statistical mechanics - Langevin Equation - Stochastic Differential Equation. What are the subtleties?

I am trying to find out the motion of a particle in 3D governed by the Langevin equation, numerically.

Anyway, the Langevin equation is given by

$$m \ddot{x} = -(6\pi a\nu) \dot{x} + F_b$$

where $F_b$ is due to random fluctuations.

From various sources I've read, $F_b$ is treated as stochastic term. According to wiki http://en.wikipedia.org/wiki/Langevin_dynamics it translates to

$$m \ddot{x} = -(6\pi a\nu) \dot{x} + \sqrt{2\gamma K_BTm}R(t)$$

where $\gamma$ is a friction term, and $R(t)$ is a delta-correlated stationary Gaussian process with zero-mean.

If my guess is correct, there isn't an explicit form of $R(t)$?

So I'm wondering if anyone can explain what $R(t)$ means and how I can go about trying to implement it into a simulation for example.

Answer

The exact solution of your equation can be written as

$$x(t) = x(0) + \frac{m}{6 \pi a \nu} \dot{x}(0) - \frac{m}{6 \pi a \nu} \dot{x}(0) \, \text{e}^{-\frac{6\pi a\nu}{m}t} + \frac{1}{m}\int_0^t \text{d}\tau_1 \int_0^{\tau_1} \text{d}\tau_2 \text{e}^{\frac{6\pi a\nu}{m} \left(\tau_2-\tau_1\right)} F(\tau_2) \, .$$

$x(0)$ and $\dot{x}(0)$ are the initial conditions. Then whatever you want to compute, you can write it as an expression that depends on $F(t)$ and take its average over the fluctuations of $F(t)$.

For example

$$\begin{align*} \langle x(t) \rangle = & \langle x(0) \rangle + \frac{m}{6 \pi a \nu} \langle \dot{x}(0) \rangle- \frac{m}{6 \pi a \nu} \langle \dot{x}(0) \rangle \, \text{e}^{-\frac{6\pi a\nu}{m}t} \\ & + \frac{1}{m}\int_0^t \text{d}\tau_1 \int_0^{\tau_1} \text{d}\tau_2 \text{e}^{\frac{6\pi a\nu}{m} \left(\tau_2-\tau_1\right)} \langle F(\tau_2) \rangle \, .\end{align*}$$

If your particle starts at rest and at the origin,

$$\langle x(0) \rangle = 0 \, \qquad \langle \dot{x}(0) \rangle = 0 \, ,$$

and if it experiences a constant force,

$$\langle F(t) \rangle = F \, ,$$

then you find

$$\begin{align*}\langle x(t) \rangle & = \frac{F}{m}\int_0^t \text{d}\tau_1 \int_0^{\tau_1} \text{d}\tau_2 \text{e}^{\frac{6\pi a\nu}{m} \left(\tau_2-\tau_1\right)} \\ & = \frac{F}{m} \frac{m}{6 \pi a \mu} \left[ t + \frac{m}{6 \pi a \nu} \left( \text{e}^{-\frac{6 \pi a \nu}{m}t}-1\right) \right] \, .\end{align*}$$

You see that you can choose the statistics of $F(t)$ freely. Then if you know the moments of $F(t)$, you can compute the moments of $x(t)$. It's all about computing integrals. Typically one chooses Gaussian statistics with

$$\langle F(t) \rangle = 0 \, , \qquad \langle F(t_1) F(t_2) \rangle = D \, \delta(t_1-t_2) \, .$$

If you insist on solving this problem numerically, you need to discretise time

$$t \in \left[0,\infty\right[ \rightarrow t \in \left\{t_i\right\}_{i = 1,..,N} \, .$$

Then you can sample $F(t)$ according to your favourite probability distribution

$$P\left[F(t_1),F(t_2),..,F(t_N)\right] \, .$$

For every sample you get a discretised function, $F(t_i)$ and you can switch to finite difference derivatives (for example) to solve your differential equation. Then you average at the end.