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I'm having difficulty seeing how the net torque on a surface in Stokes flow is non-zero.

For Stokes flow, we have $\nabla\cdot\sigma = 0$, where $\sigma$ is a symmetric stress tensor. The net torque on a surface $S$ is defined as $$ L = \int_S f\times x~ ds,$$ where $f = \sigma\cdot n$ ($n$ being the unit normal to $S$). Writing this in index notation gives $$L_i = \int_S \epsilon_{ijk}f_j x_k ~ds = \epsilon_{ijk}\int_S \sigma_{ij}n_i x_k~ds.$$ Applying the divergence theorem $$\int_S \sigma_{ij}n_i x_k~ds = \int_V \partial_i (\sigma_{ij} x_k)~dV = \int_V x_k\partial_i\sigma_{ij}~dV + \int_V \sigma_{ij}\delta_{ik}~dV.$$

The first volume integral is zero, since $\nabla\cdot\sigma=0$, and the second is $\int_V \sigma_{kj}~dV$. The net torque is then just $$ L_i = \epsilon_{ijk}\int_V \sigma_{kj}~dV,$$ which I think is zero because of the symmetry of $\sigma$. If $i=1$ for instance we get $L_1 = \int_V \sigma_{23} - \sigma_{32}~dV$. Have I made a mistake somewhere?

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Your application of the divergence theorem and computation of zero net torque is correct -- only assuming that $S$ is the entire surface bounding a region $V \subset \mathbb{R}^3$, on which the velocity field of a Stokes flow is defined.

A conclusion that the net torque (or force for that matter) on a body is zero in Stokes flow is incorrect.

Suppose that a solid body occupies $V'$ with fluid occupying $\mathbb{R}^3 \setminus V'.$ Let $S = \partial V'$ be the boundary of the body and let $S_{\infty}$ denote the surface of a large sphere enclosing the body. If $V$ denotes the region bounded by $S \cup S_{\infty}$, then as you showed,

$$\mathbf{L} = \int_{S \cup S_{\infty}} \mathbf{r} \times (\mathbf{\sigma} \cdot \mathbf{n}) \, dS = \int_{S } \mathbf{r} \times (\mathbf{\sigma} \cdot \mathbf{n}) \, dS + \int_{S_{\infty} } \mathbf{r} \times (\mathbf{\sigma} \cdot \mathbf{n}) \, dS = \mathbf{0},$$

where the unit normal vector points out of the region $V$.

It follows that

$$\mathbf{L}_{\text{body}} = \int_{S} \mathbf{r} \times (\mathbf{\sigma} \cdot \mathbf{n}) \, dS = -\int_{S_{\infty} } \mathbf{r} \times (\mathbf{\sigma} \cdot \mathbf{n}) \, dS, $$

and the torque exerted by the fluid on the body need not be zero. The net torque on the fluid in region $V$, however, is zero since Stokes flow is quasi-static with inertial effects neglected.

For example, if a sphere of radius $R$ is held fixed in a rotational flow where $\mathbf{u} \to \mathbf{r} \times \mathbf{\Omega_{\infty}}$ as $|\mathbf{r}| \to \infty$, then it can be shown under conditions of Stokes flow that the net torque on the sphere is

$$\mathbf{L}_{\text{sphere}} = 8\pi \mu R^3 \mathbf{\Omega}_{\infty}$$

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  • $\begingroup$ +1 for saving the day again! $\endgroup$ – Chee Han May 30 '18 at 6:03
  • $\begingroup$ @CheeHan: Thank you. $\endgroup$ – RRL May 30 '18 at 18:09

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