# Notions of simultaneity in strongly curved spacetime

by PAllen
Tags: curved, notions, simultaneity, spacetime, strongly
P: 3,179
I just have time for this, as this sub-discussion took off:
 Quote by grav-universe ["a clock kept at this place would go at the rate zero". - Einstein] Quote by PeterDonis " (1) Einstein is claiming that a clock can be kept at the horizon, and saying that it would go at rate zero. (2) Einstein is claiming that a clock *cannot* be kept at the horizon, because if it could, it would go at rate zero, and that doesn't make sense. Are you interpreting him as saying #1 or #2? If it's #1, the refutation is pretty easy: the clock would have to go at the speed of light, and no clock can do that. " Just thought I'd add my two cents, but if that is your refutation of #1, then we should also add (3) A clock can't cross the horizon, because if it could, it would have to go at the speed of light, and no clock can do that. [...]
None of them can be a correct interpretation, just as (as I mistakenly thought to have clarified,) none of the following can be a correct interpretation of "For v=c all moving objects—viewed from the “stationary” system—shrivel up into plane figures":

1. Einstein is claiming that it is *possible* for an object to move at c.
2. Einstein is claiming that an object cannot move at c, because if it could, it would shrivel up into a plane figure, and that doesn't make sense.
3. Einstein is claiming that an object cannot move at c, because if it could, it would have infinite energy, and that is impossible.
etc.

Instead, such statements simply refer to (unattainable) physical limits; and in both cases it takes infinite coordinate time to reach such limits. This is acceptable shorthand among physicists, but "forbidden" for mathematicians.
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 Quote by harrylin Instead, such statements simply refer to (unattainable) physical limits; and in both cases it takes infinite coordinate time to reach such limits.
So basically, you are saying that the correct interpretation of what Einstein said is simply:

(4) No object can reach the horizon, because that would take an infinite amount of coordinate time.

In other words, eliminate all mention of "moving at c", and just focus on the coordinate time. Correct? If so, I'm confused about where the phrase "a clock kept at this place would go at rate zero" fits in.
P: 406
 Quote by PeterDonis The more technical way of putting this is that the solution of the Einstein Field Equation is perfectly finite and continuous at the horizon; there is nothing in the solution leading up to the horizon that makes the horizon a place where the solution (i.e., spacetime) could just stop. It has to continue if the EFE is valid, and therefore the worldline of the clock has to continue as well. To claim otherwise is to claim that the EFE suddenly stops being valid at the horizon, for no apparent reason. Do you mean isotropic coordinates? As in the ones described in the "alternative formulation" section here: http://en.wikipedia.org/wiki/Schwarzschild_metric If so, you are correct that these do not cover the interior region; that's because they double cover the exterior region (outside the horizon). The range 0 < R < m/2 (where R is the isotropic radial coordinate) covers the same set of events as the range m/2 < R < infinity; each event in the exterior region maps to *two* values of R, not one. How do you figure that? Isotropic coordinates do not somehow make a point mass magically appear at the horizon.
That particular coordinate system is the one I discovered and you and I discussed in the other thread. Rather than keep calling it the 1/sqrt(1 + 2m/r1) coordinate system, let's call it GU coordinates :) . It coordinately transforms from SC with

r1 = r (1 - 2m/r), r = r1 (1 + 2m/r1) with the metric

ds^2 = c^2 dt^2 / (1 + 2 m / r1) - dr1^2 (1 + 2 m / r1) - dθ^2 r1^2 (1 + 2 m / r1)^2

I like it because it completely eliminates the event horizon and interior spacetime altogether, leaving only what external observers observe. It shrinks the boundary of the event horizon to a point, so that from the perspective of external observers applying this coordinate system, the mass lies at a point singularity in the center just as in Newtonian with infinite acceleration there, no event horizon and no interior spacetime. A clock falling to the point mass will still do so in finite time. Proper distance measured to the point mass is also finite. But we would expect these when measuring the distance to a point or the time to fall to a point. It is just as valid as SC, and the EFE's valid also, being only a coordinate transformation, with all of the same external observables, but looking at it, one would not expect any more spacetime to exist within a point. From the perspective of this coordinate system, that would be like falling out of this universe altogether into some other dimension if there were interior spacetime within a point. If Schwarzschild had happened to come up with this coordinate system rather than the one he did, each just as likely to have been derived before the other, we might not even consider that any interior spacetime exists in the first place.

