Question

Use a computer to produce a table and graph, like those in this section, for the case where one Einstein solid contains $200$ oscillators, the other contains$100$ oscillators, and there are $100$ units of energy in total. What is the most probable macrostate, and what is its probability? What is the least probable macrostate, and what is its probability?

Short answer

The most likely macrostate is when $67$energy units out of the total energy units are in solid $\mathrm{A},{\mathrm{q}}_{\mathrm{A}}=67$and ${\mathrm{q}}_{\mathrm{B}}=33$, with a probability of $0.07315$; the least likely macrostate is when all energy units are in partition${\mathrm{Bq}}_{\mathrm{A}}=0$or when ${\mathrm{q}}_{\mathrm{B}}=100$, with a probability of $2.6926\times {10}^{-37}$.

Step-by-step solution

Calculation for overall multiplicity

Probability is,

$\mathrm{P}\left({\mathrm{q}}_{\mathrm{A}}\right)=\frac{{\mathrm{\Omega }}_{\mathrm{A}}{\mathrm{\Omega }}_{\mathrm{B}}}{{\mathrm{\Omega }}_{\text{overall}}}$

Over all multiplicity is,

${\mathrm{\Omega }}_{\text{overall}}\left({\mathrm{N}}_{\text{overall}},{\mathrm{q}}_{\text{overall}}\right)=\left(\begin{array}{c}{\mathrm{q}}_{\text{overall}}+{\mathrm{N}}_{\text{overall}}-1\\ {\mathrm{q}}_{\text{overall}}\end{array}\right)$

$=\frac{\left({\mathrm{q}}_{\text{overall}}+{\mathrm{N}}_{\text{overall}}-1\right)!}{{\mathrm{q}}_{\text{overall}}!\left({\mathrm{N}}_{\text{overall}}-1\right)!}$

Where,

${\mathrm{q}}_{\text{overall}}={\mathrm{q}}_{\mathrm{A}}+{\mathrm{q}}_{\mathrm{B}}=100$

${\mathrm{N}}_{\text{overall}}={\mathrm{N}}_{\mathrm{A}}+{\mathrm{N}}_{\mathrm{B}}=300$

Then,

${\mathrm{\Omega }}_{\text{overall}}=\left(\begin{array}{c}100+300-1\\ 100\end{array}\right)$

$=\frac{(100+300-1)!}{100!(300-1)!}$

$=1.681\times {10}^{96}$

Calculation for total multiplicity and probability

Total multiplicity is,

${\mathrm{\Omega }}_{\text{total}}={\mathrm{\Omega }}_{\mathrm{A}}{\mathrm{\Omega }}_{\mathrm{B}}$

${\mathrm{N}}_{\mathrm{A}}=200$

${\mathrm{\Omega }}_{\mathrm{A}}=\left(\begin{array}{c}{\mathrm{q}}_{\mathrm{A}}+199\\ {\mathrm{q}}_{\mathrm{A}}\end{array}\right)$

$=\frac{\left({\mathrm{q}}_{\mathrm{A}}+199\right)!}{{\mathrm{q}}_{\mathrm{A}}!(199!)}$

${\mathrm{\Omega }}_{\mathrm{B}}=\left(\begin{array}{c}{\mathrm{q}}_{\mathrm{B}}+{\mathrm{N}}_{\mathrm{B}}-1\\ {\mathrm{q}}_{\mathrm{B}}\end{array}\right)$

For ${\mathrm{N}}_{\mathrm{B}}=100$,${\mathrm{q}}_{\mathrm{B}}=100-{\mathrm{q}}_{\mathrm{A}}$

${\mathrm{\Omega }}_{\mathrm{B}}=\left(\begin{array}{c}100-{\mathrm{q}}_{\mathrm{A}}+100-1\\ 100-{\mathrm{q}}_{\mathrm{A}}\end{array}\right)$

$=\frac{\left(199-{\mathrm{q}}_{\mathrm{A}}\right)!}{\left(100-{\mathrm{q}}_{\mathrm{A}}\right)!\left(99\right)!}$

probability is,

$\mathrm{P}\left({\mathrm{q}}_{\mathrm{A}}\right)=\frac{{\mathrm{\Omega }}_{\mathrm{A}}{\mathrm{\Omega }}_{\mathrm{B}}}{{\mathrm{\Omega }}_{\text{overall}}}$

$=\frac{1}{1.681\times {10}^{96}}\frac{\left({\mathrm{q}}_{\mathrm{A}}+199\right)!}{{\mathrm{q}}_{\mathrm{A}}!(199!)}\frac{\left(199-{\mathrm{q}}_{\mathrm{A}}\right)!}{\left(100-{\mathrm{q}}_{\mathrm{A}}\right)!\left(99\right)!}$

$\mathrm{P}\left({\mathrm{q}}_{\mathrm{A}}\right)=\frac{1}{1.681\times {10}^{96}}\frac{\left({\mathrm{q}}_{\mathrm{A}}+199\right)!}{{\mathrm{q}}_{\mathrm{A}}!(199!)}\frac{\left(199-{\mathrm{q}}_{\mathrm{A}}\right)!}{\left(100-{\mathrm{q}}_{\mathrm{A}}\right)!\left(99\right)!}$

Python program for creation of graph

Graph for probability versus energy