= Solution
Let $e=(\nabla\mathbf u+\nabla\mathbf u^T)/2$ be the <rate-of-strain tensor> and $\boldsymbol\sigma=-pI+2\mu e$ the <Newtonian fluid stress tensor>. The <viscous dissipation> is $D[\mathbf u]=2\mu\int_V e:e\,dV$. The <Minimum-dissipation theorem for Stokes flow> compares the <Stokes flow> with every sufficiently regular <incompressible flow> in the same domain having the same prescribed boundary velocity, with no <body force> or with a conservative <body force> absorbed into the <fluid pressure>. A trial field need not satisfy the <Stokes flow> equations. Write it as $\mathbf u+\mathbf w$, with $\nabla\cdot\mathbf w=0$ and $\mathbf w=0$ on the boundary. Integration by parts, the symmetry of the <Newtonian fluid stress tensor>, and $\nabla\cdot\boldsymbol\sigma=0$ give
$$
2\mu\int_V e(\mathbf u):e(\mathbf w)\,dV=\int_V\boldsymbol\sigma:\nabla\mathbf w\,dV=\int_{\partial V}\mathbf w\cdot\boldsymbol\sigma\mathbf n\,dS-\int_V\mathbf w\cdot\nabla\cdot\boldsymbol\sigma\,dV=0.
$$
Consequently \b[the Stokes flow minimizes dissipation]:
$$
\boxed{D[\mathbf u+\mathbf w]-D[\mathbf u]=2\mu\int_V e(\mathbf w):e(\mathbf w)\,dV\geq0.}
$$
Equality requires a <rigid body> motion of $\mathbf w$, which the prescribed boundary eliminates. Extending the flow inside the inserted particle as its <rigid body> velocity produces an admissible comparison field with zero internal <rate-of-strain tensor>; hence the <extra dissipation due to a rigid inclusion> is nonnegative.
Here and below $\mathbf n$ on $A$ points into the particle, so the <traction> $\boldsymbol\sigma\mathbf n$ represents force exerted by the particle on the fluid. Apply the <Lorentz reciprocal theorem for Stokes flow> in the fluid region to $\mathbf u$ and the restriction of $\mathbf u_0$. The equal outer boundary velocities yield
$$
\int_{\partial V}\mathbf V_0\cdot(\boldsymbol\sigma-\boldsymbol\sigma_0)\mathbf n\,dS=\int_A(\mathbf u\cdot\boldsymbol\sigma_0\mathbf n-\mathbf u_0\cdot\boldsymbol\sigma\mathbf n)\,dS.
$$
The boundary-work formula for <viscous dissipation> therefore gives $D'=\int_A[(\mathbf u-\mathbf u_0)\cdot\boldsymbol\sigma\mathbf n+\mathbf u\cdot\boldsymbol\sigma_0\mathbf n]dS$. The last integral is zero: $\boldsymbol\sigma_0$ is symmetric and divergence-free throughout the particle's original volume, so its total force and <torque> there vanish, and $\mathbf u$ on $A$ is a <rigid body> velocity. This also allows the sign of this zero-work term to be reversed, giving exactly
$$
\boxed{D'=\int_A(\mathbf u\cdot\boldsymbol\sigma-\mathbf u_0\cdot\boldsymbol\sigma-\mathbf u\cdot\boldsymbol\sigma_0)\cdot\mathbf n\,dS.}
$$
With the particle's centre as origin, define $\mathbf F=\int_A\boldsymbol\sigma\mathbf n\,dS$, $\mathbf G=\int_A\mathbf x\times(\boldsymbol\sigma\mathbf n)\,dS$ and the <particle stresslet tensor> $S=-\tfrac12\int_A[\mathbf x(\boldsymbol\sigma\mathbf n)+(\boldsymbol\sigma\mathbf n)\mathbf x]dS$. Contracting the linear background velocity with the <traction> gives
$$
\boxed{D'=\mathbf F\cdot(\mathbf U-\mathbf U_0)+\mathbf G\cdot(\boldsymbol\Omega-\boldsymbol\Omega_0)+S:E_0.}
$$
The background <rate-of-strain tensor> $E_0$ is symmetric and trace-free; the trace of this unprojected <particle stresslet tensor> does not affect the contraction.
