1
$\begingroup$

The Stokes-Einstein rotational diffusion relation tells us that we can write down a rotational diffusion coefficient for a sphere as:

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

Where $k_B$ is Boltzmann's constant, $T$ is the temperature in Kelvin, $\zeta_f \approx (8 \pi \eta)(r)^3$ is the friction, $\eta$ is the viscosity of the medium (e.g. $\approx 1 \space cP$ in pure water), and $r$ is the radius of the sphere.

For an example calculation of $D_r \approx 0.2 \space rad^2/s$ (or $Hz$) of a $\approx 1 \space \mu m$ radius sphere in pure water at room temperature, please see here: http://www.microurl.info/YOtwp (this shortened URL is going to WolframAlpha).

My question is the following: Take a point on the sphere undergoing rotational diffusion in solution, and select a random point on its surface. Define a vector $v_1$ between this random point and the point at the center of the sphere. Let $v_2$ be another randomly oriented vector. What probability distribution do we have for the time until a minimum angle $\theta$ between $v_1$ and $v_2$ is achieved?

Alternatively, if we start with the vectors $v_1$ and $v_2$ being within a minimum angle $theta$, what time distribution do we have until the minimum angle becomes larger than $\theta$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

If I have understood you correctly, your rotational diffusion problem is equivalent to the problem of a particle diffusing on the surface of a sphere (angular diffusion constant $D_r$), starting at some polar angle $\theta_0\in(0,\pi)$ and first crossing the angle $\theta\in(0,\theta_0)$ at time $t$. This is a classic problem in first-passage-times, see for example S. Redner, A guide to first-passage processes, or arXiv:1101.5043.

The mean first passage time is $$\langle t\rangle=\frac{2}{D_{r}}\ln\left(\frac{\sin(\theta_0/2)}{\sin(\theta/2)}\right).$$

If you choose the starting point at random, uniformly on the sphere, then you would perform one more average over $\theta_0$ (with weight $\sin\theta_0$), producing the average

$$\langle t\rangle=-\frac{2}{D_{r}}\left[\ln[\sin(\theta/2)]+\frac{1+\cos\theta}{4}\right].$$

The full probability distribution of the first passage time seems more problematic, I have not found it in the literature.

Improve this answer
$\endgroup$
8
  • $\begingroup$ I agree with your interpretation of the problem. However, I'm a little confused about your expression for $<t>$ provided random uniform initiation, shouldn't there for a $D_r$ term somewhere? $\endgroup$
    – Ayn
    Commented Feb 17, 2014 at 15:02
  • $\begingroup$ Also, is the expression for $<t>$ starting from $\theta_0 \in (0, \pi)$ from the linked arXiv paper? I can't seem to find the derivation for it in the 3D case section. $\endgroup$
    – Ayn
    Commented Feb 17, 2014 at 15:05
  • $\begingroup$ This answers my question! However, if you could point out where $<t>$ was derived (unless you did it yourself) that would be very helpful for me in terms of understanding. $\endgroup$
    – Ayn
    Commented Feb 17, 2014 at 15:18
  • $\begingroup$ typo (missing $D_r$) corrected; the expression for $\langle t\rangle$ is the unnumbered equation after Eq. 57 in arXiv:1101.5043, for $\omega=0$ (no bulk diffusion, only surface diffusion). $\endgroup$
    – Carlo Beenakker
    Commented Feb 17, 2014 at 15:19
  • $\begingroup$ Ok, I see. Regarding the unnumbered equation after Eq. 57 in arXiv:1101.5043: we're assuming the unit radius sphere, otherwise it's: $\langle t\rangle=\frac{2R^2}{D_{r}}...$, we're dropping the terms involving contributions from $D_2$ (related to bulk diffusion) (which seems OK to me), and we're neglected the latter summation/integral term as a small positive constant that does not scale with the diffusion coefficient. $\endgroup$
    – Ayn
    Commented Feb 17, 2014 at 15:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .