How does GR handle metric transition for a spherical mass shell?

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The discussion focuses on the transition of metrics in General Relativity (GR) for a static thin spherical mass shell, highlighting the Schwarzschild metric's application in the exterior region and the Minkowski metric in the interior. There is a consensus that the temporal component's transition is smooth, represented by a frequency redshift scale factor, while the behavior of spatial metric components remains unclear, particularly regarding the tangential scale factor. The conversation delves into the physical justification for the tangential scale factor's decrease from unity to a fraction of the gravitational potential, with references to the stress-energy tensor's role. Participants debate the significance of pressure in the shell's stability and the implications for the metric coefficients, emphasizing the need for a deeper understanding of the underlying physics. The thread concludes with a call for expert insights to clarify these complex interactions in GR.
  • #91
PeterDonis said:
So what I see from "far away" depends on how I look, and therefore it can't tell me whether the little objects "really are" packed more tightly, or "contracted". In a curved space, there simply isn't a unique answer to such questions for distant objects; to see how an object "really is", you have to get close to it. There's no alternative.
Allright, you've done good job explaining the perspective issue and need for local invariants, in my terms - thanks. Perhaps there is another angle to this worth looking at.
Suppose we have an inflatable sphere centered within a transparent and initially unstressed elastic medium. Inflating the sphere slightly creates radial compressive and tangential tensile hoop stresses, and corresponding small displacements - the medium expands non-uniformly. In polar coord terms, the perturbed changes in radial and tangential strain and displacement (the integration of strain over distance) can be expressed as factors operating on the polar ordinates. A tiny elastic being caught up in it all cannot sense this directly - only 'tidal' elastic strain is locally evident. Yet in the lab, there is a need to relate changed, stress-strain induced optical properties (e.g. light deflection) which require knowledge of the elastic perturbations - both strain and displacement.

No point asking elastic being who knows only 'tidal' effects. But having a good handle on medium properties and knowing the sphere inflating pressure, all parameters of interest are readily calculable. And it necessarily assumes definite 'before' vs 'after' relations that from the lab must be inferred. Do we agree that, regardless of the particular coordinate chart used, elastic deformation and total displacement of any given elastic element should here be considered physically meaningful, coordinate independent quantities (and recall it is perturbative, before/after differences we want)? I should think yes. Expressed in say polar coords, that in turn locks down the radial and tangent strain factors say, to definite relationships if proper, accurate calculations and predictions are to be possible. Allowing treatment of both local (stress/strain), and non-local (displacements, optical paths) phenomena.
I believe gravitational light bending, on a geometric interpretation, assumes something entirely analogous if I'm not mistaken. So what this amounts to is - is gravity really that different one cannot say equivalent things - perturbative factors precisely defined? Still have a hangup on this - sorry.
 
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  • #92
Q-reeus said:
Qualified agreeance on that. Still feel there are 'remotely determined' coordinate independent spatial relations one should be able to tie down. Will relate some more in another post.

The problem with spatial relations is they are tied to decisions about simultaneity. 3-space 'now' is global statement about simultaneity. This is well defined for inertial observers in flat spacetime. Otherwise, it is just not well defined, there a number of perfectly reasonable choices. Given different choices for distant simultaneity, you get different conclusions about spatial relationships.
 
  • #93
Q-reeus said:
Have always understood it that stress induced elastic/hydrodynamic energy contributions aught to be incorporated into the T00 source term - i.e. just an addition to mass density, and thought everyone else saw it that way. Obviously assuming incompressibility assumes zero contribution from that.

Yes, work done on the system by compression shows up in the energy density (rho, or T_00); that's why I corrected myself in my exchange with pervect about that.

Q-reeus said:
Thought we were just discussing the relative contribution of the 'pure pressure' terms p11, p22, p33 to m(r).

Not quite. As I said in a previous post, the formula for m(r) in MTW is generally applicable; it says that the pressure does *not* contribute directly to m(r) for any spherically symmetric object; m(r) is *just* an integral over the energy density. The pressure only contributes to the mass indirectly, through hydrostatic equilibrium.

Q-reeus said:
Which I here interpret as referring to 'p only' terms p11 etc.

Just to clarify terminology, the "pressure" p is just another name for the diagonal space components of the stress-energy tensor, in the case where that tensor is spatially isotropic; in other words, p = T_11 = T_22 = T_33. (The energy density rho = T_00.) Also, this assumes that we are in the rest frame of the object, so each little piece of it that is ascribed the pressure p is at rest in the coordinates we are using.

The way the pressure affects the metric in this scenario is through the r-r component of the Einstein Field Equation, G_rr = 8 pi T_rr. (T_rr is what I was calling T_11 above, if we are using spherical coordinates.) This equation leads to the Tolman-Oppenheimer-Volkoff equation, which describes hydrostatic equilibruim in GR:

http://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_equation

The derivation of the metric for the constant density case in MTW, which I quoted from earlier, makes essential use of this equation.
 
  • #94
Q-reeus said:
Suppose we have an inflatable sphere centered within a transparent and initially unstressed elastic medium. Inflating the sphere slightly creates radial compressive and tangential tensile hoop stresses, and corresponding small displacements - the medium expands non-uniformly...

...I believe gravitational light bending, on a geometric interpretation, assumes something entirely analogous if I'm not mistaken. So what this amounts to is - is gravity really that different one cannot say equivalent things - perturbative factors precisely defined? Still have a hangup on this - sorry.

I see the analogy you are trying to make, but let's dig into it a little deeper.

Viewing the non-Euclideanness of space around a black hole as an elastic distortion in the space has been tried; I believe Sakharov, for one, came up with a reformulation of relativity along these lines. I'm not saying it's an invalid analogy, but to make sense of it and see what it can and can't tell you, you have to first define what the "unstressed" state of the space is, so to speak. Is it the Euclidean state? Let's suppose it is.

The general method of dealing with elastic deformation (as described, for example, in the Greg Egan pages I linked to in an earlier post) is to label each point in the elastic object by its unstressed location, and use the label of a given point to track it as it moves, relative to other points, due to the stresses imposed. The analogous procedure for spacetime would be to label each event by its "Euclidean" coordinates, and interpret those as "unstressed" distances, and then track the actual, "stressed" distances relative to them. This is, in fact, what Schwarzschild coordinates can be viewed as doing; the Schwarzschild r coordinate can be viewed as the "Euclidean radius" of a point, and the actual distance given by the Schwarzschild metric can be viewed as the "stressed" distance, due to "elastic deformation" of the space.

