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A molecule with a red cross on it's front undergoing 3 dimensional rotational diffusion. The red cross moves erratically as the sphere is made to randomly rotate by collisions with surrounding molecules.

Rotational diffusion is the rotational movement which acts upon any object such as particles, molecules, atoms when present in a fluid, by random changes in their orientations. Whilst the directions and intensities of these changes are statistically random, they do not arise randomly and are instead the result of interactions between particles. One example occurs in colloids, where relatively large insoluble particles are suspended in a greater amount of fluid. The changes in orientation occur from collisions between the particle and the many molecules forming the fluid surrounding the particle, which each transfer kinetic energy to the particle, and as such can be considered random due to the varied speeds and amounts of fluid molecules incident on each individual particle at any given time.

The analogue to translational diffusion which determines the particles/molecules position in space, Rotational diffusion randomises the orientation of any particle it acts on. Anything in a solution will experience rotational diffusion, from the microscopic scale where individual atoms may have an effect on each other, to the macroscopic scale.

History

Peter Debye was the first to apply Einstein's theory of translational diffusion in a rotational context

Brownian motion was first described by Robert Brown in 1828 as the movement of small, solid particles in a solution . From the years 1904, 1905 and 1906, a formula for the diffusion coefficient was revealed independently by William Sutherland, Albert Einstein, and Marian Smoluchowski respectively. This formula took the form:

$D=\mu {k_{B}}T$

However, specifically rotational Brownian motion was first discussed by Peter Debye in 1913 who related Albert Einstein's theory of translational Brownian motion to the rotation of molecules with electric dipoles.

Applications

Complex 3-dimensional molecule rotating in 2-dimensions about a singular axis

Rotational diffusion has multiple applications in chemistry and physics, and is heavily involved in many biology based fields. For example, protein-protein interaction is a vital step in the communication of biological signals. In order to communicate, the proteins must both come into contact with each other and be facing the appropriate way to interact with each others binding site, which relies on the proteins ability to rotate . As well as this, the overall mass of a protein and the distribution of the mass within the structure of the protein can be approximated by measuring the rotational diffusion constants present in the system, which is a useful tool for identifying unknown particles. As an example concerning physics, rotational Brownian motion in astronomy can be used to explain the orientations of the orbital planes of binary stars, as well as the seemingly random spin axes of supermassive black holes.

Relation to translational diffusion

The standard translational model of Brownian motion

Much like translational diffusion in which particles in one area of high concentration slowly spread position through random walks until they are near-equally distributed over the entire space, in rotational diffusion, over long periods of time the directions which these particles face will spread until they follow a completely random distribution with a near-equal amount facing in all directions. As impacts from surrounding particles rarely, if ever, occur directly in the centre of mass of a 'target' particle, each impact will occur off-centre and as such it is important to note that the same collisions that cause translational diffusion cause rotational diffusion as some of the impact energy is transferred to translational kinetic energy and some is transferred into torque.

Two-dimensional rotational diffusion

A sphere rotating around a fixed central axis can be modelled as a circle rotating in 2-dimensions when viewed from the axis of rotation. Here A₀ is the starting position at t₀ and A is the position at time t when the circle has rotated by θ.

A sphere rotating around a fixed axis will rotate in two dimensions only and can be viewed from above the fixed axis as a circle. In this example, a sphere which is fixed on the vertical axis rotates around that axis only, meaning that the particle can have a θ value of 0 through 360 degrees, or 2π Radians, before having a net rotation of 0 again.

These directions can be placed onto a graph which covers the entirety of the possible positions for the face to be at relative to the starting point, through 2π radians, starting with -π radians through 0 to π radians. Assuming all particles begin with single orientation of 0, the first measurement of directions taken will resemble a delta function at 0 as all particles will be at their starting, or 0^th, position and therefore create an infinitely steep single line. Over time, the increasing amount of measurements taken will cause a spread in results; the initial measurements will see a thin peak form on the graph as the particle can only move slightly in a short time. Then as more time passes, the chance for the molecule to rotate further from it’s starting point increases which widens the peak, until enough time has passed that the measurements will be evenly distributed across all possible directions.

The distribution of orientations will reach a point where they become uniform as they all randomly disperse to be nearly equal in all directions. This can be visualized in two ways.

1. For a single particle with multiple measurements taken over time. A particle which has an area designated as it’s face pointing in the starting orientation, starting at a time t₀ will begin with an orientation distribution resembling a single line as it is the only measurement. Each successive measurement at time greater than t₀ will widen the peak as the particle will have had more time to rotate away from the starting position.
2. For multiple particles measured once long after the first measurement. The same case can be made with a large number of molecules, all starting at their respective 0^th orientation. Assuming enough time has passed to be much greater than t₀, the molecules may have fully rotated if the forces acting on them require, and a single measurement shows they are near-to-evenly distributed.

