Spagettification and a singularity

[PeterDonis]

The point I was making was that inside the hole there isn't a "radial direction" for the spaghettification to happen in, because the "radial direction" is timelike.

This is not correct. You stated earlier that the $r$ coordinate is timelike--that is true in Schwarzschild coordinates, but not others. But regardless of any choice of coordinates, it is always true that there are spacelike directions that point radially inward and outward. So there always is a radial direction for spaghettification to happen in. It's just that in Schwarzschild coordinates, that direction doesn't happen to point along the radial "grid lines" of the $r$ coordinate.

 

[PeterDonis]

I think Eddington-Finkelstein coordinates share a $t$ direction with Schwarzschild, but their radial coordinate lines are not orthogonal to that direction.

Yes, that is correct. The integral curves of $t$ are the same for Schwarzschild, Eddington-Finkelstein, and Painleve. (In the Kruskal diagram, they are the hyperbolas that are labeled with different constant values of $r$.) But only the Schwarzschild surfaces of constant $t$ are orthogonal to those curves.

 

[BoraxZ]

No. The singularity is reached in a finite proper time for both particles.

Yes. But what about the spatial component? If two particles end up at infinity they can do that spatially separated. Two points at infinity van be spatially separated.

It would involve infinite speed though.

 

[BoraxZ]

There is no "co-falling frame" for both particles together. They do not remain the same radial distance apart; the radial distance between them increases.

Of course. That's spaghettification, ain't it? In the frame falling along with the second particle, you will see the first particle end up a finite distance from you after your proper time has ended. But the same holds for all particles behind you. Isn't there a spatial infinity involved then?

 

[PeterDonis]

That's spaghettification, ain't it?

One aspect of it, yes.

In the frame falling along with the second particle

Ah, ok, that's what you meant by "co-falling frame".

you will see the first particle end up a finite distance from you after your proper time has ended.

In the sense that there is a finite "distance" along the $r = 0$ line from the event where you hit it, to the event where the first particle hit it, yes. But this "distance" is not radial. It can't be, since it is entirely along the $r = 0$ line, i.e., along a line with the same value of $r$ everywhere.

Isn't there a spatial infinity involved then?

The $r = 0$ line is infinitely long, yes. But it never reaches "spatial infinity". It stays inside the horizon the whole time. Welcome to the wonderful world of curved spacetimes.

 

[BoraxZ]

The r=0 line is infinitely long, yes. But it never reaches "spatial infinity". It stays inside the horizon the whole time. Welcome to the wonderful world of curved spacetimes.

I was just about to ask you the question, "what happens to your time if all your proper time has elapsed". Does this have anything to do with your last statement? I mean, you fall in, as a particle, you look at your watch, and then, when the proper falling time to reach the end is over, what next?
IS there a next?

 

[PeterDonis]

"what happens to your time if all your proper time has elapsed"

If you mean, what happens to you when you hit the singularity, in the model we are discussing, you are destroyed. Your worldline ends there.

Most physicists believe that this is not physically reasonable, and that some new physics will come into play at some point. But we have no good theory for what that new physics might be.

IS there a next?

Not in the model we are discussing. See above.

 

[BoraxZ]

Maybe particles have some kind of geometric structure preventing the singularity to form, like strings or branes, or whatever. Particles accumulating because the angular distances become too small. Point-like particles cause trouble.
But that's a different story...

 

[PeterDonis]

Maybe particles have some kind of geometric structure preventing the singularity to form, like strings or branes, or whatever. Particles accumulating because the angular distances become too small. Point-like particles cause trouble.
But that's a different story...

Please do not speculate. Personal speculations are off limits at PF.

 

[Ibix]

This is not correct. You stated earlier that the $r$ coordinate is timelike--that is true in Schwarzschild coordinates, but not others. But regardless of any choice of coordinates, it is always true that there are spacelike directions that point radially inward and outward. So there always is a radial direction for spaghettification to happen in.

I understand your point, but I don't agree.