For the <straight-rod resistance in a linear flow>, write $\Delta\mathbf U=\mathbf U-\mathbf U_0$, $\Delta\boldsymbol\Omega=\boldsymbol\Omega-\boldsymbol\Omega_0$, $M=I-\mathbf p\mathbf p/2$, and $\mathbf X=s\mathbf p$, $-L\leq s\leq L$. The <slender-body force density> becomes $\mathbf f=C M[\Delta\mathbf U+s(\Delta\boldsymbol\Omega\times\mathbf p-E_0\mathbf p)]$. The integrals of $s$ and $s^2$ are $0$ and $2L^3/3$, respectively, so
$$
\boxed{\mathbf F=2CLM\Delta\mathbf U,\qquad\mathbf G=\frac{2CL^3}{3}\left[(I-\mathbf p\mathbf p)\Delta\boldsymbol\Omega-\mathbf p\times E_0\mathbf p\right].}
$$
For a <force-free>, <torque-free> rod, $\Delta\mathbf U=0$ and the perpendicular component of $\Delta\boldsymbol\Omega$ is $\mathbf p\times E_0\mathbf p$. Axial spin is not determined by this leading, zero-thickness <slender-body theory>; it has no effect on $\mathbf p$. Put $a_p=\mathbf p\cdot E_0\mathbf p$. The vector identity $(\mathbf p\times E_0\mathbf p)\times\mathbf p=E_0\mathbf p-a_p\mathbf p$ gives
$$
\boxed{\mathbf f=-\frac C2s a_p\mathbf p,\qquad\dot{\mathbf p}=\boldsymbol\Omega_0\times\mathbf p+E_0\mathbf p-a_p\mathbf p.}
$$
The <force-free straight-rod orientation equation> combines the background rotation with the part of strain that turns the rod; removing $a_p\mathbf p$ preserves its unit length. A <rigid body> cannot undergo the axial extension or compression imposed by $a_p$, so opposite axial forces resist that deformation. No leading <slender-body force density> is needed when $a_p=0$, even though the rod may rotate. Its <particle stresslet tensor> and <viscous dissipation> are
$$
\boxed{S=\frac{CL^3}{3}a_p\mathbf p\mathbf p,\qquad D'=\frac{CL^3}{3}a_p^2.}
$$
In the <simple shear flow>, $\boldsymbol\Omega_0=-\gamma\mathbf e_z/2$, $(E_0)_{xy}=(E_0)_{yx}=\gamma/2$ and $a_p=\gamma\cos\theta\sin\theta$. Thus the <rod excess dissipation in shear> is
$$
\boxed{D'(\theta)=\frac{CL^3\gamma^2}{12}\sin^2(2\theta),\qquad\dot\theta=-\gamma\sin^2\theta.}
$$
The maxima at $\theta=\pi/4,3\pi/4$ correspond to strongest axial extension and compression. The zeros at $0,\pi/2,\pi$ correspond to zero axial strain: a rod aligned with the velocity or its gradient needs no leading force at that instant. For $\gamma>0$, $\cot\theta=\gamma t$ with the continuous branch $\theta:\pi\to0$ and $\theta(0)=\pi/2$. Reversing the <shear flow> reverses the traversal. Substitution gives
$$
D'(t)=\frac{CL^3\gamma^2}{3}\frac{(\gamma t)^2}{[1+(\gamma t)^2]^2},\qquad\boxed{\int_{-\infty}^{\infty}D'(t)\,dt=\frac{\pi CL^3|\gamma|}{6}.}
$$
The $t^{-2}$ tails make the total finite; the instantaneous zero at $t=0$ lies between two maxima.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-329-rod-dissipation.png]
{title=Rod excess dissipation versus orientation and time in simple shear}
For a <dilute rod suspension>, the added bulk stress is the number density times the orientation average of the <particle stresslet tensor>. In this ideal infinitely slender, non-interacting model, rods approach alignment with the <shear flow> and their excess <viscous dissipation> decays. A real finite-aspect-ratio rod continues to tumble; rotational diffusion and interactions also maintain a spread of orientations. One therefore expects increased effective <viscosity>, with alignment offering a mechanism for reduced excess <viscosity> at strong <shear flow>. This is a qualitative expectation, not a claim that a finite rod suspension has zero steady excess stress.
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