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## Notions of simultaneity in strongly curved spacetime

 Quote by PAllen The path of an outer edge infall particle has finite proper time integrated to the SC radius. If you declare it stops there, you have a hole in spacetime. You have a geodesic ending with finite 'interval', where curvature is finite.
PAllen, do you realise that in Minkowski geometry zero space-time distance (zero proper time) between two points does not mean it's the same point?
This is very different from traditional geometry where zero distance between two points does mean it's the same point.

 Quote by PAllen If you imagine the surface of such particles infalling, and you don't allow it to proceed to Sc radius, you have, geometrically, a hole: proper time on this surface is finite, area is finite, but if you stop it from continuing, it is geometrically a hole. Geometry is defined by invariants, not coordinate quantities. Consider the example I gave to harrylin several posts back of a horizontal geodesic in the plane in (u,v) coordinates. u coordinate goes to infinity on both sides of coordinate singularity, but it is still nothing but a geodesic in the flat plane.
I found your example. I just have to think what I have to say about it. It is just mock transformation when you undo all the consequences of transformation using transformed metric.

 Quote by PAllen Nope. Einstein was very clear that simultaneity is purely a convention, not an observable. There is no observation or measurement in SR that changes if you use a different one than the standard one (but you have to change the metric as well; it is no longer eg. diag(+1,-1,-1,-1) if you use a funky convention.
There is one statement in SR that gives it physical content - it is principle of relativity.
But principle of relativity applies to certain class of inertial coordinate systems. This class of inertial coordinate systems is defined using particular simultaneity convention.
So you can't really speak about SR with different simultaneity convention as this particular simultaneity convention is integral part of the theory (and it's predictions).

If you want you can say that relativity principle gives physical content to particular simultaneity convention.

 Quote by PAllen Just a quick comment here - the SC metric you've seen (as derived from the EFE) applies inside the horizon. That is, if you look at it, it works just fine for r < Rs; it only doesn't work (without limits for the finite invariants) on Rs itself. So it is not added later - it is the same solution, and EFE are telling us it applies everywhere - the derivation applies up to r=0. So, if you look back at my (u,v) versus (x,y) example, it is as if you got the full (u,v) solution; you just have to deal with technical problems at x and y axis coordinate (but not invariant) discontinuity.
And a quick comment on that quick comment: I don't see the qualitative difference with "The light speed limit doesn't exist; the tachyon space works just fine, it is the same solution. You just have to deal with technical problems to get through c".

 Quote by PAllen Having come this far, I want to emphasize a very physical reason for not being satisfied with an object's history stopping at an arbitrary time in its history. This is that a key physical foundation of GR is the principle of equivalence; one aspect of this, is that sufficiently locally, GR = SR, everywhere, every when. So we have an infall clock whose mechanics is following normal SR physics locally at all times; tidal gravity is irrelevant due to its small size (esp. for a supermassive BH); the relation between its time rate and some distant clock when sending messages is irrelevant locally. So you posit at 2:59:59 pm, it is operating the same as any similarly constructed clock; but at 3 pm it stops dead for no reason explicable with SR physics. This is a gross violation of the principle of equivalence: a free fall clock near the horizon has a behavior completely different than free fall clocks everywhere else.
So far I was following your explanations, but here I am I little confused. Why do you say that the clock will stop? Surely passing the horizon will not stop the clock and someone with the clock will see it ticking after 3:00pm, no?
 A related mathematical point is that the EFE are system of 10 equations that satisfy 4 identities.
What do you mean by this? Equations that satisfy identities!

