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- 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.

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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