Gravity on supermassive black hole's event-horizon

Question

• $M =$ black hole mass

• Gravitation is about $r^{-2}$

• Schwarzschild radius, $r_{\text{S}}$, is $\propto M$

• So, more massive black holes have weaker gravitation at their event horizon.

Consider a black hole so enormous that the gravitation on its event horizon is negligible.

Person A is 1 meter 'outside' the horizon, and Person B is inside (1 meter from the horizon as well). Person B throws a ball to Person A. Both just started accelerating towards the black hole very slowly, so why person A wont catch the ball? Why won't Person A even ever see person B granted A will somehow escape later on?

Reference: https://mathpages.com/rr/s7-03/7-03.htm

Answer 1

With the proper definition of the meaning of gravitational acceleration, the questioner is correct and the other answers that claim that the gravitational force at the event horizon is infinite are wrong.

Per Wikipedia:

In relativity, the Newtonian concept of acceleration turns out not to be clear cut. For a black hole, which must be treated relativistically, one cannot define a surface gravity as the acceleration experienced by a test body at the object's surface. This is because the acceleration of a test body at the event horizon of a black hole turns out to be infinite in relativity. Because of this, a renormalized value is used that corresponds to the Newtonian value in the non-relativistic limit. The value used is generally the local proper acceleration (which diverges at the event horizon) multiplied by the gravitational redshift factor (which goes to zero at the event horizon). For the Schwarzschild case, this value is mathematically well behaved for all non-zero values of r and M.

[...]

Therefore the surface gravity for the Schwarzschild solution with mass $M$ is $\frac{1}{4M}$

So with this definition, the OP is correct that the suitably defined surface gravity at the event horizon decreases as the mass of the black hole increases.

Now this surface gravity does not mean that a rocket engine that can produce that acceleration will enable you to hover at that distance from the black hole. It does take an infinitely powerful rocket engine to hover arbitrarily close to the horizon and, of course, no rocket engine could let you hover inside the event horizon.

However, if both observers, A and B are freely falling in from infinity, nothing at all unusual will happen as first B and then A (one meter later) crosses the event horizon. Neither will lose sight of the other at any time. What actually happens is that the photons bouncing off of B as he crosses the horizon will be frozen at the horizon waiting for A to run into them at the "speed of light". B who is inside can toss the ball to A who is falling in but is currently outside the event horizon and A will catch the ball after he crosses the event horizon. This is true since to first order A and B, when freely falling are in a common inertial reference frame and they can do whatever they could do when far from the black hole.

The problem comes if they try to hover with one person inside and one outside the horizon. That is not possible – the person inside cannot hover at all and the person outside would need a very powerful continuously firing rocket engine to try to hover. But then all the effects of time dilation etc. will be occurring for both of them and all the problems noted by the other answers will be true.

Read these questions and answers for more insight:

• How does an object falling into a plain Schwarschild black hole appear from near the black hole?

• How does the star that has collapsed to form a Schwarschild black hole appear to an observer falling into the black hole?

Answer 2

The horizon, for any static black hole, is the surface where the escape velocity is $c$. Thus, your notion that gravitation is weaker at the horizon for larger black holes is incorrect.

EDIT: Consider that the thrust required to hover goes to infinity at the horizon regardless, i.e., the proper local acceleration for a stationary observer goes to infinity at the horizon.

EDIT 2: In your thought experiment, you write that person B is $1m$ inside the horizon. But, it's crucially important to understand that, within the horizon, the radial coordinate is timelike with the future towards the singularity and the past towards the horizon. Person B can no more throw a ball towards person A outside the horizon than person A can throw a ball into the past.

Answer 3

In very simple terms, everything that enters a BH will become part of it. So, there is no question of A ever being able to catch the ball.
You could say that there is a Fundamental Law of Nature- " Thou shalt not travel faster than light"(sentence taken from a TV show, initial speaker unknown to me)
Which means that no form of matter that belongs to this Universe(Okay, certainly no one know about the matter inside of a BH ) can travel faster than the speed of light in pure vacuum .
And since the escape velocity at the Event Horizon is the speed of light, literally nothing can ever escape a BH after entering it, not even light.Thus, A cannot see the real time B after B enters the BH.

Answer 4

As @Alfred said, your notion that gravitation is weaker at the horizon for larger black holes is incorrect.

Let's $E$ - energy of a body of mass $m$, which is falling from the infinity, at 1 meter above horizon of BH with Schwarzschild radius $R_{S1}$:

$\large {E=\frac{mc^2}{\sqrt{1-R_{S1}/(R_{S1}+1)}}}$

and the same energy of the body at $x$ meters above horizon of BH with Schwarzschild radius $R_{S2}$:

$\large {E=\frac{mc^2}{\sqrt{1-R_{S2}/(R_{S2}+x)}}}$

Then

$\large {\frac{mc^2}{\sqrt{1-R_{S1}/(R_{S1}+1)}}}=\frac{mc^2}{\sqrt{1-R_{S2}/(R_{S2}+x)}}$

and

$\large {x=\frac{R_{S2}}{R_{S1}}}$

So, if $R_{S1}>R_{S2}$ then $x<1$

Answer 5

Person A will not see anybody beyond the event horizon, even a meter ahead. That is because one meter in flat coordinates (which I suppose you mean) corresponds to infinite distance in the co-moving coordinates of the observer A.

Observer A will be able to see large objects (larger than 1 meter) ahead of him which are still outside the event horizon. At the same time, an observer at infinity will see observer A shortened in the radial direction and becoming like a flat disk on the surface of the black hole.

Crossing the horizon for observer A (if it happened) would look not like crossing a spatial surface, but like crossing a moment of time: now he is before the horizon, and now he is inside. All objects around him, ahead and beyond, cross the horizon nearly simultaneously (the difference being only the time it takes for light to travel between them).

Something a meter ahead of him in flat coordinates corresponds to an object that crossed the horizon infinite time before he did, so he would not be able to see the observer B. Even if observer B is also outside the horizon, the distance between them would be so large that they hardly could see each other.

If you meant that the observers were within 1 meter of each other in co-moving coordinates, then they are both either outside the horizon or inside of it. They cannot see each other, even when a meter apart, because they are separated by the horizon which is a null surface; it is not a spatial surface.

Two friends travelling in one spaceship will cross the horizon nearly simultaneously, even if spatially separated (for a distant observer the length of their spaceship will become zero at the horizon).

Answer 6

As you approach the event horizon of a black hole, the rate of time approaches zero. The acceleration due to gravity is caused by a rate of change in the rate of time, which must also approach zero.

If a person managed to slow down to zero speed on their way to the event horizon, they would have lost just about all of their energy in doing so (presumably by hitting stuff on the way), and become very tiny indeed. Person B must not have hit quite so many things and just overshot by a metre.

Let's say that the tiny people throw their tiny ball to one another across the event horizon. It would probably be a bit like doing the same thing in deep space, except that everything has shrunk.

I should point out that this answer is slightly tongue in cheek as I don't consider the scenario to be plausible.