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.
  • #61
DaleSpam said:
By the way, the burden of proof is always on the person challenging mainstream established science.
We've discussed this before, and my response was this is a forum, not a peer-review panel of some prestigious journal, and I'm not a specialist presenting a paper for publishing.
However, I will go ahead and calculate the metric for the sphere and the sphere with gas, mostly for my own practice since I don't believe that it will make a difference to you.
Will be most interested how it is arrived at, how order of unity quantity can be shaped by order of one trillionth of a trillionth effect.
 
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  • #62
PAllen said:
Q-reeus: Can you try to succinctly state your issue(s) in the context of SC geometry fitted to interior Minkowsdie geometry, with no matter shell at all. Despite the disconinuty in metric derivatives (but continuity of metric itself) all physical observables even ai femtometer away from the 0 thickness shell are well defined. This situation is no different from the junction of ideal inclined plane with a plane. Continuity combined with discontinuity of derivative. Yet this is routinely considered a plausible idealization. So please phrase some specific objection you have to GR physics of the zero width shell.
Even with finite thickness shell wall there are discontinuities in derivatives and that I have no problem accepting. Maybe my prejudice but infinitely thin shell sounds like pure boundary matching exercise that can hide physics. So would prefer to stick with explanations involving finite thickness. Later posting wil try and summarise afresh.
 
  • #63
PeterDonis said:
But as we keep on saying, it's not the *mass* (or energy density) of the air or the shell that matters, but the spatial stress components, and you specified the scenario in such a way that those can't be negligible: you specifically said that the puff of air had enough pressure to significantly change the stress components inside the shell. That all by itself is enough to ensure that the stress-energy tensor of the puff of air is enough to make spacetime inside the shell non-flat, to the level of accuracy you are assuming...
That is the key sticking point. More later on that. Thanks for the link to Egan's site - a huge resource. Not finding the particulars wanted yet, but will keep looking.
 
  • #64
PeterDonis said:
The metric inside the static spherical object is:
ds^{2} = - \left( \frac{3}{2} \sqrt{1 - \frac{2 M}{R}} - \frac{1}{2} \sqrt{1 - \frac{2 M r^{2}}{R^{3}}} \right)^{2} dt^{2} + \frac{1}{1 - \frac{2 m(r)}{r}} dr^{2} + r^{2} d\Omega^{2}

...Note that arriving at this result, as the details in MTW show, requires taking the pressure inside the object into account as well as the density.

...So the ratio of pressure to energy density is indeed similar to the ratio of mass to radius (in geometric units); but that's still enough to have an effect on the metric.

But anything like big enough? I must be missing something basic here, because we all agree shell stresses are utterly minute compared to matter in gross effect. The exterior SM, and interior MM level, owe essentially exclusively to the matter contribution, and shell geometry. Nothing else. Microscopic effects of stress cannot be doing much, regardless of how they 'point', surely! I can only conclude, since all of you insist there is no vast order of magnitude chasm to ford here, that a changing K vs J in shell wall is some kind of mirage, a mathematical artefact of coordinate system. Is this where it's at - is the metric actually an isotropic one according to my conception? I could then believe there is no issue, but can't see it is that way.
Perhaps it best to summarize again the principle issues as I have till now seen them.

1: What SC's are really saying about SM. On a straight reading of the standard SC's

e55cd5c7e42dfd5865febb4757f96fb6.png

it is evident potential operates on the temporal, and spatial r components, but not at all on the tangent components. And that this is referenced to coordinate measure, even if for spatial measure it can only be 'inferred' not directly measured (Actually even for temporal component one must infer that clocks tick slower rather than light 'loses energy in climbing out' of potential well). Hence we find frequency slows by factor J = 1-rs/r, and for a locally undistorted ruler placed radially, inferred coordinate measure shrinks by the same factor J.

And it makes perfectly good physical sense. Direct proportionality to J means direct proportionality to depth in gravitational potential. A simple, physical linkage to relative energy level. Somehow, according to straight SC reading, tangent spatials are immune to this principle - in exterior SM region that is. To be clear about what 'immunity' means here, adding mass to the shell while compensating perfectly for any elastic strain in shell wall, a ruler horizontal on the surface will not change as viewed by a telescope looking directly down on it (negligible light bending). whereas a vertically oriented ruler will shrink by K-1 = J. That this is not evident locally is immaterial imo. Locally there is this packing ratio that will change. Fine. But is there not a 'real' transition from anisotropic to isotropic to explain? That will be evident locally - packing ratio change. And non-locally - 'inferred' ruler tangent contraction in passing through hole in the shell wall.

2: Underlying physical principle. Inferred tangent component having by SC's no metric operator in SM region, but obtains one in shell wall. And uber minute stresses explain that? Here's the problem. Diagonals in p, if added in three (isotropic pressure), are utterly puny in effect. And it's not like isotropic pressure is the difference of huge, almost cancelling terms. One just adds arithmetically. Yet take just one away - the radial component in the shell case, and lo and behold, the other two seem to acquire miraculous capabilities. Either that, or as I say, everything is 'really' isotropic and there is no 'real' transition issue to account for. That's how I see it.
Later
 
  • #65
Trying to be as succinct as possible, can you contrast where you see a problem in GR versus Newtonian gravity. In Newtonian gravity, outside the shell, there is a clear physical anisotropy - gravity points toward the shell. Across the shell, this radial force diminishes. Inside the shell there is perfect isotropy. In the weak field case, it is trivial to show GR is identical because it recovers Newtonian potential. So again, I still see no comprehensible claim about what exactly is the problem GR supposedly has.

Another take on this: it is pure mathematics that any invariant quantity computed in isotropic SC coordinates (which still, clearly, have radial anisotropy built in - redshift and coordinate lightspeed vary radially; however, coordinate lightspeed is locally isotropic) is the same as in common SC coordinates. All measurements in GR are defined as invariants constructed from the instrument (observer) world line and whatever is being measured. Thus it is a mathematical triviality that isotropic SC coordinates describe the same physics as common SC coordinates, for every conceivable measurement.

So, can you describe your objection in terms of isotropic coordinates? If you can't, your complaint is analogous to the following absurdity:

- In polar coordinates on a plane, the distance per angle varies radially. How does this effect disappear in Cartesian coordinates?
 
  • #66
Q-reeus said:
But anything like big enough? I must be missing something basic here, because we all agree shell stresses are utterly minute compared to matter in gross effect. The exterior SM, and interior MM level, owe essentially exclusively to the matter contribution, and shell geometry. Nothing else. Microscopic effects of stress cannot be doing much, regardless of how they 'point', surely!

I think I see an actual question about physics here, but you are making it far more complicated than it needs to be by mixing in coordinate-dependent concepts. See below.

Q-reeus said:
I can only conclude, since all of you insist there is no vast order of magnitude chasm to ford here, that a changing K vs J in shell wall is some kind of mirage, a mathematical artefact of coordinate system.

No, it isn't, if by K and J you mean those terms as I defined them. I specifically defined them as physical observables, so a change in their relationship is likewise a physical observable. See below.

Q-reeus said:
1: What SC's are really saying about SM. On a straight reading of the standard SC's...

Here is the problem. As I acknowledged above, you are asking a legitimate question about the physics, but you keep on thinking about it, and talking about it, in terms of things that are coordinate-dependent, which makes it very difficult to discern exactly what you are asking. The metric coefficients in Schwarzschild coordinates are *only* applicable to Schwarzschild coordinates; they don't tell you anything directly about the physics. The physics is entirely summed up in terms of the two observables, K and J, that I defined, and everything can be talked about without talking about coordinates at all.

Here's the actual physical question I think you are asking; I'll take it in steps.

(1) We have two observables, K and J, defined as follows: J is the "redshift factor" (where J = 1 at infinity and J < 1 inside a gravity well), and K is the "non-Euclideanness" of space (where K = 1 at infinity and K > 1 in the exterior Schwarzschild vacuum region).

(2) These two observables have a specific relationship in the exterior vacuum region: J = 1/K.

(3) We also have an interior vacuum region in which space is Euclidean, i.e., K = 1. However, J < 1 in this region because there is a redshift compared to infinity.

(4) Therefore, the non-vacuum "shell" region must do something to break the relationship between J and K. The question is, how does it do this?

(5) The answer DaleSpam and I have given is that, in the non-vacuum region, where the stress-energy tensor is not zero, J is affected by the time components of that tensor, while K is affected by the space components. Put another way, J is affected by the energy density--more precisely, by the energy density that is "inside" the point where J is being evaluated. K, however, is affected by the pressure.

(6) Your response is that, while this answer seems to work for J, it can't work for K, because K has to change all the way from its value at the outer surface of the shell, which is 1/J, to 1 at the inner surface. Since J at the outer surface is apparently governed by the energy density, and the change in K to bring it back to 1 at the inner surface must be of the same order of magnitude as the value of J at the outer surface, it would seem that whatever is causing that change in K must be of the same order of magnitude as the energy density. And the pressure is much, much smaller than the energy density, so it can't be causing the change.

Now that I've laid out your objection clearly and in purely physical terms, without any coordinate-dependent stuff in the way, it's easy to see what's mistaken about it. You'll notice that I bolded the word apparently. In fact, the value of J at the outer surface of the shell is *not* governed by the shell's energy density; J (or more precisely the *change* in J) is only governed by the shell's energy density *inside* the shell. At the outer surface, because of the boundary condition there, the value of J is governed by the ratio of the shell's total mass to its radius, in geometric units (or, equivalently, by the ratio of its Schwarzschild radius to its actual radius). And if you look at what I posted before, you will see that the pressure inside the shell is of the *same* order of magnitude as the ratio of the shell's mass, in geometric units, to its radius. So the pressure inside the shell is of just the right size to change K from 1/J at the outer surface of the shell, back to 1 at the inner surface of the shell.
 
  • #67
PeterDonis said:
I think I see an actual question about physics here, but you are making it far more complicated than it needs to be by mixing in coordinate-dependent concepts. See below.

No, it isn't, if by K and J you mean those terms as I defined them. I specifically defined them as physical observables, so a change in their relationship is likewise a physical observable. See below.

Here is the problem. As I acknowledged above, you are asking a legitimate question about the physics, but you keep on thinking about it, and talking about it, in terms of things that are coordinate-dependent, which makes it very difficult to discern exactly what you are asking. The metric coefficients in Schwarzschild coordinates are *only* applicable to Schwarzschild coordinates; they don't tell you anything directly about the physics. The physics is entirely summed up in terms of the two observables, K and J, that I defined, and everything can be talked about without talking about coordinates at all.

Here's the actual physical question I think you are asking; I'll take it in steps.

(1) We have two observables, K and J, defined as follows: J is the "redshift factor" (where J = 1 at infinity and J < 1 inside a gravity well), and K is the "non-Euclideanness" of space (where K = 1 at infinity and K > 1 in the exterior Schwarzschild vacuum region).

(2) These two observables have a specific relationship in the exterior vacuum region: J = 1/K.

(3) We also have an interior vacuum region in which space is Euclidean, i.e., K = 1. However, J < 1 in this region because there is a redshift compared to infinity.

(4) Therefore, the non-vacuum "shell" region must do something to break the relationship between J and K. The question is, how does it do this?

