# 物理代写|广义相对论代写General relativity代考|Null Geodesics in Kerr Spacetime

#### Doug I. Jones

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## 物理代写|广义相对论代写General relativity代考|Null Geodesics in Kerr Spacetime

We know that freely falling massless particles describe null geodesics. For the Kerr spacetime, we can construct the Lagrangian for the Boyer-Lindquist form (10.13) as
\begin{aligned} L= & \frac{\Delta}{\left(r^2+a^2 \cos ^2 \theta\right)}\left(\dot{t}-a \sin ^2 \theta \dot{\phi}\right)^2-\frac{\sin ^2 \theta}{\left(r^2+a^2 \cos ^2 \theta\right)}\left[\left(r^2+a^2\right) \dot{\phi}-a i\right]^2 \ & -\frac{\left(r^2+a^2 \cos ^2 \theta\right) \dot{r}^2}{\Delta}-\left(r^2+a^2 \cos ^2 \theta\right) \dot{\theta}^2 . \end{aligned}
[Dot indicates the derivative with respect to affine parameter $p$ ]
The corresponding Euler-Lagrange equations are
$$\frac{d}{d p}\left(\frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}=0, x=t, r, \theta, \phi .$$
We know that the first integral of geodesics equation is
$$g_{\mu \gamma} \frac{d x^\mu}{d p} \frac{d x^\gamma}{d p}=\epsilon$$
where $\epsilon=1$ for the time-like geodesics and $\epsilon=0$ for null geodesics.
Note that the Kerr metric is axially symmetric and without any loss of generality we consider the null geodesic in equatorial plane $\theta=$ constant, i.e., null geodesics lie in the hypersurface $\theta=$ constant, then $\frac{d \theta}{d p}=0$. Kerr metric coefficients do not contain $t$ and $\phi$ explicitly which means $\phi$ and $t$ are cyclic coordinates.

For the above Lagrangian, we get the following equations of motion corresponding to $x=t, \phi$ and $\theta$, respectively,
\begin{aligned} & \frac{\Delta}{\left(r^2+a^2 \cos ^2 \theta\right)}\left(i-a \sin ^2 \theta \dot{\phi}\right)+\frac{a \sin ^2 \theta}{\left(r^2+a^2 \cos ^2 \theta\right)}\left[\left(r^2+a^2\right) \dot{\phi}-a i\right]=l . \ & \frac{a \Delta \sin ^2 \theta}{\left(r^2+a^2 \cos ^2 \theta\right)}\left(i-a \sin ^2 \theta \dot{\phi}\right)+\frac{\left(r^2+a^2\right) \sin ^2 \theta}{\left(r^2+a^2 \cos ^2 \theta\right)}\left[\left(r^2+a^2\right) \dot{\phi}-a i\right]=n . \ & \frac{a^2 \Delta}{\left(r^2+a^2 \cos ^2 \theta\right)^2}\left(i-a \sin ^2 \theta \dot{\phi}\right)^2-\frac{2 a \Delta \dot{\phi}}{\left(r^2+a^2 \cos ^2 \theta\right)}(i-a \sin \theta \dot{\phi}) \ & -\frac{\left(r^2+a^2\right)}{\left(r^2+a^2 \cos ^2 \theta\right)^2}\left[\left(r^2+a^2\right) \dot{\phi}-a i\right]^2+\frac{a^2 \dot{r}^2}{\Delta}=0 . \end{aligned}

## 物理代写|广义相对论代写General relativity代考|Kerr Solution in Eddington-Finkelstein Coordinates

Now, we are interested in finding a new coordinate $(v, r, \theta, \psi)$ in which ingoing radial null geodesics becomes straight lines,
$$v=t+r^, \psi=\phi+\phi^,$$
where
\begin{aligned} r^= & \int \frac{r^2+a^2}{\Delta} d r=r+\left(\frac{m r_{+}}{\sqrt{m^2-a^2}}\right) \ln \left|\frac{r}{r_{+}}-1\right| \ & -\left(\frac{m r_{-}}{\sqrt{m^2-a^2}}\right) \ln \left|\frac{r}{r_{-}}-1\right| \ \phi^= & \int \frac{a}{\Delta} d r=\frac{a}{2 \sqrt{m^2-a^2}} \ln \left|\frac{r-r_{+}}{r-r_{-}}\right|+\text {constant. } \end{aligned}
Here, we use the transformation $t \rightarrow v, \phi \rightarrow \psi$, where
$$d v=d t+\frac{\left(r^2+a^2\right)}{\Delta} d r ; d \psi=d \phi+\frac{a}{\Delta} d r$$

For advanced time coordinate $(v, r, \theta, \psi)$ the Boyer-Lindquist line element is transformed into
\begin{aligned} d s^2= & \left(1-\frac{2 m r}{R^2}\right) d v^2-2 d v d r+\frac{4 a m r \sin ^2 \theta}{R^2} d v d \psi \ & +2 a \sin ^2 \theta d r d \psi-R^2 d \theta^2-\frac{\Sigma^2 \sin ^2 \theta}{R^2} d \psi^2 . \end{aligned}
This form is known as advanced Eddington-Finkelstein form of the Kerr solution. These coordinates produce an extension of the Kerr metric across the future horizon. Actually, the coordinates $(v, r, \theta, \psi)$ work well on the future horizon, however, not on regular, i.e., singular on the past horizon.

