 \theta d\phi)^2-\frac{sin^2\theta}{\rho^2}((r^2+a^2)d\phi-a dt)^2-\frac{\rho^2}{\Delta}dr^2-\rho^2d\phi[/tex]

Then I break out the terms [tex]dt^2, dr^2, d\theta^2, d\phi^2[/tex] and [tex]dt d\phi[/tex], this gives the metric:

[tex]g_{ab} = \left(\begin{array}{cccc} \frac{\Delta-a^2 sin^2\theta}{\rho^2} & 0 & 0 & \frac{2a sin^2\theta (r^2+a^2-\Delta)}{\rho^2} \\ 0 & -\frac{\rho^2}{\Delta} & 0 & 0 \\ 0 & 0 & \rho^2 & 0 \\ \frac{2a sin^2\theta (r^2+a^2-\Delta)}{\rho^2} & 0 & 0 & \frac{sin^2 \theta (\Delta a^2 -(r^2 + a^2)^2)}{\rho^2} \end{array} \right)[/tex]

Calculating the determinant of this matrix in Maple gives the expression:

[tex]det(g_{ab})={\frac{1}{\delta}}\left(- \left( \sin \left( \theta \right) \left( \Delta\,{a}^{2}-{r

}^{4}-2\,{r}^{2}{a}^{2}-{a}^{4} \right) \right) ^{2}\Delta+ \left(

\sin \left( \theta \right) \left( \Delta\,{a}^{2}-{r}^{4}-2\,{r}^{2}{

a}^{2}-{a}^{4} \right) \right) ^{2}{a}^{2} \left( \sin \left( \theta

\right) \right) ^{2}+4\,{a}^{2} \left( \sin \left( \theta \right)

\right) ^{4}{r}^{4}+8\,{a}^{4} \left( \sin \left( \theta \right)

\right) ^{4}{r}^{2}[/tex]

[tex]-8\,{a}^{2} \left( \sin \left( \theta \right)

\right) ^{4}{r}^{2}\Delta+4\,{a}^{6} \left( \sin \left( \theta

\right) \right) ^{4}-8\,{a}^{4} \left( \sin \left( \theta \right)

\right) ^{4}\Delta+4\,{a}^{2} \left( \sin \left( \theta \right)

\right) ^{4}{\Delta}^{2}[/tex][tex])[/tex]

Are my calculations correct? And how can I find the coordinates where it goes towards infinity in an analytical way? I'm not even sure how to plot the determinant on a computer given I'm not entirely used to Boyer-Lindquist coordinates.

Edit: Pardon the "bad" thread title, I pushed the submit button rather prematurely.

[tex]ds^2=\frac{\Delta}{\rho^2}(dt-a sin^2 \theta d\phi)^2-\frac{sin^2\theta}{\rho^2}((r^2+a^2)d\phi-a dt)^2-\frac{\rho^2}{\Delta}dr^2-\rho^2d\phi[/tex]

Then I break out the terms [tex]dt^2, dr^2, d\theta^2, d\phi^2[/tex] and [tex]dt d\phi[/tex], this gives the metric:

[tex]g_{ab} = \left(\begin{array}{cccc} \frac{\Delta-a^2 sin^2\theta}{\rho^2} & 0 & 0 & \frac{2a sin^2\theta (r^2+a^2-\Delta)}{\rho^2} \\ 0 & -\frac{\rho^2}{\Delta} & 0 & 0 \\ 0 & 0 & \rho^2 & 0 \\ \frac{2a sin^2\theta (r^2+a^2-\Delta)}{\rho^2} & 0 & 0 & \frac{sin^2 \theta (\Delta a^2 -(r^2 + a^2)^2)}{\rho^2} \end{array} \right)[/tex]

Calculating the determinant of this matrix in Maple gives the expression:

[tex]det(g_{ab})={\frac{1}{\delta}}\left(- \left( \sin \left( \theta \right) \left( \Delta\,{a}^{2}-{r

}^{4}-2\,{r}^{2}{a}^{2}-{a}^{4} \right) \right) ^{2}\Delta+ \left(

\sin \left( \theta \right) \left( \Delta\,{a}^{2}-{r}^{4}-2\,{r}^{2}{

a}^{2}-{a}^{4} \right) \right) ^{2}{a}^{2} \left( \sin \left( \theta

\right) \right) ^{2}+4\,{a}^{2} \left( \sin \left( \theta \right)

\right) ^{4}{r}^{4}+8\,{a}^{4} \left( \sin \left( \theta \right)

\right) ^{4}{r}^{2}[/tex]

[tex]-8\,{a}^{2} \left( \sin \left( \theta \right)

\right) ^{4}{r}^{2}\Delta+4\,{a}^{6} \left( \sin \left( \theta

\right) \right) ^{4}-8\,{a}^{4} \left( \sin \left( \theta \right)

\right) ^{4}\Delta+4\,{a}^{2} \left( \sin \left( \theta \right)

\right) ^{4}{\Delta}^{2}[/tex][tex])[/tex]

Are my calculations correct? And how can I find the coordinates where it goes towards infinity in an analytical way? I'm not even sure how to plot the determinant on a computer given I'm not entirely used to Boyer-Lindquist coordinates.

Edit: Pardon the "bad" thread title, I pushed the submit button rather prematurely.

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