Response and Dissipation: Part 1 of the Fluctuation-Dissipation Theorem

I’ve referred to the fluctuation-dissipation theorem many times on this blog (see here and here for instance), but I feel like it has been somewhat of an injustice that I have yet to commit a post to this topic. A specialized form of the theorem was first formulated by Einstein in a paper about Brownian motion in 1905. It was then extended to electrical circuits by Nyquist and then generalized by several authors including Callen and Welten (pdf!) and R. Kubo (pdf!). The Callen and Welton paper is a particularly superlative paper not just for its content but also for its lucid scientific writing. The fluctuation-dissipation theorem relates the fluctuations of a system (an equilibrium property) to the energy dissipated by a perturbing external source (a manifestly non-equilibrium property).

In this post, which is the first part of two, I’ll deal mostly with the non-equilibrium part. In particular, I’ll show that the response function of a system is related to the energy dissipation using the harmonic oscillator as an example. I hope that this post will provide a justification as to why it is the imaginary part of a response function that quantifies energy dissipated. I will also avoid the use of Green’s functions in these posts, which for some reason often tend to get thrown in when teaching linear response theory, but are absolutely unnecessary to understand the basic concepts.

Consider first a damped driven harmonic oscillator with the following equation (for consistency, I’ll use the conventions from my previous post about the phase change after a resonance):

$\underbrace{\ddot{x}}_{inertial} + \overbrace{b\dot{x}}^{damping} + \underbrace{\omega_0^2 x}_{restoring} = \overbrace{F(t)}^{driving}$

One way to solve this equation is to assume that the displacement, $x(t)$, responds linearly to the applied force, $F(t)$ in the following way:

$x(t) = \int_{-\infty}^{\infty} \chi(t-t')F(t') dt'$

Just in case this equation doesn’t make sense to you, you may want to reference this post about linear response.  In the Fourier domain, this equation can be written as:

$\hat{x}{}(\omega) = \hat{\chi}(\omega) \hat{F}(\omega)$

and one can solve this equation (as done in a previous post) to give:

$\hat{\chi}(\omega) = (-\omega^2 + i\omega b + \omega_0^2 )^{-1}$

It is useful to think about the response function, $\chi$, as how the harmonic oscillator responds to an external source. This can best be seen by writing the following suggestive relation:

$\chi(t-t') = \delta x(t)/\delta F(t')$

Response functions tend to measure how systems evolve after being perturbed by a point-source (i.e. a delta-function source) and therefore quantify how a system relaxes back to equilibrium after being thrown slightly off balance.

Now, look at what happens when we examine the energy dissipated by the damped harmonic oscillator. In this system the energy dissipated can be expressed as the time integral of the force multiplied by the velocity and we can write this in the Fourier domain as so:

$\Delta E \sim \int \dot{x}F(t) dt = \int d\omega d\omega'dt (-i\omega) \hat{\chi}(\omega) \hat{F}(\omega)\hat{F}(\omega') e^{i(\omega+\omega')t}$

One can write this more simply as:

$\Delta E \sim \int d\omega (-i\omega) \hat{\chi}(\omega) |\hat{F}(\omega)|^2$

Noticing that the energy dissipated has to be a real function, and that $|\hat{F}(\omega)|^2$ is also a real function, it turns out that only the imaginary part of the response function can contribute to the dissipated energy so that we can write:

$\Delta E \sim \int d \omega \omega\hat{\chi}''(\omega)|\hat{F}(\omega)|^2$

Although I try to avoid heavy mathematics on this blog, I hope that this derivation was not too difficult to follow. It turns out that only the imaginary part of the response function is related to energy dissipation.

Intuitively, one can see that the imaginary part of the response has to be related to dissipation, because it is the part of the response function that possesses a $\pi/2$ phase lag. The real part, on the other hand, is in phase with the driving force and does not possess a phase lag (i.e. $\chi = \chi' +i \chi'' = \chi' +e^{i\pi/2}\chi''$). One can see from the plot from below that damping (i.e. dissipation) is quantified by a $\pi/2$ phase lag.

Damping is usually associated with a 90 degree phase lag

Next up, I will show how the imaginary part of the response function is related to equilibrium fluctuations!

2 responses to “Response and Dissipation: Part 1 of the Fluctuation-Dissipation Theorem”

1. Rahul

Hi, wonderful explanation, where is your part 2?

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• Anshul Kogar

Hi Rahul, I haven’t written it as yet. It’s on my to-do list though, so thanks for the reminder. Originally, I thought it wouldn’t be so difficult to sketch part 2 on the blog, but it turned out to be quite a formidable task. I moved onto other things and eventually forgot… But I much appreciate you asking – it gives me some motivation to think about a good way to explain it.

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