Is a black hole's mass uniformly distributed?

Question

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.

Answer 1

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.

Answer 2

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!

Answer 3

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...

Answer 4

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.