Can large Black Holes ever spaghettify? (and more)

[.Scott]

TL;DR Summary

I am unclear on how tidal forces can substantially increase after crossing the event horizon. The crux is whether moving forward in time ever really gets you closer to the singularity.

So Alice and Bob are hanging out near a really large black hole.
It's quiet. Nothing has entered the BH is a while.
Alice tosses Bob in and then waits long enough for him to collide with the singularity.

Of course, Bob is keeping time differently than Alice - so I rather doubt that the time period Alice calculates for Bob to reach the "center" has any real meaning to Bob. But Alice's real goal is to enter the BH herself and to never seen Bob again.

So my first question: Is her strategy for avoiding Bob a good one? If she waits for Bob to get within a meter of the EH and then waits a second more than the radius of the EH divided by c, will Bob really be completely unreachable?

Alice is patient. If she needs to wait a few weeks longer than this to avoid Bob, she will.
During her wait, nothing else enters the BH.

Then she takes her dive. What is before her is a mass M that can only become less than M if she passes some of the mass on the way in. And if she does pass some of that mass, the first of it should be Bob.

If she doesn't pass any of that mass, how can the diameter of that mass ever fall below the original diameter of the event horizon? If it does, would that not violate the Bekenstein bound - an attempt to put too much information into too small a space?

So, she keeps on moving through time in the direction of this singularity - but because the "width" of her space doesn't shrink, tidal forces cannot increase. She will stay in the same geometry until she passes something.

I really can't tell if she ever reaches Bob. If she does, it would seem that from Bob's perspective, upon crossing the EH, Alice was already there - having picked up on the in-going time-line at a longer radius (therefore "earlier") than him. Isn't that what it looks like when someone catches up to you in time?

 

[Ibix]

But Alice's real goal is to enter the BH herself and to never seen Bob again.

So my first question: Is her strategy for avoiding Bob a good one?

In principle, Alice can always see Bob, unless she does something like go round the other side of the hole. In practice the light will eventually be redshifted into oblivion, so she won't actually see him. She will, however, strike the singularity before she sees Bob hit the singularity.

I really can't tell if she ever reaches Bob.

Alice and Bob can always manoeuvre to meet after they cross the horizon, in the sense that the future light cones of their horizon crossing events necessarily overlap. I think it requires teamwork, though, so Alice could evade Bob if she wished. If they do meet up it's a kind of suicidal twin paradox setup - they need not be the same age. So it's not that Alice was "there first" so much as they took different routes to get to the meeting event.

I recommend sketching a Kruskal diagram - it's all fairly clear on that.

 

[PeterDonis]

Is her strategy for avoiding Bob a good one?

As long as she knows how Bob will fall--for example, that he will free-fall all the way--then yes, she can calculate a time on her clock after which she can fall in herself and be guaranteed not to meet Bob. This time will be the time by her clock at which, if she emits a light signal inward, that signal will hit the singularity at the same event that Bob does. In other words, this signal is on the past light cone of the event at which Bob hits the singularity. At any time after that time by her clock, if she falls in there is no chance of her encountering Bob because she is outside the past light cone.

What is before her is a mass M that can only become less than M if she passes some of the mass on the way in.

No. The mass of the hole has nothing to do with the presence or absence of matter inside the horizon. It is a global geometric property of the spacetime. Inside the horizon, that property doesn't really have any meaning anyway, because the interior region is not stationary and the physical meaning of $M$ as "mass" only makes sense in a stationary region of spacetime.

In any case, if Alice has waited long enough to make sure she won't meet Bob, as above, and nothing else fell in after Bob, it is impossible for her to meet anything during her fall inside the horizon. She will see nothing but vacuum all the way to the singularity.

If she doesn't pass any of that mass, how can the diameter of that mass ever fall below the original diameter of the event horizon? If it does, would that not violate the Bekenstein bound - an attempt to put too much information into too small a space?

No. The Bekenstein bound, like the physical interpretation of $M$ as a mass, only makes sense in the exterior region, where the spacetime is stationary. In other words, it only makes sense as a limit on how much information can be contained inside a surface if spacetime everywhere exterior to that surface is stationary.

she keeps on moving through time in the direction of this singularity - but because the "width" of her space doesn't shrink, tidal forces cannot increase

I don't know what you mean by the "width" of her space, but there is no valid concept I'm aware of that has the properties you describe. The tidal gravity in Alice's vicinity will increase continuously during her fall.

 

[Ibix]

This time will be the time by her clock at which, if she emits a light signal inward, that signal will hit the singularity at the same event that Bob does.

I think you are assuming Bob is not free to manoeuvre. If he goes to Absurd Acceleration in the +r direction the instant he crosses the horizon then he can stay (on a Kruskal diagram) right at the upper edge of his future light cone and still be just inside the horizon whenever Alice crosses.

