Conditions for spacetime to have flat spatial slices

[George Jones]

Heh, good phrase. Can you give a reference? I've read a fair amount of Penrose's writing (at least his writing for the lay reader) and I haven't come across this one.

Look at pages 189-190 in the hardcover edition of Penrose's Road to Reality.

The diagram below is Schwarzschild spacetime in Painleve(-Gullstrand) coodinates \left(T,R\right), which are related to the usual Schwarzschild coordinates \left(t,r\right) by

<br /> \begin{equation*}<br /> \begin{split}<br /> T &amp;= t+4M\left( \sqrt{\frac{r}{2M}}+\frac{1}{2}\ln \left| \frac{\sqrt{\frac{r}{2M}}-1}{\sqrt{\frac{r}{2M}}+1}\right| \right)\\<br /> R &amp;= r.<br /> \end{split}<br /> \end{equation*}<br />

On the diagram, the R-axis is horizontal and the T-axis is vertical, the black line is the worldline of an observer who freely falls radially from rest at infinity, and the event horizon is the vertical line R = 1. At three events on the observer's worldline (outside the horizon, on the horizon, and inside the horizon) I have plotted forward light cones in red. In green, I have plotted lines of constant Schwarzschild t that go through the worldline events inside and outside the horizon. These green lines would be horizontal lines on a Schwarzschild \left(t,r\right) grid which has t as the vertical axis and r as the horizontal axis.The green lines both asymptotically approach the event horizon as T \rightarrow -\infty.

Of course, coordinate time in these coordinates is not actual proper time for any observer except at r=infinity. You used an interesting phrasing earlier:

"suppose we have a coordinate system where coordinate time directly represents the proper time of some family of observers"

Do you mean something other than equals for "directly represents"? Or are you thinking of some simple transform of the standard Schwarzschild coordinates that normalized t to equal tau ?

Suppose the observer's watch is set such that it reads zero when the observer "hits" the singularity R = 0. Then, the events on the observer's (black) worldline all have their T coordinates equal to the observer's watch readings.

Just to check that I've got this right, after some digestion, here's my grasp of these four items, slightly out of order:

2) The transformation between t and T doesn't "change the direction" of the integral curves; it just reparametrizes them.

Yes. Sorry, I going to be a bit pedantic. More pecisely, for every T-integral curve parametrized by T (T-integral curves don't have to be parametrized by T), there is a t-integral curve parametrized by t (t-integral curves don't have to be parametrized by t) such that the two integral curves differ by a constant shift of curve parameter. Even without shiftting parameter, however, every T-integral curve is already a t-integral curve (and vice versa).

This means that, in both coordinate systems, we can use the integral curves of "time" to uniquely label spatial points (each curve has constant values of r = R, \theta, and \varphi).

Yes.

For the rest of this, I'll leave out the angular coordinates (assume them held constant) and only talk about "time" and "radius".

1) The lines of constant T, which are integral curves of \partial_{R}, "cut at a different angle" from the lines of constant t, which are integral curves of \partial_{r}. So even though the integral curves of "time" stay the same, the "spatial slices" cut through them can be different if they're cut at different angles.

Yes.

3) It looks to me like the 4-velocity you gave, which gives us 4), also gives us 3), since it makes it obvious that the 4-velocity is *not* just \partial_{T}, so the integral curves of the 4-velocity can't be the same as the integral curves of \partial_{T}. The integral curves of the 4-velocity are "tilted inward", while the integral curves of \partial_{T} (and thus, of course, \partial_{t}) are "vertical".

Yes.

4) Since the 4-velocity "tilts inward", the integral curves of \partial_{R}, to be orthogonal to them, must "tilt downward" relative to the integral curves of \partial_{r}, which are "horizontal".

Oops, I think this should be "tilt upward", since this is spacetime, not space, so "orthogonal" works differently.

On a \left(t,r\right) diagram, the nature of an integral curve of \partial_{R} depends on which side of the horizon the curve lies. Here, I think you mean outside the horizon. On an \left(t,r\right) diagram, integral curves of \partial_{R} look like the natives of the green curves below.

5) The integral curves of \partial_{R} are always spacelike (horizontal lines outside all lightcones below). Contrast this with the integral curves of \partial_{r}, which are spacelike outside the horizon (right green line outside lightcone) and timelike inside (left green line inside lightcone).

6) \left(T,R\right) coodinates are defined on the horizon (unlike \left(t,r\right) coodintes), with T lightlike on the horizon (part of red lightcone is vertical).

7) T is timelike outside the horizon, but all (four) Painleve coordinates are spacelike inside the horizon!

[PLAIN]http://img832.imageshack.us/img832/4121/painlevegullstrand.jpg

 

[PAllen]

Suppose the observer's watch is set such that it reads zero when the observer "hits" the singularity R = 0. Then, the events on the observer's (black) worldline all have their T coordinates equal to the observer's watch readings.

You responded this way to my observation:

"Of course, coordinate time in these coordinates is not actual proper time for any observer except at r=infinity. You used an interesting phrasing earlier:"

I should have said "any constant r observer" rather than "any observer". I thought the context was clear (implied by r=infinity, for example), but perhaps not. Anyway, thanks, it is useful to know this.

 

[PeterDonis]

Look at pages 189-190 in the hardcover edition of Penrose's Road to Reality.

Hmm, I have that book but I must have missed the quote. I'll see if I can dig it out of the pile of books awaiting shelving...

Here, I think you mean outside the horizon.

Yes, I did. Thanks for the detailed response and diagram, it makes things very clear.

7) T is timelike outside the horizon, but all (four) Painleve coordinates are spacelike inside the horizon!

I hadn't thought about this before, but it looks to me like this follows from the Painleve metric, since the coefficient g_{TT} is the same as the coefficient g_{tt} in the Schwarzschild metric, and changes sign for r < 2M.

This does bring up another question, which is whether such a coordinate system really "qualifies" as a coordinate system, since there is no timelike coordinate inside the horizon. Is there some way to tell "by inspection" that the system is still "OK" (where "OK" means something like "covers everything we want it to cover"--in this case that would be the exterior and the future interior)? Or does one just need to turn the crank and look at the integral curves of the coordinates in some other system, e.g., Kruskal, that is known to cover the entire manifold?

 

[JDoolin]

Ok, good, that makes it clear what the lightcone in your model means--except that this...



...is *not* correct if what you said above is true. There is no "horizontal plane through (0, 0)" in the FWD. See my comments further below.



This is correct, because, as I just noted, there is *no* "horizontal line" in the FWD (I said this in a previous post as well). The FWD includes lines that get as close to "horizontal" as you like, but none that are exactly horizontal. Yes, that means that this...



