B Spagettification and a singularity

[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.

 

[BoraxZ]

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

I still have difficulty understanding what the r in r=0 means. It is the radial coordinate. Which is timeline behind the horizon. Which means a particle travels on, after passing the horizon, the amount of proper time it takes to reach the singularity (which is linearly dependent on M and is about the time it takes light to cross the Schwarzschild radius).

Do the particles just travel on the time coordinate behind the horizon while in space in front of the horizon? So, again, two particles end up at the singularity time at r=0 while they get separated in space in front of the horizon? Or do I have coordinates mixed up here, as they never crosses the horizon but only approach the artificial singularity at the Schwarzschild radius (which is a spatial distance, in front of the horizon)?

Can't we say, instead of saying that the spatial distance the particles fall into (or onto) the hole (which obviously isn't c times the proper time, corresponding to the Schwarzschild radius), is indeterminate, approaches infinity (which indeterminate too)? This would allow for particles being spatially separated at the singularity. Or wouldn't it?

 

[BoraxZ]

What should we think about the images of 2D black holes that are frequently shown? With a funnel inside getting thinner and thinner and reaching towards infinity in the center?

 

[PeterDonis]

I still have difficulty understanding what the r in r=0 means.

It means the notional 2-spheres on the horizon singularity [corrected] are actually points--they have zero radius. The $r$ that is used as a coordinate in many charts on Schwarzschild spacetime (but not all) is the "areal radius" of 2-spheres; a 2-sphere with coordinate $r$ has surface area $4 \pi r^2$.

It is the radial coordinate. Which is timeline behind the horizon.

Only in Schwarzschild coordinates. But whether a coordinate is timelike or not has nothing to do with the nature of a particular curve. The singularity curve is spacelike regardless of any choice of coordinates. The worldlines of infalling objects are timelike regardless of any choice of coordinates. Inside the horizon, the areal radius of 2-spheres along any infalling object's worldline decreases with increasing proper time; but that has nothing to do with whether the $r$ in some coordinate chart is timelike or not. (To really understand this requires more than a B-level discussion.)

All of the rest of your post is just confusion based on you not recognizing the above points.

 

[PeterDonis]

What should we think about the images of 2D black holes that are frequently shown? With a funnel inside getting thinner and thinner and reaching towards infinity in the center?

Those have nothing to do with what we are discussing here.

 

[Ibix]

It means the notional 2-spheres on the horizon are actually points

Singularity, I think you mean, not horizon.

 

[PeterDonis]

Singularity, I think you mean, not horizon.

Yes, thanks! Post corrected.

 

[BoraxZ]

It means the notional 2-spheres on the horizon singularity [corrected] are actually points--they have zero radius. The r that is used as a coordinate in many charts on Schwarzschild spacetime (but not all) is the "areal radius" of 2-spheres; a 2-sphere with coordinate r has surface area 4πr2.

Does this mean that particles falling in, at the same time, from different angularly separated points but at the same r, will end up crammed together at the singularity?

 

[Ibix]

Assuming they have the same initial velocity, yes.

 

[BoraxZ]

Assuming they have the same initial velocity, yes.

Even if you look at it from from the faraway outside, where they end up in the horizon (where r is not zero)?

 

[Ibix]

Even if you look at it from from the faraway outside,

You can't look at the singularity from outside.

 

[BoraxZ]

You can't look at the singularity from outside.

No, that's true. But if they cram up in the inside, don't they cram up in the outside too? Or do particles (not angularly but radially separated) cram up, seen from the outside, on the horizon?

 

[PeterDonis]

if they cram up in the inside, don't they cram up in the outside too?

No. The horizon is a true 2-sphere with a nonzero area. Different angular coordinates correspond to different points on the horizon. The reason they don't correspond to different points on the singularity is that the singularity is not a true 2-sphere; it has zero area (because $r = 0$ there).

 

[Ibix]

No, that's true. But if they cram up in the inside, don't they cram up in the outside too? Or do particles (not angularly but radially separated) cram up, seen from the outside, on the horizon?

You mean, do things falling radially inwards get closer to each other? Sure. For a stellar mass black hole the circumference of the event horizon is tens of kilometres or more, so there's a lot of space. It only goes to zero size at the singularity.

