Brownian Motion (Langevin equation) correlation function

[Tim667]

TL;DR Summary

How does one calculate the correlation function?

So the Langevin equation of Brownian motion is a stochastic differential equation defined as
$$m {d \textbf{v} \over{dt} } = - \lambda \textbf{v} + \eta(t)$$

where the noise function eta has correlation function $$\langle \eta_i(t) \eta_j(t') \rangle=2 \lambda k_B T \delta_{ij} \delta(t - t')$$.

I have two questions. How does one actually calculate a correlation function and where exactly do the constants (with temperature, the Boltzmann constants etc) proceeding the delta functions originate here? I understand that the delta functions ensure that there is no correlation at different times etc, but I don't get where the rest comes from.

Thanks

 

[vanhees71]

That's the fluctuation-dissipation theorem. It's one of Einstein's most famous results from his "miracle year" 1905. There are two ways to answer your question. One is to derive the Fokker-Planck equation for the probability distribution function $f(t,\vec{v})$ and show that the stationary solution is the Maxwell-Boltzmann distribution with temperature $T$, as it should be.

The other is to evaluate the expectation value of the kinetic energy and use the equipartition theorem for the long-time limit. In other words we have to calculate $\langle \vec{v}^2(t)$. To do that we can formally solve the stochastic differential equation, using the Green's function of the "deterministic part", i.e., we look for
$$m \dot{G}(t)+\lambda G(t)=\delta(t). \qquad (*)$$
For $t \neq 0$ we have
$$G(t)=A \exp(-\gamma t), \quad \gamma=\lambda/m.$$
Since we want a "causal Green's function" we make $G(t)=0$ for $t<0$ and determine $A$ for $t>0$ such that we get the right singularity. So we integrate (*) over a small interval $(-\epsilon,\epsilon)$ and make $\epsilon \rightarrow 0^+$ to get
$$m A=1 \; \Rightarrow \; A=1/m.$$
So we have
$$G(t)=\Theta(t) \frac{1}{m} \exp(-\gamma t).$$
Then the formal solution of the Langevin equation is (for $t>0$)
$$\vec{v}(t)=\vec{v}_0 \exp(-\gamma t) + \int_0^t \mathrm{d} t' G(t-t') \eta(t')=\vec{v}_0 \exp(-\gamma t) + \vec{v}_{\text{fluct}}(t).$$
From this we get
$$\vec{v}^2(t) = \vec{v}_0^2 \exp(-2 \gamma t) + 2 \vec{v}_0 \cdot \vec{v}_{\text{fluct}}(t) \exp(-\gamma t) + \vec{v}_{\text{fluct}}^2(t).$$
Taking the expectation value, we get because of $\langle \vec{\eta}(t)=0$
$$\langle \vec{v}^2(t) \rangle=\vec{v}_0^2 \exp(-2 \gamma t) +\langle \vec{v}_{\text{fluct}}^2(t) \rangle.$$
Now
$$\langle \vec{v}_{\text{fluct}}^2(t) \rangle=\int_0^t \mathrm{d} t_1 \int_0^t \mathrm{d} t_2 G(t-t_1) G(t-t_2) \langle{\eta_j(t_1) \eta_j(t_2)}=\int_0^t \mathrm{d} t_1 \int_0^t \mathrm{d} t_2 6 \lambda k_{\text{B}} T \delta(t_1-t_2)G(t-t_1) G(t-t_2).$$
Evaluating the integral leads finally to
$$\langle \vec{v}_{\text{fluct}}^2(t) \rangle=\frac{3 k_{\text{B}} T}{m}[1-\exp(-2 \gamma t)].$$
For $t \rightarrow \infty$ we thus get
$$\frac{m}{2} \langle \vec{v}^2(t) \rangle \rightarrow \frac{3}{2} k_{\text{B}} T,$$
as it should be in the equilibrium limit according to the equipartition theorem. So the diffusion coefficient $D=\lambda k_{\text{B}} T$ as assumed is the correct choice. That's known as dissipation-fluctuation theorem, because it connects the friction coefficient $\lambda$ ("dissipation") to the strength of the random force in the Langevin equation, which is describing (from a macroscopic point of view) the "diffusion" of the heavy particle in the heat bath made up by the light particles.

