Q:

Does Stoke's drag formula apply even in the absence of flow?

I know that Stoke's flow (creeping flow) applies in the presence of low Reynold's number situations, i.e. $Re<<1$, which results in the following formula for the drag coefficient of a spherical object placed in the flow:
$$C_d = 6\pi\eta a$$
where $\eta$ is the viscosity of the fluid and $a$ is the radius of the sphere.
My question is, does this formula also apply in situations in which there is no net fluid flow? In other words, the net velocity of the fluid, relative to the sphere, is 0 and the only motion which the sphere experiences is due to fluctuations caused by diffusive (Brownian) motion. 

A:

Yes, it totally applies if your spherical particle is much larger than the mean free path of surrounding molecules (i.e. the continuum case). This is a common way of deriving the Einstein relation. Below I demonstrate this derivation.
A spherical particle immersed into a liquid experiences two forces. The first one is a random force of molecular collisions, which forces the particle to move (the reason for Brownian motion). The second once is the drag force that is trying to stop the particle set in motion. Observe that both of this forces have the same nature --- they are just particles hitting the sphere.
Let $\boldsymbol v$ be the particles's velocity and $m$ its mass. Then the drag force is $- 6\pi \eta a \boldsymbol\, v$ and we denote the random force of the brownian motion by $\boldsymbol F_{\text{st}}$. Then Newton's second law is
$$
  m\frac{d\boldsymbol v}{dt} = - 6\pi \eta a\, \boldsymbol v + \boldsymbol F_{\text{st}}
$$
Then you take an inner product of both sides with the particle's position $\boldsymbol r$:
$$
  m\,\frac{1}2{}\biggl[\frac{d^2 \boldsymbol r^2}{dt^2} - 2\boldsymbol v^2\biggr] = - 6\pi \eta a\, \frac{1}{2} \frac{d\boldsymbol r^2}{dt} + \boldsymbol r \cdot \boldsymbol F_{\text{st}}.
$$
Now you take an average $\langle {\cdot} \rangle$ of all the terms and recall that $\frac{1}{2}m \langle \boldsymbol v^2 \rangle = \frac{3}{2}kT$ (Equipartition theorem) and that the random force and the particle's position are uncorrelated $\langle \boldsymbol r \cdot \boldsymbol F_{\text{st}} \rangle = 0$:
$$
  m\frac{d^2 \langle \boldsymbol r^2 \rangle }{dt^2} - 6kT = - 
6\pi \eta a\, \frac{d\boldsymbol \langle r^2 \rangle}{dt}.
$$
Omitting some finer details, you can see that the mean square of the displacement grows linearly with time as
$$
  \langle \boldsymbol r^2 \rangle (t) = \frac{kT}{\pi \eta a} t.
$$
In general the diffusion coefficient $D$ is defined as
$$
  \langle \boldsymbol r^2 \rangle (t) = 2d\, D\, t,
$$
where $d$ is the dimension of the problem (3 in our case). Thus we have shown that
$$
  D = \frac{kT}{6\pi\eta a}.
$$

On the other hand, I am not sure how well a microscopically derived drag coefficient (like the one you are talking about) describes the drag of the real Brownian particles. By those I mean the particles of sizes for which the Brownian motion is really pronounced, those could be too small for the continuum Stokes theory. I think I saw a statement that the effective drag is lower. But I think definitely you can take that the drag is proportional to the velocity, it's just the drag coefficient that changes.