The problem with this analogy is, as I said before, that in the spacetime case, a small object sitting at r is *not* deformed; it looks the same from every direction, just as it would in an "unstressed" flat space. The "deformation" is only visible globally, and only as a non-Euclideanness in the relationship between radial distances and tangential areas. (Note that you can't just say radial and tangential distances here, though you could say tangential *circumferences*, and some do; the key is that you can only spot the non-Euclideanness by measuring distances around an entire circle, or sphere, at "radius" r, *not* by just measuring small distances tangentially.) Also, a small object *feels* no stress just from this non-Euclideanness of space; put strain gauges in it and they will all read zero. This is *not* the case with normal elastic deformation; if I take a small spherical portion of an unstressed elastic object, label it somehow so I can see its boundary, and then stress the object, that small spherical portion will appear deformed *locally*, when I look at it from right next to it. I won't have to make global observations to spot it. And if I put strain gauges in that little spherical portion, they will register nonzero values.

Go back to the analogy with the house at the North Pole and circles around it. You can set up the same sort of "elastic" model there, where the actual surface of the Earth is "elastically deformed" from Euclidean flatness. But you can only spot the deformation by comparing complete circumferences of circles. You can't spot it by just looking locally. So what physical meaning can you ascribe to the "elastic deformation"? Since you can't spot it by looking locally, you can't ascribe any physical meaning to it locally. You can say that it's a global property of the space, but you can't tie it to anything on a local scale. And since even the observation of it from a distance depends on how you look, you're limited in the physical interpretations you can put even on the global property.
 
  • #95
PAllen said:
The problem with spatial relations is they are tied to decisions about simultaneity. 3-space 'now' is global statement about simultaneity. This is well defined for inertial observers in flat spacetime. Otherwise, it is just not well defined, there a number of perfectly reasonable choices. Given different choices for distant simultaneity, you get different conclusions about spatial relationships.
Given that the arrangement considered is static, I'm thinking simultaneity issue is referring to a gravitationally effected c as measuring stick?
 
  • #96
PeterDonis said:
The way the pressure affects the metric in this scenario is through the r-r component of the Einstein Field Equation, G_rr = 8 pi T_rr. (T_rr is what I was calling T_11 above, if we are using spherical coordinates.) This equation leads to the Tolman-Oppenheimer-Volkoff equation, which describes hydrostatic equilibruim in GR:

http://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_equation

The derivation of the metric for the constant density case in MTW, which I quoted from earlier, makes essential use of this equation.
OK no disagreement here I can see.
 
  • #97
PeterDonis said:
This is, in fact, what Schwarzschild coordinates can be viewed as doing; the Schwarzschild r coordinate can be viewed as the "Euclidean radius" of a point, and the actual distance given by the Schwarzschild metric can be viewed as the "stressed" distance, due to "elastic deformation" of the space.
Precisely the handle I've been trying to get a hold of, but not so easy.
The problem with this analogy is, as I said before, that in the spacetime case, a small object sitting at r is *not* deformed; it looks the same from every direction, just as it would in an "unstressed" flat space. The "deformation" is only visible globally, and only as a non-Euclideanness in the relationship between radial distances and tangential areas.
Sure agree and have been trying to make it clear that was understood. Obviously didn't express the analogy too clearly in #91 with
"A tiny elastic being caught up in it all cannot sense this directly - only 'tidal' elastic strain is locally evident...No point asking elastic being who knows only 'tidal' effects." Was trying to convey the analogy re local unobservability of 1st order metric effects. Basically that 'elastic being' deforms with it's surroundings, and must use a kind of 'K' factor to 'navigate' but with a limited perspective. Which answers to your later comments on that matter. Sensing only the gradients of strain, there are important properties only available - yes on an indirect inferred basis - to 'outside observer'.
(Note that you can't just say radial and tangential distances here, though you could say tangential *circumferences*, and some do; the key is that you can only spot the non-Euclideanness by measuring distances around an entire circle, or sphere, at "radius" r, *not* by just measuring small distances tangentially.)
This may be homing in on where it's at for me. My whole suspiscion has been that the spatial part of spacetime 'strains' was predicted wrongly - radial to transverse 'strain ratio' that, in passing to flat interior region with 1:1 'strain ratio', implied inconsistent functional dependence on potential. Just about ready to accept your collective wisdom on this and take a break. Had no initial idea defining spatial quantities of interest would prove so tricky. But thanks for an interesting if very circuitous ride!:zzz:
 
  • #98
Q-reeus said:
Given that the arrangement considered is static, I'm thinking simultaneity issue is referring to a gravitationally effected c as measuring stick?

No; I assume PAllen was referring to the fact that surfaces of constant Schwarzschild time, which are surfaces of constant time for static observers (who stay at the same radius r), will not be surfaces of constant time for observers that are not static. For example, observers freely falling towards the black hole will have different surfaces of simultaneity, so "space" will look different to them. I believe I brought up the fact in an earlier post that spatial slices are flat in Painleve coordinates, which is equivalent to saying that observers freely falling into the black hole will *not* see the "non-Euclideanness" of space that observers hovering at a constant radius will.
 
  • #99
Q-reeus said:
Was trying to convey the analogy re local unobservability of 1st order metric effects. Basically that 'elastic being' deforms with it's surroundings, and must use a kind of 'K' factor to 'navigate' but with a limited perspective.

I think you're twisting the analogy around here. A "local" being does *not* deform with the surroundings, in the sense you are using the term "deformation". That's the point. The K factor is *not* observable locally; it's only observable by taking measurements over an extended region. Locally, space looks Euclidean; there is no "deformation". Just as locally on Earth, its surface looks flat; we only see the non-Euclideanness of the surface by making measurements over an extended region. Furthermore, the non-Euclideanness never shows up as any kind of "strain" on individual objects. It's just a fact about the space, that it doesn't satisfy the theorems of Euclidean geometry. That's all.

I really think it's a mistake to look for a "real" physical meaning to the non-Euclideanness of space, over and above the basic facts that I described using the K factor--i.e., that there is "more distance" between two spheres of area A and A + dA, or between two circles of circumference C and C + dC, than Euclidean geometry would lead us to expect. If I start from my house at the North Pole and walk in a particular direction, I encounter circles of gradually increasing circumference. Between two such circles, of circumference C and C + dC, I walk a distance K * (dC / 2 pi), where K is the "non-Euclideanness" factor and is a function of (C / 2 pi). If space were Euclidean, I would find K = 1; but I find K > 1. So what? If I insist on ascribing the fact that K > 1 to some actual physical "strain" in the space, or anything of that sort, what is my reason for insisting on this? The only possible reason would be that I ascribe some special status to K = 1, so that when I see K > 1, I think something must have "changed" from the "natural" state of things. But why should Euclidean geometry, K = 1, be considered the "natural" state of things? What makes it special? The answer is, as far as physics is concerned, nothing does. Euclidean geometry is not special, physically. It's only special in our minds; *we* ascribe a special status to K = 1 because that's the geometry our minds evolved to comprehend. But that's a fact about our minds, not about physics.
 