...is true; that's usually how "conformal" diagrams work. (A Mercator projection of the Earth's surface, for example, is "conformal" in this sense--it maps the North and South poles to horizontal lines, not points.) But that does *not* mean that this...



is true. Consider the Mercator projection again: the lines of longitude (great circles through the poles) are mapped to vertical lines, which *appear* to meet the poles at "different places". But that's an artifact of the "infinite distortion" that the projection makes at the poles. In the same way, the *apparent* "multiple light rays" from the initial singularity in the CPD are an artifact of the "infinite distortion" that this diagram makes at the initial singularity.

How do we deal with this? The only really consistent way is to accept that these "conformal" diagrams *cannot* actually represent the singularities (just as we don't actually use the Mercator projection at the poles). What the "multiple light rays" in the conformal diagram are actually indicating is that, at very short times after the initial singularity, worldlines which emerged from that singularity "in different directions" will be causally disconnected; the more "different" the initial directions are, the longer it will take for the worldlines to become causally connected again. The light that is reaching is now from "close to the big bang" is coming from worldlines that emerged from the big bang in a direction that was "more different" from ours than light that reached us from close to the big bang some time ago.

I realize the above is a somewhat vague description; when I have more time I can try to make it more precise if needed.



Strictly speaking, this is false, although it's sometimes used colloquially to describe what I was describing above, that events very close to the big bang happened throughout the universe (in the sense of causal disconnection I gave above).



As I noted above, the FRW models do not claim that we can see light (or any other signal) "from the big bang" itself. The earliest *photons* we can see in the FRW models are, as you say, those from the time of recombination. However, the FRW model would predict that we could see other radiation from earlier times (e.g., neutrinos from the electroweak phase transition, or gravitational waves from even earlier times). We don't currently have any way of testing such predictions because of our poor ability to detect any kind of radiation other than electromagnetic.



I reached this point in the e-book today, as it happens. I still find the "infinite density" part of the model physically unreasonable, but I agree that *if* you stipulate that the density goes to infinity at the "photon shell", there would be no possibility of anything coming in from outside the shell.



This is one reason (but hardly the only reason) that I find the infinite density physically unreasonable.



Another reason I find the infinite density physically unreasonable is that it should result in infinite curvature at the "photon shell", which, aside from any other objections, would contradict the initial assumption of a flat background Minkowski spacetime. Even if spacetime *inside* the shell were flat (which it could be since that's a general result for inside a symmetrical spherical shell even in Newtonian gravity), the *complete* spacetime in which everything is embedded could not be.



This is much closer, I think, but I still have a couple of comments:

(1) According to General Relativity, there are no situations where the EFE does not apply. It always does. There are certainly a wide variety of particular *solutions* to the EFE, among which are the various spacetimes we've been discussing (Schwarzschild, Kerr, FRW, etc.), and which specific solution applies in a particular case will depend on the distribution of matter. But the EFE, as the equation to be solved, applies in every case. So if we disagree on the distribution of matter, we may well disagree on which specific solution to the EFE to apply, but if we accept GR, we *have* to agree that the EFE applies. If you don't accept that, you don't accept GR.

(The only caveat to the above is what I've said before about spacetime singularities: there the EFE itself tells us it can't apply. But we can get as close to the singularities as we like and still apply the EFE.)

(2) Specifying a "matter distribution" in order to solve the EFE can be done in a variety of ways; it can, as you say, be "general", but you may not be appreciating just how general it can be. For example, to obtain the FRW solutions, we specify: "The matter distribution is a perfect fluid, and we'll write the solution in coordinates in which that fluid is isotropic." That's all. Similarly, to obtain the Schwarzschild solution, we specify: "There is no matter--the stress-energy tensor is identically zero--and the solution must be spherically symmetric." (As you can see, often our "specification" takes the form of symmetry properties that the solution must satisfy.)

This last post comes off as though you are correcting me, rather than acknowledging that we really are talking about two distinct theories. Do you understand that I am describing TWO DIFFERENT models yet?

The Milne Model (Milne Minkowski Diagram MMD), where the Big Bang is ONE EVENT, and the horizontal plane in the FWD is mapped to the light-cone.

The Standard Model (Comoving Particle Diagram CPD), where the Big Bang is MANY EVENTS, and the horizontal plane in the FWD does not exist.

I appreciate your objection to Milne Model, that it has "infinite density" and that is a problem for you. t least I can see that you are actually familiar with the model. But infinite density is a natural extension of lorentz contraction and time dilation as rapidities go from -infinity to infinity. It's not something that is simply assumed; it is something that follows logically from Special Relativity.

I also appreciate that the EFE's start with a symmetry. The Milne Model has the following symmetry: It is the ONLY distribution of matter which is invariant under Lorentz Transformation.

Finally, I wanted to ask whether this transformation (where a single event gets mapped to multiple locations) happens in other metrics; schwarszchild, Kerr, etc.? You seem comfortable with it, as though obviously, it happens at the big bang, but is there anything other than the imperfect analogy with the Mercator Projection which makes it follow naturally or inevitably?

 

[PeterDonis]

This last post comes off as though you are correcting me, rather than acknowledging that we really are talking about two distinct theories. Do you understand that I am describing TWO DIFFERENT models yet?

I haven't been sure, because some of the things you've been saying seem to imply that there is at least a correspondence between the models; for example, this:

The Milne Model (Milne Minkowski Diagram MMD), where the Big Bang is ONE EVENT, and the horizontal plane in the FWD is mapped to the light-cone.

The Standard Model (Comoving Particle Diagram CPD), where the Big Bang is MANY EVENTS, and the horizontal plane in the FWD does not exist.

If the Milne model is really supposed to be a "different model", then where does the "horizontal plane in the FWD is mapped to the light-cone" come from? That correspondence between the FWD and a "light cone" diagram, for example the "SR" diagram on Ned Wright's page, *only* applies if we are talking about the FRW model with k = -1 and zero density (as Wright says on his web page). In fact, that correspondence is the basis for the claim that the Milne model is a special case of the FRW models (the case with k = -1 and zero density), which you reject. If you are not talking about the FRW model but about some other model, then I don't see how you're coming up with a correspondence between your "light cone" diagram and the FWD.

But infinite density is a natural extension of lorentz contraction and time dilation as rapidities go from -infinity to infinity. It's not something that is simply assumed; it is something that follows logically from Special Relativity.

Only if you also assume that spacetime can be flat with a non-zero stress-energy tensor. See next comment.

I also appreciate that the EFE's start with a symmetry. The Milne Model has the following symmetry: It is the ONLY distribution of matter which is invariant under Lorentz Transformation.

First of all, I assume you mean "globally invariant under Lorentz Transformation," since any distribution which is proportional to the metric (e.g., a cosmological constant or a scalar field) will be *locally* invariant under Lorentz Transformation.

Second, and more important, your model assumes that you can have a non-zero distribution of matter without affecting the spacetime geometry. Or, if you insist on avoiding any "geometric" interpretation, your model assumes that you can have a non-zero distribution of matter without any tidal gravity effects--that any pair of freely moving particles in your model will have the same constant relative velocity (each one as seen by the other) for all time. This is known to be false--for example, because of the curvature in the Hubble diagram that we discussed before.