You also mention Eddington's isotropic coordinates. These are also valid. But as you said, with a one to one correspondence to SC coordinates, they only map some of the interior spacetime of SC, then double back. If one were to fall past the horizon and all the way to the center of EIC, then, when transformed back to SC, it would be like falling part way past the horizon, then doubling back and travelling back out of the horizon again. Likewise, I could find coordinate systems that have more spacetime than SC or even one that cuts out part of the exterior coordinates. So arbitrary coordinate systems may be valid, but obviously they are not equal. Some map out more or less spacetime than others, and some in ways that don't make sense, like the doubling back of EIC, although it would not actually double back in EIC itself. So how are we to know which one maps it out correctly? Personally I would go with the one I found, but if you insist that there must be interior spacetime, as I'm sure you do :) , then as you stated "you are correct that these do not cover the interior region" referring to EIC as compared to SC apparently, how do you know that they do not, or that SC does, with no more interior spacetime than actually exists and no less? SC is only the first coordinate system found. Since then, many others have been determined, and infinitely many are possible, all different in terms of how much spacetime is mapped, so statistically speaking, it is unlikely that SC maps it perfectly. How much spacetime is the right amount?
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 Quote by grav-universe That particular coordinate system is the one I discovered and you and I discussed in the other thread. Rather than keep calling it the 1/sqrt(1 + 2m/r1) coordinate system, let's call it GU coordinates :) . It coordinately transforms from SC with r1 = r (1 - 2m/r), r = r1 (1 + 2m/r1) with the metric ds^2 = c^2 dt^2 / (1 + 2 m / r1) - dr1^2 (1 + 2 m / r1) - dθ^2 r1^2 (1 + 2 m / r1)^2
Can you give a reference to the "other thread" you refer to? This does not look at all familiar to me, but I may just be failing to remember a previous discussion. I'll refrain from commenting on the rest of your post until I've got the context clear.
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 Quote by PeterDonis Can you give a reference to the "other thread" you refer to? This does not look at all familiar to me, but I may just be failing to remember a previous discussion. I'll refrain from commenting on the rest of your post until I've got the context clear.
Sure, here it is. "Shrinking event horizon to point singularity"
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 Quote by grav-universe Sure, here it is. "Shrinking event horizon to point singularity"
Ah, ok, thanks, that helps to jog my memory.

Pretty much everything I would say in response has already been said in the other thread, so I don't see much point in anything more than a quick recap (and what I'm saying applies just as well to Eddington isotropic coordinates as any other chart):

(1) You can't change the physics by changing coordinate charts. You can choose coordinates such that what used to be r = 2m is now r1 = 0; but you can't change the physical nature of the spacetime at what used to be r = 2m and is now r1 = 0. Just labeling it with r1 = 0 doesn't make it a point instead of a surface.

(2) To actually talk about the physics, you have to compute invariants--quantities that don't change when you change coordinate charts. If your chart is singular at a particular place, you can't compute invariants there using the chart, so you can't say anything about the physics there using the chart. Your chart is singular at r1 = 0, so it can't say anything about the physics at that location: in particular, you can't compute any invariant in your chart that shows that what you are labeling r1 = 0 is an actual, physical point, instead of, say, a surface that your coordinates don't cover well.

These points are basic facts of differential geometry as it is used in physics. They have been stated ad nauseam, and you don't seem to be accepting them. That means we really don't have a good basis for discussion.
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Right, but that could go either way. Both being equally valid external coordinate systems, how do you know we're not making a surface out of a point? What I am asking, though, is what coordinate system you think accurately maps out the spacetime, no more and no less? You stated that EIC do not. Why not? Do SC? Why or why not?
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 Quote by grav-universe Right, but that could go either way. Both being equally valid external coordinate systems, maybe we're making a surface out of a point. How do you know that is not the case?
Because I have computed the invariants at r = 2m (using a chart that's not singular there), so I know what the actual physical quantities are there. That includes a computation of the physical area of a 2-sphere at r = 2m, *and* a computation of the causal nature of a curve with constant r = 2m (and constant theta, phi if we include the angular coordinates) to verify that it's a null curve, not a timelike curve.

 Quote by grav-universe What I am asking is what coordinate system would accurately map out the spacetime, no more and no less? You stated that EIC do not. Why not? Do SC? Why or why not?
If you want a map of the *entire* spacetime, including all regions that are mathematically possible according to the vacuum, spherically symmetric solution of the Einstein Field Equation, the only charts I'm aware of that cover it all are the Kruskal chart and the Penrose chart. (The technical term for the spacetime that the full Kruskal chart maps is the "maximally extended Schwarzschild spacetime".) However, as has been said before, nobody believes that this entire spacetime is physically reasonable, because it includes a white hole and a second exterior region.

If you want a map of a highly idealized spacetime consisting of a spherically symmetric region of collapsing matter with zero pressure, plus the vacuum region surrounding it, the only chart I'm aware of that covers it all with a single expression for the metric is the Penrose chart. There is a "Kruskal-type" chart for this spacetime, which covers it all, but the expression for the line element is different depending on whether you're in the vacuum region or the matter region. This spacetime is at least physically reasonable, though obviously it is highly idealized because of the exact spherical symmetry.

