Hydrodynamic resistance matrix

At fixed orientation, linearity of Stokes flow relates a rigid body's translational velocity $\mathbf{U}$ to the hydrodynamic force $\mathbf{F}$ byThe Lorentz reciprocal theorem makes $\mathbf{R}$ symmetric, and positive viscous dissipation makes it positive definite.

Two widely separated spheres rigidly joined by a hydrodynamically negligible connector obey the free-draining approximation. Their translational hydrodynamic resistance matrix is $6\pi \mu (a_1+a_2)I$. The displayed point is the centre about which every pure translation has zero hydrodynamic torque: ${\sum }_i{\zeta }_i({\mathbf{r}}_i-{\mathbf{r}}_c)=0$. A pure rotation about that point gives no resultant force at this order. This is an exact cancellation within the approximation, rather than an exact formula at arbitrary finite separation. It does not assert that every chiral rigid body has an origin eliminating its translation-rotation coupling.

With a consistent force convention, the mobility matrix maps applied generalized forces to rigid-body velocities; it is the inverse of the hydrodynamic resistance matrix. For a sphere with force on the fluid $\mathbf{F}$, its isolated mobility is $1/(6\pi \mu a)$. Coupling between bodies represents hydrodynamic interactions.

Here $\mathbf{F},\mathbf{G}$ are forces and torques exerted by the body on the fluid, or the external force and torque needed to maintain its motion. The Lorentz reciprocal theorem for Stokes flow makes the six-dimensional hydrodynamic resistance matrix symmetric. The power identity $\mathbf{F}\cdot \mathbf{U}+\mathbf{G}\cdot \mathbf{\Omega }=2\mu \int e:edV$ makes it a positive-definite matrix for a body moving in otherwise stationary unbounded fluid.

For a fixed rigid-body shape and origin in an unbounded stationary Stokes flow, the Lorentz reciprocal theorem makes the off-diagonal blocks of the six-dimensional hydrodynamic resistance matrix transposes. Here $\mathbf{F},\mathbf{T}$ are the fluid's force and torque on the body. If every pure translation has zero torque about the chosen origin, then $B^T=0$, hence $B=0$: every pure rotation about that origin has zero resultant force. This conditional statement does not require such an origin to exist for every body.