(5) The answer DaleSpam and I have given is that, in the non-vacuum region, where the stress-energy tensor is not zero, J is affected by the time components of that tensor, while K is affected by the space components. Put another way, J is affected by the energy density--more precisely, by the energy density that is "inside" the point where J is being evaluated. K, however, is affected by the pressure.

I haven't been following this thread in detail, but I"d like to say that there are known examples where J is affected by pressure. So it's wrong to think that J isn't affected by pressure. The right answer is that J and K are both affected by pressure.

My first attempt at a post wasn't too good, let's hope this one, after replenishing my blood sugar, is better.

If you have a stationary metric , you have a timelike Killing vector, and J has an especially useful coordinate-independent interpretation as the length of said vector, sqrt |\xi^a \xi_a|

If you analyze the case of a shell enclosing a photon gas, you'll find that J, measured just below the surface of the shell is 1 -(2G/R) \int \rho dV rather than 1 -(G/R) \int \rho dV The difference from unity is twice as large, you can think of this as "twice the surface gravity" if you care to think in those terms.

You can think of J as being congtrolled by the Komarr mass, which is the integral of rho+3P, i.e. the Komar mass depends on both pressure and energy density.

However, if you measure J outside the shell, you'll find a sudden increase in J (towards unity, which you can interpret as a REDUCTION of the surface gravity), and J outside the surface of the shell will be equal to 1 -(G/R) \int \rho dV as you might naievely suspect.

The reason for the anti-gravity effect is that the intergal of the tension in the spherical shell is negative. It's a form of exotic matter to have something with a tension higher than it's energy density (which in this case is being oversimplifed to zero, though you can un-over-simplify it to have a more realistic value if you want to bother and want to avoid exotic matter).

For a small system, where you can neglect the gravitational self-energy as a further source of gravity, you can say that the total volume intergal of the pressure cancels out, and the integral of rho+3P is just equal to the integral of rho as the later term is zero.
 
  • #68
pervect said:
I haven't been following this thread in detail, but I"d like to say that there are known examples where J is affected by pressure. So it's wrong to think that J isn't affected by pressure. The right answer is that J and K are both affected by pressure.

Hi pervect, yes, this is a good point; in this particular case the pressure contribution to J is negligible (I believe--see below), but in general it might not be.

pervect said:
For a small system, where you can neglect the gravitational self-energy as a further source of gravity, you can say that the total volume intergal of the pressure cancels out, and the integral of rho+3P is just equal to the integral of rho as the later term is zero.

In post #57, if you have time to look, I posted a metric from MTW Box 23.2 for the interior of a static spherical object that is not a "shell", i.e., it has no hollow portion inside it. The total mass M that appears in that metric is defined in MTW as

M = \int_{0}^{R} 4 \pi \rho r^{2} dr

I.e., M does not contain any contribution from the pressure inside the object. If I'm reading MTW correctly here, they don't intend this formula to be an approximation; it is supposed to be exact. They certainly are not assuming that the pressure is negligible compared to the energy density; they explicitly talk about their formulas as applying to neutron stars, for which that is certainly not the case. (The specific metric I wrote down is for a uniform density object, which would not describe a neutron star, but the mass formula above is supposed to be general.) They are, I believe, assuming that the material of the object is ordinary matter, not photons; is it just because of the different energy condition (i.e., no "exotic matter" in this case) that the pressure does not appear in the mass integral, and hence (if I'm reading right) does not contribute to J in this particular case?
 
  • #69
PeterDonis said:
In post #57, if you have time to look, I posted a metric from MTW Box 23.2 for the interior of a static spherical object ... The specific metric I wrote down is for a uniform density object

Schwarzschild's solution.
 
  • #70
George Jones said:
Schwarzschild's solution.

Yes. It's interesting that Schwarzschild was able to arrive at it, even with the idealization of uniform density, without knowing the Tolman-Oppenheimer-Volkoff equation. MTW's derivation of the metric makes essential use of that equation.
 
  • #71
PeterDonis said:
Hi pervect, yes, this is a good point; in this particular case the pressure contribution to J is negligible (I believe--see below), but in general it might not be.



In post #57, if you have time to look, I posted a metric from MTW Box 23.2 for the interior of a static spherical object that is not a "shell", i.e., it has no hollow portion inside it. The total mass M that appears in that metric is defined in MTW as

M = \int_{0}^{R} 4 \pi \rho r^{2} dr

I.e., M does not contain any contribution from the pressure inside the object. If I'm reading MTW correctly here, they don't intend this formula to be an approximation; it is supposed to be exact.

The formula is exact - but read the part in MTW that says that 4 pi r^2 dr is NOT a volume element.

It superfically looks like one at first glance, but isn't. 4 pi r^2 is ok, but dr needs a metric correction. Using the actual volume element, MTW also calculates the integral of rho* dV, dV being the volume element, and find that said integral is larger than the mass M. The quantity \int \rho dV is given a name, the "mass before assembly". Because there is no compression to worry about, (the pieces are modeled as not changing volume with pressure), the only work being done by assembly is the binding energy, which you can think of being taken out of the system as you assemble it - for instance, you might imagine cranes lowering the pieces into place, and work is made available in the process.

You'll see a chart, where they tabluate the binding energy for various sizes too, as I recall.
 
  • #72
pervect said:
...For a small system, where you can neglect the gravitational self-energy as a further source of gravity, you can say that the total volume intergal of the pressure cancels out, and the integral of rho+3P is just equal to the integral of rho as the later term is zero...
This bit I had initially forgotten re my 'pufff of air' thing, but recall now from Elers et al paper http://arxiv.org/abs/gr-qc/0505040 cited in #11. Yes, external to shell, complete cancellation of internal gas and shell hoop stress contributions applies. My only interest though was in how insignificant the effect of pressure on given arrangement is, and whether or not external cancellation is considered is effectively moot imo.
 
