I What happens to the inertia of a mass falling into a black hole?

KurtLudwig

What happens to the inertia of a mass falling into a black hole? I am not even sure if I frame the questions correctly. Will this mass reach the center or is mass distributed within the black hole? Is the singularity the whole volume of the black hole or is it a point in the center? If a large star falls into a medium-sized black hole, will the black hole move towards the star, due to gravitational attraction, or will the impact move the black hole away, due to the inertia of the star?

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PeterDonis

Mentor
What happens to the inertia of a mass falling into a black hole?
This doesn't seem like the same question as the many others you ask (quoted below). I don't even think this question is well defined; inertia isn't a thing that something can "happen" to.

Will this mass reach the center or is mass distributed within the black hole?
There is no "center" of the black hole in the usual sense. The "center" at $r = 0$ is to the future of every event inside the horizon, so anything that falls into the hole will eventually reach it since it is impossible to avoid traveling into the future.

Is the singularity the whole volume of the black hole or is it a point in the center?
Neither. See above.

If a large star falls into a medium-sized black hole, will the black hole move towards the star, due to gravitational attraction, or will the impact move the black hole away, due to the inertia of the star?
There is no impact because a black hole is not an object; it's spacetime geometry. The hole will simply increase in size as matter from the star falls in. Whether or not you view this as the star falling into the hole, or the hole moving towards the star, is a matter of choice of coordinates and does not affect the physics either way.

PAllen

Given the basic level of your questions, I will try to give beginner answers.

After an object falls into a BH, the BH horizon is larger, and the inertia is larger by the amount of mass absorbed by the BH. Note that since it is hard to push on BH, you can speak of charged BH, and a neutral infalling body, and apply an electric field, and measure inertial mass.

A black hole has no center.

The singularity is not a point. In an ideal, uncharged, nonrotating, BH it is a missing spacelike hypersurface that can be thought of as a time. It is more complex in other cases.

The mass will approach the singularity in the ideal case just referred to, as certainly as tomorrow will be approached. In other cases, it may or may not reach the singularity.

In the case of a medium BH interacting with a large star, the BH would move just like any body of the same mass. If the BH touches the star rather than orbiting, the interaction would be complex, with the BH absorbing some, but far from all the mass of the star, with some being ejected, and some forming an accretion disc.

pervect

Staff Emeritus
What happens to the inertia of a mass falling into a black hole? I am not even sure if I frame the questions correctly. Will this mass reach the center or is mass distributed within the black hole? Is the singularity the whole volume of the black hole or is it a point in the center? If a large star falls into a medium-sized black hole, will the black hole move towards the star, due to gravitational attraction, or will the impact move the black hole away, due to the inertia of the star?
It's wrong in detail to think of the black hole as being a point, but the details of what happens inside are hidden by the event horizon, so the mistake may not matter much. It's probably slightly less wrong (if there can be such a thing) to think of it as a point than to think of it has having a volume. Technically, we call it a singularity.

I can attempt an answer at the second part, though I haven't done any of the detailed calculations. If you have a large star and a black hole, initially at rest, and nothing else in the universe, they both will start to fall towards each other.

During this process, gravitational radiation will be emitted. I'm afraid I don't know the exact details, but more radiation will be emitted as the mass of the star increases. I am assuming the star has less mass than the black hole, I'm not sure of what might hapen if we imagine a small black hole and a large star.

The end result of this process will be that the black hole swallows the star. According to an observer on the star being swallowed, the infall will take finite (proper) time. For an external observer, the answer to the question is trickier and may be infinite. A more detailed answer here gets very technical.

With the somehat unrealistic assumption that there is nothing else in the universe, I would expect that the emission of gravitatioanl radiation probably would result in the resulting black hole no longer being at rest (even though the star and the black hole were initally at rest) but I don't know in what direction it might move. If the radiation was symmetrical, it might not move at all, I'm not aware of anyone doing such a calculation but there is at least a possibility that the resulting black hole might be moving.

