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:

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

and: $\displaystyle \displaystyle \frac{dx(t)}{dt}=v(t)$

I have then that for a brownian particle:

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

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

The assumption that the noise is Gaussian means that the noise is delta-correlated: $\displaystyle \displaystyle \left < \xi(t_1)\xi(t_2) \right>_{\xi}=g\delta (t_2-t_1)$

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

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

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

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 $\displaystyle v(0)=v_0$): $\displaystyle \displaystyle \left < v(t) \right>_{\xi}=v_0e^{-\frac{\gamma}{m}t}$

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

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

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

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

$\displaystyle \xi(t)$ 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: $\displaystyle \displaystyle \left <\xi(t) \right>_{\xi}=0$ and $\displaystyle \displaystyle \left < \xi(t_1)\xi(t_2) \right>_{\xi}=g\delta (t_2-t_1)$, 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.