To be clear, I am not talking here about "local" effects such as bending of light by the Sun, but about "global" effects, about the relative velocity of "freely moving" particles on a cosmological scale (the ones whose worldlines are straight lines radiating out from the Big Bang event in your diagram). In GR, such particles, even though they are freely falling (they feel no acceleration), can have relative velocities that vary with time. This shows up in our observations as a variation in the "Hubble constant"--the slope of the curve in the Hubble diagram--with time. According to the Milne model, this is impossible--this should be obvious from the fact that, as you say, the Milne model is based on logical deductions from SR, since in SR there can be no such variation with time in the relative velocity of freely falling objects (i.e., objects moving on inertial worldlines). This is why GR was necessary--because in the presence of gravity (i.e., when the effect of mass-energy on the behavior of inertial worldlines is significant), freely falling objects can change their relative velocity with time (in other words, tidal gravity is present), and SR cannot account for that.

You can see this "curvature of freely falling worldlines" in Ned Wright's diagram of the "critical density" case (the "FPD" version). Notice that in that diagram, the worldlines radiating out from the Big Bang curve inward towards each other--unlike the "zero density" diagram, where they are straight. This is the effect of non-zero mass-energy (i.e., gravity) on freely falling worldlines (or "spacetime geometry" in the usual terminology). The usual pop-science way of describing this is that "the gravitational attraction of the mass-energy in the universe causes the expansion of the universe to slow down." (This terminology was invented before we discovered that, for the last few billion years or so, the expansion has actually been "speeding up", which is why dark energy has been added to the "standard" cosmological model--Ned Wright's diagrams don't cover that case, although I believe he discusses it elsewhere on his cosmology site.)

Finally, I wanted to ask whether this transformation (where a single event gets mapped to multiple locations) happens in other metrics; schwarszchild, Kerr, etc.? You seem comfortable with it, as though obviously, it happens at the big bang, but is there anything other than the imperfect analogy with the Mercator Projection which makes it follow naturally or inevitably?

Mathematically, it's fairly easy to construct transformations that do weird things like this. For example, the transformations used to construct Penrose diagrams map various points or lines at "infinity" to finite coordinate values (see the Wikipedia page here: http://en.wikipedia.org/wiki/Penrose_diagram). There's nothing inconsistent about them; you just have to get used to how they work.

As far as other metrics are concerned, yes, there are transformations often used in GR that have similar effects. For example, in Schwarzschild coordinates, there appears to be an entire infinite line at the horizon, r = 2M, t = minus infinity to plus infinity, that actually, physically, is just a point, as you can see by transforming to Kruskal coordinates, where that entire line becomes the single point at the center of the diagram. (Here I've been ignoring the angular coordinates; when we put them back in, the "point" is actually a 2-surface.) This transformation also maps the "point" at t = infinity in Schwarzschild coordinates to an entire null line (the 45-degree line between regions I and II in the diagram with a yellow background on the Wikipedia page here: http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates); this null line, the "future horizon", is where all the interesting physics at the horizon actually happens, and it is "invisible" in Schwarzschild coordinates, which often leads to confusion if those coordinates are taken too literally.

The fact that stuff like this can happen is a big reason why physicists are hesitant to attribute too much meaning to coordinates; you always have to check the physical invariants to see what's really going on. For example, I asserted just now that the apparent "line" at the horizon in Schwarzschild coordinates is actually just a point--or, if we include the angular coordinates, what appears to be a 3-surface is actually just a 2-surface. How do I know this is right? (Put another way, how do I know that the description in Kruskal coordinates is the "right" one physically?) Because I can compute the physical 3-volume of the apparent 3-surface, using the metric, and find that it is zero (because the metric coefficient g_{tt} is zero at r = 2M in Schwarzschild coordinates). A similar computation in FRW coordinates shows me that the initial singularity is, physically, a point (because a(t) is zero there, so the spatial metric vanishes), even though it looks like a line (actually a 3-surface, if we include the angular coordinates) in the "conformal" diagram. (Here I do really mean a literal point--zero dimensions--unlike the horizon of a black hole, which is physically a 2-surface--we can compute its area and find that it's non-zero, because the spatial part of the metric doesn't vanish completely. In the FRW case, the entire spatial metric vanishes at the initial singularity.)

 

[Passionflower]

For example, in Schwarzschild coordinates, there appears to be an entire infinite line at the horizon, r = 2M, t = minus infinity to plus infinity, that actually, physically, is just a point,...

A point, please explain? Are you saying that all travelers from different times on a given radial angle meet at the same time at this point?

For example, I asserted just now that the apparent "line" at the horizon in Schwarzschild coordinates is actually just a point--or, if we include the angular coordinates, what appears to be a 3-surface is actually just a 2-surface. How do I know this is right? (Put another way, how do I know that the description in Kruskal coordinates is the "right" one physically?) Because I can compute the physical 3-volume of the apparent 3-surface, using the metric, and find that it is zero (because the metric coefficient g_{tt} is zero at r = 2M in Schwarzschild coordinates).

What volume are you computing?

Could you please explain a bit more what you mean here, I am not sure I can agree here, but I probably misunderstand.

 

[PeterDonis]

A point, please explain? Are you saying that all travelers from different times on a given radial angle meet at the same time at this point?

No, as I noted elsewhere in the post, it's actually a 2-surface when the angular coordinates are taken into account. Also, I was *not* saying that any traveler crossing the horizon passes through this point; as I noted further on, all the actual physics at the horizon is on the "future horizon" null line that runs at 45 degrees up and to the right from the center point in the Kruskal diagram. That's where worldlines crossing the horizon go, and they can cross at anyone of an infinite number of different events.

What volume are you computing?

The "3-volume" spanned by r = 2M, t = minus infinity to plus infinity, theta = 0 to pi, phi = 0 to 2 pi. Since the metric coefficient g_{tt} is zero at r = 2M, the integral corresponding to this 3-volume vanishes, indicating that what looks like a 3-volume in Schwarzschild coordinates is actually, at most, a 2-surface. (We can verify that it is, in fact, a 2-surface and not something with even fewer dimensions by, for example, integrating over the full range of angular coordinates at the "point" at the center of the Kruskal diagram, which gives the nonzero area of the horizon.)

 

[Passionflower]

That's where worldlines crossing the horizon go, and they can cross at anyone of an infinite number of different events.

If the worldlines don't cross there then there must be a dimension to separate them right? How do you explain that?

The "3-volume" spanned by r = 2M, t = minus infinity to plus infinity, theta = 0 to pi, phi = 0 to 2 pi. Since the metric coefficient g_{tt} is zero at r = 2M, the integral corresponding to this 3-volume vanishes, indicating that what looks like a 3-volume in Schwarzschild coordinates is actually, at most, a 2-surface. (We can verify that it is, in fact, a 2-surface and not something with even fewer dimensions by, for example, integrating over the full range of angular coordinates at the "point" at the center of the Kruskal diagram, which gives the nonzero area of the horizon.)

