I How Long Does it Take to Fall into a Black Hole?

[PeterDonis]

An object just hugging the horizon would still have added it's mass to some region of spacetime much closer to the hole than the outside observer.

The mass isn't a property of "some region of spacetime". It's a global property of the spacetime geometry as a whole.

An object can't "just hug the horizon" unless it has some way of "hovering" at a constant altitude; below $r = 6M$ (3 times the horizon radius), there are no stable free-fall orbits, and below $r = 3M$ (3/2 the horizon radius) there are no free-fall orbits, period. And any way of "hovering" (such as, for example, firing a rocket) will require energy, and that energy will have been part of the object when it fell past you, so it will get counted in the increased mass of the hole that you detect; and since that energy will have to be finite, at some point it will run out and the object will fall into the hole anyway. It is not possible for an object containing finite energy to "hug the horizon" indefinitely.

a change in the radius of the horizon itself by some significant mass falling in would be detectable no?

You might be able to detect it by observing a change in the behavior of light rays coming from behind the hole, yes.

 

[JLowe]

An object can't "just hug the horizon" unless it has some way of "hovering" at a constant altitude; below r=6M (3 times the horizon radius), there are no stable free-fall orbits, and below r=3M (3/2 the horizon radius) there are no free-fall orbits, period. And any way of "hovering" (such as, for example, firing a rocket) will require energy, and that energy will have been part of the object when it fell past you, so it will get counted in the increased mass of the hole that you detect; and since that energy will have to be finite, at some point it will run out and the object will fall into the hole anyway. It is not possible for an object containing finite energy to "hug the horizon" indefinitely.

Yes, I understand that. I wasn't necessarily talking about an object doing so indefinitely. Just if outside observer would be able to tell the difference between object being outside or inside the horizon from calculation of mass alone.

In any case, thanks for the replies, I don't really have the math chops for GR.

 

[Vanadium 50]

If it is large enough to do this, it is large enough to cause a detectable increase in the mass of the hole, and the outside observer will detect that increase (by, for example, detecting a change in his orbital parameters, or a change in the proper acceleration he has to maintain to keep the same altitude) at a time that depends on where the object fell in, relative to where he is.

Can you elaborate?

If I am in orbit far from both the BH and the approaching object, I am in orbit around both of them - in orbit around their center of mass which has whatever momentum it has. That's true both before and after the object crosses the horizon.

 

[PeterDonis]

If I am in orbit far from both the BH and the approaching object,

Then the approaching object has already fallen past you (more precisely, past the distance from the hole that you are), long enough ago that the effects on the spacetime geometry will have had time to propagate to you, so yes, you will count the mass of the object as part of the overall mass of the thing you are orbiting.

 

[Vanadium 50]

Let's call it a javelin. I'm in orbit around the BH + javelin system. To me, nothing special happens when the javelin crosses the horizon (or, to be a little more accurate, gets close enough to the BH for the horizon to form around the javelin).

I don't think I can do better than calculating when the javelin must have crossed the horizon.

 

[PeterDonis]

I'm in orbit around the BH + javelin system.

Yes, which means, as I said, that the javelin already, some time ago, fell close enough to the BH for you to measure the change in mass.

To me, nothing special happens when the javelin crosses the horizon

It never does. Heuristically, the "something special happens" when the change in the spacetime geometry reaches you. If the javelin fell right past you, the "something special" happens as soon as it passes you. If it falls in somewhere distant from you, the "something special" happens when the change in spacetime geometry propagates to you at the speed of light. But none of these will have any necessary connection to when the javelin crosses the horizon.

 

[DPH]

Yes, 10 Msun.
$M_s=1 Yes, 10 solar mass. With your formula and data:$1.756\times 10^{-3} ~s##

A related problem I have thought about, a 1kg mass increases the event horizon of a 10 solar mass black hole by 1.5e-27 m. It takes only a finite and relatively short observer time for a falling mass to reach this distance where it would be absorbed. Question, as the 1kg mass approaches the BH does the added mass move the horizon so we never see the mass absorbed?.

 

[meekerdb]

If it's true that a distant observer never "sees" an object reach the event horizon of a black hole, then what do you make of LIGO's observation of one black hole falling into another and the observed merger being complete in a minute. It certainly makes no sense for one black hole to be seen as "stuck" on the surface of another.

 

[PeterDonis]

as the 1kg mass approaches the BH does the added mass move the horizon so we never see the mass absorbed?

My previous posts have already addressed this type of scenario (for example, see my responses to @Vanadium 50 regarding the "javelin").

 

[PeterDonis]

If it's true that a distant observer never "sees" an object reach the event horizon of a black hole, then what do you make of LIGO's observation of one black hole falling into another and the observed merger being complete in a minute.

