Physics Asked by GGphys on June 9, 2021
Langevin equation of a free Brownian particle has the solution of the form:
$$v(t)=v(0)e^{-tgamma}+dfrac{1}{m}int_0^t e^{-gamma(t-tau)}eta(tau)dtau$$
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;
$$langle v_i(t)v_j(t’)rangle=v(0)_iv(0)_je^{-gamma(t-t’)}+int_0^t dtau int_0^t dtau’ langle eta_i(tau)eta_j(tau’)rangle e^{-gamma(t+t’-tau-tau’)} $$
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 is over multiple Brownian particles, so experimentally, you would track these particles and compute the property you want at each time step for every particle and do the mean
Answered by and85 on June 9, 2021
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: $$ int_0^t dtauint_0^{t'} dtau'langle eta_i(tau)eta_j(tau')rangle e^{-gamma(t+t'-tau-tau')} = int_0^t dtauint_0^{t'} dtau'delta_{i,j}delta(tau-tau') e^{-gamma(t+t'-tau-tau')}= delta_{i,j}theta(t'-t)int_0^t dtau e^{-gamma(t+t'-2tau)} + delta_{i,j}theta(t-t')int_0^{t'} dtau' e^{-gamma(t+t'-2tau')}, $$ 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.
Answered by Roger Vadim on June 9, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP