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)?



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.

  • $\begingroup$ "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 '18 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. $\endgroup$ – Ben Crowell Aug 27 '18 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.