Outside the hole, independent of your state of motion you can point at the hole. You'd call that a radial direction. You could also call it an axial direction because (unless you are hovering) it has a component in the Schwarzschild $t$ direction, the integral curves of which join spherically symmetric surfaces of constant areal radius. But you wouldn't usually call it an axial direction because it's the spacelike component you are interested in, and also you have a special case to deal with for hovering observers whose radial direction has no axial component.

Much the same argument applies the other way around inside the hole. Certainly there are spacelike directions that point to spherially symmetric surfaces of different radii, but I wouldn't call them "radial directions" for the same reason I wouldn't call radial directions outside the hole "axial directions".

 

[Dale]

Certainly there are spacelike directions that point to spherially symmetric surfaces of different radii, but I wouldn't call them "radial directions" for the same reason I wouldn't call radial directions outside the hole "axial directions".

Inside or outside the horizon the Schwarzschild spacetime is spherically symmetric and is foliated by 2-spheres. Each event in the spacetime belongs to exactly one such sphere, so you can unambiguously construct the “tangential” directions.

I don’t know if there is a geometric method that can be used to construct a unique “radial” direction. Simply requiring orthogonality to the “tangential” directions gives a 2D plane. That plane could be spanned by a spacelike and a timelike vector. If that could be done in a unique way then the spacelike vector could indeed be called “radial”. But I don’t know how to do it uniquely.

 

[BoraxZ]

Inside or outside the horizon the Schwarzschild spacetime is spherically symmetric and is foliated by 2-spheres. Each event in the spacetime belongs to exactly one such sphere, so you can unambiguously construct the “tangential” directions.

I don’t know if there is a geometric method that can be used to construct a unique “radial” direction. Simply requiring orthogonality to the “tangential” directions gives a 2D plane. That plane could be spanned by a spacelike and a timelike vector. If that could be done in a unique way then the spacelike vector could indeed be called “radial”. But I don’t know how to do it uniquely.

I have a question. Suppose, again, that there are two particles. The first one in a geodesic from infinite far away. The second one too, but that one is stopped in its way for, say, a second (in a hovering platform). So the second one falls in freely from the platform one second after the first one.

Now let's look at the second particle in a frame that falls along with it (so it stays at the origin of that frame), after being released from the hovering platform.

Now, two marbles released one second after one another from a fixed height above the Earth will end up at the same location on the surface below them. Such a surface is absent in a black hole. Nevertheless, they end up in the same amount of proper time on the singularity. But they don't end up, like the marbles, in a solid surface.

How far, as seen in the frame falling along with the second particle, will the first particle be removed from it, after their proper times have ended? The two marbles just end up at the surface, "on top" of each other. But the two particles seem to end up with a finite distance between them. You see the second particle move away from you during all the proper time that is left for you (I'm not sure if you can actually see the first particle ahead of you because behind the horizon, all signals travel towards the singularity).

 

[Dale]

How far, as seen in the frame falling along with the second particle, will the first particle be removed from it, after their proper times have ended? The two marbles just end up at the surface, "on top" of each other. But the two particles seem to end up with a finite distance between them. You see the second particle move away from you during all the proper time that is left for you (I'm not sure if you can actually see the first particle ahead of you because behind the horizon, all signals travel towards the singularity)

The calculation of this quantity would be fairly involved. First, you would need to calculate the equation for the two radial geodesics of interest. Then you would need to find the limit of the $t$ coordinate for each as the $r$ coordinate approaches $0$. Then you would construct the path of constant $r$ from one $t$ to the other and integrate the metric along that path. Finally, you would take the limit of that integral as $r\to 0$.

 

[BoraxZ]

The calculation of this quantity would be fairly involved.

Could you somehow use the fact that the second particle stops for, say, one second at the hovering platform? The only difference between the geodesic path of the first particle (falling freely from infinity to the singularity) and the path of the second, is the temporary acceleration (for one second) at the platform. How many meters does this one-second acceleration (to keep it on the platform, after which it continues its free fall) add to the length of the path (through spacetime) of the second particle, in comparison with the length of the first particle's geodesic path? Is it just ct meters, with t=1? Will the total proper time elapsed for the second particle be just one second longer than the total proper time "experienced" by the first one (so the particles end up with different ages, which in this case seems the inverse of the twin paradox where the accelerated observer ends up younger).