Langevin dynamics

Collisions with the surrounding fluid molecules will create a fluctuating torque on the sphere due to the varied speeds, numbers, and directions of impact. When trying to rotate a sphere via an externally applied torque, there will be a systematic drag resistance to rotation. With these two facts combined, it is possible to write the Langevin-like equation:

${\frac {dL}{dt}}={I}\,\cdot {\frac {d^{2}{\phi }}{dt^{2}}}=-{\zeta }^{r}\cdot {\frac {d{\theta }}{dt}}+TB(t)$

Where:

• L is the angular momentum.
• ${\frac {dL}{dt}}$ is is torque.
• I is the moment of inertia about the rotation axis.
• t is the time.
• t₀ is the start time.
• θ is the angle between the orientation at t₀ and any time after, t.
• ζ^r is the rotational friction coefficient.
• TB(t) is the fluctuating Brownian torque at time t.

The overall Torque on the particle will be the difference between:

$TB(t)$ and $({\zeta }^{r}\cdot {\frac {d{\theta }}{dt}})$.

This equation is the rotational version of Newtons second equation of motion. For example, in standard translational terms, a rocket will experience a boosting force from the engine whilst simultaneously experiencing a resistive force from the air it is travelling through. The same can be said for an object which is rotating.

Due to the random nature of rotation of the particle, the average Brownian torque is equal in both directions of rotation. symbolised as:

$\left\langle TB(t)\right\rangle =0$

This means the equation can be averaged to get:

${\frac {d\left\langle L\right\rangle }{dt}}=-{\zeta }^{r}\cdot \left\langle {\frac {d{\theta }}{dt}}\right\rangle =-{\frac {\zeta ^{r}}{I}}\left\langle L\right\rangle$

Which is to say that the first derivative with respect to time of the average Angular momentum is equal to the negative of the Rotational friction coefficient divided by the moment of inertia, all multiplied by the average of the angular momentum.

As ${\frac {d\left\langle L\right\rangle }{dt}}$ is the rate of change of angular momentum over time, and is equal to a negative value of a coefficient multiplied by $\left\langle L\right\rangle$, this shows that the angular momentum is decreasing over time, or decaying with a decay time of:

${\tau {_{L}}}={\frac {I}{\zeta ^{r}}}$.

For a sphere of mass m, uniform density ρ and radius a, the moment of inertia is:

$I={\frac {2ma^{2}}{5}}={\frac {8{\pi }{\rho }a^{5}}{15}}$.

As mentioned above, the rotational drag is given by the Stokes friction for rotation:

${\zeta ^{r}}=8\pi \eta a^{3}$

Combining all of the equations and formula from above, we get:

${\tau {_{L}}}={\frac {\rho a^{2}}{15\eta }}={\frac {3}{10}}\tau _{p}$ where:

• $\tau _{p}$ is the momentum relaxation time
• η is the viscosity of the Liquid the sphere is in.

Example: Spherical particle in water

Water particles (blue) and larger virus particles (pink) The impact between the virus and water, and to a lesser extent two virus molecules, will cause translational and rotational movement

Let's say there is a virus which can be modelled as a perfect sphere with the following conditions:

• Radius (a) of 100 nanometres, a = 10^-7m.
• Density ρ = 1500 kg m^-3.
• Orientation originally facing in a direction denoted by π.
• Suspended in water.
• Water has a viscosity of η = 8.9 × 10^-4 Pa·s at 25 ° C
• Assume uniform mass and density throughout the particle

First, the mass of the virus particle can be calculated:

$m={\frac {4\rho \pi a^{3}}{3}}={\frac {4\times 1500\times \pi \times (10^{-7})^{3}}{3}}=6.3\times 10^{-18}kg$

From this, we now know all the variables to calculate moment of inertia:

$I={\frac {2ma^{2}}{5}}={\frac {2\times (6.3\times 10^{-18})\times (10^{-7})^{2}}{5}}=2.5\times 10^{-32}kg\cdot m^{2}$

Simultaneous to this, we can also calculate the rotational drag:

$\zeta ^{r}=8\pi \eta a^{3}=8\times \pi \times (8.9\times 10^{-4})\times (10^{-7})^{3}=2.237\times 10^{-23}Pa\cdot s\cdot m^{3}$

Combining these equations we get:

$\tau _{L}={\frac {I}{\zeta ^{r}}}={\frac {2.5\times 10^{-32}kg\cdot m^{2}}{2.2\times 10^{-25}Pa\cdot s\cdot m^{3}}}=1.1\times 10^{-9}kg\cdot Pa^{-1}\cdot s^{-1}\cdot m^{-1}$

As the SI units for Pascal are $kg\cdot m^{-1}\cdot s^{-2}$ the units in the answer can be reduced to read:

$\tau _{L}=1.1\times 10^{-9}s$

For this example, the decay time of the virus is in the order of hundreds of nanoseconds.

