What path does matter take after entering a black hole?

[MikeeMiracle]

TL;DR Summary

When matter passes the event horizon does it continue in a spiral towards the center or does it take a more direct path to the center?

Probably a silly question I thought of last night but would appreciate some clarification. Matter falling towards a black hole joins the disk spinning around the black hole slowly inching towards the event horizon with each orbit of the black hole. When matter passes the event horizon does it continue in a spiral towards the center or does it take a more direct path to the center?

I'm making some assumptions here that I have no doubt you will correct me on.

1) Matter falling towards a black hole outside the event horizon forms into a disk due to angular momentum more than being a geodesic.
2) After crossing the event horizon the geodesic effect is all that counts and the geodesic forms a straight line path direct to the center?

So to sum up if we had an outside observer that could see inside the event horizon, they would see matter spiraling towards the event horizon and then once crossing the event horizon taking a direct path straight to the center?

Thanks in advance.

 

[Orodruin]

So to sum up if we had an outside observer that could see inside the event horizon

This is a contradiction in terms.

Also note that it is not clear what you would mean by "straight to the center". The geometry of the interior of the black hole is not Euclidean and it does not really have a "center". The singularity at $r = 0$ is more like a moment in time than a place in space as the $r$ coordinate is time-like inside the horizon.

 

[MikeeMiracle]

Hi Orodruin

I was trying to very mindful about not mentioning anything to do with what the center might be, it's outside the scope of the question and I believe it will be irrelevant to the answer.

Let me try asking again slightly differently. Let's take a 5km wide black hole with it's center 2.5km away from the event horizon in all directions. Matter crosses the event horizon and travels 500m, is that 500m a spiral towards the center or a straight line towards the center?

I'm not interested about the final destination of the matter, what it might find when it gets there, or what will happen to it when it does get there. I am just interested in the path taken down to that point for the purposes of this question.

Thanks

 

[Orodruin]

I was trying to very mindful about not mentioning anything to do with what the center might be, it's outside the scope of the question and I believe it will be irrelevant to the answer.

Your question was "does it go straight to the center?" Clearly, what is meant by "center" is highly relevant to that question.

Let's take a 5km wide black hole with it's center 2.5km away from the event horizon in all directions.

What do you mean by 5 km wide? Remember that the geometry is not Euclidean and you need to define what you mean by "center" in order for that word to hold any meaning.

 

[MikeeMiracle]

Ok, maybe I am not getting this. I am just interested in what happens as soon as you pass the event horizon.

From what I have read on this forum previously (and maybe/probably interprited incorrectly,) is that the distance from the "whatever" it is that exists at the "center" to the event horizon is determined by the mass of a black hole. So the more massive the black hole, the further away the event horiszn would be from this "center." This implies an area inside the event horizon which is not the center that must be traveled through in order to reach the center. So the question is based on what happens between crossing the horizon and reaching the center.

Perhaps the term "center" is not the correct term and I should be using "final destination" instead? Or perhaps we can assume there is indeed a singularity and I should use this instead of the word "center." What happens between crossing the event horizon and traveling towards the singularity.

Your responces are leading me to believe that I have this all wrong and there is no center/singularity and eveything inside the event horizon is/acts the same, therefore making the question irrelevant.

As I understand it a singularity is a point in space and time where everything converges into a single "point." Does that "point" occupy the entire black hole? If not then there should be "something" that you need to pass through in order to reach it.

You seem to be suggesting the geometry inside the black hole cannot be thought of this way and in effect everything inside the event horizon "is" the singularity.

Ok, that's enough from me, some clarification would be welcome.

Thanks again.

 

[256bits]

So to sum up if we had an outside observer that could see inside the event horizon, they would see matter spiraling towards the event horizon and then once crossing the event horizon taking a direct path straight to the center?

As I understand it,
For an outside observer, a singularity occurs at the event horizon, where time appears infinite.
And where an object, to the far observer, does not appear to cross the event horizon.

Yet we do know that an object can cross the event horizon ( which to the object offers no barrier to penetration ), by using other coordinates. Since spacetime acts no different just outside the horizon from just inside the horizon, ( no disjoint in physics ), for the traveler crossing through the horizon, you are asking whether the object looses its tangential velocity by virtue of the crossing. And within the black hole, how is the tangential velocity manifested, perhaps in a comparison of the trajectory of an object that fell radially directly into a black hole.

PS.As for the question, black hole questions are never silly IMO, as they are difficult to fathom, imagine, analyze,...

