# Fluctuation dissipation on a ring?

The integral fluctuation theorem is given by: $$\left< e^{-R}\right>=1\tag{0}$$ where: $$R\equiv \ln \left( \frac{p_0(\vec n_0) p[\vec n(\tau),\vec c(\tau)]}{p_f(\vec n) \cdot p[\tilde n(\tau),\tilde c(\tau)]}\right)\tag{1}$$ where my notation follows in part (arxiv:0605080) with $p[\vec n(\tau),\vec c(\tau)]$ being the trajectory weight for a given trajectory $\vec n(\tau)$ with initial state $\vec n_0$. And $p[\tilde n(\tau),\tilde c(\tau)]$ being that for the trajectory $\tilde n(\tau)\equiv n(t-\tau)$ under the time reversed protocol $\tilde c(\tau) \equiv c(t-\tau)$. Lastly $p_0(\vec n_0)$ is the initial distribution and $p_f(\vec n)$ the finial distribution.

From this on a ring it is possible to derive the equation: $$p(-\Delta s_{tot})=e^{-\Delta s_{tot}}p(\Delta S_{tot})\tag{2}$$ where $p(\Delta S_{tot})$ is the pdf for total entropy production. Following back references it appears that this relation originated in (Crooks, 1999).

My question: I understand how in general such relations are derived but I can't see where the $p(\Delta s_{tot})$'s in equation (2) come from in relation to equation (1). I.e. how do we relate $p(\Delta s_{tot})$ to probabilities in equation (1)?

The Crooks relation cannot be derive (as far as I can tell) directly from the "integral fluctuation theorem" (eq. (0) in the OP) and a generalized theorem is required - which the rest of this answer will look into.

# Notation

I will follow the notation in the reference {1} below. This is summarized here:

• $$\dagger$$ denotes quantities relating to the time-reversed process.
• $$S_\alpha[x(\tau)]$$ is a functional of the original dynamics and: $$S_\alpha^\dagger[x(\tau)^\dagger, \lambda^\dagger, F^\dagger]=\varepsilon_\alpha S_\alpha[x(\tau), \lambda, F]$$
• $$g$$ is a function depending on an arbitrary number of functionals $$S_\alpha$$

# Generalized Theorem

The generalized theorem is then given by: $$\langle g(\{ \varepsilon_\alpha S^\dagger_\alpha[x^\dagger(\tau)] \})\rangle^\dagger=\langle g(\{ \varepsilon_\alpha S_\alpha[x(\tau)] \}) \exp(-R[x(\tau)])\rangle \tag{A1}$$ This is proved on page 7 of {1} and as such I will not reproduce the proof here.

# Deriving Crooks theorem

Both reference {1} and {2} then goes into explaining how we derive crooks theorem. Let us first consider $$R[x(\tau)]$$. We start both the original and reversed dynamics in the stationary state. This means that: $$R=\frac{\Delta S_m}{k_B}+\ln \left( \frac{p_i(\vec x_0)}{ p_f(\vec x)}\right)$$ $$=\frac{\Delta S_m}{T}+\frac{ \Delta V-\Delta \mathcal{F}}{T}$$ where $$\Delta V$$ is the change in potential and $$\Delta \mathcal{F}$$ the change in free energy. Using $$W[x(\tau)]=\Delta S_m+ \Delta V$$ we get: $$R=(W[x(\tau)]-\Delta \mathcal{F})/T$$ Further more we chose that $$S_\alpha[x(\tau)]=W[x(\tau)]$$ (which corresponds to $$\varepsilon_\alpha=-1$$) and take: $$g(W[x(\tau)])=\exp(-k W[x(\tau)])$$ Thus (A1) becomes: $$\langle \exp(k W[x(\tau)]) \rangle^\dagger =\exp(\Delta \mathcal{F}/T)\langle \exp(-k W[x(\tau)]-W[x(\tau)]/T) \rangle \tag{A2}$$ Note that: $$\langle \exp(-\alpha W[x(\tau)]) \rangle=\int P(W)e^{-\alpha W} dW$$ so taking the inverse Laplace transform of (2) w.r.t $$k$$ gives us: $$P^\dagger(-W)=P(W) \exp(\Delta \mathcal{F}/T-W[x(\tau)]/T)$$ which is the Crooks theorem as given in terms of work rather then entropy as given in the OP.

# References

1. Stochastic Thermodynamics by Luka Pusovnik. Avalible from: http://mafija.fmf.uni-lj.si/seminar/files/2016_2017/luka-pusovnik-stochastic-thermodynamics.pdf

2. Täuber, U.C., 2014. Critical dynamics: a field theory approach to equilibrium and non-equilibrium scaling behavior. Cambridge University Press. (pg 338)