The equation for Brownian Motion is

\[\frac{d}{dt} v + \gamma v = R(t),\]

where \(R(t)\) is random force which requires

\[\begin{split}\avg{R(t)} &= 0 \\
\avg{R(t)R(t')} = C \delta(t-t') .\end{split}\]

This equation is the linear Langevin equation.

The formal solution, which seems to be useless, is

\[v(t) = v(0)e^{-\gamma t} + \int_0^t dt' e^{-\gamma (t-t')} R(t') .\]

Knowing that \(R(t)\) is random, or stochastic, we imediately realize that \(v(t)\) is stochastic. The thing to do is to find out the statistics.

This quantity squared is,

\[v(t)^2 = v(0)^2 e^{-2\gamma t} + \int_0^t dt'\int_0^t dt'' e^{-\gamma (t- t')} e^{-\gamma (t - t'')} R(t')R(t'') + \mathrm{Cross Terms}\]

Average of the solution is

\[\begin{split}\avg{v} &= \avg{v(0)e^{-\gamma t}} + {\color{magenta}\avg{\int_0^t dt' e^{-\gamma (t-t')} R(t') } } \\
\avg{v^2} &= \avg{v(0)^2 e^{-2\gamma t}} + \avg{\int_0^t dt'\int_0^t dt'' e^{-\gamma (t- t')} e^{-\gamma (t - t'')} R(t')R(t'')} + {\color{magenta}\avg{ \mathrm{Cross Terms}} }\end{split}\]

where these magenta colored terms are zero.

Hint

Note that the average here is ensemble average. Define probability density \(P(v,t)\) for velocity at time t. Recall that in velocity space, Liouville equation is

\[\frac{d}{dt}P(v,t) + \frac{d}{dt}j = 0.\]

It’s very important to realize that this current density is the current density in velocity space. Applying \(d_t v = -\gamma vt R(t)\),

\[\frac{d}{dt}P(v,t) + \frac{d}{dv} \left( (-\gamma v+ R(t))P(v,t) \right) = 0 .\]

Warning

Now I am curious what ENSEMBLE average is. It’s not this probability density because there is time dependence.

Warning

What about Fokker-Plank equation?

By using the definition of random motion, the ensemble average shows us very simple results.

\[\begin{split}\avg{v(t)} &= \avg{v(0)} e^{-\gamma t} \\
\avg{v(t)^2} & = \avg{v(0)^2}e^{-2\gamma t} + C \frac{1- e^{-2\gamma t}}{2\gamma}\end{split}\]

In the limit that time goes to infinity, we have

\[\begin{split}\avg{v(t\to\infty)} & = 0 \\
\avg{v(t\to \infty)^2} & = \frac{C}{2\gamma} .\end{split}\]

The legacy of Einstein is to measure Boltzmann constant using Brownian motion. The regorous derivation can be found in Physical Mathematics . The results, however, is that \(\avg{v(t)^2}\) is related to thermal energy of the particle and also related to the temperature according to equipartition theorem,

\[\frac{1}{2}m \avg{v(t\to \infty)^2} = \frac{1}{2}k_B T .\]

Equation for this process is

\[\frac{d}{dt} x = R(t).\]

Derived from this,

\[\avg{x^2} \propto t.\]

Change the velocity in Brownian motion to displacement,

\[\frac{d}{dt} x + \gamma x = R(t).\]

Suppose we have \(\avg{v(0)}=0\), \(\avg{x^2}\) becomes constant.

A damped, driven harmonic oscillator is

\[m \frac{d^2}{dt^2} x + \alpha \frac{d}{dt} x + \beta x = R(t) .\]

Divided by \(\alpha\),

\[\frac{d^2}{dt^2} x + \frac{\alpha}{m} \frac{d}{dt} x + \frac{\beta}{m} x = \frac{1}{m}R(t) .\]

We understand immediately that this reduces to a Ornstein-Uhlenbeck equation if mass is very small.

Warning

Does it ring a bell of the WKB approximation?

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