Callen problem 15.1-2:
“A system is composed of two harmonic oscillators of natural frequencies
ωo and 2ωo . If the system has total
energy E = (n + ½)ħωo,
where n is an odd integer, what is the entropy of the system?”
As in Problem 15.1-1, the
total energy constraint is used to count the total number of microstates. The
energies possessed by both oscillators must equal the total energy
(i+½)ħωo + (j+½)2ħωo = (n
+ ½ )ħωo,
giving the constraint
relating the integers i, j, and n
(i+½) + (j+½)2 = n + ½
i+½ + 2j+1
= n + ½
i + 2j
= n – 1
2j = n – 1 – i
2j = (n – i) – 1.
Since n is odd,
i must be even (or zero), so that (n – i) – 1 is even and j is an integer. This relationship is shown in
a table:
|
i |
2j |
j |
|
0 |
n
– 1 |
(n
– 1)/2 |
|
2 |
n
– 3 |
(n
– 3)/2 |
|
… |
|
|
|
n – 3 |
2 |
1 |
|
n – 1 |
0 |
0 |
Comparing this to the first
table in Problem 15.1-1, there would be (n – 1) + 1 = n states if we were counting with n = all integers. But we’re only
counting with n
= odd integers. So do we just divide n by two? That doesn’t work, since n
is odd. But I didn’t see how to avoid that at
first, until I realized that counting from zero to n –
1 is
one too many states to divide by two to get rid of the n = even numbers. We need
instead to count from i
= 1
to i
= n
– 1 and divide that by two,
then add one state for the i
= 0:
(n – 1) / 2 + 1 = (n – 1 + 2) / 2 = (n + 1) / 2
This
is the number of microstates of the system. The entropy is
S = k log [(n+1)/2]
= k log(n+1) – k log
2
= k log(n+1) –
0.693k.
When n = 1, the
entropy is zero, as
expected, since there’s only one microstate. As n increases, there’s always the constant 0.693k to be subtracted from the increasing entropy.
