- #1
JD_PM
- 1,131
- 158
- 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!
\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!