Callen problem 15.1–1. (part one) "A system is composed of two harmonic oscillators, each
of natural frequency ωo and each having permissible energies
(n
+½) ħωo, with n being any non–negative integer. Given that the total energy of
the system is E′ = , where n′ can be any positive integer, calculate the number of microstates available to the
system, and the entropy of the system." (See Chapter 15 in Thermodynamics and an Intro to Thermostatistics (1985) for background reading that
leads up to this problem.)
The first thing to note is these aren't coupled oscillators. Thus this is the simplest possible example of a Planckian oscillator system with zero-point energy. The oscillators can share the total energy available to them, but otherwise don't interact. Yeh, like photons themselves. As Peter Milonni (The Quantum Vacuum, chapter 2) says, an electromagnetic field mode is a (quantum) harmonic oscillator. Also, to be more precise, "mode" should be "normal mode".
The other thing to consider is indistinguishability versus distinguishability. Callen says nothing that I see in Chapter 15 or earlier about this, so we just rather automatically accept that the energy elements are indistinguishable. This makes it easier to count the microstates because permutations of the energy elements for a particular microstate don't need to be counted.
You can write the energy
constraint of the two oscillators as
(i+½) ħωo + (j+½)
ħωo = n′ ħωo ,
which gives
(i+½) + (j+½) = n′ ,
i + j = n′ – 1,
j = n′ – i –
1
From this expression, you
can just build a table of all possible i
and j values,
i
|
j
|
0
|
n′ – 1
|
1
|
n′ – 2
|
2
|
n′ – 3
|
…
|
…
|
n′ – 3
|
2
|
n′ – 2
|
1
|
n′ – 1
|
0
|
For any given positive integer n' this table gives all the ways the energy can be shared between the two oscillators. So, just counting the number of rows in the table gives the number of microstates. Since we have
to start counting from zero, the number of microstates is
(n′ – 1) + 1 = n′ .
The “–1” comes from having to
account for zero-point energy (ZPE) in the energy constraint, and the “+1” from starting the count at zero. So there are n′ microstates. Yeah, sort of an anticlimactic result—the number of microstates is just equal to the number of "energy elements" the system has. With only two oscillators,
whatever part of the total energy isn't in one is in the other. When there are three or more oscillators, permutations of the total
number of oscillators among the energy states must be counted. (But--not to be dealt with in this problem--is this equal
to the number of permutations of the total number of energy elements among the
oscillators? That is, we can count oscillators per energy level, or energy elements per oscillator. Is the number of microstates the same for both methods of counting?)
Also, in this two-oscillator case, the equality of # of microstates and # of energy elements is only true when both oscillators have a zero-point energy, as you can see if you leave out the ZPE in the energy constraint equation above.
How many microstates does a single oscillator have? Remember, the given energy of the system is not the highest energy available because this is the thermally-isolated-system case, the microcanonical case, and the given energy is the total energy. A single oscillator with constant total energy (with or without ZPE) has one microstate.
As an example for our two oscillators, let n′ = 5. Then i + j = n′ – 1 = 4, and there are five microstates:
i
|
j
|
0
|
4
|
1
|
3
|
2
|
2
|
3
|
1
|
4
|
0
|
Solution to part one: Since there
are n′ microstates, the entropy is
S1 =k log (n′) = k log (E′/ħωo).
where log means natural log. I’ll
come back later to this part of the problem to examine what kind of physical
system this two–oscillator model might represent.
(What if the problem
involved N oscillators without ZPE, a la Dr. Planckenstein in 1900? I’ll look at that case below.)
-----------------------------
Now for part two of the problem:
"A second system is
composed of two harmonic oscillators, each of natural frequency 2ωo. The total energy of this system is E″ = n″ħωo, where n″
is an even integer. How many microstates
are available to this system, and what is its entropy?"
The energy constraint is
now
(i+½) 2ħωo + (j+½)2ħωo = n″ħωo
,
(i+½) + (j +½) = n″/2 ,
i + j = n″/2 – 1,
j = n″/2 – i –
1
Why does Callen give us this second part of the problem? What does it tell us that’s not told in the first part of the problem? This is the same relation between i and j as in part one, but with n″/2 instead of n′. And n″/2 still gives us consecutive positive integers as possible values, since we are told n″ is an even integer. The table of available microstates is
|
i |
j |
|
0 |
n″/2
– 1 |
|
1 |
n″/2
– 2 |
|
2 |
n″/2
– 3 |
|
… |
… |
|
n″/2
– 3 |
2 |
|
n″/2
– 2 |
1 |
|
n″/2
– 1 |
0 |
There are (n″/2
– 1) + 1 = n″/2 microstates. The entropy is
S2 = k log (n″/2).
This part of the problem definitely has some Planck-original-quantum-paper similarity to it, although it's entirely superficial: Planck used different numbers of "primes" on his variables (E, E′, E″, etc., for energy, for instance) to indicate different frequencies ω, ω', ω'', etc. And that's all we have here in part two, a different frequency. But, the big thing with Planck's theory was and is that the energy of a quantum mechanical oscillator is determined by its single, perfectly monochromatic frequency (amazing!) and by integer multiples of that frequency. I think this is the point of Callen's problem. Is the number of microstates going to change just because we have a higher frequency, and thus a higher energy? What's really different about this and part one?
Consider this:
E″ = n″ħωo = (n″/2) 2ħωo,
where the consecutive energy levels for n″/2 are 1, 2, 3, ... , since n″ is even. But this is just like counting the number of energy levels for part one. What we have is n″/2 = n′, and that means
E″ = (n″/2) 2ħωo= n′ 2ħωo
Now the energy constraint
is
(i+½) 2ħωo + (j+½)2ħωo = n′ 2ħωo ,
(i+½) + (j +½) = n′ ,
i + j = n′ – 1,
j =n′ – i –
1
This is the same relation between i and j as in part one. The number of microstates is the same, n′ , so the entropy is the same.
Now for part three of the problem: What
is the entropy of the system composed of the two preceding subsystems if they
are separated and enclosed by a totally restrictive wall?
Since
the subsystems are noninteracting in the most restrictive way possible--they can't exchange energy or particles (oscillators) with each other--their entropies just add together to give
S1 + S2 =
k log (n′) + k log(n″/2)
=
k log (E′/ħωo) + k log(E″/2ħωo)
=
k log (E′ E″/2ħ2ωo2
).
S1 + S2 = k log (n′) + k log(n″/2)
= k log (n′) + k log(n′)
= 2k log(n′)
=2k log(E′/ħωo).
Now that the problem is solved, I'll just relate it to the general problem of N oscillators by using Planck's (and before him, Boltzmann's) combinatorial formula for
(P+N–1)! / P!(N–1)!
In part one of this problem we have N = 2 oscillators and P = n′ energy elements, giving the number of complexions as
(n′ + 2 – 1)!/ n'!1! = (n′ + 1)! / n′! = n′ + 1.
We don't have the 1/2 + 1/2 ZPE terms here. With the zero-point energy included, the minus one as shown above in part one of the problem would cancel the plus one here, and we'd again have n′ microstates.
Now let N=3 oscillators and keep P=n′, no ZPE:
(n′ + 3 – 1)!/ n'!2! = (n′ + 2)! / 2n′! = (n′ + 2)(n′ + 1)/2.
Now let N=4 oscillators:
(n′ + 4 – 1)!/ n'!3! = (n′ + 3)! / 6n′! = (n′ + 3)(n′ + 2)(n′ + 1)/6.
(To be continued in a
later post using Planck’s example with ZPE from his Theory of Heat Radiation, 2nd
edition, 1912.)
Planck included or postulated or uncovered the zero-point energy in a paper published in 1911, and included this derivation in his 1912 book The Theory of Heat Radiation (the 2nd edition, which is very different from the 1st edition of 1906, with ZPE being one of the big differences). Einstein and Stern in their 1913 paper discussed some consequences of ZPE, although its physical meaning caused confusion theoretically until the 1920s when quantum non-commutativity of certain operators representing physical quantities was discovered.