  • #100
PeterDonis said:
Originally Posted by Q-reeus:
"Given that the arrangement considered is static, I'm thinking simultaneity issue is referring to a gravitationally effected c as measuring stick?"

No; I assume PAllen was referring to the fact that surfaces of constant Schwarzschild time, which are surfaces of constant time for static observers (who stay at the same radius r), will not be surfaces of constant time for observers that are not static. For example, observers freely falling towards the black hole will have different surfaces of simultaneity, so "space" will look different to them. I believe I brought up the fact in an earlier post that spatial slices are flat in Painleve coordinates, which is equivalent to saying that observers freely falling into the black hole will *not* see the "non-Euclideanness" of space that observers hovering at a constant radius will.
Interesting point of difference for free-fall, but my query was all about stationary source and observer, which is why non-simultaneity wasn't making sense to me in that context. So a more general comment was being made.
 
  • #101
PeterDonis said:
I think you're twisting the analogy around here. A "local" being does *not* deform with the surroundings, in the sense you are using the term "deformation". That's the point. The K factor is *not* observable locally; it's only observable by taking measurements over an extended region. Locally, space looks Euclidean; there is no "deformation". Just as locally on Earth, its surface looks flat; we only see the non-Euclideanness of the surface by making measurements over an extended region. Furthermore, the non-Euclideanness never shows up as any kind of "strain" on individual objects. It's just a fact about the space, that it doesn't satisfy the theorems of Euclidean geometry. That's all.
Umm...'not observable locally' in which strict sense here? There is not something here roughly analogous with 3- divergence - unobservable at a point, but locally observable in the sense of finite observables ('excess' field lines in that case) over a small but finite volume/bounding surface? In other words, is it actually necessary to do a series of complete circuits around the globe to detect the geometry? Specific example - fill a standard container of local volume V with tiny, uniform hard spheres and count them as N. Take another container that by linear ruler measuring has volume nV, and do a refill/recount, and the numbers would always be nN? And this would ccontinue to hold at any r (within reason - not for BH obviously)? Seems strange, but if that's the case, think that's provided just the handle I want!
[EDIT: Better pin down the matter of measuring that container. If one measured with ruler always held either radial or in tangent plane, and rotated the container to make say, length vs diameter measurements for a cylindrical container, it would or wouldn't matter if instead one held container fixed and reoriented the ruler in measuring?]
EDIT 2: Occurred to me now this is probably more like the situation of Ehrenfest paradox - so any 'divergence' probably of such high order as to be nearly unobservable locally. So - a good example of where effect of 'non-euclideanness' can only be appreciated by non-local (or in this instance, non-rotating) observer.
 
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  • #102
Q-reeus said:
Specific example - fill a standard container of local volume V with tiny, uniform hard spheres and count them as N. Take another container that by linear ruler measuring has volume nV, and do a refill/recount, and the numbers would always be nN? And this would continue to hold at any r (within reason - not for BH obviously)?

Yes, if the respective counts of little objects are N and nN, then local measurements of volume will find V and nV. (My personal preference would be to phrase it the way I just did: take two containers and fill them with little identical objects, and arrive at the counts N and nN. Then measure the volume of each by local measurements, and the volumes will come out to be V and nV.)

Q-reeus said:
[EDIT: Better pin down the matter of measuring that container.

Good instinct. :wink:

Q-reeus said:
If one measured with ruler always held either radial or in tangent plane, and rotated the container to make say, length vs diameter measurements for a cylindrical container, it would or wouldn't matter if instead one held container fixed and reoriented the ruler in measuring?]

No, it wouldn't matter.
 
  • #103
PeterDonis said:
Originally Posted by Q-reeus:
"If one measured with ruler always held either radial or in tangent plane, and rotated the container to make say, length vs diameter measurements for a cylindrical container, it would or wouldn't matter if instead one held container fixed and reoriented the ruler in measuring?]"

No, it wouldn't matter.
Right, and I sort of realized late it was a bit of a lame question given I had already argued that 'co-stretching' couldn't allow such. Just wondered if differential gradients of 'co-stretching' might come in somehow.
Yes, if the respective counts of little objects are N and nN, then local measurements of volume will find V and nV. (My personal preference would be to phrase it the way I just did: take two containers and fill them with little identical objects, and arrive at the counts N and nN. Then measure the volume of each by local measurements, and the volumes will come out to be V and nV.)
Interesting. Hopefully one more clue here will fill the puzzle to my level of satisfaction. [STRIKE]Taken to extreme, as V grows it will eventually engulf the source and something has to give with that invariance, surely. So could this be analogous with the case of say the straight current carrying wire. Line integral of B is zero unless enclosing wire? In shell case, K effect is only non-zero (K > 1) if shell is enclosed within any 'counting volume/areas'?[/STRIKE]
[Latest take on that. If fractional excess volumetric particle count between two concentric shells is a function of radius r, this 'must' be true for subdivided portions - conic sections through the shells say. So I'm under the strong impression it really boils down to a kind of spatial divergence - the small counting spheres are only capable of being a reference if their relative volumetric expansion is negligible compared to much larger container volume. However it's not just relative volume that matters. Expanding volume in tangent directions (wider conic solid angle) makes no change, but expanding in radial direction will. A directed non-euclidean effect that must to some extent be 'locally' observable. What to call this beast apart from 'delta K effect' I don't know but certainly imo physics not just coordinate peculiarity. My take on what's fundamentally going on, but bound to be shot down s'pose.]
 
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  • #104
I did some digging and found a paper on the metric of a photon gas star (without the shell).

http://arxiv.org/abs/gr-qc/9903044

The general solution is numerical, but there's one solution that's simple that's an "attractor" to the numerical solutions:

<br /> \frac{7}{4}\, dr^2 + r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \sqrt{\frac{7}{3}}\,r\,dt^2<br />

This corresponds to a photon gas with a density per unit proper volume of 3 / (7 r^2) (the density has to depend on r), and a pressure per unit volume in each direction of one third of that. (This later was calculated by me to confirm it was a photon gas solution, it wasn't in the paper).

As usual, you need to specify an orthonormal co-frame basis to see that the actual density is in fact constant. The long way of doing it is to say that you transform the metric so it's locally Minkowskian, and take the density in that locally Minkowskian transformed space.
 
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  • #105
pervect said:
As usual, you need to specify an orthonormal co-frame basis to see that the actual density is in fact constant. The long way of doing it is to say that you transform the metric so it's locally Minkowskian, and take the density in that locally Minkowskian transformed space.
Is this saying that there are no locally measurable physical consequences? As a further elaboration on what I concluded in #103, one should be able to notice the following: Make the container shape a slender tube, fluid filled and with a fine 'breather' capillary tube sticking out one end. Orienting the tube axis in the tangent plane will give some reading for height of fluid in the capillary (think of old style mercury thermometer). Orient tube along radial direction, at same mean radial position r, and the level in capillary will drop - differential rate of 'volume expansion/contraction' along r direction is such that 'expanded volume' in tube portion nearest source of gravity wins over opposite effect in portion furtherest from source. this is just a reinterpretation of physical implications of K factor imo.

