# Gravity on supermassive black hole's event horizon

M = black hole mass Gravitation is about 1/r^2 Schwarzschild radius r is ~ to M

Greater BH -> weaker gravitation on its horizon. Lets take black hole so enormous that the gravitation on its horizon is negligible.

Person A is 1 meter 'outside' the horizon, person B is inside (1 meter from the horizon aswell) person B throws a ball to person A. Both just started accelerating towards BH very slowly, so why person A wont catch the ball? Why A wont even ever see person B granted A will somehow escape later on?

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Person A will not see anybody beyond the even horizon, even in metre ahead. That is because one meter in a 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 still outside the even horizon. At the same time, ofserver at infinity will see observer A shortened in radial direction and becoming like a flat disk on the surface of black hole.

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

Something in meter ahead him in flat coordinates corresponds to a thing 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 1 meter of each other in co-moving coordinates, then they both either outside the horizon or inside it. They cannot see each other in a meter but be separated by a horizon, because the horizon is null surface, it is not spatial surface.

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

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Thank you, are there any equations which would help? I don't think I understood it well... – darksticko Sep 12 '12 at 18:10
@darksticko, perhaps this will help: mathpages.com/rr/s7-03/7-03.htm – Alfred Centauri Sep 12 '12 at 18:34

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.

...snip...

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.

and

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

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The surface gravity, as defined, obviously varies inversely with M. The point that's been lost here is this: the surface gravity has nothing to do with this problem. In fact, it is what has confused the OP into thinking that B can toss a ball to A. The fact is this: According to an accelerometer on a rocket, the thrust required to hover goes to infinity at the horizon. The fact that a distant observer see this differently is completely irrelevant to this problem. What's relevant is that Person B cannot toss a ball towards the horizon. – Alfred Centauri Sep 13 '12 at 20:37
If they are both freely falling into the black hole, 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. What I said for light also applies to balls. Since to first order A and B, when freely falling are in a common inertial reference frame they can do whatever they could do when far from the BH. – FrankH Sep 13 '12 at 22:26
Don't drop the context of the problem. They're not freely falling: "granted A will somehow escape later on". Remember, my points are related to the context of the problem as (ill) posed – Alfred Centauri Sep 13 '12 at 23:09

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.

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But 2x heavier black hole would have en.wikipedia.org/wiki/Schwarzschild_radius 2x longer so gravity on its horizon would be 2x weaker. What am I missing? if nothing, then gravity on BH's horizon can be as little as we want – darksticko Sep 12 '12 at 16:29
@user1666076 The local acceleration due to gravity is lower. In that Alfred is simple wrong. The issue is that the nature of space-time actually changes at the horizon---the radial dimension of space becomes time-like and one can no more move outward once inside the horizon then you can move backward in time while sitting in front of your computer. I had hopes to avoid writing about this because it has been a long time since I worked through the math. Perhaps one of our theorists will say a few words. – dmckee Sep 12 '12 at 16:50
@dmckee, it's not the local acceleration I'm referring to, it's the escape velocity. The OP seems, to me, to indicate that his notion of "strength of gravity" is related to escape velocity thus his idea that B could throw a ball through the horizon. Perhaps I'm misunderstanding his question. – Alfred Centauri Sep 12 '12 at 16:59
OK, so I spoke too strongly. Please accept my apology. As I read the question @user1666076 asks why the two observer separated--as they can be--by little gravitational potential can't pass the ball back and forth. And he is right that it is not because gravity is locally too strong: it is because space time doesn't behave in the way we're used to once you cross the invisible locus of the event horizon. – dmckee Sep 12 '12 at 17:02
I think he is talking about gravitational acceleration at the BH surface which is indeed small for a massive BH. – Anixx Sep 12 '12 at 17:02

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.

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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 $x<1$

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