1

A Kerr Black Hole (BH) is a spinning BH. There is an Event Horizon (EH) which is $$r_H^\pm =\frac{r_{S} \pm \sqrt{r_{S}^2 - 4a^2}}{2},$$ where $a = \frac{J}{Mc}$ and $r_{S}$ is the Schwarzschild radius. My question is, suppose I'm in a spacecraft, not in orbit, but stationary at a distance $r$. I want to have a comprehensive understanding of the effects of that BH on the spacecraft, and by extension on me. I would like to know, if I'm subjected to the effects of:

  1. Time Dilation: If this is present, the formula should be derived from the $g_{tt}$ component of the Kerr metric, which is $$\gamma = \frac{1}{\sqrt{1-\frac{r_{S}r}{r^2+a^2\cos^{2}\theta}}}.$$
  2. Tidal Force: What is the formula for this in a Kerr metric? Normally the equation is $$\frac{2GMd}{r^3},$$ with $d$ being in the instance of a human being, about 2 meters.
  3. Acceleration due to the gravitational attraction of the BH at that distance (surface gravity from arbitrary distance). Normally the formula is $$\frac{GM}{r^2}.$$ But I've read here that this should be multiplied by $\gamma$
  4. Escape velocity I need for my spacecraft from this distance $r$; normally the formula is $$\sqrt{\frac{2GM}{r}}.$$

Can the above four quantities be derived from the metric?

To prevent ambiguity, let me edit the question with the following:

  1. Kerr BH mass is: $10^8$ times that of Solar mass
  2. Spin parameter: $0.998$
  3. Distance $r$: $2.5208\cdot 10^{11}$ m
Vick
  • 123

1 Answers1

1

The time dilation $\rm dt/d\tau$ for a ZAMO is $\sqrt{g^{\rm tt}}$ and for a stationary observer $1/\sqrt{g_{\rm tt}}$.

Your equation for the proper acceleration is wrong though even for the Schwarzschild case, the force is ${\rm F/m=a=}\sqrt{|\sum_{\mu, \nu} \ g_{\mu \nu} \rm \ a^{\mu} \ a^{\nu}|}$ with $\rm a^{\kappa}=\ddot{x}^{\kappa}+\sum_{\mu, \nu} \ \Gamma^{\kappa}_{\mu \nu} \ \dot{x}^{\mu} \ \dot{x}^{\nu} $, which is $\rm a=G M/r^2/\sqrt{1-r_s/r}$ for a stationary Fido in Schwarzschild spacetime.

With Kerr you have to choose between the corotating ZAMO (the required force gets infinite at the horizon) and the stationary oberserver (infinite force required to stay stationary at the ergosphere). For the tidal force take the difference between two points.

If you are looking for the coordinate acceleration $\rm \ddot{x}$ instead of the proper acceleration $\rm a$ see here at Output 9.

The escape velocity to or the free fall velocity from infinity relative to a local ZAMO is ${\rm v_{esc}}=\pm {\rm c} \ \sqrt{1-1/g^{\rm tt}}$ (which is $\rm c$ at the horizon).

Yukterez
  • 14,655