B Spagettification and a singularity

[HansH]

TL;DR Summary

is spagettification and a singularity in contradiction with eachother?

a singularity is assume to be a point without dimensions. When 2 mass particles fall into a black hole in the direction of the centre and separated by small distance, the particle closest to the center will have an acceleration as seen from the other particle as I understood. so distance between the particles will increase. Iassume you can do that thought experiment for any moment during the path towards the singularity, so place 2 particles close to eachother in free fall. But the singuarity has no dimensions and both particles will finally arrive there. So how can it be the on te one hand the distance increases and on the other hand they will finally arrive at the same point (the singularity). To me this looks as if dimensions relatively scale down towards 0 the closer you come to the singularity. So how does this work? (I posted it as basic as I am looking for a high level explanation if possible)

 

[Ibix]

a singularity is assume to be a point without dimensions.

This isn't correct. The Schwarzschild singularity is a line - it is more like a moment in time than a place in space.

When 2 mass particles fall into a black hole in the direction of the centre and separated by small distance, the particle closest to the center will have an acceleration as seen from the other particle as I understood.

Yes, although it can be stated more rigorously avoiding terms like "center" that don't really apply.

So how can it be the on te one hand the distance increases and on the other hand they will finally arrive at the same point (the singularity). To me this looks as if dimensions relatively scale down towards 0 the closer you come to the singularity. So how does this work? (I posted it as basic as I am looking for a high level explanation if possible)

I don't understand the issue. Even if the singularity were a point, two things arriving at a point at different times is in no way contradictory. As it is, once you're in the black hole the singularity is more like Monday morning than anything else. Tidal forces may tear you apart over the weekend, but you will still arrive at Monday morning.

 

[HansH]

perhaps some clarification to make my issue more clear. First of all I only used the word 'center' to make clear in which direction the curvature gets stronger, in order to be able to indicate which of the 2 particles does what with respect to the other particle. so perhaps speaking in terms of curvature would be better?

But the point where it is really about is that I have some trouble with understanding that we talk about 'distance' first (as seen from the first point the distance to the second point becomes larger and larger in time) and arriving at the singularity we talk only about a place in time, so term 'distance or size' then does not make sense for the singularity if I understand you correctly. Of course this solves my problem, as we dont have to bother about distance getting larger and larger. But Is there some simple way to explain a singularity in terms of distance and size having no meaning then?

 

[Ibix]

The objects just strike different places on the singularity. (Again, that can be phrased more rigorously).

Wikipedia has a Kruskal diagram which shows a maximally extended Schwarzschild spacetime. The right hand region is the exterior and the upper region is the interior. The singularity is the hyperbola across the top, and the $r$ coordinate lines are marked. I've pasted it below (thanks to @DrGreg) sketched in a couple of worldlines of infalling observers in black (probably not completely correctly, but roughly ok). You can see the separation between them growing until they strike the singularity.

 

[malawi_glenn]

Isn't spagettification just due to a strong tidal effect? That the gravitational pull is vastly stronger on one side compared to the other.

 

[Ibix]

Isn't spagettification just due to a strong tidal effect? That the gravitational pull is vastly stronger on one side compared to the other.

Well, you have to get a bit more technical than that if you want to see gravity as spacetime curvature because it doesn't really "pull". But yes, it's due to gravity causing different accelerations on opposite sides of the object.

 

[HansH]

for the Kruskal diagram. Just from a beginners view: Do I see it correctly that depending on where you start falling into the black hole you end up at the singularity at a certain time and position (x?) so starting at r=1.4 following that line you end up at r=0 but x is still approx 1.25. so what is then the meaning of the x coordinate ending up for example at the point (t=2,r=0) of the right side line in the figure?

 

[phinds]

is spagettification and a singularity in contradiction with each other?

Since spaghettification happens outside the EH of small BH's, I don't see how it can be a contradiction. Spacetime curvature (the cause of sphagettification) doesn't change in type, just degree (and that only slowly) as the EH is passed, so there's really no difference inside or outside as far as sphagettification is concerned (other than due to the size of the BH)

 

[DrGreg]

for the Kruskal diagram. Just from a beginners view: Do I see it correctly that depending on where you start falling into the black hole you end up at the singularity at a certain time and position (x?) so starting at r=1.4 following that line you end up at r=0 but x is still approx 1.25. so what is then the meaning of the x coordinate ending up for example at the point (t=2,r=0) of the right side line in the figure?

