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Random Force and Its Applications

Stochastic Equation

We wrote kinetic equation for the average x as

\frac{d\left\langle x\right\rangle}{dt}=-\lambda\left\langle x\right\rangle\end{displaymath}

What happens if we want to measure x for one system in the ensemble? Then we write  
 \frac{dx}{dt}=-\lambda x + f(t)\end{displaymath} (6)
with random variable x. The function f(t) is random force.

Properties of Random Force

We can speak only about average properties of f(t). We will calculate $\left\langle f(t)\right\rangle$ and $\left\langle f(0)f(t)\right\rangle$. We will average over equilibrium ensemble.

Averaging (6), we obtain:

\frac{d\left\langle x\right\rangle}{dt}=-\lambda\left\langle x\right\rangle+\left\langle f(t)\right\rangle\end{displaymath}

 \left\langle f(t)\right\rangle=0\end{displaymath} (7)
Equation (6) can be solved. The solution is  
 x(t)=e^{-\lambda t} \int_{-\infty}^t f(\tau)e^{\lambda\tau}\,d\tau\end{displaymath} (8)
Correlation function:

\left\langle x(0)x(t)\right\rangle = e^{-\lambda t} \int_{-\...
 ...eft\langle f(\tau)f(\tau')\right\rangle e^{\lambda(\tau+\tau')}\end{displaymath}

But we know $\left\langle x(0)x(t)\right\rangle$! Result:

\int_{-\infty}^t d\tau
 \int_{-\infty}^0 d\tau'
 ...\tau')\right\rangle e^{\lambda(\tau+\tau')} = \frac{kT}{\gamma}\end{displaymath}

This could be rewritten as

\int_{-\infty}^t F(\tau)\,d\tau = \mathit{const}, \quad t\gt \end{displaymath}

This is possible only if $F(\tau)=0$ for $\tau\gt$But

F(\tau) = \int_{-\infty}^0 d\tau'
 \left\langle f(\tau)f(\tau')\right\rangle e^{\lambda(\tau+\tau')} \end{displaymath}

and $\left\langle f(\tau)f(\tau')\right\rangle$ depends only on $\tau-\tau'$. We obtain:

\left\langle f(\tau)f(\tau')\right\rangle=0,\quad \tau\gt\tau'\end{displaymath}

From symmetry

\left\langle f(\tau)f(\tau')\right\rangle=0,\quad \tau<\tau'\end{displaymath}

Since $\left\langle f(\tau)f(\tau')\right\rangle$ is not zero, it is a $\delta$-function:

\left\langle f(\tau)f(\tau')\right\rangle = C\delta(\tau-\tau')\end{displaymath}

Then we have

 &\int_{-\infty}^t d\tau
 \int_{-\infty}^0 d\t...
 ...ty}^0 e^{2\lambda\tau'}\,d\tau'=\frac{C}{2\lambda}

and since this is $kT/\gamma$,

C=\frac{2\lambda kT}{\gamma}\end{displaymath}

We obtained:  
 \left\langle f(t)f(t')\right\rangle = \frac{2\lambda kT}{\gamma}\delta(t-t')\end{displaymath} (9)
or, after Fourier transform  
 \left\langle f(t)f(t')\right\rangle_{\omega}\propto\mathit{const}\end{displaymath} (10)

White Noise and Colored Noise

Sometimes the function f(t) is called random noise. Equation (9) shows that random noise is $\delta$-correlated. In other words, all harmonics are present in its Fourier transform with equal weights (equation (10)). Such noise is called white noise (remember rainbow?).

Colored noise: not all harmonics are equal[*]:

\left\langle f(t)f(t')\right\rangle_{\omega}\ne\mathit{const}\end{displaymath}

and $\left\langle f(t)f(t')\right\rangle\not\propto\delta(t-t')$.

How can it be so? What's wrong in our derivation?

We started from the equation  
 \left\langle x(0)x(t)\right\rangle\propto e^{-\lambda t}\end{displaymath} (11)
This is a consequence of the fact that x is the slowest mode in the system and we averaged out everything else! If this is not true, equation (11) does not work, and noise is no longer white!

Another interpretation: the $\delta$-function in equation (9) is the $\delta$-function only approximately. In fact it is a sharp peak with the width about the characteristic time of the molecular processes.

next up previous
Up: Time Correlation Functions. Random Previous: Spectral Representation

© 1997 Boris Veytsman and Michael Kotelyanskii
Tue Oct 28 22:16:24 EST 1997