Spagettification and a singularity

[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?

 

[jbriggs444]

Isn't that c times the number of seconds between now and next Tuesday?

The distance between two objects is the space-time separation between the event of the one object "now" and the event of the other object at the "same time" according to some chosen foliation.

Next Tuesday is not at the "same time" as "now".

 

[PeterDonis]

Isn't that c times the number of seconds between now and next Tuesday?

No. The interval between now and next Tuesday noon on your worldline is timelike, not spacelike. It's not a "distance" at all. It's a time.

You can of course measure that time in meters instead of seconds, by multiplying the number of seconds by 299,792,458. But that doesn't make the interval a distance. It just changes the units of time you are using.

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)

You don't. There is no way for you to say how long after me you arrive. Taking "according to my proper time" doesn't help any, because your proper time only applies along your worldline, it says nothing about times on my worldline.

Your basic problem here is that you are still thinking of the singularity as a place instead of a time. That doesn't work. The singularity is spacelike, not timelike. Only something described by a timelike curve can be thought of as a "place", and only for that kind of thing does it make sense to ask how long after person A person B arrives.

The singularity, being spacelike, has to be thought of as a time. Does it make sense to ask how long after me you arrive at next Tuesday noon? Of course not. So asking how long after me you would arrive at the singularity doesn't make sense either. The only question that makes sense is how far (i.e., what distance) apart our arrival points are.

above the horizon we are separated (about) one lightsecond? Is that what you said?

No.

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

No.

 

[BoraxZ]

The distance between two objects is the space-time separation between the event of the one object "now" and the event of the other object at the "same time".

Next Tuesday is not at the "same time" as "now".

But how can there be a separation in time if we don't consider different times? If you are standing still, the separation between you now and you tomorrow is the number of seconds between now and tomorrow times c. In lightseconds (one lightsecond being 300 000 meter). Or do I misunderstand you?

 

[PeterDonis]

how can there be a separation in time if we don't consider different times?

We were considering different times: now and next Tuesday noon.

f you are standing still, the separation between you now and you tomorrow is the number of seconds between now and tomorrow times c.

See my first response to you in post #92.

 

[BoraxZ]

It's not a "distance" at all. It's a time.

But when you label the t-axis ct doesn't it hold? Don't you have to express the time-position fourvector with equal components?

All wordlines entering the hole have their unique proper time. But they all end at the future singularity time. The difference in their proper times is not an indication of how far apart they end up wrt each other at the singularity. But what DOES determine that separation? If they fall from a platform it is the time that elapses on the platform between the first particle falling and the second.

 

[PeterDonis]

when you label the t-axis ct doesn't it hold?

This issue has nothing to do with coordinates or labeling. The singularity is a spacelike line. That is an invariant, independent of any choice of coordinates or labeling. That invariant fact is what makes the singularity like a time (like next Tuesday noon) and not a place.

All wordlines entering the hole have their unique proper time.

Which tells you nothing about anything off those worldlines, as I've already said.

The difference in their proper times is not an indication of how far apart they end up wrt each other at the singularity.

That's correct.

what DOES determine that separation? If they fall from a platform it is the time that elapses on the platform between the first particle falling and the second.

Yes. The separation in time, according to the clock on the platform, between the instants when they start to fall does tell you something about how far apart in distance their arrival points on the singularity will be. See my posts #86 and #87.

 

[BoraxZ]

We were considering different times: now and next Tuesday noon.

Yes, but I wrote this in response to a comment which said:

The distance between two objects is the space-time separation between the event of the one object "now" and the event of the other object at the "same time".

When a black hole has formed isn't the time on the inside the time of the constituting particles?

 

[PeterDonis]

I wrote this in response to a comment which said

That comment specifically talked about spacelike distances, i.e., distances along a spacelike curve from one event to another. But the worldlines of objects are timelike, not spacelike.

When a black hole has formed isn't the time on the inside the time of the constituting particles?

No. There is no such thing as "the" time on the inside.

 

[BoraxZ]

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

No.

But if we consider the particles constituting the hole, don't their proper times (since the big bang) end in the hole? Which, for example, means that when I jump into the hole a million years after its formation, my proper time ends with a million years difference (between my proper time and that of the hole particles) at the singularity? What will be then the spatial distance between me and the hole particles on the r=0 line?

