Shear stress in Energy-momentum Stress Tensor

Q-reeus

Hi all - first post at PF. As a 'science enthusiast' with no training in the tensor math of GR, was initially bewildered by the common assertion that still hypothetical 'dark energy' would act as a source of 'negative gravity' despite having positive energy density. Finally grasped that pressure in GR acts as a source of gravity 'all by itself' - quite additional to any hydrostatic energy density associated with that pressure. As 'dark energy' is supposed to exert negative pressure, that term wins out as source of gravity (space-time curvature).

All well and good, but when visiting a Wikipedia article http://en.wikipedia.org/wiki/Stress-energy_tensor noticed that energy density and pressure are just some of the terms contributing to the energy-momentum stress tensor T^ab, which in GR is the total source of gravity. There is also energy flux, momentum flux, and most curiously, shear stress terms (the off-diagonals).

My question is this: what role do shear stresses play as source of gravity? What seems so strange is that it is well known that shear stress can be resolved into orthogonal acting tensile (negative pressure) and compressive (positive pressure) components of equal amplitude. As the diagonal pressure terms are in the first power only of p, to my mind shear stress should therefore make NO net contribution! Remember these stresses are supposed to contribute 'all by themselves' - NOT as shorthand for the associated elastic/hydrostatic energy density. So, are the shear stress terms just 'padding' or can someone explain what a chunk of matter under static shear stress contributes to the gravity of that chunk?

pervect

If you take a sufficiently small region of static space-time, it is only the diagonal terms that are needed to calculate the Komar mass, i.e. rho + Px + Py + Pz, rho + 3P if you have an isotropic gravitating fluid.

The Komar mass isn't directly used to calculate the gravitational field, either, but you can get the "far field" at infinity knowing just the Komar mass, which I hope is sufficient for your question.

But I'm not sure you can even have shear stresses in a static space-time, I'd have to think about the issue more. It's possible my calculations reduce to saying that "if the terms are zero, you don't have to worry about them" :-).

You can apply some of the concepts to a stationary space-time, which basically allows constant rotation. (I'm being a bit loose here, but I don't have time to dot all the t's - I mean dot all the i's and cross the t's). But I'd have to think about how to do it more and I'm out of time at the moment.

Q-reeus

pervect - appreciate your feedback but need some clarification:
"If you take a sufficiently small region of static space-time, it is only the diagonal terms that are needed to calculate the Komar mass, i.e. rho + Px + Py + Pz, rho + 3P if you have an isotropic gravitating fluid.

The Komar mass isn't directly used to calculate the gravitational field, either, but you can get the "far field" at infinity knowing just the Komar mass, which I hope is sufficient for your question."

Well indirectly maybe. Firstly, what do you refer to by "If you take a sufficiently small region of static space-time..."? Further down you add "But I'm not sure you can even have shear stresses in a static space-time...." Are you referring in either or both cases to the source or the field from that source? The latter in particular seems to be talking about the possibility of "shear space-time curvature" in vacuum, which has nothing to do with my query.
I asked about the contribution from 'pure' static shear stresses (apart from the associated elastic energy) in the source. Maybe I should have specifically said this implies a solid under shear stress - eg. a shaft under static torsion. Your first remarks imply the off-diagonals give zero contribution, which was my own conclusion. It is not clear if the context relates only to a static fluid though.

The point is, for a solid under static shear, the off-diagonals are non-zero but by my reasoning contribute nothing, unlike the case for the diagonal terms - compression & tension does contribute, but with opposite signs. So mathematically, how do you operate on non-zero diagonal terms to get zero result? As I remarked earlier, are they just treated as 'padding' on an ad hoc contextual basis and ignored? One could surmise that because of the symmetric nature of shear stress then for every 'plus' off-diagonal term there is an equal and opposite 'minus' term so it always somehow adds to zero. However recall that when resolved into orthogonal tensile/compressive stress components, EACH TERM SEPARATELY has zero net effect - as I see it anyways. Which raises the question of why the terms are there at all - pure formality?

