Crossing the Bekenstein Bound at Black Hole Event Horizon

[.Scott]

TL;DR Summary

I'm very unclear on how the Bekenstein Bound is enforced during EH crossings.

The Bekenstein Bound places a upper limit on the amount of entropy that a given volume of space may contain.
This limit was described by Jacob Bekenstein who tied it quite closely to the Black Hole Event Horizon.

Put simply, black holes hold the maximum entropy allowed for their volume. If you drop more stuff into the BH, it gets a little bigger - just enough to accommodate that additional information burden.

So I pick a black hole that's big enough to allow me to comfortably cross its event horizon while momentarily avoiding that sinking feeling of tidal spaghettification. And I ponder the mass of the black hole that remains below me.

I do not doubt that I will fall "towards the center", but as I fall "towards the center" how can the diameter of the space around me and the diameter of the object below me shrink unless I fall pass other falling mass in the process?

What is different about the interior of a black hole that allows it to ignore the Bekenstein Bound?

 

[PeterDonis]

I ponder the mass of the black hole that remains below me.

There is no such thing. Once you are inside the hole, there is no sense in which some of its mass is "above" you and some is "below" you.

how can the diameter of the space around me and the diameter of the object below me shrink

There are no such things; your ordinary concepts of "the space around me" and "the object" do not work inside a black hole. Once you are inside the hole, it has no well-defined "size" and, as above, there is no sense in which some of it is "above" you while the rest of it is "below" you.

 

[PeterDonis]

The Bekenstein Bound places a upper limit on the amount of entropy that a given volume of space may contain.

While this is a common heuristic description, it's actually not correct. The bound is actually an upper bound on the entropy that can be associated with a given amount of energy enclosed by a 2-sphere with a given area. (In the usual formula, a "radius" $R$ appears, but this is an "areal radius", i.e., it is defined by the area of a 2-sphere enclosing the object, not its volume.)

For ordinary objects the difference is not a big deal, since there is a well-defined connection between area and volume (in GR you have to allow for the non-Euclideanness of space when deriving the connection, but that's easy to do). But there is no such connection in the interior of a black hole; as soon as you are dealing with 2-spheres inside a black hole with areas smaller than the horizon area, those 2-spheres do not enclose any well-defined volume. That is the geometric reason behind the statements I made in post #2 above.

 

[.Scott]

Once you are inside the hole, there is no sense in which some of its mass is "above" you and some is "below" you.

With the help of friends, I will recast my description below and get rid of "above" and "below".

There are no such things; your ordinary concepts of "the space around me" and "the object" do not work inside a black hole. Once you are inside the hole, it has no well-defined "size"...

When I recast my description, I will still be using "distances". But I think that is fair. After all, I am supposedly closing in on the singularity (getting "closer").

The bound is actually an upper bound on the entropy that can be associated with a given amount of energy enclosed by a 2-sphere with a given area. (In the usual formula, a "radius" $R$ appears, but this is an "areal radius", i.e., it is defined by the area of a 2-sphere enclosing the object, not its volume.)

I am actually depending on this.

So here is my recast:
I get a thousand friends (or perhaps enemies) and we spread ourselves out evenly around the event horizon and allow ourselves to drop through the event horizon at the same time (from the appropriate reference frame) and at the same speed. As we all cross through the even horizon, we keep track of the distances between pairs of adjacent friends.
At the time we cross through the event horizon, that distance will work out to approximately the minimum 2-sphere area for the singularity energy, thus this distance can never get smaller until we reach the singularity. But it's unclear to me whether there would really be a singularity. If my friends are keeping their distance, why the crunch?

 

[PeterDonis]

I am supposedly closing in on the singularity (getting "closer").

Not in the sense of spatial distance. The singularity is a moment of time, not a place in space.

I get a thousand friends (or perhaps enemies) and we spread ourselves out evenly around the event horizon and allow ourselves to drop through the event horizon at the same time (from the appropriate reference frame)

What reference frame? How are you going to ensure this?

and at the same speed.

Speed relative to what?

As we all cross through the even horizon, we keep track of the distances between pairs of adjacent friends.

How are you going to keep track of these distances?

At the time we cross through the event horizon, that distance will work out to approximately the minimum 2-sphere area for the singularity energy

The mass of the black hole is not "the singularity energy". It is not "located" at the singularity. (As above, the singularity is not even a place in space.)

thus this distance can never get smaller until we reach the singularity.

You have given no valid basis for this claim.

 

[.Scott]

Not in the sense of spatial distance. The singularity is a moment of time, not a place in space.

Okay. I'm guessing it's only a moment in time after you cross the event horizon. Before that, one could say it's at the center of the event horizon, but I'm not sure how meaningful that would be.

How are you going to keep track of these distances?

I am still able to exchange information with some of these friends. For objects that are adjacent to my fall, can't I just use RADAR?

