I am reading Molecular Driving Forces, 2nd ed., by Dill & Bromberg.

On page 53, example 3.9, we consider why energy exchanges between two systems from the point of view of the 2nd law.

We consider two separate systems. Each has ten particles, and each particle has two possible energy states. System A has total energy $U_{a}$ = 2, and $U_{b}$ = 4. Thus binomial statistics predicts the multiplicities of these systems:

$W(U_{a})$ = $\frac{10!}{8!2!} = 45$

$W(U_{b})$ = $\frac{10!}{6!4!} = 210$

Now the confusing part, to me, is the math when these two systems come into thermal contact. Then the author asserts that the initial multiplicity is

$W(U_\text{total})$ = $\frac{10!}{8!2!}\frac{10!}{6!4!}$

And that maximum multiplicity is found at

$W(U_\text{total})$ = $\frac{10!}{7!3!}\frac{10!}{7!3!}$ = 14,400

But why consider the systems in this way, as opposed to thinking of a new system, with 20 particles, having $U_{a}$+$U_{b}$ = 6?

Then we get

$W(U_{a+b})$ = $\frac{20!}{14!6!}$ = 38760 ≠ $W(U_\text{total})$

I'm trying to develop a sense of the difference, I suppose, between two systems in thermal contact and one system. After they've equilibrated, how are they not treatable as one system? They clearly aren't, because if they were, then the total multiplicity of that one system must = the total multiplicity of the two systems A and B.

  • $\begingroup$ Put a different way, why couldn't I take any system and arbitrarily partition it up and consider it as several systems in thermal contact with each other? What we've just seen here is that the math is different between "one system" and "two systems in thermal contact". $\endgroup$
    – masonk
    Commented Jan 15, 2013 at 0:23

1 Answer 1


In this example both the systems are of the same type of particles (with two energy states) and same number of particles. Therefore thermal equilibrium is defined when energy is equally shared between the systems, but the particles are still not allowed to be exchanged. The particles, although of the same kind, are distinguished as being in system A or B. If you allowed particles to be exchanged then you are allowing swapping of particles between the two systems effectively raising the possibility to 20 particles with 6 energy units.

  • $\begingroup$ By introducing rigid partitions you are "distinguishing" particles between each partition. While the complete system doesn't distinguish between the particles. If you allow leaky partitions, i.e., both matter and energy can move between partition you get back to the same answer for the whole system $\endgroup$
    – Sankaran
    Commented Jan 15, 2013 at 0:36

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