A thermodynamic system doesn’t have to be big. Although thermodynamics was originally concerned with very large objects like steam engines for pumping out coal mines, thermodynamic thinking can equally well be applied to very small systems consisting of say, just a few atoms.

Of course, we know that very small systems play by different rules – namely quantum rules – but that’s ok. The rules are known and can be applied. So let’s imagine that our thermodynamic system is an idealized solid consisting of three atoms, each distinguishable from the others by its unique position in space, and each able to perform simple harmonic oscillations independently of the others. At the absolute zero of temperature, the system will have no thermal energy, one microstate and zero entropy, with each atom in its vibrational ground state.

Harmonic motion is quantized, such that if the energy of the ground state is taken as zero and the energy of the first excited state as ε, then 2ε is the energy of the second excited state, 3ε is the energy of the third excited state, and so on. Suppose that from its thermal surroundings our 3-atom system absorbs one unit of energy ε, sufficient to set one of the atoms oscillating. Clearly, one unit of energy can be distributed among three atoms in three different ways – 100, 010, 001 – or in more compact notation [100|3].

Now let’s consider 2ε of absorbed energy. Our system can do this in two ways, either by promoting one oscillator to its second excited state, or two oscillators to their first excited state. Each of these energy distributions can be achieved in three ways, which we can write [200|3], [110|3]. For 3ε of absorbed energy, there are three distributions: [300|3], [210|6], [111|1].

Summarizing the above information

Energy E (in units of ε) | Total microstates W | Ratio of successive W’s |

0 | 1 | |

1 | 3 | 3 |

2 | 6 | 2 |

3 | 10 | 1⅔ |

The summary shows that as E increases, so does W. This is to be expected, since as W increases, the entropy S (= k log W) increases. In other words E and S increase or decrease together; the ratio ∂E/∂S is always positive. Since ∂E/∂S = T, the finding that E and S increase or decrease together is equivalent to saying that the absolute temperature of the system is always positive.

**– – – –**

**Adding an extra particle**

It is instructive to compare the distribution of energy among three oscillators (N =3)*

E = 0: [000|1]

E = 1: [100|3]

E = 2: [200|3], [110|3]

E = 3: [300|3], [210|6], [111|1]

with the distribution among four oscillators (N = 4)*

E = 0: [0000|1]

E = 1: [1000|4]

E = 2: [2000|4], [1100|6]

E = 3: [3000|4], [2100|12], [1110|4]

*For any single distribution among N oscillators where n_{0}, n_{1},n_{2} … represent the number of oscillators in the ground state, first excited state, second excited state etc, the number of microstates is given by

It is understood that 0! = 1. Derivation of the formula is given in Appendix I.

For both the 3-oscillator and 4-oscillator systems, the first excited state is never less populated than the second, and the second excited state is never less populated than the third. Population is graded downward and the ratios n_{1}/n_{0} > n_{2}/n_{1} > n_{3}/n_{2} are less than unity.

Example calculations for N = 4, E = 3:

Comparisons can also be made of a single ratio across distributions and between systems. For example the values of n_{1}/n_{0} for E = 0, 1, 2, 3 are

(N = 4) : 0, ⅓, ½, ⅗

(N = 3) : 0, ½, ⅔, ¾

Since for a macroscopic system

this implies that for a given value of E the 4-oscillator system is colder than the 3-oscillator system. The same conclusion can be reached by looking at the ratio of successive W’s for the 4-oscillator system sharing 0 to 3 units of thermal energy

Energy E (in units of ε) | Total microstates W | Ratio of successive W’s |

0 | 1 | |

1 | 4 | 4 |

2 | 10 | 2½ |

3 | 20 | 2 |

For the 4-oscillator system the ratios of successive W’s are larger than the corresponding ratios for the 3-oscillator system. The logarithms of these ratios are inversely proportional to the absolute temperature, so the larger the ratio the lower the temperature.

**– – – –**

**Finite differences**

The differences between successive W’s for a 4-oscillator system are the values for a 3-oscillator system

W for (N =4) : 1, 4, 10, 20

Differences : 3, 6, 10

Likewise the differences between successive W’s for a 3-oscillator system and a 2-oscillator system

W for (N =3) : 1, 3, 6, 10

Differences : 2, 3, 4

Likewise for the differences between successive W’s for a 2-oscillator system and a 1-oscillator system

W for (N =2) : 1, 2, 3, 4

Differences : 1, 1, 1

This implies that W for the 4-particle system can be expressed as a cubic in n, and that W for the 3-particle system can be expressed as a quadratic in n etc. Evaluation of coefficients leads to the following formula progression

For N = 1

For N = 2

For N = 3

For N = 4

It appears that in general

Since n = E/ε and ε = hν, the above equation can be written

For a system of oscillators this formula describes the functional dependence of W microstates on the size of the particle ensemble (N), its energy (E), the mechanical frequency of its oscillators (ν) and Planck’s constant (h).

**– – – –**

**Appendix I**

**Formula to be derived**

For any single distribution among N oscillators where n_{0}, n_{1},n_{2} … represent the number of oscillators in the ground state, first excited state, second excited state etc, the number of microstates is given by

**Derivation**

In combinatorial analysis, the above comes into the category of permutations of sets with the possible occurrence of indistinguishable elements.

Consider the distribution of 3 units of energy across 4 oscillators such that one oscillator has two units, another has the remaining one unit, and the other two oscillators are in the ground state: {2100}

If each of the four numbers was distinct, there would be 4! possible ways to arrange them. But the two zeros are indistinguishable, so the number of ways is reduced by a factor of 2! The number of ways to arrange {2100} is therefore 4!/2! = 12.

1 and 2 occur only once in the above set, and the occurrence of 3 is zero. This does not result in a reduction in the number of possible ways to arrange {2100} since 1! = 1 and 0! = 1. Their presence in the denominator will have no effect, but for completeness we can write

4!/2!1!1!0!

to compute the number of microstates for the single distribution E = 3, N = 4, {2100} where n_{0} = 2, n_{1} = 1, n_{2} = 1 and n_{3} = 0.

In the general case, the formula for the number of microstates for a single energy distribution of E among N oscillators is

where the terms in the denominator are as defined above.

**– – – –**

P Mander April 2016

Lower states are never less populated than higher states because there are always more ways to distribute a given amount of quantized energy in smaller packets than in bigger packets. It’s a statistical thing I guess.

Question: Why is it that the first excited state has to be at least as populated as the second and so on?