Relation of translational and rotational diffusion of a spherical particle The mean squared displacement in time $\Delta t$ of a particle undergoing Brownian motion (translational diffusion) in $d$ dimensions is
$\langle x^2\rangle = 2 d D \Delta t$, where the diffusion coefficient $D=\frac{k_B T}{6\pi\,\eta\,r}$ (per Stokes–Einstein equation),
where''η'' is the dynamic viscosity, ''r'' is the radius of the spherical particle, ''kB'' is Boltzmann's constant and ''T'' is the absolute temperature.
Similarly, the mean-square angular deviation for rotational diffusion is given by $\langle\theta^2\rangle = 2 D_r \Delta t$, where the rotational diffusion coefficient $D_r = \frac{k_B T}{8 \pi \eta r^3}$.
Equipped with this, essentially, wikipedia knowledge, I noticed that there are many common factors that cancel when dividing one by the other, leading me to
$\frac{\langle x^2\rangle}{\langle\theta^2\rangle} = \frac{4d}{3}r^2$,
which I would interpret to mean, for example, "in three dimensions, a spherical particle on average rotates by about 1 rad in the time it is displaced by the distance equal to its diameter". That sounds like a catchy conclusion that I have not seen spelled out anywhere.
Is this a reasonable interpretation or is there a reason why calculating the quotient of $\langle x^2\rangle$ and $\langle\theta^2\rangle$ is not meaningful?
(And am I missing something else, e.g. an adjustment of the angular deviation for the number of dimensions?)
 A: Yes, your interpretation is correct. This surprising coincidence can be understood with the fluctuation-dissipation theorem, which describes the Brownian motion for any degree of freedom in a thermodynamic system. 
The theorem relates Brownian motion for a coordinate $x$ to the energy dissipation caused by changes in $x$. In particular, if two coordinates $x_1$ and $x_2$ cause the same energy dissipation when varied, then they will have the same diffusion coefficient.
How does this relate to our problem? For a rotating sphere, we have six degrees of freedom - three of them translational and three rotational. When the sphere is translated or rotated, there will be a flow induced in the surrounding fluid, causing energy to be dissipated through viscosity. This is precisely the "energy dissipation" that the theorem deals with!
Let's compare a translational degree of freedom $x$ to a rotational degree of freedom $\theta$ in terms of their energy dissipation. If we vary $x$, the fluid will flow with the Stokes flow past a sphere (link), dissipating energy at a rate $P_{x} = 6 \pi \mu r \dot{x}^2$. If we vary $\theta$, the fluid will flow azimuthally, dissipating energy at a rate $P_{\theta} = 8 \pi \mu r^3 \dot{\theta}^2$ (derivation, page 6). In other words, moving through the liquid at speed $v = \dot{x}$ dissipates $\frac{3}{4}$ as much energy as rotating the surface at speed $v = r \dot{\theta}$.
With this fluid dynamics factoid, we can go back to the fluctuation-dissipation theorem to see the diffusivity in $x$ must be $\frac{4}{3}$ of the diffusivity in $r \theta$. The origin of this result is now clear - it basically comes down to the similarity between two fluid flows.
A: I think its important to note that
$$
\frac{\langle x^2 \rangle}{\langle \theta^2 \rangle}
\neq
\langle \frac{x^2}{\theta^2} \rangle
$$
In general expectation values of ratios are more meaningful than ratios of expectation values. As an example, due to rotational symmetry, if you let $y$ and $z$ be two coordinates in the d-dimensions available to diffusion, $\langle y^2 \rangle = \langle z^2 \rangle$, so their ratio is one. OTOH, the mean-square value of their ratio is $\langle y^2/z^2 \rangle = \langle \tan^2 \phi_{yz} \rangle$ is likely not equal to one. In fact, due to symmetry $\phi_{yz}$ is uniformly distributed between $0$ and $2\pi$, so
$$
\langle \frac{y^2}{z^2} \rangle = \frac{1}{2\pi} \int_0^{2\pi}\tan^2 \phi \, d \phi = \mbox{does not converge}
$$
Similarly, the quantity $\langle x^2 / \theta^2 \rangle$ could be interpreted as the mean-square ratio of radial displacement to angular displacement. However, I have doubts that this is well-defined, and it might be infinite because $\theta$ can be 0.
Finally, your interpretation, 

in three dimensions, a spherical particle on average rotates by about 1 rad in the time it is displaced by the distance equal to its diameter

doesn't seem to exactly correspond to either of these ratio cases. The most literally transcription of your interpretation is
$$
\langle \theta \mid x=x' \rangle
$$
i.e. the average rotational displacement when the radial displacement is $x'$ (in particular, the diameter of the particle). This is probably something that could be calculated:
$$
\langle \theta \mid x=x' \rangle = \frac{\int_\Omega \theta p(\theta,x') \, d\Omega}{\int_\Omega p(\theta,x') \, d\Omega}
$$
I think this is probably the best way to phrase your concept.