Further, one could take a fluid filled spherical container (again with a capillary tube sticking out of it), and find that for inwardly directed radially displacement, fluid level in capillary will drop. This might be interpreted as a weird volumetric expansion of containment vessel - one without explanation in terms of any mechanical stress/strain. We assume here a notionally incompressible fluid and containment vessel such that the ever present tidal forces have no appreciable mechanical strain influence. So I would maintain purely metric distortions are locally observable - as gradient 'stretching' phenomena.
 
  • #106
Q-reeus said:
Orienting the tube axis in the tangent plane will give some reading for height of fluid in the capillary (think of old style mercury thermometer). Orient tube along radial direction, at same mean radial position r, and the level in capillary will drop - differential rate of 'volume expansion/contraction' along r direction is such that 'expanded volume' in tube portion nearest source of gravity wins over opposite effect in portion furtherest from source. this is just a reinterpretation of physical implications of K factor imo.

No, the K factor does not imply this. Remember that a spherical object (more precisely, an object that in flat spacetime, under zero stress, is spherical) will still be spherical if placed at radial coordinate r; the K factor does not cause any distortion in the object. There is no "distortion" in the effect on the capillary tube either, for the same reason.

Q-reeus said:
Further, one could take a fluid filled spherical container (again with a capillary tube sticking out of it), and find that for inwardly directed radially displacement, fluid level in capillary will drop.

No, it won't. See above.

Q-reeus said:
This might be interpreted as a weird volumetric expansion of containment vessel - one without explanation in terms of any mechanical stress/strain.

This is not possible; if the physical volume of the container were expanded, the containment vessel would *have* to show strain. That's part of what "physical volume" means.

Think again about what the K factor means. It does not mean that "the physical volume of a particular piece of space is expanded". That's impossible. It means that there are *more* "pieces of space", more physical volume, per unit radial coordinate than Euclidean geometry would lead one to expect. But as I said in a previous post, to view this as somehow a "distortion of space" implies that the Euclidean state is the "natural" state, so any variation from it is a "distortion" and requires some physical manifestation. That's wrong. There is nothing privileged about Euclidean geometry in physics, and the fact that the geometry of space is non-Euclidean along the radial dimension in the spacetime surrounding a gravitating object is just that: a fact about the geometry of that spacetime. Just as the fact that, in my "house at the North Pole" scenario, there is "more distance" along a given unit of the radial coordinate I defined than Euclidean geometry would lead one to expect is simply that: a fact about the geometry of the surface of the Earth. None of these facts change the behavior of physical objects locally; they only change the global structure of the geometry.
 
  • #107
I'll add one more bit to what Peter said. Your stated goal of having tidal effects ignorable guarantees you can't detect Euclidean deviations. Tidal effects are the first order influence of curvature, thus they define the minimum scale needed to detect curvature. However, if you are willing to span a relatively large distance, and have near mathematically ideal measuring devices, you can detect Euclidean deviation as follows:

You pick a configuration of 5 points in space (e.g. the vertices of the figure made by joining two tetrahedra). You set up distances and angles between them per Euclidean predictions (e.g. using round trip laser time to define distance, and laser path the define straight lines). Then, at the very end, with all angles and all but one edge length set up, the last edge will be the wrong length.

J.L. Synge, in his 1960 book, develops this 5 point curvature detector. He shows that 5 points is the minimum needed to make this work (because, for example, flat Euclidean planes can be embedded in general 4-manifolds).

[EDIT: as for scale, if you use a 10 meter device near earth, your final deviation would be 10^-20 centimers or so. Less than a millionth the radius of a proton. ]
 
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  • #108
PeterDonis said:
Originally Posted by Q-reeus:
"Orienting the tube axis in the tangent plane will give some reading for height of fluid in the capillary (think of old style mercury thermometer). Orient tube along radial direction, at same mean radial position r, and the level in capillary will drop - differential rate of 'volume expansion/contraction' along r direction is such that 'expanded volume' in tube portion nearest source of gravity wins over opposite effect in portion furtherest from source. this is just a reinterpretation of physical implications of K factor imo."

No, the K factor does not imply this. Remember that a spherical object (more precisely, an object that in flat spacetime, under zero stress, is spherical) will still be spherical if placed at radial coordinate r; the K factor does not cause any distortion in the object. There is no "distortion" in the effect on the capillary tube either, for the same reason.
Didn't really expect this to go down quietly.:rolleyes: Still having great difficulty reconciling that bit (and the remainder of your comments) with just this excised bit of mine from #103:
"If fractional excess volumetric particle count between two concentric shells is a function of radius r, this 'must' be true for subdivided portions - conic sections through the shells say. So I'm under the strong impression it really boils down to a kind of spatial divergence - the small counting spheres are only capable of being a reference if their relative volumetric expansion is negligible compared to much larger container volume..."

Thought I had it conceptually pinned down there. Do we agree that if K factor applies to excess volume between complete concentric shells, it must apply to partitioned portions. Apply a soccer-ball style tesselation over shell surface and cut through radially at the boundaries.That defines intimately joined volume segments. An observer in each segment does a count. How could the excess count by each observer not add to give just that for the whole shells? Ergo - there is an non-euclidean effect observable in a 'container'. No?!

Let's take your analogy of north pole - or anywhere on a curved spherical surface. Instead of concentric circles, just take a hoop, fill it with tiny marbles. We know that non-euclidean surface curvature means being able to fit more marbles inside the hoop than would be the case on a flat surface. But the analogy is flawed - we can move the hoop anywhere over a spherical surface and marbles fit the same. The proper analogy is more like a surface in the shape of an egg - with pointy end corresponding to the source of gravity in 'real' case. We note now that our hoop, despite having a fixed locally measured perimeter, fits more and more marbles within upon approach to the pointy end. Do you still say there will be no observable 'delta K factor'?
Originally Posted by Q-reeus:
Orienting the tube axis in the tangent plane will give some reading for height of fluid in the capillary (think of old style mercury thermometer). Orient tube along radial direction, at same mean radial position r, and the level in capillary will drop - differential rate of 'volume expansion/contraction' along r direction is such that 'expanded volume' in tube portion nearest source of gravity wins over opposite effect in portion furtherest from source. this is just a reinterpretation of physical implications of K factor imo.

No, the K factor does not imply this. Remember that a spherical object (more precisely, an object that in flat spacetime, under zero stress, is spherical) will still be spherical if placed at radial coordinate r; the K factor does not cause any distortion in the object. There is no "distortion" in the effect on the capillary tube either, for the same reason.
I see what you are saying but re my previous argument, something, albeit exceedingly tiny, is physically happening here.
Originally Posted by Q-reeus:
"This might be interpreted as a weird volumetric expansion of containment vessel - one without explanation in terms of any mechanical stress/strain."