  • #73
PAllen said:
Trying to be as succinct as possible, can you contrast where you see a problem in GR versus Newtonian gravity. In Newtonian gravity, outside the shell,

there is a clear physical anisotropy - gravity points toward the shell. Across the shell, this radial force diminishes.
Sure and as indicated in #62 that kind of thing is not an issue because such are gradient functions of potential, and disappearing in the shell interior is a simple consequence of that. Ditto for tidal effects. Not where it's at for me.
Inside the shell there is perfect isotropy. In the weak field case, it is trivial to show GR is identical because it recovers Newtonian potential.
Yes and not an issue for the same reason.
...So again, I still see no comprehensible claim about what exactly is the problem GR supposedly has.
As I've tried real hard but am obviously failing to get it through, there is by the SC's this direct dependence on potential alone for certain metric components but not others that is not so trivial imo - anisotropy of metric itself that leads to 'jump' issues at a shell boundary. Gets down to my belief 'remote' view of spatial component jumps is both real and has physical significance, whereas it seems everyone else here thinks only locally observable physics - 'tidal' effects in essence, matters as far as spatial components go. Can appreciate the latter will have a reasonable smoothness across shell, so sure, none or at least a lesser problem from that perspective.
Another take on this: it is pure mathematics that any invariant quantity computed in isotropic SC coordinates (which still, clearly, have radial anisotropy built in - redshift and coordinate lightspeed vary radially; however, coordinate lightspeed is locally isotropic) is the same as in common SC coordinates. All measurements in GR are defined as invariants constructed from the instrument (observer) world line and whatever is being measured. Thus it is a mathematical triviality that isotropic SC coordinates describe the same physics as common SC coordinates, for every conceivable measurement.
So, can you describe your objection in terms of isotropic coordinates? If you can't, your complaint is analogous to the following absurdity:
- In polar coordinates on a plane, the distance per angle varies radially. How does this effect disappear in Cartesian coordinates?
You might have forgotten an earlier post where I acknowleged ISC's are just a reformulation of standard SC's without deep significance. They do not imply underlying isotropy of metric. It's all a question of whether standard SC's are just a trivial cartesian to polar mapping kind of thing as your last example alludes to, or accurately reflect that SM has this anisotropy implied by the J factor *not* operating on all spatials. Isn't that a statement about the properties curved spacetime surrounding a spherically symmetric mass has? Put it this way, do you agree that if J factor were also applicable to tangent components (with no redefinition of r as per isotropic ISC's), it would be *implying* different physics? Of course so. Will cover what I think is a 'crunch' issue in another posting.
 
  • #74
PeterDonis said:
Now that I've laid out your objection clearly and in purely physical terms, without any coordinate-dependent stuff in the way, it's easy to see what's mistaken about it. You'll notice that I bolded the word apparently. In fact, the value of J at the outer surface of the shell is *not* governed by the shell's energy density; J (or more precisely the *change* in J) is only governed by the shell's energy density *inside* the shell. At the outer surface, because of the boundary condition there, the value of J is governed by the ratio of the shell's total mass to its radius, in geometric units (or, equivalently, by the ratio of its Schwarzschild radius to its actual radius). And if you look at what I posted before, you will see that the pressure inside the shell is of the *same* order of magnitude as the ratio of the shell's mass, in geometric units, to its radius. So the pressure inside the shell is of just the right size to change K from 1/J at the outer surface of the shell, back to 1 at the inner surface of the shell.
You seem to have indeed grasped the essence of my objection well. There has been a minor revelation for me that clears one aspect up, but first some comments on above. Externally, J = (1-rs/r)1/2 = (1-2GM/(rc2))1/2, and without any confusion, M is the shell total mass, with pressure an insignificant contribution. And it is understood from the first expression shown in #57 that m(r) is what replaces M on descent through the shell wall re variation in J - outer layers become successively equipotential regions until at inner shell radius it is all equipotential. No confusion there - I think.

Have now come to understand the nature of K used here differently. Had looked at it as equivalent to scale factor for r, but see that only coincidentally applies in external SM region. From your descriptions earlier, I think it can be roughly expressed like K ~ ∂/∂r(V(r)/A(r)3/2), (V the volume, A the area, locally measured) and now see the divergent relationship within the matter region as necessary because of how K is actually defined. Yes see now it has to go to unity inside the equipotential region by this reckoning, unlike J. My continued problem is believing that K can be seriously influenced by shell wall pressure that is physically minute. The numbers just aren't there - one cannot have an ant physically lifting a mountain. More likely surely it's just the different functional dependence on mass as function of radius and nothing more. I'm confident that explicit general expressions for J and K, valid everywhere, would show that only energy density mattered in shell case. Will post on another aspect still far from satisfied about.
 
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  • #75
Firstly, correction for a silly slip-up: clear through from #1 to #64 have at times written J as J = 1-rs/r, whereas that should have been J = (1-rs/r)1/2 (doesn't however alter relations like J=K-1 in SM region, or the general problem as I see it).
Anyway let's try and settle an aspect I've tried to have pinned down but without prior success. The oft repeated claim is that unlike redshift, spatial components of the metric have no unambiguous meaning or means of measure on a relational 'down there' to 'out here' basis. So here's a scenario:
Planet X orbits at a well defined distance, and astronauts on it's surface have placed both a calibrated frequency source, and self-illuminated ruler. Mass of planet X is known from orbital mechanics, and redshift of frequency source pins down precisely the J factor. We know that astronomers have confirmed gravitational lensing effects predicted by astrophysicists - therefore optical corrections owing purely to gravitational bending of light can and have been accounted for. And similarly local mechanical distortion from gravitational gradient effects can be largely eliminated and/or otherwise completely accounted for. What's left must be entirely owing to the metric itself. Hence viewing the ruler via a telescope presents no inherent difficulty re obtaining corrected-for-all-but-pure-metric-distortion 'true' readings. This can be repeated for another planet Y, differing only in redshift factor J. That allows cross-checking and the capacity to completely eliminate all extraneous influences. We have left just direct metric influence on coordinate measured length scale.

What then will be the correction factor, if:
1: ruler lies in a plane tangent to planet surface, at various heights, and thus potentials, above the planet surface.
2: same as for 1:, but 'limb' readings of ruler alligned radially.

Repeating here my naive expectation: correction factor for case 1 will be unity for any elevation, but J for case 2. Anyone else have a different take? If it is claimed this experiment cannot be done or has no meaning, please explain why not.
 
  • #76
pervect said:
Because there is no compression to worry about, (the pieces are modeled as not changing volume with pressure), the only work being done by assembly is the binding energy, which you can think of being taken out of the system as you assemble it - for instance, you might imagine cranes lowering the pieces into place, and work is made available in the process.

Ah, that makes sense. So for the more general case where the density is not constant, and the individual pieces can't be modeled as not changing volume with pressure, there *would* be a pressure contribution to the mass integral, because the "assembly" process would have to do work to compress the pieces, and that would offset some of the gravitational binding energy being taken out; or, put another way, some of the gravitational potential energy in the system when the pieces were very far apart would be converted to compression work, and would therefore show up in the object's final mass, instead of being radiated away as the object was "assembled". So the final mass would be larger--i.e., there would not be as much net binding energy subtracted.
 
  • #77
Q-reeus said:
And it is understood from the first expression shown in #57 that m(r) is what replaces M on descent through the shell wall re variation in J

No, you didn't read that expression carefully enough. m(r) appears in the g_rr metric component, but *not* in the g_tt metric component; in the latter, only the total mass M appears. So in physical terms, K becomes dependent on m(r), but J does *not*; it remains dependent only on the total mass M and on the radial coordinate r (but *not* m(r)).

Q-reeus said:
My continued problem is believing that K can be seriously influenced by shell wall pressure that is physically minute.

Did you look at the actual calculations, which show that the pressure is of the right order of magnitude compared to the value of J at the outer surface of the shell?
 
  • #78
Q-reeus said:
What then will be the correction factor, if:
1: ruler lies in a plane tangent to planet surface, at various heights, and thus potentials, above the planet surface.
2: same as for 1:, but 'limb' readings of ruler alligned radially.

I can't give an answer to the above because, as I said, it requires calculating the paths of light rays, and that will take some time. However, I do have a couple of questions/comments:

(1) The astronauts themselves, who are next to the ruler, will see no difference if they just look at the ruler; the ruler's length will look the same whether it is placed tangentially or radially. So let's suppose that the observations by telescope from far away *do* show a difference, in accordance with your naive expectation. What will that prove?

(2) If you are correcting for optical distortion due to gravitational bending of light rays, that in itself may change the answer, because if the ruler's length looks the same both ways to astronauts next to it, the only way it can look different when viewed through a telescope far away, it seems to me, is via some sort of distortion of the light rays when traveling through the intervening spacetime with its changing curvature. After all, the astronauts are seeing the ruler via light rays too, and they don't see any difference. So if you assume that all distortions due to how light moves in curved spacetime are eliminated, how can the corrected view through the telescope possibly be any different from the view seen by the astronauts next to the ruler?
 
  • #79
Q-reeus said:
Firstly, correction for a silly slip-up: clear through from #1 to #64 have at times written J as J = 1-rs/r, whereas that should have been J = (1-rs/r)1/2 (doesn't however alter relations like J=K-1 in SM region, or the general problem as I see it).
Anyway let's try and settle an aspect I've tried to have pinned down but without prior success. The oft repeated claim is that unlike redshift, spatial components of the metric have no unambiguous meaning or means of measure on a relational 'down there' to 'out here' basis. So here's a scenario:
Planet X orbits at a well defined distance, and astronauts on it's surface have placed both a calibrated frequency source, and self-illuminated ruler. Mass of planet X is known from orbital mechanics, and redshift of frequency source pins down precisely the J factor. We know that astronomers have confirmed gravitational lensing effects predicted by astrophysicists - therefore optical corrections owing purely to gravitational bending of light can and have been accounted for. And similarly local mechanical distortion from gravitational gradient effects can be largely eliminated and/or otherwise completely accounted for. What's left must be entirely owing to the metric itself. Hence viewing the ruler via a telescope presents no inherent difficulty re obtaining corrected-for-all-but-pure-metric-distortion 'true' readings. This can be repeated for another planet Y, differing only in redshift factor J. That allows cross-checking and the capacity to completely eliminate all extraneous influences. We have left just direct metric influence on coordinate measured length scale.

What then will be the correction factor, if:
1: ruler lies in a plane tangent to planet surface, at various heights, and thus potentials, above the planet surface.
2: same as for 1:, but 'limb' readings of ruler alligned radially.

Repeating here my naive expectation: correction factor for case 1 will be unity for any elevation, but J for case 2. Anyone else have a different take? If it is claimed this experiment cannot be done or has no meaning, please explain why not.

Trying to fill in the missing information in your measurement proposal will explain the difficulties, and also why different reasonable methods will yield different answers. A fundamental observation on SR vs. GR (flat vs. curved spacetime) is that for SR, there is a unique, natural, global system of measurements for inertial observers; this is simply a consequence of the fact that any reasonable way of performing a measurement comes out the same. In GR, this remains true for inertial observers only locally. Once you are making global measurements in GR, different reasonable methods yield different answers.

