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?

 

[PeterDonis]

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 anybody 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]

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]

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]

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 spatial 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 spatial 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.

 

[PeterDonis]

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]

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]

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]

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]

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]

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]

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]

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]

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:

https://www.physicsforums.com/threa...lar-momentum-in-gr.492449/page-4#post-3269586
https://arxiv.org/abs/1012.2028

 

[PeterDonis]

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.

 

[Orodruin]

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).

The easy way to see that this has no real physical meaning is to note that $\partial_t$ is a Killing field also in the interior region (although it is no longer time-like). Thus, if you go along a coordinate line of $t$, you will end up in exactly the same situation as you were before. If I am not mistaken, the transformation generated by $\partial_t$ is just a hyperbolic rotation of the Kruskal-Szekeres coordinates. Since the metric in those coordinates is on the form (with $T$ and $X$ as the KS coordinates)
$$
ds^2 = f(r) (dT^2 - dX^2) - r^2 d\Omega^2,
$$
and $r$ is constant along the hyperbolae of constant $T^2 - X^2$, this is invariant under hyperbolic rotations between $X$ and $T$.

 

[PeterDonis]

the transformation generated by $\partial_t$ is just a hyperbolic rotation of the Kruskal-Szekeres coordinates

That's my understanding as well, similarly to the way that the Rindler $\partial_t$ transformation is a hyperbolic rotation of Minkowski coordinates.

 

[PAllen]

[

The easy way to see that this has no real physical meaning is to note that $\partial_t$ is a Killing field also in the interior region (although it is no longer time-like). Thus, if you go along a coordinate line of $t$, you will end up in exactly the same situation as you were before. If I am not mistaken, the transformation generated by $\partial_t$ is just a hyperbolic rotation of the Kruskal-Szekeres coordinates. Since the metric in those coordinates is on the form (with $T$ and $X$ as the KS coordinates)
$$
ds^2 = f(r) (dT^2 - dX^2) - r^2 d\Omega^2,
$$
and $r$ is constant along the hyperbolae of constant $T^2 - X^2$, this is invariant under hyperbolic rotations between $X$ and $T$.

Great point, but I am more convinced by the simple SR example I gave that, in general, there is a fundamental ambiguity talking about distance from an event to a light like surface.

 

[PAllen]

On further thought, I claim that there is a meaningful way to argue that t going to infinity along an r=k surface represents approach to the horizon, even if you can’t pick an unambiguous distance:

Every spacelike path across the horizon will encounter unbounded t values as its affine parameter approaches the horizon crossing value from below.

 

[PeterDonis]

Every spacelike path across the horizon will encounter unbounded t values as its affine parameter approaches the horizon crossing value from below.

I don't think this argument works, because "unbounded t values" does not name a particular event or even a particular "distance" from the horizon. Curves of constant $t$ go all the way from the horizon to the singularity, for all values of $t$ no matter how large. So just saying "I encountered unbounded t values" says nothing about how close to the horizon you are; you could equally well have encountered unbounded t values close to the singularity.

 

[PeterDonis]

I claim that there is a meaningful way to argue that t going to infinity along an r=k surface represents approach to the horizon

To phrase my objection a different way, t going to infinity along a surface of constant $r$ is not the same as t going to infinity along a spacelike path that intersects the horizon.

 

[PAllen]

To phrase my objection a different way, t going to infinity along a surface of constant $r$ is not the same as t going to infinity along a spacelike path that intersects the horizon.

True, but conversely, there is no way to approach the horizon without t approaching infinity. (There is a tiny exception - the 2 sphere connecting the WH and BH regions, where the t coordinate is degenerate and doesn’t exist in the same sense as the longitude of the North Pole doesn’t exist).

 

[PAllen]

Keeping the water muddy, I also note (because I just found one) that there exists a valid foliation of (a large portion of) the Kruskal geometry where the horizon distance along an r=k surface increases with increasing t!

