0
$\begingroup$

By a simple dynamical model one can derive the equilibrium distribution of $N = n_1 + n_2$ identical two-level particles which are subject to Boltzmann-like thermal fluctuations.

enter image description here

Consider a particle being picked at random and given an extra amount of energy $\Delta E > 0$ with probability

$$p_1(\Delta E) \sim e^{-\Delta E/kT}$$

If $\Delta E$ is large enough to lift the particle over the energy barrier $E_3$ it may fall back with equal probability $0.5$ to state $E_1$ or $E_2$. Otherwise it will remain in its state.

Equilibrium is reached when particles go with equal probability from $E_1$ to $E_2$ as from $E_2$ to $E_1$:

$$p_2(E_1 \rightarrow E_2) = p_2(E_2 \rightarrow E_1).$$

We have (by definition)

$$p_2(E_1 \rightarrow E_2) = \frac{n_1}{2}\ p_3(\Delta E > E_3-E_1)$$

and

$$p_2(E_2 \rightarrow E_1) = \frac{n_2}{2}\ p_3(\Delta E > E_3-E_2).$$

with

$$p_3(\Delta E > X) = \int_X^\infty p_1(E) dE $$

Regardless of the specific form of $p_1$ and $p_3$ we have equilibrium at

$$\frac{n_1}{n_2} = \frac{p_3(\Delta E > E_3-E_2)}{p_3(\Delta E > E_3-E_1)}$$

With $p_1$ as above (Boltzmann distribution) we have

$$p_3(\Delta E > X) \sim \int_X^\infty e^{-E/kT} dE \sim e^{-X/kT}$$

and equilibrium at

$$\frac{n_1}{n_2} = \frac{p_3(\Delta E > E_3-E_2)}{p_3(\Delta E > E_3-E_1)} = \frac{e^{-(E_3-E_2)/kT}}{e^{-(E_3-E_1)/kT}} = e^{-(E_1-E_2)/kT}$$

This (correct) result looks both like an insight (with a rather elaborate derivation) and like a sleight (which seems to help to understand the Boltzmann factor – but heavily relies on the Boltzmann distribution itself).

So what can be learned from this model?

Maybe not in terms of physics but in terms of didactics?


Added (1): I wonder if $p_1(E) = e^{-E/kT}$ might be the only distribution that "reproduces" itself in this way (via $p_3$).

Added (2): The Wikipedia article on the Boltzmann distribution confirms this asssumption. (Compare also @Pieter's comment that "it may be a proof that the distribution must be a negative exponential".)

$\endgroup$
1
$\begingroup$

This argument looks a bit circular and is not immediately convincing, but it may be a proof that the distribution must be a negative exponential. However, it does not really give a link to temperature.

In an earlier answer, I wrote about temperature, how one can show that the fractional change of the multiplicity with energy is the same when there is thermal equilibrium: $\frac{1}{\Omega}\frac{{\rm d}\Omega}{{\rm d}E} = \beta = \frac{1}{kT}$. Now consider a small system in contact with a large reservoir. The small system could be a single oscillator with equally spaced energies. These energies are so small that they would not affect the temperature of the large reservoir. Now, for each small amount of energy $E$ transferred to the oscillator, the multiplicity of the large system goes down with the same fractional amount. And that multiplicity is also proportional to the probability of the combined system. So the probability of finding the oscillator in a state with energy $E$ is proportional to $e^{-\beta E}$.

Didactically, a numerical example may be useful. At room temperature, the thermodynamic beta is $ \beta \approx 0.04$ meV$^{-1}$, or about 4 % per meV. So for each meV transferred to the oscillator, the multiplicity of the reservoir is reduced by a factor 0.96. At 25 meV this is $0.96^{25} \approx 0.36 \approx 1/e$, at 50 meV it is $0.96^{50} \approx 1/e^2$.

$\endgroup$
  • $\begingroup$ I agree that it does not give a link to temperature (i.e. explain it). But at least the given temperature of the heat bath (that produces the thermal fluctuation) is reflected in the population numbers of the two-level system. $\endgroup$ – Hans-Peter Stricker Jan 21 '18 at 19:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.