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Q. 2.11

Expert-verifiedFound in: Page 60

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

Use a computer to produce a table and graph, like those in this section, for two interacting two-state paramagnets, each containing $100$ elementary magnetic dipoles. Take a "unit" of energy to be the amount needed to flip a single dipole from the "up" state (parallel to the external field) to the "down" state (antiparallel). Suppose that the total number of units of energy, relative to the state with all dipoles pointing up, is$80$; this energy can be shared in any way between the two paramagnets. What is the most probable macrostate, and what is its probability? What is the least probable macrostate, and what is its probability?

The most likely macrostate is when the energy units are evenly distributed,${\mathrm{q}}_{\mathrm{A}}={\mathrm{q}}_{\mathrm{B}}=40$ , with a probability of $0.07513$. The least likely state is when all the energy units are in partition $\mathrm{B}$or$\mathrm{A},{\mathrm{q}}_{\mathrm{A}}=40$ , or when ${\mathrm{q}}_{\mathrm{B}}=40$, with a chance of $7.8726\times {10}^{-20}$.

The probability of $\mathrm{P}\left({\mathrm{q}}_{\mathrm{A}}\right)$ is,

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

The overall multiplicity is,

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

The total multiplicity is ,

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

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

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

Multiplicity is,

${\mathrm{\Omega}}_{\text{overall}}=\frac{\left({\mathrm{q}}_{\text{overall}}+{\mathrm{N}}_{\text{overall}}-1\right)!}{{\mathrm{q}}_{\text{overall}}!\left({\mathrm{N}}_{\text{overall}}-1\right)!}$${\mathrm{q}}_{\text{overall}}={\mathrm{q}}_{\mathrm{A}}+{\mathrm{q}}_{\mathrm{B}}=80$

${\mathrm{N}}_{\text{overall}}={\mathrm{N}}_{\mathrm{A}}+{\mathrm{N}}_{\mathrm{B}}=200$

So,

role="math" localid="1650306409339" ${\mathrm{\Omega}}_{\text{overall}}=\frac{(80+200-1)!}{80!(200-1)!}$

$=2.1225\times {10}^{71}$

Multiplicity of $\mathrm{A}$ is,

${\mathrm{\Omega}}_{\mathrm{A}}=\left(\begin{array}{c}{\mathrm{q}}_{\mathrm{A}}+99\\ {\mathrm{q}}_{\mathrm{A}}\end{array}\right)=\frac{\left({\mathrm{q}}_{\mathrm{A}}+99\right)!}{{\mathrm{q}}_{\mathrm{A}}!\left(99\right)!}$

Multiplicity of $\mathrm{B}$is,

${\mathrm{\Omega}}_{\mathrm{B}}=\left(\begin{array}{c}{\mathrm{q}}_{\mathrm{B}}+99\\ {\mathrm{q}}_{\mathrm{B}}\end{array}\right)$

Substitute ${\mathrm{q}}_{\mathrm{B}}=80-\mathrm{qA}$

so,

role="math" localid="1650306390399" ${\mathrm{\Omega}}_{\mathrm{B}}=\left(\begin{array}{c}179-{\mathrm{q}}_{\mathrm{A}}\\ 80-{\mathrm{q}}_{\mathrm{A}}\end{array}\right)$

$=\frac{\left(179-{\mathrm{q}}_{\mathrm{A}}\right)!}{\left(80-{\mathrm{q}}_{\mathrm{A}}\right)!\left(99\right)!}$

Probability is,

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

$=\frac{1}{2.1225\times {10}^{71}}\frac{\left({\mathrm{q}}_{\mathrm{A}}+99\right)!}{{\mathrm{q}}_{\mathrm{A}}!\left(99\right)!}\frac{\left(179-{\mathrm{q}}_{\mathrm{A}}\right)!}{\left(80-{\mathrm{q}}_{\mathrm{A}}\right)!\left(99\right)!}$

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