The Stokes's law states that a sphere of radius $R$ dragged at a constant speed $U$ in a very viscous fluid of viscosity $\eta$ creates a force $F = \mu U$ with $\mu=6\pi\eta R$.

Similarly, if the same sphere is rotated at an angular velocity $\Omega$ in the same fluid, I would expect that it creates a torque $\Gamma$ such that $\Gamma = \nu\Omega$. Dimensional analysis gives $\nu\propto \eta R^3$. Does anybody know what is the prefactor (the equivalent of the $6\pi$) and how to compute it? (I'm pretty sure one can find a reference in which it is done.)

EDIT: As I am working in systems that are constrained to 2d, I would be satisfied by the value of $\nu$ for a rotating cylinder.

Reason: The Einstein relation states that the (translational) diffusion coefficient is $D_{trans} = k_B T/\mu$. Similarly, one expects a rotational diffusion coefficient $D_{rot} = k_BT/\nu$. I want to know what is the numerical value of $D_{rot}R^2/D_{trans} = \mu R^2/\nu = 6\pi/(\dots)$

This article (page 2, end of first column) states that $D_{rot} = 3 D_{trans}/R^2$, so do we have $\nu = 2\pi\eta R^3$?

  • $\begingroup$ This seems to apply to laminar flow only. Are you sure this assumption is valid? $\endgroup$ Sep 21, 2018 at 14:05
  • 1
    $\begingroup$ For a velocity (or an angular velocity) small enough, in a fluid viscous enough, the flow should not be turbulent. That's the ony limit in which I am interested. $\endgroup$
    – styko
    Sep 21, 2018 at 14:11

2 Answers 2


For Stokes flow (i.e. a flow in which there is no inertia and pressure gradient balances viscous forces) the torque on a sphere of radius $a$ rotating with angular velocity $\Omega$ in a fluid of viscosity $\mu$ is: $$T=8\pi\mu a^3\Omega$$

Quick derivation:

Stokes equation for incompressible fluid is: $$\nabla\cdot\mathbf{u}=0\\ \nabla p=\mu\nabla^2\mathbf{u}$$ in which symbols have the usual meaning. Vectors are denoted by bold face font. The boundary condition is: $$\mathbf{u}|_{r=a}=\mathbf{\Omega}\times\mathbf{r}_s$$ in which $\mathbf{r}_s$ is the position vector of a point on the sphere's surface w.r.t. sphere centre.

Taking divergence of $\nabla p=\mu\nabla^2\mathbf{u}$ and using continuity equation we see that $\nabla^2 p=0$. Solutions of the 3-d Laplace equation that decay at infinity are the following so-called spherical solid harmonics: $$\frac{1}{r},~\frac{\mathbf{r}}{r^3},~\frac{\mathbf{I}}{r^3}-\frac{3\mathbf{r}\mathbf{r}}{r^5},~\ldots \textrm{(solutions of higher tensorial order)}$$ in which $\mathbf{r}$ is the position vector of a point in the fluid (w.r.t. sphere centre), $r=|\mathbf{r}|$, and $\mathbf{I}$ is the identity tensor (Kronecker delta in indicial notation).

The Laplace equation is linear, and the boundary condition is linear in the driving term $\mathbf{\Omega}$. Therefore pressure field must be a linear function of $\mathbf{\Omega}$ while also being proportional to one or more of the spherical solid harmonics. Thus the only possible combination is: $$p=c_1\mathbf{\Omega}\cdot\frac{\mathbf{r}}{r^3}$$ in which $c_1$ is a constant. However $\mathbf{\Omega}$ is a pseudo-vector because its direction depends on whether you adopt the right-handed or left-handed coordinate system. This implies that $\mathbf{\Omega}\cdot\mathbf{r}$ changes sign when we change from right-handed to left-handed coordinate system or vice versa. But pressure $p$ is a scalar which shouldn't depend on our choice of coordinate system, therefore we must have $c_1=0$. This means that there is no disturbance pressure field caused by the rotating sphere.

Similar argument for velocity field (which may be shown to be biharmonic, i.e. $\nabla^2\nabla^2\mathbf{u}=0$) shows that it must be of the form: $$\mathbf{u}=c_2\mathbf{\Omega}\times\frac{\mathbf{r}}{r^3}$$

Here cross-product also flips sign when we change from right-handed to left-handed coordinate system (or vice versa) and compensates for the sign flip of $\mathbf{\Omega}$, to give a true-vector field $\mathbf{u}$. The boundary condition shows that $c_2=a^3$.

Once you know the velocity field you may calculate the strain rate tensor $\mathbf{e}$ and thus the stress tensor $\mathbf{\sigma}=2\mu\mathbf{e}$. The torque on the sphere is found by computing the surface integral (on sphere's surface): $$\int_{S}\mathbf{r}\times\sigma\cdot\mathbf{dS}$$

Reference: Guazzelli & Morris, A physical introduction to Suspension Dynamics, Chapter 2.