Then again I can see how there can be non-zero effects in the case of a viscous fluid under shear (eg. heat generation from turbulence). But then surely the off diagonal terms would have a totally different status and functional form to the diagonals? It has been frustrating doing a Google search using the thread title, only to come up with definitions, never explanations. Perhaps someone knows of a section in some tome like MTW that clearly explains it all, covering the stressed solid case. If so, please quote chapter, heading. page numbers etc!

Mentz114

Your question is very interesting. I recently started reading this paper, 'Relativistic Elastodynamics' by Michael Wernig-Pichler ( arXiv:gr-qc/0605025v1) but I haven't got far enough to know if it answers your questions.

It's probably not relevant, but the SET of the EM field has spatial off-diagonal elements in the form of the Maxwell stress tensor,

[tex]
\sigma_{ij}=\epsilon_0 E_iE_j + B_iB_j/\mu_0,\ \ \ i\ne j
[/tex]

Q-reeus

Mentz114:
Thanks for the reference. Ploughed through as much of that paper by Wernig-Pichler as I could, but the maths is too dense and notation too arcane for me to distill anything relevant from. Looked at quite a few similar papers but all have similar style - best left to specialists!

"It's probably not relevant, but the SET of the EM field has spatial off-diagonal elements in the form of the Maxwell stress tensor,"

Actually it seems to be quite relevant. Looking at http://en.wikipedia.org/wiki/Electro...-energy_tensor, the point is made there the entire EM ST acts as source in the GR ST. There is evidently a one-to-one correspondence in the terms. In the EM case it becomes clear the off-diagonals are really an 'artifact' of the coordinate orientation. That is, for a uniform field the coordinates can always be chosen so as to only have diagonal terms for a given field (E or B), but not generally both. The same will I think apply in the GR case of uniform stress (tension/compression and/or shear). And since there is really only an 'E' field equivalent, a tensor with zero off-diagonals is always possible. So it seems to be just a 'perspective' thing so to speak. Am I right? Nonuniform stress distributions are another matter of course.

pervect

Well, one of the places the pressure terms make a difference is if you imagine having a relativistic gas enclosed in a sturdy shell. But I'm not sure I want to go into that now, it's interesting but it's probably not the root of the problem, it's a bit of a digression.

I'm not familiar with your identification of the off-diagonal terms of the stress-energy tensor as being shear stress, and perhaps that's the problem. It could be a difference between physics and engineering usage.

For instance, if you imagine having a pressurized sphere, and analyze the spherically symmetric problem in cartesian coordinates, you'll see off-diagonal terms in the stress energy tensor at the boundry, where the containing sphere is a shell under tension, but the tension is not along any of the principle axes.

But if you analyze the same problem in spherical coordinates, you'll see only diagonal terms.

It seems to me wrong to say that you have shear stress when you analyze the problem in cartesian coordinates and no shear stress when you analyze the problem in spherical coordinates unless I'm misunderstanding the meaning of shear stress.

Q-reeus

I think you are referring to one of the so-called Tolman Paradoxes - annihilate electron/positron pairs within a perfectly reflecting vessel to create a 'photon gas' and the total mass will apparently double. The resolution is to take into account the negative pressure (tension) within the containing vessel's walls; there is then no net change in gravitational mass. http://arxiv.org/abs/gr-qc/0505040
http://arxiv.org/abs/gr-qc/0510041

Just a moment of madness on my part. A little more thought and it became obvious that you cannot eliminate off-diagonals merely by a coordinate rotation (nothing to do with changing from Cartesian to spherical - stick with Cartesian!). All you achieve is redistributing among the diagonals, and likewise redistributing among the off-diagonals. That is, as long as the principal stresses (assumed to lie along the diagonals) are unequal, there MUST be non-zero off-diagonal terms present. Further, the analogy with EM breaks down early on as even for the simplest case of a single uniform E or B field, the Maxwell 'stresses' are always biaxial - tension along the field lines, pressure orthogonal to the lines. Uni-axial mechanical stress in a solid by contrast (eg. tension in a rod) has only a single component.

So the original problem remains - assuming the inherently non-zero off-diagonal terms in T^ab for a solid under static shear stress make no contribution, how is that arrived at on a consistent mathematical basis?

Again - it is emphasized we are dealing with 'stress only' contributions proportional to the first power in p or tau (shear), NOT the strain energies associated, which for linear elasticity go as p² etc and are by definition part of the 'time-time' component T⁰⁰.

