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Is stress a source of gravity?

  1. Mar 10, 2012 #1
    In GR normal stresses as per the three lower diagonal terms T11, T22, T33 in the SET (stress-energy tensor) (e.g. http://en.wikipedia.org/wiki/Stress–energy_tensor) are source terms for gravitating mass. And afaik true for any similar theories of gravity. Owing to their being resolved into equal and oppositely signed normal stresses, the off-diagonal shear stress components cannot even formally be a source of gravitation in solids and so are of no relevance here. Elastic/hydrostatic energy with typically quadratic dependence on stress/pressure is formally part of the T00 rest-energy term, and that role for stress is not in doubt.

    My contention is that if normal stresses truly are a source for gravitating mass m, it implies the following:
    [1] Existence of monopole GW's, which represents an internal inconsistency in GR.
    [2] Generation of GW's violating the conservation of energy in general.

    First, a situation illustrating [1]

    Komar mass is considered a valid definition of gravitating mass m in a stationary metric setting. From Wikipedia article at http://en.wikipedia.org/wiki/Komar_mass#Komar_mass_as_volume_integral_-_general_stationary_metric
    With these coordinate choices, we can write our Komar integral as
    9d928b3328fb2da0e1eb7ed1b6a03c93.png
    ...Essentially, both energy and pressure contribute to the Komar mass. Furthermore, the contribution of local energy and mass to the system mass is multiplied by the local "red shift" factor aaae79e5eeb85b11af0941939fb1e9f4.png

    Evidently m above applies for any matter distribution assumed non-rotating and having a stationary center of energy. Apply that to the case of a perfectly elastic thin spherical shell vibrating at natural frequency f in the n = 0 'fundamental' membrane breathing mode (uniform sinusoidal oscillation in radius R). Spherical symmetry allows no mass dipole or quadrupole moments P, Q. Periodic exchange at frequency 2f between KE of radial motion and elastic energy in circumferential uniform biaxial stress/strain leaves total energy (integration over the T00 term) time invariant. [Assuming for the moment a possible tiny monopole GW drain is entirely absent!] The off-diagonal SET shear terms play no role in determining m. For the momentum-energy flux terms Ti0 = -T0i, having radial acting velocity vector character, spherical symmetry implies net cancellation. Even for other configurations such as a straight bar vibrating freely in fundamental axial mode, these 'magnetic' terms, although then non-zero, scale very differently as functions of say material elesticity and density to that for the stress terms considered below.

    Which just leaves the Tii stress terms, that are not time invariant. At minimum radial excursion there is positive (compressive) circumferential stress, and negative (tensile) stress at maximum excursion. For the biaxially stressed shell, let's say we have |T11|max = |T22|max = p0 (c=1), with radial component |T33| negligible. Choose time t = 0 when R is undergoing maximum inward motion. If the shell has a thickness δ << R there will be a harmonic monopole moment ms = 4∏R2δp0sin(2∏ft), owing solely to the almost purely biaxial stress. Implying radial acting monopole GW radiation owing to d/dt(ms) = 8∏2R2δp0fcos(2∏ft). (see e.g. http://www.tapir.caltech.edu/~teviet/Waves/gwave.html - with the g' monopole series there continued to 1/r radiative term). [Note that adding in a non-negligible radial T33 contribution (thick shell case not considered) merely acts to redistribute the stress contributions in the Komar expression. All that matters is that pressure is the sole time varying net contribution to m] Contrary to the GR claim that the lowest possible GW mode is pure transverse quadrupolar. As to whether monopole GW generation is a conservative process here requires detailed calculations. It does seem to scale correctly wrt the relevant parameters. Not so for the next example involving forced vibrations.

    Now, a situation illustrating [2], first introduced here: https://www.physicsforums.com/showpost.php?p=3790816&postcount=65 , necessarily cleaned up below:

    Suppose two 'G'-clamps are welded back-to-back, and by means of say electric motors & batteries, the screws are periodically tightened and loosened. In this forced oscillation regime, frequency assumed well below mechanical self-resonance, inertial forces play no important role. By inspection periodic stresses in the assembly having a quadrupolar type distribution Qs arise - compression in the screwed arms coinciding with tension in the opposite arms, and vice versa. If the screwed legs are taken as verticallly inclined, the stress moment Qs would be linear and horizontal in orientation. Bending and shear stresses also present are self-cancelling wrt net pressure. This dominantly quadrupolar stress distribution acts as a source of quadrupolar GW's whose amplitude for a given driving frequency is directly proportional to the stresses (as Tii source terms for m).

