When a black hole increases in size by engulfing matter, does this matter fall into the event horizon or does it just rotate around the accretion disk?
EDIT: Clarified the beginning of the question.
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1$\begingroup$ When matter falls into a black hole implies that the matter has fallen through the event horizon. $\endgroup$– HDE 226868Commented Jul 14, 2015 at 18:04
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2 Answers
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It doesn't just rotate around the accretion disk. As to what actually happens to it isn't clear-cut. Most people will say the infalling matter goes right through the event horizon all the way to the central point-singularity in finite proper time. But check out The Formation and Growth of Black Holes on mathspages:
"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'..."
Most people just aren't aware that there's this other GR interpretation. And some people who are, such as the author of the article, tend to discount it saying things like this:
"Therefore, if the 'frozen star' interpretation gave equivalent predictions for all externally observable phenomena, and was logically consistent, it might be the preferred view. However, we saw above that the idea of a frozen star as an empty region around which matter 'bunches up' outside an event horizon isn't viable".
There's an error here, which you can appreciate by thinking about a hailstone. You're a water molecule, and you alight upon the hailstone's surface. You can't pass through this surface. However other water molecules can settle around you and bury you, so the surface passes through you. So you can effectively pass through the surface. In similar vein the black and its event horizon can grow. I should add that on top of all this there are "firewall" issues, but that's one for another day.
answered Jul 14, 2015 at 18:38
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$\begingroup$ While the mathpages link is fine, John Duffield's comments here are wrong (from previous experience, he's a nonphysicist has an axe to grind against the existence of black holes). The "field interpretation" doesn't pick out a preferred definition of simultaneity, presumably the mathpages author is just talking about what happens if you use the field interpretation for Schwarzschild coordinates or some other system where things take an infinite time to reach the horizon, but you could just as well apply the field interpretation to another coordinate system where they cross in finite coor. time. $\endgroup$ Commented Jan 12, 2016 at 0:14
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$\begingroup$ My comments aren't wrong, I'm an IT guy with a lifelong interest in physics, and I have no axe to grind against the existence of black holes. In fact I'm on record as saying the frozen-star black hole is even more of a black hole than the point-singularity version. $\endgroup$ Commented Jan 12, 2016 at 13:23
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$\begingroup$ It's wrong in the sense of disagreeing with current mainstream physics--and the point of this site is to teach people about mainstream physics, not promote alternatives even if history someday proves them true. In addition, even if it were true that a falling observer would never experience crossing the horizon and entering an interior region, your argument for this position above is definitely wrong--the field interpretation is quite compatible with the idea that the observer will experience this, since the field interpretation does not pick out a preferred definition of simultaneity, $\endgroup$ Commented Jan 12, 2016 at 18:51
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$\begingroup$ (cont) so it must be usable with any of the various simultaneity definitions used in different coordinate systems which the geometric interpretation can be applied to, including simultaneity definitions where events on the outside are simultaneous with the event of falling objects crossing the horizon, like Eddington-Finkelstein coordinates or Kruskal-Szekeres coordinates. $\endgroup$ Commented Jan 12, 2016 at 18:53
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Matter in the accretion disk is not (yet) inside the black hole. It is orbiting, and it can even escape. In fact, some of it must escape for accretion to happen at all: The disk has too much angular momentum to be accreted, and so some material must be ejected, carrying off excess angular momentum, in order for the rest to fall into the hole.
The general idea is that matter in the accretion disk slowly makes its way inward (dependent entirely on how well angular momentum can be transported outward with help from magnetic fields). Once it reaches the innermost stable circular orbit or thereabouts, it plunges toward the black hole rather quickly. This defines the inner edge of the disk, and is somewhat above the event horizon.
Now whether or not the matter ever does fall in is another question, but it's as much a question of semantics as physics.