In Schwarzschild coordinates $(t,r,\phi,\theta)$

• outside the event horizon (region I)
  • $r$ represents a distorted distance
  • $t$ represents a distorted time

• inside the event horizon (region II)
  • $r$ represents a distorted time
  • $t$ represents a distorted distance

In Kruskal coordinates $(T,X,\phi,\theta)$, in all regions,

• $X$ represents a distorted distance
• $T$ represents a distorted time

(To be pedantic, when I say "represents a distorted distance" I really mean "is a spacelike coordinate", and when I say "represents a distorted time" I really mean "is a timelike coordinate".)

 

[vanhees71]

"coordinate" is better than "distorted something". In which sense is it "distorted"?

 

[DrGreg]

"coordinate" is better than "distorted something". In which sense is it "distorted"?

$r$ "represents a distorted distance" (outside the event horizon) in the sense that, if you hold all other coordinates constant, you can plot a graph of distance versus $r$, which isn't a straight line. The scale ratio of $\Delta r$ to distance is variable. In that sense it's "distorted". It seems to me to be a reasonable simple description for a B-level thread where we don't need to get into technicalities like the definition of "spacelike".

 

[Ibix]

Do I see it correctly that depending on where you start falling into the black hole you end up at the singularity at a certain time and position

There are all sorts of circumstances that lead to you arriving at different "places" on the singularity. Dropped simultaneously from different heights, or from the same height at different times or with different velocities. I drew two things dropped from different heights because you were talking about an elongated object that gets further stretched. Think of those lines as the worldlines of the opposite ends of a rod. You can see the length growing all the way to the singularity.

 

[HansH]

The objects just strike different places on the singularity. (Again, that can be phrased more rigorously).

in #2 you say: 'the Schwarzschild singularity is a line - it is more like a moment in time than a place in space.'
but if I interpret the picture in #4 then I see the singularity as a line with r=0 and time having different values. so based on that I would conclude that the singulatity is a point with zero dimensions, but with all moments in time so not one moment in time.

so for me there are still some mysteries left:
1)
I understood that for points in region 1 time flows forwards (as this is our 'normal' world), but crossing the event horizon time goes to infinity, so that would be as seen from the opserver outside the black hole looking at the particle falling in I suppose. But the orginal question was about how the picture looks as seen from the observer free falling into the black hole towards the singularity including a mist of particlea around him in freefall with him. And that observer following particles that are for example 1cm in front of him (1 cm compared to his measuring device) in the path following together.

so then distances cannot be interpreted in the way as represented in the picture I suppose, so for me not sure how to interpret the distance between the 3 sketched lines in the picture. so How should I understand that for every particle 1 cm away, the particle is acellerating away from the observer and how that looks like just before hitting the singularity?

2)
does time stop at the singularity? so is it possible to followthe blue line going to the right? (time is plottet represented by straight lines with time related to the angle of the line, so I then would expect that that also holds for the singularity?)

 

[timmdeeg]

The Finkelstein diagram (time in z-direction) could be helpful:

It shows the horizon and the singularity "as a line" as mentioned by @Ibix in #2, where the second particle arrives later. You can assume the worldline of the second particle roughly as being parallel to that of the first one. That's all.

P.S. You can find this and other simple but very enlightening diagrams in Robert Geroch's General Relativity from A to B.

 

[Ibix]

r=0

It's important to realise that $r$ is the timelike coordinate inside the black hole. You can see that by considering the sign of the Schwarzschild coordinates $dr^2$ and $dt^2$ terms in the metric for $r$ greater than and less than the Schwarzschild radius. Thus the singularity has 1d spatial extent (the circumference of a circle in the $\theta$ and $\phi$ directions is zero at the singularity, so it's 1d) but no timelike extent. That's more like an instant in time than anything else.

The diagram is a very distorted representation of the spacetime and obeys some of the same rules as a Minkowski diagram. Specifically, light travels on 45° lines, which is why the event horizon is one such, and any definition of "time at one place" is a line above 45° (one of my worldlines probably dips below 45°, but that's bad drawing on my part).

But the orginal question was about how the picture looks as seen from the observer free falling into the black hole towards the singularity including a mist of particlea around him in freefall with him.

Your original question was about a contradiction you seemed to see between things getting further apart but hitting the singularity. The Kruskal diagram shows both that you misunderstood what the singularity was, and shows how things can reach it at different places.

If you want to know what happens to a cloud of free particles it's the same as in Newtonian gravity - it lengthens in one direction and shortens in the other two. The lengthening direction just doesn't relate to the $r$ coordinate inside the black hole because that's not a spacelike direction.

does time stop at the singularity?

