1
$\begingroup$

The surface gravity (acceleration at event horizon, as measured by an observer at infinity, since the proper acceleration is infinite) of a Kerr-Newman black hole is given (e.g. here) as:$$ \kappa = \frac{r_+ - r_-}{2 \cdot (r_+^2 + a^2)} $$What is the corresponding generalization of this formula for any distance from the event horizon (i.e., what's $\kappa(r)$)?

$\endgroup$

1 Answer 1

0
$\begingroup$

This would be $\rm \rm {\rm d}^2r/{\rm d}\tau^2 = \ddot{r}$, so you evaluate that for a corotating particle by solving the geodesic equation (see here) and take the limit for $\rm r \to r_{+}$ in case of the horizon, or any $\rm r$ and $\theta$ since $\rm \ddot{r}$ is not only a function of the radial but also of the poloidial coordinate.

The general equation for a black hole of charge $\rm Q$ and spin $\rm a$ for a particle with the charge $\rm q$ is in natural units and in Boyer Lindquist coordinates

$ \rm \ddot{r} =$ $\rm (a^2 \dot{\theta} \sin (2 \theta ) \ \dot{r})/(a^2 \cos ^2 \theta +r^2)+\dot{r}^2 ((r-1)/(a^2+(r-2) \ r+Q ^2)-r/(a^2 \cos ^2 \theta +r^2))+ $ $ \rm ((a^2+(r-2) \ r+Q ^2) (8 a \sin ^2 \theta \ \dot{\phi} \ (a^2 \cos ^2 \theta \ (q \ Q -2 \dot{t})+r (2 (r-Q ^2) \dot{t}-q \ Q \ r))+ $ $\rm 8 \dot{t} \ (a^2 \cos ^2 \theta \ (\dot{t}-q \ Q )+r \ (q \ Q \ r+(Q ^2-r) \ \dot{t}))+8 r \ \dot{\theta}^2 \ (a^2 \cos ^2 \theta +r^2)^2+ $ $ \rm \sin ^2 \theta \ \dot{\phi}^2 \ (2 a^4 \sin ^2(2 \theta )+r \ (a^2 (a^2 \cos (4 \theta )+3 a^2+4 \ (a-Q ) (a+Q ) \cos (2 \theta )+4 Q ^2)+ $ \rm $ \rm 8 r \ (-a^2 \sin ^2 \theta +2 a^2 r \cos ^2 \theta +r^3)))))/(8 \ (a^2 \cos ^2 \theta +r^2)^3) $

so you have to set $\rm \dot{r}=\dot{\theta}=q=0, \ \dot{\phi}=|g_{t \phi}/g_{\phi \phi}|/\sqrt{|g^{t t}|}$ and $\rm \dot{t}=\sqrt{|g^{t t}|}$ to get $\rm \ddot{r}$ for a locally stationary particle with zero angular momentum (a so called ZAMO in his so called LNRF) at any given $\rm r$ and $\theta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.