Spagettification and a singularity

[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.

 

[PeterDonis]

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

Ok, so to you, "pointing away from the center" and "pointing at spatial infinity" are two distinct things. Fair enough. The first is the one you pick out by the outgoing side of the spacelike exterior of the light cone in the 2D $t$-$r$ subspace. The second you gave no specific definition for since I'm the one that brought it up, but I think "defines a spacelike geodesic that reaches spatial infinity" is reasonable.

 

[PeterDonis]

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

I agree. Another advantage of that restriction is that it makes it easier to justify the term "away from the center" for all spacelike vectors in the outgoing half of the light cone in the 2D $t$-$r$ subspace, even the ones that define spacelike geodesics that end on the singularity: they all do obviously point away from the timelike part of the $r = 0$ locus, the worldline of the center of the collapsing matter that formed the hole. But that locus is only there in the realistic restriction.

 

[BoraxZ]

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.

Does this mean that different points on r=0 correspond to different times, like in GP coordinates? So, for example, one particle ends up at r=0 on Friday, and the other on Monday (so to speak)? If so, does this mean that the tidal forces that initially separated two particles spatially (together with an acceleration on the hovering platform), will somehow change into separating them temporarily, while spatially they get on top of each other again (like they started at infinity, after which one of them is accelerated at the platform before continuing)?

 

[PeterDonis]

Does this mean that different points on r=0 correspond to different times, like in GP coordinates?

They correspond to different coordinate times in some charts, such as Painleve. But in those charts, the "time" coordinate is not timelike inside the horizon (in Painleve, it's spacelike), so you cannot conclude anything about "times" in a physical sense from them. To look at the physics, you have to look at invariants. The relevant invariant in this case is that the singularity is spacelike. That means it is physically meaningful to talk about arriving at the singularity at different places, but not at different times.

 

[BoraxZ]

That means it is physically meaningful to talk about arriving at the singularity at different places, but not at different times.

Yes, it would be rather odd if two particles end up at the same point on different times. So they end up spatially separated while the imaginary clocks n their backs show different times?
But then again, if every particle falling in falls the same distance to the singularity, does that imply the singularity lies infinitely far away (for else they could not end up spatially separated).

 

[PeterDonis]

So they end up spatially separated while the imaginary clocks n their backs show different times?

In the scenario you specified previously, yes.

if every particle falling in falls the same distance to the singularity

The particle doesn't fall "the same distance". It takes the same time by its clock to fall to the singularity (if we start the clock when it begins to fall, and every particle falls from the same altitude).

The concept of what "distance" it falls doesn't even make sense, any more than the concept of what "distance" it is from now to next Tuesday noon makes sense. The singularity is a moment of time, not a place.

does that imply the singularity lies infinitely far away

No. See above.

 

[BoraxZ]

Okay. So, just for clarity, two particles on a hovering platform, are pushed off from the platform one second after one another. Say it takes 10 seconds proper time to reach the singularity from the platform (with zero initial velocity). The first one hits the singularity after ten seconds proper time while the second hits the singularity (which for both particles is the same moment in time, though their clocks show one second difference) spatially separated from the first. (which is the reason it makes no sense to ask about the distance fallen?). Is this right (more or less)?

Can we say what that moment in time (the singularity) is? Is it the time when the hole was formed?

 

[vanhees71]

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.

For that you simply need some other coordinate chart covering the Schwarzschild horizon, i.e., coordinates that are non-singular at the Schwarzschild horizon (like Kruskal coordinates etc.).

 

[jbriggs444]

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 fact that all signals proceed toward the singularity does not prevent you from receiving them on your own trajectory toward the singularity. For instance, for a sufficiently massive black hole, you can fall feet-first through the horizon and still see your feet. Your eyes will have passed through the horizon by then, of course. Or, perhaps more aptly, the horizon will have passed through you at the speed of light.

 

[BoraxZ]

The fact that all signals proceed toward the singularity does not prevent you from receiving them on your own trajectory toward the singularity.

Yes, indeed. You can see the light that comes from the particles in front of you. Locally the space, in the freely falling frame, is flat. But there will come a point when the tidal forces prevents the light from reaching you.

 

[BoraxZ]

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.

But suppose the tape is made of tachyonic matter instead of tardyonic. What will it show? I'm still under the impression you fall an infinite amount of space in the proper time you are inside the hole. Which makes it possible for all particles to end up with a finite mutual distance. If all particles fall to infinity they all can end up at finite distances wrt each other. All at the same time.
So you end up at infinity at the "endtime" T, and me too. We will have a spatial distance between us, and our clocks show different times. My clock, if I jumped in after you, will show a later time. So in a sense, accelerating outside the hole (in a platform) is a kind of "reverse" twin paradox (where you end up younger after accelleration).

