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 &lt; v(t) \right &gt;=\left &lt; v_0e^{-\frac{\gamma}{m}t}+\frac{1}{m}\int_0^{t}dse^{-\frac{\gamma}{m}(t-s)}\xi(s) \right &gt;=\left &lt; v_0e^{-\frac{\gamma}{m}t} \right &gt; + \left &lt; \frac{1}{m}\int_0^{t}dse^{-\frac{\gamma}{m}(t-s)}\xi(s) \right &gt;

If then I take that: \displaystyle\left &lt; v_0e^{-\frac{\gamma}{m}t} \right &gt; = v_0e^{-\frac{\gamma}{m}t}

And: \displaystyle \left &lt; \frac{1}{m}\int_0^{t}dse^{-\frac{\gamma}{m}(t-s)}\xi(s) \right &gt;=\frac{1}{m}\int_0^{t}dse^{-\frac{\gamma}{m}(t-s)}\left &lt; \xi(s) \right &gt;=0

I get the desired result from the given first moment: \displaystyle \left &lt; v(t) \right &gt;= 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).