Shear viscosity in an acoustic plane wave

Question

In the absence of bulk viscosity, $\eta_b = 0$, in Landau and Lifshitz book and many other places, the viscous stress tensor is defined as:

\begin{equation} \sigma'_{ik} = \eta_s\left(\frac{\partial v_i}{\partial x_k} + \frac{\partial v_k}{\partial x_i}-\frac{2}{3}\delta_{ik}\frac{\partial v_l}{\partial x_l}\right) \end{equation}

where $\eta_s$ is the shear viscosity and $v_i$ the fluid velocity. If we consider an acoustic plane wave propagating in the $x_1$ direction in a gas at rest, we have:

\begin{equation} v_1 =v\cdot\mathrm{e}^{i(Kx_1-\Omega t)}, \quad v_2 = v_3 = 0 \end{equation}

Then, we find the the non-zero viscous stress coefficients to be:

\begin{equation} \sigma'_{11} = \frac{4}{3}i\eta_s K v_1, \end{equation}

\begin{equation} \sigma'_{22} = \sigma'_{33} = -\frac{2}{3}i\eta_s K v_1 \end{equation}

First, I thought that the shear viscosity would only apply forces parallel to a surface element (i.e. $\sigma'_{ii} = 0$), but it is not the case. Second, since we have a plane acoustic wave, the only velocity gradient is along the velocity direction itself, so I intuitively see this effect as a compression/relaxation of the fluid, for which only the bulk viscosity is concerned, as I thought.

My question is then: is the results I obtained correct? And if yes, how can we intuitively visualize what happens ?

Answer 1

You can think of uniaxial compression as a sum of bulk compression + volume preserving deformation. If it's hard to recognize a pure shear on the second picture, consider rotating coordinates 45°.

Finally, shear viscosity can apply forces perpendicular to element. The only forces that are always tangent are rotational (and the corresponding tensor is antisymmetric).