Neither time nor space is defined at the singularity. If we were being rigorous, most claims about what happens "at" the singularity would be replaced by claims about limiting behaviour as one approaches the singularity.

 

[Ibix]

The Finkelstein diagram (time in z-direction) could be helpful:

I think in this diagram the z direction is Schwarzschild t coordinate, so again calling that "time" is something of a misnomer inside the horizon. On this diagram I believe that "the future" generally points to the right inside the horizon, which is why the singularity is a vertical line in this representation.

 

[BoraxZ]

But still, when falling as a point particle behind another one, it is stretched away from you in the finite proper time time left for you. Or does the particle ahead of you stops in the radial space direction? It's different from two marbles falling after one another from a fixed height. The marbles literally hit solid earth which is absent in a black hole.
Can't we say all particles end up separated in radial space (as seen from a freely falling frame) and separated in time?

If one particle falls in freely 1 year after another, will they both up at infinity but separated in radial space (there is no change in space and time coordinates), and about 1 year separation in time?

 

[Ibix]

But still, when falling as a point particle behind another one, it is stretched away from you in the finite proper time time left for you.

Yes.

Or does the particle ahead of you stops in the radial space direction?

There isn't a radial direction once you're inside the hole - the direction of changing radius is timelike there.

It's different from two marbles falling after one another from a fixed height. The marbles literally hit solid earth which is absent in a black hole.

It's qualitatively similar, up until the marbles hit the ground. The particles just get further and further apart. This process just doesn't get interrupted by ground because there isn't any; the marbles instead reach the singularity.

Can't we say all particles end up separated in radial space (as seen from a freely falling frame) and separated in time?

There isn't a radial direction, really, any more. And "separated in time" doesn't really make sense for objects. For events on their worldline, sure they can be separated in time (or not, there's always some ambiguity).

If one particle falls in freely 1 year after another, will they both up at infinity but separated in radial space (there is no change in space and time coordinates), and about 1 year separation in time?

This question makes no sense to me.

 

[BoraxZ]

Yes.

There isn't a radial direction once you're inside the hole - the direction of changing radius is timelike there.

It's qualitatively similar, up until the marbles hit the ground. The particles just get further and further apart. This process just doesn't get interrupted by ground because there isn't any; the marbles instead reach the singularity.

There isn't a radial direction, really, any more. And "separated in time" doesn't really make sense for objects. For events on their worldline, sure they can be separated in time (or not, there's always some ambiguity).
This question makes no sense to me.

Suppose you fall freely along with a particle. How do space and time interchange? That change is seen only for a faraway observer. Why should the space ahead of you, when falling in, change into time and what does that even mean?

Imagine you fall in one year after another particle. You will never reach the place where the other particle ends his inward voyage in a short proper time. There is not an end of space like the surface of the Earth. Can't we say both end up at infinity, so there can be a finite distance between them at the end of their journey?

Insofar time is concerned, the clock on the back of a particle in front of us shows a different time than ours. If we fall one year after the other particle (dat we both jump in from a huivering platform at sufficient distance from the EH and we wait one year before jumping after the first particle), won't our clicks end up with about a year difference?

Which means that both particles end up separated in space, as well in time? I just can't grasp the meaning of space and time changing roles behind the horizon as seen from a frame falling along.

 

[timmdeeg]

I think in this diagram the z direction is Schwarzschild t coordinate, so again calling that "time" is something of a misnomer inside the horizon. On this diagram I believe that "the future" generally points to the right inside the horizon, which is why the singularity is a vertical line in this representation.

Thanks and that's why the lightcones are tipped towards the black hole. Robert Geroch doesn't mention the coordinates, which I think are Eddington-Finkelstein coordinates. He mentions that the light cones become "vertical" in the distant external region (meaning asymptotically flat).

 

[Ibix]

Suppose you fall freely along with a particle. How do space and time interchange?

They don't, not for anybody. As you say, what would it even mean? What changes is the relationship between your local directions and global measures. It's loosely analogous to travelling north. North is forward and south is backward, right? But at some point you pass through the coordinate singularity at the pole and suddenly north is backward and south is forward. Have "north and south changed meaning"? Or did you just pass through a place where your coordinates don't work well and had a discontinuous change in how your notions of direction relate to your global coordinate system?

