Schwarzschild metric with angular momentum

[Mr rabbit]

Homework Statement

Given the Schwarzschild metric generalisation for a mass M rotating with angular momentum J

$ds^2 = -(1-\frac{2 M}{r}) \; dt^2 +(1-\frac{2 M}{r})^{-1} \;(dr^2 +r^2 \;d\theta ^2 +r^2 \sin ^2 \theta \; d\phi ^2) -\frac{4J}{r} \sin ^2 \theta \; dt d\phi$

a) Write the associated conserved magnitudes $E, L$ and clear $p^t = p^t(E,L)$, $p^{\phi}=p^{\phi} (E,L)$
b) Consider a test particle in a stable circular orbit of radius $r=R$ on the equatorial plane $\theta = \pi /2$. Staying at first order for $M/R$ and $(L/R)^2$, show that for a test particle of mass m=1 E and L are given by

$E^2=1-M/R$
$L=\sqrt{M R}$

Homework Equations

$E= - \vec{T} \cdot \vec{u}$
$L=\vec{R} \cdot \vec{u}$

where $T^{\mu} = \delta ^{\mu} _t =(1,0,0,0)$ and $R^{\mu} = \delta ^{\mu} _{\phi} = (0,0,0,1)$ are the Killing vectors

$u^{\mu} u_{\mu}= u^{\mu} g_{\mu \nu} u^{\nu} = -1$

The Attempt at a Solution

a)
Calling $B(r)=(1-\frac{2 M}{r})$, the conserved quantities are

$E=-T^{\mu} g_{\mu \nu} u^{\nu} = -T^t g_{t t} u^t -T^t g_{t \phi} u^{\phi} = B(r) \dot{t} + \frac{2 J}{r} \sin ^2 \theta \; \dot{\phi}$

$L= R^{\mu} g_{\mu \nu} u^{\nu} = R^{\phi} g_{\phi \phi} u^{\phi} + R^{\phi} g_{\phi t} u^t= B^{-1} (r) r^2 \sin ^2 \theta \; \dot{\phi} - \frac{2 J}{r} \sin ^2 \theta \; \dot{t}$

From this equations we can obtain the 4-momentum (which it will be equal to the 4-velocity because m=1)

$\displaystyle \dot{t} = \frac{E r^2 B^{-1} (r) - \frac{2JL}{r}}{r^2+\frac{4 J^2}{r^2}\sin^2 \theta}$ $\; \; \; \; \;\displaystyle \dot{\phi} = \frac{B(r) L+\frac{2JE}{r}\sin^2\theta}{r^2 \sin^2\theta + \frac{4J^2}{r^2}\sin^4\theta}$

b) Once this is done, we have to use $u^{\mu} u_{\mu}= u^{\mu} g_{\mu \nu} u^{\nu} = -1$ for getting the effective potential. Due to the difficulty of expressions, we are forced to make some approximations... and here is my problem. How to make it? I tried (setting $r=R$ and $\theta = \pi /2$)

$\displaystyle \dot{t} = \frac{ R^4 E B^{-1} - 2JLR}{R^4+4J^2} \approx \frac{ R^4 E B^{-1} - 2JLR}{R^4} = E B^{-1}-2JL/R^3 \approx E B^{-1} = \frac{E}{1-\frac{2M}{R}} \approx E (1+\frac{2M}{R})$

$\displaystyle \dot{\phi} = \frac{R^2 B L+2JER}{R^4+4J^2} \approx \frac{R^2 B L+2JER}{R^4} = \frac{BL}{R^2}+\frac{2JE}{R^3} \approx \frac{BL}{R^2} =\frac{L}{R^2} (1-\frac{2M}{R}) \approx \frac{L}{R^2}$

but I don't get the correct effective potential. Rather, doing $\frac{dV}{dr} (r=R)=0$ I don't get the expressions for E and L. Any ideas?

 

[vacuum]

One approach to solving this problem is to first consider the equations of motion for a test particle in the equatorial plane. These can be written as:

$\ddot{r} = -\frac{1}{2} \frac{d}{dr} (g_{tt} + 2g_{t\phi} \dot{\phi} + g_{\phi\phi} \dot{\phi}^2)$
$\ddot{\phi} = -\frac{1}{2} \frac{d}{dr} (g_{t\phi} + g_{\phi\phi} \dot{r}^2)$

Substituting in the Schwarzschild metric and setting $\theta = \pi/2$, we get:

$\ddot{r} = -\frac{M}{r^2} + \frac{J^2}{r^3}$
$\ddot{\phi} = \frac{2J}{r^3}$

Now, for a test particle in a stable circular orbit at radius $r = R$, we know that $\ddot{r} = 0$ and $\dot{\phi} = \omega$, where $\omega$ is the orbital frequency. Plugging this in and solving for $E$ and $L$, we get:

$E = \sqrt{1 - \frac{2M}{R}}$
$L = \frac{J}{\sqrt{1 - \frac{2M}{R}}}$

We can also use the equations of motion to find the effective potential, which is given by:

$V_{eff} = \frac{1}{2} \left( \frac{L^2}{r^2} + \frac{J^2}{r^4} \right) - \frac{M}{r}$

Plugging in our expressions for $E$ and $L$ and expanding to first order in $M/R$ and $(J/R)^2$, we get:

$V_{eff} \approx \frac{1}{2} \left( 1 - \frac{2M}{R} \right) - \frac{M}{R} = \frac{1}{2} - \frac{3M}{2R}$

Setting $dV_{eff}/dr = 0$ and solving