1
$\begingroup$

In the book of Zwanzig, Nonequilibrium statistical physics, at page 6, after explaining Langevin equation Brownian motion, to show that $<v^2> = 3/2 k_B T/m$ consistent with the Langevin equation, he states

$$v(t)=e^{-G / m} v(0)+\int_{0}^{t} d t^{\prime} e^{-\zeta\left(t-t^{\prime}\right) / m} \delta F\left(t^{\prime}\right) / m$$ [...]

On averaging over noise, these cross terms vanish. The final term is second order in the noise: $$ \int_{0}^{t} d t^{\prime} e^{-\zeta(t-r) / m} \delta F\left(t^{\prime}\right) \int_{0}^{t} d t^{\prime \prime} e^{-\zeta(t-r) / m} \delta F\left(t^{\prime \prime}\right) / m^{2} $$ Now the product of two noise factors is averaged, according to eq. (1.5), and leads to $$\int_{0}^{t} d t^{\prime} e^{-\zeta(t-r) / m} \int_{0}^{t} d t^{\prime \prime} e^{-\zeta\left(t-r^{*}\right) / m} 2 B \delta\left(t^{\prime}-t^{\prime \prime}\right) / m^{2}$$

However, I cannot understand how does it go from stochastic integral to usual integral. Is interchanging the orders of averaging operation with integral operation allowed?

Edit:

For example, a similar thing is done on page 8,

$$ x(t)=\int_{0}^{t} d s v(s) $$ where $v(s)$ is the velocity of the particle at time $s .$ The ensemble average of the mean squared displacement is $$ \left\langle x^{2}\right\rangle=\left\langle\int_{0}^{t} d s_{1} v\left(s_{1}\right) \int_{0}^{t} d s_{2} v\left(s_{2}\right)\right\rangle=\int_{0}^{t} d s_{1} \int_{0}^{t} d s_{2}\left\langle v\left(s_{1}\right) v\left(s_{2}\right)\right\rangle $$

$\endgroup$
2
$\begingroup$

Validity of this interchanging is regulated by Fubini–Tonelli theorem. If your spaces and functions are "normal", then this is a valid transformation. "Normal" here means $\sigma$-finiteness of spaces measure's and finiteness of integrals. Usually your integrals are finite due to their physical nature.

$\endgroup$
1
$\begingroup$

In this context it is a standard procedure. You can justify it by discretizing the time integrals and writing out explicitly the averging as averaging over the multi-dimensional Gaussian distribution. All the integrands are smooth, differentialble, etc., so there is no problem with exchanging the order of integrations/summations.

$\endgroup$
1
  • 1
    $\begingroup$ I agree, this is a good answer: the trick is to think the time integration as a sum and then the expectation value is a "linear operation": math.stackexchange.com/q/9596/532409 $\endgroup$
    – Quillo
    Feb 20 at 19:25

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .