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If you were to fly around a black hole, would the gravitational pull be uniform and centered on the singularity, regardless of your relative location?

If yes, how can this be consistent with models in which an object can never pass the event horizon from an outside observer as time slows down as you get closer to the event horizon? If the object you're feeding into the black hole is massive enough, shouldn't you be able to detect the mass 'snapping' from where the object passes the event horizon to the singularity?

EDIT: for simplicity you can assume a non-rotating black hole.

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    $\begingroup$ How familiar are you with general relativity? $\endgroup$ – Kyle Kanos Jul 29 '15 at 2:36
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    $\begingroup$ @KyleKanos I'm afraid my knowledge is limited to the concept of relativity. My entire understanding of it is based on (popular) thought experiments, simulations, and imagination. I have no experience whatsoever with the actual formulae. But please do not feel refrained from posting on in-depth answer - I'm not the only reader here :) $\endgroup$ – orlp Jul 29 '15 at 2:41
  • $\begingroup$ I'm not quite qualified for an answer, but I'm fairly certain that, for your first question, it depends on whether the BH rotates or not (rotating BH are not spherically symmetric). For your second question, you may be interested in this Physics.SE post. $\endgroup$ – Kyle Kanos Jul 29 '15 at 2:48
  • $\begingroup$ orlp : I thought Nathaniel's answer was pretty good in the post KK linked to. But note this: if you're a water molecule alighting on the surface of a hailstone, you can't pass through the surface. However you can get buried by other water molecules. So whilst you can't pass through the surface, the surface can pass through you. In similar vein you could say that falling matter can't pass through the event horizon, but the event horizon can pass through it. $\endgroup$ – John Duffield Jul 29 '15 at 15:38
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There are two issues.

One issue is that some people try to oversell the "no hair" theorem to areas beyond where it applies, they apply a long term analysis of final states as if it applies in the short term.

The second issue is that in the short term we have two objects, the black hole and the incoming object. We have to think about both.

I'll go into more detail, first about the long term analysis. The no hair theorem is basically about waiting a long time, if you wait a long time without new things coming in (but letting gravitational waves travel out) then the system will start to look like a very simple charged rotating black hole with possibly zero charge and/or possibly zero rotation. So if you put your massive body in at just one location into your initially not rotating black hole and then wait a huge amount of time it will end up looking very close to a simple rotating black hole (with possibly zero rotation). For some reason it is popular to ignore the part where you have to wait a long time and you only get approximately the simple solution. Really it is just that there are only a small number of rather simple back holes that things can approach over time. There are many ways it can be as it approaches the final state. It's like if you leave some hot metal out in a room, there is just one final temperature it approaches, an equilibrium temperature, but it takes forever to reach it, so it is always a bit hotter than the room but starts to get really really really close. Your set up will start to get really really really close to the final state of a simple rotating black hole (with possibly zero rotation) if you throw your mass thing into the side of a black hole.

If you throw it straight towards the exact center of the black hole, then the final state might be a nonrotating black hole.

OK, so that's about wrong things people say about the long term as if it applies to the short term. So now let's talk about what really happens in the short term.

In the short term we have two objects, the black hole and the incoming object. Each produces gravitational effects. You feel the total gravitational effects including the effect of their interactions.

In the long term the system of original hole plus object starts to become like one object, the long term black hole. So you might be approaching this long term gravitational pull of that long term larger black hole, but there isn't going to be a snap. To be fair if you have inspiral there is a transition where it goes from a slow orbit to an faster inspiral but that faster inspiral just means more gravitational waves emitted so it starts to approach the final state faster which is different than when you leave the hot metal out where it slows the rate of cooling as it gets closer to the equilibrium temperature. But it is still continuous in space and time. The gravitational waves are emitted as it approaches the final configuration.

So ... you feel a pull that starts out like the pull from two objects. So more pull near the incoming object. But over time you continuously start to feel a pull that is more and more like the pull from one larger black hole.

The change happens because the system evolves (and possibly rotates if the black hole plus object together have rotation) and it emits gravitational waves that remove bit by bit everything except the total charge, total surviving energy, and total surviving rotation.

If this contradicts stories you've heard, it is likely that people oversold their story.

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A non-rotating black hole can be treated as spherically symmetric using the Schwarzschild metric.