 

[Tim667]

That's the fluctuation-dissipation theorem. It's one of Einstein's most famous results from his "miracle year" 1905. There are two ways to answer your question. One is to derive the Fokker-Planck equation for the probability distribution function $f(t,\vec{v})$ and show that the stationary solution is the Maxwell-Boltzmann distribution with temperature $T$, as it should be.

The other is to evaluate the expectation value of the kinetic energy and use the equipartition theorem for the long-time limit. In other words we have to calculate $\langle \vec{v}^2(t)$. To do that we can formally solve the stochastic differential equation, using the Green's function of the "deterministic part", i.e., we look for
$$m \dot{G}(t)+\lambda G(t)=\delta(t). \qquad (*)$$
For $t \neq 0$ we have
$$G(t)=A \exp(-\gamma t), \quad \gamma=\lambda/m.$$
Since we want a "causal Green's function" we make $G(t)=0$ for $t<0$ and determine $A$ for $t>0$ such that we get the right singularity. So we integrate (*) over a small interval $(-\epsilon,\epsilon)$ and make $\epsilon \rightarrow 0^+$ to get
$$m A=1 \; \Rightarrow \; A=1/m.$$
So we have
$$G(t)=\Theta(t) \frac{1}{m} \exp(-\gamma t).$$
Then the formal solution of the Langevin equation is (for $t>0$)
$$\vec{v}(t)=\vec{v}_0 \exp(-\gamma t) + \int_0^t \mathrm{d} t' G(t-t') \eta(t')=\vec{v}_0 \exp(-\gamma t) + \vec{v}_{\text{fluct}}(t).$$
From this we get
$$\vec{v}^2(t) = \vec{v}_0^2 \exp(-2 \gamma t) + 2 \vec{v}_0 \cdot \vec{v}_{\text{fluct}}(t) \exp(-\gamma t) + \vec{v}_{\text{fluct}}^2(t).$$
Taking the expectation value, we get because of $\langle \vec{\eta}(t)=0$
$$\langle \vec{v}^2(t) \rangle=\vec{v}_0^2 \exp(-2 \gamma t) +\langle \vec{v}_{\text{fluct}}^2(t) \rangle.$$
Now
$$\langle \vec{v}_{\text{fluct}}^2(t) \rangle=\int_0^t \mathrm{d} t_1 \int_0^t \mathrm{d} t_2 G(t-t_1) G(t-t_2) \langle{\eta_j(t_1) \eta_j(t_2)}=\int_0^t \mathrm{d} t_1 \int_0^t \mathrm{d} t_2 6 \lambda k_{\text{B}} T \delta(t_1-t_2)G(t-t_1) G(t-t_2).$$
Evaluating the integral leads finally to
$$\langle \vec{v}_{\text{fluct}}^2(t) \rangle=\frac{3 k_{\text{B}} T}{m}[1-\exp(-2 \gamma t)].$$
For $t \rightarrow \infty$ we thus get
$$\frac{m}{2} \langle \vec{v}^2(t) \rangle \rightarrow \frac{3}{2} k_{\text{B}} T,$$
as it should be in the equilibrium limit according to the equipartition theorem. So the diffusion coefficient $D=\lambda k_{\text{B}} T$ as assumed is the correct choice. That's known as dissipation-fluctuation theorem, because it connects the friction coefficient $\lambda$ ("dissipation") to the strength of the random force in the Langevin equation, which is describing (from a macroscopic point of view) the "diffusion" of the heavy particle in the heat bath made up by the light particles.

Thank you! Brilliant answer