We can describe the outgoing radial null geodesics by using the following coordinate $(u, r, \theta, \chi)$, where
\begin{aligned} u= & t-r^, \chi=\phi-\phi^, \ d s^2= & \left(1-\frac{2 m r}{R^2}\right) d u^2+2 d u d r+\frac{4 m r a \sin ^2 \theta}{R^2} d u d \chi-2 a \sin ^2 \theta d r d \chi \ & -R^2 d \theta^2-\frac{\Sigma^2 \sin ^2 \theta}{R^2} d \chi^2 . \end{aligned}
This form is known as retarded Eddington-Finkelstein form of the Kerr solution.
Here, the coordinates $(v, r, \theta, \chi)$ are regular on the past horizon but singular on the future horizon.

# 广义相对论代考

## 物理代写|广义相对论代写General relativity代考|Null Geodesics in Kerr Spacetime

\begin{aligned} L= & \frac{\Delta}{\left(r^2+a^2 \cos ^ 2 \theta\right)}\left(\dot{t}-a \sin ^2 \theta \dot{\phi}\right)^2-\frac{\sin ^2 \theta}{\left(r ^2+a^2 \cos ^2 \theta\right)}\left[\left(r^2+a^2\right) \dot{\phi}-ai\right]^2 \ & -\ frac {\left(r^2+a^2 \cos ^2 \theta\right) \dot{r}^2}{\Delta}-\left(r^2+a^2 \cos ^2 \theta\右）\dot{\theta}^2 。 \end{aligned}
[点表示对仿射参数$p$的导数]

$$\frac{d}{dp}\left(\frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}=0, x=t, r, \theta, \phi 。$$

$$g_{\mu \gamma} \frac{dx^\mu}{dp} \frac{dx^\gamma}{dp}=\epsilon$$

\begin{aligned} & \frac{\Delta}{\left(r^ 2+a^2 \cos ^2 \theta\right)}\left(ia \sin ^2 \theta \dot{\phi}\right)+\frac{a \sin ^2 \theta}{\left( r^2+a^2 \cos ^2 \theta\right)}\left[\left(r^2+a^2\right) \dot{\phi}-ai\right]=l 。\ & \frac{a \Delta \sin ^2 \theta}{\left(r^2+a^2 \cos ^2 \theta\right)}\left(ia \sin ^2 \theta \dot{\ phi}\right)+\frac{\left(r^2+a^2\right) \sin ^2 \theta}{\left(r^2+a^2 \cos ^2 \theta\right)} \left[\left(r^2+a^2\right) \dot{\phi}-ai\right]=n 。\ & \frac{a^2 \Delta}{\left(r^2+a^2 \cos ^2 \theta\right)^2}\left(ia \sin ^2 \theta \dot{\phi}\右)^2-\frac{2 a \Delta \dot{\phi}}{\left(r^2+a^2 \cos ^2 \theta\right)}(ia \sin \theta \dot{\ phi}) \ & -\frac{\left(r^2+a^2\right)}{\left(r^2+a^2 \cos ^2 \theta\right)^2}\left[\左(r^2+a^2\right) \dot{\phi}-ai\right]^2+\frac{a^2 \dot{r}^2}{\Delta}=0 。 \结束{对齐}

## 物理代写|广义相对论代写General relativity代考|Kerr Solution in Eddington-Finkelstein Coordinates

$v=t+r \wedge$, , psi $=$ Iphi+|phi^,
$$v=t+r^{\prime} \psi=\phi+\phi$$

$$r^{=} \int \frac{r^2+a^2}{\Delta} d r=r+\left(\frac{m r_{+}}{\sqrt{m^2-a^2}}\right) \ln \left|\frac{r}{r_{+}}-1\right|$$

$$d v=d t+\frac{\left(r^2+a^2\right)}{\Delta} d r ; d \psi=d \phi+\frac{a}{\Delta} d r$$

$$d s^2=\left(1-\frac{2 m r}{R^2}\right) d v^2-2 d v d r+\frac{4 a m r \sin ^2 \theta}{R^2} d v d \psi$$

$$u=t-r^{\prime} \chi=\phi-\phi^{\prime} d s^2=\quad\left(1-\frac{2 m r}{R^2}\right) d u^2$$

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