 

[Dale]

But suppose the tape is made of tachyonic matter instead of tardyonic.

Tachyonic matter doesn’t exist, and we don’t even have a good theory of how it would behave if it did hypothetically exist. This part of your question is unanswerable.

A note to anyone who would like to answer it: I may be wrong that there is no theory of tachyonic matter. If you would like to answer this then it is mandatory to provide a peer reviewed reference for the theory of tachyonic matter you are using. Note, this is an tachyonic tape measure, not just a tachyonic particle. Any answers without such a reference will be deleted.

 

[BoraxZ]

Tachyonic matter doesn’t exist, and we don’t even have a good theory of how it would behave if it did hypothetically exist. This part of your question is unanswerable.

A note to anyone who would like to answer it: I may be wrong that there is no theory of tachyonic matter. If you would like to answer this then it is mandatory to provide a peer reviewed reference for the theory of tachyonic matter you are using. Note, this is an tachyonic tape measure, not just a tachyonic particle. Any answers without such a reference will be deleted.

Okay, sorry for that. But don't we fall a distance through space? How else can it be? If I fall freely to the hole don't I fall through space? Do we fall through time inside the hole? The time/space reversal appears to a faraway observer only who coordinates the spacetime. manifold with Schwarzschild coordinates.

To put it differently, suppose a lot of particles are thrown in after one another. They all end up spatially separated from one another, but at the same time of the singularity. Will not the spatial distances between the first and last particle grow with the number of particles?

 

[BoraxZ]

Can we say that In Gullstrand-Painlevé coordinates the space inside is flat but a particle in front of you accelerates away from you due to "space expansion" (or the tidal effect)? The horizon behind you idem dito?

And say we place, again on the GP coordinates when we fall in freely, pendulae everywhere in front of us and behind us. What will they show?

 

[PeterDonis]

The first one hits the singularity after ten seconds proper time while the second hits the singularity (which for both particles is the same moment in time, though their clocks show one second difference) spatially separated from the first.

Yes.

Can we say what that moment in time (the singularity) is?

It is a moment of time that is to the future of all other moments inside the horizon. In general it has no well-defined relationship in time to events outside the horizon; to put it into your future, you have to fall into the hole.

 

[PeterDonis]

don't we fall a distance through space?

You are continually "falling" into the future. What is the distance you fall between now and next Tuesday noon?

The natural way to answer this question is in terms of time, not distance. The same is true for the singularity.

suppose a lot of particles are thrown in after one another. They all end up spatially separated from one another, but at the same time of the singularity. Will not the spatial distances between the first and last particle grow with the number of particles?

You use the phrase "the" spatial distance. There is no such thing. There is no one, unique, well-defined "spatial distance" between the particles once they are inside the horizon. So there is no one, unique, well-defined answer to your question.

As far as the "spacing" on the horizon of the events at which the particles arrive there, I have already answered that: it will be the same as the spacing in time between the particles starting to fall, provided they all fall from the same altitude above the horizon. So if they each fall one second after the previous one, each one will arrive at the singularity one light-second away from the previous one.

 

[PeterDonis]

As far as the "spacing" on the horizon of the events at which the particles arrive there, I have already answered that: it will be the same as the spacing in time between the particles starting to fall, provided they all fall from the same altitude above the horizon. So if they each fall one second after the previous one, each one will arrive at the singularity one light-second away from the previous one.

Actually, I'm not sure this is quite true. I think an additional factor appears involving the ratio of the altitude above the horizon to the horizon radius. I'm not sure of the exact factor, though.

 

[BoraxZ]

There is no one, unique, well-defined "spatial distance" between the particles once they are inside the horizon. So there is no one, unique, well-defined answer to your question.

Do you look at the hole from the faraway outside when saying this? Can't we assign a distance to a first particle in front of a second particle in the frame co-falling with the second?

 

[PeterDonis]

Do you look at the hole from the faraway outside when saying this?

No. I explicitly said "once they are inside the horizon".

Can't we assign a distance to a first particle in front of a second particle in the frame co-falling with the second?

Not for the entire infall process, no. The frame co-falling with the second particle is limited in extent; there will come a point where the worldline of the first particle will not be included in the second particle's co-falling frame (because the curvature of spacetime will be too large).

Also, the notion of "space" defined by the frame co-falling with the second particle is not unique. The frame co-falling with the first particle defines a different notion of "space"; and neither of those notions are global (both are limited to the extent of their respective co-falling frames).

 

[BoraxZ]

What is the distance you fall between now and next Tuesday noon?

Isn't that c times the number of seconds between now and next Tuesday?
Again, If you fall in your proper time to the future time of singularity, and I arrive one second after you (according to my proper time), the above the horizon we are separated (about) one lightsecond? Is that what you said?

Can we say the future time towards which we inevitably fall in the hole, is the time that registered when the hole formed?