 Quote by PeterDonis A clarification about terminology: there are several concepts/objects that can be referred to as "Schwarzschild's solution", or "the Schwarzschild metric" [..] My reason for making these distinctions will be evident in a moment.
As I elaborate in a parallel thread, I make a similar distinction between different "flavours" of GR.
 [..] this is not a statement about any actual physics; it's only a statement about a particular coordinate chart, with no real physical content.
In fact that chart is an equation. You next suggest that another equation has more physical content than the equation which was derived from it, using reasonable physical assumptions. I suspect that the one who derived that equation would disagree with you for reasons that I will briefly* mention.
 [..] when you actually work through the solution of the Einstein Field Equation, in either case #1 above ([..] physically unreasonable) or case #2 above (more complex [..] but physically more reasonable, though still highly idealized), you find that any solution that only includes the region outside the horizon is incomplete; the EFE itself tells you that that region cannot be the entire spacetime. [..] These all seem like innocuous statements about coordinate charts on geometric manifolds to me. Apparently they seem like huge issues to you; I'm not sure why. Why is it such a big deal that some charts can cover points or regions that others can't?
 Quote by PeterDonis [..] technically the Schwarzschild observer has to change charts to make some of the predictions I'm going to give; but there is nothing stopping him from doing that. [..] why does your worldline stop? No reason. That's not physically reasonable. [..]
 Quote by PeterDonis [..] Based on the answer to that question given by the Einstein Field Equation, we are taking the incomplete O-S map and adding a new region, and an expanded set of directions, onto it to make it physically complete. [..] Understood; and the answer is yes, it is "really GR".
If I correctly understood the explanations, those equations lead to white holes when blindly followed through without physical concerns; following your arguments, white holes are "really GR". Is it a big deal to you?

However, and as we discussed earlier, we agree that considerations of what makes physical sense should play a role. Most theories do contain more than mere equations, and GR is no exception. For me a theory consists of its physical foundations (both those postulated and those clearly mentioned). Those foundations are all kept except if they lead to contradictions; and an equation is only physically valid insofar as those physical foundations are not violated. An obvious one is the Einstein equivalence principle, but there are also the physical reality of gravitational fields and the requirement that theorems must describe the relation between measurable bodies and clocks.

Consequently I expect that Einstein would reject Peter's argument and say that Peter denies the physical reality of gravitational fields. Einstein would argue that the clock's worldline never really stops, but does not reach beyond 42 due to the physical reality of the gravitational field. I think that Kruskal's white holes and his inside solution for black holes are not compatible with Einstein's GR.
 Quote by PAllen Having come this far, I want to emphasize a very physical reason for not being satisfied with an object's history stopping at an arbitrary time in its history. This is that a key physical foundation of GR is the principle of equivalence; one aspect of this, is that sufficiently locally, GR = SR, everywhere, every when. So we have an infall clock whose mechanics is following normal SR physics locally at all times; tidal gravity is irrelevant due to its small size (esp. for a supermassive BH); the relation between its time rate and some distant clock when sending messages is irrelevant locally. So you posit at 2:59:59 pm, it is operating the same as any similarly constructed clock; but at 3 pm it stops dead for no reason explicable with SR physics. This is a gross violation of the principle of equivalence: a free fall clock near the horizon has a behavior completely different than free fall clocks everywhere else.
PAllen, it looks to me that you are mixing up reference frames. As far as I can see, in no valid GR reference system is the clock suddenly stopping dead.
As described from S, the clock never stops ticking. I guess that for such an extreme case the validity of SR probably shrinks to nothing. And as described from S', dramatic things happen upto 3 pm but no stopping of clocks is observed.

[ADDENDUM: It may look a little weird if you believe that the universe is eternal. But in case you believe that the universe is not eternal, as is commonly thought, then the universe ends at for example 2:59:58.]

And don't you think that Einstein would have exclaimed the same, if it was really a "gross violation of the principle of equivalence"? Instead he commented that "there arises the question whether it is possible to build up a field containing such singularities". (E. 1939.)

The Einstein principle of equivalence:
"K' [..] has a uniformly accelerated motion relative to K [..] [This] can be explained in as good a manner in the following way. The reference-system K' has no acceleration. In the space-time region considered there is a gravitation-field which generates the accelerated motion relative to K'."
- https://en.wikisource.org/wiki/The_F..._of_Relativity

*Regretfully this forum has been stripped from philosophy on the grounds that the mentors don't want to spend time on monitoring such discussions; I will respect that by not elaborating much on philosophy of science.

 Quote by martinbn [..] Surely passing the horizon will not stop the clock and someone with the clock will see it ticking after 3:00pm, no? [..]
According to the Schwartzschild equations as used by Einstein and Oppenheimer, that clock will only reach the horizon (and indicate 3:00pm) at t=∞. In common physics that means "never"; however some smart person (in fact, who?) invented a different interpretation!

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 Quote by harrylin In fact that chart is an equation. You next suggest that another equation has more physical content than the equation which was derived from it, using reasonable physical assumptions.
No, I don't. I suggest that the first chart/equation (exterior Schwarzschild) does not cover a particular portion of the spacetime that the second chart/equation (Kruskal) does.

