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Both, tensile loading and shear stress are represented in the Stress-Energy Tensor and hence are sources of gravity. As I understand it (please correct if wrong) tensile loading implies forces perpenticular to the surface, so that e.g. x-momentum flows in the x-direction across the unit surface. The same direction means that $T^{11}$ is pressure (negative or positiv depending on tension or compression). Whereas the non-diagonal elements are shear stress in case the force parallel to the surface.

In the FRW context negative pressure acts as repulsive gravity. Regarding a solid body, does negative pressure decrease the effective gravitational mass?

Can the non-diagonal shear stress components be negativ and positiv? What would a good example to show that, e.g. the deformation of a elastic ball if a gravitational wave passes by?

How does shear stress depending on it‘s sign act on gravity (in the sense of attractive vs. repulsive)?

Thanks

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2 Answers 2

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The statement that pressure exists is not frame-independent. For example, if you describe dust in its own rest frame, the only nonvanishing component of the stress-energy tensor is $T^{00}$, but if you do a Lorentz boost, now you can have a nonzero $T^{11}$ as well.

It is not possible in general to describe the effect of pressure and stress using plus or minus signs, as corrections to be added to the mass. This is what would happen if the source of gravitational fields in GR were a scalar such as mass. The source is not a scalar, it's a tensor.

However, there are certain cases where the effect reduces to scalar addition. One such example is the pressure inside atomic nuclei due to electrical repulsion. In this example, the effect is equivalent to adding a positive correction onto the scalar mass. (Conceptually, it's similar to a box full of photons.) This was tested by Will in 1976, in a reinterpretation of a 1966 experimental test of the equivalence principle by Kreuzer. A more sensitive test is provided by lunar laser ranging. I have a somewhat more detailed description of all this in section 8.1 of my GR book.

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  • $\begingroup$ "The statement that pressure exists is not frame-independent." Ok, the dust example is convincing, thanks. But how about the negative pressure due to the cosmological constant, is it frame dependent too? Talking about vacuum energy I can't imagine that, but perhaps I'm wrong. $\endgroup$
    – timm
    Aug 27, 2018 at 7:35
  • $\begingroup$ @timm: The only reason that the word "pressure" needs to be there is in order to make contact with our Newtonian intuition and experience. Pressure is not a universal or fundamental category in relativity. The pressure of water in a swimming pool is isotropic, and that makes up part of our Newtonian experience: pressure acts like a scalar. In relativity, it is not true in general that $T^{11}=T^{22}=T^{33}$. This is only true for a perfect fluid, considered in the fluid's rest frame. Dark energy happens to behave as a perfect fluid. $\endgroup$
    – user4552
    Aug 27, 2018 at 21:31
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If the shear stress is within a tensor that transforms with the metric, it is somewhat likely that the metric would shear. The metric and stress energy tensors are tensors. negative Pressure repels. Changing the coordinates with jacobian [[1,a],[0,1]], that transforms to [[1,a],[0,1]]M[[1,0],[a,1]]. If the SE is full of pressure, now we have negative pressure plus negative shear stress (both regions off the diagonal), (negative being the value), repels in a metric tensor that is 're-created' by a sheared jacobian. Simply put if pressure repels outward, then shear stress plus pressure repels outward in a sheared jacobian spacetime. So that means (approximately), [that something to do] with the ratio of shear stress to pressure when pressure is negative, pushes outward in a sheared frame of reference. We know that pressure minus shear then becomes pressure (in the shear) which therefore we know shear_spacetime_op^-1 * pressure_op = pressure_minus_shear_op. That means that positive shear stress with negative pressure behaves 'opposite' to the shear when observed while pushing outward. This becomes (shearing both Metric and SE tensor), negative pressure pushes outward. (negative pressure pushes outward)<->(shear stress with negative pressure behaves 'opposite' to the shear when observed while pushing outward). So shear versus pressure does the opposite in observed spacetime. That may make the object more straight. In observer independent viewpoints, adding shear stress to pressure is much like doing that same thing (operator) to the metric tensor in certain situations. This assumes the metric and SE tensor are trace-only and we modify both sides.

By the very definition of a tensor, gravity often 'resists' deformation. For example, near a spinning black hole one is in lockstep with the spinning event horizon. They won't notice it locally. Another example is frame-dragging. In your case adding pressure to shear stress gravitationally resists that near the object. This is because of a coordinate transformation. Altering the tidal effects in many locations doesn't not count as a coordinate transformation, so we notice it and since we are not points we would eventually feel gravity. On earth the gravity is slamming us down. But that isn't exactly what it 'feels' like since we are just accelerating. If everybody falls the same gravity is unnoticed. But when gravity is different in two nearby places it becomes felt. The closer those two locations and the bigger the difference the larger - and with many locations - the crazier the gravity. So your shear stress could possibly warp spacetime to where it is unnoticed. Two different shear stresses depending on the situation may bend the object to align itself to the stress of the objects. This would likely be the same in any spacetime where coordinates don't matter. For almost any kind of classical 'gravity' it may tend to resist being observed at every point independently. Using your knowledge of what pressure does, you can now see that certain constant linear transformations that result in the shear stress being included [for example:a new metric transformed by [[0,1],[1,0]], which requires a 45 degree four dimensional set of rotations and boosts. Why: because nxy=k isa 45 degree hyperbola, while n2x^2-n3*y^2=k is a 0 degree hyperbola. Your metric therefore needs to be rotated 45 degrees for pure shear stress in the space dimension.], are not measured. A constant shear stress from a constant pressure does nothing. A variable shear stress likely requires something like around a 45 degree rotation (with some stretching). If your object is creating a 45 degree tilted well that is pressure on the inside pushing outward given a perfect fluid (possibly with two dimensions of time inside it), then locally that is shear stress (recorded by an outside observer). That means shear stress could tend to rotate spacetime with a lot of pressure observed on the inside. Rotated Negative pressure of a perfect fluid (and altering time as we will see), is equivalent to shear stress. But this applies in hyperbolic dimensions. If we make one of the dimensions hyperbolic, then rotate it 45 degrees in 4d, then we have 2 perfect fluid pressure components turned directly into two shear stress components. Doing this 1 time for x and y (rotate 45 deg, (dx)^2->-((dx)^2)), and then rotating the original dimension of time ((z,t)->(z+t,z-t)) with So this transformation would be tend to be resisted by the gravitational field (as in harder to measure as you get closer, except for how one travels through space compared to light). If one wants to check the exact effects with variable pressure and shear stress on non-local macroscopic scales try using the Einstein field equation. (These effects would be a tendency for any [tensorial] classical theory of gravity not just Einstein's, but it could be different over different distances).

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  • $\begingroup$ Read the definition of a tensor. $\endgroup$
    – Misha
    Aug 27, 2022 at 15:53
  • $\begingroup$ Note: hyperbolic means becoming the opposing distance component. $\endgroup$
    – Misha
    Aug 27, 2022 at 18:47
  • $\begingroup$ A theory of gravity is how 'coordinate' transformations spread out. Quantum physics is somewhat how 'matter' spreads out. Gravity is a smearing of the coordinate transformations of stress-energy. They are very similar. $\endgroup$
    – Misha
    Aug 27, 2022 at 19:13
  • $\begingroup$ I wonder if quantum gravity is the smearing of integral transforms and coordinate transforms? $\endgroup$
    – Misha
    Aug 27, 2022 at 19:16
  • $\begingroup$ I would guess one needs a quantity that is independent of each. $\endgroup$
    – Misha
    Aug 27, 2022 at 19:16

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