 

[Dale]

Could you somehow use the fact that the second particle stops for, say, one second at the hovering platform?

Yes. You could (you would need to) use that fact. It wouldn’t make anything simpler from what I described. The calculation can be done, it just is not trivial.

 

[Ibix]

Inside or outside the horizon the Schwarzschild spacetime is spherically symmetric and is foliated by 2-spheres. Each event in the spacetime belongs to exactly one such sphere, so you can unambiguously construct the “tangential” directions.

I don’t know if there is a geometric method that can be used to construct a unique “radial” direction. Simply requiring orthogonality to the “tangential” directions gives a 2D plane. That plane could be spanned by a spacelike and a timelike vector. If that could be done in a unique way then the spacelike vector could indeed be called “radial”. But I don’t know how to do it uniquely.

There's a Killing vector field, which is the Schwarzschild $t$ basis vector field outside the horizon and the "axial" direction inside. It lies in the 2d plane to which you refer, and the orthogonal direction is radial. (Doesn't work on the horizon where the Killing field is null.)

 

[Dale]

There's a Killing vector field, which is the Schwarzschild $t$ basis vector field outside the horizon and the "axial" direction inside. It lies in the 2d plane to which you refer, and the orthogonal direction is radial. (Doesn't work on the horizon where the Killing field is null.)

You are right, but outside the horizon $\partial_r$, the direction orthogonal to the Killing field, is the spatial direction away from the center, whereas inside the horizon $\partial_t$, the direction of the Killing field itself, is the spatial direction away from the center.

I mean, I guess we can do a piecewise construction, but I couldn’t come up with one geometric rule that works everywhere.

 

[BoraxZ]

Suppose that when you enter the hole, just before passing the horizon, you attach a measuring tape to a station hovering above the horizon. What will the tape, rolling off from some magical device, show when you hit the singularity?

Will this even be possible (since no information can leave in the direction of the event horizon behind you)?

When you fall freely behind another particle, will you actually see that particle? Or will the light coming from it never reach you, because that radiation can only flow towards the singularity, its light one being inwardly directed directed? Or are the lightcones, in a freely falling frame, normal, since it's locally flat?

 

[Dale]

You may find this page helpful

https://www.gregegan.net/SCIENCE/Rindler/RindlerHorizon.html

 

[timmdeeg]

When you fall freely behind another particle, will you actually see that particle?

Post #14 show the future light cones of an infalling particle. If you add the complementing past light cones you will recognize that a second particle can see the first one crossing the horizon and for some time within the horizon but can't see it reaching the singularity.

 

[PeterDonis]

Outside the hole, independent of your state of motion you can point at the hole.

Yes. And you can also point exactly opposite from that direction and point away from the hole.

And inside the hole, independent of your state of motion, you can point away from the hole. So you can also point exactly opposite from that direction and point towards the singularity. (Note that this is true even though the singularity is in your future--there are always timelike, null, and spacelike directions that point towards the singularity.) As I said, there is always a spacelike radial direction you can point in.

You could also call it an axial direction because (unless you are hovering) it has a component in the Schwarzschild $t$ direction

Now you are restricting your definition to spacelike vectors that are orthogonal to your worldline. But I did not claim that there is always a spacelike radial direction orthogonal to your worldline. That depends on your worldline. My claim was weaker: that there is always some spacelike vector at whatever event on your worldline you pick that points in a purely radial direction (i.e., no $\theta$ or $\phi$ components).

That said, for realistic worldlines for actual observers free-falling radially into the hole, I think you will always be able to find a spacelike radial vector that is orthogonal to your worldline. For example, this is easy to prove for a Painleve observer, free-falling radially from rest at infinity.

Here's a heuristic argument for the claim I just made (which is stronger than the claim I made before): pick any event on the infalling observer's worldline that is inside the horizon. Take the ingoing radial null curve that intersects that point (there will always be exactly one). This can be thought of as light coming in from some distant object that is "directly overhead" for the radial infaller. Convert that null direction to a spacelike direction in the observer's orthonormal frame at that event. That spacelike direction is "radially outward". (And note that, for a radial infall, the same distant object can serve as the reference for "directly overhead" during the entire infall process.)