However, underlying all of this is just one equation, the EFE. That equation is what's really at issue here. See below.

 Quote by harrylin If I correctly understood the explanations, those equations lead to white holes when blindly followed through without physical concerns; following your arguments, white holes are "really GR".
You didn't correctly understand the explanations. The EFE leads to white holes only if we assume the spacetime is vacuum everywhere (and spherically symmetric, but that's a minor point for this discussion). Nobody thinks that assumption is physically reasonable. If the spacetime is not vacuum everywhere--for example, if there is collapsing matter present--then the EFE does *not* predict white holes. So white holes are part of the set of all possible mathematical solutions of the EFE, but they are not part of the set of physically reasonable solutions of the EFE.

Just an "equation" isn't enough; you have to add constraints--initial/boundary conditions--to get a particular solution. Which solution of the equation you get--i.e., which spacetime geometry models the physical situation you're interested in--depends on the constraints.

 Quote by harrylin However, and as we discussed earlier, we agree that considerations of what makes physical sense should play a role. Most theories do contain more than mere equations, and GR is no exception. For me a theory consists of its physical foundations (both those postulated and those clearly mentioned).
Of course. See above.

 Quote by harrylin Those foundations are all kept except if they lead to contradictions; and an equation is only physically valid insofar as those physical foundations are not violated. An obvious one is the Einstein equivalence principle, but there are also the physical reality of gravitational fields and the requirement that theorems must describe the relation between measurable bodies and clocks.
Sure.

 Quote by harrylin Consequently I expect that Einstein would reject Peter's argument and say that Peter denies the physical reality of gravitational fields.
Einstein *did* reject arguments of this type. Einstein was wrong.

 Quote by harrylin Einstein would argue that the clock's worldline never really stops, but does not reach beyond 42 due to the physical reality of the gravitational field.
What is "the gravitational field"? What mathematical object in the theory does it correspond to? Before we can even evaluate this claim, we have to know what it refers to. But let's try it with some examples:

(1) The "gravitational field" is the metric. The metric (the coordinate-free geometric object, not its expression in particular coordinates) is perfectly finite and continuous at the horizon, and for reasons that both PAllen and I have explained, it can't "just stop" at the horizon without violating the EFE.

(2) The "gravitational field" is the Riemann curvature tensor. Like the metric, this is perfectly finite and continuous at the horizon.

(3) The "gravitational field" is the proper acceleration experienced by a "hovering" observer (an observer who stays at the same radius and does not move at all in a tangential direction). This *does* increase without bound as you get closer and closer to the horizon. However, there is *no* "hovering" observer *at* the horizon, because the horizon is a null surface: i.e., a line with constant r = 2M and constant theta, phi is not a timelike line; it's a null line (the path of a light ray--a radially outgoing light ray). So there is no observer who experiences infinite proper acceleration, and this definition of "gravitational field" simply doesn't apply at or inside the horizon.

As far as I can see, the only possible basis you could have for claiming that "the physical reality of the gravitational field" means that the clock's worldline stops as tau->42, would be #3. However, #3 doesn't apply to infalling observers; it only applies to accelerated, "hovering" observers. Infalling observers don't feel any acceleration, so there's nothing stopping them from falling through the horizon. The "gravitational field" in the sense of #3 is simply not felt by them at all.

Note that in all these cases, the physical "field" has to correspond to something invariant in the mathematical model, *not* something that only exists in a particular coordinate chart. That is something Einstein would have *agreed* with. Note also that none of the definitions of "gravitational field" I gave above used Schwarzschild coordinate time, or the fact that t->infinity as you approach the horizon. Einstein simply didn't understand that claims about t->infinity as you approach the horizon were claims about something that only exists in a particular coordinate chart.

 Quote by harrylin I think that Kruskal's white holes and his inside solution for black holes are not compatible with Einstein's GR.
See above. You are equivocating on different meanings of "Einstein's GR". White holes are mathematically compatible, but not physically reasonable. Black hole interiors are both mathematically compatible *and* physically reasonable.

 Quote by harrylin And don't you think that Einstein would have exclaimed the same, if it was really a "gross violation of the principle of equivalence"? Instead he commented that "there arises the question whether it is possible to build up a field containing such singularities". (E. 1939.)
As I've said before, Einstein's paper only considered the stationary case--i.e., he only considered systems of matter in stable equilibrium. All his paper proves is that *if* a system has a radius less than 9/8 of the Schwarzschild radius corresponding to its mass, the matter can't be in stable equilibrium. A collapsing object that forms a black hole meets this criterion: the collapsing matter is not in stable equilibrium. So Einstein's conclusion doesn't apply to it.

