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

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:
$$
\begin{aligned}
d s^{2}=&\left(1-\frac{2 m r}{\Sigma}\right) c^{2} d t^{2}-\frac{\Sigma}{\Delta} d r^{2}-\Sigma d \theta^{2}-\left(r^{2}+a^{2}+\frac{2 m a^{2} r \sin ^{2} \theta}{\Sigma}\right) \sin ^{2} \theta d \phi^{2} \
&+\frac{4 m r a \sin ^{2} \theta}{\Sigma} c d t d \phi,
\end{aligned}
$$
116
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
117
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.

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

物理代考

6.6.1 度量和视野
从天体物理学的角度来看,纯粹的非旋转黑洞是不可能的。一般来说,黑洞会拥有自旋或非零角动量。引力波天文台LIGO和Virgo这几年发现的黑洞都是一般的,自转具有不可忽略的角动量。只有当发现角动量很小时,Schwarzschild 解才近似适用。在本节中,我们将简要描述旋转黑洞,它体现在克尔解(1963 年)中,描述了旋转质量之外的时空。这可能是爱因斯坦方程除了史瓦西方程之外最重要的精确解。

我们在 BoyerLindquist 坐标中写下具有角动量 $J$ 的质量 $M$ 的度量:
$$
\开始{对齐}
ds^{2}=&\left(1-\frac{2 mr}{\Sigma}\right) c^{2} dt^{2}-\frac{\Sigma}{\Delta} dr^{2 }-\Sigma d \theta^{2}-\left(r^{2}+a^{2}+\frac{2 ma^{2} r \sin ^{2} \theta}{\Sigma} \right) \sin ^{2} \theta d \phi^{2} \
&+\frac{4 m r a \sin ^{2} \theta}{\Sigma} c d t d \phi,
\end{对齐}
$$
116
6 史瓦西解和黑洞
其中$a=J / c M$是黑洞每单位质量的角动量,$m=G M / c^{2}$。 $a$ 和 $m$ 都有长度维度。数量 $\Delta$ 和 $\Sigma$ 由下式定义,
$$
\Delta=r^{2}-2 m r+a^{2}, \quad \Sigma=r^{2}+a^{2} \cos ^{2} \theta
$$
这是一个轴对称的解决方案。它不像 Schwarzschild 解那样是球对称的。如果我们将 $a=0$ 放入方程式描述的 Kerr 度量中。 (6.6.1) 和 (6.6.2),然后度量减少到 Schwarzschild 度量。

地平线由方程 $\Delta=0$ 的根给出。由于这是一个二次方程,它有两个根 $r_{\pm}$。较大的根 $r_{+}=m+\sqrt{m^{2}-a^{2}}$ 定义了地平线。另一个根 $r=r_{-}$ 也定义了一个地平线,但由于 $r_{-}2 \mathrm{~m }$。这些被称为静态观察者。现在我们对 Kerr 解决方案提出同样的问题。只有当 $g_{00}>0$ 时,观察者才能保持在固定的 $(r, \theta, \phi)$。注意下标0代表$x^{0}=c t$。因此,当 $g_{00}=0$ 或当:
其中点表示关于适当时间 $\tau$ 的导数(注意我们已经吸收了 $c$ 的因子,因此我们有常数的正确维度)和
$6.6$ 克尔时空和旋转黑洞
117
我们也没有明确写出指标。取极限为 $r \longrightarrow \infty$,我们发现 $g_{0 \phi} \longrightarrow 0$ 而 $g_{\phi \phi} \longrightarrow-r^{2} \sin ^{2} \theta $ 因此 $g_{0 \phi} c \dot{t}+g_{\phi \phi} \dot{\phi} \longrightarrow-r^{2} \sin ^{2} \theta \dot{\ phi} \equiv$ $-L$,其中 $L$ 是测试粒子在无穷远处的每单位质量的角动量。因此,方程中的常数是 $-L$。 (6.6.4)。从无穷大的静止开始的粒子有 $L=0$(为什么?),因此:
$$
\frac{d \phi}{dt}=\frac{\dot{\phi}}{\dot{t}}=-\frac{c g_{0 \phi}}{g_{\phi \phi}} =\omega(r, \theta)>0
$$
因此,落入黑洞的粒子必然与黑洞共同旋转。事实上,如果观察者的角速度 $d \phi / d t=\omega(r, \theta)$ 如公式 1 所示。 $(6.6 .5)$,那么这样的观察者被称为零角动量观察者或 ZAMO。它们也被称为局部非旋转观察者或 LNRO。从物理的角度来看,这些观察者起着重要的作用。见米斯纳等人。 (1973 年)。所有这些效果都称为帧拖动。

最后,正如彭罗斯所说,能层可用于从黑洞中提取能量。考虑该区域中坐标 $x^{0}=c t$ 的切向量 $\xi$。它是类空间的,因为 $\xi \cdot \xi=g_{00}<0$。存在与 $\xi$ 的标量积为负的类时向量(参见练习 1)。因此,可以设想一个动量向量 $\mathbf{p}$(类未来),其与 $\xi$ 的标量积为负,$\mathbf{p} \cdot \xi<0$。但这只是无穷大的能量 $\mathcal{E}$ (Misner et al. (1973))。因此,在能层内可以有动量为 $\mathbf{p}$ 的粒子,使得 $\mathbf{p} \cdot \xi=\mathcal{E}<0$。然后我们可以安排一个能量为 $\mathcal{E}$ 的粒子从无穷大进入并在能层内分裂成两个粒子,其中一个粒子的 $\mathcal{E}_{2}<0$。因为能量是守恒的,我们

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