Much the same argument applies the other way around inside the hole. Certainly there are spacelike directions that point to spherially symmetric surfaces of different radii, but I wouldn't call them "radial directions" for the same reason I wouldn't call radial directions outside the hole "axial directions".

The "axial" direction inside the hole is the direction of integral curves of $\partial_t$ (the fourth Killing field, which is timelike outside the horizon, null on the horizon, and spacelike inside the horizon). Yes, that direction is not aptly referred to as "radial". But the timelike "radial" direction orthogonal to it is not the worldline of any observer that free-falls in from outside the hole: it can't be, because it comes from the bifurcation point (the intersection of the axes on the Kruskal diagram), which is not a point that can be traversed by any timelike observer falling into the hole from outside. The only way for a timelike observer to reach that point is to come from the white hole region on the diagram. So calling that direction "radial" is also not very apt. But it's also not a direction that is relevant if you're thinking of an observer who fell into the hole from outside.

 

[PeterDonis]

I don’t know if there is a geometric method that can be used to construct a unique “radial” direction.

See the ingoing light ray construction I described in post #56.

 

[PeterDonis]

two marbles released one second after one another from a fixed height above the Earth will end up at the same location on the surface below them. Such a surface is absent in a black hole. Nevertheless, they end up in the same amount of proper time on the singularity.

They take the same amount of proper time to fall, yes. But they don't end up at the same point on the singularity, because they didn't start from the same point on the worldline of the platform they fell from. One started at a point one second later in proper time along that worldline than the other.

 

[PeterDonis]

Will the total proper time elapsed for the second particle be just one second longer than the total proper time "experienced" by the first one

The total proper time during the fall will be the same for both. But the total proper time starting from the time when the first one is dropped will be one second longer for the second particle--because it waits for one second before being dropped itself.

 

[PeterDonis]

The calculation of this quantity would be fairly involved.

The actual numerical value would be (since the infalling worldlines after being dropped would be the worldlines of Novikov observers, who fall from rest at a finite altitude, which are more complicated to compute than those of Painleve observers, who fall from rest at infinity). But the comparisons @BoraxZ is asking about don't require the full computation. See my previous posts in response to him.

 

[PeterDonis]

There's a Killing vector field, which is the Schwarzschild $t$ basis vector field outside the horizon and the "axial" direction inside. It lies in the 2d plane to which you refer, and the orthogonal direction is radial.

The orthogonal direction is unproblematically radial outside the horizon. Inside, I don't think that description works well, for the reason I gave in the last part of post #56.

whereas inside the horizon $\partial_t$, the direction of the Killing field itself, is the spatial direction away from the center.

I don't think this is true. The integral curves of $\partial_t$ stay at a constant $r$ coordinate; they do not point in a direction of increasing $r$. A past-directed tangent vector along an integral curve of $\partial_r$ (Schwarzschild) points in a direction of increasing $r$, but of course this direction is timelike, not spacelike (and it doesn't point outside the hole, for the reason I gave in the last part of post #56--which also explains why this timelike direction is the wrong one to think of if you are considering observers free-falling into the hole from outside).

 

[PeterDonis]

Suppose that when you enter the hole, just before passing the horizon, you attach a measuring tape to a station hovering above the horizon. What will the tape, rolling off from some magical device, show when you hit the singularity?

It won't show anything. It will break. To stay attached to you once you fall below the horizon, the tape would have to unroll faster than the speed of light, which is impossible.

When you fall freely behind another particle, will you actually see that particle?

Yes.

Or will the light coming from it never reach you, because that radiation can only flow towards the singularity

No. It's true that the light can only move in the direction of decreasing $r$, even though it's radially outgoing--but, heuristically, it does that more slowly than you do, so you catch up to it as you fall and can see it.

are the lightcones, in a freely falling frame, normal, since it's locally flat?

Yes. This is a good way to avoid confusion when considering local behavior in scenarios like this.