 

[PeroK]

Ok, maybe I am not getting this. I am just interested in what happens as soon as you pass the event horizon.

Try this to start:

 

[PeroK]

Or, this one:

 

[Ibix]

The singularity isn't the "centre" of a black hole - as @Orodruin notes, it's a point in time in your future (once you've crossed the horizon). That's one reason you can't avoid it - how would you avoid Tuesday morning?

The time you measure from crossing the horizon to striking the singularity (for a non-rotating uncharged black hole) depends on your trajectory, yes. The maximum, for a black hole of $K$ solar masses, is about $15K\ \mu \text{s}$. Trajectories (powered or otherwise) with a shorter survival time are possible.

In practice, tidal forces will necessarily kill you before that. Or you enter a regime where quantum gravity is important and all bets are off.

 

[PeroK]

The singularity isn't the "centre" of a black hole - as @Orodruin notes, it's a point in time in your future (once you've crossed the horizon). That's one reason you can't avoid it - how would you avoid Tuesday morning?

I know someone who regularly avoids mornings!

 

[Ibix]

I know someone who regularly avoids mornings!

Spoiler: Black humour

I want to die like my grandfather did, peacefully in his sleep. Not screaming in terror like his passengers.

 

[Vanadium 50]

The singularity isn't the "centre" of a black hole - as @Orodruin notes, it's a point in time in your future (once you've crossed the horizon).

Which is also why discussions of length or radius run into trouble here. How wide is Thursday?

This is a "what do the laws of physics say would happen if we break the laws of physics" question. We can't see beyond the horizon, so "what would we see if we could" is not a well-defined question. Maybe a well-defined question could be asked, but this isn't it.

My answer - "there is no such path. Anything crossing the horizon turns into invisible pink unicorns". Prove me wrong. More importantly, describe an experiment that would prove me wrong. That might lead to an answerable question.

 

[MikeeMiracle]

So...according to the video a black hole is a series of events all happening at exactly the same time at different spatial cordinates inside the event horizon. So a larger black hole just contains more events. It also suggests we do not understand why or how a black hole has mass, it's like the events "project" mass somehow or we do not really fully understand what mass is.

I'm still confused about the density of the black hole and why a smaller black hole will spaghetify you but a larger one would not. I don't understand why they do not all have the same density and cause the same effects.

Going back to my original question though, after watching this video my question is basically meaningless and irrelevant.

 

[PeroK]

Which is also why discussions of length or radius run into trouble here. How wide is Thursday?

This is a "what do the laws of physics say would happen if we break the laws of physics" question. We can't see beyond the horizon, so "what would we see if we could" is not a well-defined question. Maybe a well-defined question could be asked, but this isn't it.

My answer - "there is no such path. Anything crossing the horizon turns into invisible pink unicorns". Prove me wrong. More importantly, describe an experiment that would prove me wrong. That might lead to an answerable question.

You could, of course, go beyond the event horizon yourself and do a experiment (if you are quick). The experiment itself can be done, but publishing the results is a different matter!

Does this mean that the predictions of GR inside the event horizon are well defined? I would lean towards saying "yes".

 

[Ibix]

Maybe a well-defined question could be asked, but this isn't it.

You can meaningfully ask what GR says about whether particles with a non-zero tangential component to their velocity as they approach the horizon continue to have a non-zero tangential component to their velocity after crossing the horizon. Yes they do. Or are there distinct trajectories inside the hole depending on your infall. Again, yes there are.

Whether or not this is testable is a nice point. It doesn't seem impossible that you could insert an object into an evaporating black hole and get it out as the hole evaporates. Whether or not it's actually possible depends on what quantum gravity actually looks like, and we don't know. And an evaporating black hole isn't a Schwarzschild black hole anyway.

 

[Ibix]

The experiment itself can be done, but publishing the results is a different matter!

Publish or perish, taken literally.

 

[MikeeMiracle]

The second video linked was very useful. I have at leave some basic understand about what a Penrose diagram is now.

I'm trying to understand as much as is possible without being an academic and have already come further than I thought possible thanks to this forum.

Thank you all for your continuing patience and knowledge transfer, it is very much appreciated.

 

[timmdeeg]

I'm still confused about the density of the black hole and why a smaller black hole will spaghetify you but a larger one would not.

This is because "spaghettification" goes with $M/R^3$. In case of a black hole the Schwarzschild radius is $R=2M$ and thus "spaghettification" goes with $1/M^2$.

