Is this a type of Horizon?

In summary: I disagree. Two particles dropped one after the other from a given height will get further apart as they fall but this...In summary, a thought experiment about a spherical shell of matter may help introduce the concepts of black hole, Unruh radiation and cosmological horizons.
  • #1
Naty1
5,606
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We have talked about black hole, Unruh and cosmological horizons in these forums...I am wondering if a simple example [thought experiment] might usher in an introduction to such rather abstract horizons:

the sphercial hollow shell of matter...from the outside, a test particle is attracted just as if the shell were a point charge; but after passing thru the shell suddenly there is no gravity...the particle is suddenly no longer subject to any gravitational attractions.

For an infinitesimally thin shell, there is a discontinuous jump in gravitational potential.

This seems to have SOME similarities to the first three horizons I mentioned...for example, as a test particle approaches the hollow shell from the outside would it not appear to slow when viewed from a distant intertial frame since some time dilation occurs...increased gravitational potential. What happens to that particle's "aging" when viewed from that same distant inertial frame inside the shell? Does time continue to "slow" for that particle, so that for example, maybe a radioactive decay would appear delayed from the distant frame?? Seems like it must...Or does it resume zero gravitational behavior??

I fear I have missed something because I have never seen such a simple example discussed as an illustration of "horizons"...

Comments,insights,criticisms appreciated.
 
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  • #2
Naty1 said:
...the sphercial hollow shell of matter...from the outside, a test particle is attracted just as if the shell were a point charge; but after passing thru the shell suddenly there is no gravity...the particle is suddenly no longer subject to any gravitational attractions.
For an infinitesimally then shell, there is a discontinuous jump in gravitational potential..
Shouldn't that be discontinuous jump in gravitational gradient? Potential G00 (referenced to [URL]http://upload.wikimedia.org/math/3/1/8/318a357012f7ad029e2d8781eed017df.png[/URL] - hope that's correct usage) will be uniform within, but depressed wrt infinity so time dilation and length contraction apply just as at the exterior surface of the shell, I would say.
 
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  • #3
If correct, does it not seem like the distant observer in a zero gravitational potential observes time passing more slowly elsewhere, also at zero gravitational potential.

I'm simply making an observation, not making any right versus wrong conclusions.
 
  • #4
Q-reeus said:
Shouldn't that be discontinuous jump in gravitational gradient? Potential G00 (referenced to [URL]http://upload.wikimedia.org/math/3/1/8/318a357012f7ad029e2d8781eed017df.png[/URL] - hope that's correct usage) will be uniform within, but depressed wrt infinity so time dilation and length contraction apply just as at the exterior surface of the shell, I would say.
What length contraction?

In the Schwarzschild solution the volume inside a given area is greater than in an Euclidean space.

[tex]

V >1/6\,{\frac {{A}^{3/2}}{\sqrt {\pi }}}

[/tex]
 
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  • #5
Naty1 said:
If correct, does it not seem like the distant observer in a zero gravitational potential observes time passing more slowly elsewhere, also at zero gravitational potential.

I'm simply making an observation, not making any right versus wrong conclusions.
My limited understanding is that if the potentials are equal, any difference in time passing could only arise from that twin paradox thing of differing relative velocities - ie SR not GR. The 'real' time dilation would then be for say, one observer moving in a circle, the other 'stationary'. At very large separations I guess expanding space comes into play and that's for a cosmologist to explain!:rolleyes:
 
  • #6
Passionflower said:
What length contraction?

In the Schwarzschild solution the volume inside a given area is greater than in an Euclidean space.

[tex]

V >1/6\,{\frac {{A}^{3/2}}{\sqrt {\pi }}}

[/tex]
I was referring to the size of say an in-falling astronaut as viewed from afar. Are you saying the distant observer would notice the astronaut grow larger, not smaller as he/she approached then passed through the shell surface?
 
