# Combinatorics and new Bose-Einstein Statistics Problem

Can you check my solution? Is this answer correct?

As there are four indistinct particles, and their sum must add 4, I am considering that exist the following partitions:

(4,0,0,0) implies C(17,1)= 17

(3,1,0,0) implies C(10,1)*C(2,1)= 20

(2,2,0,0) implies C(5,1)*C(5,1)= 25

(2,1,1,0) implies C(5,1)*C(2,1)*C(2,1)= 20

(1,1,1,1) implies C(2,1)*C(2,1)*C(2,1)*C(2,1)= 16

Finally, 17 + 20 + 25 + 20 + 16 = 98.

Winther, who is a postdoc in theoretical astrophysics at the ICG in Portsmouth, gave me a "translation" of the problem:

Translating the problem to a more real world scenario might help: there are four wallets in a bag and each wallet contains from zero to four $1$ notes. The total money in the wallets combined is $4$.

Your questions in terms of this analogy: "Is the energy of each particle the same?" = "Do we have the same money in all wallets?" Not necessarily.

"some of the energy of the particles can cancel" = "some of notes cancel each other" No they can't. "Are we applying Pauli's principle?" No, bosons do not satisfy this principle.

Here is the problem: Bose Statistics Problem

5) A physical system consists of four identical particles. The total energy of the system is $4E_0>0$. Each of the particles can have an energy level equal to $kE_0$ for $k\in\{0,1,2,3,4\}$. A particle with energy $kE_0$ can occupy any one of the $k^2+1$ distinct energy states at that energy level. How many distinct energy configurations can the system have? [In statistical mechanics the particles are said to obey Bose-Einstein statistics.]

After my "research" in many physics websites, I found this:

Consider an energy level $\varepsilon_i$ with degeneracy $g_i$, containing $n_i$ bosons. The states may be represented by $g_i-1$ lines, and the bosons by $n_i$ circles; distinguishable microstates correspond to different orderings of the lines and circles. For example, with $9$ particles in the $8$ states corresponding to a particular energy, a particular microstate might be: The number of distinct orderings of lines and circles is: $$t_i = \frac{(n_i+g_i-1)!}{n_i!(g_i-1)!}.$$

A particular distribution has a specified number of particle $n_i$ within each of the possible energy levels $\varepsilon_i$. The total number of microstates for a given distribution is therefore: $$t(\{n_i\}) = \prod_i \frac{(n_i+g_i-1)!}{n_i!(g_i-1)!}.\tag 8$$

I also found a couple of Bose-Einstein Statistics problems:

A combinatorics problem related to Bose-Einstein statistics

Indistinguishable particles: obeing "Bose-Einstein Statistics"

• Which part are you lost with? Which parts do you understand? – Mark May 28 '17 at 21:15
• This is the stars and bars problem. Have a look here: en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) – jvdhooft May 28 '17 at 21:15
• Please type out images and use MathJax. – Em. May 28 '17 at 21:39

This is a guide of what you have regarding the binomial coefficient $t_i$.

The binomial in $t_i$ is basically the problem of arranging a word of $n_i+g_i-1$ letters, from which $n_i$ letters are, say A (or bosons $b$), and $g_i-1$ letters are, say B (or boxes $b$). What you have then is a Mississippi problem (very well worked out in the early part of combinatorics courses and books). Martin's book Counting: the art of enumerative combinatorics can help you to understand these concepts.

Now to the problem. You can form $N!$ words from $N$ distinct letters. Consider now that the letters are not all different, and in fact that you have only two types of letters, say $n_A$ $A$'s and $n_b$ $B$'s such that $n_A+n_B=N$. Then $N!$ is an overcounting of all the possible distinct words that you can form (try this with, say $3$ $A$'s and $2$ $B$'s so that you can convince yourself of this fact). By the division principle, the total number of words of length $N$ that you can form with $n_A$ $A$'s and $n_b$ $B$'s is

$$\frac{(n_A+n_B)!}{n_A!n_B!}.$$

Now it is easy to say that there are $n_A$ bosons and $n_B+1$ boxes (you need essentially $n_B$ "bars" to form $n_b+1$ boxes).