 

[PeterDonis]

do particles (not angularly but radially separated) cram up, seen from the outside, on the horizon?

If they're at the same angular coordinates and are falling purely radially, then yes, from the outside, they will in principle appear to pile up at the same point on the horizon (though the light coming from them will also be increasingly redshifted so it will quickly become undetectable in a practical sense).

 

[PeterDonis]

do things falling radially inwards get closer to each other? Sure.

No, they don't. Their radial separation will increase due to spacetime curvature (tidal gravity). Here the OP is asking about particles separated only radially, not tangentially. (Particles falling from the same radius but separated tangentially will get closer to each other due to tidal gravity--I think this is what you were thinking of, but it's not what the OP asked about in that particular post.)

They will appear to a distant observer to pile up at the horizon, but this is an optical effect due to the behavior of outgoing light rays and can easily be shown not to contradict the statement I made above (which is invariant).

 

[Ibix]

No, they don't.

Ah yes, sorry, misread the question. Same radius, different angular coordinates will bunch up because there's less area. Same angular coordinates different radius will separate just like in Newtonian gravity (qualitatively).

 

[Dale]

No, that's true. But if they cram up in the inside, don't they cram up in the outside too? Or do particles (not angularly but radially separated) cram up, seen from the outside, on the horizon?

This is what “spaghettification” means: Things get crammed up in the horizontal directions and stretched out in the vertical direction. This is basically the familiar form of tidal forces from Newtonian gravity.

 

[BoraxZ]

This is what “spaghettification” means: Things get crammed up in the horizontal directions and stretched out in the vertical direction. This is basically the familiar form of tidal forces from Newtonian gravity.

Yes. But on the inside particles get crammed in the angular directions and seen from faraway on the outside the particles seem to get crammed in the radial direction.

 

[PeterDonis]

on the inside particles get crammed in the angular directions

Which, as noted, is part of spaghettification.

seen from faraway on the outside the particles seem to get crammed in the radial direction.

Which has nothing whatever to do with spaghettification so it's off topic for this thread.

 

[Dale]

on the inside particles get crammed in the angular directions and seen from faraway on the outside the particles seem to get crammed in the radial direction.

I am not sure why you believe that, but it is untrue. Tidal forces always stretch in the vertical direction and compress in the horizontal directions, both inside and outside the horizon.

Perhaps you are thinking of some coordinate-based description. If so it is yet another reason not to take coordinates too seriously.

 

[BoraxZ]

I am not sure why you believe that, but it is untrue. Tidal forces always stretch in the vertical direction and compress in the horizontal directions, both inside and outside the horizon.

Yes, but on the outside the cramming seems to take place spatially on the horizon for radially separated particles. In the inside these particles all collect at the singularity time but get spatially separated.

Particles that are angular separated (and not radially) don't get (totally) crammed on the horizon, while in the inside they get totally crammed angularly.

 

[PeterDonis]

I am not sure why you believe that

I think he is referring to what I described in post #120. Which, as I noted, is only an optical effect. And which in any case has nothing to do with spaghettification.

 

[BoraxZ]

Perhaps you are thinking of some coordinate-based description. If so it is yet another reason not to take coordinates too seriously

Can we describe the hole without coordinates? With invariants?

 

[PeterDonis]

Can we describe the hole without coordinates? With invariants?

Of course. Many of the statements that have been made in this thread are about invariants.

 

[Dale]

I think he is referring to what I described in post #120. Which, as I noted, is only an optical effect. And which in any case has nothing to do with spaghettification.

All this obsessing over what distant observers visually see is rather pointless. But at least it isn’t obsessing over coordinates which is pointlesser

 

[BoraxZ]

All this obsessing over what distant observers visually see is rather pointless. But at least it isn’t obsessing over coordinates which is pointlesser

You got a point there!

Somehow, what we see from faraway is the opposite of what happens inside. From afar the particles seem to cram up on the Schwarzschild radius, while on the inside they cram up on the singularity time and angular directions, while their radial distances from each other have grown. Very loosely speaking.

 

[PeterDonis]

pointlesser

Is that a word? I guess it is now, since you've used it.