  • #7
Q-reeus said:
I was referring to the size of say an in-falling astronaut as viewed from afar. Are you saying the distant observer would notice the astronaut grow larger, not smaller as he/she approached then passed through the shell surface?
No, the astronaut would not change size at all, assuming the tidal forces do not stretch him.

A remote observer knowing general relativity would obviously conclude the same, however he would measure something different. Based on the light signals he receives he would see the astronaut getting longer assuming the astronaut travels head first towards the center.

And all observers would obviously agree that the volume inside the hollow sphere is larger than if the space was Euclidean for the given area of the shell.
 
  • #8
Passionflower said:
No, the astronaut would not change size at all, assuming the tidal forces do not stretch him.

A remote observer knowing general relativity would obviously conclude the same, however he would measure something different. Based on the light signals he receives he would see the astronaut getting longer assuming the astronaut travels head first towards the center.

I disagree. Two particles dropped one after the other from a given height will get further apart as they fall but this is a tidal effect which we have said we will ignore. If the falling astronaut maintains his proper length then not all parts of the astronaut are free falling so this is not the same case as the two free falling particles. An stationary astronaut standing on the shell would be length contracted by the gravitational length contraction factor sqrt(1-2m/r). Locally on the shell spacetime is Minkowskian and the stationary observer would measure the falling astronaut to be length contracted by the SR velocity gamma factor of sqrt(1-v^2) as he passed at a velocity of v. The falling velocity for an object dropped from infinity happens to a terminal velocity that has has a gamma factor that is exactly equal to the gravitational gamma factor at any point, so the observer at infinity sees the falling astronaut length contracted by the gravitational gamma factor squared (1-2m/r).

Passionflower said:
And all observers would obviously agree that the volume inside the hollow sphere is larger than if the space was Euclidean for the given area of the shell.
Yes, I would agree with this. You can gets lots more of those gravitationally length contracted astronauts in there than the Euclidean expectation. :wink:
 
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  • #9
Passionflower said:
No, the astronaut would not change size at all, assuming the tidal forces do not stretch him.
A remote observer knowing general relativity would obviously conclude the same, however he would measure something different. Based on the light signals he receives he would see the astronaut getting longer assuming the astronaut travels head first towards the center...
I'm assuming SR contraction has been excised out of this picture, as well as any effects of gravitational mechanical stressing - ie just the spatial part of the metric? Let's treat the situation on that basis. If I recall correctly, first got the idea that purely gravitational time dilation and (isotropic)length contraction were in equal measure was in 'Gravitation & Relativity' by M.G.Bowler. Made an intuitive derivation of light bending as a geometric diffraction effect exactly half length contraction and half time dilation (gradients). Just did some searching and this forum had an interesting post on this topic here: https://www.physicsforums.com/showthread.php?p=3047988#post3047988 - particularly entry #9! That entry is my understanding. Since the context there was Schwarzschild co-ords, a stationary object is being discussed, not in-falling. That is what I originally was thinking of in response to the OP - an object 'floating' inside the shell. So there are two cases of interest.

1: The astronaut A is lowered gently to the surface. Will a distant observer see A smaller, unchanged, or larger wrt there being no gravitating shell? My understanding is the first option - A will appear shrunk, as per entry #9 in referenced thread. Size will then stay the same if A hops through a manhole and floats around anywhere inside.

2: Astronaut A free-falls right through the manhole. You say the distant observer will perceive - neglecting tidal forces & SR contraction, a length increase. Can this square with the perception of a second astronaut B standing next to the manhole and waving hello-goodbye? Will not B notice only the SR contraction of A and vice versa? So if a distant observer see's A shrunk and B extended, that is purely an 'optical illusion'?
 
  • #10
Naty1 said:
the sphercial hollow shell of matter...from the outside, a test particle is attracted just as if the shell were a point charge; but after passing thru the shell suddenly there is no gravity...the particle is suddenly no longer subject to any gravitational attractions.