If you are willing to settle for a map that only covers the vacuum region exterior to a spherically symmetric collapsing body, there are two additional charts that will cover the entirety of this region: the ingoing Eddington-Finkelstein chart and the ingoing Painleve chart.

The common feature of all these charts is that they are nonsingular over the entire spacetime (or over the entire vacuum region, in the case of the last two), *and* the full range of their coordinates spans the full range of the region they cover. Both the SC chart and the EIC chart fail on at least one of these properties:

* The coordinate singularity at the horizon means that the SC chart can't accurately map the spacetime there, and it also means that the interior SC chart (with r < 2m) is a different, disconnected chart from the exterior SC chart (with r > 2m).

* The EIC chart is nonsingular at the horizon (actually, technically the inverse metric is singular there, but opinions differ on whether that counts as a "coordinate singularity" so I'm giving it the benefit of the doubt). However, the full range of the EIC "r" coordinate doesn't cover anything inside the horizon--instead, as I've said before, it double covers the region outside the horizon. Another way of putting this is that the area of a 2-sphere at radius "r" in EIC coordinates is not monotonic in r; it has a minimum at r = m/2, and increases both for r > m/2 *and* r < m/2. So there are two values of "r" that both map to the same physical 2-sphere (except at the horizon, r = m/2). This makes it pretty obvious that the EIC chart's coverage is incomplete: where are the 2-spheres with smaller area?

[Edit: btw, it's worth noting that the computation of invariants at the horizon that I referred to above can actually be done in the EIC chart, since the line element is not singular there. To compute the area of the 2-sphere at the horizon, plug in r = m/2 and dt = dr = 0, and integrate ds^2 over the full range of theta and phi. You should get 16 pi m^2. To compute the causal nature of a curve with constant r at the horizon, plug in r = m/2 and dr = dtheta = dphi = 0. You should find ds^2 = 0, indicating that a line element with constant r, theta, phi at the horizon (but nonzero dt) is null.]
 PF Patron Sci Advisor P: 4,772 I realized I should add an additional comment to my last post about which charts cover which regions. In the case of a spherically symmetric collapsing body, one can cover the interior of the body (the region containing the matter) with a collapsing FRW-type chart (the time reverse of the expanding FRW-type chart that is used in cosmology to model the universe). MTW does this in their treatment of this model, for example. One can also construct a chart for the vacuum region that matches up with this chart at the boundary (the surface of the collapsing matter); IIRC this chart for the vacuum region is not the same as any of the ones I named. I believe the treatment of this model in MTW uses this type of chart for the vacuum region; if I get a chance I'll check my copy to see.
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 Quote by PeterDonis So basically, you are saying that the correct interpretation of what Einstein said is simply: (4) No object can reach the horizon, because that would take an infinite amount of coordinate time. In other words, eliminate all mention of "moving at c", and just focus on the coordinate time. Correct? If so, I'm confused about where the phrase "a clock kept at this place would go at rate zero" fits in.
No. With that phrase he merely explains the meaning of Schwartzschild's solution (he even says so!). Here's a last attempt to clarify this.

Einstein: "a clock kept at this place would go at rate zero".
My translation attempt for mathematicians: setting dr/dt=0, dτ/dt->0 for r->μ/2

Compare Einstein in 1905: "For v=c all moving objects—viewed from the “stationary” system—shrivel up into plane figures".
My translation attempt for mathematicians: L->0 for v->c

And then I get to what might be "the mother" of all bugs (any further discussion on this topic is useless as long as this has not been fixed) - and this issue is perfectly on topic:
 Quote by PeterDonis Eve *is* accelerating; she feels a nonzero acceleration, an accelerometer attached to her reads nonzero, if she stood on a scale it would register weight, etc. There is no way in which she "thinks she is not accelerating". The term "co-moving inertial reference frame" is more precisely stated as "momentarily co-moving inertial reference frame" (MCIF).
What may be confusing is that Egan says to "use only SR", but as you see his discussion is an application of Einstein's equivalence principle and that is pure GR. For this illustration one only needs SR math. Now your claims:

1. Peter: There is no way in which Eve "thinks she is not accelerating", as "an accelerometer attached to her reads nonzero".

According to the Einstein Equivalence Principle, when she is sitting in her chair Eve can think that she is not accelerating; she may think that instead the force that she feels is due her being in rest in a gravitational field. As you noticed yourself, "if she stood on a scale it would register weight".

2. Peter: The term "co-moving inertial reference frame" is more precisely stated as "momentarily co-moving inertial reference frame".