 

[Ibix]

pointlesser

Is that a word? I guess it is now, since you've used it.

Should be "eventlesser" in a discussion about spacetime.

 

[HansH]

I meant what vanhees71 showed in his comment. Two particles falling together freely, after which one of them (or both) accelerates away from the other to meet up again later in. Can they meet up again behind the horizon?

I think tis is still the essential part of my original question (but not answered yet I suppose, or did I mis something? could of course be after 140 answers). So constant accelarating away from eachother, how can they ever meet again?

 

[PeterDonis]

I think tis is still the essential part of my original question

Your original question was about spaghettification. That has nothing to do with the twin paradox. The twin paradox subthread is a hijack that will shortly be moved to its own thread.

 

[PeterDonis]

constant accelarating away from each other, how can they ever meet again?

Any particles involved in the spaghettification of an object as it approaches the singularity, if the spaghettification is moving them apart, will not meet again.

 

[PeterDonis]

The twin paradox subthread is a hijack that will shortly be moved to its own thread.

And now it has been:

https://www.physicsforums.com/threads/the-twin-paradox-and-black-holes.1050610/

 

[HansH]

ok thanks. my original question was indeed about spagettification, but the reason was that I was wondering if this spagettification process continues no matter how close you are to the singularity. (as I would expect and now asume you confirm that in #157, but could be that I do not understand you exactly)

 

[HansH]

probably I am a bit confused about the term 'size' and where we can speak of a 'size' of something. outside the event horizon we can speak for example of the size of an atom. With a person in freefall into a black hole Iassume the atoms still are there so also the size can be measured an also the size between 2 atoms separating due to the spagettification process. but how close can we be to the singularity before size becomes a useless definition so when is the spagettification process not able anymore to be followed by an observer (being very close nearby)?

 

[BoraxZ]

I think tis is still the essential part of my original question (but not answered yet I suppose, or did I mis something? could of course be after 140 answers). So constant accelarating away from eachother, how can they ever meet again?

All particles end up in the r=0 singularity. Spacelike separated. The r is timelike. So two particles falling in after one another end up at the same time but spatially separated. Their clocks showing different times (when synchronized before falling in).

 

[PeterDonis]

I was wondering if this spagettification process continues no matter how close you are to the singularity.

Yes, it does. Your reading of the previous post about this is correct.

probably I am a bit confused about the term 'size' and where we can speak of a 'size' of something.

Whether we can do this, and what "size" means, depends on the something and the circumstances.

At some point during the spaghettification process any kind of internal structure that determines a "size" of anything will be destroyed. For example, at some point atoms will get torn apart by it, so we can no longer speak of the size of atoms since there aren't any.

 

[PeterDonis]

two particles falling in after one another

The general topic in this thread is spaghettification, i.e., an object falls in, initially, as a single object, and the discussion is about what happens to it as it approaches the singularity--it gradually gets torn apart by increasing spacetime curvature. Two originally separate particles falling in one after the other is not quite the same scenario.

 

[BoraxZ]

The general topic in this thread is spaghettification, i.e., an object falls in, initially, as a single object, and the discussion is about what happens to it as it approaches the singularity--it gradually gets torn apart by increasing spacetime curvature. Two originally separate particles falling in one after the other is not quite the same scenario.

Yes. But aren't two particles starting at the same event, while holding one back, after which you release it, the same as an extended object?

 

[PeterDonis]

ren't two particles starting at the same event, while holding one back, after which you release it, the same as an extended object?

No. An extended object has interactions between its particles.

 

[BoraxZ]

What time is it inside the hole?

 

[BoraxZ]

No. An extended object has interactions between its particles.

Yes, that's true. But these are overcome inside so they end up spatially separated still. But less indeed.

 

[BoraxZ]

No. An extended object has interactions between its particles.

Will even a proton get ripped apart? Or maybe even quarks (when there are preons inside it)?

 

[HansH]

two particles falling in after one another end up at the same time

which time?

 

[PeterDonis]

What time is it inside the hole?

This question is not answerable. Please do not clutter the thread with pointless questions.

 

[PeterDonis]

which time?

He means the singularity, which is a moment of time.