# Proof of Classical Fluctuation-Dissipation Theorem

• A
• Twigg
In summary, the conversation discusses a proof from Wikipedia involving the probability density of finding a system at a given state at a given time. The proof is based on a perturbation expression and the autocorrelation function. The speaker is having trouble evaluating an integral and doubts the validity of the form of the Fluctuation-Dissipation Theorem stated in Wikipedia. They are open to alternative proofs that may be easier to remember.
Twigg
Gold Member
TL;DR Summary
Found a proof of the classical version of the fluctuation-dissipation theorem, couldn't figure out one step. Involves classical transition probabilities for a stochastic, dissipative system. How does one evaluate the integral ##\int dx \text{ } x \int dx' P(x,t|x',0) W_0 (x') x'(0)##.
Sorry if there's latex errors. My internet connection is so bad I can't preview.

Here's the wikipedia proof I'm referring to. I'm fine with the steps up to $$W(x,0) = W_0 (x) [1 + \beta f_0 (x(0) - \langle x \rangle_0) ]$$ where ##W(x,t)## is the probability density of finding the system at state ##x## at time ##t##, ##f_0## is the magnitude of a perturbation ##f_0 << kT## on the Hamiltonian H of the form ##H = H_0 - x f_0 \Theta (-t) ## where ##\Theta (t)## is the heaviside step function, ##\langle \cdot \rangle_0## is expectation value in the unperturbed equilibrium distribution (called ##W_0 (x)##), and where ##\beta = \frac{1}{kT}##.

I was able to reproduce the approximate, first order perturbation expression for ##W(x,0)## just fine using taylor/binomial series. It's the next step, using this distribution to prove $$\langle x(t) \rangle = \langle x \rangle_0 + \beta f_0 A(t)$$ where ##A(t) = \langle [x(t) - \langle x \rangle_0][x(0) - \langle x \rangle_0] \rangle_0## is the autocorrelation function.

I'm confident that the general procedure is to write $$\langle x(t) \rangle = \int dx \text{ } x \int dx' P(x,t|x',0) W(x',0)$$ where ##P(x,t|x',0)## is the conditional probability for state x at time t given the system is in state x' at time 0. I can see why ##\int dx' P(x,t|x',0) W_0 (x') = W_0(x)## since ##W_0## is an equilibrium distribution. I also have a hunch that ##\lim_{t \rightarrow \infty} \int dx' P(x,t|x',0) W(x',0) = W_0(x)## for any initial distribution since it's a dissipative system and will reach equilibrium eventually.

However, given all this, what I get is $$\langle x(t) \rangle = \int dx \text{ } x \int dx' P(x,t|x',0) W_0 (x') [1 + \beta f_0 (x'(0) - \langle x \rangle_0) ] = \langle x \rangle_0 + \beta f_0 \left[ \int dx \text{ } x \int dx' P(x,t|x',0) W_0 (x') x'(0) - \int dx \text{ } x W_0(x) \langle x \rangle_0 \right] = \langle x \rangle_0 + \beta f_0 \left[ \int dx \text{ } x \int dx' P(x,t|x',0) W_0 (x') x'(0) - \langle x \rangle_0^2 \right]$$ It's the integral $$I = \int dx \text{ } x \int dx' P(x,t|x',0) W_0 (x') x'(0)$$ that gets me. I have no idea how to evaluate this, because the thing that ##P(x,t|x',0)## is multiplied by isn't a probability distribution. Can anyone point me in the right direction?

Alternatively, if you have a different concise proof, I'm all ears. I was interested in this proof because it's (relatively) short and might be easier to remember.

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Update: I'm having serious doubts with the form of the FDT stated in the wikipedia article. They claim $$S_x (\omega) = \frac{2kT}{\omega} \mathrm{Im}[\chi(\omega)]$$ However, for the vanilla Langevin equation $$m\ddot{x} = -\gamma \dot{x} + \eta$$ I am fairly confident that $$\lim_{t \rightarrow \infty} \langle \dot{x}(t+\tau) \dot{x}(t) \rangle \propto e^{-\gamma \tau}$$ In other words $$\lim_{t \rightarrow \infty} \langle \dot{x}(t+\tau) \dot{x}(t) \rangle \propto \chi(\tau)$$ and that's inconsistent with the wikipedia article's claim since the Fourier transform of ##e^{-\gamma \tau}## has non-zero real part.

## 1. What is the Classical Fluctuation-Dissipation Theorem?

The Classical Fluctuation-Dissipation Theorem is a fundamental principle in statistical mechanics that relates the fluctuations of a system to its response to external perturbations. It states that the strength of the fluctuations in a system is directly proportional to its dissipation, or the rate at which the system loses energy.

## 2. What is the significance of the Classical Fluctuation-Dissipation Theorem?

The Classical Fluctuation-Dissipation Theorem is important because it provides a link between the microscopic behavior of a system and its macroscopic properties. It allows us to understand how fluctuations in a system affect its overall behavior and how these fluctuations are related to the dissipation of energy.

## 3. How is the Classical Fluctuation-Dissipation Theorem derived?

The Classical Fluctuation-Dissipation Theorem can be derived from the fundamental principles of statistical mechanics, such as the Boltzmann distribution and the equipartition theorem. It can also be derived using the linear response theory, which relates the response of a system to an external perturbation to its equilibrium fluctuations.

## 4. Can the Classical Fluctuation-Dissipation Theorem be applied to all systems?

The Classical Fluctuation-Dissipation Theorem is applicable to a wide range of systems, including classical gases, liquids, and solids. However, it is important to note that it is only valid for systems that are in thermal equilibrium, where the energy distribution is described by the Boltzmann distribution.

## 5. How is the Classical Fluctuation-Dissipation Theorem related to other fluctuation theorems?

The Classical Fluctuation-Dissipation Theorem is a special case of more general fluctuation theorems, such as the Jarzynski equality and the Crooks fluctuation theorem. These theorems extend the concept of equilibrium fluctuations to non-equilibrium systems and provide a deeper understanding of the relationship between fluctuations and dissipation.

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