Ok, so we are looking at an illuminated, distant ruler, say, 1 meter long measured locally. Through the telescope, what we measure is that it subtends some angle. How do we convert that to our claim of 'length at a distance'? Oh, we need the distance to the ruler. How do we get that? Well we can measure light bounce time, but oh, speed of light will vary along the path (or not) depending on conventions chosen. The answer here depends on coordinate choice. Pick the most common SC coordinates, for example. Now propose instead measuring distance with a long ruler; not possible in practice, but can be modeled mathematically - given a choice of simultaneity. Which one to use? One common one amounts to extending a spacelike geodesic 4-orthogonal to your world line, and measuring its ruler length (which is equivalent to saying one end of ruler doesn't look like it is moving relative to the other). Guess what, you will get a different answer for distance than a radar based approach. Ok, pick some set of answers for all these choices.

Now we face distant measurement of the end on ruler. Well, angle subtended is irrelevant. Nothing you can see in the telescope is relevant. So, maybe have a half slivered mirror on the closer end of the ruler and regular mirror on the other, and compare round trip light time difference. But what speed of light to use? Now for consistency, you better use distance measured by light travel time for the perpendicular ruler.

So, you can think this through a lot more and come up with a well defined set of measurements that you will interpret to be length of the ruler in two orientations at a distance. If your procedure is fully defined, there will be a unique, answer given all the choices you have made. However, you would have a hard time defending your choices against other reasonable ones that would yield a different result.
 
  • #80
PeterDonis said:
No, you didn't read that expression carefully enough. m(r) appears in the g_rr metric component, but *not* in the g_tt metric component; in the latter, only the total mass M appears. So in physical terms, K becomes dependent on m(r), but J does *not*; it remains dependent only on the total mass M and on the radial coordinate r (but *not* m(r)).
Oops, yes my mea culpa re m(r). But what I have not got there is how p is incorporated into m(r)? I would expect we just have a net source density in the shell wall given by ρt = ρm + 2p, where ρt, ρm are the total and matter only contributions. So dm = ρtdx3. Now while you have given the relationship p(0)=1/2ρM/R, applying at I think the center of a fluid sphere of uniform density, this would probably be quite a deal larger than for a thin shell, but I guess specifics are in the pipeline on that. From http://en.wikipedia.org/wiki/Schwarzschild_radius we have
M = Gm/c2, and rs = 2M, so for that p(0) expression, one gets as you said p(0) = ρrs/r, which is exceedingly small. The fractional p modification to m(r) is then of the order 2rs/r as upper limit (center of sphere). Thats my reading of it anyway.
 
  • #81
PeterDonis said:
...I can't give an answer to the above because, as I said, it requires calculating the paths of light rays...
No it won't. I specified we have accounted for gravitational *bending* of light - it is taken to be subtracted out, either by computer or clever lens optics. If astrophysicists can calculate distortion effects of 'g lensing', those effects must in principle be able to be subtracted out. We just want the 'raw' effects of spatial components of spacetime curvature 'right there at the ruler'.
...However, I do have a couple of questions/comments:

(1) The astronauts themselves, who are next to the ruler, will see no difference if they just look at the ruler; the ruler's length will look the same whether it is placed tangentially or radially. So let's suppose that the observations by telescope from far away *do* show a difference, in accordance with your naive expectation. What will that prove?
We will have established to what degree SC's tell us the 'true' values of gravitational length changes - on a coordinate basis.
(2) If you are correcting for optical distortion due to gravitational bending of light rays, that in itself may change the answer, because if the ruler's length looks the same both ways to astronauts next to it, the only way it can look different when viewed through a telescope far away, it seems to me, is via some sort of distortion of the light rays when traveling through the intervening spacetime with its changing curvature. After all, the astronauts are seeing the ruler via light rays too, and they don't see any difference. So if you assume that all distortions due to how light moves in curved spacetime are eliminated, how can the corrected view through the telescope possibly be any different from the view seen by the astronauts next to the ruler?

There must be a sense in which we can cleanly separate gravitational 'lensing' distortions which are an accumulated effect of transverse bending of light rays, from the metric spatial contractions that are just 'there'. Can't be all smoke and mirrors - SM must be telling us something definite via SC's, or whatever coordinate scheme is deemed relevant. I want to be clear this is not some arduous exercise I'm imposing. Not asking anyone to actually perform all those lensing correction calcs etc an astronomer might need. We just note such 'extraneous' influences exist and claim it's possible in principle to completely factor them out. This kind of thing is done all the time in other arenas - 'corrective optics' is fact. What I'm asking is, are SC's (or equivalent) implicitly predicting anisotropy of spatial components - on a 'down there' vs 'out here' basis? Just read PAllen's comments and wonder if there is any agreed sense of anything that can be got here.
 
  • #82
PAllen said:
So, you can think this through a lot more and come up with a well defined set of measurements that you will interpret to be length of the ruler in two orientations at a distance. If your procedure is fully defined, there will be a unique, answer given all the choices you have made. However, you would have a hard time defending your choices against other reasonable ones that would yield a different result.
Yikes! How on Earth will astronomers ever get a handle on testing contending theories for GBH candidates!? Look I recognize what you are saying about practical issues, but is there no sense of what's 'actually' going on at the ruler? If we say mass distorts spacetime, is there no sense that we can use SC's to simply predict the spatial part 'down there'? I'm astonished to be reading that it seems an in principle impossibility. Good grief! :zzz:
 
  • #83
Q-reeus said:
Yikes! How on Earth will astronomers ever get a handle on testing contending theories for GBH candidates!? Look I recognize what you are saying about practical issues, but is there no sense of what's 'actually' going on at the ruler? If we say mass distorts spacetime, is there no sense that we can use SC's to simply predict the spatial part 'down there'? I'm astonished to be reading that it seems an in principle impossibility. Good grief! :zzz:

The curvature of spacetime affects only measurements over some span of time or distance. A defining feature of (semi)Riemannian geometry is that sufficiently locally, spacetime is identical to flat Minkowski space. This is not different that the tangent to a curve approximates it arbitrarily well over a sufficiently small length.

