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Kerr metric

Note

To keep things clean we will assume \(r_g \equiv GM_\bullet/c^2 = 1\) and \(a\equiv J_\bullet c/(GM_\bullet^2)\). Dots "\(\cdot\)" will be used instead of zeros "0" for visual clarity.

Kerr-Schild coordinates

Kerr metric in Kerr-Schild coordinates is given by the following covariant tensor (and its inverse contravariant counterpart):

\[ g_{\mu \nu}=\begin{bmatrix}-(1-z) & z & \cdot & -az\sin^2{\theta}\\z & 1+z & \cdot & -a(1+z)\sin^2{\theta} \\\cdot & \cdot & \Sigma & \cdot \\-az\sin^2{\theta} & -a(1+z)\sin^2{\theta} & \cdot & \frac{A\sin^2{\theta}}{\Sigma}\end{bmatrix}~~~ g^{\mu\nu}=\begin{bmatrix}-(1+z) & z & \cdot & \cdot\\z & \frac{\Delta}{\Sigma} & \cdot & \frac{a}{\Sigma}\\\cdot & \cdot & \frac{1}{\Sigma} & \cdot\\\cdot & \frac{a}{\Sigma} & \cdot & \frac{1}{\Sigma\sin^2{\theta}}\end{bmatrix} \]
\[\begin{align*} &\Sigma=r^2+a^2\cos^2{\theta} & &\Delta = r^2-2r+a^2\\ &A=(r^2+a^2)^2-a^2\Delta \sin^2{\theta} & &z=\frac{2r}{\Sigma}\\ &\sqrt{-g}=\Sigma\sin{\theta} \end{align*}\]

3+1 formulation

To rewrite the metric in 3+1 formulation we find the lag function and the shift vector to be:

\[\begin{align*} & \beta^i=\begin{bmatrix}\frac{z}{1+z} \\ \cdot \\ \cdot\end{bmatrix} & \alpha^2=\frac{1}{1+z} \\ & \beta_i=\left[z~~ \cdot~~ -az\sin^2{\theta}\right] & \beta_i\beta^i=\frac{z^2}{1+z} \end{align*}\]

Then we find

\[ h_{ij}=\begin{bmatrix}1+z & \cdot & -a(1+z)\sin^2{\theta} \\ \cdot & \Sigma & \cdot \\ -a(1+z)\sin^2{\theta} & \cdot & \frac{A\sin^2{\theta}}{\Sigma}\end{bmatrix}~~~ h^{ij}=\begin{bmatrix}\frac{A}{\Sigma(\Sigma + 2r)} & \cdot & \frac{a}{\Sigma} \\ \cdot & \frac{1}{\Sigma} & \cdot \\ \frac{a}{\Sigma} & \cdot & \frac{1}{\Sigma\sin^2{\theta}}\end{bmatrix} \]

and \(\sqrt{h}=\Sigma\sin{\theta}/\alpha\).

It is also useful to define a locally orthonormal (tetrad) basis. This special frame is defined with the following vectors:

\[ e^i_{\hat{i}}=\begin{bmatrix}\sqrt{h^{rr}} & \cdot & \cdot \\ \cdot & 1/\sqrt{h_{\theta\theta}} & \cdot \\ -\sqrt{h^{rr}}h_{r\varphi}/h_{\varphi\varphi} & \cdot & 1/\sqrt{h_{\varphi\varphi}}\end{bmatrix}~~~ e_i^{\hat{i}}=\begin{bmatrix} 1/\sqrt{h^{rr}} & \cdot & h_{r\varphi}/\sqrt{h_{\varphi\varphi}}\\ \cdot & \sqrt{h_{\theta\theta}} & \cdot \\ \cdot & \cdot & \sqrt{h_{\varphi\varphi}}\end{bmatrix} \]

Transformations to and from the tetrad basis can be performed via \(A^{\hat{i}}=e^{\hat{i}}_j A^j\) \(A^{i}=e_{\hat{j}}^i A^{\hat{j}}\) \(a_{\hat{i}}=e_{\hat{i}}^j a_{j}\) and \(a_{i}=e_{i}^{\hat{j}} a_{\hat{j}}\). In the more explicit simplified form this reads:

\[\begin{aligned} & A^{\hat{r}}=\frac{1}{\sqrt{h^{rr}}}A^r & & a_{\hat{r}}=\sqrt{h^{rr}}a_r-\sqrt{h^{rr}}\frac{h_{r\varphi}}{h_{\varphi\varphi}}a_\varphi \\ & A^{\hat{\theta}}=\sqrt{h_{\theta\theta}}A^\theta & & a_{\hat{\theta}}=\frac{1}{\sqrt{h_{\theta\theta}}}a_\theta \\ & A^{\hat{\varphi}}=\frac{h_{r\varphi}}{\sqrt{h_{\varphi\varphi}}}A^r+\sqrt{h_{\varphi\varphi}}A^\varphi & & a_{\hat{\varphi}}=\frac{1}{\sqrt{h_{\varphi\varphi}}}a_\varphi \end{aligned}\]