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 Quote by zonde I found your example. I just have to think what I have to say about it. It is just mock transformation when you undo all the consequences of transformation using transformed metric.
I only have time for one quick comment:

But that is the way all coordinate transforms work in differential geometry! That is the whole point! The the metric is transformed along with coordinates so all geometric properties (lengths, angle, intervals, etc.) are preserved as mathematical identities.

 Quote by harrylin According to the Schwartzschild equations as used by Einstein and Oppenheimer, that clock will only reach the horizon (and indicate 3:00pm) at t=∞. In common physics that means "never"; however some smart person (in fact, who?) invented a different interpretation!
That is not true, the clock will pass the horizon, and its proper time will go after 3:00pm.

 Quote by PeterDonis [..] Einstein *did* reject arguments of this type. Einstein was wrong.
Just give me the physics paper that proves that your philosophy is right, and Einstein's was wrong.
 What is "the gravitational field"? [..]
Perhaps your beef with Einstein could be summarized as follows:

Peter: What is "the gravitational field"? It is not a real mathematical object
Einstein: What is a "region of spacetime"? It is not a real physical object.

In my experience it can be interesting to poll opinions, and to inform onlookers about different points of view; however discussions of that type are useless.

 Quote by martinbn That is not true, the clock will pass the horizon, and its proper time will go after 3:00pm.
Not in "Schwartzschild" time t of the Schwartzschild equation. Did you calculate it?
Once more: According to the Schwartzschild equations as used by Einstein and Oppenheimer, that clock will only reach the horizon (and indicate 3:00pm) at t=∞. In common physics that means "never"; however some smart person (in fact, who?) invented a different interpretation. In that different interpretation, which I still don't fully understand, the clock will pass the horizon despite Schwartzschild's t=∞.

For details, see the ongoing discussion: http://www.physicsforums.com/showthread.php?t=651362
incl. an extract of Oppenheimer-Snyder: http://www.physicsforums.com/showpos...5&postcount=50

 Quote by harrylin Not in "Schwartzschild" time t of the Schwartzschild equation. Did you calculate it? For details, see the ongoing discussion: http://www.physicsforums.com/showthread.php?t=651362
This makes no sense. The clock shows its proper time, nothing else, and it will not stop when it reaches 3:00pm.

 Quote by martinbn This makes no sense. The clock shows its proper time, nothing else, and it will not stop when it reaches 3:00pm.
A so-called "asymptotic observer" predicts that it will slow down so much that it will not reach 3:00pm before the end of this universe. However, a "Kruskal observer" says that that is true from the viewpoint of the asymptotic observer but predicts that the clock will nevertheless continue to tick beyond 3:00pm.

PeterDonis and PAllen say that these predictions do not contradict each other. And that still makes no sense to me, despite lengthy efforts of them to explain this to me.

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 Quote by harrylin And a quick comment on that quick comment: I don't see the qualitative difference with "The light speed limit doesn't exist; the tachyon space works just fine, it is the same solution. You just have to deal with technical problems to get through c".
It does require closer inspection to see if the apparent singularity in the equations of motion is removable or not.

What do I mean by a removable singularity?

[quote]
http://en.wikipedia.org/w/index.php?...ldid=507006469

 n complex analysis, a removable singularity (sometimes called a cosmetic singularity) of a holomorphic function is a point at which the function is undefined, but it is possible to define the function at that point in such a way that the function is regular in a neighbourhood of that point. For instance, the function $$f(z) = \frac{\sin z}{z}$$ has a singularity at z = 0. This singularity can be removed by defining f(0) := 1, which is the limit of f as z tends to 0. The resulting function is holomorphic.
It's been known for a very long time that in the black hole case that the singularity is removable.

IT does takes a bit of work to decide if the apparent singularity is the result of a poor coordinate choice , or is an inherent feature of the equations.

It might be helpful to give a quick example of how this happens. Consider the equations for spatial geodesics on the surface of the Earth. (Why geodesics? Because that's how GR determines equation of motion. So this is an easy-to-understand application of the issues involved in finding geodesics).