The statement that, for global measurements, there is no unique way to factor curvature to effects on distance, time, and light speed is no more surprising than the statement that there are many useful projections for representing a globe on flat map. Further, for any such projection, you can choose where the biggest distortions are (you can pick any two antipodes to function as the poles).

Of course, you can relate measurements to SC coordinates; you can also relate them to Isotropic SC coordinates; or any of several other popular choices. The choice doesn't affect predictions of actual measured values of anything, but it definitely affects how you interpret what those measurements say about distant events - down to the most basic question of how far away they are.
 
  • #84
Q-reeus said:
But what I have not got there is how p is incorporated into m(r)?

Read pervect's exchange with me a few posts ago. For the particular case I posted the metric for, constant density, there is no pressure contribution to m(r), because the process of "assembling" the object doesn't do any compression work (because constant density implies that the individual pieces of the object are not compressible). This is obviously an idealization. For a real object, m(r) will include a contribution for the work required to compress the pieces of the object that are inside the radial coordinate r, from their size "at infinity" to their (smaller) size when they are part of the object. That will be a function of the pressure at r.

Also note that, as pervect pointed out, I should not have implied that J is determined by the energy density and K by the pressure. In fact J and K are *both* affected by the energy density *and* the pressure, in a real object (where m(r), and hence the total mass M, include a contribution from the pressure). However, there is an additional factor to consider: the solution for the "constant density" metric that I gave depends on the pressure, because deriving the form of the metric components requires solving the Tolman-Oppenheimer-Volkoff equation, which is the relativistic equation for hydrostatic equilibrium (i.e., the balance between pressure and gravity). So even if, in the idealized case I posted, the pressure does not appear to contribute to J and K (because the constant density assumption implies no compression work when "assembling" the object), the pressure is still essential because the form of the metric is determined by the balance of pressure and gravity within the object.
 
  • #85
Q-reeus said:
What I'm asking is, are SC's (or equivalent) implicitly predicting anisotropy of spatial components - on a 'down there' vs 'out here' basis?

I know it seems to you that you are asking a genuine question here, but I don't think it's actually a well-defined question at all. Consider the following analogy:

Suppose I live at the North Pole, and I set up a coordinate grid to label points near my home. We'll idealize the Earth as a perfect sphere to avoid any complications from oblateness. I draw a series of circles with gradually increasing circumference around my home, and label each circle with a "radial coordinate" r, defined such that the circumference of a circle with radial coordinate r is 2 \pi r. (My house is then at r = 0.) I then define an angular coordinate \phi to label the different directions I can look in from my house, so I can label any point with a pair of coordinates (r, \phi).

If I then start measuring physical distances between points, what will the metric for my little coordinate grid look like? It will look like this:

ds^{2} = \frac{1}{1 - \frac{r^{2}}{R^{2}}} dr^{2} + r^{2} d\phi^{2}

where R is a particular constant number that I find popping up in all my distance calculations, which happens to have dimensions of a length. (If I try to determine R's exact value by really accurate measurements, I will find it to be 6.378 x 10^6 meters.) (I should also note, by the way, that this thought experiment is only intended to cover the region near the North Pole; if I were to extend my measurements down past the Arctic Circle, I would start to see errors in the formula above and would have to add additional terms in the denominator of g_rr, with higher powers of the ratio r / R. We won't cover that here.)

You will notice, of course, that this metric has the same general property as the metric for the spacetime around a black hole, what I have called the "non-Euclideanness" of space. Suppose I first measure the circumference of a circle at radial coordinate r very precisely by lining up little identical objects around it. Then I measure the circumference of a slightly larger circle at r + dr the same way. Then I measure the distance between the two circles by lining up the same little identical objects between them. I will find that there is *more* distance between the circles than Euclidean geometry would lead me to expect, based on their circumferences. I should emphasize that, even though we discovered this property by looking at a specific expression for the metric in this "space", in a specific set of coordinates, the property itself is an actual physical observable, just as it is in the spacetime around a black hole. The number of little identical objects that can be lined up between two nearby circles, relative to the number of little identical objects that can be lined up around the circles' circumferences, is independent of the coordinates I use to describe the space.

What should we make of the "anisotropy" I have just described? We might wonder if it is a sign of a genuine "anisotropy of space", but we can quickly dispense with that by going to various circles and verifying that, locally, objects appear the same to us, with no distortion, regardless of which circle we are on. But what about "down here vs. up there"? Could it not be that, "from far away", there is a genuine "distortion" in this space?

I am wondering about all this one day when a friend shows up with a helicopter and offers to give me a view of the area from above. It just so happens I have little identical objects laid out all over the place, and I tell my friend to fly me around to give me a look at them from the air. Consider again two circles at r and r + dr. If I am looking down on these circles from directly above a point on one of them, I will see no distortion. But if I hover directly over my house, and look down on the circles from that angle, I will see that the little identical objects appear to be packed more tightly the further away the circles are from my house; they appear to be "contracted" in the radial direction while maintaining the same size in the tangential direction.

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.

Postscript (added by edit): Suppose that while I am hovering in the helicopter and looking at the distorted objects, I have an idea: what if I apply an optical "correction" to the image I see, to correct for the fact that I am looking at the circles "at an angle"? Well, what correction should I apply? I can certainly apply an image transformation that converts the image I see from above my house, at r = 0, to the image I would see from above a circle at some other radius r. As we've seen, that would remove the apparent distortion for the little objects at radius r. Does that mean they're not "really" distorted? There is no unique answer to this question. I can figure out how an object at r would appear from any vantage point I want, but there's nothing that singles out any particular vantage point as the "real" one, the one that determines how things "really are"--except, as I said before, the *local* vantage point, the one as seen by an observer right there, at the circle at r (not even hovering above it, but *at* it).
 
