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

by Q-reeus
Tags: handle, mass, metric, shell, spherical, transition
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Q-reeus
#91
Oct23-11, 11:03 AM
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Quote Quote by PeterDonis View Post
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.
PAllen
#92
Oct23-11, 11:36 AM
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Quote Quote by Q-reeus View Post
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.
PeterDonis
#93
Oct23-11, 01:34 PM
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Quote Quote by Q-reeus View Post
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.

Quote Quote by Q-reeus View Post
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.

Quote Quote by Q-reeus View Post
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%...lkoff_equation

The derivation of the metric for the constant density case in MTW, which I quoted from earlier, makes essential use of this equation.
PeterDonis
#94
Oct23-11, 01:51 PM
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Quote Quote by Q-reeus View Post
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.
Q-reeus
#95
Oct23-11, 03:38 PM
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Quote Quote by PAllen View 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.
Given that the arrangement considered is static, I'm thinking simultaneity issue is referring to a gravitationally effected c as measuring stick?
Q-reeus
#96
Oct23-11, 03:41 PM
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Quote Quote by PeterDonis View Post
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%...lkoff_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.
Q-reeus
#97
Oct23-11, 03:50 PM
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Quote Quote by PeterDonis View Post
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!
PeterDonis
#98
Oct23-11, 04:49 PM
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Quote Quote by Q-reeus View Post
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.
PeterDonis
#99
Oct23-11, 05:01 PM
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Quote Quote by Q-reeus View Post
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.
Q-reeus
#100
Oct24-11, 07:15 AM
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Quote Quote by PeterDonis View Post
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.
Q-reeus
#101
Oct24-11, 07:18 AM
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Quote Quote by PeterDonis View Post
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.
PeterDonis
#102
Oct24-11, 09:00 AM
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Quote Quote by Q-reeus View Post
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.)

Quote Quote by Q-reeus View Post
[EDIT: Better pin down the matter of measuring that container.
Good instinct.

Quote Quote by Q-reeus View Post
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.
Q-reeus
#103
Oct24-11, 09:56 AM
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Quote Quote by PeterDonis View Post
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. 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'?
[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.]
pervect
#104
Oct24-11, 08:52 PM
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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:

[tex]
\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
[/tex]

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.
Q-reeus
#105
Oct25-11, 05:23 AM
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Quote Quote by pervect View Post
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.
PeterDonis
#106
Oct25-11, 03:22 PM
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Quote Quote by Q-reeus View Post
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.

Quote Quote by Q-reeus View Post
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.

Quote Quote by Q-reeus View Post
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.
PAllen
#107
Oct25-11, 04:36 PM
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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. ]
Q-reeus
#108
Oct25-11, 05:24 PM
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Quote Quote by PeterDonis View Post
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. 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.


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