Getting a conserved charge out of the Kerr metric

[JD_PM]

Homework Statement

I am struggling to evaluate certain limits while computing the Komar integral

Relevant Equations

N/A

Compute the Komar integral for the Kerr metric

\begin{equation*}
J=-\frac{1}{8 \pi G} \int_{\partial \Sigma} d^2 x \sqrt{\gamma^{(2)}} n_{\mu} \sigma_{\nu} \nabla^{\mu} R^{\nu}
\end{equation*}

The Kerr metric is given by

\begin{align*}

(ds)^2 &= -\left(1-\frac{2GMr}{\rho^2} \right)(dt)^2 - \frac{2GMar \sin^2 \theta}{\rho^2}(dt d\phi + d\phi dt) \\

&+ \frac{\rho^2}{\Delta}(dr)^2 + \rho^2 (d \theta)^2 + \frac{\sin^2 \theta}{\rho^2} \left[ (r^2+a^2)^2-a^2 \Delta \sin^2 \theta \right] (d \phi)^2

\end{align*}

Where

\begin{equation*}

\Delta = r^2 -2GMr+a^2

\end{equation*}

\begin{equation*}

\rho^2 = r^2+a^2 \cos^2 \theta

\end{equation*}The idea is to compute the conserved charge associated to the Killing vector $R=\partial_{\phi}$ via the Komar integral.

First off, we need to compute the inverse metric:

\begin{equation*}
g^{rr}= \frac{\Delta}{\rho^2}
\end{equation*}

\begin{equation*}
g^{\theta \theta}= \frac{1}{\rho^2}
\end{equation*}

\begin{equation*}
\begin{pmatrix}
g^{tt} & g^{t\phi} \\
g^{tt} & g^{t\phi} \\
\end{pmatrix}=
\begin{pmatrix}
g^{tt} & g^{t\phi} \\
g^{\phi t} & g^{\phi\phi} \\
\end{pmatrix}^{-1}=\frac{1}{g_{tt}g_{\phi\phi}-g_{t \phi}^2}\begin{pmatrix}
g_{\phi \phi} & -g_{t\phi} \\
-g_{\phi t} & g_{t t} \\
\end{pmatrix}
\end{equation*}

Where the following is particularly tedious to compute

\begin{equation*}
g_{tt}g_{\phi\phi}-g_{t \phi}^2 = -\Delta \sin^2 \theta
\end{equation*}

I got that the spacelike-hypersurface $\Sigma$ has the following unit normal vector $n_{\mu}$ associated to it

\begin{equation*}
n_{\mu} = \left( -\Delta^{1/2} \rho\left[ (r^2+a^2)^2 - a^2 \Delta \sin^2 \theta\right]^{-1/2},0,0,0 \right) \tag{1}
\end{equation*}

The boundary of $\Sigma$ i.e. $\partial \Sigma$ has the following unit normal vector $\sigma_{\mu}$ associated to it

\begin{equation*}
\sigma_{\mu} = \left(0,\frac{\rho}{\sqrt{\Delta}},0,0 \right) \tag{2}
\end{equation*}

I checked $(1)$ and $(2)$ and they are OK. My doubts come later.

The Killing vector $R$ can be expressed in component form

\begin{equation*}
R^{\mu} = (0,0,0,1)
\end{equation*}

Thus $\partial_{\mu} R^{\nu}=0$. Now, let us compute $n_{\mu}\sigma_{\nu} \nabla^{\mu} R^{\nu}$

\begin{align*}
n_{\mu}\sigma_{\nu} \nabla^{\mu} R^{\nu} &= n_{\mu}\sigma_{\nu}g^{\mu \rho} \nabla_{\rho} R^{\nu} \\
&= n_{\mu}\sigma_{\nu}g^{\mu \rho} \left( \partial_{\rho} R^{\nu} + \Gamma_{\rho \sigma}^{\nu} R^{\sigma}\right) \\
&= n_{\mu}\sigma_{\nu}g^{\mu \rho}\Gamma_{\rho \phi}^{\nu} R^{\phi} \\
&= n_{t}\sigma_{r}g^{t \rho}\Gamma_{\rho \phi}^{r} R^{\phi} \\
&= n_{t}\sigma_{r}\left( g^{tt}\Gamma_{t \phi}^{r} + g^{t \phi}\Gamma_{\phi \phi}^{r}\right)
\end{align*}

OK so far.