 

[PeterDonis]

In principle, Alice can always see Bob

She can see Bob in the past, yes. But she cannot see Bob all the way down to the singularity. There will be a point on Bob's worldline where the future light cone of that point no longer contains the event on the singularity where Alice ends up. Alice can never see anything that happens to Bob after that point. (I am assuming here that Bob free-falls all the way, since that is how I understand the scenario to have been specified.)

In practice the light will eventually be redshifted into oblivion.

No, it won't. You are confusing what is observed by a static observer who remains outside the horizon, with what is observed by a free-falling observer who falls inside the horizon. They're not the same.

I believe that the light Alice sees coming from Bob as she herself falls in will always be blueshifted, but I haven't done the detailed math. However, it will certainly not be "eventually redshifted into oblivion".

She will, however, strike the singularity before she sees Bob hit the singularity.

In fact, she will strike the singularity before she ever sees anything happening to Bob after the point I described above, which is well before Bob hits the singularity.

Alice and Bob can always manoeuvre to meet after they cross the horizon

Not if Alice waits long enough (and Bob does not have unbounded amounts of proper acceleration available--see below). See my response to @.Scott.

the future light cones of their horizon crossing events necessarily overlap

This is true in the vacuous sense that any point on the horizon contains all later points on the horizon in its future light cone (since the horizon itself is the outward boundary of that future light cone), but by itself it does not entail the consequence you describe. You would need to add the proviso that Alice cannot wait too long before falling in herself (or that Bob has unbounded amounts of proper acceleration available).

 

[Ibix]

She can see Bob in the past, yes. But she cannot see Bob all the way down to the singularity. There will be a point on Bob's worldline where the future light cone of that point no longer contains the event on the singularity where Alice ends up. Alice can never see anything that happens to Bob after that point.

Agreed.

No, it won't.

I disagree. Imagine that I freefall from rest at a hovering space station and you wait a short time them drop the same way. When you cross the horizon you should receive light from me that I emitted as I crossed, and as you note there's a final pulse I emit that you receive as you strike the singularity, which I emitted before I hit the singularity. Since we both followed identical worldlines our watches should show the same elapsed time from horizon to singularity, so in your horizon-to-singularity journey you see less time elapse on my clock. Light I emit must be redshifted.

or that Bob has unbounded amounts of proper acceleration available

Fair point. I should have said "in principle". Given a finite maximum acceleration available to Bob, Alice can wait long enough.

 

[PeterDonis]

Since we both followed identical worldlines our watches should show the same elapsed time from horizon to singularity, so in your horizon-to-singularity journey you see less time elapse on my clock. Light I emit must be redshifted.

Hm, yes, you're right. The redshift effect of Alice's fall towards the singularity is greater than the blueshift effect of Bob's fall on Bob's light emitted to Alice.

 

[.Scott]

No. The mass of the hole has nothing to do with the presence or absence of matter inside the horizon. It is a global geometric property of the spacetime. Inside the horizon, that property doesn't really have any meaning anyway, because the interior region is not stationary and the physical meaning of $M$ as "mass" only makes sense in a stationary region of spacetime.

Given what you are saying, Alice will always be falling to a mass $M$ until she hit the singularity. And all of the information will always be in the direction of the singularity.

Is it fair to say that once Alice crosses the EH, her world is the 3D surface of a 4H sphere with a decreasing radius? In time, that radius reaches 0 and the singularity is reached. But I really wonder if that radius can get anywhere near zero. And if it cannot, there will be no singularity.

 

[PAllen]

She can see Bob in the past, yes. But she cannot see Bob all the way down to the singularity. There will be a point on Bob's worldline where the future light cone of that point no longer contains the event on the singularity where Alice ends up. Alice can never see anything that happens to Bob after that point. (I am assuming here that Bob free-falls all the way, since that is how I understand the scenario to have been

But she can see Bob until she hits the singularity. This is how I interpreted @Ibix statement.

...

Not if Alice waits long enough (and Bob does not have unbounded amounts of proper acceleration available--see below). See my response to @.Scott.

But I understood this to be the key point @Ibix implied. That, if Alice waits around hovering (even for a million years of her time), then free falls into the BH, as long as Bob knows Alice's intent, Bob can pick a timelike trajectory inside the horizon that can still meet Alice before they both hit the singularity. Of course this requires proper acceleration, but only one pulse is strictly needed. A single course correction after crossing, chosen with knowledge of Alice intent, is sufficient.

 

[PeterDonis]

Given what you are saying, Alice will always be falling to a mass $M$ until she hit the singularity

No, that's not what I'm saying. What I'm saying is that the concept of "the mass of the black hole" makes no sense inside the horizon. There is no meaningful concept of "how much mass Alice is falling towards" inside the horizon.