    We have not so far included the usual contributions:

    a) Gross matter motion under mechanical strain. This is expected to be overwhelmingly the dominant source of GW's. One also expects dominantly vertical strain motion, generating a net vertical linear quadrupole moment Qm, orthogonal to that for stress generated Qs. Hence little if any cross-coupling between the two. Further striking differences between Qm and Qs is the scaling wrt elastic constant E (Young's modulus), and material density ρ. Given a specified driving stress amplitude, strain is inversely proportional to E. Hence gross matter motion and thus Qm scales accordingly. Additionally, Qm is directly proportional to material density ρ. As Qs is in this setting independent of both E and ρ, it is not possible for cross-coupling between these two GW sources to cancel anything in general.

    b) Relativistic energy-momentum flux owing to redistributions of energy between the driving power source (battery etc.) and elastic strain energy in the clamps. Suppose this gives rise to a quadrupole moment Qe. The same scaling feature wrt E mentioned above applies here also. As well, by careful arrangement of power sources one could eliminate any quadrupole moment term Qe, leaving only insignificant higher-order terms.

    It is this independence from E and ρ of stress contribution Qs to GW amplitude, in this forced oscillation regime, that is critical. Plastic will flex far more than say steel. It follows back reaction from stress generated GW's must induce far greater power drain in the plastic clamps case than for the steel ones. Much longer 'stroke' for the same retarding 'force'. And given the E and ρ parameter dependence of all non-stress GW contributions, there is no way they can in general nullify the conclusion GW's owing to Qs trend to 'for free' as E trends upward. There cannot be in general a conservative power balance. And importantly, this setting is in arbitrarily flat background metric - so Noether's theorem appears to be in serious trouble!

    Of course as hinted in the title there is one possible ready cure for all this - pressure is in fact *not* a source of gravity. Assuming no fatal blunders in the foregoing, seems to me a stark choice has to be made. Has pressure as source ever been derived from first principles - as in direct calculation of motion contributed gravitating mass generated by a 'gas' of colliding particles? That might prove to be interesting. If anyone knows of such a study, please provide a reference to the literature. Why was stress inserted into the SET in the first place? Symmetry considerations perhaps - all the SET slots have to mean something physical? My suspicion as complete GR outsider is it was a carry over from SR, where pressure applied to a flowing fluid does exhibit inertial properties as a consequence of non-simultaneity. I believe it can be shown this inertial behaviour, implying a sort of 'mass' to pressure, is really a type of pseudo quantity that fails when stretched just a bit, but this is not the place to expand on that.

    Don't expect all this to be taken lying down, so await breathlessly for sensible and constructive critiques.
     
    Last edited: Mar 10, 2012
  2. jcsd
  3. Mar 10, 2012 #2

    Dale

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    I am not sure what you mean by this. The 3D stress tensor at any single point can always be rotated into a coordinate system where the off-diagonal terms are 0 and the diagonal terms are called the principal stresses. I assume that the same is true of the 4D stress energy tensor at a point. But I am not at all sure that it can be done globally.

    If you have GW's then the metric is, by definition, not stationary, so the Komar mass is not defined.

    This claim here requires much more than a hand-waving argument like the one above. You need to actually derive some metric and show that it:
    A) Is a monopole source
    B) Is a solution to the EFE
    C) Exhibits GW's

    Sure, in non static spacetimes energy is not generally globally conserved in GR.
    http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
     
  4. Mar 10, 2012 #3
    Sure rotation is possible as you say, but since in the SET off-diagonals represent shear stresses, we must simultaneously have Tik + Tki present as matching pair. Always then this resolves into orthogonal normal stresses of equal and opposite amplitude. A basic property of shear stress. Disagree?
    Is this anything more than a trivial point? The GW's for given scenarios represent an extremely weak perturbation. Are you seriously suggesting that will throw out the validity of Komar mass used here? Surely non-stationary as significant factor re coe implies something like being in an FLRW setting or whatever where parallel transport issues etc. - 'counting difficulties' - lies at the heart of presumed failure of coe.
    You are well aware I'm not some GR pro capable of doing that math. The 'hand-waving' is though sufficient imo to establish in principle what I have claimed. Let's see what others have to say. Meanwhile, what specific points do you find to be obviously in error?
     
