B What path does matter take after entering a black hole?

[PeterDonis]

All the angular momentum of the object would be transferred to the black hole before it falls in.

"Transferred" is a misleading word here. Once the object is inside the horizon, it is part of the hole as far as outside observers are concerned, so its angular momentum can be part of the hole and part of the object. The object does not "give up" its angular momentum magically at the horizon and change its trajectory.

So it seems that the "trajectory" of two objects, one originally with angular momentum and one without ( as I mentioned in post 5 ), as viewed from a distant observer, actually ends up the same.

No, it isn't. See above.

 

[Ibix]

The answer [...] won't be expressible as a "trajectory" in the usual sense of the word since spacetime inside the horizon lacks some relevant properties (for example, it isn't stationary).

In what way does spacetine being non-stationary prevent the path being a trajectory? It's certainly expressible, at least in principle, as $(t(r),r,\theta(r),\phi(r))$. Is there some technical meaning of the word trajectory that I'm not aware of?

 

[256bits]

An infalling object has linear and angular momentum, and while an outside observer can't see the post horizon trajectory, she can see that the linear and angular momentum has been transferred to the hole.

So you should not expect any drastic change in the "trajectory" of the thing (to the extent that term is well-defined) when it crosses the horizon.

All the angular momentum of the object would be transferred to the black hole before it falls in. Stands to reason since whatever happens within cannot be observed outside the hole.
The non-rotating hole then becomes rotating. As well, any object outside the hole then has angular momentum transferred to it from the hole. Which again has its momentum transferred back to the hole.

so it seems that the "trajectory"

"Transferred" is a misleading word here. Once the object is inside the horizon, it is part of the hole as far as outside observers are concerned, so its angular momentum can be part of the hole and part of the object. The object does not "give up" its angular momentum magically at the horizon and change its trajectory.
No, it isn't. See above.

I know. Words are difficult to describe black holes. Not sure what terms are appropriate at times
The influence of the gravitational field of the black hole extends to infinity, so momentum transfer I agree does not magically happen at the horizon, but does becomes more pronounced the closer the object moves toward the black hole. Due to frame dragging, ergosphere, as the object approaches the hole - it is falling the same time as having a tangential velocity ( wrt an far outside observer ) -
Then Question: the outside observer records the same rotational velocity for both black hole and object at the event horizon?

 

[PeterDonis]

In what way does spacetine being non-stationary prevent the path being a trajectory?

The fact that "trajectory" usually means a path in space, not spacetime, and in a non-stationary spacetime there is no invariant notion of "space". Of course you can choose coordinates and define "space" according to your chosen coordinates, but there is no geometric property of the spacetime involved. In a stationary spacetime the integral curves of the timelike Killing vector field mark out "points in space" independent of any choice of coordinates.

 

[PeterDonis]

All the angular momentum of the object would be transferred to the black hole before it falls in.

I have already explained why this is wrong. If you keep repeating wrong statements after they have been corrected, you will receive a warning.

The non-rotating hole then becomes rotating.

Which, as I have already explained, does not mean the object with the angular momentum somehow "stops rotating" and magically changes its trajectory at the horizon. It doesn't. I have already explained what does happen.

momentum transfer I agree does not magically happen at the horizon, but does becomes more pronounced the closer the object moves toward the black hole

Wrong. There is no "momentum transfer". That's not what is happening.

Due to frame dragging, ergosphere, as the object approaches the hole - it is falling the same time as having a tangential velocity

If the hole was non-rotating before the object fell in, there is no frame dragging.

If the hole was rotating to start with, what you describe is correct (for geodesic motion), but it does not change the object's angular momentum. In Kerr spacetime, angular momentum and angular velocity do not have the simple relationship you are assuming.

the outside observer records the same rotational velocity for both black hole and object at the event horizon?

The outside observer has no way of measuring the "rotational velocity" of the hole independently of objects falling in.

 

[DaveC426913]

The best answer to your underlying question is the one I gave in a previous post a little bit ago: none of the relevant properties of the thing that is falling in change when it crosses the horizon. Locally the thing can't even tell that it's crossed the horizon. So you should not expect any drastic change in the "trajectory" of the thing (to the extent that term is well-defined) when it crosses the horizon.

...and the one I gave waaayy back in post 19...

"...there is nothing special about the event horizon as regards to trajectory. It's not a barrier or boundary; it is simply a geometrically-defined surface below which light (et al) cannot escape.

An infalling observer will not experience anything unusual passing through the EH; in fact he won't even know without doing some measurements and calculations (quickly!)."

 

[Ibix]

The fact that "trajectory" usually means a path in space, not spacetime, and in a non-stationary spacetime there is no invariant notion of "space".

Ah - I see your point. I think you are using a fairly strict definition of "trajectory", but fair enough.

 

[PeterDonis]

I think you are using a fairly strict definition of "trajectory"

The obvious alternative meaning would be "worldline", i.e., the curve in spacetime that describes the object's history. The object certainly does continue to have a worldline when it falls through the horizon, and that worldline continues to be a curve in spacetime, yes.

