Getting a conserved charge out of the Kerr metric

In summary, the conversation discusses the computation of the Komar integral for the Kerr metric, which is a conserved charge associated with the Killing vector R=∂φ. The Kerr metric is given by a complicated equation and involves calculating the inverse metric. The equation for the Komar integral involves the unit normal vectors and the induced metric on the boundary of the spacelike-hypersurface, which lies at infinity. The equations for the normal vectors and induced metric are given, but there are difficulties in evaluating them at infinity. The summary concludes with a request for help and gratitude for any assistance.
  • #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! :biggrin:
 
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  • #2
JD_PM said:
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##.
 
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  • #3
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
 
  • #4
I was overcomplicating things! Actually it was pretty easy, I feel a bit ashamed! :doh:

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*}
 
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Related to Getting a conserved charge out of the Kerr metric

1. What is the Kerr metric?

The Kerr metric is a mathematical solution to Einstein's field equations in general relativity that describes the spacetime around a rotating black hole.

2. What is a conserved charge in the context of the Kerr metric?

A conserved charge in the Kerr metric refers to a physical quantity, such as energy or angular momentum, that remains constant as a particle moves along a geodesic in the spacetime described by the Kerr metric.

3. How do you calculate a conserved charge from the Kerr metric?

To calculate a conserved charge from the Kerr metric, you need to use the Killing vector associated with the specific symmetry of the metric. This vector represents the conserved quantity and can be used to calculate its value at any point in the spacetime.

4. Can conserved charges be measured in real-world scenarios?

Yes, conserved charges can be measured in real-world scenarios, such as in astrophysical observations of black holes. By studying the motion of particles around a black hole, scientists can calculate the conserved charges and use them to understand the properties of the black hole and its surrounding spacetime.

5. What are the practical applications of understanding conserved charges in the Kerr metric?

Understanding conserved charges in the Kerr metric can help us better understand the behavior of rotating black holes and their effects on surrounding matter and spacetime. This knowledge can also have practical applications in fields such as astrophysics, where the study of black holes is crucial for understanding the universe.

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