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From this answer and from the Stokes-Einstein equation the diffusivity of a particle of radius $R$ in a fluid of viscosity $\eta$ is

$$D=\frac{k_B T}{6 \pi \eta R}$$

where $\xi=6 \pi \eta R$ is a coefficient of friction Stokes' law such that for velocity $v$ the viscous drag force is

$$F_D=\xi v.$$

The diffusion in 1 dimension would then be given as

$$\langle x^2\rangle = 2Dt. $$

As suggested in this answer to that question, the diffusivity is usually so strongly limited by the drag force that it does not depend much on the density of the particle.

Question: Is there an orientation analog to the positional diffusion? For example, if the particle were a long thin rod, would the direction of its axis move in a random-walk type process? If so, is there an analogous orientation diffusivity, perhaps something like $D_{rot}$ where $\langle\theta^2\rangle = 2D_{rot}t$ based on the length of the thin rod and likewise independent of particle density?

update: As pointed out by @KyleKanos there is indeed the concept of "rotational diffusivity". This question mentions that for a spherical particle:

$$D_{rot} \approx \frac{k_B T}{\zeta_f} \approx \frac{k_B T}{(8 \pi \eta)(r)^3}$$

and @KyleKanos's answer begins to describe how it can be used.

What I'm looking for here instead is $D_{rot}$ for a long thin rod.

uhoh
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    The diffusion coefficient is generally a rank-2 tensor. I also mention rotational analog here. – Kyle Kanos Apr 02 '19 at 11:46
  • @KyleKanos thanks for that! Yes, that expression for $D_r$ in the question (for a spherical particle) is just the kind of thing I'm looking for, except for a thin rod. I'll have a look at expression in your answer, I'm hoping that for small times it will result in something that reduces to $\langle\theta^2\rangle = 2D_{r}t$ or similar. – uhoh Apr 02 '19 at 11:56

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Rotational diffusion was solved by Edwardes / Perrin / Langevin at their times. It is normally made for particles of Ellipsoid shape, since it is easier to solve analytically. A long and thin ellipsoid is your rod. Try Google "ellipsoid brownian motion". The model is for macroscopic particles immersed in a viscous fluid, but should work also for long molecules in a gas.

Solution, thin ellipsoid of axis $a$ (roughly equivalent to $2a$ rod) and (small) radius $b$:

$D = \frac{kT}{C_r} $

Where $C_r$ is the friction for the ellipsoid rotating in the fluid; for our thin rod,

$ C_r \simeq \frac{16 \pi \eta}{3 P_r} $

Where again, the $P_r$ is an elliptic integral that represent the particle size, at the moment I don't manage to calculate:

$ P_r = \int ^{\infty} _0 \frac{dx}{(a^2 +x)^{\frac{3}{2}} (b^2 + x) } $

All this discussion work for orientation the mayor axis of a thin rod (rotation around the two perpendicular axes); if the particle has a stranger shape, I guess we must consider all components and tensors.

Here some paper where the model is discussed in English: Paper Wegener et al, A general ellipsoid cannot always .. Koening, Brownian motion of an ellipsoid And the book The Langevin equation by Coffey has a chapter. Wikipedia Perrin friction factors

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