With the unrealistic assumption that there is nothing else in the universe, we are able to assign masses and momentum (what you mean by inertial) to the initial black-hole and star, and to the end result, the black hole after it swallowed the starr and the emitted gravitational radiation. The technical wording for this assumption that there is nothing else in the universe is that we have an "asymptotically flat space-time".

Detailed calculations for the amount of gravitational radiation have been calculated for the more usual case of a non-direct infall, where the star slowly spirals into the black hole. In this case, I know that the amount of energy and momentum carried away by the gravitational radiation can be significant. For instance in the Ligo observations of one black hole falling into another in a circular inspiral, <<link>>, the initial masses of the black holes were 36 and 29 solar masses, a total of 65 solar masses. the end result was a 62 solar mass black hole, with the three solar mass being carried away by gravitational radiation.

This isn't the case you asked about, but it may be close enough to be of interest. There are a bunch of other Ligo papers, some of them seem to involve two neutron stars merging to form a black hole, but I don't see any observations of a black hole swallowing a star in my quick glance at the Ligo webiste for detection papers <<here>>.

PeterDonis

Mentor
During this process, gravitational radiation will be emitted.
Actually, it won't in the idealized case you describe, where the hole and the mass are initially at rest relative to each other. For this idealized case, the motion is purely radial, so there is no time-varying quadrupole moment, only a time-varying dipole moment. A time-varying quadrupole moment is necessary for GW emission.

In a more realistic case, where there is some sideways motion initially, the hole and the mass will spiral in towards each other, and in this case, yes, GWs will be emitted.

Buzz Bloom

Gold Member
There is no "center" of the black hole in the usual sense. The "center" at r=0r = 0 is to the future of every event inside the horizon, so anything that falls into the hole will eventually reach it since it is impossible to avoid traveling into the future.
Hi Peter:

In several other threads I have seen the cencept of the above quote as a description of the nature of the r=0 "point" inside a black hole. I have also seen some references cited on this topic, but I have not yet had the time to study them. The concept is apparently something other than a geometric spacial point, but as I am tring to understand it, it is some sort of point in the time dimension. The following are some guesses of plausible possibilities to my uneducated mind. I hope you can help me improve my understanding by correcting these guesses.

1. The z=0 point represents the entire interior space at infinite time in the future.
2. The z=0 point represents the entire interior space as the limit of all future time. (Thus is of cource very similar to #1 but avoids the ambiguous term "infinity".)
3. The z=0 point represents the entire interior space some finite time in the future corresponding to the mass of the black hole.
4. The z=0 point represents the limit of all future time, but not any spacial point or points.

Regards,
Buzz

PAllen

Every r=k hypersurface inside the horizon is a 3 manifold of topology S2 x R1, and it is a simultaneity hypersurface per the Schwarzschild foliation. Note that very different foliations are also possible. r=0 is not part of the manifold, but can be approached arbitrarily closely. In the limit of such hypersurfaces, the S2 radius approaches 0.

Any interior timelike worldline will approach r=0 in finite proper time. Further, a world line of constant t, theta, and phi, is a geodesic timelike world line, with r being a timelike parameter along it.

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PeterDonis

Mentor
it is some sort of point in the time dimension
It's a spacelike line. More precisely, it's the limit of a series of spacelike 3-surfaces that each contain an infinite continuum of 2-spheres with radius $r$, as $r \rightarrow 0$. This can be viewed as an instant of time, which might be what you are trying to get at with "point in the time dimension"; if you view it this way, it is an instant of time that is in the future for any observer inside the hole's horizon.

The following are some guesses of plausible possibilities
All of these are wrong. See above.

PAllen

I guess one other observation about the interior is that for an r=k hypersurface, the horizon is approached at axial infinity, i.e. as t goes to infinity.

While Schwarzschild interior and exterior coordinates are two completely disjoint coordinate patches, neither of which can include the horizon, many other coordinates do include the horizon. These show there are smooth spacelike or timelike paths across the horizon.

