Adopted finite reversible bath rates and field coupling

Use the finite state set, energies $e_a$, rates $w_{ab}$, Gibbs weights ${\pi }_a$, and row generator $Q$ defined in Finite reversible stochastic rate model. Assign the state energies the physical meaning of coarse-state energies in joules. Use the primitive SI Boltzmann constant $k_B$ in joules per kelvin from Adopted inertial velocity Langevin bath, and set $\beta =(k_B{T}_b{)}^{-1}$ in joules${}^{-1}$ for the positive bath-temperature parameter $T_b$ in kelvins. Adopt the baseline transition rates $w_{ab}$ in seconds${}^{-1}$ and the initial law $p_0=\pi$ for the equilibrium-response branch. These choices define a finite phenomenological thermal rate model; they do not say that a real material has these rates.

Fix a finite time horizon $t_f>0$. For a fixed state observable $B:E\to \mathbb{R}$ with units denoted $[B]$, a dimensionless control amplitude $ϵ$, and a continuous field $h:[0,t_f]\to \mathbb{R}$ with units joules per $[B]$, adopt the controlled rates

$w_{ab}^{ϵ}(t)=w_{ab}\exp \text{\hspace{0.17em}⁣}(\frac{\beta ϵh(t)}{2}(B_b-B_a)),a\mathrm{\ne }b\mathrm{.}$

Their units remain seconds${}^{-1}$ because the exponent is dimensionless. This is one explicit symmetric allocation of the field-dependent rate factor. For constant $h$ and each positive-rate edge ($w_{ab}>0$, equivalently $w_{ba}>0$), the rate ratio obeys

$\frac{w_{ab}^{ϵ}}{w_{ba}^{ϵ}}=\exp \text{\hspace{0.17em}⁣}(-\beta [(e_b-ϵh{B}_b)-(e_a-ϵh{B}_a)]),$

so the Gibbs weights of the shifted energies $e_a-ϵh{B}_a$ satisfy detailed balance by direct multiplication on every positive-rate edge. For an absent edge, both controlled rates remain zero, so the detailed-balance product identity holds there as well; no rate ratio is defined there. A rate ratio alone leaves state-pair prefactors undetermined; the symmetric splitting above is an additional modeling choice and is the specific perturbation analyzed in the consuming theorem. No experiment or empirical rate estimate is asserted.