\(a=0\) (Schwarzschild solution)

\[ g_{\mu \nu}=\begin{bmatrix}-(1-z) & z & \cdot & \cdot\\z & 1+z & \cdot & \cdot \\\cdot & \cdot & \Sigma & \cdot \\\cdot & \cdot & \cdot & \frac{A\sin^2{\theta}}{\Sigma}\end{bmatrix}=\begin{bmatrix}-\left(1-\frac{2}{r}\right) & \frac{2}{r} & \cdot & \cdot\\\frac{2}{r} & 1 + \frac{2}{r} & \cdot & \cdot \\\cdot & \cdot & r^2 & \cdot \\\cdot & \cdot & \cdot & r^2\sin^2{\theta}\end{bmatrix} \]
\[ g^{\mu\nu}=\begin{bmatrix}-(1+z) & z & \cdot & \cdot\\z & \frac{\Delta}{\Sigma} & \cdot & \cdot\\\cdot & \cdot & \frac{1}{\Sigma} & \cdot\\\cdot & \cdot & \cdot & \frac{1}{\Sigma\sin^2{\theta}}\end{bmatrix}=\begin{bmatrix}-\left(1+\frac{2}{r}\right) & \frac{2}{r} & \cdot & \cdot\\\frac{2}{r} & 1-\frac{2}{r} & \cdot & \cdot \\\cdot & \cdot & \frac{1}{r^2} & \cdot \\\cdot & \cdot & \cdot & \frac{1}{r^2\sin^2{\theta}}\end{bmatrix} \]
\[\begin{align*} & \Sigma=r^2 & & \Delta = r^2-2r \\ & A=r^4 & & z=\frac{2}{r} \end{align*}\]

And the 3+1 components reduce to the following:

\[\begin{align*} & \beta^i=\begin{bmatrix}\frac{2}{r+2} \\ \cdot \\ \cdot\end{bmatrix} & & \alpha^2=\frac{r}{r+2} \\ & \beta_i=\left[2/r ~~ \cdot ~~ -2(a/r)\sin^2{\theta}\right] & & \beta_i\beta^i=\frac{4}{r^2+2r} \end{align*}\]
\[ h_{ij}=\begin{bmatrix}1+\frac{2}{r} & \cdot & \cdot \\ \cdot & r^2 & \cdot \\ \cdot & \cdot & r^2\sin^2{\theta}\end{bmatrix}~~~ h^{ij}=\begin{bmatrix}\frac{r}{r+2} & \cdot & \cdot \\ \cdot & \frac{1}{r^2} & \cdot \\ \cdot & \cdot & \frac{1}{r^2\sin^2{\theta}}\end{bmatrix} \]

Boyer-Lindquist coordinates

\[ g_{\mu \nu}=\begin{bmatrix}-\left(1-\frac{2r}{\Sigma}\right) & \cdot & \cdot & -\frac{2ar\sin^2{\theta}}{\Sigma}\\ \cdot & \frac{\Sigma}{\Delta} & \cdot & \cdot \\ \cdot & \cdot & \Sigma & \cdot \\-\frac{2ar\sin^2{\theta}}{\Sigma} & \cdot & \cdot & \frac{A\sin^2{\theta}}{\Sigma}\end{bmatrix}~~~ g^{\mu\nu}=\begin{bmatrix}-\frac{A}{\Delta\Sigma} & \cdot & \cdot & -\frac{2ar}{\Delta\Sigma}\\ \cdot & \frac{\Delta}{\Sigma} & \cdot & \cdot \\ \cdot & \cdot & \frac{1}{\Sigma} & \cdot \\-\frac{2ar}{\Delta\Sigma} & \cdot & \cdot & \frac{\Delta -a^2\sin^2{\theta}}{\Delta\Sigma\sin^2{\theta}}\end{bmatrix} \]
\[\begin{align*} & \Sigma=r^2+a^2\cos^2{\theta} & & \Delta = r^2-2r+a^2 \\ & A=(r^2+a^2)^2-a^2\Delta \sin^2{\theta} \\ & \sqrt{-g}=\Sigma\sin{\theta} \end{align*}\]