Mentz114

Suppose we have a cubic lump of material as the source of a gravitational field. We can introduce shear in the material by using a clamp that squeezes the cube so two opposite faces 'slide'. Using an argument raised in an earlier post - would the shear stress contribution be cancelled by the stress in the clamp. In this way nature will ensure that shear stress does not contribute to the field.

So, for the cube we will have off-diagonal elements that are exactly cancelled by those in the clamp. I.e. T_F^ab+T_C^ab is diagonal.

I can't think of a way that shear stress could be induced other than by static forces , which cancel the shear stress.

Q-reeus

Without a picture of your arrangement hard to know how shear is generated. Agree with your general direction here though - by Newton's law of equal and opposite reaction, internal forces and thus integrated stresses must all balance out to zero overall (net 'monopole' contribution = 0). This does not per se preclude higher moment terms ('dipole', quadrupole' etc); the very notion of stress distribution implies a separation between stressed regions. Symmetry considerations alone can sometimes cancel all moment contributions. An example would be the typical scenario of a spherical pressure vessel used in the Tolman Paradox cited in a previous posting. While harder to see, 'dipole' terms are probably not possible, though I thought briefly they could be. Take the case of a G-clamp screwed down on itself. The screw side will be under compression. To balance this the opposite side must take on a tensile stress, conveyed via shear and bending moments in the top and bottom sides. At first this looks like a 'diople' distribution, but neglects the existence of a substantial bending moment in the tensioned side. Taken together, there is no stress dipole. By the same token, quadrupole and higher moments are allowed (think of two G-clamps welded back-to-back).

In the case of shear the resolved orthogonal tensile/compressive stresses are spatially coincident, hence no contribution of any kind - zero 'monopole' or any higher terms seem possible. As no-one has come forward to say otherwise, I proceed on that basis.

Think of a motor spinning up a flywheel via a torsion shaft. Or just a shaft by itself undergoing free-standing torsional oscillations (apply equal and opposite torque to each end, then suddenly let go). Or just whack a rigid body (eg. a bell) with a hammer and set in train a huge number of vibrating modes that would include shear as well as tension/compression waves.

There is I think something else hidden in the foregoing that suggests the possibility of a 'genuine paradox', but that should be for a new topic.

Mentz114

I can't argue with that.

That's interesting. Certainly while torque is present a rotating disc will get concentric circles of equal shear ( ideally) , but what gets me is that when I construct the EMT of a spinning disc using basic mechanics or field theory, I always get off-diagonal terms, even in equilibrium ( i.e. free spinning). I can undertand some tension along the diagonals but not this 'shear'.

As for ringing bells, etc, I suppose two plane compression waves crossing paths would generate shear.

This is food for thought, so I'm glad you raised it.

I doubt it !

Q-reeus

I failed to elaborate on that one - the focus was meant to be on the pure torsion (shear) in the shaft. The flywheel was intended as just inertial 'ballast' along for the ride so to speak. But now that you've raised it, it is an interesting digression. Im very much an 'un-expert' in these things but here's my take.

Recall that earlier I had stated whenever the diagonal terms are unequal there must be non-zero off-diagonals. Basic physical reasoning suggests the radial and circumferential stresses cannot be uniform throughout the disk. For instance at the outer rim radial tension must be zero but circumferential tension will be substantial. I think it makes sense that radial acting tension in a spinning disc decreases monotonically as a function of radius, while circumferential tension mostly increases. Hence in general the orthogonal components of tension must be different. By superposition you then finish up in general with shear stress plus uniform bi-axial stress plus uni-axial stress.

Well I intend to pose the problem anyhow and see where the dust settles!

pervect

Exactly what I had in mind - thanks for the references! It makes it a lot more concise to talk about.

Note that even without thinking of it as a paradox, the internal field in this case is affected by the pressure. "Field" is a vague term, in this case I'm thinking of what an accelerometer would measure held just inside the shell, i.e. I'm thinking of "field" in a quasi-Newtonian sense. The "field" or acceleration is twice what you would expect from a Newtonian analysis So in this case the pressure terms make an observable difference to the field configuration, though they don't make a difference to the total mass or to the field at infinity.

