Stokes’ law describes the size of the drag force
at low Reynolds number
experienced by a spherical object in a steady, uniform flow at velocity .
Imagine a sphere with radius sinking in a viscous liquid.
To model this situation, let us pretend that the sphere is fixed instead,
and the fluid comes from infinity at velocity along the -axis,
flows past the sphere, and continues to infinity at the same .
The Reynolds number is:
We assume that , in which case
the incompressible Navier-Stokes equations
are reduced to the steady Stokes equations:
The goal is to solve for and .
We make the following ansatz in
spherical coordinates ,
where , and are unknown functions:
The fluid hits the sphere head on,
so the solution is taken to be -independent due to symmetry.
Note that is the angle to the positive -axis,
which is the direction of .
Moreover, note that
where and are basis vectors.
To begin with, we insert this ansatz into the incompressibility condition,
The parenthesized expression must be zero for all ,
leading us to the following relation:
Next, we take the divergence of the first Stokes equation,
and insert incompressibility:
This is simply the Laplace equation,
which is as follows for our ansatz :
Again, the parenthesized expression must be zero for all ,
meaning it is an ODE for ,
whose solution is straightforwardly found to be:
Where and are linearity constants ( and appear later).
The pressure is therefore:
Consequently, its gradient in spherical coordinates is as follows:
According to the Stokes equation, this equals .
Let us look at the -component of :
Substituting for the expression we found from incompressibility lets us simplify this:
The Stokes equation says that this must be equal to the -component of :
Where we have inserted .
Dividing out leaves an ODE for ,
Then, thanks to our earlier relation again,
we know that is as follows:
So what about , , and ?
For , we expect that ,
meaning that and .
This implies that and , leaving:
Furthermore, the viscous no-slip condition demands
that at the sphere’s surface , so there.
Inserting into and , setting them to zero,
and solving the resulting system of equations
yields and .
Therefore the full solution is:
From the definition of viscosity,
we know that there must be shear stresses at the sphere surface,
described by the fluid’s Cauchy stress tensor .
The drag force on the surface is:
Where is the sphere’s surface normal vector.
The integrand can be expanded as follows:
To calculate this, we start by taking the gradient of the velocity field :
Some of these terms are necessary to calculate the stress elements and :
At the sphere’s surface we set , so these expressions reduce to the following:
Now we can finally calculate the effective stress on the surface,
by converting the basis vectors and to Cartesian coordinates:
Remarkably, the stress at every point on the sphere is purely in the -direction!
This is not entirely unexpected though: symmetry cancels out all other components.
With this, we can do the integrals for ,
which reduce to a surface area factor :
At last, we arrive at Stokes’ law,
which simply expresses the magnitude of :
To arrive at this result,
we assumed that the sphere was fixed, and the fluid was flowing past it.
We can equally well let the fluid be at rest,
with the sphere falling through it at .
The force of gravity then exerts the following force on it,
Where and are the sphere’s and fluid’s densities,
and is the gravitational acceleration.
Since acts in the opposite sense of ,
after some time, they cancel out:
This is an equation for the terminal velocity ,
which we find to be as follows:
The falling sphere will accelerate until ,
and then continue falling at constant speed.
- B. Lautrup,
Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition,