June 2022 – Physics Notes

Despite some rather dismissive responses from other experts I think that this is a very good question which deserves a real answer.

It is indeed reasonable for a student to ask why the equal distribution of energy between degrees of freedom in subsystems that appears to result from equipartition and heat flow does not extend down to the level of individual molecules in a gas. And the answer does not require a complete calculus-based derivation of the Maxwell-Boltzmann distribution.

All that is required is some clarification of the distinction betwee microstates and macrostates, where the former are specified in complete detail but the latter always involve some unknown details which have to be dealt with by probabilities.

For a “gas” of just two identical spherical atoms confined by perfectly elastic collisions with the walls of a massive container, the actual energy distribution of the atoms never changes (because during a collision between two atoms conservation of energy and momentum requires that both ${v_{1}}^{2} + {v_{2}}^{2}$ and $v_{1} + v_{2}$ remain unchanged which allows only that $v_{1}$ and $v_{2}$ either stay the same or exchange values). But if the interactions with the walls allow for energy transfers to or from the atoms then the distribution may change over time. And if the transfers are small and random in such a way that we end up not knowing the individual particle energies but do know that there is negligible accumulated change in the total energy of the system, then it seems reasonable to make the assumption that all of the microstates (ie actual $v_{1}$ and $v_{2}$ values) corresponding to the macrostate defined just by total energy are equally likely.

For a simpler analysis (avoiding the need to discuss a continuous range of energy values – which would require calculus) let’s just imagine that the two atoms are distinguishable, that each of them can have energy values corresponding to whole numbers 0,1,2,3… , and that the macrostate is defined by having a total energy of 2. Then there are just three equally likely microstates with the actual values of $(e_{1}, e_{2})$ being (0,2), (1,1), or (2,0). In this situation the case of equal energies is actually less likely than that of their being unequal, but if the atoms were indistinguishable then the cases (0,2) and (2,0) would be the same and the probabilities would be equal. (And the fact that identical particles really are indistinguishable does make an experimentally significant difference in statistical mechanics.)

For the case of N distinguishable atoms in the macrostate with total energy E, the possible microstates are $(e_{1}, e_{2}, \dots e_{N})$ with $e_{1} + e_{2} + \dots + e_{N} = E$ .

With $N = 3$ and $E = 6$ , this allows $(e_{1}, e_{2}, e_{3})$ =(6,0,0),(5,1,0),(5,0,1),(4,2,0),(4,1,1),(4,0,2),(3,3,0),(3,0,3),(3,2,1),(3,1,2),(2,4,0),(2,0,4),(2,3,1),(2,1,3),(2,2,2),(1,5,0),(1,0,5),(1,4,1),(1,1,4),(1,3,2),(1,2,3),(0,6,0),(0,0,6),(0,5,1),(0,1,5),(0,4,2),(0,2,4),(0,3,3). For each particle, $p$ , this includes one microstate with $e_{p} = 6$ , two with $e_{p} = 5$ , three with $e_{p} = 4$ , four with $e_{p} = 3$ , five with $e_{p} = 2$ , six with $e_{p} = 1$ , and seven with $e_{p} = 6$ , (and in only one of the 28 cases are all three energies equal).

When we combine two such systems with different temperatures (amounts of energy per degree of freedom) then at first the distribution of energies will have two peaks (with one corresponding to each of the combined populations). But assuming that the interactions with the external world allow (perhaps indirect) energy exchange in the usual way, then eventually we’ll reach a situation in which the probability of a microstate depends only on its total energy and every particle has the same probability distribution for its possible energy values.

And even for the case in which the two combined systems each consists of a single particle, the evolution towards a situation in which the two particles both have the same energy distribution does not lead to one in which they each actually have the same energy.

(This simple model was not intended to be realistic but just to show why the process of approaching thermal equilibrium does not lead to all of the molecules having the same velocity.)

Source: (254) Alan Cooper’s answer to Why does the Boltzmann distribution of gas particles not collapse to equal velocities (thermal equilibrium)? I would like help explaining it to high school chemistry students. – Quora