3+1 formulation

\[\begin{align*} & \beta^i=\begin{bmatrix} \cdot \\ \cdot \\ -2ar/A\end{bmatrix} & & \alpha^2=\frac{\Delta \Sigma}{A} \\ & \beta_i=\left[~~ \cdot ~~ \cdot ~~ -2ar\sin^2{\theta}/\Sigma\right] & & \beta_i\beta^i=\frac{4a^2r^2\sin^2{\theta}}{A} \end{align*}\]
\[ h_{ij}=\begin{bmatrix}\frac{\Sigma}{\Delta} & \cdot & \cdot \\ \cdot & \Sigma & \cdot \\ \cdot & \cdot & \frac{A\sin^2{\theta}}{\Sigma}\end{bmatrix}~~~ h^{ij}=\begin{bmatrix} \frac{A}{\Delta\Sigma} & \cdot & \cdot \\ \cdot & \frac{1}{\Sigma} & \cdot \\ \cdot & \cdot & \frac{1}{\Sigma\sin^2{\theta}} + \frac{(4r^2-1)a^2}{\Delta\Sigma} \end{bmatrix} \]

\(a=0\) (Schwarzschild solution)

\[ g_{\mu \nu}=\begin{bmatrix}-\left(1-\frac{2}{r}\right) & \cdot & \cdot & \cdot \\ \cdot & \left(1-\frac{2}{r}\right)^{-1} & \cdot & \cdot \\ \cdot & \cdot & r^2 & \cdot \\ \cdot & \cdot & \cdot & r^2 \sin^2{\theta}\end{bmatrix}~~~ g^{\mu\nu}=\begin{bmatrix}-\left(1-\frac{2}{r}\right)^{-1} & \cdot & \cdot & \cdot \\ \cdot & 1-\frac{2}{r} & \cdot & \cdot \\ \cdot & \cdot & \frac{1}{r^2} & \cdot \\ \cdot & \cdot & \cdot & \frac{1}{r^2\sin^2{\theta}}\end{bmatrix} \]
\[\begin{align*} & \beta^i=\begin{bmatrix} \cdot \\ \cdot \\ \cdot \end{bmatrix} & & \alpha^2=1-\frac{2}{r} \\ & \beta_i=\left[~~\cdot ~~ \cdot ~~ \cdot ~~\right] & & \beta_i\beta^i=0 \end{align*}\]
\[ h_{ij}=\begin{bmatrix} \left(1-\frac{2}{r}\right)^{-1} & \cdot & \cdot \\ \cdot & r^2 & \cdot \\ \cdot & \cdot & r^2 \sin^2{\theta}\end{bmatrix}~~~ h^{ij}=\begin{bmatrix}1-\frac{2}{r} & \cdot & \cdot \\ \cdot & \frac{1}{r^2} & \cdot \\ \cdot & \cdot & \frac{1}{r^2\sin^2{\theta}}\end{bmatrix} \]

Conversion from KS to BL

Assuming \(g_{\mu\nu}\) is the metric in KS coordinates, and \(g_{\tilde{\mu}\tilde{\nu}}\) -- in BL.

\[ J_{\tilde{\nu}}^{\mu}=\begin{bmatrix} 1 & 2r/\Delta & \cdot& .\\ \cdot& 1 & \cdot& .\\ \cdot& \cdot& 1 & .\\ \cdot& a/\Delta & \cdot& 1 \end{bmatrix}~~~ J_{\nu}^{\tilde{\mu}}=\begin{bmatrix} 1 & -2r/\Delta & \cdot& .\\ \cdot& 1 & \cdot& .\\ \cdot& \cdot& 1 & .\\ \cdot& -a/\Delta & \cdot& 1 \end{bmatrix} \]
\[\begin{align*} & x^{\tilde{\mu}}=J^{\tilde{\mu}}_\nu x^\nu & & x^{\mu}=J^{\mu}_{\tilde{\nu}} x^{\tilde{\nu}}\\ & x_{\tilde{\mu}}=J_{\tilde{\mu}}^{\nu} x_{\nu} & & x_{\mu}=J_{\mu}^{\tilde{\nu}} x_{\tilde{\nu}} \end{align*}\]