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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
#55
Oct20-11, 01:48 PM
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Quote Quote by PeterDonis View Post
Ok, that makes it clearer where you're coming from.
Seriously, you didn't get that till now? How about entry #1! Or my posts over there in the blackhole thread which lead to this one. Now come on.
If you doubt the EFE, then the whole discussion in this thread is pretty much useless, because everything everybody else has been saying assumes the EFE is valid.
Last bit obviously true but at the same time everyone here has understood my sceptical stance. The challenge, and my opinions were clearly set out to all you GR buffs from the start. If you feel an explicit and definitive resolution has been given, I'd love a recap because must have missed it.
The puff of air makes the interior region not vacuum; it has a non-zero stress-energy tensor. If the pressure of the air is enough to change the stresses in the shell, then its stress-energy tensor certainly can't be neglected; and a non-vacuum interior region changes the entire problem, because the spacetime in the interior region is no longer flat. Now you're talking about something more like a static model of a planet or a star, just with a weird density profile. (Though again, the density profile can't be that weird, because if the puff of air has enough pressure to significantly affect stresses in the shell, and the air is non-relativistic, then its pressure has to be much less than its energy density, so the energy density of the "air" would be pretty large.) We'll see more specifics when DaleSpam runs the numbers, but these considerations strongly suggest to me that what you have proposed does not have any significant bearing on the original shell problem, where the interior region is vacuum and spacetime there is flat.)
It was evidently all about contrasting vanishingly small self-gravity contribution to shell stress, with what a tiny mass of air would greatly overwhelm - a mass in turn many orders of magnitude less than that of the 8 ton shell. True I didn't run specific figures because seemed quite evident there was no need given the scenario. One gets a pretty good feel for orders of magnitude with that kind of thing (but perfectly happy to put figures to it if challenged to do so). And that bit of air would matter one hoot re non-flat interior spacetime? But it's ok I think I get what's going on and why. Kind of sad but guess this is the end of your participation. So be it - but thanks anyway.

One last thing though. I think it important to know the nature of contribution to metric that in particular uniaxial stress (being the extreme of stress anisotropy), or biaxial if you like, in some element of stressed matter makes. For instance, element having uniaxial stress axis along polar axis - how different are the radial and tangent SC's generated compared to equivalent element of unstressed mass. Would be fascinating to know that, given it's ability to solve the shell issue. You may not wish to bother with a personal contribution, but pointing to a good resource explaining it in a way a layman can grasp would be appreciated. Cheers.
PeterDonis
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Oct20-11, 03:21 PM
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Quote Quote by Q-reeus View Post
It was evidently all about contrasting vanishingly small self-gravity contribution to shell stress, with what a tiny mass of air would greatly overwhelm - a mass in turn many orders of magnitude less than that of the 8 ton shell.
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. (Obviously the puff of air doesn't affect the flatness of spacetime in a way our normal senses or even fairly accurate instruments can perceive, but then again our normal senses and even fairly accurate instruments can't perceive the self-gravity of a steel ball either. So the whole scenario obviously assumes a much higher level of accuracy than we can currently achieve.)

Quote Quote by Q-reeus View Post
One last thing though. I think it important to know the nature of contribution to metric that in particular uniaxial stress (being the extreme of stress anisotropy), or biaxial if you like, in some element of stressed matter makes. For instance, element having uniaxial stress axis along polar axis - how different are the radial and tangent SC's generated compared to equivalent element of unstressed mass. Would be fascinating to know that, given it's ability to solve the shell issue. You may not wish to bother with a personal contribution, but pointing to a good resource explaining it in a way a layman can grasp would be appreciated. Cheers.
If I can find a good resource on this specific topic I'll post it. You might try Greg Egan's science pages for some good notes that are at least somewhat related:

http://gregegan.customer.netspace.ne....html#CONTENTS

Try in particular the pages on "Rotating Elastic Rings, Disks, and Hoops", since he specifically discusses stress-energy tensors there. All these examples are in flat spacetime, but they might still be at least somewhat relevant to your example.
PeterDonis
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Oct20-11, 05:28 PM
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After digging around in Chapter 23 of MTW, which discusses stellar structure, I found enough info to write down a metric for a static, spherically symmetric object with uniform density. This is not quite the same as the "shell" scenario, but it's close, and may even be close enough to use. (The specific case of a uniform density star is in Box 23.2 of MTW; I'm taking the g_tt and g_rr metric coefficient expressions from equations (6) and (3), respectively, in that box.)

The metric inside the static spherical object is:

[tex]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}[/tex]

Here R is the radius of the object (i.e., its surface radius, which is constant); r is the Schwarzschild r coordinate (i.e., a 2-sphere at r has physical area 4 pi r^2); M is the total mass of the object; and m(r) is the mass inside radial coordinate r. I have not written out the angular part of the metric in detail since it's the standard spherical form.

At the surface of the object, where r = R, the above expression is identical to the exterior Schwarzschild metric at r = R; so the above is completely consistent with the metric being Schwarzschild in the exterior vacuum region.