This is not possible; if the physical volume of the container were expanded, the containment vessel would *have* to show strain. That's part of what "physical volume" means.
Not if one accepts a physical gradient of length scale operates - gradient non-euclidean is necessary if any non-euclidean at all, yes? I have given the 2-D example above - hoop+marbles on egg re seemingly impossible effects.
Think again about what the K factor means. It does not mean that "the physical volume of a particular piece of space is expanded". That's impossible. It means that there are *more* "pieces of space", more physical volume, per unit radial coordinate than Euclidean geometry would lead one to expect. But as I said in a previous post, to view this as somehow a "distortion of space" implies that the Euclidean state is the "natural" state, so any variation from it is a "distortion" and requires some physical manifestation. That's wrong...
Hope this doesn't bog down into arguing over meaning of things. Would you say that an apple falling to the ground represents physics, or 'just' an expression of non-euclidean geometry? For me, excess counts owing to non-euclidean geometry manifest as physical phenomena. Taking your example of North-pole (or anywhere on a spherical surface), surface curvature means more marbles between concentric circles than on flat ground. Yes I can call that just geometry, but since it is mass that causes the 3-space curvature in gravitational situation, that's physics to me. Stubborn me.
 
  • #109
PAllen said:
I'll add one more bit to what Peter said. Your stated goal of having tidal effects ignorable guarantees you can't detect Euclidean deviations. Tidal effects are the first order influence of curvature, thus they define the minimum scale needed to detect curvature.
My stated aim was simply to say 'we know tidal effects will matter in practice - but let's fully account for them and just look at what remains'.
...However, if you are willing to span a relatively large distance, and have near mathematically ideal measuring devices, you can detect Euclidean deviation as follows:...[EDIT: as for scale, if you use a 10 meter device near earth, your final deviation would be 10^-20 centimers or so...]
Interesting and ingenious but I'm struggling here over principle - goes without saying nothing practical to patent. (caught your post after preparing response to Peter, so some things are repetitive - sorry)
 
  • #110
Q-reeus, first of all, did you read my post in the other thread about a similar issue?

https://www.physicsforums.com/showpost.php?p=3580102&postcount=6

I think it might be relevant here.

I'm going to comment on this particular thing you say first because it may be the key to the issue:

Q-reeus said:
Not if one accepts a physical gradient of length scale operates - gradient non-euclidean is necessary if any non-euclidean at all, yes?

The K factor is not a "gradient of length scale". One meter is one meter, physically, regardless of what radial r coordinate you are at. (Or one nanometer, or one width of an atomic nucleus.) The only thing the non-Euclideanness of space affects is *how many meters* are between two concentric spheres; if K > 1, there are more meters between the spheres than the Euclidean formula predicts. That's all. It doesn't change what a meter is at all.

It may be worth thinking about this for a bit before reading the rest of what I have to say below. I'll be drawing on it.

Q-reeus said:
Do we agree that if K factor applies to excess volume between complete concentric shells, it must apply to partitioned portions. Apply a soccer-ball style tesselation over shell surface and cut through radially at the boundaries.That defines intimately joined volume segments. An observer in each segment does a count. How could the excess count by each observer not add to give just that for the whole shells? Ergo - there is an non-euclidean effect observable in a 'container'. No?!

Not if you specify the size of the container in meters. (Or in some unit of distance, anyway.) If you specify it in coordinate units, that's different; for example, if you specify it in units of radial coordinate r. But in your original container scenario, you didn't; you specified the size of the container in physical distance units. I suspect you didn't because you realized, subconsciously, that the "physical" definition of size is in meters, not coordinate units. If K > 1, there are more meters in a given unit of radial coordinate r, but each meter itself is still the same.

Q-reeus said:
Let's take your analogy of north pole - or anywhere on a curved spherical surface. Instead of concentric circles, just take a hoop, fill it with tiny marbles. We know that non-euclidean surface curvature means being able to fit more marbles inside the hoop than would be the case on a flat surface. But the analogy is flawed - we can move the hoop anywhere over a spherical surface and marbles fit the same. The proper analogy is more like a surface in the shape of an egg - with pointy end corresponding to the source of gravity in 'real' case. We note now that our hoop, despite having a fixed locally measured perimeter, fits more and more marbles within upon approach to the pointy end. Do you still say there will be no observable 'delta K factor'?

Of course there will be an observable change in the K factor; in fact, if you go back and read my "North Pole" post carefully, you will see that K varies even in that scenario, because of the way I defined the r coordinate. Constant curvature does not necessarily imply constant K.

Also, once again, there will be a "non-Euclideanness" in the number of marbles that can fit between a pair of hoops, but each marble itself remains the same size. Marbles, in this scenario, are like meters; they are the physical measure of distance. They themselves don't change, but how many of them fit between a pair of hoops does.

Q-reeus said:
Hope this doesn't bog down into arguing over meaning of things. Would you say that an apple falling to the ground represents physics, or 'just' an expression of non-euclidean geometry?

It represents physics, and the non-Euclidean geometry is one way of modeling the physics.

Q-reeus said:
For me, excess counts owing to non-euclidean geometry manifest as physical phenomena. Taking your example of North-pole (or anywhere on a spherical surface), surface curvature means more marbles between concentric circles than on flat ground. Yes I can call that just geometry, but since it is mass that causes the 3-space curvature in gravitational situation, that's physics to me. Stubborn me.

I never said it was *just* geometry. I said explicitly, when I defined the K factor, that it was a physical observable. But you have to be careful when you think about *which* physical observable it is.
 
  • #111
I'm not really following the philosophical end of this discussion, but I think I can write a bit how to enclose the photon gas metric I previously presented in a shell to join the two together properly, which should serve as an actual concrete example.

Going back to the very basics, we can use Wald's metric

-f(r)*dt^2 + h(r)*dr^2 + r^2 (d theta^2 + sin(theta) dphi^2)

and Wald's results 6.2.3, 6.2.4

(6.2.3) 8 pi rho = (r h^2)^-1 dh/dr + (1-1/h) / r^2
(6.2.4) 8 pi P = (r f h)^-1 df/dr -(1-1/h) / r^2

rho and P are not the 'coordinate' density and pressure, but the densities in the orthonormal basis given by Wald in 6.1.6, i.e. they represent the "physical" density and pressure seen by an observer in a local Minkowskii frame.

6.2.3 can be written as 6.2.6

8 pi rho = (1/r^2) d/dr [r (1-1/h) ]

If we envision a thick shell where rho=0, this immediately implies that r(1-1/h) is constant through the shell.