Perhaps you could show me the formulas you use, let's say we take the "volume" (I am still not sure what physical volume you are talking about) of r=4M and r=2M so we can all see the difference between a surface at r=4M and r=2M.

Just to remind everybody, the r-coordinate is not a radius or a measure of distance, the r-coordinate is instead a function of an area!

So r is defined as:

<br /> r = {1 \over 2}\,\sqrt {{\frac {A}{\pi }}}<br />

Let's keep that in mind.

 

[JDoolin]

I haven't been sure, because some of the things you've been saying seem to imply that there is at least a correspondence between the models; for example, this:



If the Milne model is really supposed to be a "different model", then where does the "horizontal plane in the FWD is mapped to the light-cone" come from? That correspondence between the FWD and a "light cone" diagram, for example the "SR" diagram on Ned Wright's page, *only* applies if we are talking about the FRW model with k = -1 and zero density (as Wright says on his web page). In fact, that correspondence is the basis for the claim that the Milne model is a special case of the FRW models (the case with k = -1 and zero density), which you reject. If you are not talking about the FRW model but about some other model, then I don't see how you're coming up with a correspondence between your "light cone" diagram and the FWD.

If I recall correctly, yes, the Milne model has k=0, a=1. It is Minkowski Spacetime with a density that approaches infinity as you go out from the center toward r=c*t, and it has a density that approaches infinity as to go toward t=0. That is not at all compatible with a zero density.

If anyone thinks they have correctly modeled the Milne universe with a zero density, they are just fooling themselves, and using some kind of circular reasoning or a straw-man argument.

There are two variables in the Friedmann Walker Diagram, the horizontal variable is a "space-like" variable, and the vertical is a "time-like" variable. To map to the Comoving Particle Diagram, I'm not sure exactly how it is done, but I think the vertical "time component" is just mapped straight over, while the horizontal "space component" is some form of velocity * distance. I may be wrong, but I *think* it is the integral of the changing scale factor with respect to the "cosmological time."

The horizontal variable is SPACE, and the vertical variable is TIME; some kind of "Absolute" or "cosmological" time, which really doesn't exist in the realm of Special Relativity.

What the Milne model does is treats the horizontal variable in the FWD as sort of a "rapidity-space" The mapping from the FWD to the MMD assumes that the meaning of the FWD "space-like" variable is distance = rapidity * proper time. For a set of particles all coming from an event (0,0), giving the rapidity and proper time for a particle uniquely defines its position in space and time. To map from the FWD to the MMD, you are simply mapping: (d'=rapidity*proper time,t'=proper time) to (d=space, t=time).

To map from the FWD to the CPD, you are mapping (d'="Stretchy" Velocity * Cosmological Time, t'=Cosmological Time) to (d=Space,t=Cosmological Time).

I'll see if I can express this as mathematically and unambiguously as I can, so that if I'm wrong it can be corrected.

\begin{matrix} FWD \mapsto CPD \text{ as }(d\int a(\tau)d\tau,\tau)\mapsto(d,\tau) \\ d=Proper Distance = Cosmological Distance \\ \tau=Proper Time=CosmologicalTime \\ a(\tau)=ScaleFactor \end{matrix}​

On the other hand, the Milne mapping looks like this:

\begin{matrix} FWD \mapsto MMD \text{ as }(\varphi \cdot\tau,\tau)\mapsto(v \cdot t,t) \\ \varphi=rapidity \\ \tau=proper time \\ v = velocity \\ t = time \end{matrix}​

As you can see, the Milne mapping is linear; there's no changing scale factor. The relation between rapidity and velocity and distance, time, and proper time is the same as is usually given in Special Relativity.

Rapidities between -infinity and +infinity map to velocities between -c and +c. So the horizontal plane (representing infinite rapidity) in the Friedmann Walker Diagram maps to the light-cone in the Milne Minkowski Diagram.

So after any corrections to the FWD to CPD mapping, I wonder if you are yet convinced that we are talking about two different mappings? Can you see that the rapidity=infinity line is included in the Milne Model? Can you see that the distance vs. time relation is fundamentally different? Can you see how an infinite density naturally results from this mapping? Can you see how the Milne Model has only one event at x=0,t=0, while the CPD has an infinite number of events at x=0,t=0?

Only if you also assume that spacetime can be flat with a non-zero stress-energy tensor. See next comment.

First of all, I assume you mean "globally invariant under Lorentz Transformation," since any distribution which is proportional to the metric (e.g., a cosmological constant or a scalar field) will be *locally* invariant under Lorentz Transformation.

Yes, I mean "globally invariant"

Second, and more important, your model assumes that you can have a non-zero distribution of matter without affecting the spacetime geometry. Or, if you insist on avoiding any "geometric" interpretation, your model assumes that you can have a non-zero distribution of matter without any tidal gravity effects--that any pair of freely moving particles in your model will have the same constant relative velocity (each one as seen by the other) for all time. This is known to be false--for example, because of the curvature in the Hubble diagram that we discussed before.

I may have implied that "any pair" of freely moving particles would have the same constant relative velocity, but I must back off on that. But it is not "any pair." It is a specific set of particles which are completely undisturbed after the moment of the Big Bang which will maintain a constant relative velocity.

You have the "Relativity, Gravitation, and World Structure" e-book; check section 112, the list of properties, item 10, Milne says:

"Every particle of the system is in uniform radial motion outward from any arbitrary particle O of the system, and the acceleration of every particle in the system is zero. But the acceleration of a freely projected particle, other than the given particles, is not zero."

I think what is happening is that every particle that is at the center of mass remains at the center of mass, but if you move away from that center of mass, you'll be attracted toward it.

The unusual thing is that accelerating toward the center of mass actually causes the center to move away from you rather than toward you. That results in inflation.

To be clear, I am not talking here about "local" effects such as bending of light by the Sun, but about "global" effects, about the relative velocity of "freely moving" particles on a cosmological scale (the ones whose worldlines are straight lines radiating out from the Big Bang event in your diagram). In GR, such particles, even though they are freely falling (they feel no acceleration), can have relative velocities that vary with time. This shows up in our observations as a variation in the "Hubble constant"--the slope of the curve in the Hubble diagram--with time. According to the Milne model, this is impossible--this should be obvious from the fact that, as you say, the Milne model is based on logical deductions from SR, since in SR there can be no such variation with time in the relative velocity of freely falling objects (i.e., objects moving on inertial worldlines).

This is the elephant in the room that I wanted to talk about earlier. Acceleration.
The Big Bang is one explosion, but particle decay should lead to secondary explosions. If you have secondary (and tertiary, etc.) explosions of matter, this naturally leads to variations in the "Hubble Constant" based on logical deductions from SR. Our local Hubble Constant is based not on the age of the universe, but on the time since the most recent explosion. Only the most distant Hubble Constant tells you the age of the universe.