Evans evidently means constantly co-moving inertial reference frame, and I will explain why. According to you, Evans means that according to Eve the force she feels is due to acceleration; so that she thinks that she is one moment at rest in one inertial frame, and the next moment she is at rest in a different inertial frame. Consequently she would use the same set of inertial frames as Adam - that is standard SR. In any such reference frame there is a time for Eve when Adam passes through the horizon. It would be just an SR simultaneity disagreement.

To the contrary, according to Egan there is no time for Eve when, in her co-moving inertial reference frame, Adam passes through the horizon.
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 Quote by harrylin Einstein: "a clock kept at this place would go at rate zero". My translation attempt for mathematicians: setting dr/dt=0, dτ/dt->0 for r->μ/2
Don't you mean r -> 2μ? (Or 2m in the more usual symbols.)

Anyway, I wasn't asking about how you would translate Einstein's statement into mathematics. I was asking what you thought it meant physically.

 Quote by harrylin What may be confusing is that Egan says to "use only SR", but as you see his discussion is an application of Einstein's equivalence principle and that is pure GR.
Egan's discussion is about a scenario in flat spacetime, which can be handled using only SR. That's why he says to "use only SR". The reason the scenario is relevant for our discussion here is that, as Egan says, the scenario he decribes in flat spacetime is equivalent to "a first-order approximation of the Schwarzschild metric near a black hole's horizon". Whether you call that "pure GR" or not is a matter of words, not physics.

 Quote by harrylin 1. Peter: There is no way in which Eve "thinks she is not accelerating", as "an accelerometer attached to her reads nonzero". According to the Einstein Equivalence Principle, when she is sitting in her chair Eve can think that she is not accelerating; she may think that instead the force that she feels is due her being in rest in a gravitational field. As you noticed yourself, "if she stood on a scale it would register weight".
Sigh. I should have clarified (again) the distinction between coordinate acceleration and proper acceleration. Yes, Eve can think that she experiences no *coordinate* acceleration; she can view herself as at rest in a gravitational field. But she cannot think that she experiences no *proper* acceleration, because she feels weight, and that is the *definition* of proper acceleration. Proper acceleration is an invariant; it's a direct observable, so it's there regardless of which coordinates Eve uses (Rindler or Minkowski). Coordinate acceleration is *not* an invariant; Eve can make it disappear by viewing herself as at rest in Rindler coordinates (in which a "gravitational field" is present) rather than as accelerating in Minkowski coordinates (where there is no "gravitational field"). I thought you understood the difference between coordinate and proper acceleration, since it's been discussed enough times, but apparently not, so I'll try to be more careful about qualifying the term "acceleration", as in "feeling acceleration", which makes it clear that I'm referring to the direct physical observable, that Eve feels weight.

 Quote by harrylin 2. Peter: The term "co-moving inertial reference frame" is more precisely stated as "momentarily co-moving inertial reference frame". Evans evidently means constantly co-moving inertial reference frame
If you left out the word "inertial", this would be fine. But with it included, it's false. There is no such thing as a "constantly co-moving inertial reference frame" for Eve (and Egan certainly isn't claiming any such thing). Eve feels proper acceleration; nobody at rest in an inertial frame (for more than an instant) can feel proper acceleration. That's the *definition* of an inertial frame: that any observer at rest in it (for more than an instant) is weightless, in free fall. Again, I thought you understood this, but apparently not. Sigh.

 Quote by harrylin according to Eve the force she feels is due to acceleration; so that she thinks that she is one moment at rest in one inertial frame, and the next moment she is at rest in a different inertial frame.
This is not due to her "thinking" that the force is due to "acceleration": it's due to her actually *feeling* acceleration, i.e,. feeling weight.

 Quote by harrylin Consequently she would use the same set of inertial frames as Adam - that is standard SR.
Adam only uses one inertial frame, since he is in free fall. Only Eve has to use a "set" of inertial frames if she wants to use inertial frames to describe her motion.

 Quote by harrylin In any such reference frame there is a time for Eve when Adam passes through the horizon. It would be just an SR simultaneity disagreement.
And this is true; in any of the inertial frames in which Eve is momentarily at rest, there *is* a finite time at which Adam crosses the horizon. But Eve doesn't *stay* at rest in any of these frames, because she feels acceleration, i.e., she feels weight.

