A Statistical View of Entropy — Question 5

Question 5

Most probable and mean energy shares need not coincide in a small system

Two weakly coupled model solids have $N_A=2$ and $N_B=3$ distinguishable oscillators. They share four indistinguishable energy quanta. For $N$ oscillators and $q$ quanta, the supplied multiplicity is $\Omega(N,q)=\binom{q+N-1}{q}$. All joint microstates at total energy four quanta are equally likely.

1. Compute joint multiplicities for $q_A=0,1,2,3,4$, normalize them and identify the most probable share.

2. Calculate $\langle q_A\rangle$ and compare it with the most probable value. Verify the total multiplicity by treating all five oscillators together.

Question 5 - Solution

Most probable and mean energy shares need not coincide in a small system

(a) Method and calculation

$$w(q_A)=\Omega(2,q_A)\Omega(3,4-q_A).$$$$(w_0,w_1,w_2,w_3,w_4)=(15,20,18,12,5),\quad\sum w_q=70.$$$$P(q_A)=w(q_A)/70,\qquad q_{A,\mathrm{mode}}=1.$$

(b) Interpretation and further calculation

$$\langle q_A\rangle=\frac{0(15)+1(20)+2(18)+3(12)+4(5)}{70}=1.60.$$$$\Omega(5,4)=\binom84=70.$$The mean is an ensemble average and need not be an allowed integer value. It differs from the mode because this small-system distribution is broad and asymmetric.

Physical check

The total number of microstates, not the number of possible energy shares, sets the normalization.