One can split spherically symmetric spacetimes up into spheres. If you do that in Schwarzschild spacetime the event horizon is where it stops making sense to describe smaller spheres as being "inside" larger ones. Rather, the next smaller sphere is in the future of the current one. To put it another way, if you slice spacetime into a set of spheres, those spheres will be indexed by two coordinates, one timelike and one spacelike. If you fix the timelike coordinate then changing the spacelike coordinate takes you to smaller or larger spheres. Fixing the spacelike coordinate and changing timelike coordinate you remain on spheres of the same area. Inside the hole, though, it's vice versa with the spheres getting smaller in the timelike direction and remaining constant area in the spacelike direction. And there's a surface between the two regions, the event horizon, where the same size sphere's are found in a null direction.

I think it's a mistake to label the spatial coordinate $t$ because it confuses people. Labelling the time coordinate $r$ is also confusing, but it is related to the radius of the spheres making up your current spatoal slice.

Imagine you fall in one year after another particle. You will never reach the place where the other particle ends his inward voyage in a short proper time

Depends what you mean. Given sufficient delta v it's possible to meet anyone who falls in after you, and a lot of people who fell in before you. The proper time you experience on any such route will be less than 15$\mu$s per solar mass of black hole. So I would say it's either possible to reach somebody's impact on the singularity in short proper time or else it's not possible to reach it at all.

Can't we say both end up at infinity, so there can be a finite distance between them at the end of their journey?

Infinity of what? When we talk about "infinity" in the context of black holes we usually mean $r\rightarrow\infty$, which is the very opposite of what's happening here.

Which means that both particles end up separated in space, as well in time?

Particles can be separated in space, but I don't know what you mean by "separated in time".

 

[Ibix]

Thanks and that's why the lightcones are tipped towards the black hole. Robert Geroch doesn't mention the coordinates, which I think are Eddington-Finkelstein coordinates. He mentions that the light cones become "vertical" in the distant external region (meaning asymptotically flat).

I think Eddington-Finkelstein coordinates share a $t$ direction with Schwarzschild, but their radial coordinate lines are not orthogonal to that direction. I'd need to double check - I don't know my way around them so well as Schwarzschild and Kruskal-Szekeres ones. Something like that has to be true to make sense of the light cone tipping and where the singularity is, though.

 

[BoraxZ]

They don't, not for anybody. As you say, what would it even mean? What changes is the relationship between your local directions and global measures. It's loosely analogous to travelling north. North is forward and south is backward, right? But at some point you pass through the coordinate singularity at the pole and suddenly north is backward and south is forward. Have "north and south changed meaning"? Or did you just pass through a place where your coordinates don't work well and had a discontinuous change in how your notions of direction relate to your global coordinate system?

One can split spherically symmetric spacetimes up into spheres. If you do that in Schwarzschild spacetime the event horizon is where it stops making sense to describe smaller spheres as being "inside" larger ones. Rather, the next smaller sphere is in the future of the current one. To put it another way, if you slice spacetime into a set of spheres, those spheres will be indexed by two coordinates, one timelike and one spacelike. If you fix the timelike coordinate then changing the spacelike coordinate takes you to smaller or larger spheres. Fixing the spacelike coordinate and changing timelike coordinate you remain on spheres of the same area. Inside the hole, though, it's vice versa with the spheres getting smaller in the timelike direction and remaining constant area in the spacelike direction. And there's a surface between the two regions, the event horizon, where the same size sphere's are found in a null direction.

I think it's a mistake to label the spatial coordinate $t$ because it confuses people. Labelling the time coordinate $r$ is also confusing, but it is related to the radius of the spheres making up your current spatoal slice.

Depends what you mean. Given sufficient delta v it's possible to meet anyone who falls in after you, and a lot of people who fell in before you. The proper time you experience on any such route will be less than 15$\mu$s per solar mass of black hole. So I would say it's either possible to reach somebody's impact on the singularity in short proper time or else it's not possible to reach it at all.

Infinity of what? When we talk about "infinity" in the context of black holes we usually mean $r\rightarrow\infty$, which is the very opposite of what's happening here.

Particles can be separated in space, but I don't know what you mean by "separated in time".

If they end up with clicks that show different times then they are separated by an amount of time. Which corresponds to ct meters.

In other words, our incomplete geodesic is ct longer in time than the geodesic of the particle before us. Or not?

 

[vanhees71]

Yes.

There isn't a radial direction once you're inside the hole - the direction of changing radius is timelike there.