 Quote by harrylin To the contrary, according to Egan there is no time for Eve when, in her co-moving inertial reference frame, Adam passes through the horizon.
As you state it, this is false; you need to leave out the phrase "in her co-moving inertial reference frame" (which Egan does *not* use, and your attributing it to him is mistaken). The "time for Eve" that Egan refers to is Rindler coordinate time, which is the same as proper time along her worldline. Since she feels acceleration, i.e., feels weight, that proper time is *not* the same as the time in *any* inertial frame, even inertial frames in which she is momentarily at rest. Egan's statement simply means that there is no Rindler coordinate time at which Adam crosses the horizon; it's not referring to the time in *any* inertial frame.
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 Quote by PeterDonis Don't you mean r -> 2μ? (Or 2m in the more usual symbols.) Anyway, I wasn't asking about how you would translate Einstein's statement into mathematics. I was asking what you thought it meant physically.
I directly used the notation of Einstein, in order not to mix up my own interpretation with my translation. And I already told you, it has no physical meaning without context, just as "v=c" has no physical meaning in itself. The context (incl. other papers) suggests to me that Einstein held both extremes for impossible in physical reality.
 Egan's discussion is about a scenario in flat spacetime, which can be handled using only SR. That's why he says to "use only SR". The reason the scenario is relevant for our discussion here is that, as Egan says, the scenario he decribes in flat spacetime is equivalent to "a first-order approximation of the Schwarzschild metric near a black hole's horizon". Whether you call that "pure GR" or not is a matter of words, not physics.
I almost fully agree; but regretfully our differences are mostly a matter of words, which obscures eventual differences in physical models. For example, my notion of "flat space-time" means SR with reference systems that relate to each other by means of the Lorentz transformations - such a space-time lacks a Rindler horizon.

Apart of simple mistakes, we certainly come from different "schools" (even literally) that teach definitions which are incompatible with each other (Sigh indeed!). I will come back to the issue of definitions in a next thread.
Now, it will be a waste of time start a text exegesis of the meaning of words as used by Egan, with speculations of the school of thought that he is following; we don't really need him (except if we want to discuss "Egan's theory). And in the context of this thread you appear to agree with me on the simple point that I tried to make, and also the different "notions of simultaneity" are not in question:
 [..]Yes, Eve can think that she experiences no *coordinate* acceleration; she can view herself as at rest in a gravitational field. [..] in any of the inertial frames in which Eve is momentarily at rest, there *is* a finite time at which Adam crosses the horizon. [..] Egan's statement simply means that there is no Rindler coordinate time at which Adam crosses the horizon [..]
That looks to me a reasonable summary of different "notions of simultaneity in strongly curved spacetime"; perhaps PAllen would like to clarify how his first post relates to this example (if indeed it does).

And from an earlier post, you seem to agree that at the moment that Adam "falls away" according to Eve, she ascribes the frequency difference from two clocks to the effect of a gravitational field which makes her clocks go at different rates; and that in contrast, for Adam the frequency difference that Eve observes is almost completely due to "classical" Doppler. My point was that in GR much more than in SR the different views relate to a disagreement about physical reality. However, that is a bit off-topic in this thread; and now that I decided to start my own thread I'll include further elaboration there.
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 Quote by harrylin my notion of "flat space-time" means SR with reference systems that relate to each other by means of the Lorentz transformations - such a space-time lacks a Rindler horizon.
No, it doesn't. You may think it does, but that's because you don't fully understand the implications of "reference systems that relate to each other by means of the Lorentz transformations". Such a spacetime includes hyperbolas such as the worldline that Eve travels on, and it also includes the fact that a light ray emitted from the origin will never cross such a hyperbola (since the light ray is an asymptote of the hyperbola). That is the definition of a Rindler horizon, so your notion of flat space-time includes a Rindler horizon, whether you think so or not. If you didn't realize that all that was already included in your notion of a flat space-time, well, then you need to think more carefully about the implications of your notions.

 Quote by harrylin Apart of simple mistakes, we certainly come from different "schools" (even literally) that teach definitions which are incompatible with each other (Sigh indeed!).
I don't know what "school" you come from, so I can't really evaluate this statement. I'm not aware of any definitions from my "school", i.e., standard GR, which are incompatible with each other. I don't really see a problem with incompatible definitions in our discussion; unclear definitions, yes, but that can be fixed by making them clear. When we've managed to do that, I don't see any incompatibility.

 Quote by harrylin And from an earlier post, you seem to agree that at the moment that Adam "falls away" according to Eve, she ascribes the frequency difference from two clocks to the effect of a gravitational field which makes her clocks go at different rates; and that in contrast, for Adam the frequency difference that Eve observes is almost completely due to "classical" Doppler.
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Quote by harrylin

