B Spagettification and a singularity

[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.

 

[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).