A rotating black hole has an axis of symmetry and can be represented with the Kerr metric.

Treatments of black holes using either of these would make the assumption that the "test particle" you are considering does not influence the metric (is much less massive). If this were not the case then the metric symmetry would be broken and problems become much harder!

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  • $\begingroup$ I don't feel this answers my other question. You say that "A non-rotating black hole can be treated as spherically symmetric using the Schwarzschild metric." I don't understand how that's consistent in a model where objects never pass the horizon from an outside observer - if this is the case you will have a non-uniform distribution of mass, from the perspective of an outside observer, no? $\endgroup$ – orlp Jul 29 '15 at 15:16
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    $\begingroup$ @orlp Correct - if you allow the black hole to interact with an object that has a mass that is not negligible, then you cannot use the Schwarzschild metric to describe that situation. $\endgroup$ – Rob Jeffries Jul 29 '15 at 15:23
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In the mathematical models of black holes that we use, there is a parameter M which is related to the mass-energy of the black hole. There is NO hint of distribution of mass-energy inside the black hole. In particular:

a non-rotanting BH is spherically symmetric (these BH probably do not exist in nature, since all astronomical objects rotate and pick up angular momentum as they swallow other bodies). the singularity is pointlike.

a rotating BH is axially symmetric and it has a ring-like singularity.

When you fall toward a BH, you will pass through the event horizon. There is no question about that. An external observer will see your image getting more and more red-shifted until the photons will have so little energy that they will be undetectable. This will take a very long time. But this is just your image. You will pass through the event horizon instantaneously (it is a 2D surface) and you will never be able to come back.
What do you feel when you pass the horizon? Depends. If the BH is very massive, the curvature at the event horizon is not too strong, so you won't even notice the gravitational effects on your body, if the BH is not massive (just few solar masses or less) the curvature is very high and different parts of your body will feel different pulls/pushes. These are called tidal forces and might rip you apart. Actually, if they are strong enough, they will rip apart the individual atoms of your body.

An excellent non technical book describing this is Thorne's Black Holes...

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  • $\begingroup$ I don't feel this answers my other question. You say that "There is NO hint of distribution of mass-energy inside the black hole." I don't understand how that's consistent in a model where objects never pass the horizon from an outside observer - if this is the case you will have a non-uniform distribution of mass, from the perspective of an outside observer, no? $\endgroup$ – orlp Jul 29 '15 at 15:15
  • $\begingroup$ Why would that be the case? The image of the falling object is not part of the black hole mass, and the object itself contributes to the position of the center of mass (and the location of the event horizon) in the same was as in any other two-body system. $\endgroup$ – Asher Jul 29 '15 at 15:34
  • $\begingroup$ I do not understand why you keep thinking that objects never pass the horizon. I explained that they do pass the horizon instantaneously, just the images of them approaching the horizon appear to move slower and slower and redder and redder. $\endgroup$ – magma Jul 29 '15 at 23:15
  • $\begingroup$ There are many things wrong about this answer: 1) There exist stable orbits around black holes under a variety of conditions, just because you fall towards a black hole doesn't mean you will pass the event horizon. 2) If you mean "fall directly at the black hole," it is still incorrect as until you cross the event horizon you will still theoretically be able to accelerate away from the black hole and escape. 3) Since you are a 3d object you don't pass the event horizon instantly. 4) You don't really explain the actual answer of "NO hint of distribution of mass-energy inside the black hole" $\endgroup$ – Benjamin Horowitz Nov 10 '15 at 23:59
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Is a black hole's mass uniformly distributed?

One hears conflicting answers to this. One article I rather like is the mathspages Formation and Growth of Black Holes. See this bit:

"Historically the two most common conceptual models for general relativity have been the 'geometric interpretation' (as originally conceived by Einstein) and the 'field interpretation' (patterned after the quantum field theories of the other fundamental interactions). These two views are operationally equivalent outside event horizons, but they tend to lead to different conceptions of the limit of gravitational collapse. According to the field interpretation, a clock runs increasingly slowly as it approaches the event horizon (due to the strength of the field), and the natural 'limit' of this process is that the clock asymptotically approaches 'full stop' (i.e., running at a rate of zero). It continues to exist for the rest of time, but it's 'frozen'..."