The particle is suddenly no longer subject to gravitational attractions, but it is still subject to gravitational time dilation. Stationary clocks inside the cavity are all running at the same rate, but they are all running slower than stationary clocks outside the shell.

Naty1 said:
For an infinitesimally thin shell, there is a discontinuous jump in gravitational potential.

There is a rapid change in gravitational force but it is not discontinuous because it changes smoothly as you pass through the shell. It only appears discontinuous in the limit that the shell has no thickness but in that case the shell would have no mass. It is the same for the gravitational potential gradient and gravitational time dilation. The gradient changes smoothly as you pass through the shell and becomes zero (flat) everywhere inside the shell. The potential (and time dilation factor) is equal everywhere inside the shell, but lower than outside the shell.

Q-reeus : You have linked the wrong thread in post #9.
 
  • #11
Q-reeus said:
If I recall correctly, first got the idea that purely gravitational time dilation and (isotropic)length contraction were in equal measure was in 'Gravitation & Relativity' by M.G.Bowler.
That is correct. The vertical coordinate speed of light is c*(1-2m/r). A local observer measures the local speed of light to be c because his rulers are length contracted by a factor of sqrt(1-2m/r) and his clocks are also time dilated by the same factor. It is also fairly easy to calculate the gravitational length contraction factor directly from the Schwarzschild metric.

Q-reeus said:
Since the context there was Schwarzschild co-ords, a stationary object is being discussed, not in-falling. That is what I originally was thinking of in response to the OP - an object 'floating' inside the shell. So there are two cases of interest.

1: The astronaut A is lowered gently to the surface. Will a distant observer see A smaller, unchanged, or larger wrt there being no gravitating shell? My understanding is the first option - A will appear shrunk, as per entry #9 in referenced thread. Size will then stay the same if A hops through a manhole and floats around anywhere inside.
This is correct.
Q-reeus said:
2: Astronaut A free-falls right through the manhole. You say the distant observer will perceive - neglecting tidal forces & SR contraction, a length increase. Can this square with the perception of a second astronaut B standing next to the manhole and waving hello-goodbye?
Not correct.
Q-reeus said:
Will not B notice only the SR contraction of A and vice versa?
Correct.
Q-reeus said:
So if a distant observer see's A shrunk and B extended, that is purely an 'optical illusion'?
A distant observer does not see B extended. He sees B length contracted by sqrt(1-2m/r) and A as length contracted by (1-2m/r) as A falls past B.
 
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  • #12
yuiop said:
I disagree. Two particles dropped one after the other from a given height will get further apart as they fall but this is a tidal effect which we have said we will ignore.
I don't think that is the point.

The light on the front will take a longer time to reach the remote observer than the light on the back even if we take into account the proper distance between the front and the back. So the radar distance increases and this is what the remote observer actually measures.

yuiop said:
Yes, I would agree with this. You can gets lots more of those gravitationally length contracted astronauts in there than the Euclidean expectation. :wink:
So what are you saying, that AND the volume increases AND the lengths get contracted? You are talking about proper lengths right?

I believe it when you can demonstrate it mathematically.
 
  • #13
Passionflower said:
The light on the front will take a longer time to reach the remote observer than the light on the back even if we take into account the proper distance between the front and the back. So the radar distance increases and this is what the remote observer actually measures.
In relativity, when we talk about measurements we normally allow for light travel times. If a car is coming towards you and you measure the length using your radar method, the car will appear shorter when it coming towards you and longer when it going away from you even at sub-relativistic speeds. This is an optical illusion due to light travel times and I am sure you will agree that the length of a car does not physically change as it goes past you at constant velocity under normal circumstances. Of course we often informally say "see" something get shorter in relativity when we really should say "measure" if we are not specifically talking about optical illusions.