This is why GR was necessary--because in the presence of gravity (i.e., when the effect of mass-energy on the behavior of inertial worldlines is significant), freely falling objects can change their relative velocity with time (in other words, tidal gravity is present), and SR cannot account for that.

You can see this "curvature of freely falling worldlines" in Ned Wright's diagram of the "critical density" case (the "FPD" version). Notice that in that diagram, the worldlines radiating out from the Big Bang curve inward towards each other--unlike the "zero density" diagram, where they are straight. This is the effect of non-zero mass-energy (i.e., gravity) on freely falling worldlines (or "spacetime geometry" in the usual terminology). The usual pop-science way of describing this is that "the gravitational attraction of the mass-energy in the universe causes the expansion of the universe to slow down." (This terminology was invented before we discovered that, for the last few billion years or so, the expansion has actually been "speeding up", which is why dark energy has been added to the "standard" cosmological model--Ned Wright's diagrams don't cover that case, although I believe he discusses it elsewhere on his cosmology site.)

The pop-science way of describing these things is not nearly sufficient to give any hint to me about what they are actually measuring. I'm interested in whatever data they've gathered to determine that the "expansion of the universe is speeding up" but I think it more likely that whatever the effect, the cause is much more likely to be our galaxy's acceleration toward the receding center of mass, rather than some universal cosmological scale factor increasing at a faster rate.

Mathematically, it's fairly easy to construct transformations that do weird things like this. For example, the transformations used to construct Penrose diagrams map various points or lines at "infinity" to finite coordinate values (see the Wikipedia page here: http://en.wikipedia.org/wiki/Penrose_diagram). There's nothing inconsistent about them; you just have to get used to how they work.

As far as other metrics are concerned, yes, there are transformations often used in GR that have similar effects. For example, in Schwarzschild coordinates, there appears to be an entire infinite line at the horizon, r = 2M, t = minus infinity to plus infinity, that actually, physically, is just a point, as you can see by transforming to Kruskal coordinates, where that entire line becomes the single point at the center of the diagram. (Here I've been ignoring the angular coordinates; when we put them back in, the "point" is actually a 2-surface.) This transformation also maps the "point" at t = infinity in Schwarzschild coordinates to an entire null line (the 45-degree line between regions I and II in the diagram with a yellow background on the Wikipedia page here: http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates); this null line, the "future horizon", is where all the interesting physics at the horizon actually happens, and it is "invisible" in Schwarzschild coordinates, which often leads to confusion if those coordinates are taken too literally.

The fact that stuff like this can happen is a big reason why physicists are hesitant to attribute too much meaning to coordinates; you always have to check the physical invariants to see what's really going on. For example, I asserted just now that the apparent "line" at the horizon in Schwarzschild coordinates is actually just a point--or, if we include the angular coordinates, what appears to be a 3-surface is actually just a 2-surface. How do I know this is right? (Put another way, how do I know that the description in Kruskal coordinates is the "right" one physically?) Because I can compute the physical 3-volume of the apparent 3-surface, using the metric, and find that it is zero (because the metric coefficient g_{tt} is zero at r = 2M in Schwarzschild coordinates). A similar computation in FRW coordinates shows me that the initial singularity is, physically, a point (because a(t) is zero there, so the spatial metric vanishes), even though it looks like a line (actually a 3-surface, if we include the angular coordinates) in the "conformal" diagram. (Here I do really mean a literal point--zero dimensions--unlike the horizon of a black hole, which is physically a 2-surface--we can compute its area and find that it's non-zero, because the spatial part of the metric doesn't vanish completely. In the FRW case, the entire spatial metric vanishes at the initial singularity.)

I'm afraid I'm at a loss for following most of this. I'm not entirely convinced you have anywhere a single event which is mapped to multiple places (except for the FRW case where a(t) shrinks to zero).

If we could focus on the nature of the Schwarzschild coordinates; I'm afraid you'll have to go completely remedial to explain it to me, because I don't know the equations for the Schwarzschild coordinates or the reasoning. All I know is that time slows down as you get near the surface, and this causes light to turn toward the planet and get a higher frequency. When you're talking about the "line" at the horizon which is actually just a point, or the three-surface which is really just a two-surface, I'm not sure precisely what is meant.

 

[PeterDonis]

If the worldlines don't cross there then there must be a dimension to separate them right? How do you explain that?

There is--it's the Kruskal V coordinate (in the null U, V formulation of Kruskal coordinates, which uses V to label events along the future horizon).

Perhaps you could show me the formulas you use, let's say we take the "volume" (I am still not sure what physical volume you are talking about) of r=4M and r=2M so we can all see the difference between a surface at r=4M and r=2M.

The 3-volume is a 3-volume in 4-dimensional spacetime; the only difference between it and a "normal" spatial 3-volume is that one of the dimensions is the time coordinate. All you do is integrate the appropriate volume measure over the appropriate range of coordinates. The volume measure is just the product of the distance measures along each coordinate; each distance measure is the square root of the appropriate metric coefficient times the coordinate differential. (Some of this may be simpler because the Schwarzschild metric is diagonal; I'd have to go back and do some review to remind myself of what, if any, complications arise when dealing with a non-diagonal metric such as the Kerr metric.) So for a constant value of r, the "volume integral" looks like:

V = \int_{t_{1}}^{t_{2}} \sqrt{- g_{tt}} dt \int_{0}^{\pi} \sqrt{g_{\theta \theta}} d\theta \int_{0}^{2 \pi} \sqrt{g_{\varphi \varphi}} d\varphi = 4 \pi r^{2} \sqrt{1 - \frac{2M}{r}} \left( t_{2} - t_{1} \right)

where I've included the minus sign in front of {g_{tt}} because of the opposite sign of that metric coefficient. At r = 4M, this integral gives a positive value (how large a value depends on the range of t we choose--we can make it infinite by letting t cover its full range of - \infty to \infty), but at r = 2M, the integral vanishes identically.

Just to remind everybody, the r-coordinate is not a radius or a measure of distance, the r-coordinate is instead a function of an area!

Yes, it is, which makes the integral over the angular coordinates very easy to do at a constant value of r (that's why there's just 4 \pi r^{2} in the integral above, instead of some more complicated function of r). But the r coordinate is still being used to *label* events along the radial dimension, which is different from the two angular dimensions. It's not a direct measure of radial distance because the metric coefficient g_{rr} varies, but it's still a coordinate in the radial direction.

 

[Passionflower]

There is--it's the Kruskal V coordinate (in the null U, V formulation of Kruskal coordinates, which uses V to label events along the future horizon).

So then if you agree that different observers at different times can pass the event horizon at the same physical location then clearly there must be a line and not a point for a given theta and phi?

I have to get back on your volume calculation, I am not very encouraged by what I see (but clearly that must be my shortcoming).

 

[PeterDonis]

If I recall correctly, yes, the Milne model has k=0, a=1.