According to that interpretation, the black hole's mass is uniformly distributed, and it grows rather like a hailstone. But according to "the other" interpretation, it isn't distributed at all. It's all at some central point singularity. However there are some issues with this:

"In contrast, according to the geometric interpretation, all clocks run at the same rate, measuring out real distances along worldlines in curved spacetime. This leads us to think that, rather than slowing down as it approaches the event horizon, the clock is following a shorter and shorter path to the future time coordinates. In fact, the path gets shorter at such a rate that it actually reaches the future infinity of Schwarzschild coordinate time in finite proper time..."

The infalling mass goes to future infinity of Schwarzschild coordinate time. In plain English, that's the end of time. I don't like it, or the way this sort of thing tends to get glossed over in popscience books and articles.

If you were to fly around a black hole, would the gravitational pull be uniform and centered on the singularity, regardless of your relative location?

Notwithstanding what I said above, I think the answer is broadly yes, whichever interpretation you use.

If yes, how can this be consistent with models in which an object can never pass the event horizon from an outside observer as time slows down as you get closer to the event horizon?

Because the black hole is a place where the coordinate speed of light is zero, and the force of gravity relates to the local gradient in the coordinate speed of light at your location. And a black hole is a massive thing, it's round. You can of course read about the Kerr black hole which is flattened because it's spinning. And you can read about black holes spinning at half the speed of light. But at the event horizon the coordinate speed of light is zero, so there's issues there too. There's more issues to do with black hole firewalls, wherein matter is said to be unable to survive falling into a black hole. IMHO things are less certain than some people say.

If the object you're feeding into the black hole is massive enough, shouldn't you be able to detect the mass 'snapping' from where the object passes the event horizon to the singularity?

No, because at the event horizon the coordinate speed of light is zero, and it can't go lower than that. The gradient in the coordinate speed of light where you are isn't going to change.

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    $\begingroup$ Actually, your statement there explicitly says that the gravitational pull would be uniform; this is only true for non-rotating BHs. $\endgroup$ – Kyle Kanos Jul 29 '15 at 15:54
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    $\begingroup$ @JohnDuffield I mean, the nice thing about the Einstein equations is that they let you choose other coordinates. A continuously accelerating observer sees a wall of death (horizon plane) at a fixed distance behind them; do you also think they should assume that half of the universe is not real? It's mostly philosophical: but if there's a better coordinate system that lets you describe more, why be religiously attached to another one? $\endgroup$ – CR Drost Jul 29 '15 at 23:05
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    $\begingroup$ "As for coordinate systems, perhaps you've missed the crucial point. When light doesn't move you have no way to measure distance and time." I think maybe you've missed my crucial point too: this is one way of looking at the situation, but not necessarily the way to look at it, because it's not true for all coordinate systems. When you know that other coordinate systems exist which disagree with yours, you should hesitate to pronounce that objective reality is your coordinates and everyone else is wrong. Infalling coordinates pass the event horizon like it's nothing. Why dismiss them? $\endgroup$ – CR Drost Jul 30 '15 at 14:42
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    $\begingroup$ @JohnDuffield The heart of the issue seems to be that you don't particularly like the mathematical framework of general relativity, even though I'm pretty sure that you don't disagree with its physical predictions. (That is, I can only read your last comment as "I don't like this mathematical framework for philosophical reasons," not "This mathematical framework is wrong on an experimental basis.") The only caveat I'd add here is that not everyone will agree with your philosophy and Physics.SE may not be the best place to preach it. Grad students will just use whatever is simple and works, no? $\endgroup$ – CR Drost Jul 30 '15 at 16:15
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    $\begingroup$ @JohnDuffield I don't need to defend "Chicken-Little gravity"; both infalling and infinity-based coordinates are mathematically equivalent for the entire region outside the black hole, hence there is no physical difference assuming that the Einstein equations hold. Sorry for using the word "preach"; it's just that you keep saying philosophical/preachy things like "X just doesn't square with Y" but not actual physical things like "here is a mathematical discrepancy between these two leading to a real physical difference that we can test." Your coordinates make you comfortable. Good for you. $\endgroup$ – CR Drost Jul 30 '15 at 17:07

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