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  • #86
PeterDonis said:
Ah, that makes sense. So for the more general case where the density is not constant, and the individual pieces can't be modeled as not changing volume with pressure, there *would* be a pressure contribution to the mass integral, because the "assembly" process would have to do work to compress the pieces, and that would offset some of the gravitational binding energy being taken out; or, put another way, some of the gravitational potential energy in the system when the pieces were very far apart would be converted to compression work, and would therefore show up in the object's final mass, instead of being radiated away as the object was "assembled". So the final mass would be larger--i.e., there would not be as much net binding energy subtracted.

There is a pressure contribution even when the pieces don't change volume. Even though the pressure doesn't do any work, it alters (increases) the gravitational field.

To be specific, if you look at the gravitational field of a contained photon gas, by measuring the gravity field just inside the outer edge of the container so that the container doesn't contribute, you find that it generates more gravity than you'd expect if gravity were due to E/c^2. (Which is not the case, and this example illustrates why).

For static gravity, you can think of (rho+3P) as the source of gravity. So in simple terms, for static systems (and only for static systems) you can think of gravity as being caused by a scalar, but the scalar is not the energy, relativistic mass, invariant mass, or anything else from special relativity.

The important quantity (for static systems) is rho+3P. The pressure doesn't cause gravity by doing work and contributing to the energy density. The pressure causes gravity just be existing.

The tension in the container doesn't change it's special relativistic mass if the container does not expand. There's no work done on the container if you pressurize the interior.

It does, however, change the gravitational field that the container produces, even when the container does not expand. You can't really quite test this directly, because in order for the container to be in tension, it has to have some contents which cause the tension, pure radiation being the thing that will produce the most tension for the least amount of added mass-energy.

However, you can make the container spherically symmetrical, and measure the surface gravity inside and out. When you do this, you find that the container under tension adds less to the gravity than it would if it were not under tension. If you idealize the container to having zero mass, while still being under tension, it will actually subtract from the gravitational field.
 
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  • #87
pervect said:
There is a pressure contribution even when the pieces don't change volume. Even though the pressure doesn't do any work, it alters (increases) the gravitational field...

I understand and agree that the pressure contributes to the stress-energy tensor, and hence to the Einstein tensor (or Ricci tensor). That's where the \rho + 3p comes from. I also understand and agree that the pressure doesn't have to do any work to appear in the Ricci tensor.

But in post #71 you agreed that, for the particular idealized case I was talking about, a spherical object with constant density, the formula I quoted from MTW for the total mass M was exact. That formula only contains \rho; it does not contain p. So in this particular case, it appears to me that the pressure does not contribute to the mass M that appears in the metric. The pressure still affects the object's internal structure through the equation of hydrostatic equilibrium (and this also affects the form of the metric, i.e., where and how the mass M appears in it); but it doesn't, in this idealized case, contribute to M.

(Actually, looking again at MTW, they seem to be saying that the equation for the mass inside radius r, m(r), applies even when the density isn't constant. So it looks like they're saying the pressure doesn't contribute to the total mass M that appears in the metric for any spherical object whose stress-energy tensor is of the form of a perfect fluid. That means I was wrong a few posts ago when I said pressure would contribute to the mass integral when the density wasn't constant. The only contribution the pressure could make to the mass of the object would be indirect, by affecting the density profile of the object; for example, any gravitational potential energy that got converted to compression work instead of being radiated away during the assembly process would show up as increased density, and would increase the mass that way.)
 
  • #88
Q-reeus said:
We just want the 'raw' effects of spatial components of spacetime curvature 'right there at the ruler'.

Just re-read the thread and saw this phrase, which I must have missed before. I've given the answer to this one several times: to an observer "right there at the ruler", the ruler will look the same whether it's placed radially or tangentially. There will be no distortion.
 
  • #89
PAllen said:
Of course, you can relate measurements to SC coordinates; you can also relate them to Isotropic SC coordinates; or any of several other popular choices. The choice doesn't affect predictions of actual measured values of anything, but it definitely affects how you interpret what those measurements say about distant events - down to the most basic question of how far away they are.
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.
 
  • #90
PeterDonis said:
Read pervect's exchange with me a few posts ago. For the particular case I posted the metric for, constant density, there is no pressure contribution to m(r), because the process of "assembling" the object doesn't do any compression work (because constant density implies that the individual pieces of the object are not compressible). This is obviously an idealization. For a real object, m(r) will include a contribution for the work required to compress the pieces of the object that are inside the radial coordinate r, from their size "at infinity" to their (smaller) size when they are part of the object. That will be a function of the pressure at r.
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. Thought we were just discussing the relative contribution of the 'pure pressure' terms p11, p22, p33 to m(r).
...In fact J and K are *both* affected by the energy density *and* the pressure, in a real object (where m(r), and hence the total mass M, include a contribution from the pressure).
Agreed, but as having insisted since it was first raised in #3, relative contribution of pressure (whether 'pure' pressure or via elastic energy density) is essentially zero in shell case, and it's time to give that one a quiet burial. We all know what results of a certain undertaking will show. The pressure thing has really become a sidetracking issue, and it was my confusion in seeing K-1 as entirely equivalent to J that sustained my interest given insistence that p terms entirely determined K's evolution within the shell. Once I understood that JK-1 = 1 is a coincidental thing - 'a perversity of spherical symmetry' in exterior region, this no longer matters to me.
...the pressure is still essential because the form of the metric is determined by the balance of pressure and gravity within the object.
Which I here interpret as referring to 'p only' terms p11 etc.
 

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