Heat fluctuations in overdamped non-isothermal processes

Abstract

In stochastic thermodynamics, non-isothermal processes are essential for most Brownian thermal machines. In this paper, we investigate how the inclusion of a temperature protocol affects heat and its fluctuations. The protocol needs to satisfy certain constraints to allow the particle to be stationary at the end of the process. We investigate the fluctuations of heat when there is a non-isothermal process for the non-stationary final velocity case, and the overdamped harmonic and quartic potential. Our results are consistent with thermodynamics and demonstrates the importance of adding a kinetic energy term to the equations in overdamped systems.

Appendix A: Path integral calculation

Since the path integrals considered are Gaussian. We only need to calculate the stochastic action in its extremum. Both the harmonic and the non-stationary cases are very similar mathematically. Therefore, we will only show one of the cases: the harmonic potential.

The stochastic action with the temperature protocol is [32]

$$\begin{aligned} \mathcal {A}[x] = \frac{1}{4\gamma }\int _0^\tau \frac{\left( \gamma \dot{x}(t)+k x(t)\right) ^2}{T(t)} \mathrm{{d}}t, \end{aligned}$$

(A1)

extremizing this action we obtain the differential equation

$$\begin{aligned} \dot{u}(t) = \frac{k}{\gamma } u(t), \end{aligned}$$

(A2)

where $u(t)=\frac{\left( \gamma \dot{x}(t)+k x(t)\right) }{T(t)}$, solving for u(t) we find the differential equation for the extremum of the position.

$$\begin{aligned} \gamma \dot{x} + k x = T(t) C e^{\frac{k}{\gamma }t}, \end{aligned}$$

(A3)

where C is an unknown constant. Solving for x(t) we have

$$\begin{aligned} x(t) = x_0 e^{-\frac{k}{\gamma }t}+ C \frac{e^{-\frac{k}{\gamma }t}}{\gamma }\int _0^t T(t')e^{2\frac{k}{\gamma }t'} \mathrm{{d}}t' \end{aligned}$$

(A4)

To find C we only need the boundary condition $x(\tau ) = x_\tau$, which gives

$$\begin{aligned} C = -\gamma \frac{\left( x_0-x_\tau e^{\frac{k\tau }{\gamma }}\right) }{\mathcal {T}_k(\tau )}. \end{aligned}$$

(A5)

Finally, to find the stochastic action in its extremum $\mathcal {A}_e[x]$, we only need to substitute u(t) since

$$\begin{aligned} \mathcal {A}_e[x] = \frac{C^2}{4\gamma }\int _0^\tau T(t) e^{\frac{2kt}{\gamma }} \mathrm{{d}}t = \frac{\gamma }{4 \mathcal {T}_k(\tau )}\left( x_0-x_\tau e^{\frac{k\tau }{\gamma }}\right) ^2. \end{aligned}$$

(A6)

Thus, the transitional probability will be given by $P[...] \sim \exp \left( - \mathcal {A}_e[x]\right)$, with a normalization constant that can be calculated a posteriori.