Passionflower said:
So what are you saying, that AND the volume increases AND the lengths get contracted? You are talking about proper lengths right?
Length contraction in SR does not imply that the proper length of the object is getting shorter, just the coordinate length. Imagine we had an empty hollow shell with radius r=10 astronauts and a circumference of 20pi astronauts and added lots of mass to the centre. With sufficient mass we might be able to fit 20 astronauts along the radius but as far as we can tell the circumference of the shell has not changed either by the measurements of a local observer on the shell, or according to a very distant observer. If we had another much larger shell that is subject to insignificant gravity effects (so we can be fairly sure this outer shell does not change significantly) then ruler measurements from the outer shell to the inner shell will be longer implying that if anything the inner shell has contracted rather than grown. The ruler measurements inside the inner shell suggest the shell has expanded while the ruler measurements outside the shell suggest it has shrunk and the circumference measurements suggest it has not changed. It only makes sense if we consider the rulers to have length contracted physically and then everyone agrees the inner shell has not shrunk or expanded and we have a consistent picture. Has the inner shell "really" got bigger in a way that no external observer can measure or have the astronauts "really" shrunk so that you can fit more of them inside a shell of fixed circumference? I am not sure there is any way to determine what has "really" happened.
 
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  • #14
yuiop said:
Length contraction in SR does not imply that the proper length of the object is getting shorter, just the coordinate length.
Of course I agree with that.

But in a Schwarzschild solution the physical volume between two shells with two given areas is larger than if the space were Euclidean. However an extended object in this solution does not change size except for size changes due to tidal effects.
 
  • #15
yuiop said:
That is correct. The vertical coordinate speed of light is c*(1-2m/r). A local observer measures the local speed of light to be c because his rulers are length contracted by a factor of sqrt(1-2m/r) and his clocks are also time dilated by the same factor. It is also fairly easy to calculate the gravitational length contraction factor directly from the Schwarzschild metric.


This is correct.
Not correct.
Correct.

A distant observer does not see B extended. He sees B length contracted by sqrt(1-2m/r) and A as length contracted by (1-2m/r) as A falls past B.
Nice to get a few 'ticks' yuiop - glad we basically agree on this one!:smile: To be fair to passionflower, he may be right about the perceived length increase, but it needed to be specified the distant observer was aligned with both the in-falling object and the shell center. If on the other hand the distant observer, in-falling object and shell center form a right-angle (or near enough allowing for curvature), I would expect the 'optics' to be quite different, and length contraction as per your sqrt(1-2m/r) would apply. It's a matter of specifying things carefully. What bugs me is how to reconcile the Schwarzschild situation near a BH event horizon with the free-falling observer who experiences typically only a brief moment before oblivion at the singularity. How much outside time has passed during this one-way trip? I would think forever, in which case will the BH still be there - for the infaller at any distance past the event horizon in fact? But that's for another thread I guess.
 
  • #16
yuiop said:
It only makes sense if we consider the rulers to have length contracted physically and then everyone agrees the inner shell has not shrunk or expanded and we have a consistent picture. Has the inner shell "really" got bigger in a way that no external observer can measure or have the astronauts "really" shrunk so that you can fit more of them inside a shell of fixed circumference? I am not sure there is any way to determine what has "really" happened.
The space in the inner shell has expanded that is the whole point!

The space is no longer Euclidean. We have more volume than the area justifies in Euclidean space. It is not due some shortening of rods that 'fool' us, no the space inside is in fact no longer Euclidean.
 
  • #17
Passionflower said:
The space in the inner shell has expanded that is the whole point!

The space is no longer Euclidean. We have more volume than the area justifies in Euclidean space. It is not due some shortening of rods that 'fool' us, no the space inside is in fact no longer Euclidean.
In what context is it non-Euclidean? I'm thinking because the potential is constant within, space-time is everywhere flat within. So an astronaut floating around with a ruler will measure diameter and circumference as 'normal'. The space and time units will be uniformly changed wrt a distant observer, yes?
 