    Last edited: Mar 10, 2012
  5. Mar 10, 2012 #4
    As a possible first principle and definition of matter, pressure is a very nice idea. After all, that's what a star seems to use (gravity/pressure) to create elements, according to the main stream view. And matter is indeed coupled to gravity. Are you saying that pressure should be equivalent to gravity? And the 'stress' then would be the geometry?
     
  6. Mar 10, 2012 #5

    PeterDonis

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    This is way too oversimplified. The radial components are not the same as the tangential ones, and the components vary radially (they are different on the inner surface of the shell than on the outer). This basically invalidates all of your reasoning about your case #1.
     
  7. Mar 10, 2012 #6
    And where did I suggest they were? Use of 'way too' suggests there are important factors completely overlooked. What are they exactly?
    No it doesn't. Notice I specified thin shell. For which circumferential biaxial shell stresses become arbitrarily close to uniform as thickness declines. Is it not clear to you radial stress in this thin shell situation is negligible? And anyway, what if one stupidly picked a thick shell scenario where stresses did vary significantly with radius, just to throw in unnecessary complication? In what way would that invalidate the essential argument? Would it invalidate in any way whatsoever the fact of a net sinusoidal (strictly - near sinusoidal) fluctuating pressure contribution - uncancelled by the other SET terms? Please, if there is some basic flaw, argue it on important principle, not by blowing up inessential details into major flaws. But despite that bit, glad to see you involved.
     
  8. Mar 10, 2012 #7
    Not sure your question is directed to me. If so, my argument goes like this: If pressure adds to gravity as assumed in GR, it causes problems listed. If it doesn't, that's a major problem for how GR is formulated.
     
  9. Mar 10, 2012 #8

    PeterDonis

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    See below.

    No, it isn't. It can't be, because without radial stress, the shell will collapse under its own gravity. And the radial stress has to change from inner to outer surface. That means that, for the EFE to hold at every event in the shell, the tangential stresses must vary too. For a thin shell, the variation becomes very sharp (i.e., very large spatial derivatives).

    You have already admitted that you can't write down an actual mathematical description of your scenario. That makes it very difficult to even understand what the scenario actually is, because you are using imprecise English to describe it. The only help I have in trying to make the imprecise English more precise is the details; hence, they are not "inessential". Without them I can't even analyze your scenario at all.
     
    Last edited: Mar 10, 2012
  10. Mar 10, 2012 #9
    Huh? What? Did I say above the radial stress was exactly zero? Negligible wrt tangent stresses - that's what I said, and it's obviously the case, although you're welcome to argue otherwise. But I ask again - in what way would the addition of a significant radial stress (thick shell case) invalidate the general argument? But I'm thinking you probably will duck that question. Can sense where this is going below - and it's a pity.
    What can I say to that. In essence you deny my arguments because it is not expressed in some full blown, complex mathematical model? I'm not seeing you deal with others on that basis, so why here? I have in fact used some math, just enough imo to help put the first scenario in clear enough terms. There really wasn't any need for that much - the idea was to apply a process of elimination there. And you couldn't follow it?! The second scenario is inherently beyond analytic solution if one demands 'the full maths'. In my opinion that would be a wholly unreasonable stance.

    Honestly, there are truckloads of gedanken experiments accepted as valid that regularly fail to include every single possible factor and detail. How could Einstein get away with his use of trains and lights in SR setting when 'clearly' the masses involved are warping spacetime thus invalidating the flat spacetime postulated in SR. But of course we use reasonableness and accept such warping is of no real consequence. Anyway, better if you just come out and say plainly "I reject your arguments out of hand because they don't line up with established consensus opinion", if that is so and it seems to me to be so.
    If it's not so and you genuinely can't fathom what #1 is all about, ask for help on any part therein and trust me I'll do my best to clarify any grey areas.
    [EDIT: Have gone back and specifically added commentary re radial stress contribution, in light of your criticisms]
     
    Last edited: Mar 10, 2012
  11. Mar 10, 2012 #10

    Jonathan Scott

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    I certainly find it hard to believe that Komar mass makes a real contribution to gravity.

    If you have a static configuration of masses then the Komar mass stress contribution on the diagonal is positive and matches the internal (negative) potential energy of the system, and the mass of the set of particles making up the system is effectively decreased by twice the internal potential energy because of time dilation effects of all of the particles on one another. That seems to make a lot of sense and match the Newtonian model, as the overall energy of the system is simply decreased by the potential energy within the system.