Not if it's a different model from the FRW models; the k and a parameters apply to those models. The FRW model that is claimed to be an alternate formulation of the Milne model has k = -1, a(t) = t (linear variation, starting with a = 0 at t = 0, the Big Bang) in the FRW parameter system; but you don't agree that that model actually is an alternate formulation of the Milne model.

It is Minkowski Spacetime with a density that approaches infinity as you go out from the center toward r=c*t, and it has a density that approaches infinity as to go toward t=0. That is not at all compatible with a zero density.

If anyone thinks they have correctly modeled the Milne universe with a zero density, they are just fooling themselves, and using some kind of circular reasoning or a straw-man argument.

No, they are pointing out the same physical issue that I pointed out in my last post, that a nonzero density will curve freely falling worldlines, so you can't have Minkowski spacetime with a non-zero density. See below.

So after any corrections to the FWD to CPD mapping, I wonder if you are yet convinced that we are talking about two different mappings? Can you see that the rapidity=infinity line is included in the Milne Model? Can you see that the distance vs. time relation is fundamentally different? Can you see how an infinite density naturally results from this mapping? Can you see how the Milne Model has only one event at x=0,t=0, while the CPD has an infinite number of events at x=0,t=0?

I understand everything you're saying about your diagram of the Milne model. You're not correct that the CPD has an infinite number of events at x = 0, t = 0; it's only one event., and it has coordinates t = 0, but x, y, z undefined (in the same way the longitude of the South Pole is undefined), and it looks like a line in the diagram only because we need to have a reference for the "limit point" of all the vertical worldlines. As you'll see from my comments near the end of this post, I'm afraid I can't really do justice to the mathematical side of this, but I'll try to briefly explain what I think it's supposed to mean physically.

The "conformal" diagram of the FRW model (your "CPD") is intended to make it easy to see the causal structure of the spacetime, and that's all--it highly distorts everything else in order to ensure that light rays always travel on 45 degree lines, so it's easy to see the boundaries of light cones. Pick two worldlines that are widely separated in the conformal diagram, and follow them back towards the "initial singularity" line at the bottom. You'll see that the past light cones of those two worldlines stop overlapping while they're still a fair distance away from the initial singularity. If you pick two worldlines that are closer, their past light cones stop overlapping closer to the singularity. As worldlines get closer and closer, their past light cones get closer and closer to the singularity before they stop overlapping. But for any time after the initial singularity, there will be some "conformal separation" (separation horizontally in the conformal diagram) of worldlines for which the past light cones have stopped overlapping. The conformal diagram was constructed to make all this visually obvious, but the price you pay is making the single event at t = 0 (all space coordinates undefined) look like a line. Actually, the boundary line itself is *not* part of the spacetime; it's just a convenient way of showing, visually, that the vertical worldlines "end" at the initial singularity (which, technically, is not part of the spacetime either, as I've said before--the density and curvature are infinite there, so physically we expect new physics, perhaps quantum gravity, to take over before that point is reached).

I may have implied that "any pair" of freely moving particles would have the same constant relative velocity, but I must back off on that. But it is not "any pair." It is a specific set of particles which are completely undisturbed after the moment of the Big Bang which will maintain a constant relative velocity.

You have the "Relativity, Gravitation, and World Structure" e-book; check section 112, the list of properties, item 10, Milne says:

"Every particle of the system is in uniform radial motion outward from any arbitrary particle O of the system, and the acceleration of every particle in the system is zero. But the acceleration of a freely projected particle, other than the given particles, is not zero."

Yes, thanks for including that quote as it clarifies the point I was trying to make. The "specific set of particles" you mention is the set of "given particles" in Milne's terminology. In your diagram, they appear as straight lines radiating out from the big bang event, all within the "envelope" of the light cone. The point I am making is that in the presence of gravity (i.e., with a non-zero density), GR predicts that those lines cannot be straight lines! As I noted before, in the diagram of the "critical density" FRW model on Ned Wright's page, you see the corresponding lines curve inward towards each other, whereas in the "zero density" model, they are straight.

So here we have a clear case where standard GR and the Milne model make different physical predictions: Milne's model says that we can have a non-zero density in the universe but still have a family of "given particles" that maintain constant relative velocity for all time; GR says that's not possible and predicts that with non-zero density we will see the relative velocities even of "comoving" particles (the ones that correspond to Milne's "given" particles) vary with time.

The Hubble diagram provides a way to test this; but since you commented on that with regard to "inflation", I'll comment further on it below.

The unusual thing is that accelerating toward the center of mass actually causes the center to move away from you rather than toward you. That results in inflation.

This is the elephant in the room that I wanted to talk about earlier. Acceleration. The Big Bang is one explosion, but particle decay should lead to secondary explosions. If you have secondary (and tertiary, etc.) explosions of matter, this naturally leads to variations in the "Hubble Constant" based on logical deductions from SR. Our local Hubble Constant is based not on the age of the universe, but on the time since the most recent explosion. Only the most distant Hubble Constant tells you the age of the universe.

Saying that "our local Hubble constant is based not on the age of the universe, but on the time since the most recent explosion" is not based on direct observation but on a conclusion from your model; you need to give me some way of directly testing this assertion vs. the contrary assertion of standard GR, which is that the "Hubble Constant" at any distance indicates the general "expansion of the universe" at the epoch corresponding to that distance. Since we observe variation in the observed Hubble constant with distance (the curvature in the Hubble diagram), the standard GR model (the FRW model) infers that the relative velocity of "comoving" observers can change with time.

It seems as though you're saying that the Milne model explains the curvature in the Hubble diagram by "secondary explosions". The main problem I see with this is that it should not produce curvature in the diagram; it should produce anisotropy--different slopes of the Hubble plot in different directions (corresponding to different epochs for each "secondary explosion"), but each plot should be a straight line, no curvature (because once the explosion happens, the debris, so to speak, travels in straight lines, with each piece having a constant relative velocity with respect to all other pieces--each "piece" being one of the set of "given particles" created by that particular explosion). We don't see anything like this; we see curvature in the diagram, but it's isotropic.

Even if you push all the secondary explosions back into the "inflation" era, they should still produce anisotropies in, for example, the CMBR, that would have to be larger than those we've observed--at least it seems that way to me, but since your model isn't quantified, it's hard to tell what it predicts for this. But in any case, if all the secondary explosions occurred during the inflation era, they wouldn't affect the Hubble diagram at all, since that only goes back as far as we can see galaxies, quasars, and other objects that have measurable redshifts. So the Hubble diagram, in this case, would indicate the general expansion of the universe, and should *not* have any curvature according to the Milne model.

I'm afraid I'm at a loss for following most of this. I'm not entirely convinced you have anywhere a single event which is mapped to multiple places (except for the FRW case where a(t) shrinks to zero).

If we could focus on the nature of the Schwarzschild coordinates; I'm afraid you'll have to go completely remedial to explain it to me, because I don't know the equations for the Schwarzschild coordinates or the reasoning. All I know is that time slows down as you get near the surface, and this causes light to turn toward the planet and get a higher frequency.