  • #18
Q-reeus said:
How much outside time has passed during this one-way trip? I would think forever, in which case will the BH still be there - for the infaller at any distance past the event horizon in fact? But that's for another thread I guess.

That "other" thread would be https://www.physicsforums.com/showthread.php?t=337236 but its already a long thread so it may be hard to catch up.
 
  • #19
Q-reeus said:
In what context is it non-Euclidean?
In a Euclidean space we have:

[tex]
V =1/6\,{\frac {{A}^{3/2}}{\sqrt {\pi }}}
[/tex]

In a Schwarzschild solution the volume is larger. For instance we can calculate the volume between two shells of a given area if we have the mass of the solution.
 
  • #20
Q-reeus said:
In what context is it non-Euclidean? I'm thinking because the potential is constant within, space-time is everywhere flat within. So an astronaut floating around with a ruler will measure diameter and circumference as 'normal'. The space and time units will be uniformly changed wrt a distant observer, yes?
Sorry, that was my fault introducing a shell with mass inside in a discussion with Passionflower, but the shell of the OP has no mass inside. In the original hollow shell, spacetime is indeed flat everywhere inside and there Euclidean, but the also as you say, the gravitational gamma factor is uniformly greater inside the shell than outside, so clocks run slower and rulers are shorter than outside the shell. The radius measured inside the shell will be greater than the shell circumference suggests. It is difficult to justify the reason for the shell radius or volume being greater than the Euclidean expectation is due to spacetime being non-Euclidean, when the spacetime everywhere inside the shell IS Euclidean!
 
  • #21
yuiop said:
Sorry, that was my fault introducing a shell with mass inside in a discussion with Passionflower, but the shell of the OP has no mass inside.
Thanks for clearing that one up, should have followed the threads more closely.
In the original hollow shell, spacetime is indeed flat everywhere inside and there Euclidean, but the also as you say, the gravitational gamma factor is uniformly greater inside the shell than outside, so clocks run slower and rulers are shorter than outside the shell. The radius measured inside the shell will be greater than the shell circumference suggests. It is difficult to justify the reason for the shell radius or volume being greater than the Euclidean expectation is due to spacetime being non-Euclidean, when the spacetime everywhere inside the shell IS Euclidean!
And I thought Doctor Who was just sci-fi rubbish. Silly me!:confused:
 
  • #22
yuiop said:
... when the spacetime everywhere inside the shell IS Euclidean!
Oops! Just worked out that the space inside a hollow shell is NOT Euclidean. :redface: If an observer inside the hollow shell measures radii and circumferences, he will find that Circ < 2*pi*r which is a bit weird in flat space!

Q-reeus said:
And I thought Doctor Who was just sci-fi rubbish. Silly me!:confused:
Yup, spacetime in highly curved space is just like the Tardis, much bigger on the inside than on the outside!
 
  • #23
Passionflower said:
In a Euclidean space we have:

[tex]
V =1/6\,{\frac {{A}^{3/2}}{\sqrt {\pi }}}
[/tex]

In a Schwarzschild solution the volume is larger. For instance we can calculate the volume between two shells of a given area if we have the mass of the solution.

You can calculate an invariant volume or just a Scharzchild metric volume? Does it become negative?
 
  • #24
Phrak said:
You can calculate an invariant volume or just a Scharzchild metric volume? Does it become negative?
I think you are hinting at something here. Only just realized that Schwarzschild metric is anisotropic and that may explain the strange relations that seem non-Euclidean within the shell but are maybe just an artifact of that system. At http://en.wikipedia.org/wiki/Schwarzschild_metric it reads:
"Alternative (isotropic) formulations of the Schwarzschild metric

The original form of the Schwarzschild metric involves anisotropic coordinates, in terms of which the velocity of light is not the same for the radial and transverse directions (pointed out by A S Eddington).[3] Eddington gave alternative formulations of the Schwarzschild metric in terms of isotropic coordinates (provided r ≥ 2GM/c2 [4]).
In isotropic spherical coordinates, one uses a different radial coordinate, r1, instead of r. They are related by..."
The considerably more complicated expressions follow. My hunch is using these would remove the non-Euclidean findings interior to shell.
 