If you let lattitude be represented by $\psi$ and longitude by $\phi$, then you can write the metrc $ds^2 = R^2 (d \psi^2 + cos^2(\psi) d\phi^2)$ and come up with the equations for the geodesic (which we know SHOULD be a great circle) for $\psi(t)$ and $\phi(t)$

$$\frac{d^2 \psi}{dt^2} + \frac{1}{2} \sin 2 \psi \left( \frac{d \phi}{dt} \right)^2$$$$\frac{d^2 \phi}{dt^2} - 2 \tan \psi \left(\frac{d\phi}{dt}\right) \left( \frac{d\psi}{dt} \right) = 0$$

Now, one solution of these equations is $\phi$ = constant. It makes both equations zero. It's also half of a great circle. But, if we look more closely, we see that we have a term of the form 0*infinity in the second equation as we approach the north pole, because of the presence of $\tan \psi$ when $\psi$ reaches 90 degrees.

THis apparent singularity is mathematical, not physical. If you're drawing a great circle around a sphere, there's no physical reason to stop at the north pole.

Of course we already know what the answer is - we need to join two half circles together. In particular, we know we need to splice together two half circles, 180 degrees apart in lattitude, though as far as I know all the solution techniques (change of variable, etc) are equivalent to not using lattitude and longitude coordinates at the north pole, because the coordinates are ill-behaved there.

The same is in the black hole case, though to justify it you need to either do the math yourself, or read a textbook where someone else has. Note that you probably won't find this sort of thing in papers so much, it's assumed everyone knows it in the literature. Where you're more likely to find an explanation in a textbook or lecture notes.

Which brings me to the next point.

We don't have textooks online, but we've got several good sets of lecture notes.

What does Carroll's lecture notes have to say on the topic?
He defines the geodesic equation of motion - they're pretty complex looking, and I wouldn't be surprised if you didn't want to solve them yourself. But what does Caroll have to say about solving them?

I'll give you a link http://preposterousuniverse.com/grno...otes-seven.pdf, and a page reference (pg 182) in that link.

Then I'll give you some question

1) Does Carroll support your thesis? Or does he disagree with it?
2) What do other textbooks and online lecture notes have to say?

And for my own information
3) Do you think you know the difference between "absolute time" and "non-absolute time"
4) Do you think your argument about "time slowing down at the event horizon" depends on the existence of "absolute" time?

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 Quote by harrylin A so-called "asymptotic observer" predicts that it will slow down so much that it will not reach 3:00pm before the end of this universe. However, a "Kruskal observer" says that that is true from the viewpoint of the asymptotic observer but predicts that the clock will nevertheless continue to tick beyond 3:00pm. PeterDonis and PAllen say that these predictions do not contradict each other. And that still makes no sense to me, despite lengthy efforts of them to explain this to me.
Here's an analogy that may help it make a bit more sense.
Consider an ordinary boring constant-velocity special relativity problem: You are rest and you watch me passing by at some reasonable fraction of the speed of light, so you observe that my clock is ticking more slowly than yours. If the universe has a a finite age, it is certainly possible for me to observe a time on my clock that you will claim will never be reached - all that necessary is that:
1) I get to read my clock on my worldline before it terminates at the end of the universe.
2) Your worldline terminates at the end of the universe before it intersects the line of (your) simultaneity through the event of me reading my clock.

But, you will say, I'm cheating by introducing this arbitrary "end of the universe" to cut off your worldline (actually, you introduced it - I'm just abusing it )before it can intersect the relevant line of simultaneity. If I didn't do that, then no matter how much of my time passes before I read my clock, you'd be able to extend your worldline to intersect the line of simultaneity. That is true enough, but then again the entire concept of "line of simultaneity" only really makes sense in flat space.

The bit about a "Kruskal observer" is a red herring. The geometry around a static non-charged non-rotating mass is the Schwarzchild geometry, no matter what coordinates we use, and the only meaningful notion of time that we have is proper time along a time-like worldline. The Kruskal coordinates allow us to calculate the proper time along the infalling clock's worldline as it crosses the Schwarzchild radius, whereas the the Schwarzchild coordinates (as opposed to geometry) do not. So it's not that the "Schwarzchild observer" and the "Kruskal observer" are producing conflicting observations, it is that the Kruskal coordinates are producing a prediction for the infalling observer's worldline and the Schwarzchild coordinates are not.

