# Shear viscosity in an acoustic plane wave

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:

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

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:

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

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

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

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

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 ?