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I derived a formula for the time constant $\tau$ by which a toy-system of identical particles having two energy levels $E_1$ and $E_2$ approaches equilibrium. I'd like to ask if this derivation may be considered essentially correct, and if both result and derivation are of any theoretical or didactical interest.

The system (red) may be in thermal contact with a one-dimensional ideal gas of pressure $p$ and temperature $T$ (yellow). Let there be initially arbitrary $n_1(0)$ and $n_2(0)$ particles with energy $E_1$ and $E_2$. Let $N = n_1(0) + n_2(0)$. The kinetic energies of the atoms of the gas are distributed like $p(E) \propto e^{-E/kT}$. (See also here and here.)

enter image description here

Assume that in the time intervall $dt \ll 1$ an average number of $\mathcal{P}Ndt$ particles are hit, $\mathcal{P}$ being proportional to the pressure $p$ of the gas.

The number of particles with energy $E > E_3$ then is

$$\mathcal{P}Ndt\Big(\frac{n_1(t)}{N}e^{-\Delta E_1/kT} + \frac{n_2(t)}{N}e^{-\Delta E_2/kT}\Big)$$

with $\Delta E_i = E_3 - E_i$.

Then mixture occurs: Half of these go back to $E_1$, the other half to $E_2$.

So we have

$$\dot{n_1} = \frac{\mathcal{P}}{2}\Big(-n_1e^{-\Delta E_1/kT} + n_2e^{-\Delta E_2/kT}\Big) $$

$$\dot{n_2} = \frac{\mathcal{P}}{2}\Big(+n_1e^{-\Delta E_1/kT} - n_2e^{-\Delta E_2/kT}\Big) $$

A little rewriting gives

$$\dot{n_1} = \alpha (-n_1 + \beta n_2) $$

$$\dot{n_2} = \alpha (+n_1 - \beta n_2) $$

with

$$\alpha = \frac{\mathcal{P}}{2}e^{-\Delta E_1/kT}$$

$$\beta = e^{-(\Delta E_2 - \Delta E_1)/kT} = e^{(E_2 - E_1)/kT}$$

Rescaling time by $dt' = \alpha dt$ we finally get

$$\dot{n_1} =-n_1 + \beta n_2 $$

$$\dot{n_2} = +n_1 - \beta n_2 $$

This system of differential equations has the general solution

$$n_1(t)=C_1 (\beta +e^{-(1+\beta )t})+C_2 (\beta -\beta e^{-(1+\beta )t}) \\ n_2(t)=C_1 (1-e^{-(1+\beta )t})+C_2 (1+\beta e^{-(1+\beta )t})$$

which fulfills $N = n_1(t) + n_2(t) = (C_1+C_2)(\beta + 1)$ for all $t$. (For a derivation see here.)

From the initial conditions we get

$$C_1 = \frac{n_1^0}{\beta + 1}$$

$$C_2 = \frac{n_2^0}{\beta + 1}$$

and for $N_i = \lim_{t \rightarrow \infty}n_i(t)$

$$N_1 = \beta(C_1 + C_2)$$

$$N_2 = C_1 + C_2$$

With the definition of $\beta = e^{(E_2 - E_1)/kT}$ we find - as expected for the thermal equilibrium the system has approached - that

$$\frac{N_1}{N_2} = \beta = e^{(E_2-E_1)/kT}$$

Scaling the time constant $\tau' =1/(1+\beta)$ back to the original time $t = t'/\alpha$ we have

$$\tau = \frac{2}{\color{green}{\mathcal{P}}}\frac{e^{(\color{red}{E_3}-E1)/k\color{blue}T}}{(1 + e^{(E_2-E_1)/k\color{blue}T})}$$

This is how $\tau$ depends qualitatively on $T$ (blue), $\mathcal{P}$ (green) and the energy barrier $E_3$ (red) when keeping the other two fixed:

enter image description here (see here)

As suggested by user @JalfredP I tried to relate this result to Kramers problem (as exposed here), but I didn't manage.

A side question concerns the proportional factor between $\mathcal{P}$ and pressure $p$. This should not depend on any of the other parameters of the system, so it should be composed of natural constants only.

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  • $\begingroup$ You will never be able to get time constants from equilibrium statistical mechanics, since equilibrium means "no time evolution". You will maybe be able to get the scaling behavior of typical time scales, but never an explicit expression for them. $\endgroup$ – valerio Jan 30 '18 at 1:01
  • $\begingroup$ I definitely did not get time constants from equilibrium statistical mechanics: the equilibrium distribution for $N_1/N_2$ is a by-product of the calculation (see above). And what do you say to Kramers problem: "The Kramers problem is to find the rate (!) at which a Brownian particle escapes from a potential well over a potential barrier." $\endgroup$ – Hans-Peter Stricker Jan 30 '18 at 6:56
  • $\begingroup$ You are assuming that the particles are distributed according to the Boltzmann distribution in every one of your posts. So how is that a "by-product of the calculation"? The Kramers problem uses a non-equilibrium approach, and that is surely what you need if you want to say something about time constants. $\endgroup$ – valerio Jan 30 '18 at 8:39
  • $\begingroup$ It's the heat bath (the ideal gas below, yellow) that is supposed to be in equilibrium, that's right. But the question concerns the other system (red) and how this approaches equilibrium (with $N_1/N_2$ like above), starting in non-equilibrium with arbitrary $n_1(0)$, $n_2(0)$. How could I have made this clearer? $\endgroup$ – Hans-Peter Stricker Jan 30 '18 at 8:48

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