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

pervect

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.

PeterDonis

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 [itex]\rho + 3p[/itex] 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 [itex]\rho[/itex]; it does not contain [itex]p[/itex]. 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.)

PeterDonis

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.

Q-reeus

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.

Q-reeus

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).
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.
Which I here interpret as referring to 'p only' terms p11 etc.

Q-reeus

Allright, you've done good job explaining the perspective issue and need for local invariants, in my terms - thanks. Perhaps there is another angle to this worth looking at.
Suppose we have an inflatable sphere centered within a transparent and initially unstressed elastic medium. Inflating the sphere slightly creates radial compressive and tangential tensile hoop stresses, and corresponding small displacements - the medium expands non-uniformly. In polar coord terms, the perturbed changes in radial and tangential strain and displacement (the integration of strain over distance) can be expressed as factors operating on the polar ordinates. A tiny elastic being caught up in it all cannot sense this directly - only 'tidal' elastic strain is locally evident. Yet in the lab, there is a need to relate changed, stress-strain induced optical properties (e.g. light deflection) which require knowledge of the elastic perturbations - both strain and displacement.

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

PAllen

The problem with spatial relations is they are tied to decisions about simultaneity. 3-space 'now' is global statement about simultaneity. This is well defined for inertial observers in flat spacetime. Otherwise, it is just not well defined, there a number of perfectly reasonable choices. Given different choices for distant simultaneity, you get different conclusions about spatial relationships.

PeterDonis

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

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

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

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

http://en.wikipedia.org/wiki/Tolman%...lkoff_equation

The derivation of the metric for the constant density case in MTW, which I quoted from earlier, makes essential use of this equation.

PeterDonis

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

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

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

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

Go back to the analogy with the house at the North Pole and circles around it. You can set up the same sort of "elastic" model there, where the actual surface of the Earth is "elastically deformed" from Euclidean flatness. But you can only spot the deformation by comparing complete circumferences of circles. You can't spot it by just looking locally. So what physical meaning can you ascribe to the "elastic deformation"? Since you can't spot it by looking locally, you can't ascribe any physical meaning to it locally. You can say that it's a global property of the space, but you can't tie it to anything on a local scale. And since even the observation of it from a distance depends on how you look, you're limited in the physical interpretations you can put even on the global property.

Q-reeus

Given that the arrangement considered is static, I'm thinking simultaneity issue is referring to a gravitationally effected c as measuring stick?

Q-reeus

OK no disagreement here I can see.

Q-reeus

Precisely the handle I've been trying to get a hold of, but not so easy.
Sure agree and have been trying to make it clear that was understood. Obviously didn't express the analogy too clearly in #91 with
"A tiny elastic being caught up in it all cannot sense this directly - only 'tidal' elastic strain is locally evident....No point asking elastic being who knows only 'tidal' effects." Was trying to convey the analogy re local unobservability of 1st order metric effects. Basically that 'elastic being' deforms with it's surroundings, and must use a kind of 'K' factor to 'navigate' but with a limited perspective. Which answers to your later comments on that matter. Sensing only the gradients of strain, there are important properties only available - yes on an indirect inferred basis - to 'outside observer'.
This may be homing in on where it's at for me. My whole suspiscion has been that the spatial part of spacetime 'strains' was predicted wrongly - radial to transverse 'strain ratio' that, in passing to flat interior region with 1:1 'strain ratio', implied inconsistent functional dependence on potential. Just about ready to accept your collective wisdom on this and take a break. Had no initial idea defining spatial quantities of interest would prove so tricky. But thanks for an interesting if very circuitous ride!

PeterDonis

No; I assume PAllen was referring to the fact that surfaces of constant Schwarzschild time, which are surfaces of constant time for static observers (who stay at the same radius r), will not be surfaces of constant time for observers that are not static. For example, observers freely falling towards the black hole will have different surfaces of simultaneity, so "space" will look different to them. I believe I brought up the fact in an earlier post that spatial slices are flat in Painleve coordinates, which is equivalent to saying that observers freely falling into the black hole will *not* see the "non-Euclideanness" of space that observers hovering at a constant radius will.

PeterDonis

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

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

Q-reeus

Interesting point of difference for free-fall, but my query was all about stationary source and observer, which is why non-simultaneity wasn't making sense to me in that context. So a more general comment was being made.

Q-reeus

Umm...'not observable locally' in which strict sense here? There is not something here roughly analogous with 3- divergence - unobservable at a point, but locally observable in the sense of finite observables ('excess' field lines in that case) over a small but finite volume/bounding surface? In other words, is it actually necessary to do a series of complete circuits around the globe to detect the geometry? Specific example - fill a standard container of local volume V with tiny, uniform hard spheres and count them as N. Take another container that by linear ruler measuring has volume nV, and do a refill/recount, and the numbers would always be nN? And this would ccontinue to hold at any r (within reason - not for BH obviously)? Seems strange, but if that's the case, think that's provided just the handle I want!
[EDIT: Better pin down the matter of measuring that container. If one measured with ruler always held either radial or in tangent plane, and rotated the container to make say, length vs diameter measurements for a cylindrical container, it would or wouldn't matter if instead one held container fixed and reoriented the ruler in measuring?]
EDIT 2: Occurred to me now this is probably more like the situation of Ehrenfest paradox - so any 'divergence' probably of such high order as to be nearly unobservable locally. So - a good example of where effect of 'non-euclideanness' can only be appreciated by non-local (or in this instance, non-rotating) observer.

PeterDonis

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

Good instinct.

No, it wouldn't matter.