Black holes don't emit light of their own, so LIGO is not observing light emitted by one black hole as it falls into another.

What LIGO is observing is gravitational waves emitted as black holes merge. These waves are emitted from outside the combined horizon of the two holes, so LIGO can see them; but they also carry identifiable "signatures" of the spacetime geometry of the holes as they merge, signatures which would not be present (according to our best current understanding of GR) if it were any other kind of object besides black holes that were merging. The duration of the LIGO observations corresponds to the time when detectable gravitational waves are emitted from the merger.

 

[meekerdb]

There's a subtlety here. The outside observer will observe the increase in the size of the black hole and this happens in a finite time, but that's not the same thing as seeing the infalling object cross the horizon - light from that event is still trapped at the horizon.

There's an older thread where this was discussed at more length - I'll see if I can find it.

(And "their own gravitational field" isn't right, it's their mass/energy that matters. That's a bit of a digression in this context)

Is the increase in the size of the black hole observable at a great distance in a finite time? If so, I'd say that's an observation that the object has crossed the event horizon...although of course one can never "see" that.

 

[meekerdb]

In a spacetime containing a black hole merger, there are not multiple event horizons; there is only one. If visualized in a spacetime diagram with time going upwards, the horizon just looks like a pair of trousers instead of a cylinder: the legs of the trousers are the two holes that merge, and the top of the trousers is the merged hole. (The legs are also twisted around each other because the holes orbit each other before merging.) So it is indeed never the case that one hole is "stuck" just outside the horizon of the other.

Ok, now consider the limit in which one black hole is much smaller than the other. It's never "stuck" on the event horizon, but how is it different from some other extended bit of mass-energy falling in? You say there is only one event horizon, but it must change and it's change can only propagate at the local speed of light. So there is a wave of change in the event horizon spreading out from the entry point of the infalling object, as well as some radiated gravitational waves. I agree that one can never see something actually reach the event horizon. But as these gravitational waves are radiated from the merger event, so can photons be emitted and LIGO shows these can be detected at great distance and they cease sharply; so in that sense they have "seen" the infall event.

 

[PeterDonis]

now consider the limit in which one black hole is much smaller than the other.

In that limit I would not expect gravitational waves to be detectable from the merger. However, my earlier posts in this thread, regarding when an observer orbiting the larger hole at some distance would observe a change in their orbital parameters, indicating that something had fallen into the larger hole, would still apply.

how is it different from some other extended bit of mass-energy falling in?

It isn't.

You say there is only one event horizon, but it must change

Bear in mind that the event horizon is a surface in spacetime, not space. The geometry of this surface in spacetime already includes all the information about all the things that will fall into the black hole over its entire future history. When we talk about the horizon "changing", what we really mean is that we have adopted some convention about what events on the horizon--the surface in spacetime--are happening "at the same time" as some particular reading on our own clock. Any such convention is not a matter of physics, it's a matter of how we humans choose to describe the physics.

there is a wave of change in the event horizon spreading out from the entry point of the infalling object

No. This is not how it works. The event horizon is not that kind of thing. See above.

as well as some radiated gravitational waves

See my comment above about this for the limiting case you describe.

as these gravitational waves are radiated from the merger event, so can photons be emitted

Whether or not photons are emitted will depend on what kinds of matter are present when the merger takes place. If there is no matter present at all, we would not expect anything but gravitational waves to be emitted. But for most real-world black hole mergers, we expect that the holes are surrounded by matter which is falling in at the same time the merger takes place, and that matter will emit radiation as it falls in, and the radiation will be affected by the merger.

 

[Nugatory]

Is the increase in the size of the black hole observable at a great distance in a finite time?

Yes, with clarification that the “size” (by which you seem to mean the Schwarzschild radius) is not something that we measure directly. We measure the mass, by observing the gravitational effects produced by the mass. But…

If so, I'd say that's an observation that the object has crossed the event horizon.

No, because these gravitational effects are not interestingly different when the infalling mass is just outside the horizon and just inside. In the most extreme case, when the distribution of the incoming mass is spherically symmetric there is no difference at all.

But the problem with identifying the moment that the object crosses the horizon goes deeper than that. The (necessarily timelike) worldline of an infalling test body intersects the event horizon (a lightlike surface) at a particular point in spacetime and there’s no ambiguity about which one. That allows us to make statements about how much time passes for the infalling object, what an infalling clock reads as it falls through the horizon, how much time an infalling observer experiences before they die at the singularity. However, there’s no non-arbitrary way of matching events on the infaller’s worldline to our clock readings as we sit outside the black hole watching the proceedings.