  • $\begingroup$ @styko You are welcome. $\endgroup$
    – Deep
    Sep 22, 2018 at 4:06

This is a partial answer for the case of a cylinder.

The value for a rotating cylinder can be computed by solving the Navier-Stokes equation for the right boundary values and in the stationary limit by assuming the symmetry of the problem carries over to the solution (this does not tell us the solution is actually stable against perturbations, if it is not there may will be turbulence). The Navier-Stokes equation for and incompressible fluid is given by: $$ \rho \partial_t \vec v + \rho (\vec v \cdot \nabla) \vec v - \eta \Delta \vec v = -\nabla p + \vec f$$ where $\rho$ is the density of the fluid, $\vec v$ its velocity field, $\eta$ its viscosity, $p$ its pressure and $\vec f$ the acting external forces (e.g. gravity).

Now, we a have rotating cylinder of radius $R$ in our fluid, this means under no-slip boundary conditions we have the boundary condition $$ \vec v = \vec e_\varphi \Omega R$$ for $\left|\vec r\right| = R$. Further, $\left|\vec v\right| \to 0$ as $\left|\vec r \right| \to \infty$.

Sine we want the stationary solution we can set $\partial_t u$ to zero, we assume no external force (gravity would simply be countered by static pressure). Further, we guess that the solution will respect the cylinder symmetry of the problem and write the stationary Navier-Stokes equations in cylinder coordinates and drop all terms that are zero if the symmetry and the boundary conditions are respected: $$ \rho\left(v_\varrho \partial_\varrho + \frac{1}{\varrho} v_\varphi \partial_\varphi + v_z \partial_z\right)\vec v - \eta \left(\frac 1 \varrho \partial_\varrho \varrho \partial_\varrho + \frac 1 {\varrho^2} \partial_\varphi^2 + \partial_z^2\right) \vec v = 0$$ $v_\varrho$ must be zero in the stationary case (since the fluid is incompressible and we want cylinder symmetry, so the volume inside any cylinder coaxial to our coordinates would change if it were not). $v_z$ must be zero for the velocity to vanish at infinity. Further, $\partial_zv_\varphi$ and $\partial_\varphi v_\varphi$ must vanish for the solution to have the full symmetry of the problem. This leaves us with an ordinary differential equation for $v_\varphi(\varrho)$ (careful $\vec e_\varphi$ depends on the coordinates, $\partial_\varphi^2 \vec e_\varphi = -\vec e_\varphi$): $$ \eta \frac 1 \varrho \partial_\varrho \rho \partial_\varrho v_\varphi - \frac 1 {\varrho^2} v_\varphi = 0.$$ Equivalently: $$ \left(\varrho^2 \partial_\varrho^2 + \varrho \partial_\varrho - 1 \right)v_\varphi = 0$$ This form of differential equations allows a power function ansatz $v_\varphi = C \varrho^a$ reducing it to an algebraic equation for the exponent: $$ C\rho^2 a(a-1)\rho^{a-2} + Ca\rho \rho^{a-1} - C\rho^{a} = 0, $$ $$ a^2 - 1 = 0$$ So independently of $\eta$ the general solution is: $$ v_\varphi(\varrho) = \frac{C_1}{\varrho} + C_2 \varrho. $$ Adapted to the boundary conditions $v_\varphi(R) = \Omega R$ and $v_\varphi \to 0$ as $\varrho \to \infty$ this results in: $$ v_\varphi(\varrho) = \frac{\Omega R^2}{\varrho}.$$

The torque per unit length can now be calculated. From Newton's definition of the viscosity we have $$ F = \frac{\eta A \Delta v}{d}, $$ for parallel plates. This equation holds for an infinitesimally thin layer of the fluid so the total force acting on a small patch of area $A$ of the cylinder is $$ F = \eta A \partial_\varrho v(R).$$ This results in a torque per unit length of (the torque of each area is $\tau_A = R \eta A \partial_\varrho v(R)$, so in the total torque the area must just be replaced by the total area) $$ \frac{\tau}{l} = -2\pi R^2 \eta \Omega \frac{R^2}{R^2} = -2\pi\eta R^2 \Omega.$$

Similarly, albeit requiring more work due to the lower symmetry, the case of the sphere can be handled. In this case it helps to develop the solution in terms of spherical harmonics. To do make the equation solvable one uses the reduced Stokes equation (which holds in the limit of low Reynold numbers), which neglects the convective forces $\rho(\vec v \cdot \nabla)\vec v$ which leaves the Laplace equation in the steady state. (For more, see https://en.wikipedia.org/wiki/Stokes_flow).


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