Tensors for Physics

180 10 Multipole Potentials

According to (8.91), the force acting on a solid body is given by the surface integral
−

∮

∂Vnνpνμd

(^2) s. For the case of the sphere, the outer normalnis equal toˆr, and
d^2 s=R^2 d^2 rˆ. As used before, cf. (7.55), the pressure tensorpνμis the sum of the
isotropic partpδμνand the symmetric traceless partpνμ=−ηR−^1 (∇νvμ+∇μvν).
The part of the force associated with the distorted pressure field is

F

Stp

μ ≡

∮

AXνVνr̂μR^2 d^2 rˆ=

4 π

3

R^2 AVμ= 2 πRηVμ. (10.64)

Here

∮

r̂μr̂νd^2 rˆ=^43 πδμνwas used.

The contribution of the force associated withvrequires the computation of

̂rν∇νvμ+r̂ν∇μvν.

The first of these two terms is

r̂ν∇νvμ=

∂

∂r

vμ=a′Vμ+(b′− 3 r−^1 b)XμνVν.

On the other hand, one has

∇μvν=a′r̂μVν+b′r̂μXνλVλ−bXμνλ,

and

r̂ν∇μvν=a′r̂μr̂νVν+b′r̂μ 2 r−^3 r̂λVλ−br̂μXμνλ.

Due to the compressibility condition∇μvμ=0, one has anda′+r−^3 b′=0, and

r̂ν∇μvν=−br̂μXμνλ.

The term involvingr̂μXμνλ∼Xνλ, however, vanishes in the integration over the
surface of the sphere.
Thus the additional force is

FμStv≡ηR−^1

∮

r̂ν∇νvμR^2 d^2 rˆ= 4 πηRVμ. (10.65)

The term involvingXμνgives no contribution to the integral over the surface of the
sphere.
Notice, in the calculation, the sphere was at rest and the fluid, far away from
the sphere, moved with the velocityV. As mentioned before, this corresponds to a
situation, where the sphere moves with the velocity−Vand the fluid is at rest, far
away from the obstacle. This explains the opposite sign in the expression for the
friction force.