The key point, though, is that for r < R, the g_tt term continues to get more negative (note that M, the total mass, appears in g_tt, not m(r)), meaning the "potential" continues to decrease; while the g_rr term gets less positive, closer to 1, finally becoming equal to 1 at the center of the object, r = 0. (Strictly speaking, we have to take the limit as r -> 0 since 1/r appears in the expression; but for uniform density, m(r) goes like r^3, so the expression as a whole goes to zero like r^2.) Since this is the same sort of thing we expected to happen in the "shell" case, it looks to me like the above is basically a degenerate case of the "shell" scenario, where the flat "interior" vacuum region shrinks to zero size (the single point r = 0--note that the spatial metric at r = 0 is flat and the "distortion" is zero).

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. Equation (7) in Box 23.2 gives the central pressure (at r = 0) as a function of the density:

[tex]p(0) = \rho \left( \frac{1 - \sqrt{1 - \frac{2M}{R}}}{3 \sqrt{1 - \frac{2M}{R}} - 1} \right)[/tex]

For the extreme non-relativistic case, M << R, this approximates to:

[tex]p(0) = \frac{1}{2} \rho \frac{M}{R}[/tex]

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.

In view of the all this, it looks to me like the metric for the "shell" scenario with an interior vacuum region, in the non-vacuum region, ought to look similar to the above; the major changes would be that m(r) would go to zero at some r_i > 0 instead of at r = 0 (so g_rr = 1 at that radius), and that the potential would stop changing at r = r_i, so the g_tt expression might have to change some (maybe replace the r^2 with something that equals R^2 at r = R but goes to zero at r_i). The only thing I'm not sure of is how the different pressure profile required (which has been discussed before) would affect things.

Of course, for the "puff of air inside the shell" scenario, the metric would be very similar to the above; the only difference would be having two uniform-density regions with differing densities (steel, then air), but matching the pressure at the boundary between them. That would mean that g_rr would be very close to 1 at the boundary between steel and air, because m(r) would be close to zero there (most of the mass is in the steel, not the air). It would also mean, I think, that the r^2 in the expression for g_tt would need to be replaced by something that decreased faster in the steel region and slower in the air region, still going to zero at r = 0 (and of course still being equal to R^2 at r = R).
PAllen
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Oct20-11, 06:35 PM
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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.
pervect
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Oct20-11, 06:57 PM
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There's a section in MTW about "boundary" or "junction" conditions in MTW, which shows how to handle spherical shells.

In my copy it's pg 551,the section title is $21.13, look for "Junction conditions" in the index

Quote Quote by MTW
The intrinsic and extrinsic curvatures of a hypersurface, which played such fundamental roles in the initial-value formalism, are also powerful tools in the analysis of "junction conditions."
Recall the junction conditions of electrodynamics: across any surface (e.g., a Junction conditions for capacitor plate), the tangential part of the electric field, En, and the normal part electrodynamics of the magnetic field, Bą, must be continuous

....

Similar junction conditions, derivable in a similar manner, apply to the gravitational field (spacetime curvature), and to the stress-energy that generates it.
DaleSpam
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Oct20-11, 08:10 PM
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Quote Quote by Q-reeus View Post
Was meant as good advice, based on what I wrote in #40
"Oh, here's a possible fly in the ointment. Add the tiniest puff of fresh, pure mountain air inside the shell. Just a touch. Just enough to reverse the sign of shell hoop stresses and blow the amplitude up by, say, a mere factor of one million."
If you choose to reject the basic logic of that bit, then recall - you have committed to proving me wrong by calculations I consider doomed to failure - but go ahead and show that I'm the mistaken one.
Meaning that you cannot prove it.

By the way, the burden of proof is always on the person challenging mainstream established science. 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.
Q-reeus
#61
Oct21-11, 12:11 PM
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Quote Quote by DaleSpam View Post
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.
Q-reeus
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Oct21-11, 12:13 PM
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Quote Quote by PAllen View Post
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.
Q-reeus
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Oct21-11, 12:15 PM
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Quote Quote by PeterDonis View Post
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.
Q-reeus
#64
Oct21-11, 12:23 PM
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Quote Quote by PeterDonis View Post
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


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
PAllen
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Oct21-11, 12:56 PM
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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?
PeterDonis
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Oct21-11, 01:33 PM
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Quote Quote by Q-reeus View Post
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.

Quote Quote by Q-reeus View Post
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.

Quote Quote by Q-reeus View Post
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.
pervect
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Oct21-11, 07:14 PM
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Quote Quote by PeterDonis View Post
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, [itex]sqrt |\xi^a \xi_a| [/itex]

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 [itex]1 -(2G/R) \int \rho dV [/itex] rather than [itex]1 -(G/R) \int \rho dV [/itex] 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 [itex]1 -(G/R) \int \rho dV [/itex] 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.
PeterDonis
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Oct21-11, 08:58 PM
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Quote Quote by pervect View Post
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.

Quote Quote by pervect View Post
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

[tex]M = \int_{0}^{R} 4 \pi \rho r^{2} dr[/tex]

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?
George Jones
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Oct21-11, 10:13 PM
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Quote Quote by PeterDonis View Post
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.
PeterDonis
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Oct21-11, 11:03 PM
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Quote Quote by George Jones View Post
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.
pervect
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Oct22-11, 04:12 AM
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Quote Quote by PeterDonis View Post
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

[tex]M = \int_{0}^{R} 4 \pi \rho r^{2} dr[/tex]

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 [itex]\int \rho dV[/itex] is given a name, the "mass before assembly". Because there is no compression to worry about, (the pieces are modelled 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.
Q-reeus
#72
Oct22-11, 10:30 AM
P: 1,115
Quote Quote by pervect View Post
...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.


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