 To the contrary, according to Egan there is no time for Eve when, in her co-moving inertial reference frame, Adam passes through the horizon.
 Quote by PeterDonis As you state it, this is false; you need to leave out the phrase "in her co-moving inertial reference frame" (which Egan does *not* use, and your attributing it to him is mistaken). The "time for Eve" that Egan refers to is Rindler coordinate time, which is the same as proper time along her worldline. Since she feels acceleration, i.e., feels weight, that proper time is *not* the same as the time in *any* inertial frame, even inertial frames in which she is momentarily at rest. Egan's statement simply means that there is no Rindler coordinate time at which Adam crosses the horizon; it's not referring to the time in *any* inertial frame.
I don't know what Egan had to say but I think you are quite mistaken regarding Rindler coordinates and the horizon.
I don't think Rindler has anything to do with it. It is a coordinate artifact due to the dynamic metric in any accelerating system. This applies just as well to momentarily co-moving inertial frames. It happens because the distance to a point towards the rear shrinks due to contraction comparable to the increase in length due to system motion. SO the system asymptotically stops moving relative to points nearing the horizon as calculated . from a point within the system.
So harrylin is correct that Adam never crosses the horizon in any MCRF that Eve is at rest in.
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 Quote by PeterDonis Because I have computed the invariants at r = 2m (using a chart that's not singular there), so I know what the actual physical quantities are there. That includes a computation of the physical area of a 2-sphere at r = 2m, *and* a computation of the causal nature of a curve with constant r = 2m (and constant theta, phi if we include the angular coordinates) to verify that it's a null curve, not a timelike curve.
Right. Most of those invariants, some of which you mentioned in the other thread, don't really demonstrate much as far as I can see, as they are mostly measured externally to the horizon, so don't say a lot about what happens at the horizon itself. For instance, with the finite proper time of a freefalling clock, we could just as easily see that time showing the clock striking a point mass singularity in GUC as crossing a surface in SC.

Those invariants may make sense if taken all together as you said in the other thread, but on the flip side, there are also a few things that don't make sense, such as a clock travelling at c to a *hypothetical* static observer (which actually doesn't exist at the horizon, I know), infinite acceleration applied at a finite surface, and charts such as SC mapping out the physical space between the center and the horizon, but being unable to say anything at all about the spacetime there or events that occur there without referring to a different chart altogether. But these don't demonstrate anything definite either.

The one and only thing so far that I can see that does demonstrate anything substantial is what you just mentioned, the invariant locally measured surface area. In SC, there is only radial contraction and no tangent contraction of static rulers as inferred by a distant observer, so if the distant observer measures A = 4 pi r^2 = 4 pi (2 m)^2 = 16 pi m^2 at the horizon, then with no tangent contraction in SC all the way down to the horizon, so presumably at the horizon as well, a *hypothetical* static observer there should also. And regardless of how we change the coordinate system, that local measurement is invariant.

Even in GUC, the distant observer measures A = 0, but the tangent length contraction is 1 / (1 + 2 m / r1) = 0 at r1 = 0, so a static observer at the horizon measures A' = A (1 + 2 m / r1)^2 = 0 / 0^2 = any real number, including zero. Being invariant, however, it should agree with the local measurement made in SC, which is finite and non-zero. So there's that. Of course, however, since there can be no static observer at the horizon anyway, though, the surface cannot actually be measured there, which negates this result (lol jk). We could perhaps instead consider what an observer measures that just begins to freefall from rest (or near rest?) at the horizon, although that would already be assuming that a surface exists there that one could fall through, or if falling from rest just before the horizon, he could not reach it at less than c, so still very far from measuring its surface while static. Hmm, I'm actually not sure how that surface would be measured locally.

By the way, you said that some charts are not singular there, but how could that be? The local acceleration there is infinite, that is an invariant. Are you not defining a singularity as a place with infinite local acceleration? Also, due to the infinite acceleration, static observers cannot exist there, so that is also an invariant. Surely the chart you are referring to does not allow static observers there, right? Wouldn't that define the horizon, a place where static observers cannot exist and observers can never accelerate at a large enough rate to escape once there?