Is it fair to say that once Alice crosses the EH, her world is the 3D surface of a 4H sphere with a decreasing radius?

No. The 4-volume inside the horizon is infinite. There are spacelike 3-surfaces inside it that are infinite in extent. There are also spacelike 3-surfaces inside it that are finite in extent, These statements are also true if we replace "the horizon" with any surface of constant Schwarzschild radial coordinate $r$ for $r > 0$. Pretty much all of your intuitions about how "space" works are not valid inside a black hole horizon.

 

[PeterDonis]

she can see Bob until she hits the singularity.

Yes, agreed.

if Alice waits around hovering (even for a million years of her time), then free falls into the BH, as long as Bob knows Alice's intent, Bob can pick a timelike trajectory inside the horizon that can still meet Alice before they both hit the singularity. Of course this requires proper acceleration, but only one pulse is strictly needed.

Only one pulse, but of unbounded magnitude--that is, for any time that Bob crosses the horizon, we can find an Alice trajectory that requires his single proper acceleration pulse to be of any desired magnitude, no matter how large.

 

[PAllen]

Only one pulse, but of unbounded magnitude--that is, for any time that Bob crosses the horizon, we can find an Alice trajectory that requires his single proper acceleration pulse to be of any desired magnitude, no matter how large.

Agreed, but I assumed this is what was meant by cooperating, and that it wasn't meant to be realistic. Note this is the same as the problem of maximizing your proper time to the singularity. To do this, you want an arbitrarily large thrust arbitrarily soon after horizon crossing to put you on a t=constant geodesic as close as possible to t=infinity.

 

[PeterDonis]

maximizing your proper time to the singularity. To do this, you want an arbitrarily large thrust arbitrarily soon after horizon crossing to put you on a t=constant geodesic as close as possible to t=infinity.

I thought it was to put you on the same trajectory as a Painleve observer (i.e., one who is "free falling from rest at infinity") passing through the event at which you apply the impulse.

 

[PAllen]

I thought it was to put you on the same trajectory as a Painleve observer (i.e., one who is "free falling from rest at infinity") passing through the event at which you apply the impulse.

No. The t=constant geodesics all have the same proper time from the beginning of the future horizon to the singularity. From any point inside the horizon, any path to the singularity different from one of these geodesics passing through through that point necessarily has less proper time. This can almost just be read from the metric. Thus, the issue isn't that there is something better about a larger t value geodesic, but simply that you want to be on one of these as soon as possible after horizon crossing. The sooner you do this, and the greater your acceleration available, the higher t value geodesic you happen to end up on.

 

[PeterDonis]

The t=constant geodesics all have the same proper time from the beginning of the future horizon to the singularity. From any point inside the horizon, any path to the singularity different from one of these geodesics passing through through that point necessarily has less proper time.

Ah, yes, you're right. To put it another way, from any event on an $r = \text{const}$ hyperbola inside the horizon, the geodesic with maximal proper time to $r = 0$ is the $t = \text{const}$ one passing through that event. If you've just fallen through the horizon, you're at an $r$ only a little less than $2M$, and you're on an event on that $r = \text{const}$ hyperbola close to the $t = \infty$ curve, so the $t = \text{const}$ geodesic passing through the event you're at will have $t$ only a little less than $\infty$.

 

[.Scott]

For those that are looking to get away from it all, Black Holes are less effective than I thought.

 

[.Scott]

She can see Bob in the past, yes. But she cannot see Bob all the way down to the singularity. There will be a point on Bob's worldline where the future light cone of that point no longer contains the event on the singularity where Alice ends up. Alice can never see anything that happens to Bob after that point. (I am assuming here that Bob free-falls all the way, since that is how I understand the scenario to have been specified.)

I'm not sure whether I have a good picture of this or not ...

Let's say that both Alice and Bob are within the event horizon and they are both dropping towards r=0. Let's also assume that the share the same non-R coordinates and neither has any non-R component to their velocity.

So, before reading your post (quoted above), this was the story as I understood it:

Story A: Bob is leaving photons behind - and as Alice traverses Bob's path, she will continuously run into these photons and therefore "see" Bob - though perhaps very red-shifted. Finally, photons from Bob will turn back towards Bob at a rate faster that Alice is approaching r=0. So before Alice reaches the singularity, she will see Bob red-shifted to the extreme as he reaches an Alice-relative Event Horizon. All of Bob and all of Alice hit the singularity but spaghettified to the point where only one "bit" enters at a time. When the real Alice bit reaches the singularity, it (she) will reach it's own event horizon just as Bob's final ($\lambda = 0$) photon does.

After reading your post:

I don't think that you are contradicting "Story A", but I'm not sure. In Story A, the singularity has a time dimension only. But separate world lines that cannot intersect could suggest that this singularity has some de facto acreage.

