- #1

Telemachus

- 835

- 30

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.