Well, one can show that the off-diagonal terms don't contribute to the overall Komar mass, there's an intergal for the Komar mass in terms of the stress-energy tensor in Wald. I could dig up the reference , it's mentioned in a recent thread though, complete with the formula.

http://www.physicsforums.com/showpos...1&postcount=19
has the formula, there's some related discussion in previous posts

Also, there's an interesting paper by Baez that shows that the second derivative of the volume of a sphere of coffee grounds is proportional to rho+Px+Py+Pz, i.e. again it doesn't depend on the diagonal terms.

I would guess that the off-diagonal terms do have some effect - just not on the mass, or the field at infinity. Also the analysis I have in mind only works for a stationary system (among other reasons, the Komar mass is only defined if you have a time-like Killing vector), I don't have any definite proof of whether they would have an effect on the mass (ADM or Bondi) if you had a time-varying system.

Q-reeus

Just so. While it all balances out formally, I have real difficulty seeing the physical origin of p contributions to active gravitational mass M_a in terms of particle dynamics. Take for instance a dilute highly relativistic gas, where nearly all the energy is kinetic. Increased passive gravitational mass M_p (and by direct analogy, inertial mass M_i) owing to twice the Newtonion curvature in an applied gravitational field is easy to grasp in principle. Formally M_a follows by appealing to the EP, but whence the physical basis from 'first principles'? I had always assumed it was taken care of by the SR 'gamma factor', but that only accounts for the T⁰⁰ 'rho' part. The 'p' part (diagonals) presumably involves consideration of 'apparent present position' and maybe acceleration terms akin to use in EM theory. Somewhere I guess direct calculations have been done to show this. The complex balance of kinetic, Coulomb, and QM factors in a solid makes it a much harder task so won't go there.

Thanks for the reference. One part I reproduce here:

The Komar mass integral as usually stated (Wald, pg 289, 11.2.10 is)



being a "unit future" of the volume, being the time-like killing vector. When the two are aligned, and you have a Minkowski space-time with g_00 = -1 the mass integral basically reduces to one of (rho + Px + Py + Pz)

Have no idea what "When the two are aligned" implies if they are not aligned, but this would not effect the off-diagonals either way?

Coffee grinds!!? So Nescafe funded that study!? Seriously, is shear even possible in a pile of coffee grinds? "..second derivative of the volume of a sphere of..." - just guessing but this is about free-fall gravitational collapse of 'dust'?

I have in mind an interesting system involving motion but doubt there would be any linkage.

I had better add something to my last reply to Mentz114. My statement that circumferential stress would mostly increase as a function of radius may not be accurate. At the disk center (assuming no shaft there to complicate things), 'pure radial' stress is 'pure bi-axial' and can therefore be interpreted as circumferential too. Without going into detailed calculations which for a real solid would need to take into account Poisson's ratio (material dependent!), all I should have ventured was that the two components are in general different at different radii, and thus shear must be present.

Mentz114

I too had a problem with this. In the case of a large gas cloud in near equilibrium, 'pressure' can be defined as the momentum flux passing through 3 orthogonal planes. In terms of particle dynamics this is related to the velocity distribution and mean free path - i.e. number of collisions and momentum exchanged per collision. The dimensions of pressure are the same as energy density, which I ( perhaps alone) find significant.

There was a frank discussion about this in another topic but I can't track it down right now.

The coffee grounds are freely-falling non-interacting dust. The term shear is used for velocity fields that are orthogonal to the time direction causing the ball to change shape. Not like shear stress.

If the ball was solid, there would be shear caused by tidal forces and kinematic effects.

Would this be true for a solid spinning disc with no torque being applied ?

Q-reeus

Agreed about the origin of pressure in terms of particle momentum flux, and the dimensional consistency with energy density. My problem is how one gets from first principles the p contribution to M_a, over and above the usual relativistic mass term (ie. T⁰⁰) for a moving particle.

Fair enough. Seems clear the shear terms have quite different application to the normal pressure terms.

