# Three detailed fluctuation theorem

## Recap the fluctuation theorems

### Weak system-resevoir coupling

Crooks relation $\frac{P(W)}{\bar{P}(-W)}=\exp\left[\beta(W-\Delta F)\right]$

Jaryzynski equality

$\langle \exp(-\beta W)\rangle =\exp(-\beta \Delta F)$

## detailed and integral fluctuation theorem

For a random variable $R$ it satisfy integral fluctuation theorem if

$\langle e^R\rangle=1$

and the detailed fluctuation theorem

$\exist\bar{P}\quad \frac{P(R)}{\bar{P}(-R)}=e^{R}\quad\forall R$

Kullback-Leibler distance characterize the difference between two probability distribution, for a set of random variable(s) $\mathbf{m}$ and two probability distribution $P $\mathcal{D}(P||\mathcal{\tilde{P}})=\sum_mP_\mathbf{m}\ln\frac{P_\mathbf{m}}{\tilde{P}_\mathbf{m}}$ For single trajectory, the distance between two trajectory is defined as $r_\mathbf{m}=\ln\frac{P_\mathbf{m}}{\tilde{P}_\mathbf{m}}$ One can verify that the random variable$r_\mathbf{m}$satisfy fluctuation theorem ## A physical system characterized by master equation Master equation $\dot{p}_m(t)=\sum_{m'}W_{m.m'}(\lambda_t)p_{m'}(t)$ Probability of tranjectory Now suppose we have a trajectory$\mathbf{m}={m_t,t\in [0,T]}$, discretized the time into$Nsteps, we can write down the probability of hte trajectory explicitly \begin{aligned} P_\mathbf{m} &=\prod_{j=0}^Np_{m_j}(\tau_j,\tau_{j+1})\\ &=p_{m_0}(0)\left[\prod_{j=1}^N S_{m_{j-1}}(\tau_{j-1},\tau_j,\lambda_\tau)W_{m_j,m_{j-1}}(\tau_j)\right]S_{m_N}(\tau_{N},T,\lambda_\tau)\\ \end{aligned} whereS_{m_{j-1}}(\tau_{j-1},\tau_j)is the survival probability $S_{m_{j-1}}(\tau_{j-1},\tau_j,\lambda_\tau)=\exp\left[\int_{\tau_{j-1}}^{\tau_j}d\tau'\ W_{m_{j-1},m_{j-1}}(\lambda_{\tau'})\right]$ we can also write down the probability of the reversed process $\bar{\lambda}_t=\lambda_{T-t},\ \bar{m}_j=m_{N-j}$ \begin{aligned} \bar{P}_{\bar{m}} &=\prod_{j=0}^Np_{\bar{m}_{j}}(\tau_{j},\tau_{j+1})\\ &=p_{m_N}(T)\left[\prod_{j=1}^N S_{m_j,m_j}(\tau_{j+1},\tau_j,\bar{\lambda}_{T-\tau})W_{m_{j-1},m_j}\right]S_{m_0}(\tau,0,\bar{\lambda}_{T-\tau}) \end{aligned} Total entroy production $\Delta S_{tot}[\mathbf{m}]\equiv \ln\frac{P_{\mathbf{m}}}{\bar{P}_\mathbf{\bar{m}}}$ system+reservoir picture $\Delta S_{tot}[\mathbf{m}]=\underbrace{\ln\frac{p_{m_0}(0)}{p_{m_N}(T)}}_{\text{system entropy}}+\underbrace{\sum_{j=1}^N\ln\frac{W_{m_j,m_{j-1}}(\lambda_{\tau_j})}{W_{m_{j-1},m_j}(\lambda_{\tau_j})}}_{\text{reservoir entropy}}$ adiabatic+non-adiabatic picture Using the steady state distributionp_m^{st}(\lambda_t)$$\sum_{m'}W_{m,m'}(\lambda_t)p_m^{st}(\lambda_t)=0$ we can split total entropy production to two parts, the adiabatic entropy production and the non-adiabatic entropy production $\Delta S_{tot}[\mathbf{m}]=\underbrace{\ln\frac{p_{m_0}(0)}{p_{m_N}(T)}+\sum_{j=1}^N\ln\frac{p_{m_j}^{st}(\lambda_{\tau_j})}{p_{m_{j-1}}^{st}(\lambda_{\tau_j})}}_{\text{adiabatic }\Delta S_a[\mathbf{m}]} +\underbrace{\sum_{j=1}^N\ln\frac{W_{m_j,m_{j-1}}(\lambda_{\tau_j})p_{m_{j-1}}^{st}(\lambda_{\tau_j})}{W_{m_{j-1},m_{j}}(\lambda_{\tau_j})p_{m_{j}}^{st}(\lambda_{\tau_j})}}_{\text{non-adiabatic }\Delta S_{na}[\mathbf{m}]}$ If the system’s steady state satisfy detailed balance condition $W_{m,m'}(\lambda_t)p_{m'}^{st}(\lambda_t)=W_{m',m}(\lambda_t)p_m^{st}(\lambda_t)\quad\forall m,m'$ the adbiatic contribution is zero $\Delta S_a[\mathbf{m}]=0$ If the relaxiation process is much faster than the time scale of$\lambda_t$$\langle\Delta S_{na}[\mathbf{m}]\rangle=0$ ## Three fluctuation theorems By introducing the adjoint transition matrix$\mathbf{W}^+\$

$W_{m,m'}^+=\frac{W_{m',m}(\lambda_t)p_m^{st}(\lambda_t)}{p_{m'}^{st}(\lambda_t)}$

one can obtain the following relation

$\Delta S_{na}[\mathbf{m}]\equiv\ln\frac{P_\mathbf{m}}{\bar{P}_{\mathbf{\bar{m}}}^+},\quad \Delta S_a[\mathbf{m}]\equiv\frac{P_{\mathcal{m}}}{P_{\mathbf{m}}^+}$

Using the mathematical relation we introduced before, one can obtain three detailed fluctuation theorems

$\frac{P(\Delta S_{tot})}{\bar{P}(-\Delta S_{tot})}=e^{\Delta S_{tot}}$ $\frac{P(\Delta S_{na})}{\bar{P}^+(-\Delta S_{na})}=e^{\Delta S_{na}},\quad \frac{P(\Delta S_{a})}{P^+(-\Delta S_{a})}=e^{\Delta S_{a}}$