If we shrink the shell to zero width, (a thin shell) then we say simply that h is the same inside the shell and outside, h being the spatial coefficient of the metric. So h must match where we join together the vacuum Schwarzschild metric with our photon gas metric.

If we add together 6.2.3 and 6.2.4 we can write

8 pi (rho + P) = (dh/dr) / r h^2 + (df/dr) / rfh

which we can re-write as

d/dr (f h) / (r f h^2) = 8 pi (rho +P)

Because this is NONZERO, we can say definitely that the product of f and h is not constant. We know that h is constant. Therefore we know that f changes. With a thick shell, f changes as we progress through the shell. As we shrink the shell to zero thickness, in the limit, know that f must 'jump' suddenly, because f*h can't be constant.

This doesn't tell us "how much" the jump is, and it's rather inconvenient to use this approach to actually match the metrics, but it does tell us something important, it tells us to expect 'f' to jump suddenly.

What we can do instead is say that the mass function for our metric must equal the Schwarzschild mass paramter M

i.e.

m(r) = \int 4 pi r^2 rho(r)

Furthermore, we know what M is, because we know that the h coefficents must match, and h has the value 7/4 in our photon gas metric This implies that M/r = 3/14, as

h = 1 / (1 - 2M/r)

So if we look at the photon gas metric

<br /> \frac{7}{4}\, dr^2 + r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \sqrt{\frac{7}{3}}\,r\,dt^2<br />

We see that (1/(1-2M/r)) must be 7/4, since the value of h must match. This implies that (M/r) must be 3/14.But we know that 8 pi rho(r) = 3 /( 7 r^2) from Wald's 6.2.3, and
we just have to solve for r such that m(r) = (3/14) r, where m(r) is given by the integral of 4 pi r^2 rho(r) dr , which is just the intergal of (3/14) dr.

Unless I'm mistaken, this is satisfied for all values of r, so we pick any value of r we like, set M = (3/14) r, and use that for the exterior solution.

IF we chose r = 1

we have the original metric for r<1

<br /> \frac{7}{4}\, dr^2 + r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \sqrt{\frac{7}{3}}\,r\,dt^2<br />

and for r>1

<br /> \frac{dr^2}{1-\frac{3}{7r}}+ r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \left( 1-\frac{3}{7\,r} \right) dt^2<br />
 
  • #112
pervect said:
I'm not really following the philosophical end of this discussion, but I think I can write a bit how to enclose the photon gas metric I previously presented in a shell to join the two together properly, which should serve as an actual concrete example...
pervect: thanks for all your work there, but of what I can follow, this is throwing me:
"If we envision a thick shell where rho=0, this immediately implies that r(1-1/h) is constant through the shell." Which states that h is a function of r. But later: "We know that h is constant." I'm reading the latter to merely follow as a limit of imposing zero shell thickness - ie dr -> 0. No doubt that is missing it somehow, but can't see where. I can say nothing about the derivation of the orthonormal basis stuff - whether there are any subtle assumptions that 'chop off' higher order gradient effects for instance. You will have read my reply to Peter, so perhaps let me know where you think it all comes apart, because I maintain there must be physical effects as described earlier.
 
  • #113
PeterDonis said:
Q-reeus, first of all, did you read my post in the other thread about a similar issue?

https://www.physicsforums.com/showpost.ph...02&postcount=6
I have now - and you can read my comments on that in turn!:-p
The K factor is not a "gradient of length scale". One meter is one meter, physically, regardless of what radial r coordinate you are at. (Or one nanometer, or one width of an atomic nucleus.) The only thing the non-Euclideanness of space affects is *how many meters* are between two concentric spheres; if K > 1, there are more meters between the spheres than the Euclidean formula predicts. That's all. It doesn't change what a meter is at all.
Strictly locally, I agree that 1 meter = 1 meter. My sense is though, to explain real effects, there must be a sense to '1 meter here looks different to 1 meter there'. I call that a gradient effect. Without it, there is magic - meters just slip in somehow.
Originally Posted by Q-reeus:
"Let's take your analogy of north pole - or anywhere on a curved spherical surface. Instead of concentric circles, just take a hoop, fill it with tiny marbles. We know that non-euclidean surface curvature means being able to fit more marbles inside the hoop than would be the case on a flat surface. But the analogy is flawed - we can move the hoop anywhere over a spherical surface and marbles fit the same. The proper analogy is more like a surface in the shape of an egg - with pointy end corresponding to the source of gravity in 'real' case. We note now that our hoop, despite having a fixed locally measured perimeter, fits more and more marbles within upon approach to the pointy end. Do you still say there will be no observable 'delta K factor'?"

Of course there will be an observable change in the K factor; in fact, if you go back and read my "North Pole" post carefully, you will see that K varies even in that scenario, because of the way I defined the r coordinate. Constant curvature does not necessarily imply constant K.

Also, once again, there will be a "non-Euclideanness" in the number of marbles that can fit between a pair of hoops, but each marble itself remains the same size. Marbles, in this scenario, are like meters; they are the physical measure of distance. They themselves don't change, but how many of them fit between a pair of hoops does.
You refer to a pair of hoops here, which seems to imply this will be a peculiarity of using polar coords - 'north pole effect'. But note carefully my example was using just one hoop - 2D counterpart of the 3D container earlier referenced to re fluid levels in capillary. As you seem to agree that marble count for that single enclosing hoop will be a function of surface curvature - *the coord system independent geometric object*, how can you then argue there will be no counterpart in 3D container, influenced by coord system independent spatial 3-curvature? What applies between two concentric hoops must apply within one container hoop. Likewise for concentric shells vs a 3D container.

You say 'meter hasn't changed, there are just more meters there than expected by Euclidean measure'. But how did those extra meters slip in exactly? Just because we are using SC's? Surely it's got to be the geometric object at work, totally independent of any coords used. Take that single bounding hoop example again. To keep it consistently 2D, rather than marbles, fill it with identically shaped tiny circular rings (mini-hoops), and demand that the ring count, for fixed packing density, remain constant irrespective of surface curvature. Only means to gaurantee that is one of two ways. Stress the containing hoop in compression, or stress the rings in tension, as surface curvature increases. Notice the manifestation of curvature now is stresses - and corresponding strains - rather than perceived perimeter expansion of containing hoop (or alternately, shrinking diameters of rings), from the pov of local observer, who just notices 'weirdness'.

One may wish to argue the interpretation as to what's behind it all ('more meters' vs non-uniform meter'), but for sure, there are physical effects - as I maintain there must be. And for me, that 'constant meter' idea is the problem here. Can't see other than a length scale gradient effect at work - not observable 'at a point'. This is hand-wavy, but can one not see an analogue with the well known example of triangles in curved space. Angles add to more than 1800 in positively curved space, but the effect is a non-linear function of triangle size - virtually non-existent in the small. And by analogy, those marbles/rings/water molecules act as 'standard meters' for the same basic reason - their distortion by spatial metric non-uniformity is miniscule. In my opinion that is.
 
