Consider a stationary random variable $F(t)$ representing the random force on a Brownian particle in a fluid. Suppose the autocorrelation function is given by $$\langle F(0)F(t)\rangle=Ce^{-\gamma|t|}$$ where $C$ is a constant of appropriate dimension and $\gamma$ is a real positive constant. Then the Power spectrum $S_F(\omega)$, defined as $$S_F(\omega)=\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty}dt~ e^{i\omega t}\langle F(0)F(t)\rangle$$ turns out as $$S_F(\omega)=\frac{C}{\gamma^2+\omega^2}$$ which decreases as $\omega$ increases.

On the other hand, the power spectral density of a delta correlated noise $\xi(t)$ i.e., $\langle\xi(0)\xi(t)\rangle=K\delta(t)$ turns out to be independent of frequency $\omega$.

What physical information does the power spectrum $S_F(\omega)$ provide about the nature of the random variable $F(t)$ in this case? In particular, what does $S_F(\omega)$ tell about the "frequency content" of the random variable $F(t)$? Since it is a Lorentzian (the spectral density drops as $\omega$ drops) large frequency components of the noise contributes lesser than small frequency components?

And what about constant power spectrum of the delta-correlated noise $\xi(t)$? Does it mean all frequency components contribute equally to make up $\xi$?

  • 1
    $\begingroup$ Statistical, not physical: it is just an Ornstein–Uhlenbeck process. $\endgroup$ Commented Jan 4, 2018 at 20:19
  • $\begingroup$ The power spectrum being a Lorentzian, i.e. a Cauchy (Breit-Wigner!) distribution, it means it has no frequency moments; that is, the FT autocorrelation function is highly peaked. It models the velocity of a massive Brownian particle under the influence of friction. $\endgroup$ Commented Jan 4, 2018 at 20:31

1 Answer 1


Note: I'm using $$S_F(\omega) \equiv \int_{-\infty}^\infty dt e^{i \omega t} \left \langle F(0)F(t) \right \rangle$$ because I believe it is considerably more standard than the definition given in the question post.

What physical information does the power spectrum $S_F(\omega)$ provide about the nature of the random variable $F(t)$ in this case?

The power spectral density tells you exactly what it sounds like: the amount of fluctuations at a given frequency. There are a few ways to see the meaning of "amount of fluctuations at a given frequency", the most direct and possibly intuitive is to directly relate how much the particle moves around to its power spectral density.

Mean square motion of particle

Let us ask a concrete physical question: given that the particle starts at $x=0$, what is it the probability distribution of its position after a time $t$? Since the motion of the particle is presumably due to a sum of a large number of small fluctuations, the central limit theorem guarantees that the probability distribution is Gaussian. Therefore, it's enough for us to compute the variance. We can always write $x(t)$ as an integral of its velocity, i.e. $$x(t) = \int_0^t \dot{x}(t') \, dt' \,.$$ Therefore, \begin{align} \left \langle x(t)^2 \right \rangle &= \int_0^t dt' \int_0^t dt'' \left \langle \dot{x}(t') \dot{x}(t'') \right \rangle \\ &= \int_0^t dt' \int_0^t dt'' \int_{-\infty}^\infty \frac{d\omega}{2\pi} S_{\dot{x}}(\omega) e^{-i \omega (t' - t'')} \\ &= \int_{-\infty}^\infty \frac{d\omega}{2\pi} S_{\dot{x}}(\omega) \int_0^t dt' \int_0^t dt'' e^{-i \omega (t' - t'')} \\ &= 2 t^2 \int_{-\infty}^\infty \frac{d\omega}{2\pi} S_{\dot{x}} (\omega) \underbrace{\left( \frac{1 - \cos(\omega t)}{(\omega t)^2} \right)}_{W(\omega,t)} \, . \tag{$\star$} \end{align} This result can be interpreted as a weighting function $W(\omega, t)$ integrated against the spectral density.

enter image description here Figure 1: $W(\omega, t)$ for the free particle case.

The exact form of this weighting function is particular to the case where we let the particle drift freely. If we were to filter the particle's motion (e.g. if this were an electrical signal instead of a physical particle) or change the boundary conditions in some way, we'd have a different form for $W$.

In the case of strong damping, which I believe is the case for Brownian motion, we have a large drag force $F_\text{drag} = - \gamma \dot{x}$ and Newton's law goes like this \begin{align} F_\text{total} &= m\ddot{x} \\ F_\text{external} - \gamma \dot{x} &= m\ddot{x} \leftarrow \text{(small compared to other side)} \\ F_\text{external}/\gamma &= \dot{x} \, . \end{align} Therefore, force and velocity are proportional to each other.

The $W$ we got in this case is large for small $\omega$. This means that the diffusion of the free particle is dominated by force fluctuations at low frequency. This isn't surprising: if the spectral density is large at low frequency then the force has a large roughly constant component, which pushes the particle in one direction. Spectral density at higher frequencies corresponds to force fluctuations that oscillate in time, and contribute less to the particle's total movement in a single direction.

Power spectral density as "power per bandwidth"

This section is a work in progress. I've saved it in an incomplete state so that I can work on it from another computer.

At the beginning of this post, we said that the power spectral density tells you the amount of fluctuation at a given frequency. In fact, the power spectral density tells us the power per frequency bandwidth. Let's see what that means quantitatively.

First, rewrite $(\star)$ with the variable change $z \equiv \omega t$: $$\langle x(t)^2 \rangle =2 t \int_{-\infty}^\infty \frac{dz}{2 \pi} S_{\dot{x}}(z/t) \underbrace{\left( \frac{1 - \cos(z)}{z^2} \right)}_{W(z)} \, .$$ If $S_\dot{x}(z/t)$ is a constant $S_\dot{x}$ over the range where $W(z)$ has substantial weight, then we can factor $S$ out of the integral, do the integral, and arrive at \begin{align} \langle x(t)^2 \rangle &= t S_\dot{x} \\ \text{or} \quad \langle x(t)^2 \rangle / t &= S_\dot{x} \, . \end{align} Thinking of $\langle x(t)^2 \rangle$ as "energy", the left hand side is energy per time or power. Therefore, $S_\dot{x}$ is the

  • $\begingroup$ It would be helpful for me if you could address the specific example I mentioned. What does the power spectrum of the random force tell us? In terms of the force, it may be better to understand first. @DanielSank $\endgroup$
    – SRS
    Commented Jan 4, 2018 at 20:45
  • $\begingroup$ @SRS I did the calculation in terms of $\dot{x}$ because I think that in Brownian motion the particle's velocity is proportional to the applied force via a drag coefficient. That being the case, the calculation I showed and the physical interpretation carry over directly to what you asked. Does that answer it? $\endgroup$
    – DanielSank
    Commented Jan 4, 2018 at 23:33
  • $\begingroup$ @DanielSank - When to use the factor of $2 \pi$ is kind of a matter of choice, is it not? $\endgroup$ Commented Jan 5, 2018 at 14:02
  • $\begingroup$ @SRS I'm still hoping to answer your question. Let me know what else I can add. $\endgroup$
    – DanielSank
    Commented Jan 5, 2018 at 19:30
  • 2
    $\begingroup$ @SRS yes the power spectral density $S(\omega)$ is just the power per bandwidth that would remain in the signal if it were filtered to a band centered at $\omega$. In other words, if the filter width is $\Delta \omega$, then the power is $P=S(\omega) \Delta \omega$. Is that what you were looking for? $\endgroup$
    – DanielSank
    Commented Mar 26, 2018 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.