  • #25
The particle is suddenly no longer subject to gravitational attractions, but it is still subject to gravitational time dilation. Stationary clocks inside the cavity are all running at the same rate, but they are all running slower than stationary clocks outside the shell.

That's what I decided after my original post.

Passion flower:
The space in the inner shell has expanded that is the whole point!

The space is no longer Euclidean.

I did not even think of that when posting initially...I don't know all the math involved, but it seems this would be expected as spacetime within the hollow spherical shell must be different from that of the distant (inifinite,inertial) observer since the time, for example, passes at different rates. That IS interesting...
 
  • #26
So is this a type of horizon??
 
  • #27
Naty1 said:
"The particle is suddenly no longer subject to gravitational attractions, but it is still subject to gravitational time dilation. Stationary clocks inside the cavity are all running at the same rate, but they are all running slower than stationary clocks outside the shell. "
That's what I decided after my original post.

Passion flower:
"The space in the inner shell has expanded that is the whole point! The space is no longer Euclidean."
I did not even think of that when posting initially...I don't know all the math involved, but it seems this would be expected as spacetime within the hollow spherical shell must be different from that of the distant (infinite,inertial) observer since the time, for example, passes at different rates. That IS interesting...
Apart from my comments in entry #24, came across the following which to me casts further doubt on the 'non-Euclidean within' business:
Birkhoff's theorem (relativity) http://en.wikipedia.org/wiki/Birkhoff%27s_theorem_%28relativity%29" [Broken] I'm no expert and may be wrong on this - will be interesting to see what washes up.
 
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  • #29
Q-reeus said:
Apart from my comments in entry #24, came across the following which to me casts further doubt on the 'non-Euclidean within' business:
Birkhoff's theorem (relativity) http://en.wikipedia.org/wiki/Birkhoff%27s_theorem_%28relativity%29" [Broken] I'm no expert and may be wrong on this - will be interesting to see what washes up.
At first I assumed that the geometry inside an empty cavity would be Euclidean then later decided it was not and now I am not sure which is true but Birkhoff's theorem suggest the first conclusion. If it is true that the space inside an empty cavity is Euclidean then this would suggest that gravitational time dilation is purely a function of gravitational potential at a point and that "gravitational length contraction" is a function of the potential gradient over a finite distance, possibly explaining why horizontal rulers do not length contract (zero potential gradient) and vertical rulers do (non-zero potential gradient).

Q-reeus said:
It was while trying to put some numbers and facts to your question that I stumbled across the interesting thought experiment that became the topic of the new thread :wink:.
 
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  • #30
yuiop said:
At first I assumed that the geometry inside an empty cavity would be Euclidean then later decided it was not and now I am not sure which is true but Birkhoff's theorem suggest the first conclusion. If it is true that the space inside an empty cavity is Euclidean then this would suggest that gravitational time dilation is purely a function of gravitational potential at a point and that "gravitational length contraction" is a function of the potential gradient over a finite distance, possibly explaining why horizontal rulers do not length contract (zero potential gradient) and vertical rulers do (non-zero potential gradient).

It was while trying to put some numbers and facts to your question that I stumbled across the interesting thought experiment that became the topic of the new thread :wink:.
My understanding was the ruler would uniformly contract as a function of potential only (neglecting 'weight' effects, and tidal forces which are second derivative of potential) just as time would dilate. However your remarks in the other thread about normal vs isotropic Scwarzschild coordinates leaves me wondering how to interpret 'uniform' - which system represents what a distant observer will actually observe? I think one thing here we can agree on, since potential within the shell is uniform, both clocks and rulers will behave independent of location within that shell. Which is why I found it so hard to accept that one could obtain a non-Euclidean result where volume and area (or diameter and circumference) had other than the usual relations. That's why I thought isotropic Schwarzschild held the key, but maybe not. Even Doctor Who would balk at the notion that there can be curved and flat at the same place and time (on second thoughts, the last few Doctors have been hyper-ventilating half-bonkers types)!:rolleyes:
 