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  • #114
Q-reeus said:
I have now - and you can read my comments on that in turn!:-p

I did, and I responded to them there. :wink:

Q-reeus said:
Strictly locally, I agree that 1 meter = 1 meter. My sense is though, to explain real effects, there must be a sense to '1 meter here looks different to 1 meter there'.

What is wrong with putting it the way I did in my last post? That is: 1 meter here = 1 meter there, but how many meters fit between two spheres with area A and A + dA is different here than it is there.

Q-reeus said:
I call that a gradient effect. Without it, there is magic - meters just slip in somehow.

If you think extra meters "slipping in" is "magic", why is that? The only reason I can see, is that you think the Euclidean relationship between areas of concentric spheres and the volume enclosed between them is somehow privileged. It isn't. So no "magic" is required for extra meters to be present. I don't see why this is so hard to grasp.

Q-reeus said:
You refer to a pair of hoops here, which seems to imply this will be a peculiarity of using polar coords - 'north pole effect'.

No, it isn't. I have said all along that the K factor is an invariant physical observable. You could adopt Cartesian coordinates in my north pole scenario and it would still be there; it would just be a lot harder to express in those coordinates.

Q-reeus said:
But note carefully my example was using just one hoop - 2D counterpart of the 3D container earlier referenced to re fluid levels in capillary. As you seem to agree that marble count for that single enclosing hoop will be a function of surface curvature - *the coord system independent geometric object*, how can you then argue there will be no counterpart in 3D container, influenced by coord system independent spatial 3-curvature? What applies between two concentric hoops must apply within one container hoop. Likewise for concentric shells vs a 3D container.

No, you didn't just use one hoop. You used hoops placed at different points on an egg-shaped surface, instead of on a spherical surface. (At least, that's the example I think you're referring to; if it's another, please re-state it or give a direct reference.) Did you read my response? I said constant curvature doesn't necessarily imply constant K. That means the K factor is *not* the same as curvature. It may be *related* to curvature, but it is not the same thing.

Q-reeus said:
You say 'meter hasn't changed, there are just more meters there than expected by Euclidean measure'. But how did those extra meters slip in exactly?

See above. You are assuming the Euclidean expectation is privileged. It isn't.
 
  • #115
PeterDonis said:
Originally Posted by Q-reeus:
"Strictly locally, I agree that 1 meter = 1 meter. My sense is though, to explain real effects, there must be a sense to '1 meter here looks different to 1 meter there'."

What is wrong with putting it the way I did in my last post? That is: 1 meter here = 1 meter there, but how many meters fit between two spheres with area A and A + dA is different here than it is there.
Maybe it is semantics getting in the way on that one. I think your '1 meter here = 1 meter there' statements are continually trying to clear up a non-existent conception on my part - that one would/could *locally* observe a meter changing just by moving around in a gravitational potential. No, have tried to make it abundantly clear I have never believed in such an absurdity. Rather, that move the meter rod over there into a lower grav potential, and in general it will look smaller than if done in flat spacetime. Yes, no doubt just calculating distance moved is a complication, but one that can be taken into account. So maybe we really are on the same (redshifted?) wavelength.
Originally Posted by Q-reeus:
"But note carefully my example was using just one hoop - 2D counterpart of the 3D container earlier referenced to re fluid levels in capillary. As you seem to agree that marble count for that single enclosing hoop will be a function of surface curvature - *the coord system independent geometric object*, how can you then argue there will be no counterpart in 3D container, influenced by coord system independent spatial 3-curvature? What applies between two concentric hoops must apply within one container hoop. Likewise for concentric shells vs a 3D container."

No, you didn't just use one hoop. You used hoops placed at different points on an egg-shaped surface, instead of on a spherical surface. (At least, that's the example I think you're referring to; if it's another, please re-state it or give a direct reference.)
Yes I did - where do you find me using more than one? Check my 3rd passage in #108. But suppose I used different hoops, provided they were standardized with reference to some particular starting location, how would that effect the argument? Crux of the matter is, within a single enclosing perimeter of locally measured invariant shape and size (the hoop), marble stacking density varies with surface curvature. Again, how can this not carry over more or less directly to the case of fluid-filled (water molecules = nano size marbles) container re 3-curvature effect? A rising or lowering capillary level is measurable physics. You say it won't, but where is this breakdown of analogy occurring? That is deeply baffling.
Did you read my response? I said constant curvature doesn't necessarily imply constant K. That means the K factor is *not* the same as curvature. It may be *related* to curvature, but it is not the same thing.
Accept that there is no direct relation. Having gone and re-read your #99, I see what you are driving at, but to me that situation, where 'the marbles' just sit there static on the ground, is hiding potential physics. In order for K > 1 there must be curvature - so curvature is the key operator that non-euclidean K factor manifests. I'm going to keep coming back to this same point - if there is no potential physics going on, explain or refute the matter of a locally invariant loop experiencing a varying marble area density, just by moving said hoop+marbles to a region of higher curvature. Then go back to the fluid filled container analogue, and tell me why level in the capillary would not vary with change of radial location r for container. And note, it has nothing to do with 'distortion' of the capillary tube, which is merely an indicator of change. :zzz:
 
  • #116
Q-reeus said:
Maybe it is semantics getting in the way on that one. I think your '1 meter here = 1 meter there' statements are continually trying to clear up a non-existent conception on my part - that one would/could *locally* observe a meter changing just by moving around in a gravitational potential. No, have tried to make it abundantly clear I have never believed in such an absurdity. Rather, that move the meter rod over there into a lower grav potential, and in general it will look smaller than if done in flat spacetime.

What does "look smaller" mean? Part of the issue may be the continual temptation to use ordinary English words that have imprecise or ambiguous meanings. It's very important to resist that temptation, and to phrase things carefully in terms of actual observables. (For one thing, "look smaller" involves light and how light paths are changed by gravity in the intervening space, and you've ruled all that out of bounds--we're supposed to assume that all that has been corrected for.)

Q-reeus said:
Check my 3rd passage in #108.

Do you mean the following?

Q-reeus said:
Instead of concentric circles, just take a hoop, fill it with tiny marbles. We know that non-euclidean surface curvature means being able to fit more marbles inside the hoop than would be the case on a flat surface

If so, I must have missed it previously, because I should have objected, or at least clarified. I see how it refers to one hoop, but I also think it's false, unless I'm misunderstanding how you're defining a "hoop". I was assuming "hoop" meant a *single* line of marbles going around the circumference of a circle centered on the "North Pole"; combined with the assumption that the marbles themselves are so small that they can be used as little identical objects to measure distances to any accuracy we need for the problem, then the number of marbles that fit inside a "hoop" is determined by the hoop's circumference and nothing else. Since the circumference is tangential, it is unaffected by any "spatial distortion", regardless of anything else; that's the fixed point of departure that we both agree on. So the quoted sentences above are false if "hoop" means what I think it means.