    However, if the configuration is not of minimal gravitational energy, for example held apart by poles, and one of the poles breaks or slips off its support, then the stress in that pole due to the gravitational force vanishes essentially instantly (probably at the speed of sound in the material of the pole). However, I would expect the overall mass-energy of the system and resulting gravitational field to remain constant at least initially, before anything starts moving.

    I therefore think that the Komar mass adjustment is equal to the appropriate correction to make the potential energy come out right in the static case, but it is not the actual energy, as can easily be illustrated as follows.

    If you consider the gravitational force between any two particles in the system on either side of a given surface and integrate that perpendicularly to the surface as the surface sweeps through the system, summing it for all pairs of particles you get the same (negative) totals as you get for the Komar (positive) mass diagonal integrals for each of the three axes, equal to the potential energy (assuming you only take each pair once). In Newtonian terms, this works as follows for each pair of particles:
    $$
    \int_0^r \frac{- G m_1 m_2}{r^2} dx = \frac{- G m_1 m_2}{r^2} r = \frac{-G m_1 m_2}{r}
    $$
    This means that if the system is static, the internal forces resisting the gravitational forces are equal to the gravitational forces, so the Komar mass integral is equal and opposite to the potential energy. However, even if the system is not static, the integral of the gravitational forces still gives the potential energy, so if that quantity is switched in sign and added to the time-dilated energy of the individual particles then that sum would still give the same overall energy without the requirement for being static.

    This still leaves a question of how that potential energy contribution appears in the stress-energy tensor. My feeling is that it actually effectively adjusts the normal energy term, and has nothing to do with the other diagonal terms, but as usual for anything to do with gravitational energy, I don't really know.
     
  12. Mar 10, 2012 #11

    PeterDonis

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    Why is it obviously the case? Have you done an equilibrium analysis that shows that the radial stress needed to support the shell against its own weight is negligible compared to the tangential stresses? I, for one, do not think it's at all obvious that that will be true.

    I'm not even sure I understand your general argument yet. That's why I keep asking questions about details. See below.

    I'm not necessarily asking for a full blown, complex mathematical model. What I *am* asking for is sufficient precision for me to be able to make some sort of estimate of what GR predicts for your scenario. With what you've given so far, I'm not sure I can do that. If I can't do that, I can't analyze your model; I can't say whether I think it's right or wrong at all. That's not denying your arguments; that's just saying I can't render a judgment on them one way or the other.

    But yes, if I'm in that situation and you insist on a judgment from me, my judgment will be that GR is right, and if your argument is giving answers that are not consistent with GR, then there must be some subtle flaw in your argument that I'm not smart enough to see. That seems much more probable to me than the hypothesis that you actually have discovered a basic flaw in GR.

    I'll go back and read through it again and see if anything else strikes me. But I think we're going to end up in the same place we've ended up in previous threads: your idea of what constitutes a sufficiently specified scenario for analysis is apparently much less stringent than mine, so I simply won't be able to say anything useful.
     
  13. Mar 10, 2012 #12

    PeterDonis

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    Remember that the Komar mass integral has a "redshift factor" term in it; so for a non-static (or non-stationary) system, the contribution to the total mass of an infinitesimal bit of stress-energy changes as the curvature changes. In the example of two static masses held apart by a pole, if the pole breaks, the curvature change caused by the pole breaking will propagate, as you say, at the speed of sound in the pole; hence, so will the change in the Komar mass contribution of each little bit of stress-energy in the pole. (And in the two masses themselves, once the pole is no longer holding them up and they begin to fall--by that time, the curvature in their immediate vicinity has changed, and therefore so has the effective "redshift factor".)

    So "potential energy" (or any other type of "gravitational energy") doesn't change the stress-energy tensor at all; what it changes is the "redshift factor" in the mass integral.
     
  14. Mar 10, 2012 #13
    I could scrounge up a specific study on shell stresses, but let's apply KISS to this one. Ever blown up a balloon? How much air pressure can your lungs supply - maybe a fraction of a psi. That's the differential between inside and outside radial pressure. Keep blowing and the balloon will burst. Tensile strength of rubber several thousand psi. Point made?
    BTW was in the process of editing my #9 as I had somehow overlooked this from your #8:
    In light of balloon example here, won't bother.

