Brownian motion, correlation functions

[Telemachus]

Hi. I have a doubt on the derivation of the correlation function for the velocity and position in Brownian motion from the Langevin equation.

I have that for a brownian particle:

$\displaystyle v(t)=v_0e^{-\frac{\gamma}{m}t}+\frac{1}{m}\int_0^{t}dse^{-\frac{\gamma}{m}(t-s)}\xi(s)$ (1)

$\displaystyle \xi(t)$ is a Gaussian white noice with zero mean, such that $\displaystyle \left <\xi(t) \right>_{\xi}=0$.

The assumption that the noise is Gaussian means that the noise is delta-correlated:

$\displaystyle \left < \xi(t_1)\xi(t_2) \right>_{\xi}=g\delta (t_2-t_1)$

Now, the book makes use of the fact that $\displaystyle \left < v_0\xi(t) \right>_{\xi}=0$, and gives for the correlation function:

$\displaystyle \left < v(t_2)v(t_1) \right>_{\xi}=v_0^2e^{-\frac{\gamma}{m}(t_2+t_1)}+\frac{g}{m}\int_0^{t_1}ds_1 \int_0^{t_2} ds_2\delta(s_2-s_1)e^{\frac{\gamma}{m}(s_1-t_1)}e^{\frac{\gamma}{m}(s_2-t_2)}$ (2)

I think that this must be used: $\displaystyle \left < y_1(t_1)y_2(t_2) \right>=\int dy_1 \int dy_2 y_1y_2 P_2(y_1,t_1;y_2,t_2)$

But I'm not sure of it, and I don't know what are the intermediate steps between (1) and (2).

Thanks in advance.

 

[Telemachus]

The thing is that the integrals I get confuses me. For a simpler case I was considering the average of velocity.

The book gives (subject to the condition that $v(0)=v_0$): $\displaystyle \left < v(t) \right>_{\xi}=v_0e^{-\frac{\gamma}{m}t}$

So I think that this should mean: $\displaystyle \left < v(t) \right>_{\xi}=\int_0^{t} v(t_1)\xi(t_1)dv(t_1)$

I can put all that in terms of an integral of time using the equations for brownian motion: $\displaystyle \frac{dv(t)}{dt}=-\frac{\gamma}{m}v(t)+\frac{1}{m}\xi(t)$

Then $\displaystyle \left < v(t) \right>_{\xi}=-\frac{\gamma}{m}\int_0^t v^2(t_1)\xi(t_1)dt_1+\frac{1}{m}\int_0^t v(t_1)\xi^2(t_1)dt_1$

And using (1): $\displaystyle \left < v(t) \right>_{\xi}=-\frac{\gamma}{m}\int_0^t \left ( v_0e^{-\frac{\gamma}{m}t_1}+\frac{1}{m}\int_0^{t_1}dse^{-\frac{\gamma}{m}(t_1-s)}\xi(s) \right )^2 \xi(t_1) dt_1+\frac{1}{m}\int_0^t \left ( v_0e^{-\frac{\gamma}{m}t_1}+\frac{1}{m}\int_0^{t_1}dse^{-\frac{\gamma}{m}(t_1-s)}\xi(s) \right )\xi^2(t_1)dt_1$Anyone?

BTW, I'm using the book a modern course in statistical physics, by Linda Reichl.

 

[Telemachus]

$\xi(t)$ is clearly a Gaussian, but I don't know what "shape" it has. Anyway, I think the integrals can be computed just using the given facts: $\displaystyle \left <\xi(t) \right>_{\xi}=0$ and $\displaystyle \left < \xi(t_1)\xi(t_2) \right>_{\xi}=g\delta (t_2-t_1)$, and that's the point that concerns me. How to use those facts to compute the integrals.

 

[Telemachus]

I can get the result just by taking average on both sides, and then getting the average inside the integral. Thats what the book does, but what's the justification for doing that?

This is what I mean:

$$\displaystyle v(t)=v_0e^{-\frac{\gamma}{m}t}+\frac{1}{m}\int_0^{t}dse^{-\frac{\gamma}{m}(t-s)}\xi(s)$$

$$\displaystyle \left < v(t) \right >=\left < v_0e^{-\frac{\gamma}{m}t}+\frac{1}{m}\int_0^{t}dse^{-\frac{\gamma}{m}(t-s)}\xi(s) \right >=\left < v_0e^{-\frac{\gamma}{m}t} \right > + \left < \frac{1}{m}\int_0^{t}dse^{-\frac{\gamma}{m}(t-s)}\xi(s) \right >$$

If then I take that: $$\displaystyle\left < v_0e^{-\frac{\gamma}{m}t} \right > = v_0e^{-\frac{\gamma}{m}t}$$

And: $$\displaystyle \left < \frac{1}{m}\int_0^{t}dse^{-\frac{\gamma}{m}(t-s)}\xi(s) \right >=\frac{1}{m}\int_0^{t}dse^{-\frac{\gamma}{m}(t-s)}\left < \xi(s) \right >=0$$

I get the desired result from the given first moment: $$\displaystyle \left < v(t) \right >= v_0e^{-\frac{\gamma}{m}t}$$

But I would like to see a justification on those steps, i.e. how the first average gives the result it gives, and how the average can be taken inside the integral from the formal definition given for the average. I think it can be demonstrated from the definition of the average (in the same way that trivially can be demonstrated that the average of a sum is the sum of the averages).

 

[paranormal]

Hi there,

Thank you for your question. I understand your confusion regarding the derivation of the correlation function for Brownian motion. Let me try to clarify some of the concepts and steps involved.

First, let's start with the Langevin equation. This equation describes the motion of a Brownian particle in a fluid, taking into account the effects of random collisions with the molecules of the fluid. The equation you have provided (equation 1) is a solution to this equation, where the first term on the right-hand side represents the initial velocity of the particle and the second term represents the contribution from the random collisions.

Next, let's consider the noise term, $\xi(t)$. As you correctly mentioned, this noise is assumed to be Gaussian, meaning that its probability distribution follows a Gaussian distribution. This is a common assumption in many physical systems, and it simplifies the calculations. The delta-correlated property means that the noise at different times is uncorrelated, except for when the times are the same (i.e. $\left < \xi(t_1)\xi(t_2) \right>_{\xi}=g\delta (t_2-t_1)$).

Now, to derive the correlation function, we need to consider the joint probability distribution of the velocity at two different times, $P_2(v_1,t_1;v_2,t_2)$. This distribution can be written as a product of two single-time distributions, as you have correctly mentioned in your equation. However, there is one important step missing in your derivation. We need to use the fact that the noise term is uncorrelated with the initial velocity of the particle (i.e. $\left < v_0\xi(t) \right>_{\xi}=0$). This allows us to simplify the expression for the correlation function, as shown in equation 2.

I hope this helps to clarify the derivation for you. Please let me know if you have any further questions.