Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 this site (this shortened URL is going to WolframAlpha).

However, as we can directly see in the calculation, the units for $D_r$ are clearly expressible as rad/s or Hz. Why are the units for $D_r$ always reported as rad$^2$/s in the literature?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

The diffusion equation takes the form $$ \frac{\partial f}{\partial t}=D\frac{\partial^2 f}{\partial x^2} $$ where $f$ is some function. For simplicity, let's let $f=\rho$ the mass-density. In this case, then we have, units-wise, $$ \frac{{\rm kg/m^3}}{\rm s} = \left[D\right]\frac{\rm kg/m^3}{\rm m^2} $$ Thus, in order to have the correct units on both sides, $\left[D\right]={\rm m^2/s}$. In the case of rotational diffusion, the diffusion equation takes the form $$ \frac{\partial f}{\partial t}=D_r\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial f}{\partial\theta}\right)+D_r\frac{1}{\sin^2\theta}\frac{\partial^2f}{\partial\phi^2} $$ Since $\theta$ and $\phi$ are measured in radians and $\sin\theta$ is unitless, then we must have that $\left[D_r\right]={\rm rad^2/s}$.

share|improve this answer
    
I see why it must be so, but why is this not reflected then in the (common) expression for $D_r$? –  user47046 Feb 17 at 19:01
    
I'm not sure what you mean. –  Kyle Kanos Feb 17 at 19:01
    
If you run the calculation based on the expression from Wikipedia (or elsewhere) you see rad/s or Hz, not rad^2/s? See the result of the calculation in the WolframAlpha link, for example. –  user47046 Feb 17 at 19:02
    
My question is less "why must this be so" than why I always see the expression for $D_r$ given in my posting, where a calculation will evaluate to rad/s or Hz instead of rad^2/s. –  user47046 Feb 17 at 19:05
    
Well radian isn't exactly a unit like the meter, it is a dimensionless quantity. You seem to be inserting a single radian via dimensional grounds of the components when looking at the diffusion equation suggests squaring the radian –  Kyle Kanos Feb 17 at 19:05

A rotational diffusion constant has units of $1/s$. Radians are dimensionless numbers and can not be included in a formal dimensional analysis.

The link to your calculation is broken, but here's mine (probably the same):

$k_B T$ is an energy $[kg m^2/s^2]$, $\eta$ is dynamic viscosity $[kg/ms]$. Thus

$$ \mathcal D_r \propto \frac{k_B T}{\eta r^3} \mathrm{~has~units~} [\frac{kg m^2}{s^2}\frac{ms}{kg m^3}]=[\frac{1}{s}] $$

Informally, we may think of a diffusion constant as a mean squared displacement divided by a time. And since the variable displaced is an angle, the diffusion constant is often written $\mathrm{rad}^2/s$. However, a radian is defined as a length divided by another length, and is therefore indistinguishable from $1$ in a dimensional analysis.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.