Calculation of the Gallavotti-Cohen fluctuation theorem made by Lebowitz I have a problem understanding a calculation in this paper (another form of the theorem an be found here at equation 11). For those who want to read the paper, I have difficulties with formula 2.14 in particular with the first passage in which the Authors moves from the derivative to that expression with sums.
For those who don't want to read the paper, I give a summary in the following: they take into account a Markov Jump process moving between state $\sigma$ with rates $k\left(\sigma,\sigma'\right)$ with $r\left(\sigma\right)=\sum_{\sigma'}k\left(\sigma,\sigma'\right)$
The generator is defined as:
\begin{equation}
Lf\left(\sigma\right)=\sum_{\sigma'}k\left(\sigma,\sigma'\right)\left[f\left(\sigma'\right)-f\left(\sigma\right)\right]
\end{equation}
$\mu\left(\sigma,t\right)$ is probability distribution of $\sigma$ at time $t$ so that $\left\langle f\right\rangle_{\mu\left(t\right)}=\sum_{\sigma}\mu\left(\sigma,t\right)f\left(\sigma\right)$. We then have:
\begin{equation}
\frac{d}{dt}\left\langle f\right\rangle_{\mu\left(t\right)}=\left\langle L f\right\rangle_{\mu\left(t\right)}
\end{equation}
If the function $f$ is taken as a Kronecker delta at $\sigma$ we have that:
\begin{equation}
\frac{\partial\mu\left(\sigma,t\right)}{\partial t}=\sum_{\sigma'}k\left(\sigma',\sigma\right)\mu\left(\sigma',t\right)-r\left(\sigma\right)\mu\left(\sigma,t\right)=L^*\mu\left(\sigma,t\right)
\end{equation} 
Now they define this quantity:
\begin{equation}
g\left(\sigma,t\right)=\mathbb{E}_\sigma\left[e^{-\lambda W\left(t\right)}\right]
\end{equation}
where 
-$\mathbb{E}_{\sigma}$ is the expectation value conditioned on the system being in state $σ$ at time $t=0$ (I think this is a different average with respect to the previous one but they must be somehow related)
-$W\left(t,\left\{\sigma_s,0\leqslant s\leqslant t\right\}\right)=\int_{0}^{t}\sum_{\sigma,\sigma'}\omega_{\sigma,\sigma'}\left(s\right)ds$ is an artificial quantity they define in which: $\sigma_s$ is a trajectory of the jump process and $\omega_{\sigma,\sigma'}\left(s\right)$ is a sequence of $δ$-functions, located exactly at those times $s$ when $\sigma_s$ jumps from $\sigma$ to $\sigma'$ with weight $\omega\left(\sigma,\sigma'\right)$.
Now here comes my problem: they say that,
\begin{alignat}{2}
\frac{d}{dt}g\left(\sigma,t\right)=&\sum_{\sigma'}k\left(\sigma,\sigma'\right)e^{-\lambda \omega\left(\sigma,\sigma'\right)}g\left(\sigma',t\right)-r\left(\sigma\right)g\left(\sigma,t\right)\\
=&\sum_{\sigma'}k\left(\sigma,\sigma'\right)^{1-\lambda}k\left(\sigma,\sigma'\right)^{\lambda}g\left(\sigma',t\right)-r\left(\sigma\right)g\left(\sigma,t\right)\\
=&L_{\lambda}g\left(\sigma,t\right)\\
\end{alignat}
I didn't understand the definition of $\mathbb{E}_\sigma$ together with the first passage of the $\frac{d}{dt}g\left(\sigma,t\right)$.
I hope to have stated things in a clear and complete way! Also references are appreciated!
Thanks in advance!!
 A: As often with physicists' notations, everything works fine if one knows what is going on, but otherwise...
Here one considers a jump process $(\sigma_t)_{t\geqslant0}$, and, for every $t$, the random variable
$$
W_t=\sum_{0\lt s\lt t}\omega(\sigma_{s^-},\sigma_s)\,\mathbf 1_{\sigma_{s^-}\ne\sigma_s},
$$
where, for every positive $s$,
$$\sigma_{s^-}=\lim\limits_{r\to s,r\lt s}\sigma_r.$$
The evolution of $g(\sigma,t)$ defined in the post is as follows. Fix some small positive $\tau$ and consider the situation at time $\tau$ with respect to $E_\sigma$, that is, conditionally on $\sigma_0=\sigma$. One of the following occurs.


