# Schwarzschild metric with angular momentum

1. Apr 14, 2017

### Mr rabbit

1. The problem statement, all variables and given/known data

Given the Schwarzchild 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

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

2. Relevant 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

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

3. 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?

2. Apr 19, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.