As $\partial \Sigma$ lies at infinity, we only need their leading-order behavior as $r \to \infty$

\begin{equation*}
\Gamma_{t \phi}^{r} \to -\frac{GMa \sin^2 \theta}{r^2} \tag{3}
\end{equation*}

\begin{equation*}
\Gamma_{\phi \phi}^{r} \to -r \sin^2 \theta \tag{4}
\end{equation*}

\begin{equation*}
n_{t} \to -1 \tag{5}
\end{equation*}

\begin{equation*}
\sigma_{r} \to 1 \tag{6}
\end{equation*}

\begin{equation*}
g^{tt} \to -1 \tag{7}
\end{equation*}

\begin{equation*}
g^{t \phi} \to -\frac{2GMa}{r^3} \tag{8}
\end{equation*}

Which leads to

\begin{equation*}
n_{\mu}\sigma_{\nu} \nabla^{\mu} R^{\nu} \to -\frac{3GMa \sin^2 \theta}{r^2} \tag{9}
\end{equation*}

The induced metric on $\partial \Sigma$ is given by

\begin{equation*}
\gamma^{(2)}=\gamma^{(2)}_{\theta \theta} \gamma^{(2)}_{\theta \theta}= \sin^2 \theta \left[ (r^2+a^2)^2 - a^2 \Delta \sin^2 \theta\right]
\end{equation*}

Whose asymptotic behavior is

\begin{equation*}
\sqrt{\gamma^{(2)}} \to r^2 \sin \theta \tag{10}
\end{equation*}

I am aimed at checking $(3),(4),(5),(6),(7),(8),(9)$ and $(10)$

Checking (3)

Applying the Christoffel symbol formula

\begin{align*}
\Gamma_{t \phi}^{r} &= \frac 1 2 g^{r \sigma} \left( \partial_t g_{\phi \sigma} + \partial_{\phi} g_{t \sigma} - \partial_{\sigma} g_{t \phi}\right) \\
&= \frac{\Delta}{2 \rho^2} \left[ \partial_r \left( \frac{4GMar \sin^2 \theta}{\rho^2} \right)\right]
\end{align*}

At this point, I thought of computing all brute force and the applying $\lim_{r \to \infty}$ via Hopital. But I do not get $(3)$. I get something of the form

\begin{equation*}
\Gamma_{t \phi}^{r} \sim -\frac{r^2-r}{r^4+2r^2}\frac{3r^4+r^2}{(r^4+r^2)^2} \cancel{\to} -\frac{1}{r^2}
\end{equation*}

Mmm... what am I missing? should I approach it differently?

Checking (4)

I encounter the exact same issue as above.

Checking (5)

Similar issue: how to evaluate

\begin{equation*}
\lim_{r\to\infty} \left( -\Delta^{1/2} \rho\left[ (r^2+a^2)^2 - a^2\Delta \sin^2 \theta \right]^{-1/2}\right)
\end{equation*}

?

I am having really similar difficulties regarding $(6),(7),(8),(9)$

Regarding $(10)$; as $r^4 >> 2r^2$ at $r \to \infty$ we get

\begin{equation*}
\sqrt{\gamma^{(2)}} = \sqrt{\sin^2 \theta \left[ r^4+2(ra)^2+a^4 - a^2 \Delta \sin^2 \theta\right]} \to r^2 \sin \theta \tag{*}
\end{equation*}

Is this OK? If yes at least I would have got one!

I appreciate your help.

Thank you!

 

[TSny]

Checking (3)

Applying the Christoffel symbol formula

$$ \Gamma_{t \phi}^{r} = \frac 1 2 g^{r \sigma} \left( \partial_t g_{\phi \sigma} + \partial_{\phi} g_{t \sigma} - \partial_{\sigma} g_{t \phi}\right)
= \frac{\Delta}{2 \rho^2} \left[ \partial_r \left( \frac{4GMar \sin^2 \theta}{\rho^2} \right)\right]$$

I think the factor of 4 should just be a factor of 2.

Note that both $\Delta$ and $\rho^2$ approach $r^2$ for large $r$. So, $\large \frac{\Delta}{\rho^2}$ $\to 1$ and $\large \frac{r}{\rho^2} \to$ $1/r$.

 

[JD_PM]

Oh so I think we do not need to use l'Hôpital to evaluate the asymptotic behavior of
$(3),(4),(5),(6),(7),(8),(9),(10)$

I will check it again and post what I get

 

[JD_PM]

I was overcomplicating things! Actually it was pretty easy, I feel a bit ashamed!

