We claim that the famous exponential decrease of probability with energy is natural, the vastly most probable and disordered state given the constraints on total energy and number of particles. It should be a state of maximum entropy ! The proof involves mathematical techniques beyond the scope of the text, but finding support is good exercise and not difficult. Consider a system of$\mathbf{11}$oscillators sharing a total energy of just$\mathbf{5}{\mathbf{h\omega }}_{\mathbf{0}}$ . In the symbols of Section 9.3. $\mathbf{N}\mathbf{=}\mathbf{11}$and$\mathbf{M}\mathbf{=}\mathbf{5}$ .

1. Using equation$\mathbf{(}\mathbf{9}\mathbf{-}\mathbf{9}\mathbf{)}$ , calculate the probabilities of$\mathbf{n}$ , being$\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}$ and$\mathbf{3}$ .
2. How many particles${\mathbf{N}}_{\mathbf{n}}$ , would be expected in each level? Round each to the nearest integer. (Happily. the number is still $\mathbf{11}$. and the energy still$\mathbf{5}{\mathbf{h\omega }}_{\mathbf{0}}$ .) What you have is a distribution of the energy that is as close to expectations is possible. given that numbers at each level in a real case are integers.
3. Entropy is related to the number of microscopic ways the macro state can be obtained. and the number of ways of permuting particle labels with${\mathbf{N}}_{\mathbf{0}}$ ,${\mathbf{N}}_{\mathbf{1}}\mathbf{,}{\mathbf{N}}_{\mathbf{2}}$ , and ${\mathbf{N}}_3$fixed and totaling$\mathbf{11}$ is$\frac{\mathbf{11}\mathbf{!}}{\mathbf{(}{\mathbf{N}}_{\mathbf{0}}\mathbf{!}{\mathbf{N}}_{\mathbf{1}}\mathbf{!}{\mathbf{N}}_{\mathbf{2}}\mathbf{!}{\mathbf{N}}_{\mathbf{3}}\mathbf{!}\mathbf{)}}$ . (See Appendix J for the proof.) Calculate the number of ways for your distribution.
4. Calculate the number of ways if there were$\mathbf{6}$ particles in$\mathbf{n}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}$ in$\mathbf{n}\mathbf{=}\mathbf{1}$ and none higher. Note that this also has the same total energy.
5. Find at least one other distribution in which the $\mathbf{11}$ oscillators share the same energy, and calculate the number of ways.

6. What do your finding suggests?

Consider a system of $\mathbf{11}$ one-dimensional harmonic oscillators whose total energies are$\mathbf{5}{\mathbf{h\omega }}_{\mathbf{0}}$ . It should commensurate to a system with $\mathbf{N}\mathbf{=}\mathbf{11}$and$\mathbf{M}\mathbf{=}\mathbf{5}$ .

$\mathrm{P}\left(\mathrm{n}\right)=\frac{(\mathrm{M}-\mathrm{n}+\mathrm{N}-2)!}{(\mathrm{M}-\mathrm{n})!(\mathrm{N}-2)!}\frac{(\mathrm{M}+\mathrm{N}-1)!}{\left(\mathrm{M}\right)!(\mathrm{N}-1)!}$ ….(1)

For $\mathrm{n}=0$ , The probability is

$\mathrm{P}(\mathrm{n}=0)=\frac{(5-0+11-2)!}{(5-0)!(11-2)!}\frac{(5+11-1)!}{5!(11-1)!}$

$=\frac{14!}{5!9!}\frac{15!}{5!10!}$

$$\begin{gathered} =\frac{2002}{3003} \\ =0.67 \end{gathered}$$

For $\mathrm{n}=1$, The probability is

$$\begin{gathered} \mathrm{P}(\mathrm{n}=1)=\frac{(5-1+11-2)!}{(5-1)!(11-2)!}\frac{(5+11-1)!}{5!(11-1)!} \\ =\frac{13!}{4!9!}\frac{15!}{5!10!} \\ =\frac{715}{3003} \\ =0.24 \end{gathered}$$

For $\mathrm{n}=2$, The probability is

$$\begin{gathered} \mathrm{P}(\mathrm{n}=2)=\frac{(5-2+11-2)!}{(5-2)!(11-2)!}\frac{(5+11-1)!}{5!(11-1)!} \\ =\frac{12!}{3!9!}\frac{15!}{5!10!} \\ =\frac{220}{3003} \\ =0.07 \end{gathered}$$

For $\mathrm{n}=3$ , The probability is

$$\begin{gathered} \mathrm{P}(\mathrm{n}=3)=\frac{(5-3+11-2)!}{(5-3)!(11-2)!}\frac{(5+11-1)!}{5!(11-1)!} \\ =\frac{11!}{2!9!}\frac{15!}{5!10!} \\ =\frac{55}{3003} \\ =0.018 \end{gathered}$$

b.

$\mathrm{N}\left(\mathrm{n}\right)=\mathrm{NP}\left(\mathrm{n}\right)$ ……(2)

For $\mathrm{n}=0$, the expected number of particles is

$$\begin{gathered} \mathrm{N}(\mathrm{n}=0)=11(0.67) \\ =7.37 \\ \approx 7 \end{gathered}$$

.

For $\mathrm{n}=1$, the expected number of particles is

$\mathrm{N}(\mathrm{n}=1)=11(0.24)$

$=2.64$

$\approx 3$

For $\mathrm{n}=2$, the expected number of particles is

$\mathrm{N}(\mathrm{n}=2)=11(0.07)$

$=0.77$

$\approx 1$

.

For $\mathrm{n}=3$ , the expected number of particles is

$\mathrm{N}(\mathrm{n}=3)=11(0.018)$

$=0.198$

$\approx 0$

C.

$$\begin{gathered} \mathrm{W}=\frac{\mathrm{N}!}{{\mathrm{N}}_0!{\mathrm{N}}_1!{\mathrm{N}}_2!{\mathrm{N}}_3!} \\ \mathrm{W}=\frac{11!}{7!3!1!0!} \\ =1320 \end{gathered}$$

(3), the number of ways to obtain such a distribution must be

$\begin{array}{rcl}\mathrm{W}& =& \frac{11!}{6!5!0!0!}\\ & =& 462\end{array}$

$$\begin{gathered} \mathrm{W}=\frac{11!}{7!4!0!0!} \\ =330 \end{gathered}$$

Another possible distribution.

${\mathrm{N}}_0=5,{\mathrm{N}}_1=4,{\mathrm{N}}_2=2,{\mathrm{N}}_3=0$.

The number of ways is:$$\begin{gathered} \mathrm{W}=\frac{11!}{5!4!2!0!} \\ =6930 \end{gathered}$$

Let us now attempt to evenly fill all levels. A possible distribution is ${\mathrm{N}}_0=4,{\mathrm{N}}_1=3,$${\mathrm{N}}_2=2,{\mathrm{N}}_3=2$. Although it does not appear to be possible for the $\mathrm{n}=2$ and $\mathrm{n}=3$energy levels to have identical probabilities, let us just do so for the sake of finding a trend between and how the energy can be distributed.

The number of ways is:$\begin{array}{c}\mathrm{W}=\frac{11!}{4!3!2!2!}\\ =6.93\times {10}^4\end{array}$

From our results, we can observe that the probability is decreasing with the increase in the value and also the numbers of particles are decreasing with the increase in the value of n.

Since the entropy is inversely proportional to the number of particle, with the decrease in the number of particle, the entropy will increase. So, the claim that the famous exponential decrease of probability with energy is natural, the vastly most probable and disordered state given the constraints on total energy and number of particles is justified.