 

[DaveC426913]

It might be helpful, in an oblique way, to understand that 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!).

 

[Vanadium 50]

It also suggests we do not understand why or how a black hole has mass, it's like the events "project" mass somehow or we do not really fully understand what mass is.

Presumably you are talking about gravitational mass. The mass of the hole is the mass of whatever has fallen in the hole.

 

[Ibix]

Presumably you are talking about gravitational mass. The mass of the hole is the mass of whatever has fallen in the hole.

...with a few caveats about exactly what you mean by "mass", before anyone takes that a bit too literally.

 

[Vanadium 50]

why a smaller black hole will spaghetify you but a larger one would not.

That is a property of the inverse square force and not anything specific to black holes. Tides on Earth are an example of "spaghettification" on a less extreme scale.

 

[Orodruin]

A larger mass black hole will also spaghettify you according to the GR predictions - just not as you pass the event horizon or earlier.

 

[Vanadium 50]

Whether or not this is testable is a nice point.

I don't think this is testable - your suggestions of quantum effects (Hawking radiation etc.) worry me that if you get enough down this path to see inside a horizon, you probably are far enough down this path that you can no longer think of classical trajectories. In any event, I think quantumizing the question makes it harder and not easier and so is best avoided.

Because of that, I still don't like the "looking inside the horizon" aspect for the reasons above. There are, however, some things you can say. 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.

 

[DaveC426913]

That is a property of the inverse square force and not anything specific to black holes. Tides on Earth are an example of "spaghettification" on a less extreme scale.

I've often wondered about this.

Is the lateral compression aspect of spaghettification due to the fact that there is a non-zero angle formed between the two lateral extents of any object and the point centre of mass? i.e. my left arm being pulled to centre, as is my right arm?

 

[PeterDonis]

Matter falling towards a black hole joins the disk spinning around the black hole slowly inching towards the event horizon with each orbit of the black hole.

This is for a rotating black hole; it doesn't happen for a non-rotating black hole (matter mostly just falls straight in).

When matter passes the event horizon does it continue in a spiral towards the center or does it take a more direct path to the center?

The "center" is not what you're thinking it is, so this question doesn't really make sense.

For a non-rotating black hole, there is no "center"--the singularity at $r = 0$ is spacelike and in the future; it's a moment of time, not a place in space.

For a rotating black hole, while there is a singularity at $r = 0$ that is timelike, so it can be thought of as "somewhere in space", it's a ring, not a point, and you can pass through it.

Matter falling towards a black hole outside the event horizon forms into a disk due to angular momentum more than being a geodesic.

Angular momentum is a property of geodesic motion; they aren't two different things.

It's true that matter in accretion discs around a rotating black hole is largely not in geodesic motion, but that's not because of "angular momentum"; it's because the pieces of matter in the different parts of the disc are interacting with each other, so the disc effectively has pressure and the different parts of it exert forces on each other, so they aren't moving on geodesics.

After crossing the event horizon the geodesic effect is all that counts and the geodesic forms a straight line path direct to the center?

None of the relevant properties of the matter change when it crosses the horizon. So your statement here is not correct.

 

[PeterDonis]

The singularity at $r = 0$ is more like a moment in time than a place in space as the $r$ coordinate is time-like inside the horizon.

This is strictly true only for a non-rotating black hole in Schwarzschild coordinates.

For a rotating black hole in the corresponding coordinates (Boyer-Lindquist), $r$ is timelike between the outer and inner horizons, but becomes spacelike again inside the inner horizon. Correspondingly, the $r = 0$ singularity is timelike (a ring), not spacelike.

 

[PeterDonis]

in effect everything inside the event horizon "is" the singularity.

No, this is not correct.

What is correct is that the region inside the horizon does not have all of the properties your intuition normally assigns to "space" and "time", and you cannot think of "trajectories" the way you normally would.

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.

 

[PeterDonis]

This is a "what do the laws of physics say would happen if we break the laws of physics" question.

I think this is too pessimistic.

The OP's question about "what if we could see inside the horizon" is a "what if we break the laws of physics" question, yes, because the laws of physics say we can't see inside the horizon.

But questions like "what happens to the object after it falls through the horizon" are not questions that break the laws of physics; the laws of physics can answer them perfectly well. The answer just won't have all of the intuitive properties that the OP is thinking of; it 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).

 

[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" 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.

 

[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.