Checking (3)

\begin{align*}
\Gamma_{t \phi}^{r} &= \frac 1 2 g^{r \sigma} \left( \cancel{\partial_t g_{\phi \sigma}} + \cancel{\partial_{\phi} g_{t \sigma}} - \partial_{\sigma} g_{t \phi}\right) \\
&= \frac{\Delta}{2 \rho^2} \left[ \partial_r \left( \frac{2GMar \sin^2 \theta}{\rho^2} \right)\right] \\
&\to \frac 1 2 \left[ \partial_r \left( \frac{2GMa \sin^2 \theta}{r} \right)\right] \\
&= -\frac{GMa \sin^2 \theta}{r^2}
\end{align*}

Where we used $\large \frac{\Delta}{\rho^2} \to 1$ and $\large \frac{r}{\rho^2} \to 1/r$

Checking (4)

\begin{align*}
\Gamma_{\phi \phi}^{r} &= \frac 1 2 g^{r \sigma} \left( \cancel{\partial_{\phi} g_{\phi \sigma}} + \cancel{\partial_{\phi} g_{\phi \sigma}} - \partial_{\sigma} g_{\phi \phi}\right) \\
&= -\frac{\Delta}{2 \rho^2} \left[ \partial_r \left( \frac{\sin^2 \theta \left( (r^2+a^2)^2-a^2 \Delta \sin^2 \theta\right)}{\rho^2} \right) \right] \\
&\to -\frac{\sin^2 \theta}{2} \left[ \partial_r \left( \frac{r^4}{r^2} \right)\right]=-\frac{\sin^2 \theta}{2} \left[ \partial_r r^2 \right] \\
&= -r \sin^2 \theta
\end{align*}

Where we used $\large \frac{\Delta}{\rho^2} \to 1$ and $\large \frac{\left( (r^2+a^2)^2-a^2 \Delta \sin^2 \theta\right)}{\rho^2} \to r^2$

Checking (5)

\begin{align*}
n_t &= -\Delta^{1/2} \rho\left[ (r^2+a^2)^2 - a^2 \Delta \sin^2 \theta\right]^{-1/2} \\
&\to -r^2\left( r^{-2} \right) \\
&= -1
\end{align*}

Where we used $\Delta^{1/2} \rho \to r^2$ and $\left[ (r^2+a^2)^2 - a^2 \Delta \sin^2 \theta\right]^{-1/2} \to r^{-2}$

Checking (6)

\begin{equation*}
\sigma_r = \frac{\rho}{\sqrt{\Delta}} \to 1
\end{equation*}

Checking (7)

\begin{align*}
g^{tt} &= \frac{g_{\phi \phi}}{g_{tt}g_{\phi \phi} - g^2_{t\phi}} \\
&\to \frac{-r^2 \sin^2 \theta}{r^2 \sin^2 \theta} \\
&= -1
\end{align*}

Where we used $g_{\phi \phi} \to r^2 \sin^2 \theta$ and $\Delta \to r^2$

Checking (8)

\begin{align*}
g^{t \phi} &= -\frac{g_{t \phi}}{g_{tt}g_{\phi \phi} - g^2_{t\phi}} \\
&= \frac{2GM a r}{-\Delta \rho^2} \\
&\to - \frac{2GMa}{r^3}
\end{align*}

Thus Checking (9) is straightforward

\begin{align*}
n_{\mu} \sigma_{\nu} \nabla^{\mu} R^{\nu} &= n_{t}\sigma_{r}\left( g^{tt}\Gamma_{t \phi}^{r} + g^{t\phi}\Gamma_{\phi \phi}^{r}\right) \\
&\to -\left(-\left(-\frac{GMa \sin^2 \theta}{r^2} \right) + \left(-\frac{2GMa}{r^3} \right) \left(-r \sin^2 \theta \right) \right) \\
&= -\frac{3GM a \sin^2 \theta}{r^2}
\end{align*}

(10) was OK

Thus the Komar integral yields

\begin{align*}
J&=-\frac{1}{8 \pi G} \int_{\partial \Sigma} d^2 x \sqrt{\gamma^{(2)}} n_{\mu} \sigma_{\nu} \nabla^{\mu} R^{\nu} \\
&= -\frac{1}{8 \pi G} \int_{0}^{2\pi} d \phi \int_{0}^{\pi} d \theta (r^2 \sin \theta) \left( -\frac{3GM a \sin^2 \theta}{r^2} \right) \\
&= \frac{3Ma}{8 \pi} \int_{0}^{2 \pi} d \phi \int_{0}^{\pi} d \theta \sin^3 \theta \\
&= Ma
\end{align*}