  • #31
Q-reeus said:
I think one thing here we can agree on, since potential within the shell is uniform, both clocks and rulers will behave independent of location within that shell. Which is why I found it so hard to accept that one could obtain a non-Euclidean result where volume and area (or diameter and circumference) had other than the usual relations. That's why I thought isotropic Schwarzschild held the key, but maybe not.
I agree with your sentiments here and the same thought occurred to me. In a hollow cavity with equal gravitational potential everywhere and no gravitational force, a ruler would have no sense of orientation and vertical and horizontal would be indistinguishable, so it would be reasonable to conclude that the geometry inside the cavity must be Euclidean.

On the topic of isotropic Schwarzschild coordinates, these are simply the regular metric with the r coordinate re-labelled but it changes nothing physically and measurements using local rulers would still conclude that the region outside a gravitational body is not Euclidean.
Q-reeus said:
My understanding was the ruler would uniformly contract as a function of potential only (neglecting 'weight' effects, and tidal forces which are second derivative of potential) just as time would dilate.
That has always been my assumption too, but our conclusion that length contraction does not occur inside a hollow cavity, forces the conclusion that gravitational length contraction must be a function of gradient because the potential inside the cavity is everywhere lower inside the cavity and yet length contraction does not occur. It is also worth bearing in mind that spatial distance intervals are not measured at a single point except in the trivial case of zero distance where length contraction is not relevant.
 
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  • #32
yuiop :

"... came across the following which to me casts further doubt on the 'non-Euclidean within' business:
Birkhoff's theorem (relativity) http://en.wikipedia.org/wiki/Birkhof...8relativity%29 [Broken] , states: "Another interesting consequence of Birkhoff's theorem is that for a spherically symmetric thin shell, the interior solution must be given by the Minkowski metric; in other words, the gravitational field must vanish inside a spherically symmetric shell. This agrees with what happens in Newtonian gravitation." ..

I had forgotten all about that theorem...just like the electrostatic field inside a hollow shell which is how I first came across the idea,,

So I'm concluding now that my post #25 is wrong...if there is no gravitational field potential inside the hollow shell there is no time dilation...time must pass the same inside a thin hollow (mass or energy) shell as distantly outside the shell...

and from that I conclude that while it's an interesting test situation, it's probably not a horizon...
 
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  • #33
yuiop said:
..On the topic of isotropic Schwarzschild coordinates, these are simply the regular metric with the r coordinate re-labelled but it changes nothing physically and measurements using local rulers would still conclude that the region outside a gravitational body is not Euclidean.
Yes you are correct - I had assumed spatial components were changed relatively. On going back to http://en.wikipedia.org/wiki/Schwarzschild_metric" [Broken] it is now evident spatial measure is uniformly rescaled, and the only relative change is between spatial and temporal intervals. Which leaves the claim in #4, #19 enigmatic.:redface:
That has always been my assumption too, but our conclusion that length contraction does not occur inside a hollow cavity, forces the conclusion that gravitational length contraction must be a function of gradient because the potential inside the cavity is everywhere lower inside the cavity and yet length contraction does not occur. It is also worth bearing in mind that spatial distance intervals are not measured at a single point except in the trivial case of zero distance where length contraction is not relevant.
Can't agree here. My conclusion has been that uniform length contraction and time dilation occur with equal measure inside the shell. The key word is uniform - and that's the Euclidean bit. I think of it as a uniform 3D Cartesian grid that everywhere within is simply uniformly shrunk wrt 'infinity'. Outside the shell the grid gradually expands with radius until normal at 'infinity'.
 
