Solution

For an incompressible flow, write the Newtonian fluid stress tensor as

σ = - p I + 2 μ e

, where

e = (\nabla u + \nabla u^{T}) /2

is the rate-of-strain tensor. The Stokes equation gives

\partial_{j} σ_{ij} = 0

. Symmetry of the Cauchy stress tensor therefore gives

\partial_{j} (u_{i} σ_{ij}) = σ_{ij} \partial_{j} u_{i} = σ : e = 2 μ e : e .

(1)

Integrating and applying the divergence theorem expresses the viscous dissipation as boundary power:

D = \int_{\partial V} u \cdot σn d S .

(2)

At a moving rigid body, the surface velocity is

U + Ω \times (x - x_{c})

. Its boundary-power contribution is

U \cdot F + Ω \cdot G

, with the force and torque evaluated using the fluid's outward normal vector. Both resultants vanish for a force-free, torque-free inclusion. Thus the new viscous dissipation comes entirely from the unchanged outer boundary velocity. Subtracting the particle-free boundary power gives

D^{'} = \int_{\partial V_{0}} U_{0} \cdot σ^{'} n d S .

(3)

Apply the Lorentz reciprocal theorem to

(u_{0}, σ_{0})

and

(u^{'}, σ^{'})

in the fluid outside the inclusion. On the outer boundary

u^{'} = 0

, so

0 = D^{'} + \int_{A} (u_{0} \cdot σ^{'} - u^{'} \cdot σ_{0}) n d S .

(4)

Writing

n_{p} = - n

for the particle's outward normal vector gives the required extra dissipation due to a rigid inclusion:

D^{'} = \int_{A} (u_{0} \cdot σ^{'} - u^{'} \cdot σ_{0}) n_{p} d S .

(5)

This subtraction already accounts for the fluid volume displaced by the inclusion; it is not just the integral of the disturbance's local viscous dissipation.

For the sphere, the ambient rate-of-strain tensor

E

is symmetric and trace free. The sphere in a uniform straining Stokes flow has zero translational and rotational velocity: inversion symmetry eliminates its force, and symmetry of

E

eliminates its torque. To derive the disturbance, put

r = ∣ x ∣

Q = x \cdot E x

and seek

u^{'} = f (r) E x + g (r) Q x, p^{'} = μ j (r) Q .

(6)

The incompressible flow condition and Stokes equation reduce to

\frac{f ^{'}}{r} + r g^{'} + 5 g = 0, f^{''} + \frac{4}{r} f^{'} + 4 g = 2 j, g^{''} + \frac{8}{r} g^{'} = \frac{j ^{'}}{r} .

(7)

A decaying family satisfying these equations is

f = A r^{- 5}

g = B r^{- 5} - 5 A r^{- 7} /2

j = 2 B r^{- 5}

. The no-slip boundary condition requires

f (a) = - 1

and

g (a) = 0

, so

A = - a^{5}

and

B = - 5 a^{3} /2

. Consequently

u^{'} = - \frac{a ^{5}}{r ^{5}} E x - \frac{5 a ^{3}}{2 r ^{5}} (1 - \frac{a ^{2}}{r ^{2}}) (x \cdot E x) x, p^{'} = - \frac{5 μ a ^{3}}{r ^{5}} (x \cdot E x) .

(8)

The total velocity is zero at

r = a

, and the disturbance decays as

r^{- 2}

. The pressure constant has been set to zero. These boundary and far-field conditions, together with Uniqueness of Stokes flow, establish the solution. There is no rotational background in a pure strain flow.

On the sphere

u^{'} = - u_{0} = - a E n_{p}

. Contracting the supplied Newtonian fluid stress tensor with

n_{p}

makes its two terms proportional to

(n_{p} \cdot E n_{p}) n_{p}

cancel, leaving

σ n_{p} = 5 μ E n_{p}

. Hence the extra dissipation due to a rigid inclusion is

D^{'} = 5 μ a \int_{A} ∣ E n_{p} ∣^{2} d S = 5 μ a^{3} E_{ij} E_{ik} \frac{4 π}{3} δ_{jk} = \frac{20 π}{3} μ a^{3} E : E .

(9)

Here

\int_{S^{2}} n_{j} n_{k} d Ω = 4 π δ_{jk} /3

. Taking the large outer boundary limit only after using the fixed-boundary power identity avoids replacing that prescribed boundary by an uncontrolled boundary at infinity.