That would take us pretty far afield, and there's a lot of material out there already about Schwarzschild spacetime. (Sorry to punt, more or less, but I just don't think I can do justice to the topic or give you a proper treatment of it in a discussion thread; it really does need to be studied carefully offline, at your own pace and following your own thread of inquiry.) Instead, let me just post a couple of links:

http://casa.colorado.edu/~ajsh/schwp.html

http://www.phy.syr.edu/courses/modules/LIGHTCONE/schwarzschild.html

Unfortunately, the PhysicsForums library doesn't appear to have anything on Schwarzschild spacetime (at least, not anything I could find with a quick search).

Also, all the textbooks on GR spend a fair bit of time talking about Schwarzschild spacetime because it's used so much.

(I'm also pretty much punting at this point, I'm afraid, on the general question of transformations that do weird things like map points to lines, lines to points, etc. As I said in the previous post you quoted, I think it's a mistake to get too wrapped up in what a geometry "looks like" in a particular coordinate system. I'd rather focus on the physics and what is actually observed, which is what I was trying to do with my comments about tidal gravity, curvature in the Hubble diagram, etc. There is a vast body of mathematics behind these various coordinate transformations, but again, I just don't think I can do justice to it here; it's something that really has to be studied at your own pace. Unfortunately I can't point you to any particular texts in this area--the GR textbooks go into this somewhat, but they don't really do it with the sort of rigor that mathematicians do.)

When you're talking about the "line" at the horizon which is actually just a point, or the three-surface which is really just a two-surface, I'm not sure precisely what is meant.

See my previous post in response to Passionflower for a little more info about this point.

 

[PeterDonis]

So then if you agree that different observers at different times can pass the event horizon at the same physical location then clearly there must be a line and not a point for a given theta and phi?

Yes--it's the 45 degree line that goes up and to the right from the center point of the Kruskal diagram. (This line has U = constant, V > 0 in the null Kruskal coordinates.) All worldlines that cross the horizon into the black hole cross somewhere on that line, at some positive V coordinate.

 

[Passionflower]

Yes--it's the 45 degree line that goes up and to the right from the center point of the Kruskal diagram. (This line has U = constant, V > 0 in the null Kruskal coordinates.) All worldlines that cross the horizon into the black hole cross somewhere on that line, at some positive V coordinate.

Ok so you are saying that for a given theta and phi that an observer A can cross the EH before observer an B. That implies that r=rs, phi=0, theta=0 must be a line not a point. How else would you explain it?

 

[PeterDonis]

Ok so you are saying that for a given theta and phi that an observer A can cross the EH before observer an B. That implies that r=rs, phi=0, theta=0 must be a line not a point. How else would you explain it?

It is a line, but the line does not appear in the Schwarzschild chart (either exterior or interior). The line is the one I described; U = 0, V > 0 in the null Kruskal coordinates. (The full "horizon" is actually the pair of crossing lines U = 0 and V = 0, but the portion I've described is where all the worldlines going into the black hole from "region I", the "normal" exterior, cross.) This line also has the r coordinate 2M, as can be seen from the implicit equation for r in terms of U and V, as given for example on the Wikipedia page:

http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates.

I'm using U, V here to refer to what the Wiki page calls the "light cone variant" of the Kruskal coordinates (this is its term for null coordinates, apparently--they put tildes over U, V for these coordinates and use U, V without tildes to denote the spacelike and timelike Kruskal coordinates, which I'm more used to seeing as X, T, or sometimes R, T, as the Wiki page briefly comments). In these coordinates, r is given implicitly by

UV = \left( 1 - \frac{r}{2M} \right) e^{\frac{r}{2M}}

So if either U = 0 or V = 0, r = 2M. But this pair of crossing lines (U = 0 and V = 0) does *not* have a well-defined Schwarzschild t coordinate. The equation for t is either

tanh \left( \frac{t}{4M} \right) = \frac{V + U}{V - U}

outside the horizon, or

tanh \left( \frac{t}{4M} \right) = \frac{V - U}{V + U}

inside the horizon; but if either U = 0 or V = 0 (i.e., on the horizon), both of these equations have no solution (because the tanh function would have to have a value of +/- 1, and it never does, it only asymptotes to those values).

 

[Passionflower]

It is a line

Well good I agree with that. So now we extend this line in the direction of phi and theta what do we get?

Now earlier you wrote:

For example, in Schwarzschild coordinates, there appears to be an entire infinite line at the horizon, r = 2M, t = minus infinity to plus infinity, that actually, physically, is just a point,

So do you agree it is a line or do I misunderstand here?

 

[PeterDonis]

if either U = 0 or V = 0 (i.e., on the horizon), both of these equations have no solution (because the tanh function would have to have a value of +/- 1, and it never does, it only asymptotes to those values).

I should add, to clarify, that if both U = 0 *and* V = 0 (i.e, at the "center point" of the diagram"), both expressions give 0 / 0 on the RHS, and we would have to take a limit somehow to see if the tanh function could satisfy either equation. The usual assumption seems to be that this can somehow be done in such a way as to be able to assign *any* Schwarzschild t value we like to the center point, from - infinity to infinity (but still at r = 2M).

I haven't seen anything specific in the books I've read on how this could be done, but here's one possible way I can think of: if we assume that either V is some function of U that's linear in U, or U is some function of V that's linear in V, then we can apply L'Hopital's rule and differentiate top and bottom to obtain some constant value on the RHS; and by suitable definition of the functions we can make that constant value basically be anything from -1 to 1, so in effect the center point can map to *any* Schwarzschild t value from - infinity to infinity (at r = 2M).

 

[PeterDonis]

Well good I agree with that. So now we extend this line in the direction of phi and theta what do we get?

A spacetime 3-volume. But again, this 3-volume does not appear in the Schwarzschild chart; only a single 2-surface picked out of it does (the one corresponding to the "center point" of the Kruskal diagram). To see the full 3-volume you have to use a chart that covers the "future horizon" line, such as the Kruskal chart. (The ingoing Eddington-Finkelstein chart or the Painleve chart would also work.)

 

[Passionflower]

A spacetime 3-volume. But again, this 3-volume does not appear in the Schwarzschild chart; only a single 2-surface picked out of it does (the one corresponding to the "center point" of the Kruskal diagram).

Yes, but did you claim just the opposite?

 

[PeterDonis]

Yes, but did you claim just the opposite?

No. Here's everything I said in my post #127 that first responded to this question from you:

No, as I noted elsewhere in the post, it's actually a 2-surface when the angular coordinates are taken into account. Also, I was *not* saying that any traveler crossing the horizon passes through this point; as I noted further on, all the actual physics at the horizon is on the "future horizon" null line that runs at 45 degrees up and to the right from the center point in the Kruskal diagram. That's where worldlines crossing the horizon go, and they can cross at anyone of an infinite number of different events.