*

*Either no jump occured during the time interval $[0,\tau]$, then $W_{t+\tau}$ collects the contributions of the jumps from time $\tau$ to time $t+\tau$, starting at $\sigma_\tau=\sigma$. Thus $W_{t+\tau}$ is distributed as  $W_{t}$, starting from $\sigma_0=\sigma$, unconditionally. This happens with probability $1-r(\sigma)\tau+o(\tau)$.

*Or, for some $\sigma'\ne\sigma$, a jump from $\sigma$ to $\sigma'$  occurred during the time interval $[0,\tau]$, then $W_{t+\tau}$ collects the contribution $\omega(\sigma,\sigma')$ of this first jump, plus the contributions of the jumps from time $\tau$ to time $t+\tau$, starting at $\sigma_\tau=\sigma'$. The sum of these contributions is distributed as $W_{t}$, starting from $\sigma_0=\sigma'$, unconditionally.  This happens with probability $k(\sigma,\sigma')\tau+o(\tau)$. 

*Or, several jumps occured during the time interval $[0,\tau]$.   This happens with probability $o(\tau)$. 


Summing up, one sees that, for every bounded measurable function $u$, neglecting some $o(\tau)$ terms, 
$E_\sigma(u(W_{t+\tau}))$ is 
$$
(1-r(\sigma)\tau)E_\sigma(u(W_t))+
\sum_{\sigma'\ne\sigma}k(\sigma,\sigma')\tau E_{\sigma'}(u(\omega(\sigma,\sigma')+W_t)).
$$
Substracting $E_\sigma(u(W_{t}))$ from both sides, dividing everything by $\tau$ and considering the limit when $\tau\to0^+$, one gets
$$
\frac{\mathrm d}{\mathrm dt}E_\sigma(u(W_{t}))=-r(\sigma)E_\sigma(u(W_{t}))+
\sum_{\sigma'\ne\sigma}k(\sigma,\sigma')E_\sigma(u(\omega(\sigma,\sigma')+W_{t})).
$$
When considering the function $u:x\mathrm e^{-\lambda x}$, this is the formula in the post.
Nota: In the post, the measure $\mu(t)$ depends on the initial distribution $\mu(0)$ of $\varsigma$, thus $\langle f\rangle_{\mu(t)}$ is not uniquely defined. However, for each distribution $\nu$, conditionally on $\mu(0)=\nu$, the function $v:t\mapsto\langle f\rangle_{\mu(t)}$ indeed solves the differential equation
$$
v'(t)=\langle Lf\rangle_{\mu(t)},
$$
with initial condition
$$
v(0)=\langle f\rangle_{\nu}.
$$
A: If the derivation looks anything like the proof of the Crooks fluctuation theorem, which it appears to resemble at first sight, then the expected value $\mathbb{E}_\sigma$ is carried out over all possible trajectories starting at $\sigma$, and not all possible states, which is what the angle brackets denote. The quantity $\omega(\sigma,\sigma')$ represents a path-dependent function, that is, it does not just depend on the end points $\sigma$ and $\sigma'$, but on the path connecting them.
Now, the first step when calculating the derivative of $g(\sigma,t)$ mirrors the calculation of the derivative $\frac{\partial\mu(\sigma,t)}{\partial t}$, but the weighting factor differs. Instead of being simply $k(\sigma,\sigma')$, which represents the probability of a direct transition, it includes the extra exponenital factor, to take into account the different paths going from $\sigma$ to $\sigma'$.
