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How does GR handle metric transition for a spherical mass shell? 
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#55
Oct2011, 01:48 PM

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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. 


#56
Oct2011, 03:21 PM

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http://gregegan.customer.netspace.ne....html#CONTENTS Try in particular the pages on "Rotating Elastic Rings, Disks, and Hoops", since he specifically discusses stressenergy tensors there. All these examples are in flat spacetime, but they might still be at least somewhat relevant to your example. 


#57
Oct2011, 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 2sphere 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 = 0note 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 nonrelativistic 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 nonvacuum 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 uniformdensity 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). 


#58
Oct2011, 06:35 PM

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Qreeus: 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.



#59
Oct2011, 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 


#60
Oct2011, 08:10 PM

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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. 


#61
Oct2111, 12:11 PM

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#62
Oct2111, 12:13 PM

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#63
Oct2111, 12:15 PM

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#64
Oct2111, 12:23 PM

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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 = 1r_{s}/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 nonlocally  '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
Oct2111, 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? 


#66
Oct2111, 01:33 PM

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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 "nonEuclideanness" 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 nonvacuum "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 nonvacuum region, where the stressenergy 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 densitymore 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 coordinatedependent 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
Oct2111, 07:14 PM

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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 coordinateindependent 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 antigravity 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 unoversimplify 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 selfenergy 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
Oct2111, 08:58 PM

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[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? 


#70
Oct2111, 11:03 PM

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#71
Oct2211, 04:12 AM

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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. 


#72
Oct2211, 10:30 AM

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