Question

(a) In Example, $\mathbf{5}$a mathematical model is discussed which claims to give a distribution of identical balls into boxes in such a way that all distinguishable arrangements are equally probable (Bose-Einstein statistics). Prove this by showing that the probability of a distribution of N balls into n boxes (according to this model) with ${\mathit{N}}_{\mathbf{1}}$ balls in the first box, ${\mathit{N}}_{\mathbf{2}}$in the second, ··· , ${\mathit{N}}_{\mathbf{n}}$ in the${\mathit{n}}^{\mathbf{t}\mathbf{h}}$ , is$\frac{\mathbf{1}}{\mathbf{C}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{N}\mathbf{,}\mathbf{N}\mathbf{)}}$ for any set of numbers Ni such that${\mathit{N}}_{\mathbf{i}}\sum _{i=1}^n{N}_i=N$.

b) Show that the model in (a) leads to Maxwell-Boltzmann statistics if the drawn card is replaced (but no extra card added) and to Fermi-Dirac statistics if the drawn card is not replaced. Hint: Calculate in each case the number of possible arrangements of the balls in the boxes. First do the problem of $\mathbf{4}$particles in $\mathbf{6}$boxes as in the example, and then do N particles in n boxes ($\mathit{n}\mathbf{>}\mathit{N}$ ) to get the results in Problem$\mathbf{19}$ .

Short answer

The required values are given below.

$\begin{array}{c}P=\frac{1}{C\left(n+N-1\right)}\\ P_{MB}=\frac{N!}{n^N\cdot N_1!N_2!\cdots N_n!}\\ P_{FD}=\frac{1}{C(n,N)}\end{array}$

Step-by-step solution

Given Information

Distribution of identical balls into boxes.

 Step 2: Definition of uniform sample spaces.

If a given experiment's sample space is known to be uniform, the probability of an event can be calculated using the event sizes and the sample space.

Find the values of part(a). 

There are N balls that need to be put in n boxes.

It can be done in $C\left(n+N-1\right)$ways.

Hence the probability is mentioned below.

$P=\frac{1}{C\left(n+N-1\right)}$

Step 3: Prove part(b).

The probability that first ball has$N_1$ cards and second ball has$N_2$cardsand so on is.

$\frac{N!}{N_1!N_2!\cdots N_n!}$

The Maxwell Boltzmann equation becomes as follows.

$\begin{array}{c}{P}_{MB}={\left(\frac{1}{n}\cdots \frac{1}{n}\right)}^N\cdot \frac{N!}{N_1!N_2!\cdots N_n!}\\ =\frac{N!}{n^N\cdot N_1!N_2!\cdots N_n!}\end{array}$

The Fermi-dericequation becomes as follows.

$\begin{array}{c}{P}_{FD}=({\left.\frac{1}{n}\cdot \frac{1}{n-1}\cdots \frac{1}{n-N+1}\right)}^N\cdot \frac{N!}{N_1!N_2!\cdots N_n!}\\ =\frac{N!}{n^N\cdot N_1!N_2!\cdots N_n!}\\ =\frac{1}{C(n,N)}\end{array}$

Hence, the required valuesare given below.

$\begin{array}{c}P=\frac{1}{C\left(n+N-1\right)}\\ P_{MB}=\frac{N!}{n^N\cdot N_1!N_2!\cdots N_n!}\\ P_{FD}=\frac{1}{C(n,N)}\end{array}$