 If you want a map of the *entire* spacetime, including all regions that are mathematically possible according to the vacuum, spherically symmetric solution of the Einstein Field Equation, the only charts I'm aware of that cover it all are the Kruskal chart and the Penrose chart. (The technical term for the spacetime that the full Kruskal chart maps is the "maximally extended Schwarzschild spacetime".) However, as has been said before, nobody believes that this entire spacetime is physically reasonable, because it includes a white hole and a second exterior region. If you want a map of a highly idealized spacetime consisting of a spherically symmetric region of collapsing matter with zero pressure, plus the vacuum region surrounding it, the only chart I'm aware of that covers it all with a single expression for the metric is the Penrose chart. There is a "Kruskal-type" chart for this spacetime, which covers it all, but the expression for the line element is different depending on whether you're in the vacuum region or the matter region. This spacetime is at least physically reasonable, though obviously it is highly idealized because of the exact spherical symmetry. If you are willing to settle for a map that only covers the vacuum region exterior to a spherically symmetric collapsing body, there are two additional charts that will cover the entirety of this region: the ingoing Eddington-Finkelstein chart and the ingoing Painleve chart. The common feature of all these charts is that they are nonsingular over the entire spacetime (or over the entire vacuum region, in the case of the last two), *and* the full range of their coordinates spans the full range of the region they cover. Both the SC chart and the EIC chart fail on at least one of these properties: * The coordinate singularity at the horizon means that the SC chart can't accurately map the spacetime there, and it also means that the interior SC chart (with r < 2m) is a different, disconnected chart from the exterior SC chart (with r > 2m). * The EIC chart is nonsingular at the horizon (actually, technically the inverse metric is singular there, but opinions differ on whether that counts as a "coordinate singularity" so I'm giving it the benefit of the doubt). However, the full range of the EIC "r" coordinate doesn't cover anything inside the horizon--instead, as I've said before, it double covers the region outside the horizon. Another way of putting this is that the area of a 2-sphere at radius "r" in EIC coordinates is not monotonic in r; it has a minimum at r = m/2, and increases both for r > m/2 *and* r < m/2. So there are two values of "r" that both map to the same physical 2-sphere (except at the horizon, r = m/2). This makes it pretty obvious that the EIC chart's coverage is incomplete: where are the 2-spheres with smaller area? [Edit: btw, it's worth noting that the computation of invariants at the horizon that I referred to above can actually be done in the EIC chart, since the line element is not singular there. To compute the area of the 2-sphere at the horizon, plug in r = m/2 and dt = dr = 0, and integrate ds^2 over the full range of theta and phi. You should get 16 pi m^2. To compute the causal nature of a curve with constant r at the horizon, plug in r = m/2 and dr = dtheta = dphi = 0. You should find ds^2 = 0, indicating that a line element with constant r, theta, phi at the horizon (but nonzero dt) is null.]
Um, wow, good post. Very detailed. That is a lot to look into. Thanks. :)
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 Quote by Austin0 It is a coordinate artifact due to the dynamic metric in any accelerating system. This applies just as well to momentarily co-moving inertial frames.
No, it doesn't; at least, not in flat spacetime. In flat spacetime, any inertial frame covers the entire spacetime, including the portion of Adam's worldline at and beyond the Rindler horizon. That's a basic fact about inertial frames in flat spacetime. An MCIF is an inertial frame, so this fact applies to MCIFs in flat spacetime. Another way of saying this is that in flat spacetime, every inertial frame is global.

In curved spacetime, there are *no* global inertial frames; *any* inertial frame can only cover a small patch of the spacetime. So in curved spacetime, you are correct that an MCIF at some event on an accelerated observer's worldline might not cover the horizon. But Egan's scenario is entirely set in flat spacetime, so the restrictions on inertial frames, including MCIF's, in curved spacetime doesn't apply.

Also, a word about "coordinate artifact". The fact that you can't assign a finite Rindler time coordinate to events at and beyond the Rindler horizon is an artifact of Rindler coordinates. But the fact that a light ray at the Rindler horizon will never intersect any of the "Rindler hyperbolas"--the curves with constant Rindler space coordinates--is not a coordinate artifact; you can express the same fact in any coordinate chart, because the curves themselves are geometric objects, not coordinate artifacts. So the existence of a "Rindler horizon" is not a coordinate artifact; there is something real and physical going on.
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 Quote by grav-universe with the finite proper time of a freefalling clock, we could just as easily see that time showing the clock striking a point mass singularity in GUC as crossing a surface in SC.
No, you can't, because the presence of a "point mass singularity", if it were true, would itself be an invariant; and computing invariants tells you that that whatever is there at that place in spacetime, it isn't a point mass singularity. See further comments below on the definition of a singularity.

 Quote by grav-universe Those invariants may make sense if taken all together as you said in the other thread, but on the flip side, there are also a few things that don't make sense, such as a clock travelling at c to a *hypothetical* static observer (which actually doesn't exist at the horizon, I know)
Not only does the hypothetical static observer at the horizon not exist; one *can't* exist, because the horizon is not a timelike surface. It's a null surface. That's the key fact you keep on missing, and it is one of the invariants I recommended that you compute. As I said, compute ds^2 for a line element at the horizon where dr = dtheta = dphi is zero. Technically you can't do it in SC coordinates because the line element is singular there, but PAllen said a while back that you can get around that even in SC coordinates by taking a limit as r -> 2m. Or you could do the computation in a chart that's not singular at the horizon, such as EIC. You will find that ds^2 = 0, and this is an invariant.

What is this invariant telling you? Well, look at similar line elements for r > 2m; i.e., pick some constant r > 2m, and plug in that r, plus dr = dtheta = dphi = 0, into the Schwarzschild line element. What do you get? You get ds^2 < 0 (with the usual sign convention), indicating that the line element is timelike; i.e., it's a possible worldline for an observer (a static observer, in this case). But when r = 2m, the corresponding line element is null; i.e., it's a possible worldline for a *light ray*, rather than a possible worldline for an observer.