But I think inside the Schwarzschild radius the Schwarzschild coordinates are just fine. It's only that what's labelled $t$ becomes a "spacelike coordinate" and $r$ a "timelike coordinate":
$$\mathrm{d}s^2=(1-r_S/r) \mathrm{d}t^2-(1-r_S/r)^{-1} \mathrm{d} r^2 - r^2 \mathrm{d} \Omega^2.$$
For $r>r_S$, the variable $t$ is timelike and $r$ is spacelike. At $r_S=r$ you have a coordinate singularity (but no true physical singularity). For $r<r_S$ then $r$ becomes timelike and $t$ spacelike, and you have a coordinate chart for the $0<r<r_S$, and the fundamental form has also the proper (1,3) signature. So I don't see, why one shouldnt use Schwarzschild coordinates inside the Schwarzschild horizon. Of course, for dealing with situations, where something falls from the exterior, $r>r_S$, to the interior, $r<r_S$ one should use other coordinates, which have no coordinate singularities there, or is there something making this point of view incorrect?

 

[Ibix]

But I think inside the Schwarzschild radius the Schwarzschild coordinates are just fine. It's only that what's labelled $t$ becomes a "spacelike coordinate" and $r$ a "timelike coordinate"

Sure. 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. The spaghettification happens in the (misleadingly labelled) $t$ direction, which is an axial coordinate rather than a radial one. At least some sources use $t$ and $z$ for the interior coordinates rather than $r$ and $t$ to avoid confusion (although then people try to equate the interior $t$ with the exterior one which doesn't make sense either, so you can't win).

I think it's probably worth clarifying that there are essentially two separate coordinate systems both called "Schwarzschild coordinates", one set covering the interior of the hole and one set covering the exterior. The metric components have the same functional form, and there is still meaning to the quantity usually labelled $r$, in that it is $\sqrt{A/4\pi}$ where $A$ is the area of the spherically symmetric surfaces. But the direction of changing $r$ is timelike in the interior and spacelike in the exterior. And the two coordinate patches do not overlap; one applies where $r>R_S$ and one applies where $r<R_S$ but neither applies where $r=R_S$. Ignoring that fact, comparing the functional forms of the metric components, and equating the interior $t$ with the exterior $r$ and the interior $z$ with the exterior $t$ is how silly claims like "time and space have swapped" end up appearing in popsci.

is there something making this point of view incorrect?

No, it's perfectly fine as far as I know. It's just potentially confusing to use $t$ to label a spacelike coordinate, IMO, especially for newbies.

 

[Ibix]

If they end up with clicks that show different times then they are separated by an amount of time. Which corresponds to ct meters.

I don't normally comment on language, but in both post #19 and #23 you've used "clicks" in a way I didn't follow. Did you mean clocks rather than clicks? I'm going to assume you do in what follows.

What a person's individual clock shows has absolutely no bearing on how far apart in spacetime two events are, which is a lesson from the twin paradox. It's perfectly possible for two clocks showing the same time to cross the event horizon together, separate and return so that they show different times just like in the twin paradox, and then collide with the singularity at the same event. There's no separation between the clocks any more than there is separation between two cars that leave the factory together but follow different routes to a garage - their odometers show different readings, but so what? Their odometer readings don't show the distance between them, just the length of the route they have taken. Likewise their clocks show the proper time along the route they have taken, not the time separation between them.

In other words, our incomplete geodesic is ct longer in time than the geodesic of the particle before us. Or not?

There are different geodesics that reach the singularity at the same event and ones that reach it at different events. They can have many different "lengths" from the horizon, and which is longer depends on the path taken. The "longest" available route is, as I recall, is the geodesic that corresponds to free-fall from infinity, so if you find yourself on the inside of a black hole you maximise your remaining proper time by matching speeds with an object passing you that fell from infinity.

 

[vanhees71]

Sure. 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. The spaghettification happens in the (misleadingly labelled) $t$ direction, which is an axial coordinate rather than a radial one. At least some sources use $t$ and $z$ for the interior coordinates rather than $r$ and $t$ to avoid confusion (although then people try to equate the interior $t$ with the exterior one which doesn't make sense either, so you can't win).

Sure. In the "less confusing notation" it's
$$\mathrm{d} s^2=(r_S/r-1)^{-1} \mathrm{d} \tilde{t}^2 - (r_S/r-1) \mathrm{d} z^2 - \tilde{t}^2 \mathrm{d}\Omega^2,$$
where I've written $\tilde{t}$ to avoid confusion of $t$ in the exterior Schwarzschild solution.