PeterDonis

Mentor
one other observation about the interior is that for an r=k hypersurface, the horizon is approached at axial infinity, i.e. as t goes to infinity.
This doesn't look right. An $r = k$ hypersurface, where $0 < k < 2M$, does not intersect the horizon at all, and for all values of $t$ it is equally "far" from the horizon (in terms of arc length along a curve of constant $t$).

PAllen

This doesn't look right. An $r = k$ hypersurface, where $0 < k < 2M$, does not intersect the horizon at all, and for all values of $t$ it is equally "far" from the horizon (in terms of arc length along a curve of constant $t$).
A curve of constant t is timelike, thus irrelevant. Imagine an r=k surface represented in Kruskal coordinates. Then, as t goes to infinity, you get ever closer the horizon measured along Kruskal simultaneity line (you must go outside SC coordinates to measure distance to the horizon).

What you are doing is measuring time back to the WH BH boundary on the horizon.

PeterDonis

Mentor
Imagine an r=k surface represented in Kruskal coordinates. Then, as t goes to infinity, you get ever closer the horizon measured along Kruskal simultaneity line
Ah, I see. I'm not sure how physically relevant Kruskal simultaneity lines are, but given that they are being used, yes, arc length along those lines from the horizon to any $r = k$ surface decreases as you go farther and farther around the hyperbola of the surface (i.e., as $t$ gets more and more positive or negative).

PeterDonis

Mentor
you must go outside SC coordinates to measure distance to the horizon
More precisely, trying to evaluate a spacelike arc length to the horizon from an event inside the horizon is extremely difficult in SC coordinates since there is no single coordinate you can integrate along (whereas outside the horizon, purely radial curves are spacelike and you can just integrate along them and take a limit as $r \rightarrow 2M$).

PAllen

Ah, I see. I'm not sure how physically relevant Kruskal simultaneity lines are, but given that they are being used, yes, arc length along those lines from the horizon to any $r = k$ surface decreases as you go farther and farther around the hyperbola of the surface (i.e., as $t$ gets more and more positive or negative).
Well, measuring distance is certainly more relevant than time when speaking of distance to the horizon. Also, note that distance from an event to the horizon would normally be understood as length along the spacelike geodesic that minimizes this. The Kruskal distance must be greater than or equal to this. Thus my statement remains true for the optimal definition of horizon proximity.

PeterDonis

Mentor
measuring distance is certainly more relevant than time when speaking of distance to the horizon.
Meaning, measuring along spacelike geodesics; yes, I agree that "distance" usually has that implication.

The primary counterintuitive property of spacetime inside the horizon is that it is not stationary, so there is no notion of "space" that is picked out by any symmetry of the spacetime (whereas outside the horizon the "space" orthogonal to the timelike KVF is so picked out). So the best we can do would be to look for some "natural" space that was orthogonal to the worldline of some observer of interest. I'm not sure that curves of Kruskal simultaneity correspond to such a "space", which is why I expressed reservations about how physically relevant that simultaneity is.

distance from an event to the horizon would normally be understood as length along the spacelike geodesic that minimizes this.
I think you mean "maximize" (subject to appropriate constraints), not "minimize", since it seems to me that one could always pick out spacelike geodesics that are closer and closer to the null geodesic from any chosen event to the horizon, and therefore have lengths arbitrarily close to zero.

PAllen

I think you mean "maximize" (subject to appropriate constraints), not "minimize", since it seems to me that one could always pick out spacelike geodesics that are closer and closer to the null geodesic from any chosen event to the horizon, and therefore have lengths arbitrarily close to zero.
Hmm, yes, minimize doesn’t work in this case (Minkowskian manifold). Maximize sounds plausible given the constraint that you are speaking of geodesics. It still seems to me that my claim remains true - that t approaching infinite along an r=k hypersurface approaches the horizon.

PAllen

Actually, maybe the problem has no well defined answer. Consider the simple case of distance from an event to a light path in flat spacetime restricted to x,t. Then there are spacelike geodesic distances from the event to the light path ranging from (0,infinite). Thus, at least in some cases (e.g. the simplest!) there is no well defined meaning for distance from an event to light like surface.

