Choosing the variance for a Gaussian Distribution for the Langevin Equation

Question

I am trying to numerically integrate the Langevin equation which is given by

$$ \ddot x= -\frac{6\pi\eta a}{m} \dot x + \frac{\xi(t)}{m} $$

where $\xi(t)$ is a probability distribution.

I'm assuming a gaussian distribution

So $$<\xi(t)>_\xi= 0 \space$$ $$< \xi(t_1)\xi(t_2)>_\xi= g\delta(t_1-t_2) $$

where $g$ is the strength of the fluctuation

According to this link, the fluctuation strength is given by

$$g=2\gamma k_BT $$

where $\gamma, \space k_B, t$ are just some constants

So I will need to generate a set of points that follows the gaussian distribution to pick for $\xi(t_i)$

Now I know that the mean is $\mu=0$

So my question is, how should I decide on what to use for my variance to generate the gaussian distribution?

Would it just be just $g$ directly?

Answer 1

Yes, you take $g$ for the variance.

The equation

$$\langle\xi(t_1),\xi(t_2)\rangle =g\delta(t_1-t_2)$$

gives the correlation between the samples $\xi(t_1),\xi(t_2)$ at times $t_1,t_2$, and the delta function tells you that they are assumed to be not correlated at different times. The only non-zero case of the right hand side is when $t_1=t_2$. In that case you get on the left hand side the correlation of $\xi(t_1)$ with itself, which is of course the variance of $\xi(t_1)$ at time $t_1$. So, yes, you just take $g$ for the variance.