 

[vanhees71]

I'm not aware of calculations in textbooks, where one consideres the general motion of a test mass around a (say Schwarzschild) black hole except the most simple case of radially falling towards the center. For that example you realize that nothing special happens at the event horizon, because there's nothing special their. The singularities are only coordinate singularities at the Schwarzschild radius. The motion of the particle is just smooth across the event horizon. There's only the singularity at the origin of the black hole.

 

[timmdeeg]

The object certainly does continue to have a worldline when it falls through the horizon, and that worldline continues to be a curve in spacetime, yes.

Does the worldline of an object inside a Schwarzschild black hole depend on any deviations from a radial path outside?

 

[timmdeeg]

There's only the singularity at the origin of the black hole.

Or instead something else because physics breaks down there.

 

[vanhees71]

Yes indeed, I think singularities are a clear manifestation of our ignorance, and classical point-like objects always make trouble and the theories inconsistent (like in the less complicated case of electrodynamics, where classical point particles simply don't work, i.e., there's no fully consistent dynamics of classical point charges and the em. field).

 

[timmdeeg]

there's no fully consistent dynamics of classical point charges and the em. field).

Do you think that a future theory of quantum gravity (if we ever will have it) will shed light also onto this inconsistency (besides singularity)?

 

[lavinia]

A naive question: If light cannot escape from the interior of a black hole then doesn't that say something about the light cones? Wouldn't they define bounded trajectories for light? Wouldn't a body with mass have to stay within these cones and would also follow a bounded trajectory? If there is a inevitable singularity would the light cones have to converge to it somehow?

 

[PeroK]

A naive question: If light cannot escape from the interior of a black hole then doesn't that say something about the light cones? Wouldn't they define bounded trajectories for light? Wouldn't a body with mass have to stay within these cones and would also follow a bounded trajectory? If there is a inevitable singularity would the light cones have to converge to it somehow?

Try this:

 

[lavinia]

Try this:

I found this.
https://physics.stackexchange.com/q...ght-behave-within-a-black-holes-event-horizon

 

[PeroK]

I found this.
https://physics.stackexchange.com/q...ght-behave-within-a-black-holes-event-horizon

Which is just another question based on the misapprehension that $r$ is a spacelike coordinate inside the event horizon. Inside the horizon, you still have three spatial dimensions to move around in. The inevitable singularity is not in any specific spatial direction. And it's your time evolution, as it were, that leads you towards it.

Try the PBS video.

 

[Nugatory]

A naive question: If light cannot escape from the interior of a black hole then doesn't that say something about the light cones?

Yes

Wouldn't they define bounded trajectories for light?

Yes

Wouldn't a body with mass have to stay within these cones and would also follow a bounded trajectory?

Yes

If there is a inevitable singularity would the light cones have to converge to it somehow?

No, because the singularity isn’t a place is space so there’s nothing to converge on. If you draw a spacetime diagram on a piece of paper, the singularity will be a place on the paper where worldlines just end - whether it’s represented as a point on the paper or a larger region that cuts across an entire light cone depends on how you map points in spacetime to points on the sheet of paper.

In a traditional Minkowski SR spacetime diagram we plot $t$ on the y-axis of our sheet of graph paper and $x$ on the x-axis. This doesn’t work in curved spacetime of course, but we can instead use Schwarzschild coordinates and plot the $r$ coordinate on the x-axis of our graph. When we do this the singularity appears as a vertical line at the origin and the light cones are tipped sideways so that both sides of any light cone with its base inside the horizon will intersect that line. So everything ends up reaching the singularity, but that doesn’t imply convergence.

A spacetime diagram drawn using Kruskal coordinates is particularly helpful in visualizing how light cones relate to the event horizon and the singularity because in those coordinates lightlike worldlines are drawn at the same 45-degrees-from-vertical angle as on the Minkowski diagrams that know from special relativity. The singularity is drawn across the entire top of the sheet of graph paper.

(You may be wondering why I’m making such an effort to distinguish between points in spacetime and points on the surface of the sheet of graph paper. The reason is that the singularity is not part of spacetime at all. No matter what coordinate system I use to map points in spacetime to points on the sheet of paper, there will be points on the sheet of paper that do not correspond to any point in spacetime, and those represent the singularity. A worldline that encounters the singularity... just ends because there’s no “next” point to advance to).

 

[PeterDonis]

Does the worldline of an object inside a Schwarzschild black hole depend on any deviations from a radial path outside?

Of course it does; why wouldn't it? There is nothing magical at the horizon that erases tangential motion.

 

[lavinia]

Which is just another question based on the misapprehension that $r$ is a spacelike coordinate inside the event horizon. Inside the horizon, you still have three spatial dimensions to move around in. The inevitable singularity is not in any specific spatial direction. And it's your time evolution, as it were, that leads you towards it.

