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

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
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

Mentor
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

Mentor
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

Mentor
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).

Last edited:

#### 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!

Last edited:

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
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

Staff Emeritus
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

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

Mentor

#### pervect

Staff Emeritus
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

Some extreme topics in physics are even stranger than I thought.
I believe that software modelling will be become an important tool in cosmology.

#### 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.

Mentor

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving