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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
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  • $\begingroup$ Is the spacecraft in the plane of rotation of the BH? Is it stationary relative to a distant observer? Or is it stationary relative to the frame dragging induced by the Kerr metric? $\endgroup$
    – PM 2Ring
    Commented Aug 29, 2023 at 7:44
  • $\begingroup$ @PM 2Ring I edited the original question to answer your questions. It is stationary relative to a distant observer. $\endgroup$
    – Vick
    Commented Aug 29, 2023 at 8:44

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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).

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  • $\begingroup$ @Vick - g is a force and therefore what you measure with an accelerometer. for a free faller it is 0 and for a stationary oberserver infinite at the horizon. in the posting you linked the answer gives the same g as in my answer. $\endgroup$
    – Yukterez
    Commented Aug 30, 2023 at 6:39
  • $\begingroup$ @Vick - No, that would be $\rm \ddot{x}$, I'm talking about the $\rm a$ in my answer above. $\endgroup$
    – Yukterez
    Commented Aug 30, 2023 at 8:28
  • $\begingroup$ The time dilation of a ZAMO at the equator is ṫ=3.23599 and at the pole ṫ=2.8352. For a stationary particle there is no solution at the equator since your radius is already inside the ergosphere, at the pole it's one and the same since the ZAMO there is stationary as well. The escape velocity relative to the local ZAMO is v=0.9510541 at the equator and v=0.93573335 at the pole. $\endgroup$
    – Yukterez
    Commented Aug 30, 2023 at 9:39
  • $\begingroup$ I took the numbers from your question above. The EH=[1+√(1-a²)]GM/c² is 1.5790106e11 m though and the equatorial ergosphere at r=2GM/c²=2.9702594e11 m, but maybe you rounded some masses differently, I took Msol=2e30kg. The chat here is indeed not really suitable, since if we chat too much a moderator will move it to the Chatroom where the Latex does not work. You could open a thread at physicsforums and send me the link, that is probably better suited for back and forth. $\endgroup$
    – Yukterez
    Commented Aug 30, 2023 at 10:07
  • $\begingroup$ @Vick - maybe the discrepancy is due to the spin parameter, here you used a=0.998 and there 0.999999999999986673399999999999954 and also θ=45°, I'll check that later in the evening or tomorrow if it was just that or something else. The ergosphere at θ=90° at the equator is always r=2 though. $\endgroup$
    – Yukterez
    Commented Aug 30, 2023 at 12:28

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