The "3-volume" spanned by r = 2M, t = minus infinity to plus infinity, theta = 0 to pi, phi = 0 to 2 pi. Since the metric coefficient g_{tt} is zero at r = 2M, the integral corresponding to this 3-volume vanishes, indicating that what looks like a 3-volume in Schwarzschild coordinates is actually, at most, a 2-surface. (We can verify that it is, in fact, a 2-surface and not something with even fewer dimensions by, for example, integrating over the full range of angular coordinates at the "point" at the center of the Kruskal diagram, which gives the nonzero area of the horizon.)

And I've been repeating ever since that yes, there is a line, but the line is *not* covered by the Schwarzschild chart--what looks like a line in the Schwarzschild chart is actually just a single point, which is at the center of the Kruskal diagram. The line, the "real" one that worldlines going into the hole actually cross, is the line going up and to the right at 45 degrees from the center point. And when you add in the angular coordinates, "point" becomes "2-surface" and "line" becomes "3-volume".

 

[Passionflower]

Here is your original posting (I boldfaced the relevant parts):

No, as I noted elsewhere in the post, it's actually a 2-surface when the angular coordinates are taken into account. Also, I was *not* saying that any traveler crossing the horizon passes through this point; as I noted further on, all the actual physics at the horizon is on the "future horizon" null line that runs at 45 degrees up and to the right from the center point in the Kruskal diagram. That's where worldlines crossing the horizon go, and they can cross at anyone of an infinite number of different events.

The "3-volume" spanned by r = 2M, t = minus infinity to plus infinity, theta = 0 to pi, phi = 0 to 2 pi. Since the metric coefficient g_{tt} is zero at r = 2M, the integral corresponding to this 3-volume vanishes, indicating that what looks like a 3-volume in Schwarzschild coordinates is actually, at most, a 2-surface. (We can verify that it is, in fact, a 2-surface and not something with even fewer dimensions by, for example, integrating over the full range of angular coordinates at the "point" at the center of the Kruskal diagram, which gives the nonzero area of the horizon.)

In both parts you say it is actually a 2-surface and not a 3-surface seems indicated by Schw. coordinates. But one posting ago you agreed it is actually a 3-surface? I must be seriously mistaken.

 

[PeterDonis]

In both parts you say it is actually a 2-surface and not a 3-surface seems indicated by Schw. coordinates. But one posting ago you agreed it is actually a 3-surface? I must be seriously mistaken.

You are. Here's the full relevant part of my post #125, which originally prompted your question (but you only quoted part of it in your question):

For example, in Schwarzschild coordinates, there appears to be an entire infinite line at the horizon, r = 2M, t = minus infinity to plus infinity, that actually, physically, is just a point, as you can see by transforming to Kruskal coordinates, where that entire line becomes the single point at the center of the diagram. (Here I've been ignoring the angular coordinates; when we put them back in, the "point" is actually a 2-surface.) This transformation also maps the "point" at t = infinity in Schwarzschild coordinates to an entire null line (the 45-degree line between regions I and II in the diagram with a yellow background on the Wikipedia page here: http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates); this null line, the "future horizon", is where all the interesting physics at the horizon actually happens, and it is "invisible" in Schwarzschild coordinates, which often leads to confusion if those coordinates are taken too literally.

I've stated from the very beginning that "point" and "2-surface" refer only to the "apparent line" from t = - infinity to infinity in *Schwarzschild* coordinates, which is physically just a point (or 2-surface if angular coordinates are included) and appears that way at the center of the Kruskal diagram, and that there *is* a "real" line (or 3-volume if angular coordinates are included) at the horizon that is covered by Kruskal coordinates but which does *not* appear in Schwarzschild coordinates.

 

[Passionflower]

which is physically just a point

Didn't you just agree that, for instance when two observers passing it after each other, it is a line?

 

[PeterDonis]

Didn't you just agree that, for instance when two observers passing it after each other, it is a line?

No. Go back and read my posts again, carefully, and see exactly what I've referred to as a "point" and what I've referred to as a "line", and look at a Kruskal diagram and see how the two relate to each other. Then fill in a couple of timelike worldlines on the diagram, crossing the future horizon line at different events. Where are those events placed, in relation to the point at the center? Do they pass through that point?

 

[Passionflower]

No.

Ok, so you are saying it is a point?

So then I ask you again, observer A passes this point before observer B, clearly their worldlines do not cross there, how do you explain that if there is no line but a point?

 

[PeterDonis]

Ok, so you are saying it is a point?

So then I ask you again, observer A passes this point before observer B, clearly their worldlines do not cross there, how do you explain that if there is no line but a point?

Did you do what I asked in my last post, looking at the Kruskal diagram? You will find that you can draw two timelike worldlines that cross the future horizon line at two different events (points), but neither of those events are the center point of the diagram (the "point" which appears as a line in Schwarzschild coordinates). Please do that before asking this question again.

 

[Passionflower]

Did you do what I asked in my last post, looking at the Kruskal diagram? You will find that you can draw two timelike worldlines that cross the future horizon line at two different events (points), but neither of those events are the center point of the diagram (the "point" which appears as a line in Schwarzschild coordinates). Please do that before asking this question again.

I am not talking about coordinate charts but about physics. An observer A passes the EH at the same spatial location before an observer B, what does that imply?

Too bad Peter, I think you are trying to avoid my questions.

 

[PeterDonis]

I am not talking about coordinate charts but about physics. An observer A passes the EH at the same spatial location before an observer B, what does that imply?

Um, that spacetime has a time dimension? I'm not trying to avoid anything, I just don't understand what point you're trying to make. I've already agreed that observer A and observer B, in your example just above, would cross the horizon (the line U = 0, V > 0 that goes up and to the right at a 45 degree angle from the center of the Kruskal diagram) at two different events. Neither of those events is the point at the center of the diagram.

If it's the bit about "the same spatial location" that's bothering you, I've also already said that the "future horizon" line I described has r = 2M all along it. I gave the function relating r to the Kruskal U and V that makes that clear. It's true that on the Kruskal diagram, the two events where A and B cross the horizon have different *Kruskal* coordinates (different V, or different Kruskal spatial coordinate X or R, if you use those coordinates instead of U and V to label events in the diagram). That's because the Kruskal chart is a different chart that labels events differently. What's the problem?

 

[Passionflower]

What's the problem?

Do you think r=rs, phi=0, theta=0 is a line or a point in spacetime?

 

[PeterDonis]

Do you think r=rs, phi=0, theta=0 is a line or a point in spacetime?

You said you weren't talking about coordinate charts, but about physics. Physically, I've agreed several times that there is a line at the horizon (for a given angular direction in space), so that different infalling observers (or ingoing light rays) can cross the horizon at different events. Whether or not that line is covered by a given coordinate chart depends on the chart, so if you want an answer to your question exactly as you posed it, you'll need to specify which chart the coordinates you gave relate to.