Brownian Motion (Langevin equation) correlation function

In summary, 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 $$\langle \eta_i(t) \eta_j(t') \rangle=2 \lambda k_B T \delta_{ij} \delta(t - t')$$. It follows the fluctuation-dissipation theorem, which connects the friction coefficient ##\lambda## to the strength of the random force in the Langevin equation. This is derived either by solving the Fokker-Planck equation for the probability distribution function or by evaluating the expectation value
  • #1
Tim667
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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
 
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  • #2
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.
 
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  • #3
vanhees71 said:
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
 

Related to Brownian Motion (Langevin equation) correlation function

1. What is Brownian Motion?

Brownian Motion is a physical phenomenon in which particles suspended in a fluid undergo random, erratic movements due to collisions with the molecules of the fluid.

2. What is the Langevin equation?

The Langevin equation is a mathematical model that describes the motion of a particle in a fluid under the influence of random forces. It takes into account both the deterministic forces acting on the particle and the random forces due to collisions with the fluid molecules.

3. What is the correlation function in the context of Brownian Motion?

The correlation function in Brownian Motion is a measure of the relationship between the position of a particle at a given time and its position at a later time. It is used to study the behavior and dynamics of particles undergoing Brownian Motion.

4. How is the correlation function related to the Langevin equation?

The Langevin equation can be used to derive the correlation function of a particle undergoing Brownian Motion. The correlation function can then be used to analyze the behavior of the particle and make predictions about its future movements.

5. What factors affect the correlation function in Brownian Motion?

The correlation function in Brownian Motion is affected by various factors such as the size and shape of the particle, the properties of the fluid, and the temperature. It can also be influenced by external forces and interactions with other particles in the fluid.

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