If, on the other hand, by "hoop" you mean "two circles of slightly different circumferences, C and C + dC, plus the space between them", then we've been using "hoop" to mean different things. In the following quote, it looks like you're using "hoop" in this other sense, but if you were doing that in post #108 I didn't understand that. I was using the word "circle" to avoid such ambiguity. However, I'm not entirely sure, because in the following quote you still seem to equivocate about how "local" a hoop is. See below.

Q-reeus said:
Crux of the matter is, within a single enclosing perimeter of locally measured invariant shape and size (the hoop), marble stacking density varies with surface curvature.

If the hoop has a "locally measured invariant shape and size", then it *must* be a "hoop" in the sense I was thinking--a *single* circle, with circumference C, and that's all. As I noted above, such a "hoop" must always contain the *same* number of marbles for a given "size" of hoop (i.e., circumference). If the number of marbles placed within a "hoop" of a given "size" can vary, then a "hoop" *cannot* be a single circle--it must be, as I noted above, two circles of slightly different circumferences, C and C + dC, plus the space between them. In this case, yes, the number of marbles placed within a hoop can vary, even if dC is held constant. But that just means the hoop does *not* have a "locally measured invariant shape and size".

Q-reeus said:
In order for K > 1 there must be curvature - so curvature is the key operator that non-euclidean K factor manifests.

This is basically correct, with the proviso that K does not *equal* the curvature; it is *related* to it, but not the same.

Q-reeus said:
if there is no potential physics going on, explain or refute the matter of a locally invariant loop experiencing a varying marble area density, just by moving said hoop+marbles to a region of higher curvature.

I did, by refuting your assumption that it is a "locally invariant loop". For a varying number of marbles to be seen in a "loop", the "loop" (you keep on changing words, and it doesn't help with clarity) cannot be "locally invariant"; that's obvious. More precisely, a region between two spheres of areas A and A + dA, or between two circles of circumference C and C + dC can vary in size as A or C change, even if dA or dC are held constant. That's the definition of the K factor, and I've said all along that it's a physical observable and represents "real physics" going on.

The only thing I am disagreeing with you about is that you are expecting this real physics to show up in a way that it does not, in fact, show up. The reason it does not show up the way you are expecting it to is that your expectation is based on giving a privileged status to the predictions of Euclidean geometry. In fact, there is no such privileged status. I've said that repeatedly, too, and you haven't picked up on it, or if you have, it hasn't shown in your posts. There's no point in continuing to wonder if this is about semantics, or if I think there's real physics going on. I've made all that clear multiple times. The thing to focus in on is why you believe Euclidean geometry has a privileged status, so that any departure from Euclidean geometry, meaning any K factor that is not equal to 1, requires some special manifestation over and above what I've already defined as the observable K factor.
 
  • #117
Q-reeus said:
Let's take your analogy of north pole - or anywhere on a curved spherical surface. Instead of concentric circles, just take a hoop, fill it with tiny marbles. We know that non-euclidean surface curvature means being able to fit more marbles inside the hoop than would be the case on a flat surface. But the analogy is flawed - we can move the hoop anywhere over a spherical surface and marbles fit the same. The proper analogy is more like a surface in the shape of an egg - with pointy end corresponding to the source of gravity in 'real' case. We note now that our hoop, despite having a fixed locally measured perimeter, fits more and more marbles within upon approach to the pointy end. Do you still say there will be no observable 'delta K factor'?

PeterDonis said:
I was assuming "hoop" meant a *single* line of marbles going around the circumference of a circle centered on the "North Pole"; combined with the assumption that the marbles themselves are so small that they can be used as little identical objects to measure distances to any accuracy we need for the problem, then the number of marbles that fit inside a "hoop" is determined by the hoop's circumference and nothing else. Since the circumference is tangential, it is unaffected by any "spatial distortion", regardless of anything else; that's the fixed point of departure that we both agree on. So the quoted sentences above are false if "hoop" means what I think it means.

If, on the other hand, by "hoop" you mean "two circles of slightly different circumferences, C and C + dC, plus the space between them", then we've been using "hoop" to mean different things. In the following quote, it looks like you're using "hoop" in this other sense, but if you were doing that in post #108 I didn't understand that. I was using the word "circle" to avoid such ambiguity. However, I'm not entirely sure, because in the following quote you still seem to equivocate about how "local" a hoop is.
I thought Q-reeus was completely filling the interior of a hoop with marbles -- in other words he is comparing the area of a circular disk with its circumference.

Any deviation from the expected 2D Euclidean value would be apparent only for relatively large (i.e. "non-local") hoops. Hoops that are small enough to be regarded as "local" would be too small for the deviation to be measurable -- that's pretty much what we mean by "local" as used in the Equivalence Principle.
 
  • #118
Unless Synge is wrong, even over a large region, you cannot detect Euclidean deviation on a plane. You need something 3-d, like Peter's concentric spheres (not circles). If Synge is right, the even over large regions, you can construct a Euclidean tetrahedron. You need one more vertex than a tetrahedron to detect Euclidean deviation.
 
  • #119
I haven't seen Synge's derivation (I think I tried to find it once, but I know I never managed to get a hold of it), but I'm confident you can detect intrinsic curvature of a plane with 4 points, so I find it logical to believe you can detect space curvature with 5.

Whether you use my method for detecting the intrinsic curvature of a plane (which involves comparing the ratio of the diagonals of a square to the side of a square, the square by definition having four equal sides and two equal diagonals), or the marble packing method suggested by Q, you can detect the intrinsic curvature of a plane embedded in a 3d space.

What you can't do is tell if said intrinsic curvature is due to the way the plane is embedded. Thus information on the intrinsic curvature of a single plane embedded in a higher 3d space-time doesn't directly tell you anything about the intrinsic curvature of the space it's embedded in, the intrinsic curvature of the plane could result from the way it's embedded.

A trivial example: The Earth's surface is curved, and not flat. In fact, thinking about ways to detect the curvature of the Earth's surface (while staying on the surface) is a good way to get comfortable with the concepts and properties of curvature.

But the fact that the Earth's surface is curved (has an intrinsic curvature) doesn't tell you anything about whether or not space or space-time the Earth is in is curved.

I've heard that you can decompose the Rieman into "sectional curvatures" of planes, but I'm a bit hazy about the details. Clearly, though, you need information on the intrinsic curvature of a lot of planar slices of your space-time, not just one.
 
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  • #120
A little side note on Synge is that well before Kruskal and Szekeres, Synge was the first to untangle the fully extended topology of SC geometry (though he didn't come up with K-S coordinates). MTW makes note of his clear priority as the first ever to work it all out.
 

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