Smoluchowski description of rotation

To write the Smoluchowski equation for a particle rotating in two dimensions, we introduce a probability density P(φ, t) to find the vector u at an angle θ and time t. This can be done by writing a continuity equation:

${\partial P(\theta ,t) \over \partial t}=-{\partial j(\theta ,t) \over \partial \theta }$

where the current can be written as:

$j(\theta ,t)=-D^{r}{\partial P(\theta ,t) \over \partial \theta }$

Which can be combined to give the rotational diffusion equation:

${\partial P(\theta ,t) \over \partial t}=D^{r}{\partial ^{2}P(\theta ,t) \over \partial \theta ^{2}}=D^{r}P(\theta ,t)$

We can express the current in terms of an angular velocity which is a result of Brownian torque T_B through a rotational mobility with the equation:

$j_{B}(\theta ,t)={\dot {\theta }}_{B}P(\theta ,t)$

Where:

• ${\dot {\theta }}_{B}=\mu ^{r}T_{B}$
• $T_{B}=-{\partial V_{B} \over \partial \theta }$
• $V_{B}(\theta ,t)=k_{B}T\ln P(\theta ,t)$

The only difference between rotational and translational diffusion in this case is that in the rotational diffusion, we have periodicity in the angle θ. As the particle is modelled as a sphere rotating in two dimensions, the space the particle can take is compact and finite, as the particle can rotate a distance of 2π before returning to its original position

$P(\theta +2\pi ,t)={P(\theta ,t)}$

We can create a conditional probability density, which is the probability of finding the vector u at the angle φ and time t given that it was at angle φ₀ at time t=0 This is written as such:

$P(\theta ,0\mid \theta _{0})=\delta (\theta -\theta _{0})$

The solution to this equation can be found through a Fourier series:

$P(\theta ,t\mid \theta _{0})={\frac {1}{2\pi }}\left[1+2\sum _{m=1}^{\infty }e^{-D^{r}m^{2}t}cosm(\theta -\theta _{0})\right]={\frac {1}{2\pi }}\Theta _{3}({\frac {1}{2}}(\theta -\theta _{0}),e^{-D^{r}t})$

Where $\Theta _{3}(z,\tau )$ is the Jacobian theta function of the third kind.

By using the equation

$\Theta _{3}(z,\tau )=(-i\tau )^{-1/2}exp{\biggl (}{\frac {z^{2}}{i\pi \tau }}{\biggl )}\Theta _{3}{\biggl (}{\frac {z}{\tau }},-{\frac {1}{\tau }}{\biggl )}$

The conditional probability density function can be written as :

$P(\theta ,t\mid \theta _{0})={\frac {1}{\sqrt {4\pi D^{r}t}}}\sum _{n=-\infty }^{\infty }exp\left[-{\frac {(\theta -\theta _{0}-2n\pi )^{2}}{4D^{r}t}}\right]$

For short times after the starting point where t ≈ t₀ and φ ≈ φ₀, the formula becomes:

$P(\theta ,t\mid \theta _{0})\approx {\frac {1}{\sqrt {4\pi D^{r}t}}}exp\left[-{\frac {(\theta -\theta _{0})^{2}}{4D^{r}t}}\right]+\cdots$

The terms included in the are exponentially small and make little enough difference to not be included here. This means that at short times the conditional probability looks similar to translational diffusion, as both show extremely small perturbations near t₀. However at long times, t » t₀ , the behaviour of rotational diffusion is different to translational diffusion:

$P(\theta ,t\mid \theta _{0})\approx {\frac {1}{2\pi }},t\rightarrow \infty$

The main difference between rotational diffusion and translational diffusion is that rotational diffusion has a periodicity of $\theta +(2\pi )=\theta$, meaning that these two angles are identical. This is because a circle can rotate entirely once before being at the same angle as it was in the beginning, meaning that all the possible orientations can be mapped within the space of $2\pi$. This is opposed to translational diffusion, which has no such periodicity.

The conditional probability of having the angle be φ is approximately ${\frac {1}{2\pi }}$ .

This is because over long periods of time, the particle has had time rotate throughout the entire range of angles possible and as such, the angle θ could be any amount between θ₀ and θ₀ + 2 π. The probability is near-evenly distributed through each angle as at large enough times. This can be proven through summing the probability of all possible angles. As there are 2π possible angles, each with the probability of , the total probability sums to 1, which means there is a certainty of finding the angle at some point on the circle.

3D rotational diffusion

A model of 3-dimensional rotation, now dependent on the change of two angles, both around perpendicular axes to each other

When all three of the dimensions in which molecules exist in are taken into account, the equations governing the movement of the particle now depend on an extra factor. Whereas phi was used to model 2 dimensional rotation around a single axis, theta is also used to model rotation around another axis which is perpendicular to the first. Through this spherical polar coordinate* method, rotation can be mapped to any point in the 3d surface.

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