    To get back to the essence of what #1 is all about, let's just concentrate on scenario [1] there. Simple really.
    1) Do you agree that total energy is constant there (thus net T00 contribution to m)?
    2) Do you agree or not on off-diagonals being zero contributors to time-varying m? (DaleSpam hasn't responded to my points on that issue in #3, but no sweat)
    3) Do you agree or not that Ti0, T0i energy-momentum flux terms cancel to zero by reason of spherical symmetry?
    4) Do you agree or not that all that's left is the diagonal pressure terms Tii, and that these are clearly non-zero time varying?

    As I said in #9, apply a process of elimination to above, and what does one find? Can't see the need of having to run complex numerical GR code through a supercomputer. And it would pay imo to avoid circular, self-referential arguments like demanding Birkhoff's theorem apply. The whole idea of a counterexample is to look for holes in such, imho. :zzz:
     
  15. Mar 10, 2012 #14

    Dale

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    I am just letting you know what is required in order to make a major theoretical breakthrough like the one you claim to have made. If you want to demonstrate that GR predicts monopolar GW's then you need to find a solution to the EFE's which both radiates GW's and is monopolar. Anything less will not accomplish the breakthrough you have claimed.

    The fact that you cannot yet do the math does not change the requirements.

    It is hardly trivial to point out that the key equation used to make an argument doesn't even apply. Any argument which centers around a formula that doesn't even apply to the scenario being analyzed is a fundamentally flawed argument. Major theoretical advances shouldn't be based on obviously flawed arguments.

    "Extraordinary claims require extraordinary evidence", and so far you haven't produced any.
     
    Last edited: Mar 10, 2012
  16. Mar 10, 2012 #15
    I find it hard to see how you think. On the other hand I'm not familiar with the concept of Komar mass.The stress-energy tensor includes energy density, energy flux, momentum density, and momentum flux.

    The pressure causing 'gravity' in the stress energy tensor I understand to be a result of the internal momentum flux?

    You wrote "If pressure adds to gravity as assumed in GR, it causes problems listed. If it doesn't, that's a major problem for how GR is formulated."

    Can you describe the problem in a simple way?
     
  17. Mar 10, 2012 #16

    Jonathan Scott

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    As far as I know, the "redshift" factor is exactly what I mentioned previously as the time-dilation effect on the particles of their own potential. This causes the effective energy of the system to be decreased relative to the local energy of the original components by twice the potential energy (as each interaction works both ways). I don't see why this should change in any way if, say, two parts of a supporting pole are pushed out of alignment with one another just enough to cause them to start to fall past each other. Any significant change in velocities and configurations is going to be far slower than the change in stress, and of course the fall could be stopped again a moment later.

    Clearly in the Newtonian sense there has been no immediate change in the overall potential energy nor the kinetic energy of the system caused by the support being removed, yet for example if we consider two small masses held apart by a single light pole, the Komar mass term for the stress in that pole was previously equal in magnitude to the potential energy of the pair of masses relative to one another, but if the pole is disconnected it suddenly drops to zero. It does not seem plausible that this internal change could abruptly affect the overall energy of the system, or its strength as a gravitational source.
     
  18. Mar 10, 2012 #17

    Jonathan Scott

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    I don't know about the original poster, but I see the problem as the fact that pressure can come and go almost instantly, and it doesn't seem to make sense that the shape of space-time could be directly affected by a quantity which doesn't seem to be subject to any conservation law.

    The total force perpendicular to any plane slicing through a static system is zero. The integral of the pressure over a plane gives the total force other than gravitational forces, so for a static system that force must be equal and opposite to the gravitational force. If this is integrated over three perpendicular planes which move through the system, the integral of the gravitational force is the potential energy for reasons mentioned in my previous post and for a static system this is opposite and equal to the stress part of the Komar mass. However, if the system is allowed to change, the pressure can immediately vanish, long before there is any visible change in the Newtonian potential energy or kinetic energy, and this appears to violate conservation of a form of gravitational source, regardless of whether it is actually "energy" or not.
     
  19. Mar 10, 2012 #18

    pervect

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    No. The requirement for having a stationary metric is that a coordinate system exists in which none of the metric coefficients are functions of time.

    A coordinate-free description of the requirements is that there is a time-like Killing vector.

    This explicitly rules out gravitational waves, and oscillating shells.

    Basically the Komar mass approach takes advantage of these special symmetries, so it doesn't give a general solution to Einstein's field equations. It does make it easy to calculate gravitational fields in those systems that have the requisite symmetry. Unfortunately those systems that the approach applies to cannot include gravitational waves.
     
  20. Mar 10, 2012 #19
    Interesting Johnathan. Haven't thought of pressure that way. Although it's a very theoretical description you give me here, with a lot of really difficult words in it :) I would expect gravity to obey 'c' myself?