 

[Orodruin]

Regardless of what $t$-value you choose it is going to be related to $t = 0$ by a hyperbolic rotation as long as you remain on the same constant $r$ surface. The situation is going to look exactly the same due to the symmetry of the spacetime. I would have thought that would be sufficient to rule out any sort of meaningfulness of "being closer to the horizon" for large $t$.

 

[pervect]

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.

If we consider the third time derivative of $k^2(t)$, I get:

$$ \frac{d^3 }{dt^3} k^2(t) = 6 \, \dot{k} \, \ddot{k} + 2 \, k \, \dddot{k}$$

I don't see why this should vanish. The first term, in particular, should be proportional to the velocity multiplied by the acceleration. In general temrs, the idea is that we consider Newtonian motion in flat Minkowskii space-time, and ask if the third time derivative of the quadrupole moment vanishes. The terms of the quadrupole moment tensor in these coordinates are all proportional to $k^2$, so we just need to ask if the third time derivative of $k^2(t)$ vanishes.

I'm not getting it to vanish. I could have made a mistake, but I'd need to see a reference before I was convinced that it did vanish, my attempts to calculate it don't make it vanish.

To show some of the intermediate steps in the calculation as an afterthought

$$\dot{k^2} = 2\,k\,\dot{k} \quad \ddot{k^2} = 2 \dot{k}^2 + 2 \,k \, \ddot{k}$$

 

[PeterDonis]

I don't see why this should vanish.

I was mistaken. See my response to @PAllen in post #25.

 

[pervect]

I was mistaken. See my response to @PAllen in post #25.

I did see those posts - after I made my reply, so it came out awkwards. I did find another term in addition to the third time derivative of acceleration , though, which I thought was interesting. It may tie into MTW"s "power flow" idea, that I mentioned in a different post, since mass*velocity*acceleration would represent power. That would only apply to the first term I found, though - I'm not sure how one could justify ignoring the second term, the one proportional to the third time derivative of acceleration.

The advantage of tying gravitational radiation to power flow rather than the quadrupole formula is that it's easier to communicate to the lay audience. While I can cite the appropriate reference from MTW, it's clear that they were making some assumptions in their derivation, so it's not clear how general their formula is and that makes me hesitant to present it to the lay audience who frequently wants to analyze somewhat novel scenarios.

Another minor concern is that since we do apparently have gravitational waves being emitted, to get an accurate figure for the amount of radiation emitted we need to be concerned about back reaction forces modifying the trajectory, as was done on the Hulse-Taylor binary. For the benefit of those who may not be familiar with this (I'm sure Peter is, this is for the benefit of other readers who hopefully haven't been scared off), the observation of the decay of the orbital period of this binary won the Nobel prize for it's agreement with the calculations done by General relativity. <<link>>.

Going back to the original poster's problem, while I don't have any detailed calculations I would think that the gravitational radiation would be axis-symmetric, so I don't think it would carry momentum away from the system in the center-of-mass frame. This however, is an intuition, not a hard calculation.

Assuming this is correct, what I'd expect to happen is that the black hole and the infalling star would merge into one larger black hole, but, as in the inspiral cases that Ligo analyzes, the mass of the resulting black hole would be lower than the sum of the initial black hole mass and the star mass, the difference being carried away by the emitted gravitational waves.

 

[KurtLudwig]

I defer to your knowledge.
Some extreme topics in physics are even stranger than I thought.
I believe that software modelling will be become an important tool in cosmology.
Thank you for your explanations.

 

[Lars278]

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?

1. The mass will be concentrated at the singularity. 2. Simple Newtonian mechanics apply. If the mass of the black hole is a million times the mass of the star, the black hole will hardly move at all. If it's three times the mass of the star it will move substantially.

 

[PAllen]

Please note the end of my post #3. It is a much more complete answer to BH star interaction than the last few posts. If needed, I can provide technical references.

 

[PeterDonis]

The OP question has been answered. Thread closed.