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  • #34
Naty1 said:
yuiop :

"... came across the following which to me casts further doubt on the 'non-Euclidean within' business:
Birkhoff's theorem (relativity) http://en.wikipedia.org/wiki/Birkhof...8relativity%29 [Broken] , states: "Another interesting consequence of Birkhoff's theorem is that for a spherically symmetric thin shell, the interior solution must be given by the Minkowski metric; in other words, the gravitational field must vanish inside a spherically symmetric shell. This agrees with what happens in Newtonian gravitation." ..

I had forgotten all about that theorem...just like the electrostatic field inside a hollow shell which is how I first came across the idea,,

So I'm concluding now that my post #25 is wrong...if there is no gravitational field potential inside the hollow shell there is no time dilation...time must pass the same inside a thin hollow (mass or energy) shell as distantly outside the shell...

and from that I conclude that while it's an interesting test situation, it's probably not a horizon...
Naty1 - Not all bad! See previous post. There is a uniformly depressed potential inside the shell and that means time does slow and distance does shrink (that one needs sorting out with yuiop, but we will all get there I'm sure). It's just that you can't properly call it a horizon - which carries the idea of a location where light signals cannot reach.
 
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  • #35
Q-reeus said:
Can't agree here. My conclusion has been that uniform length contraction and time dilation occur with equal measure inside the shell. The key word is uniform - and that's the Euclidean bit. I think of it as a uniform 3D Cartesian grid that everywhere within is simply uniformly shrunk wrt 'infinity'. Outside the shell the grid gradually expands with radius until normal at 'infinity'.
Up to now I have been trying to reason it out with intuition but I guess the best thing is to actually do the maths.

The interior metric using units of G=c=1 is:

[tex]
d\tau^2 = \left(\frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2M_ r}{r}}}\right)^2 dt^2 - \left(1-\frac{2M_r}{r}\right)^{-1}dr^2 -r^2(d\theta^2+r^2\sin(\theta)^2d\phi^2)
[/tex]

Where [tex]M_r[/tex] is the mass contained within a radius of r and M is the mass contained within a radius of R and R>r.

For a vacuum cavity of radius r the enclosed mass is zero and the metric for a hollow cavity enclosed in an outer shell of mass M and radius R is:

[tex]
d\tau^2 = \left(\frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\right)^2 dt^2 - dr^2 -r^2(d\theta^2+r^2\sin(\theta)^2d\phi^2)
[/tex]

The time dilation factor within the cavity is obtained by setting [tex]dr=d\theta=d\phi=0[/tex] so that:

[tex]
d\tau = \left(\frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\right) dt
[/tex]

The radial spatial interval dS is equivalent to [tex]\sqrt{-d\tau^2}[/tex] and the radial length contraction factor is found by setting [tex]dt=d\theta=d\phi=0[/tex] giving:

[tex]
dS = dr
[/tex]

i.e. no length contraction in the radial direction and the same is true for circumferential lengths [tex]d\theta [/tex] and [tex]d\phi[/tex]. The fact that circumferential lengths are the same for the interior metric and the cavity metric is a requirement for the metrics to match up at their mutual boundary otherwise there would be a contradiction in the length of the boundary circumference.

The coordinate radial speed of light in the cavity is found by setting [tex]d\tau=d\theta=d\phi=0[/tex] and solving for dr/dt and the result is:

[tex]
\frac{dr}{dt} = \left(\frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\right)
[/tex]

The coordinate speed of light is the same in all the other directions so the coordinate (and local) speed of light in the cavity is isotropic.

The conclusion is that inside the cavity the geometry is completely Euclidean in both local and coordinate terms and appears Minkowskian to local observers where “local” extends to the full extent of the cavity. Clocks run slower inside the cavity than outside the cavity and rulers are the same length as they are at infinity in any orientation.

Can you have an event horizon at the boundary of the cavity? Yes. Is it stable? That is is what I am trying to find out in the other thread.
 
Last edited:

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