 

[Dale]

I don't think this is true. The integral curves of $\partial_t$ stay at a constant $r$ coordinate; they do not point in a direction of increasing $r$. A past-directed tangent vector along an integral curve of $\partial_r$ (Schwarzschild) points in a direction of increasing $r$, but of course this direction is timelike, not spacelike (and it doesn't point outside the hole, for the reason I gave in the last part of post #56--which also explains why this timelike direction is the wrong one to think of if you are considering observers free-falling into the hole from outside).

I certainly could be wrong, but I don’t think I am. Inside the horizon you still have the foliation of the spacetime as 2-spheres. Because the signature is (-+++) and that 2D foliation is spacelike, that means that there is one spacelike and one timelike direction remaining. We are looking (or at least I am looking) for the spacelike direction that is orthogonal to the 2-sphere. That is $\partial_t$.

Now, you said “The integral curves of $\partial_t$ … do not point in a direction of increasing $r$”, which is true but not what I said. I said “inside the horizon $\partial_t$, the direction of the Killing field itself, is the spatial direction away from the center.”

The 2-spheres partition spacetime into two regions, one of which contains spacelike infinity and the other which does not. If you have a vector which is both spacelike and orthogonal to the vectors in the sphere then that vector must either point into the region which contains spacelike infinity or into the region which does not. If it points to the region with spacelike infinity then it is the spatial direction away from the center. I believe that inside the horizon $\partial_t$ is in the spacelike direction orthogonal to the 2-sphere and pointing away from the center.

It is true that the integral curves of $\partial_t$ do not point in a direction of increasing $r$, but that is not the same as pointing away from the center. If spacetime were flat, then that would be the case. However, because spacetime is curved, the relationship between going away from the center and changing the $r$ coordinate (the area-radius) differs from what you would expect. Inside the horizon the spatial direction away from the center (which I think is $\partial_t$) leads to another 2-sphere of the same surface area (same $r$).

I could still easily be wrong. But if I am then I don’t think it is for the reason you gave. The reason you gave confounds a statement that I actually said with a different statement that is equivalent in flat spacetime, but I believe is not equivalent here.

 

[PeterDonis]

nside the horizon you still have the foliation of the spacetime as 2-spheres.

Yes, that's true everywhere. Each point in the Kruskal diagram labels a 2-sphere.

Because the signature is (-+++) and that 2D foliation is spacelike, that means that there is one spacelike and one timelike direction remaining.

More precisely, it is always possible to choose an orthonormal basis for the remaining 2D subspace (the one that the Kruskal diagram represents) with one timelike and one spacelike vector; but it is not possible to choose an orthonormal basis with two spacelike vectors.

We are looking (or at least I am looking) for the spacelike direction that is orthogonal to the 2-sphere.

Every direction in the 2D subspace that the Kruskal diagram represents is orthogonal to every 2-sphere. That is an elementary consequence of the hairy ball theorem (any such vector that was not orthogonal to every 2-sphere would create an everywhere non-vanishing vector field on a 2-sphere, which cannot exist). So this criterion does not help at all in picking out one direction in the 2D subspace vs. another.

I'll refrain from responding to the rest of your post because I think it needs to be rethought in the light of the above.

 

[Dale]

Every direction in the 2D subspace that the Kruskal diagram represents is orthogonal to every 2-sphere.

That is a good point, and it does bring me back to what I said earlier to @Ibix about not having a good way to pick a specific direction.

I'll refrain from responding to the rest of your post because I think it needs to be rethought in the light of the above.

I don’t think that it requires much revision. $\partial_t$ remains an orthogonal spacelike direction pointing away from the center as I said. It simply is not unique in that regard, which was the main motivation for my comment to @Ibix.

 

[PeterDonis]

$\partial_t$ remains an orthogonal spacelike direction pointing away from the center as I said.

I don't think "pointing away from the center" is a good description, since integral curves of $\partial_t$ point in a direction of constant $r$. "Away from the center" IMO implies increasing $r$. I don't see how the spacetime being curved changes that. I would ask what "center" you think $\partial_t$ inside the horizon points away from, since it doesn't seem to be pointing away from $r = 0$. [Edit: but see a subtlety I describe in my next post.]

