What is the physical explanation for the attenuation of sound in Newtonian fluid?

Question

My question refers to Stokes's law of sound attenuation in viscous medium. At this point I don't try to understand the mathematical form of the law - I simply don't understand how viscosity effects planar waves - this law states that planar waves decay exponentially, and I don't understand how a shear stress emerges in the propogation of perfectly planar waves. So what do shear stresses and sound propogation have in common?

Edit (22.06.2025)

I want to clarify what I am actually asking here. As I wrote in my comment to @ChetMiller, the derivation of the right side of Navier-Stokes equations (the stress part) usually starts from Cauchy stress tensor as a fundamental assumption, although this tensor was originally constructed to describe the stresses a solid material is subjected to. As this tensor is symmetric, one can always diagonalize it, which physically means finding principal planes such that the stresses act perpendiculary on them.

For example, if we apply pure shear stress on a cube, we can imagine the cube as two triangular prisms attached, and than the shear forces induce normal stress on the plane (which is inclined at 45 degrees) seperating them. The two prisms act on each other like a mass on an inclined plane (normal force is formed).

So this construction is definitely valid for solids, but fluids are different - they obey Pascal law while solids do not, a fact which which manifests itself in that fluids assume the shape of their container. In particular, constant shear stress does not induce additional pressure in fluids, while it does for solids. I understand the 3D version of Newton's viscosity law speaks about rate of deformation and not deformation itself, but it is still very unclear to me.

In addition, the microscopic explanation of how shear arises as a result of compression in fluids is not clear to me; for two adjacent fluid layers with slightly different velocities, the shear force arises as a result of diffusion of particles from one layer to another, which on average transports linear momentum. Such a microscopic explanation is missing in the case of converging/diverging flow; the term $(\mu/3+\zeta)\nabla(\nabla\cdot v)$, which describes volume viscosity, refers to both the second viscosity $\zeta$ and the usual shear viscosity $\mu$.

The microscopic origins of $\zeta$ are relatively clear to me - there is a finite time till the fluid reach thermodynamic equilibrium when it undergoes volumetric change, and during this time the energy transfered to the fluid is distributed among all the internal degrees of freedom of the fluid molecoles. When the fluid undergoes very fast compression/expansion, there is no time for transfering the molecoles translational kinetic energy into their internal structure, and as a result the usual thermodynamical equations (such as $PV^{\gamma} = \text{const}$, for adiabatic process) are not valid. This results in higher pressure increase per volume change, which manifests itself as the term $\zeta\nabla(\nabla\cdot v)$. For example, monoatomic gases, who lack internal molecolar structure, have $\zeta\approx 0$.

However, the term $(\mu/3)\nabla(\nabla\cdot v)$, which refers to the first viscosity (the usual shear viscosity), is not neglible for monoatomic gases, so this explanation does not apply. So in short, I want to understand the origin of the term $(\mu/3)\nabla(\nabla\cdot v)$ from a physical/microscopic point of view.

Answer 1

Newton's law of viscosity, when properly extended using mathematical rigor to 3 dimensions, results in a linear relationship between the 3D stress tensor and the 3D rate of deformation tensor, and includes not only viscous shear stresses and shear rates, but also normal stresses and normal strain rates. These normal stresses and strains are operative in the situation(s) that you are describing. The relationship between the stress tensor and the rate of deformation tensor reduces to the usual equation relating shear stress to shear rate for the special case of pure shear (with only one velocity gradient).

Even for the case of pure shear, if the stresses and strains are resolved into their principal components (say by transforming using a coordinate system rotation), the only stress components present will be normal stresses and the only strain rate components present will be normal strain rates.

Answer 2

After much thinking/reading, I think I finally understand the comments of @ChetMiller. All my misunderstandings stemmed from a basic misconception of pressure in a moving, and viscous, fluid. One can apply the construction of Cauchy stress tensor $\tau_{ij}$ - which was originally defined for solids but can be applied to fluids when strain is replaced by strain-rate - in order to find principal normal stresses in a point in the fluid domain, which amounts to diagonalizing the (symmetric) stress tensor. Now, the definition of scalar pressure $p$ for such a fluid must be independent of the choice of the coordinate system. The only invariant (under coordinates transformation) of the stress tensor that has dimensions of pressure is its trace:

$tr(\tau_{ij}) = \tau_{11}+\tau_{22}+\tau_{33}$

and since this expression must be consistent with the definition of pressure in a static fluid $p=\tau_{11}=\tau_{22}=\tau_{33}$ (per Pascal's law), one must add a factor of $\frac{1}{3}$;that is, $p=\frac{\tau_{11}+\tau_{22}+\tau_{33}}{3}$.

The pressure in a moving and viscous flow is therefore not transfered equally in all directions, and one can give a consistent definition of it only by averaging normal stresses (this means that Pascal law of isotropic pressure is correct only for inviscid flows). The rest of the algebra (the derivation of Navier-Stokes equation) is now easily understandable.