That immediately tells us two things. First, it explains why the infalling observer moves at c relative to the horizon: the horizon is a light ray moving in the opposite direction to the observer (he's moving inward and the horizon is moving outward), so of course their relative velocity will be c. It's the *horizon* that's "moving at c", not the infalling observer; his worldline remains timelike, as it must.

Second, the fact that the horizon is null, rather than timelike, means that the horizon is not a "place" or "spatial location" in the way that the places occupied by static observers outside the horizon are. A "spatial location" requires a timelike curve going through it that has the same spatial coordinates everywhere. Curves of constant r > 2m (and constant theta, phi if we include the angular coordinates) meet that requirement; but a curve of constant r = 2m does not.

If you go back and look closely at the arguments you've made for why the infalling observer can't reach the horizon, you'll see that you were implicitly assuming that the horizon was a spatial location, a "place". It isn't. That's why your arguments don't show that an infalling observer can't reach the horizon.

 Quote by grav-universe infinite acceleration applied at a finite surface
There is no infinite acceleration, because there is no "place" where the infinite acceleration would exist. There is no "acceleration" along the path of a light ray. See above.

 Quote by grav-universe charts such as SC mapping out the physical space between the center and the horizon, but being unable to say anything at all about the spacetime there or events that occur there without referring to a different chart altogether.
I don't understand what you mean by this. The interior SC chart (i.e,. the SC chart with r < 2m) works perfectly well as a "map" of the interior of the black hole (the region inside the horizon). It's not a map that matches up with our intuitions very well, but so what? It's a perfectly valid map. It's also disconnected from the exterior SC chart, which maps the region outside the horizon, but again, so what? There's no requirement that any valid map has to continuously cover the entire spacetime.

 Quote by grav-universe The one and only thing so far that I can see that does demonstrate anything substantial is what you just mentioned, the invariant locally measured surface area.
As I've shown above, the invariant ds^2 = 0 for a line element at the horizon demonstrates something substantial as well.

 Quote by grav-universe Even in GUC, the distant observer measures A = 0
No, the distant observer can't infer anything from this chart, because it's singular at the horizon--by which I mean *really* singular; you can't even compute the physical area of the 2-sphere at the horizon at all, because the line element is mathematically undefined. That doesn't allow you to conclude A = 0. It doesn't allow you to conclude anything.

 Quote by grav-universe Of course, however, since there can be no static observer at the horizon anyway, though, the surface cannot actually be measured there, which negates this result
It's true that one can't measure the area of the horizon in the obvious way, by having observers who are static at that radius lay down rulers. But one can measure it indirectly, by having static observers on 2-spheres closer and closer to the horizon measure areas, and taking the limit as r -> 2m. There may be other more ingenious ways of doing it as well. In any case, every coordinate chart which is not singular at the horizon will give you the same answer for the value of the invariant area of the horizon.

 Quote by grav-universe We could perhaps instead consider what an observer measures that just begins to freefall from rest (or near rest?) at the horizon
Near rest is the best you can do. The fact that the horizon is a null surface means that no observer can be at rest there even for an instant.

 Quote by grav-universe although that would already be assuming that a surface exists there that one could fall through, or if falling from rest just before the horizon, he could not reach it at less than c
See above for what the relative velocity of c actually means.

 Quote by grav-universe By the way, you said that some charts are not singular there, but how could that be? The local acceleration there is infinite, that is an invariant.
No, it isn't. There is no "local acceleration" at the horizon. The formula for "local acceleration" is only valid if the curve along which it is computed is timelike. As I showed above, the corresponding curve at the horizon is not timelike, it's null. So the formula fails. Again, there is *no* invariant that is not finite at the horizon.

 Quote by grav-universe Are you not defining a singularity as a place with infinite local acceleration?
No. The usual definition of a singularity is a place where the spacetime curvature becomes infinite. The only place in Schwarzschild spacetime where that happens is r = 0.

Strictly speaking, having *any* valid invariant (as I noted above, "local acceleration" isn't valid at the horizon because it only applies along timelike curves) become infinite is sufficient for a singularity, but when you work through the math you find that if any invariant is infinite, at least one of the invariants associated with curvature is infinite, so the usual definition in terms of curvature turns out to work fine.

 Quote by grav-universe Also, due to the infinite acceleration, static observers cannot exist there, so that is also an invariant.
You're correct that static observers can't exist at the horizon, and that's an invariant, but it's not due to "infinite acceleration". See above.

 Quote by grav-universe Surely the chart you are referring to does not allow static observers there, right?
Right.

 Quote by grav-universe Wouldn't that define the horizon, a place where static observers cannot exist and observers can never accelerate at a large enough rate to escape once there?
The usual definition is that the horizon is the surface at which radially outgoing light can no longer escape to infinity. But that also implies the things you state here, so they are valid ways of describing the horizon as well.
 P: 406 What is your definition of singular?

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