PeterDonis

Mentor
maybe the problem has no well defined answer
For inside the horizon, yes, I think that's the case. Outside the horizon, as I said, distance along a radial spacelike geodesic in the hypersurface orthogonal to the timelike KVF is picked out by symmetry, so there is at least a plausible case for that being "the" distance to the horizon. But the corresponding "distance" inside the horizon is, as you pointed out, actually the time from the point at the center of the Kruskal diagram (where the WH and BH horizons meet) to the chosen event.

KurtLudwig

Physics is sure not what it seems. Our everyday physical experiences will not help us understand black holes or particle physics. (I just completed reading a textbook "The ideas of particle physics, an introduction for scientists" by Coughlan, Dodd, Gripaios of Cambridge University.) It seems physics cannot be invented, only observations and mathematical modelling will further cosmology and physics.
I have read that some massive black holes contain the mass of billions of stars. How can gravity keep the black hole from exploding? The internal pressures inside a black hole must be enormous. As a black hole accumulates more and more mass, will it ever explode? Have astronomers ever observed a black hole exploding, as massive stars do?

PAllen

Physics is sure not what it seems. Our everyday physical experiences will not help us understand black holes or particle physics. (I just completed reading a textbook "The ideas of particle physics, an introduction for scientists" by Coughlan, Dodd, Gripaios of Cambridge University.) It seems physics cannot be invented, only observations and mathematical modelling will further cosmology and physics.
I have read that some massive black holes contain the mass of billions of stars. How can gravity keep the black hole from exploding? The internal pressures inside a black hole must be enormous. As a black hole accumulates more and more mass, will it ever explode? Have astronomers ever observed a black hole exploding, as massive stars do?
Classical Theory: If classical GR is true, an exploding BH is trivially excluded because matter would have overtake light for this to be possible.

Quantum gravity theory: in the inconceivable future, all BH will become tiny due to Hawking radiation, and in the last moment explode.

Observation: No, there is no observational hint of an exploding BH, at all. I think people are looking for possible signatures of last moment BH explosions, but this is very challenging because the energy released is modest, and the likely distance large.

PeterDonis

Mentor
The internal pressures inside a black hole must be enormous.
There are no internal pressures inside a black hole. A black hole is vacuum; there is no matter inside. It is not an ordinary object with matter inside that is supported against gravity by pressure.

pervect

Staff Emeritus
Actually, it won't in the idealized case you describe, where the hole and the mass are initially at rest relative to each other. For this idealized case, the motion is purely radial, so there is no time-varying quadrupole moment, only a time-varying dipole moment. A time-varying quadrupole moment is necessary for GW emission.

In a more realistic case, where there is some sideways motion initially, the hole and the mass will spiral in towards each other, and in this case, yes, GWs will be emitted.
How so?

It's been a while, but suppose we have two unit point masses at (k,0,0) and (-k,0,0) in some locally Minksowskii frame. For the trace-free quadropole moment, which is the source of gravitational radiation, I get:

$$\begin{bmatrix} \frac{4k^2}{3} & 0 & 0 \\ 0 & -\frac{2k^2}{3} & 0 \\ 0 & 0 & -\frac{2k^2}{3} \end{bmatrix}$$

This uses MTW's formulation, as the text mention various other textbooks use different conventions.

If this is correct (and I don't see an error), then as k varies the quadrupole moment will varry.

PeterDonis

Mentor
then as k varies the quadrupole moment will varry
Yes, but (I think) the third time derivative of the $k$ will vanish, and it's actually the third time derivative of the quadrupole moment that drives GW emission. I should have been more specific.

PAllen

Yes, but (I think) the third time derivative of the $k$ will vanish, and it's actually the third time derivative of the quadrupole moment that drives GW emission. I should have been more specific.
The third time derivative will vanish only if the motion is exactly uniform acceleration. That is not even true in Newtonian gravity for this case.

Here are a couple of links from prior discussions of this:

PeterDonis

Mentor
The third time derivative will vanish only if the motion is exactly uniform acceleration. That is not even true in Newtonian gravity for this case.
Hm, yes, you're right. That's what I get for posting too late at night.

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