In a fluid, why are the shear stresses $\tau_{xy}$ and $\tau_{yx}$ equal?

Question

In solid bodies, $\tau_{xy}=\tau_{yx}$ makes sense to me because the volume elements "hold together" and can not spin against each other and therefore the resulting torque from the shear stresses has to be zero.

However, in fluids I imagine that the volume elements are able to spin against each other, and I was really surprised when I learned in my fluid mechanics lecture that $\tau_{xy}=\tau_{yx}$ holds for fluids too.

I realize that in this case a resulting torque would lead to an acceleration of the rotational speed of the volume elements, but I can't see anything which prevents that.

---

^{I unsuccessfully googled around a lot about this issue. I also asked my professor and several assistants, but neither of them was able to provide a satisfying explanation. Therefore I'm guessing my question doesn't really make sense in this way and/or is based on a very basic lack of understanding.}

Answer 1

This follows from rotational invariance. If the stress tensor is not symmetric then the angular momentum of the fluid is not conserved. More explicitly, momentum conservation is the equation $$ \frac{\partial}{\partial t}\pi_i + \nabla_j\tau_{ij} = 0 $$ where $\pi_i=\rho v_i$ is the momentum density. The density of angular momentum (about the origin) is $l_i=\epsilon_{ijk}x_j\pi_k$ and $l_i$ is conserved if $\epsilon_{ijk}\tau_{jk}=0$ (that is, if $\tau_{ij}$ is symmetric). We get $$ \frac{\partial}{\partial t}l_i + \nabla_j m_{ij} = 0 $$ where $m_{ij}=\epsilon_{ikl}x_k\tau_{lj}$ is the angular momentum flux.

Of course, the angular momentum of the fluid can change because of external torques, and the angular momentum of a fluid cell can change because of surface stresses. (That is, I can integrate the conservation law over a volume inside the fluid, and the angular momentum of the fluid volume changes because of surface torques. Of course, the total angular momentum of the fluid is conserved.)

All of this applies to any fluid that is described by a rotationally invariant Hamiltonian, that means any fluid made of atoms, electrons, quarks, gluons, etc.

An interesting question arises if rotational invariance is broken spontaneously, for example in the case of a liquid crystal. In that case, angular momentum is still conserved, and the stress tensor is symmetric, but the hydrodynamic description of the fluid (and the stress tensor itself) depend on an extra vector field $n_i$, which arises from the order parameter.

Answer 2

I have encountered the same question with exactly the same thought. But I found a way to explain. Consider a fluid element:

The most important point here is that $\tau_{xy}=\tau_{yx}$ holds only when the size of fluid elements approaches zero, or its mass approaches zero, or dx = dy = dz = a (a is the length of one size of the cube, a approaches zero).

So let's see how $\tau_{xy}$ and $\tau_{yx}$ are when a approaches zero: the idea here is that even when the size or mass of the fluid element approach zero, its angular acceleration must be a finite value. So we will try calculating the angular acceleration of the fluid element when its size approaches zero.

Consider the angular acceleration for z axis (aacz):

$$ accz = \lim_{a\to0}{\frac{\tau_{yx}a^2(a/2)-\tau_{xy}a^2(a/2)}{\frac{1}{6}(\rho a^3) a^2}} $$ where the numerator is simply the tourque applied ($a^2$ is the area of one face, a/2 is the distance of the forces to z axis), the denominator is the moment of inertia of the fluid element ($\rho a^3$ is the mass). Because a aprroaches zero so it is safe to use these simple terms and fomulars, we can assume the z axis goes through center of the cube as well. Continue calculating: $$ accz = \lim_{a\to0}{\frac{3(\tau_{yx}-\tau_{xy})}{\rho a^2}} $$ We see that when a $\to$ 0, we still have $\rho \to $ finte value. So if $\tau_{yx} \neq \tau_{xy}$, accz would go to infinity. So we must have $\tau_{yx} = \tau_{xy}$ for infinitesimally small fluid element