• First one comment, the total energy must be $4E_0>0$. This means that the particles cannot have $k=0$ (there must be at least one with $k=1$). It doesn't mean that the energy must add up to $4$. Where did you get this problem? Was it from a book? – user2820579 May 30 '17 at 17:21
• I believe that the problem says that the the total energy must add 4, this I think is incorrect. It is different to say that the energy is $4E_0$, as you state it, than to say that it must be $4E_0>0$, or $E_0>0$. Suppose each of the particles has energy $E_0$. The total energy is thus $E_{total}=E_0+E_0+E_0+E_0=4E_0$, thus $E_{total}=4E>0$ (one of the cases you stated it). Now suppose one particle has enegy $2E_0$, and all the others has energy zero, therefor $E_{total}=2E_0>0$, a valid option also. – user2820579 May 30 '17 at 18:05
• Perhaps you can ask if $E_{total}=4E_0$, or that $E_{total}>0$. These are different conditions. Can you notice the difference? – user2820579 May 30 '17 at 18:10
• If you keep $E_{total}=4E_0$, then it's ok what you are doing. – user2820579 May 30 '17 at 18:16
• Sorry, I had to correct the last comment. – user2820579 May 30 '17 at 18:17

The reasoning is almost correct, however it seems that somewhere in between occurred a confusion, which spoiled the whole calculation.

Let's investigate the problem systematically.

1. Every particle, which resides on some particular energy level, has the energy of that level.
2. The energy of every $i-$th energy level is: $i \cdot E_0$.
3. The total energy of the system is a sum of energies of all particles.
4. Let's consider different distributions of particles between different energy levels (let's call such distributions "macrostates"). In the context of the given problem each macrostate can be defined by a tuple of five numbers: $(k_0, k_1, k_2, k_3, k_4)$, where $k_i$ represents an amount of particles on the $i-$th energy level. All macrostates have to satisfy the following conditions $\label{eq:1}$: $$\sum_{i=0}^{4} k_i = 4 \\ \sum_{i=0}^{4} k_i \cdot (i \cdot E_0) = 4E_0$$
5. Every energy level contains some amount of distinct energy states (so-called "degeneration"). It means, that for every $i-$th energy level there exists a certain amount of different placements of $k_i$ particles within $g_i = (i^2 + 1)$ distinct energy states. Such variations within the macrostates are called "microstates". Let's denote $\Omega_{k_0 k_1 k_2 k_3 k_4}$ as an amount of all microstates of the macrostate $(k_0, k_1, k_2, k_3, k_4)$.
6. An amount of distinct energy configurations of the system is equal to the sum of amounts of all microstates over all macrostates: $$\sum_{(k_0, k_1, k_2, k_3, k_4)} \Omega_{k_0 k_1 k_2 k_3 k_4}$$ Where $k_0, k_1, k_2, k_3, k_4$ satisfy conditions of the problem (see equations from the paragraph 4).
7. As far as all particles are obeying Bose-Einstein statistics, the total amount of microstates of the macrostate $(k_0, k_1, k_2, k_3, k_4)$ can be calculated as follows (as already discussed in the question): $$\Omega_{k_0 k_1 k_2 k_3 k_4} = \prod_{i = 0}^{4} {k_i + g_i - 1 \choose g_i - 1} = \prod_{i = 0}^{4} {k_i + (i^2 + 1) - 1 \choose (i^2 + 1) - 1} = \prod_{i = 0}^{4} {k_i + i^2 \choose i^2}$$

There are 5 different macrostates $(k_0, k_1, k_2, k_3, k_4)$, which satisfy the conditions of the problem: $$(3, 0, 0, 0, 1), \\ (2, 1, 0, 1, 0), \\ (2, 0, 2, 0, 0), \\ (1, 2, 1, 0, 0), \\ (0, 4, 0, 0, 0)$$
Amounts of microstates of every macrostate are following: $$\Omega_{30001} = {3 + 0^2 \choose 0^2} \cdot {1 + 4^2 \choose 4^2} = 17 \\ \Omega_{21010} = {2 + 0^2 \choose 0^2} \cdot {1 + 1^2 \choose 1^2} \cdot {1 + 3^2 \choose 3^2} = 20 \\ \Omega_{20200} = {2 + 0^2 \choose 0^2} \cdot {2 + 2^2 \choose 2^2} = 15 \\ \Omega_{12100} = {1 + 0^2 \choose 0^2} \cdot {2 + 1^2 \choose 1^2} \cdot {1 + 2^2 \choose 2^2} = 15 \\ \Omega_{04000} = {4 + 1^2 \choose 1^2} = 5 \\$$
Hence the total amount of distinct energy configurations is: $$Z = \Omega_{30001} + \Omega_{21010} + \Omega_{20200} + \Omega_{12100} + \Omega_{04000} = 17 + 20 + 15 + 15 + 5 = 72$$
Also, having the presented calculations in mind, it turns out that the most probable macrostate is $(2,1,0,1,0)$, with probability: $P_{21010} = {\Omega_{21010} \over Z} = {20 \over 72}$.