# Calculation of average and correlation functions given first and second moment

• Feb 25th 2014, 05:24 PM
Ulysses
Calculation of average and correlation functions given first and second moment
Hi there. I'm stuck with this thing. I was studying Brownian motion from the book a first course on statistical physics, by Linda E. Reichl. In page 251 from her book, she treats Brownian motion from the Langevin equations.

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

The Langevin equations are:

$\displaystyle \frac{dv(t)}{dt}=-\frac{\gamma}{m}v(t)+\frac{1}{m}\xi(t)$
and: $\displaystyle \frac{dx(t)}{dt}=v(t)$

I have then 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 noise 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).

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

$\xi(t)$ is clearly a Gaussian, but I don't know what "shape" it has (I thought of using the moments to get that shape, but the deviation give something with a dirac delta in zero, which makes no sense to me). 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 thats the point that concerns me. How to use those facts to compute the integrals. I don't know if I'm giving the proper interpretation to the notation neither.

Any help will be appreciated.

PD: For a discussion on Brownian motion and Langevin equations you can check this: http://web.phys.ntnu.no/~ingves/Teac...loads/kap6.pdf

I'm interested on the formal derivation, step by step of the correlation functions, and the average velocity, because some things are not that clear to me.
• Feb 26th 2014, 10:13 AM
Ulysses
Re: Calculation of average and correlation functions given first and second moment
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 whats 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.