Lorentz reciprocal theorem for Stokes flow

Remarks

The momentum equation says $\nabla \cdot \sigma_k = 0$. Hence $\nabla \cdot (\sigma_2 u_1) = \sigma_2 : \nabla u_1 = 2 \mu \, e_2 : e_1$, where $e_k$ is the strain rate tensor of $u_k$ and the pressure term drops because $\nabla \cdot u_1 = 0$. The right side is symmetric in the two flows, so $\nabla \cdot (\sigma_2 u_1 - \sigma_1 u_2) = 0$, and the divergence theorem gives the identity. The theorem also holds in the exterior of bounded bodies if $u_k = O(1/r)$ and $\sigma_k = O(1/r^2)$ at infinity, because then the integral over a large sphere tends to $0$. Example: it shows that the resistance matrix of a rigid particle, which maps its translational and angular velocity to the force and torque, is symmetric.

Going deeper

For a student

The theorem is the Stokes-flow analogue of Green's second identity for the Laplacian, and it reflects the self-adjointness of the Stokes operator. It lets one compute the force on a particle in a complicated flow from an easy auxiliary flow. A typical auxiliary flow is the flow past a translating sphere from Stokes' law. In this way one derives Faxén's laws and the swimming speed of squirmers without solving the full problem. It requires the linear Stokes equations: the nonlinear inertial term breaks the symmetry.