But ordinary language descriptions are always vague, so I know that it's worth taking too much time to argue over them. It doesn't seem like we disagree on the math in this respect.

 

[PeterDonis]

The 2-spheres partition spacetime into two regions, one of which contains spacelike infinity and the other which does not.

In the maximally extended manifold, this is actually not true, because there are two spacelike infinities, not one, and the two are on opposite sides of each 2-sphere.

If we restrict attention to a realistic collapse spacetime, like the Oppenheimer-Snyder model, then there is only one spacelike infinity and your statement is correct.

If you have a vector which is both spacelike and orthogonal to the vectors in the sphere then that vector must either point into the region which contains spacelike infinity or into the region which does not.

Yes, but there are subtleties involved that I think are worth mentioning.

First, the integral curves of $\partial_t$ are not geodesics. So we have to distinguish two things if we look at the $\partial_t$ vector at some particular event inside the horizon: the spacelike geodesic that passes through that event with that tangent vector, and the integral curve that passes through that event with that tangent vector.

The latter of the above (the integral curve) does not go to spacelike infinity; it stays inside the horizon forever. That is how we both have been implicitly been interpreting things, when we say that this vector points in the direction of constant $r$.

The former of the above (the spacelike geodesic) should, however, go to spacelike infinity--or more precisely one end of it, the outgoing end, should (the other end, the ingoing end, should end on the $r = 0$ line somewhere inside the collapsing matter that forms the hole). I have not actually computed it to make sure, though.

The obvious next question would then be whether there are any spacelike geodesics on the outgoing side of an event inside the horizon that do not go to spacelike infinity. I think that is in fact the case--some of them will end on the singularity. (This seems obvious looking at a Penrose diagram--the Kruskal diagram is harder because the singularity is a hyperbola instead of a straight line.) So there should be some way of picking out the dividing line between the outgoing spacelike geodesics that reach spacelike infinity and the ones that don't. I'm not sure if that would have any obvious physical meaning, though.

 

[Dale]

If we restrict attention to a realistic collapse spacetime, like the Oppenheimer-Snyder model, then there is only one spacelike infinity and your statement is correct.

I am fine with that restriction.

The other possibility would be to specify one spacelike infinity as the spacelike infinity for “this” universe. Then “away from the center” would be the partition with “this” universe’s spacelike infinity and “toward the center” would be the partition with “that” universe’s spacelike infinity.

It is probably easier to just go with the realistic restriction.

First, the integral curves of ∂t are not geodesics.

So you are the one bringing in both integral curves and geodesics. I was only looking at which partition the vector is pointing in. It is a much more primitive notion I was considering.

The latter of the above (the integral curve) does not go to spacelike infinity; it stays inside the horizon forever. …

The former of the above (the spacelike geodesic) should, however, go to spacelike infinity

That is fine. I am not actually concerned about linking the vectors to any sort of curve going to spacelike infinity. I am simply identifying which region a vector is in.

I think that is in fact the case--some of them will end on the singularity. (This seems obvious looking at a Penrose diagram--the Kruskal diagram is harder because the singularity is a hyperbola instead of a straight line.)

I think you are right about that. In the Kruskal diagram you can see it by considering the light cone at an event in the interior. The outgoing light ray is parallel to the horizon (in that diagram), and since the singularity is asymptotic to the horizon the outgoing light ray must intersect it. Any event “on” the horizon beyond that point would be spacelike separated from the given event.

 

[PeterDonis]

I was only looking at which partition the vector is pointing in.

Yes, but I don't think that's enough in itself to determine whether a vector is pointing "at spatial infinity". The geodesic defined by that vector would have to actually reach spatial infinity, and I don't think all of them in the partition you have picked out do. (But again, this is a matter of choice of ordinary language, not physics. I don't think we disagree on the physics.)

 

[Dale]

Yes, but I don't think that's enough in itself to determine whether a vector is pointing "at spatial infinity".

I never claimed it was pointing at spatial infinity. I wasn’t making that claim and I don’t think that claim is necessary.