That's what I was referring to, which was your own consideration of a freely spinning disk (constant angular velocity). One could think of it this way: Divide the disk up into a large series of nested concentric hoops, initially fitting together neatly when at rest. Spin them all up to the same angular velocity, and what would happen? The outer hoops, having more mass per cross-sectional area, and being under greater centripetal acceleration will expand more than the inner hoops. The almost purely circumferential stress and strains going as r², where r is the radius of each hoop. In order recreate the original solid disk, the gap between each inner and nearest outer hoop can only be closed by pressing in on the outer hoop and out on the inner hoop, then 'gluing' them together. Remove the extra pressures, and it's obvious the end result is a tendency to more equally share the tensile circumferential stresses, while at the same time introducing a radial tension. In near the center centripetal acceleration is negligible, so the accumulated radial 'tugs' induce uniform bi-axial tension, but out at the rim circumferential stress must dominate. The two terms cannot be equal throughout. Stop press. Why didn't I think of this before, just Google it and here it is:
http://arxiv.org/abs/physics/0211004
See page 12, eq'ns 92, 93. Terms are defined on p11.

pervect

One thing I should mention - the idea that rho+3P is somehow the "source" of gravity is a very useful simplification, but only for stationary or static systems. If you try to apply it to non-stationary systems ('what happens if you have a box of relativistic gas and it explodes') it will give you misleading nonsense.

And I don't have any clue if there's a deeper reason why the theory has to be that way.
This involves issues of choice of coordinates - or you might view it as the issue of "being at rest".

If you have a stationary or static system, the metric can be expressed to be independent of time, but just because it can be, it doesn't mean it has to be. We've had some other threads about what it looks like if you don't make that choice, i.e. perhaps you want to choose the time experienced by some moving observer, some observer falling into a black hole, as your coordinate time.

The requirement that your unit future and your timelike killing vector be aligned says that you are doing the intergal "at rest". So your rho and P are measured in the static rest frame, as are your volumes, etc.

Also, the idea of integrating rho+3P to get the mass only works when your system is sufficiently small. For larger systems, g_00 is no longer constant and this has to be taken into account, in order to include gravitational self-binding energy.

Yes, exactly - "dust".

Q-reeus

OK - get the drift here, thanks.

Not sure if that last remark was meant to imply 'gravity gravitates'. Seems like a real can of worms; at http://math.ucr.edu/home/baez/physic...energy_gr.html the author states:

"One other complaint about the pseudo-tensors deserves mention. Einstein argued that all energy has mass, and all mass acts gravitationally. Does "gravitational energy" itself act as a source of gravity? Now, the Einstein field equations are

Gmu,nu = 8pi Tmu,nu

Here Gmu,nu is the Einstein curvature tensor, which encodes information about the curvature of spacetime, and Tmu,nu is the so-called stress-energy tensor, which we will meet again below. Tmu,nu represents the energy due to matter and electromagnetic fields, but includes NO contribution from "gravitational energy". So one can argue that "gravitational energy" does NOT act as a source of gravity. On the other hand, the Einstein field equations are non-linear; this implies that gravitational waves interact with each other (unlike light waves in Maxwell's (linear) theory). So one can argue that "gravitational energy" IS a source of gravity."

Well it can't be both at the same time surely. There is a lot of references to authors claiming that indeed gravity is a source of gravity.
Linked to this is the popular assertion that the total energy of the universe is zero - matter/energy + 'gravity' = 0, championed by respected heavyweights like Lawrence Krauss at http://online.wsj.com/article/SB1000...Tabs%3Darticle:
"The existence of this energy, called dark energy, has another consequence: It changes the picture so that knowing the geometry of the universe is no longer enough to determine its future. While this may be a disappointment, the existence of dark energy and a flat universe has profound implications for those of us who suspected the universe might arise from nothing.

Why? Because if you add up the total energy of a flat universe, the result is precisely zero. How can this be? When you include the effects of gravity, energy comes in two forms. Mass corresponds to positive energy, but the gravitational attraction between massive objects can correspond to negative energy. If the positive energy and the negative gravitational energy of the universe cancel out, we end up in a flat universe."

Others who also write impressive articles online pooh pooh that notion: http://www.science20.com/alpha_meme/...law_questioned
http://www.science20.com/alpha_meme/...holy_cow_urine

Who is right, and why should there be any such contested topics after 95 years of GR?