# 物理代写| Kerr Spacetime and the Rotating Black Hole 相对论代考

## 物理代写

6.6.1 The Metric and the Horizon
From the astrophysical point of view, a purely non-rotating black hole would be unlikely. Generically, a black-hole would possess spin or non-zero angular momentum. The black holes found in the past few years by the gravitational wave observatories LIGO and Virgo are all in general, spinning having non-negligible angular momentum. Only if the angular momentum is found to be small, then the Schwarzschild solution would approximately be applicable. In this section we briefly describe the rotating black hole which is embodied in the Kerr solution (1963) which describes the spacetime outside a spinning mass. This is perhaps the most important exact solution of Einstein’s equation apart from the Schwarzschild.

We write down the metric for a mass $M$ with angular momentum $J$ in BoyerLindquist coordinates:
\begin{aligned} d s^{2}=&\left(1-\frac{2 m r}{\Sigma}\right) c^{2} d t^{2}-\frac{\Sigma}{\Delta} d r^{2}-\Sigma d \theta^{2}-\left(r^{2}+a^{2}+\frac{2 m a^{2} r \sin ^{2} \theta}{\Sigma}\right) \sin ^{2} \theta d \phi^{2} \ &+\frac{4 m r a \sin ^{2} \theta}{\Sigma} c d t d \phi, \end{aligned}
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6 Schwarzschild Solution and Black Holes
where $a=J / c M$ is the angular momentum per unit mass of the black hole and $m=G M / c^{2}$. Both $a$ and $m$ have dimensions of length. The quantities $\Delta$ and $\Sigma$ are defined by,
$$\Delta=r^{2}-2 m r+a^{2}, \quad \Sigma=r^{2}+a^{2} \cos ^{2} \theta$$
This is an axially symmetric solution. It is not spherically symmmetric like the Schwarzschild solution. If we put $a=0$ in the Kerr metric described by Eqs. (6.6.1) and (6.6.2), then the metric reduces to the Schwarzschild metric.

The horizon is given by the root of the equation $\Delta=0$. Since this is a quadratic, it has two roots $r_{\pm}$. The larger root $r_{+}=m+\sqrt{m^{2}-a^{2}}$ defines the horizon. The other root $r=r_{-}$also defines a horizon but since $r_{-}2 \mathrm{~m}$. These are called static observers. Now we ask the same question for the Kerr solution. An observer can remain at a fixed $(r, \theta, \phi)$ only when $g_{00}>0$. Note that the subscript 0 represents $x^{0}=c t$. Thus the limit for such observers is obtained when $g_{00}=0$ or when:
where the dot represents derivative with respect to the proper time $\tau$ (note we have absorbed the factor of $c$ so that we have the right dimensions for the constant) and
$6.6$ Kerr Spacetime and the Rotating Black Hole
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also we have not explicitly written the metric. By taking limits as $r \longrightarrow \infty$, we find $g_{0 \phi} \longrightarrow 0$ while $g_{\phi \phi} \longrightarrow-r^{2} \sin ^{2} \theta$ and hence $g_{0 \phi} c \dot{t}+g_{\phi \phi} \dot{\phi} \longrightarrow-r^{2} \sin ^{2} \theta \dot{\phi} \equiv$ $-L$, where $L$ is the angular momentum per unit mass of the test particle at infinity. Thus the constant is $-L$ in Eq. (6.6.4). A particle starting from rest from infinity has $L=0$ (why?) and thus:
$$\frac{d \phi}{d t}=\frac{\dot{\phi}}{\dot{t}}=-\frac{c g_{0 \phi}}{g_{\phi \phi}}=\omega(r, \theta)>0$$
Therefore, such a particle as it falls into the black hole must co-rotate with the black hole. In fact if an observer has angular velocity $d \phi / d t=\omega(r, \theta)$ as given in Eq. $(6.6 .5)$, then such observers are called zero angular momentum observers or ZAMOs. They are also called locally non-rotating observers or LNROs. Such observers play an important role from the physical point of view. See Misner et al. (1973). All these effects go by the name of frame dragging.

Lastly, as Penrose argued, that the ergosphere can be used for energy extraction from the black hole. Consider the tangent vector $\xi$ to the coordinate $x^{0}=c t$ in this region. It is spacelike because $\xi \cdot \xi=g_{00}<0$. There exist time-like vectors whose scalar product with $\xi$ is negative (see Exercise 1). Thus, one can conceive of a momentum vector $\mathbf{p}$ (future time-like) whose scalar product with $\xi$ is negative, $\mathbf{p} \cdot \xi<0$. But this is just the energy at infinity $\mathcal{E}$ (Misner et al. (1973)). Therefore, one can have particles inside the ergosphere with momentum $\mathbf{p}$ such that $\mathbf{p} \cdot \xi=\mathcal{E}<0$. We can then arrange for a particle with energy $\mathcal{E}$ to come in from infinity and split into two particles inside the ergosphere with one of them having $\mathcal{E}{2}<0$. Because energy is conserved, we have $\mathcal{E}=\mathcal{E}{1}+\mathcal{E}{2}$, and so we have $\mathcal{E}{1}>\mathcal{E}$. This particle escapes to infinity with larger energy $\mathcal{E}{1}$ than the incoming particle which had energy $\mathcal{E}$. Thus we see that we can extract energy from a rotating black hole. This energy comes at the expense of the black hole losing rotational energy, because the particle with energy $\mathcal{E}{2}<0$ falls into the hole reducing its energy.

## 物理代考

6.6.1 度量和视野

$$\开始{对齐} ds^{2}=&\left(1-\frac{2 mr}{\Sigma}\right) c^{2} dt^{2}-\frac{\Sigma}{\Delta} dr^{2 }-\Sigma d \theta^{2}-\left(r^{2}+a^{2}+\frac{2 ma^{2} r \sin ^{2} \theta}{\Sigma} \right) \sin ^{2} \theta d \phi^{2} \ &+\frac{4 m r a \sin ^{2} \theta}{\Sigma} c d t d \phi, \end{对齐}$$
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6 史瓦西解和黑洞

$$\Delta=r^{2}-2 m r+a^{2}, \quad \Sigma=r^{2}+a^{2} \cos ^{2} \theta$$

$6.6$ 克尔时空和旋转黑洞
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$$\frac{d \phi}{dt}=\frac{\dot{\phi}}{\dot{t}}=-\frac{c g_{0 \phi}}{g_{\phi \phi}} =\omega(r, \theta)>0$$

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