I think it's probably worth clarifying that there are essentially two separate coordinate systems both called "Schwarzschild coordinates", one set covering the interior of the hole and one set covering the exterior. The metric components have the same functional form, and there is still meaning to the quantity usually labelled $r$, in that it is $\sqrt{A/4\pi}$ where $A$ is the area of the spherically symmetric surfaces. But the direction of changing $r$ is timelike. And the two coordinate patches do not overlap; one applies where $r>R_S$ and one applies where $r<R_S$ but neither applies where $r=R_S$. Ignoring that fact, comparing the functional forms of the metric components, and equating the interior $t$ with the exterior $r$ and the interior $z$ with the exterior $t$ is how silly claims like "time and space have swapped" end up appearing in popsci.

That's highly misleading, indeed, and I think that to deal with a particle falling into the interior one should use other coordinates without a coordinate singularity at the Schwarzschild horizon.

No, it's perfectly fine as far as I know. It's just potentially confusing to use $t$ to label a spacelike coordinate, IMO, especially for newbies.

Thanks for the clarification!

 

[BoraxZ]

I don't normally comment on language, but in both post #19 and #23 you've used "clicks" in a way I didn't follow. Did you mean clocks rather than clicks? I'm going to assume you do in what follows.

What a person's individual clock shows has absolutely no bearing on how far apart in spacetime two events are, which is a lesson from the twin paradox. It's perfectly possible for two clocks showing the same time to cross the event horizon together, separate and return so that they show different times just like in the twin paradox, and then collide with the singularity at the same event. There's no separation between the clocks any more than there is separation between two cars that leave the factory together but follow different routes to a garage - their odometers show different readings, but so what? Their odometer readings don't show the distance between them, just the length of the route they have taken. Likewise their clocks show the proper time along the route they have taken, not the time separation between them.

There are different geodesics that reach the singularity at the same event and ones that reach it at different events. They can have many different "lengths" from the horizon, and which is longer depends on the path taken. The "longest" available route is, as I recall, is the geodesic that corresponds to free-fall from infinity, so if you find yourself on the inside of a black hole you maximise your remaining proper time by matching speeds with an object passing you that fell from infinity.

Sorry for the clicks and clicks. I indeed meant clocks. I only now see the possibility to edit. I noticed it too. It's, I think a typo because my mobile typing. Small keys, big thumb!

Anyhow, imagine two geodesics, corresponding to two particles that start from the same position at infinity. Somewhere along their way we hold one of them and let it hover somewhere before reaching the EH. The particle falling along freely ends up at the singularity before the one we stopped.

Say the stop lasts one million years. After a million years it continues its downfall. Can't we say then that when it hits the singularity it's separated one million years in time from the other particle?

Suppose you and I are the particles. I fall into the singularity where my life stops. You, in the other hand, live along a million years (according to your clock). So when you fall in I will be about a million years older than you. Are we not separated then by a million years in time?

In fact, ALL particles that constitute the hole, from its beginning, will have ages that can differ as much as the particles spent their time outside the hole. All inside particles show different end-times. I.e. they end up at different times. Is this the same as today and tomorrow being separated by time? We can reach tomorrow simply by sitting still.

What happens with the hole? Your time, outside the hole, has run a million years longer than mine, inside the hole. So when you enter the hole you are obviously older than me. We are indeed separated in time in the same way that a traveler (accelerating away from you and returning to you) in the twin paradox is younger than you (remaining at rest, in an jnertial frame). But in the black hole case it's the particle without a proper acceleration that is younger after the voyage.

In the freely falling frame, you end up at a time that is different from mine. The difference can be any time, depending on the non-geodesic nature of the particles that constitute the hole. I agree that they are not separated in time like today and tomorrow are separated. But yet, they end up at different times. But they can't continue...

 

[Ibix]

Small keys, big thumb!

Happens to me too, more so with this phone than the old one, I think. This particular typo just happened to confuse me.

Can't we say then that when it hits the singularity it's separated one million years in time from the other particle?

No, because the separation between the events is spacelike. You can measure the distance between them in seconds (more usually called light seconds in this context) if you want, but it's still a spacelike separation so you're still describing something more like a distance than a time. Certainly two people falling in can take paths that mean their ages are different at death (as your example of hovering a million years), but the events of their reaching the singularity are still spacelike separated. They can always be described as simultaneous, or as either one before the other. The singularity is not a place at which you arrive, it is a time, no matter how long it took you to get there.

 

[BoraxZ]

Yeah, I see what you mean. 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?

I still don't see why spaghettification can't occur in a spatially radial direction when you fall along. The particle in front of you stretches away from you doesn't it? How fast does that happen?

 

[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.