Zon-Cohen singularity and negative inverse temperature in a trapped-particle limit

We study a Brownian particle on a moving periodic potential. We focus on the statistical properties of the work done by the potential and the heat dissipated by the particle. When the period and the depth of the potential are both large, by using a boundary layer analysis, we calculate a cumulant generating function and a biased distribution function. The result allows us to understand a Zon-Cohen singularity for an extended fluctuation theorem from a viewpoint of rare trajectories characterized by a negative inverse temperature of the biased distribution function.

Figure 1

A periodic potential that satisfies Eq. (7). We drew Eq. (8) by setting $k=2$ and $L=1$.

Figure 2

${\tilde{w}}_{h,v}\left(Y\right)$ for potentials $U_{\mathrm{harmo}}\left(y\right)=(k/2)y^2$ (left) and $U_{\mathrm{quart}}\left(y\right)=(k_4/4)y^4$ (right). Quantities are converted to dimensionless forms by setting $\gamma =T=v=1$. We fixed $k=1$, $k_4=1$, and $h=-8$. We set $L=18$ for $U_{\mathrm{harmo}}\left(y\right)$ and $L=3$ for $U_{\mathrm{quart}}\left(y\right)$. The green dashed lines and purple dotted lines are obtained from Eq. (81). These lines correspond to $14+36Y$ and $-16$ for $U_{\mathrm{harmo}}\left(y\right)$, and $14+54Y^3$ and $-16$ for $U_{\mathrm{quart}}\left(y\right)$. The red solid lines are numerical results. It can be seen that Eq. (81) agrees with the numerical results.

Figure 3

$K_{h,v}/\left(2T\right)$ for potentials $U_{\mathrm{harmo}}\left(y\right)=(k/2)y^2$, $U_{\mathrm{quart}}\left(y\right)=(k_4/4)y^4$, and $U_{\mathrm{linear}}\left(y\right)=k_1\left|y\right|$. Quantities are converted to dimensionless forms by setting $\gamma =T=v=1$. We fixed $k=1$, $k_4=1$, and $k_1=1$ and numerically evaluated $K_{h,v}/\left(2T\right)$. In each figure, dashed dotted lines (aqua blue), dotted lines (purple), and dashed lines (green) correspond to $L=6$, $L=12$, and $L=18$ for $U_{\mathrm{harmo}}$, $L=1$, $L=2$, and $L=3$ for $U_{\mathrm{quart}}$, and $L=5$, $L=25$, and $L=50$ for $U_{\mathrm{linear}}$. The red solid lines in all figures denote $h+h^2$, which is predicted by Eq. (85). As $L$ becomes large, $K_{h,v}$ approaches $h+h^2$ for $U_{\mathrm{harmo}}$ and $U_{\mathrm{quart}}$, but does not for $U_{\mathrm{linear}}$. Since $U_{\mathrm{harmo}}$ and $U_{\mathrm{quart}}$ satisfy Eq. (7) but $U_{\mathrm{linear}}$ does not, these results are consistent with our formulation.