The mass of the black hole is not "the singularity energy". It is not "located" at the singularity. (As above, the singularity is not even a place in space.)

That distance thing was one area where I thought there might be a problem with my description. It seemed to me that as I cross the event horizon, my trajectory somehow transitioned away from gravity and more to time. But I think what I was missing is that there's no real transition from a gravitational pull through space to a time singularity destiny, it's just that there is no longer any impressive consequence to location and a hugely impressive consequence to elapsed time. But it isn't on Physics to track "Impressiveness".

The "singularity energy" issue is much more of a mystery to me. Whether I cross time or distance to get there, am I not somehow colliding with whatever else fell before me?

As I approach the event horizon, the BH acts as if it is a huge mass that lies before me. As I cross the EH, it should still look that way. Does it ever not look that way? I think the answer is simply that I loose reference points to judge my approach to the presumed singularity - but not really, because we presume that tidal effects are getting more severe. So I could judge my radial acceleration that way. But do tidal effects get more severe within the EH?

You have given no valid basis for this claim.

My basis for the claim only exists if we (me and my friends) are enclosing something with mass. If we are, then our mutual distances from each other can be used to compute the area of the boundary (the boundary that started out as roughly a 2-sphere). That area cannot decrease without decreasing the topologically bounded mass. I guess the crux of the issue is whether there is a way for that mass to escape our bound. If not, then either the area (and thus the distances) cannot decrease or we need an exception to the Bekenstein Bound.

 

[PeterDonis]

I'm guessing it's only a moment in time after you cross the event horizon.

It's a moment of time, period. That is a matter of spacetime geometry. You can only reach this moment of time by falling inside the horizon.

For objects that are adjacent to my fall, can't I just use RADAR?

For a certain period of time after you cross the horizon, yes. But no matter how close to you an adjacent object is, there will be some time before you reach the singularity at which you will no longer be able to exchange light signals with them.

It seemed to me that as I cross the event horizon, my trajectory somehow transitioned away from gravity and more to time.

I don't even know what this means. But there is no "transition" in your trajectory when you cross the horizon. In fact you can't even tell when you cross the horizon from local measurements.

there is no longer any impressive consequence to location and a hugely impressive consequence to elapsed time

I don't know what this means either. I think you are focusing too much on vague ordinary language and not enough on the actual physics.

Whether I cross time or distance to get there, am I not somehow colliding with whatever else fell before me?

No. The matter that fell in before you is not reachable by you.

As I approach the event horizon, the BH acts as if it is a huge mass that lies before me.

Only if "a huge mass that lies before me" is compatible with the BH being vacuum.

As I cross the EH, it should still look that way.

What does "look that way" mean? What physical observations tell you that there is "a huge mass" before you?

do tidal effects get more severe within the EH?

Yes.

My basis for the claim only exists if we (me and my friends) are enclosing something with mass.

What does "enclosing something with mass" mean?

Again, you are focusing too much on vague ordinary language and not enough on actual physics. The actual physics is that, inside the horizon, it is simply impossible--geometrically impossible--to construct 2-spheres that "enclose mass" the way they would have to for the Bekenstein Bound reasoning to even apply. No amount of hand-waving and playing around with vague ordinary language will change that. Nor will it help you to understand why that is true. You will need to look at the actual gory details of the physics.

 

[PeterDonis]

I guess the crux of the issue is whether there is a way for that mass to escape our bound.

No, that's not the crux of the issue. The crux of the issue is that spacetime geometry inside the horizon does not work the way you think it works. The way it actually works, as I said in post #7 just now, makes it geometrically impossible for 2-spheres inside the horizon to "enclose mass" the way they would have to for the Bekenstein Bound reasoning to even apply.

 

[PeterDonis]

@.Scott here is some more food for thought: your basic intuition seems to be that we ought to be able to take an object that has some finite energy, and apply the Bekenstein Bound not just to the object as a whole, but also "divided up" between parts of the object. In other words, we ought to be able to treat the entropy contained in the object as being "spread out" among the parts of the object the way the object's energy is "spread out" among its constituents.

The problem with applying this to a black hole is that we currently do not have any model of a black hole in terms of more fundamental constituents. Our current model of a black hole, the classical GR model, only tells us that the hole itself is a self-contained entity with a mass $M$ and a horizon area $A$. It does not give any way of dividing those quantities up into smaller pieces that get added together to form the hole. That means there is no way, with our current models, to divide up the Bekenstein Bound for a black hole the way you are thinking.

We could someday come up with a quantum gravity theory that gives us a model of a black hole in terms of more fundamental constituents, that would then tell us how to account for the hole's entropy in terms of the (logarithm of the) number of ways those constituents could be assembled to form a particular hole of mass $M$ and horizon area $A$. We could then use such a model to assess the Bekenstein Bound for some subset of the hole as well as for the entire hole. But nobody currently has such a model.