I am having a problem with this form of Langevin equation:
$m d v(t)=-k v(t) d t+\sigma(v(t)) d W(t) \quad \cdots (1)$
where:
- $v(t)$ velocity of the particle
- $m$ mass of the particle
- $k$ coefficient of friction (that according to my lecture is supposed to be equal to $6 \pi \eta a$ , where :
- $a$ : radius of the particle
- $\eta$ : viscosity of the solvent
- $-k v(t)$ friction force
- $W$ Wiener process
- $\sigma$ variations amplitude (?)
By comparing it to a Langevin equation on a lecture :
$m \frac{d v(t)}{d t}=-k v(t)+\mathbf{F}(t)$
Where is the random force with the properties :
$\begin{array}{c} \langle\mathbf{F}(t)\rangle=\mathbf{0} \\ \langle\mathbf{F}(t) \mathbf{F}(0)\rangle=2 k_{B} T k \mathbf{I} \delta(t) \end{array}$
By identification I get $\sigma(v(t)) \frac{d W(t)}{dt}= F(t)$ is the random force?
Later in the lecture in a section for numerical analysis, we define a new random variable called the Wiener process :
$\int_{t_{i}}^{t_{4+1}} \mathbf{F}(t)d t \equiv \Delta \mathbf{W}_{i}$
where :
$\left\langle\Delta \mathbf{W}_{i}\right\rangle=0 \\ \left\langle\Delta \mathbf{W}_{i} \Delta \mathbf{W}_{j}\right\rangle=2 k_{B} T k \Delta t \mathrm{I} \delta_{i j}$
where :
- $I$ : is the unit tensor
- The prefactor $2k_BTk$ is determined to satisfy the fluctuation-dissipation theorem
- The angular parentheses note the average 'expectation'
but then aren't we approximating
$\frac{d W(t)}{dt} \simeq F(t)$ ( my lecture said the random force $F$ is not continuous but we assume its integral is, 'for which we introduce the new random variable $\Delta W_i$')
If so, what is the $\sigma(v(t)) $ ?
I can't see what does the $\sigma(v(t))$ mean in the equation $(1)$ and where does it come from?
Well, the first equation just shows a general version of a Langevin equation. In principle, you can make the noise term as complicated as you want. For example, if we would take $F(t,v)=vdW(t)$, then we would have $\sigma(v)=v$. However, in this case, your lecturer assumed that $F$ does not depend on $v$ and hence $\sigma$ is constant. At this point, you can choose what to do with $\sigma$. In this case $W(t)$ is defined as a Brownian motion with variance $2k_BTkI$ and $\sigma=I$. However, you could also have chosen to define $W(t)$ to be a 'standard' Brownian motion, i.e. with variance $I$. In that case, you can choose $\sigma=\sqrt{2k_B TkI}$. As you can see now by the notation, $\sigma$ informally represents the standard deviation of your noise term $F$.