I think I see the difference between staying at one position and moving in time, and moving towards the time at the singularity. With moving towards the the time at singularity you associate a spatial distance?

 

[PeterDonis]

if we consider the particles constituting the hole, don't their proper times (since the big bang) end in the hole?

Their worldlines all end on the singularity, if that's what you mean.

when I jump into the hole a million years after its formation, my proper time ends with a million years difference (between my proper time and that of the hole particles) at the singularity?

In general you have no way of making this comparison, because you don't know how big the object was that formed the hole before it started to collapse. When you say "a million years after its formation", that is not well-defined unless you can pick out a specific timelike curve that represents the "platform" from which the collapsing object started to fall (basically a hypothetical observer that sat on the surface of the object before it collapsed, and then stayed hovering at that same altitude after the object collapsed). In general you have no way of doing that.

With moving towards the the time at singularity you associate a spatial distance?

No. I have never said any such thing. I have said the opposite, that it makes no sense to associate a "spatial distance" with a timelike curve.

 

[pervect]

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.

I haven't read the entire thread, but this remark caught my interest when I was skimming it.

In order to evaluate this statement, which seems counter-intuitive, I'd need to translate it into a mathematical format. Specifically, what does "point towards" and "point away" mean in a formal, mathematical sense?

My attempt at interpreting this remark involves some context. This context is the time-like worldline some infalling observer, which I would tend to assume for simplicity is a geodesic worldline of an infalling observer that started outside the event horizon (r > r_s, r_s being the schwarzchild radius, in region I of the kruskal diagram if we introduce an extended manifold), and terminates at the singularity (r=0), using for convenience the Schwarzschild coordinates. To further simplify, I'd assume it's a radially infalling observer (theta and phi in the Schwarzschild coordinates are constant).

We can set up an orthonormal set of basis vectors in the tangent space of some event, with the local concept of "space" in the tangent space being generated by vectors that are orthogonal to the time-like vector generated by the worldline.

We are then inquiring as to whether or not the geodesic curves generated by said tangent vectors from the initial specified event intersect r=0, or r = r_s, the Schwarzschild radius.

This formalizes the concept of "pointing towards" in a formal sense, but it's unclear if that's what you meant to say.

The answer, at this point, is not particularly obvious. I'd like to know if you consider this a fair re-statement of the problem in formal terms so that it has a definite answer.

It shouldn't be too terribly difficult to set up the appropriate differential equation parameterized by some affine parameter s, I imagine one can use some of the usual references for time-like geodesics by taking into account the sign change from time-like to space-like.

 

[PeterDonis]

what does "point towards" and "point away" mean in a formal, mathematical sense?

We've had several different definitions proposed in the thread. I think the one that is closest to what I had in mind when I made the post you quoted is the one I gave in post #56, which I'll quote here:

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".

 

[PeterDonis]

We are then inquiring as to whether or not the geodesic curves generated by said tangent vectors from the initial specified event intersect r=0, or r = r_s, the Schwarzschild radius.

This is a somewhat different version of "pointing" which I did use in some other posts in the thread.

@Dale used an even simpler definition, in which every radial spacelike vector on the "outgoing" side in a Kruskal diagram points "away" from the hole and every radial spacelike vector on the "ingoing" side points "towards" it. (I think we agreed that this works better in a realistic spacetime describing the collapse of a massive object to a black hole, since then there is a timelike $r = 0$ line at the center of the collapsing matter for the "ingoing" side vectors to point to.) Note, though, that we also agreed that some of the vectors pointing "away" by this definition still determine spacelike geodesics that end up hitting the singularity instead of escaping outside the horizon and going to spacelike infinity.

 

[BoraxZ]

With moving towards the the time at singularity you associate a spatial distance?No. I have never said any such thing. I have said the opposite, that it makes no sense to associate a "spatial distance" with a timelike curve.

Yes indeed. I actually meant two different worldlines ending up in the singularity. They hit the singularity spatially separated, don't they?

 

[PeterDonis]

I actually meant two different worldlines ending up in the singularity. They hit the singularity spatially separated, don't they?

They hit the singularity at distinct points, and any two points on the singularity are spacelike separated.