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In fluid mechanics, the stress tensor writes $\sigma = -p 1 + \tau$ where the deviatoric part $\tau$ corresponds to shear. The viscous (volumic) forces are $\operatorname{div}\tau$.

For a Newtonian fluid, $\tau = \eta \dot\gamma$ where $\eta$ is the viscosity and $\dot\gamma$ is the shear rate.

As far as I understand, the shear rate is the velocity gradient $\nabla U$ which does not have to be symmetric (e.g. if vorticity is present).

I am confused with the fact that $\nabla U$ need not be symmetric while, physically, only the symmetric term contributes to viscosity/dissipation; the skew-symmetric terms of $\nabla U$ corresponds to rotation and do not "activate" viscosity. For this reasons, viscous forces are sometimes written $\eta(\nabla U + \nabla^\top U)/2$.

But then, the viscous forces for the skew-symmetric part should be 0, however the divergence of the skew-symmetric tensor is not 0 in general.

Can someone explain to me the exact relationship between the shear rate $\dot \gamma$ and the velocity gradient $\nabla U$?

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The behavior of a material must be independent of the motion of any observer. So, if an observer were rotating at the same angular velocity as the local vorticity, he would observe the same local rate of deformation tensor, but with the vorticity removed. This would have to result in the same stress tensor, because the motion of the observer can not affect the stress. This is why the anti-symmetric portion of the velocity gradient tensor must be removed in formulating a model to represent the stress tensor in terms of the kinematics of the motion.

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  • $\begingroup$ This sounds sort of right, but I am not quite sure if the argument is entirely correct. A rotating observer is non-inertial (and may see funny forces). I think the correct version of this argument is that there should not be a stress in a rigidly rotating fluid, because this in an equilibrium state. $\endgroup$ – Thomas Jan 26 '16 at 2:04
  • $\begingroup$ That's pretty much the same thing. It doesn't matter if the observer is non-inertial as long as the fluid is inertial. The stress tensor is what it is, and it doesn't change with a rotation of the observer. The Principal of Principal of Material Objectivity (aka the Principal of Material Frame Indifference) states that the mechanical response of a material is independent of the translation and/or rotation of the observer. $\endgroup$ – Chet Miller Jan 26 '16 at 2:46
  • $\begingroup$ Is that a known (valid?) principle of physics? If it is, it should follow from a symmetry of the laws of physics. $\endgroup$ – Thomas Jan 26 '16 at 14:25
  • $\begingroup$ Yes, that is a known valid principle of physics, developed by the Continuum Mechanics guys at Johns Hopkins (in particular, Walter Noll) in the early '60s. This was Clifford Truesdell's team at JH. At that time, the focus was on developing valid constitutive equations to describe the thermomechanical behavior of viscoelastic materials. Google it, and see what comes up. As far as it following from a symmetry law of physics, symmetry laws are not part of my background. $\endgroup$ – Chet Miller Jan 26 '16 at 14:42
  • $\begingroup$ That's interesting -- I was unaware of this. Some googling shows that there was (and maybe still is) an argument between Truedell, Noll, and others and various members of the physics community about whether such a principle exist. For some particularly nasty comments about Truesdell see Ingo Muller "A History of Thermodynamics." $\endgroup$ – Thomas Jan 27 '16 at 2:27
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Let me write $\Pi=\nabla\cdot u$. The shear strain is given by $$ S_{ij}=\nabla_iu_j+\nabla_ju_i-\frac{2}{3}\delta_{ij}\Pi $$ and the vorticity tensor is $$ \Omega_{ij}=\nabla_iu_j-\nabla_ju_i . $$ The most general stress tensor linear in gradients of $u$ is $$ \tau_{ij} = \eta S_{ij} + \zeta \delta_{ij} \Pi + \eta_R \Omega_{ij}. $$ We want to show that $\eta_R=0$.

1) The most general argument (I think) was given here In a fluid, why are the shear stresses $\tau_{xy}$ and $\tau_{yx}$ equal? . The stress tensor must be symmetric for angular momentum to be conserved (so we can view it as a consequence of rotational symmetry).

1b) A (somewhat) more physical argument is this: In a rigidly rotating fluid, we don't expect any dissipative stresses (as this is an equilibrium state).

2) I think the argument that $\nabla_i\omega_i=0$ is not relevant. The divergence of the vorticity vector is not the same $\nabla_i$ on $\Omega_{ij}$ (the latter is related to the curl of the curl $u$).

3) In general there is nothing wrong with $\Omega$ appearing in the stress tensor. If I go beyond linear order (non-Newtonian fluids), then terms like $\Omega_{ik}\Omega^k_j$ and $S_{ik}\Omega^k_j$ can (and do) appear.

4) Finally, microscopic expressions for $T_{ij}$ are manifestly symmetric. In kinetic theory $$T_{ij}= \frac{1}{m}\int d^3p\, p_i p_j \,f(x,p).$$ In quantum field theory the symmetry of the stress tensor is a little subtle, but once we couple to gravity, we have $$ T_{\mu\nu} = \frac{2}{\sqrt{-g}}\frac{\delta S}{\delta g_{\mu\nu}}, $$ which is manifestly symmetric.

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