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 Quote by PAllen I only have time for one quick comment: But that is the way all coordinate transforms work in differential geometry! That is the whole point! The the metric is transformed along with coordinates so all geometric properties (lengths, angle, intervals, etc.) are preserved as mathematical identities.
Okay, I have kind of working hypothesis about how this works.
We have global coordinate system where we know how to get from one place to another i.e. it provides connection, but this global coordinate system does not tell anything useful about distances and angles and such. And then we have another coordinate system that tells us distances and angles but it works only locally, meaning that if we have two adjacent patches with local coordinate systems we don't know how to glue them together.
So we take take global coordinate system with metric that will give us geometric values in accord with local coordinate systems.

Something like that. Only I don't know how to check if this is right.

 Quote by PeterDonis Since you're so insistent on doing calculations in Schwarzschild coordinates, try this one: write down the equation defining the proper time of an object freely falling radially inward from a finite radius r = R > 2M, to radius r = 2M. Write it so that the proper time is a function of r only (this is straightforward because it's easy to derive an equation relating r and the Schwarzschild coordinate time t, so you can eliminate t from the equation). This equation will be a definite integral of some function of r, from r = R to r = 2M. Evaluate the integral; you will see that it gives a finite answer. Therefore, the proper time elapsed for an infalling object is finite, even according to Schwarzschild coordinates. .
 Quote by Austin0 Correct me if I am wrong but it appears to me that the integration of proper falling time does not have a finite value..
 Quote by PeterDonis Yes, it appears that way, if you just try to intuitively guess the answer without deriving it. But when you actually derive it, you find that it *does* give a finite answer, despite your intuition.
 Quote by Austin0 It asymptotically approaches a finite limit.

 Quote by PeterDonis This is equivalent to saying the proper time integral *does* have a finite value. If you try to evaluate the integral in the most "naively obvious" way in Schwarzschild coordinates, you have to take a limit as r -> 2m, since the metric is singular at r = 2m; but the limit, when you take it, is finite..
From the statement the limit "does" have a finite value can I assume you are basing this on a mathematical theorem "proving" that such limits at 0 or infinity resolve to definite values??? While I understand the truth of such a theorem within the tautological structure of mathematics and also it's practical truth as far as, for most applications in the real world, the difference becomes vanishingly small (effectively vanishes) this does not imply that it necessarily has physical truth.

Example: Unbounded coordinate acceleration of a system under constant proper acceleration as t ---->∞

Mathematically you can say this resolves to c but in this universe as we know it or believe it to be, this is not the case.

What you are doing here seems to me to be equivalent to integrating proper time of such a system to the limit as v --->c to derive a finite value. Thus demonstrating that such a system could reach c in finite time even if it never happens according to external clocks..

The analogy is particularly apt as by assuming the free faller reaches the horizon this is also equivalent to reaching c relative to the distant static observer yes??

What difference do you see between the two cases????

In both cases it is equivalent to directly assuming reaching c or the horizon independent of determining whether they could actually arrive there. And then determining a temporal value for your assumption. Just MHO

 Quote by PeterDonis However, even if you insist on doing the integral in Schwarzschild coordinates, you can still write it in a way that doesn't even require taking a limit; as I said in the previous post you quoted, you can eliminate the t coordinate altogether and obtain an integrand that is solely a function of r and is nonsingular at r = 2m, so you can evaluate the integral directly. .
The comments above apply to any method of integration but if freefall proper time is derived from the metric how does the additional dilation factor from velocity enter into this integration??
If you are directly integrating the metric without reference to coordinate time isn't this actually integrating an infinitesimal series of static clocks between infinity and 2M???

It seems to me that either the Schwarzschild metric accurately corresponds to reality outside the hole or it doesn't. But the idea that its perfectly true up to some indeterminate pathological point "somewhere" in the vicinity of the horizon seems very shakey.
Actually the idea of a horizon as a third sector of reality between inside and outside seems like a pure abstraction. Is there a surface between air and water?

 Quote by pervect And for my own information 3) Do you think you know the difference between "absolute time" and "non-absolute time" 4) Do you think your argument about "time slowing down at the event horizon" depends on the existence of "absolute" time?
Does the returning twins age difference depend on a concept of absolute time????

What if the traveling twin hangs out close to the horizon for a time before traveling back to his distant outside brother. Does his younger age indicate time slowing down at the horizon?
Does it depend on an absolute time??? Is it a coordinate effect???