    You suspect it doesn't?
    Your idea about pressure and the conservation laws?
    That was a new angle to me, and interesting.
    ==

    Aha, rereading Pervect "Basically the Komar mass approach takes advantage of these special symmetries, so it doesn't give a general solution to Einstein's field equations. It does make it easy to calculate gravitational fields in those systems that have the requisite symmetry. . . .

    Unfortunately those systems that the approach applies to cannot include gravitational waves."

    And that would be because? There is no arrow assumed for a Komar mass? I really need to look this up.
     
    Last edited: Mar 10, 2012
  21. Mar 10, 2012 #20

    PeterDonis

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    Well, I went back and read through it again, and I was a bit pessimistic in the above quote; two things have struck me about the first of the two scenarios (the thin spherically symmetric shell). However, they don't change the verdict: GR is still right, and monpole GW radiation is still impossible.

    Here they are:

    (1) You are claiming that radial pressure is negligible and tangential pressure is not, *as a contribution to the Komar mass integral*.

    (2) You are claiming that the pressures are the only things that change significantly as the system oscillates.

    Let's take these in order. I feel like putting headings in this post, so here goes:

    Radial vs. Tangential Pressure

    The key thing you are missing here is simple: the radial pressure is positive throughout the shell (it has to be to keep the shell from collapsing under its own gravity). But the tangential pressure is *not*; it is positive (compressive) at the shell's inner surface, but *negative* (tensile) at the outer surface.

    So in the Komar mass integral, which is taken over the entire shell, the contributions from radial pressure all add up; but the contributions from tangential pressure cancel each other out since the pressure changes sign. Therefore, radial pressure can contribute significantly to the final mass of the shell, but tangential pressure will not.

    Interlude

    As a segue from my first point to my second point, consider the question: *why* does the shell oscillate? For a system to oscillate, it has to have an equilibrium configuration, and if it is perturbed by a small amount away from that equilibrium, there has to be some restoring force that acts to bring it back.

    Since we are talking about spherically symmetric oscillations, the above observation, all by itself, is enough to tell us something important: tangential forces are *irrelevant* to the dynamics of the oscillation. Only radial forces can play any role in a spherically symmetric oscillation. So again, there *must* be non-negligible radial pressure in this scenario, since the balance between radial pressure and gravity is what determines the equilibrium and the dynamics of the oscillation.

    The restoring force is then obvious: if the shell is compressed slightly (i.e., its radius gets slightly smaller), radial pressure increases and pushes it back out again; if the shell is expanded slightly (i.e., its radius gets slightly larger), the opposite happens, radial pressure decreases and the shell collapses back inward again. *That* is why the shell oscillates.

    What Varies with Time?

    Armed with the above, we can now ask: what factors in the Komar mass integral vary with time? We know radial pressure does, as we just saw in the interlude. We don't care whether tangential pressure does or not (as I said in an earlier post, I would expect it to for the EFE to continue to hold), since its contributions integrated over the shell will cancel out. But is there anything else that does?

    Yes, there is. The redshift factor also varies, because it depends on the radius of the shell. If the shell is compressed (and radial pressure rises), the redshift factor will get smaller, because the shell is more compact and so the potential within it is slightly more negative compared to "infinity". Thus, pressure gets larger but the redshift factor gets smaller, and the two effects cancel each other out to keep the Komar mass integral constant.

    If the shell is expanded, the opposite happens: radial pressure gets smaller, but the redshift factor gets larger because the shell is less compact; so again, the two effects cancel each other out and the Komar mass integral remains constant.

    Postscript

    I said the Komar mass integral remains "constant", but actually that's an approximation. As pervect pointed out, a spacetime with an oscillating shell, even if the oscillation is spherically symmetric, is not stationary. I am basically assuming that the spacetime is "almost stationary", i.e., that the oscillations are small enough that the metric can be approximated by a stationary one. To that approximation, the Komar mass integral will be constant, and I have tried to give a physical picture of how that works. But strictly speaking, the mass integral will *not* be exactly constant because the spacetime is not stationary; however, that does not mean GWs will be emitted, because we've assumed spherical symmetry (i.e., zero dipole and higher moments). Strictly speaking, I should have written "no energy is lost to GWs" instead of "Komar mass integral remains constant" in the above. But the physical reasoning remains the same.
     
    Last edited: Mar 10, 2012
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