# Langevin equation autocorrelation function

Langevin equation of a free Brownian particle has the solution of the form: $$v(t)=v(0)e^{-t\gamma}+\dfrac{1}{m}\int_0^t e^{-\gamma(t-\tau)}\eta(\tau)d\tau$$

where $$\langle \eta_i(t) \eta_j(t')\rangle=\delta_{ij}\delta(t-t')$$ and $$\langle \eta(t)\rangle=0$$.

And when we want to calculate the correlation function;

$$\langle v_i(t)v_j(t')\rangle=v(0)_iv(0)_je^{-\gamma(t-t')}+\int_0^t d\tau \int_0^t d\tau' \langle \eta_i(\tau)\eta_j(\tau')\rangle e^{-\gamma(t+t'-\tau-\tau')}$$

However, I don't understand the calculation of the correlation function. What are we averaging over? If it was $$t$$ we wouldn't have the first term as it is.

The average here is over different realizations of random process $$\eta(\tau)$$. If we measured the values of noise only at discrete time instants $$\tau_1, \tau_2, .., \tau_n$$, we could write a joint probability of these as $$w(x_1, x_2, ..., x_n),$$ which in this case should be a multivariate Gaussian distribution with diagonal covariance matrix. As we take time interval to be smaller and smaller and pass to a continuous limit, we have to average over a functional $$w[x(t)]$$ and use functional calculus. A good book covering this issues is the first volume by KLyatskin. (This is unfortunately a translation from an old Russian text. If somebody can recommend an equivalent text taht is mroe readily available, I would be glad toa dd it here.)
In the equation at hand the averaging is trivial, since the correlation function is known: $$\int_0^t d\tau\int_0^{t'} d\tau'\langle \eta_i(\tau)\eta_j(\tau')\rangle e^{-\gamma(t+t'-\tau-\tau')} = \int_0^t d\tau\int_0^{t'} d\tau'\delta_{i,j}\delta(\tau-\tau') e^{-\gamma(t+t'-\tau-\tau')}=\\ \delta_{i,j}\theta(t'-t)\int_0^t d\tau e^{-\gamma(t+t'-2\tau)} + \delta_{i,j}\theta(t-t')\int_0^{t'} d\tau' e^{-\gamma(t+t'-2\tau')},$$ where $$\theta(t)$$ is the Heaviside step function, which accounts for the overlap of integration ranges necessary for delta-function taking a non-zero value.