In deducing the zeroth law of thermodynamics in micro-canonical ensemble, there is a frequently-mentioned example. Suppose we put two isolated system, system 1 and 2, in contact and allowing them to exchange heat.

enter image description here

The total energy of the combined system is

$$E = {E_1} + {E_2}$$

The total allowed phase space is

$$\Omega (E) = \int d {E_1}{\Omega _1}({E_1}){\Omega _2}(E - {E_1}) = \int d {E_1}\exp [\frac{{{S_1}({E_1}) + {S_2}(E - {E_1})}}{{{k_B}}}]$$

By taking the maximum of the integrand,

$$\frac{{\partial {S_1}({E_1})}}{{\partial E}} + \frac{{\partial {S_2}(E - {E_1})}}{{\partial E}} = 0$$

we get the equilibrium state $(E_1^{eq},E - E_1^{eq})$, which is the macrostate with exponetially larger number of microstates, and also get the equilibrium condition

$$\frac{{\partial {S_1}({E_1})}}{{\partial {E_1}}} = \frac{{\partial {S_2}(E - {E_1})}}{{\partial {E_2}}}$$

The equal a priori probability postulate, as descripted by David Tong p4 , is

For an isolated system in equilibrium, all accessible microstates are equally likely.

So my question is, does that mean the microsates corresponds to the equilibrium macrosate $(E_1^{eq},E - E_1^{eq})$ possses equal probability or all the microstates with constant $E$ (the blue line in next figure) possess equal probability?

For me, it looks like every reference I found indicates the second statement. But for a microsate corresponds to $(E_1^{0},E - E_1^{0})$ which is not at equilibrium, does it have the same probability as a microstate of $(E_1^{eq},E - E_1^{eq})$? enter image description here


1 Answer 1


The equal a priori probability postulate says:

all microstates in the big box are equally probable

Here is what this means. If we collect all possible microstates of the big box what we will find is that the overwhelming majority of them are such that the part 1 has energy $E^{eq}$ and part 2 has energy $E-E^{eq}$. Individual microstates with different energies, say $(E', E-E')$ are equally probable with those that have $(E^{eq},E-E^{eq})$, but there are much fewer such states.

Combinatorial example Here is an example that demonstrates what is going on. Suppose we have four buckets and two balls so that a bucket can have at most one ball. All arrangements are equally probable. The buckets represent your big system, and the shaded color highlights each half of that box. Each arrangement of balls represents a microstate.

enter image description here

Out of the six arrangements, four distribute the particles in equal numbers between the two halves. This is the equilibrium distribution: $N=2$, $N_1^{eq}=1$, $N_2^{eq}=1.$ It is the distribution with the maximum number of microstates.

The microstates with both balls in one half are equally probable as the rest, but there are only 2 such microstates, compared to 4 microstates with the equilibrium distribution. With more buckets and more balls the overwhelming majority of microstates will have almost exactly equal numbers of balls in each half.

  • $\begingroup$ What is the statphysical definition of "the big box" ? The system for wich $S \sim N$ ? $\endgroup$ Commented Aug 22, 2019 at 16:38
  • 3
    $\begingroup$ I think I start to know the answer to my question. I think the key point is that equilibrium is a macroscopic property which we can't say a microscopic state is at equilirium or not. In an equilibrium ensemble with many microstates, there could be some "non-equilibrium microstates" whose energy does not corresponds to equilibrium states (likes the microscopic states with energy $(E_1^{0},E - E_1^{0})$ or the microstate that all gas atoms stay in the left box), and this is the source of fluctuation. $\endgroup$
    – FaDA
    Commented Aug 23, 2019 at 14:31
  • $\begingroup$ But since we have the equal priority probability postulate, what the system arises is the most probable microstates which has the equilibrium condition $\frac{{\partial {S_1}({E_1})}}{{\partial {E_1}}} = \frac{{\partial {S_2}(E - {E_1})}}{{\partial {E_2}}}$ $\endgroup$
    – FaDA
    Commented Aug 23, 2019 at 14:32
  • $\begingroup$ @AlekseyDruggist yes, I think the big box is a system which combine both 1 and 2. And $S~N$ is definitely a sequence due to the extensivity of entropy $\endgroup$
    – FaDA
    Commented Aug 23, 2019 at 14:34
  • 3
    $\begingroup$ @FaDA "I think the key point is that equilibrium is a macroscopic property which we can't say a microscopic state is at equilibrium or not" Exactly! $\endgroup$
    – Themis
    Commented Aug 24, 2019 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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