The derivation of Stokes’ law involves a few steps. Firstly, we shall derive the expression . Substituting eq33b in eq33, we have:

Substituting eq35b in eq47 and working out the algebra, we get:

The gradient of a function *p* in spherical coordinates is . Since fluid flow is along the z-axis, , which when compared with the above equation:

Next, we shall show that . From a previous article, , and so,

Substituting eq44 in eq49 and working out the algebra, we have:

Substituting eq50 in the first equation of eq48 and integrating, we have:

With reference to the diagram above, the stress *τ* on the sphere in the *z*-direction is given by the sum of the projections of *τ** _{rr}* and

*τ*

*on the*

_{rθ }*z*-axis:

From the articles on constitutive relation, . So,

###### Question

Show that .

###### Answer

We need to convert the term in , which is in Cartesian coordinates, to spherical coordinates. Let *i* =*1* and *j* = *2*,

In Cartesian coordinates, and . So,

Since and are scalars, we can rewrite the above as:

We can now replace the orthogonal Cartesian basis vectors with orthogonal spherical basis vectors in eq53c:

In spherical coordinates, and . So, working out the algebra and using the identities , and , the above equation becomes:

Substituting eq53d in eq53b completes the exercise.

Substitute eq45, eq51 and *r* = *a* (on the surface of the sphere) in eq53,

Substitute eq45, eq46 and *r* = *a* (on the surface of the sphere) in eq53a,

Substitute eq54 and eq55 in eq52,

The drag on the sphere *F** _{D }*is therefore the integral of the stress vector over the surface area of the sphere:

Substitute eq56 in the above integral,

Integrate by substituting *x* = *sinθ*, *dx* = *cosθdθ*, we have

where *a* is the radius of the sphere.

##### Previous article: solution of the differential equation, *E*^{2}(*E*^{2}ψ) = *0*

*E*(

^{2}*E*) =

^{2}ψ*0*