 

[PeterDonis]

Bob is leaving photons behind

No, he isn't. Even if Bob emits photons radially outward, they will still be falling towards the singularity if Bob is inside the horizon. He can't just "leave them behind".

as Alice traverses Bob's path

No, Alice does not "traverse Bob's path". Alice's worldline in spacetime is different from Bob's. And since space doesn't work the way your intuition says it does inside the horizon, there is no spatial "path" Bob follows that Alice can also follow.

photons from Bob will turn back towards Bob

No, they won't. Photons that Bob emits radially outward will always increase their proper distance from Bob.

In Story A, the singularity has a time dimension only.

This is wrong. The singularity is spacelike, not timelike.

I think you are relying way too much on intuitive reasoning instead of actually looking at the math. @Ibix suggested earlier that you look at a Kruskal diagram: I strongly support that suggestion.

 

[.Scott]

This is wrong. The singularity is spacelike, not timelike.

The Kruskal diagram certainly makes the black hole look spacelike. Does it really have acreage?
If it has acreage, wouldn't the total gravitational field be spread across it instead of concentrated at a point?
So the inverse square law would not apply - no divide by zero for the intensity of the gravitational field.

I seem to be grasping at straws in trying to get a handle on the nature of this singularity. You crash into it in the time direction - as in, 1pm is the singularity. You can maneuver to change the time you strike the singularity (as measured by your own watch). But if the singularity is "space like" does it makes sense to maneuver to strike a different location on the singularity?

 

[PeterDonis]

The Kruskal diagram certainly makes the black hole look spacelike.

Yes, that's because it is spacelike, so any proper diagram will show it as such.

Does it really have acreage?

I don't know what you mean by "acreage". The singularity is a spacelike line of infinite length. It has no surface area or volume because $r = 0$ there.

the total gravitational field

Is a meaningless concept inside the horizon.

The spacetime curvature at the singularity is infinite. Or, to be less sloppy, the singularity itself, $r = 0$, is not part of the spacetime at all. It is a limit that can only be approached, never reached. But the spacetime curvature increases without bound as it is approached. (Note, though, that the proper time along a worldline approaching the singularity does not increase without bound: it approaches a finite limit.)

I seem to be grasping at straws in trying to get a handle on the nature of this singularity.

That's because you continue to try to use intuitive concepts which are not valid inside a black hole's horizon. You need to stop relying on intuition and look at the math.

You can maneuver to change the time you strike the singularity (as measured by your own watch). But if the singularity is "space like" does it makes sense to maneuver to strike a different location on the singularity?

The two things you describe here are not two different things: they're two different ways of describing the same thing. "Maneuvering" just means changing the worldline you follow. Different worldlines hit the singularity at different places, and take different amounts of proper time to do it.

 

[Ibix]

The singularity is a spacelike line of infinite length.

Is it actually a line? It is on the Kruskal diagram, but doesn't that make it an infinite 3-volume? Or, to avoid making claims about the singularity which isn't part of the manifold, that any surface of constant $r$ inside the event horizon is an infinite 3d spacelike surface, for any $r$ as $r\rightarrow 0$.

 

[PAllen]

Is it actually a line? It is on the Kruskal diagram, but doesn't that make it an infinite 3-volume? Or, to avoid making claims about the singularity which isn't part of the manifold, that any surface of constant $r$ inside the event horizon is an infinite 3d spacelike surface, for any $r$ as $r\rightarrow 0$.

It’s the limit of S2 X R1 ( a hypercylinder) as the S2 area goes to zero. Seems reasonable to me to think of that limit as effectively R1.

 

[Ibix]

It’s the limit of S2 X R1 ( a hypercylinder) as the S2 area goes to zero. Seems reasonable to me to think of that limit as effectively R1.

Of course - even though $r$ is no longer a spacelike coordinate it's still related to the area of the sphere. So there's only one non-degenerate coordinate, $t$ at the singularity. Thanks.

 

[PeterDonis]

Is it actually a line? It is on the Kruskal diagram, but doesn't that make it an infinite 3-volume?

No, because $r = 0$ there.

 

[PeterDonis]

even though $r$ is no longer a spacelike coordinate

I should have clarified in my previous posts that by $r$ I mean the areal radius of 2-spheres, which is an invariant, not any of the coordinates named $r$ in different coordinate charts (not all of which are timelike inside the horizon--AFAIK Schwarzschild coordinates are the only common one for which that's true).

 

[PeterDonis]

So there's only one non-degenerate coordinate, $t$ at the singularity.

To be clear, this is the Schwarzschild $t$ coordinate (since you talked about $r$ not being spacelike, so you were using Schwarzschild coordinates). In other charts there are different "time" coordinates which are also non-degenerate at the singularity (I put "time" in quotes because not all of them are timelike inside the horizon).