= Finite trajectory entropy production
{title2=$\Sigma=\Delta S_{\rm tot}/k_B$}
Consider a finite-duration <Markov process> with discrete, time-even states coupled to one <thermal reservoir> at positive <temperature> $T$. Let $P_F[\gamma]$ include the initial <probability> $\rho_0(x_0)$ of the <sample trajectory> $\gamma$. Define $P_R$ by reversing the control protocol and initializing it with the actual forward final <probability distribution> $\rho_\tau$. Write $\Theta\gamma$ for the same states visited in reverse time order. Require normalized path laws with matching <probability support>: a path has positive forward weight exactly when its reverse has positive reverse weight. This excludes irreversible forbidden transitions and missing initial states.
Under <local detailed balance>, with positive <heat> $Q_{\rm bath}$ delivered to the <thermal reservoir>, the full path-weight <logarithm> is
$$
\Sigma_F[\gamma]=\ln\frac{P_F[\gamma]}{P_R[\Theta\gamma]}
=\frac{Q_{\rm bath}[\gamma]}{k_BT}+\ln\frac{\rho_0(x_0)}{\rho_\tau(x_\tau)}
=\frac{\Delta S_{\rm tot}[\gamma]}{k_B}.
$$
The endpoint term is the change in stochastic system <entropy>; the conditional path ratio alone accounts for the <thermal reservoir>. Here $k_B$ is the <Boltzmann constant>. Both terms matter even for a short observation. <Entropy production along a stochastic trajectory and an integral fluctuation theorem by Udo Seifert>[Seifert, 2005, Eqs. (14)–(18) and (30): conditional and full path weights, endpoint distributions, total entropy]{ref}
Define the conjugate reverse functional by $\Sigma_R[\eta]=\ln(P_R[\eta]/P_F[\Theta\eta])$, so $\Sigma_R[\Theta\gamma]=-\Sigma_F[\gamma]$. Summing the path identity over paths with a given value gives the exact finite <fluctuation theorem>
$$
\frac{p_F(s)}{p_R(-s)}=e^s,
$$
for nonzero denominator, using <probability masses> or corresponding <probability density functions>. This $\Sigma_R$ is defined by the paired laws: it need not be the total <entropy production> obtained by independently restarting the thermodynamic construction for the reverse experiment. A generic driven experiment therefore does not imply $p_F(s)/p_F(-s)=e^s$.
Back to article page