Try the PBS video.

read the answers

 

[timmdeeg]

Of course it does; why wouldn't it? There is nothing magical at the horizon that erases tangential motion.

Will a non-radial path be deflected towards radial (though not reaching radial at the horizon) with decreasing distance to the horizon? And is the intuition correct that an increasing non-radial worldline within the black hole will take increasing proper time to reach the singularity?

 

[pervect]

read the answers

I found this.
https://physics.stackexchange.com/q...ght-behave-within-a-black-holes-event-horizon

There are a couple of good answers in that thread from stackexchange you linked above. I only glanced at the first two, John Rennie's and Lubos' Mott's answers. Are you satisfied with those answers? Do you have any furhter qu4estions? Your responses are rather terse, I'm not quite sure if there's a need to say anything else.

 

[PeterDonis]

Will a non-radial path be deflected towards radial (though not reaching radial at the horizon) with decreasing distance to the horizon?

Why would it be deflected? As has already been pointed out multiple times, there is nothing magical at the horizon.

is the intuition correct that an increasing non-radial worldline within the black hole will take increasing proper time to reach the singularity?

What do you mean by "increasing non-radial worldline"?

 

[lavinia]

There are a couple of good answers in that thread from stackexchange you linked above. I only glanced at the first two, John Rennie's and Lubos' Mott's answers. Are you satisfied with those answers? Do you have any furhter qu4estions? Your responses are rather terse, I'm not quite sure if there's a need to say anything else.

For the moment yes. Thank you for asking.

- I was a little surprised that answers given in this thread used coordinate systems. It would seem that light cones are regions of the tangent bundle so one ought to be able to give a coordinate free description of their behavior. I later found pictures on line. One came from a book by someone named Geroch which showed the convergence of light cones to the singularity. BTW: This convergence seems real even though the singularity formally is not a part of space-time. I suppose there is a completion of the space-time manifold to include the singularity in which convergence does take place.

- The answers given here were all in terms a single point singularity and I would have thought that there is a more general answer that allows for a distribution of singular points, not just a single point. Suppose for instance that the singularity forms a two dimensional surface. This would still imply infinitely dense matter and should form a black hole.

 

[timmdeeg]

Why would it be deflected?

I was assuming that the trajectory of a non-radially infalling object is bent towards the center of mass such that it approaches but not necessarily reaches a radial path. Is that wrong?

 

[PeroK]

I was assuming that the trajectory of a non-radially infalling object is bent towards the center of mass such that it approaches but not necessarily reaches a radial path. Is that wrong?

In what way, if any, is your mental picture of a black hole inside the event horizon different from the region around a Newtonian point mass? Which is also a singularity.

 

[Ibix]

I was assuming that the trajectory of a non-radially infalling object is bent towards the center of mass such that it approaches but not necessarily reaches a radial path. Is that wrong?

It's not generally true. Qualitatively, GR orbits outside $3R_S/2$ aren't radically different from Newtonian orbits, which includes open orbits that are nearly radial, become tangential, then become nearly radial outward again. However, all free-fall orbits that graze $3R_S/2$ fall in, which is not a feature of Newtonian physics. And null trajectories are correctly handled in GR, where there isn't a clearly correct way to do it in Newtonian physics.

I'm not certain that "approaches radial" is ever a correct description of orbits nearing the centre of a spherically symmetric gravitational fields (remember that tossing a ball on Earth is the aphelion end of a very eccentric orbit whose perihelion would be deep in the core). Angular momentum is conserved, and is not affected by the field, so the tangential velocity must increase as the orbit gets lower. Certainly for open and stable orbits your characterisation is incorrect, but I haven't done the maths for orbits that fall into a black hole.

 

[timmdeeg]

In what way, if any, is your mental picture of a black hole inside the event horizon different from the region around a Newtonian point mass? Which is also a singularity.

My remark concerned the outside of the horizon and I have replaced "deflected" by "bent" hoping that then I express myself better. Inside r- and t-coordinates are changed and I think a comparison with a Newtonian trajectory doesn't make sense. But please elaborate on this if I'm wrong.

 

[timmdeeg]

Certainly for open and stable orbits your characterisation is incorrect, but I haven't done the maths for orbits that fall into a black hole.

I seem to be wrong what "orbit" means, I thought orbit means circling around.

I am talking about an object on a non-radial trajectory falling through the event horizon. Searching the web I couldn't find anything about non-radial fall into a black hole.

 

[jartsa]

Will a non-radial path be deflected towards radial (though not reaching radial at the horizon) with decreasing distance to the horizon? And is the intuition correct that an increasing non-radial worldline within the black hole will take increasing proper time to reach the singularity?

Yes sure. There's a downwards force, causing an increase of downwards speed, and a decrease of horizontal speed. It's the time dilation effect. Or the the effect that sometimes in relativity acceleration points to other direction than force.

If we are interested about momentum, downwards momentum increases, horizontal momentum stays constant.

(I'm not saying anything about the "within the black hole" -part)