B Spagettification and a singularity

[PeterDonis]

if I interpret the picture in #4 then I see the singularity as a line with r=0 and time having different values

The Kruskal "time" coordinate has different values at different points on the singularity line, but so does the Kruskal "space" coordinate, so you can't conclude anything just from looking at the coordinates. You have to look at the metric and see what kind of line the singularity line is.

When you do that, you find that the singularity line at $r = 0$ is a spacelike line, and that is the case regardless of any choice of coordinates. That means different points on the singularity line correspond to different places, not different times, if we use "places" and "times" in the ordinary way those terms are used in everyday conversation.

 

[PeterDonis]

Can we say that all stuff that will constitute the hole, so over a huge amount of time, say, from the moment the hole is formed till the last particle falling in in the future, can we say that it's all "projected" into a small time interval, speaking loosely?

Not really. You could say that all the things that fall into the hole end up at the same time (the time of the singularity) at different places. But that's just as true of you and I, for example, and a time like next Tuesday noon, if we don't meet at the same place at that time.

 

[BoraxZ]

Not really. You could say that all the things that fall into the hole end up at the same time (the time of the singularity) at different places. But that's just as true of you and I, for example, and a time like next Tuesday noon, if we don't meet at the same place at that time.

If we look at two marbles falling, say, one year after one another to the surface of the Earth, they obviously end up at the same position in spacetime, with a small differences in age (if we consider the history of the marbles identical, apart from the one year hold-up).

In a black hole there is no such material surface on which both particles end. The second particle falls in and reaches the end in the same small proper time it took for the first particle. So it falls behind the first particle (being about a year older) towards the same time but always staying behind it? Can we say, seen in the co-falling frame, the singularity is at infinity, which makes it possible for both to reach the end while having a finite spatial distance wrt to one another?

 

[PeterDonis]

In a black hole there is no such material surface on which both particles end

That's correct.

It falls behind the first particle (being about a year older) towards the same time but always staying behind it?

Basically, yes. The two particles will end up at the same time (the singularity) at different places.

say, seen in the co-falling frame

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.

the singularity is at infinity

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

 

[Vanadium 50]

B-level?

"I'll see your Kruskal and raise you and Eddington-Finkelstein!"

 

[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.