I was curious if there was an equation describing the force acting on a particle (say, sitting in a fluid) that causes it to diffuse.

If so, does it include the diffusion coefficient D?

Based on simple dimensional analysis (units for D are $length^2/time$ according to this site), I get the feeling that if such an equation existed, it would be of the form: $$F=\frac{mD}{tL}$$ where m is the mass of the particle, and t and L are a characteristic timescale and a characteristic length scale, respectively. Is this correct?

To provide some context, this is for a simulation trying to model Brownian dynamics of molecules in a solvent.


3 Answers 3


It sounds like you're trying to solve the Langevin Equation. This is a model of Brownian motion where the particle experiences stochastic kicks at discrete time intervals. Your force, in this case, is a random variable you draw from a distribution each time step (instead of being given by an explicit formula).

For the simplest case, the "kicks" are generated by thermal noise, and have a gaussian distribution.

From the Wikipedia article linked above, the equation of motion is: $$ m \frac{d^2 \vec{x}}{dt^2} = - \lambda \frac{d\vec{x}}{dt} + \vec{f(t)} $$ Here the $\lambda$ term is from viscous friction with the fluid, and $\vec{f}$ is the random force. For a particle of radius $a$ in a fluid of dynamic viscosity $\eta$, then Stokes' Law says $\lambda = 6 \pi \eta a$.

If time is continuous, the correlation function for the kicks is:

$$ \langle f_i(t) f_j(t') \rangle = 2 \lambda k_B T \delta_{ij}\delta(t-t') $$

Where $T$ is the temperature and $f_i$ is the $i$'th component of the force. To generate a force $\vec{f}(t)$ obeying this correlation function you may want to investigate gaussian processes.

Finally, there IS a relation between these (microphysical) constants and the diffusion coefficient $D$. It is called the Stokes-Einstein relation and was the first example of the Fluctuation-Dissipation Theorem. For the Langevin equation above, the relation is: $$ D = \frac{k_B T}{\lambda} \ . $$

Note: the Langevin equation is NOT an ordinary differential equation. It is a Stochastic differential equation, so the usual (e.g. Runge-Kutta) numerical methods won't apply.


The diffusion is caused by collisions between the molecules due to the random thermal motion of each molecule. So you have a blob of oxygen in a box of nitrogen then they are all moving with a random velocity from the Maxwell velocity distribution. When two molecules collide, they change velocity and so they change both speed/direction. They then collide some more. It could be an oxygen-oxygen, nitrogen-nitrogen, or oxygen-nitrogen collision. These are all random, so the tendency is to diffuse out and eventually reach a uniform distribution.

From a continuum perspective, this is described by a force that depends on the gradient of the concentrations and the pair-wise diffusion coefficient (which is a measure of how effective those collisions are at moving around the molecules). But from a microscopic perspective, there is no "diffusion" force.

There are only forces due to collisions* and these forces lead to all macroscopic phenomenon -- viscosity, thermal diffusion, mass diffusion, chemical reactions, ionization, etc..

* There's also long-range forces for polar molecules like water, which may matter, but it's still not a "diffusion" force.


Diffusion is one of several entropic phenomena the average effect of which can be described as a purely Newtonian phenomenon. This is done via the introduction of an entropic force. For diffusion the entropic force takes the shape of a repulsive radial force directed away from the starting position of the diffusion process: $$F_r=\frac{2kT}{r}$$ Here, $k$ represents Boltzmann's constant and $T$ the absolute temperature. Note that this yields a description for the average diffusion behavior only. In other words, this entropic force should be utilized to simulate the radial diffusive spreading of a cloud of particles, rather than the brownian movement of an individual particle. Details can be found in this paper.

Using the Stokes-Einstein equation, $$D=\frac{kT}{6\pi\eta R}$$ where $R$ represents the radius of the diffusing particle and $\eta$ the viscosity of the fluid in which it is Suspended, the above expression can be cast in a form that gives the entropic force in terms of the diffusion constant $D$: $$F_r=12 \pi \eta D \frac{R}{r}$$

  • 1
    $\begingroup$ The OP's question was in the context of a Brownian motion simulation. This force will make particles diffuse from the origin, but not undergo Brownian motion of any sort. $\endgroup$
    – Geoff Ryan
    Feb 27, 2015 at 3:35
  • $\begingroup$ @GeoffRyan - sure, if the objective is to simulate the diffusion of an individual particle, my answer is not relevant. If the objective is to simulate the diffusion of -say- a cloud of particles forming an ink drop, the above is highly relevant. As OP used a plural in his question "this is for a simulation trying to model Brownian dynamics of molecules in a solvent" I assumed the latter to be the case. $\endgroup$
    – Johannes
    Feb 27, 2015 at 16:13

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