## 物理代写|广义相对论代写General relativity代考|Some Elementary Properties of the Kerr Solution

2023年4月11日

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## 物理代写|广义相对论代写General relativity代考|Some Elementary Properties of the Kerr Solution

The utmost valuable configuration of Kerr metric is Boyer-Lindquist form for the study of the elementary properties of Kerr solution. One can rewrite the Boyer-Lindquist form as
$$d s^2=A d t^2-B(d \phi-\omega d t)^2-C d r^2-D d \theta^2$$
where
$$A=\frac{\triangle R^2}{\Sigma^2}, B=\frac{\Sigma^2}{R^2} \sin ^2 \theta, C=\frac{R^2}{\triangle}, D=R^2, \omega=-\frac{g_{t \phi}}{g_{\phi \phi}}=\frac{2 m r a}{\Sigma^2},$$
with
\begin{aligned} & \triangle=r^2-2 m r+a^2, R^2=r^2+a^2 \cos ^2 \theta, \ & \Sigma^2=\left(r^2+a^2\right)^2-a^2 \triangle \sin ^2 \theta=\frac{R^4 \triangle-4 a^2 m^2 r^2 \sin ^2 \theta}{\Delta-a^2 \sin ^2 \theta} . \end{aligned}
Here, the fundamental tensor takes the form
$$g_{\mu v}=\left(\begin{array}{cccc} A-\omega^2 B & 0 & 0 & \omega B \ 0 & -C & 0 & 0 \ 0 & 0 & -D & 0 \ \omega B & 0 & 0 & -B \end{array}\right) .$$
Here,
$$g_{t t}=A-\omega^2 B=1-\frac{2 m r}{R^2}, g_{t \phi}=\omega B, g_{r r}=-C, g_{\theta \theta}=-D, g_{\phi \phi}=-B .$$

The contravariant form of the fundamental tensor is
$$g^{\mu \nu}=\left(\begin{array}{cccc} \frac{1}{A} & 0 & 0 & \frac{\omega}{A} \ 0 & -\frac{1}{C} & 0 & 0 \ 0 & 0 & -\frac{1}{D} & 0 \ \frac{\omega}{A} & 0 & 0 & \frac{\omega^2 B-A}{A B} \end{array}\right) .$$
Note that the metric coefficients do not depend on $t$ and $\phi$. This implies that the solution is both stationary and axially symmetric (a solution being axially symmetric means that the solution is invariant under rotation about a fixed axis). As the Kerr metric is stationary and axially symmetric, it possesses the following Killing vectors $t^\alpha=\frac{\partial x^\alpha}{\partial t}=(1,0,0,0)$ and $\phi^\alpha=\frac{\partial x^\alpha}{\partial \phi}=(0,0,0,1)$.

This indicates that the orbits of the Killing vector field $\frac{\partial x^a}{\partial \phi}$ admit the continuous symmetries for the curves $t=$ constant, $r=$ constant, $\theta=$ constant, which are circles. This approves that a spinning source provides the Kerr field. Thus, Kerr solution represents a vacuum field exterior to a spinning source. As $a \rightarrow 0$, we reproduce the Schwarzschild line element. Also for $a \rightarrow 0$ and $r \rightarrow \infty$, we have $g_{a b} \rightarrow \eta_{a b}$ so that Kerr solution is asymptotically flat.
The determinant of the metric yields $m$ independent form as
$$\operatorname{det}\left(g_{\mu v}\right)=-\sin ^2 \theta\left(r^2+a^2 \cos ^2 \theta\right)^2$$

## 物理代写|广义相对论代写General relativity代考|Singularities and Horizons

It is obvious that the fundamental tensor of Kerr solution and its contravariant form have singularities at $\triangle=0$ and $R^2=0$. Now one needs to calculate Kretschmann $R^{a b c d} R_{a b c d}$ of the Kerr spacetime to distinguish between coordinate and curvature singularities. Here,
$$R^{a b c d} R_{a b c d}=\frac{48 m^2\left(r^2-a^2 \cos ^2 \theta\right)\left(R^4-16 a^2 r^2 \cos ^2 \theta\right)}{R^{12}} .$$
This indicates that the metric has just a coordinate singularity at $\triangle=0$ (by setting $g_{\phi t}^2-g_{t t} g_{\phi \phi}=0$ ). However, the Kerr spacetime has a real physical singularity at $R^2=0 . R^2=0$ implies
$$R^2=r^2+a^2 \cos ^2 \theta=0$$
It follows that
$$r=0 \text { and } \cos \theta=0 \text { or } \theta=\frac{\pi}{2}$$
Now, Eqs. (10.20) and (10.21) yield
$$x^2+y^2=a^2, z=0 .$$
This is a ring-type singularity having radius $a$ that lies in the $z=0$ equatorial plane.

The event horizon is a surface at which the radial coordinate $\mathrm{r}$ reverses its signature. Thus, the Kerr spacetime has horizons where the four velocity of a viewer tends to zero, or the surface $r=$ constant becomes null. The event horizon is a solution of $\triangle=0$. This yields
$$r^2-2 m r+a^2=\left(r-r_{+}\right)\left(r-r_{-}\right)=0$$
i.e.,
$$r=r_{ \pm}=m \pm \sqrt{m^2-a^2}$$
Note that $R^{a b c d} R_{a b c d}$ remains finite at $r_{ \pm}$. Thus, the points $r=r_{ \pm}$are coordinate singularity rather than a curvature singularity. Also when $a \rightarrow 0$ (nonspinning limit), we obtain $r=2 m$ or $0 . r=2 m$ is the position of the horizon in the Schwarzschild geometry. This indicates that we can recognize $r_{ \pm}$as the positions of the inner and outer horizons $\left(r=r_{+}\right.$is the outer horizon and $r=r_{-}$is the inner horizon). One can refer the region $r<r_{+}$as the interior of the Kerr black hole (see Fig. 93).

# 广义相对论代考

## 物理代写|广义相对论代写General relativity代考|Some Elementary Properties of the Kerr Solution

$$d s^2=A d t^2-B(d \phi-\omega d t)^2-C d r^2-D d \theta^2$$

$$A=\frac{\triangle R^2}{\Sigma^2}, B=\frac{\Sigma^2}{R^2} \sin ^2 \theta, C=\frac{R^2}{\triangle}, D=R^2, \omega=-$$

$$\triangle=r^2-2 m r+a^2, R^2=r^2+a^2 \cos ^2 \theta, \quad \Sigma^2=$$

$$g_{t t}=A-\omega^2 B=1-\frac{2 m r}{R^2}, g_{t \phi}=\omega B, g_{r r}=-C, g_{\theta \theta}$$

$$\operatorname{det}\left(g_{\mu v}\right)=-\sin ^2 \theta\left(r^2+a^2 \cos ^2 \theta\right)^2$$

## 物理代写|广义相对论代写General relativity代考|Singularities and Horizons

$$R^{a b c d} R_{a b c d}=\frac{48 m^2\left(r^2-a^2 \cos ^2 \theta\right)\left(R^4-16 a^2 r^2 \cos \right.}{R^{12}}$$

$$R^2=r^2+a^2 \cos ^2 \theta=0$$

$$r=0 \text { and } \cos \theta=0 \text { or } \theta=\frac{\pi}{2}$$

$$x^2+y^2=a^2, z=0 \text {. }$$

$\mathrm{IE}$
$$r=r_{ \pm}=m \pm \sqrt{m^2-a^2}$$

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