Dissipation-Time Uncertainty Relation

We show that the entropy production rate bounds the rate at which physical processes can be performed in stochastic systems far from equilibrium. In particular, we prove the fundamental tradeoff $⟨{\dot{S}}_e⟩\mathcal{T}\ge k_B$ between the entropy flow $⟨{\dot{S}}_e⟩$ into the reservoirs and the mean time $\mathcal{T}$ to complete any process whose time-reversed is exponentially rarer. This dissipation-time uncertainty relation is a novel form of speed limit: the smaller the dissipation, the larger the time to perform a process.

Figure 1

Speed limit for the dynamics Eq. (10) with process Eq. (11), obtained by numerical averages over ${10}^4$ trajectories with time steps $\mathrm{\Delta }t={10}^{-2}$, $a=1$, ${\beta }^{-1}=0.7,f=0.5,N=11$, $k_B=1$. Dashed and dotted lines correspond to the analytical value of $⟨{\dot{S}}_e⟩$ and $1/\mathcal{T}$, respectively. Inset: sketch of the dynamics Eq. (10).

Figure 2

Speed limit for the dynamics Eq. (10) with process Eq. (11), obtained by numerical averages over ${10}^4$ trajectories with time steps $\mathrm{\Delta }t={10}^{-2}$, $a=1$, ${\beta }^{-1}=0.07,f=0.6,N=2$, $k_B=1$. Dashed and dotted lines correspond to the weak-noise estimation of $⟨{\dot{S}}_e⟩$ and $r=1/\mathcal{T}$, respectively. Inset: the time averaged values of $⟨{\dot{S}}_e{⟩}_s$ (filled) and $r$ (empty) as a function of the tilt $f$. At short times $t\lesssim {10}^3$ and small tilt $f\lesssim 0.4$ the numerical estimation of $r$ is impeded by the finite statistics.

Figure 3

Speed limit for the two-state system of the main text with process Eq. (12), obtained by numerical averages over ${10}^5$ Gillespie trajectories with $1/{\beta }_h=1.5,1/{\beta }_c=0.5,{\epsilon }_2=1,{\epsilon }_1=0,E=5,\delta =6,k_B=1$. The dashed line corresponds to $⟨{\dot{S}}_e⟩$. At short times $t\lesssim 50$ the numerical estimation of $r(t)$ is impeded by the finite statistics. Inset: time-averaged value of $⟨{\dot{S}}_e{⟩}_{\tilde{s}}$ (filled) and $r$ (empty) as function of $\mathrm{\Delta }\beta ≔{\beta }_c-{\beta }_h$ at fixed average temperature $(1/{\beta }_c+1/{\beta }_h)/2=1$.