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Answers without the blur. Sign up and see all textbooks for free! Q. 2.19

Expert-verified Found in: Page 64 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # Use Stirling's approximation to find an approximate formula for the multiplicity of a two-state paramagnet. Simplify this formula in the limit ${N}_{↓}\ll N$to obtain $\Omega \approx {\left(Ne/{N}_{↓}\right)}^{{N}_{↓}}$. This result should look very similar to your answer to Problem $2.17$; explain why these two systems, in the limits considered, are essentially the same.

The two state of paramagnet and the system is formaly equivalent to an Einstein solid

$\Omega \approx {\left[\frac{Ne}{{N}_{↓}}\right]}^{{N}_{↓}}$

See the step by step solution

## Step 1 Two state of paramagnet

Consider two-state paramagnet with $N$ magnetic dipoles and ${N}_{↓}$ energy quanta, since the system is formally equivalent to an Einstein solid (we're distributing the energy quanta among dipoles rather than oscillators). The multiplicity of the paramagnet is then:

$\Omega \left({N}^{\text{'}},{N}_{↓}\right)=\left(\begin{array}{c}{N}_{↓}+N-1\\ {N}_{↓}\end{array}\right)$

Writing out the binomial coefficient:

$\Omega \left(N,{N}_{↓}\right)=\left(\begin{array}{c}{N}_{↓}+N-1\\ {N}_{↓}\end{array}\right)=\frac{\left({N}_{↓}+N-1\right)!}{{N}_{↓}!\left(N-1\right)!}$

since $N>>N↓$, and they are large numbers, so:

$\Omega \left(N,{N}_{↓}\right)=\frac{\left({N}_{↓}+N\right)!}{{N}_{↓}!N!}$

take the natural logarithms for both sides:

$\mathrm{ln}\left(\Omega \right)=\mathrm{ln}\left(\frac{\left({N}_{↓}+N\right)!}{{N}_{↓}!N!}\right)$

but, we have these logarithmic relations:

$\mathrm{ln}\left(\frac{a}{b}\right)=\mathrm{ln}\left(a\right)-\mathrm{ln}\left(b\right)\mathrm{ln}\left(ab\right)=\mathrm{ln}\left(a\right)+\mathrm{ln}\left(b\right)$

## Step 2 Logarithm

$\mathrm{ln}\left(\Omega \right)=\mathrm{ln}\left(\frac{\left({N}_{↓}+N\right)!}{{N}_{↓}!N!}\right)=\mathrm{ln}\left({N}_{↓}+N\right)!-\mathrm{ln}\left(N!\right)-\mathrm{ln}\left({N}_{↓}!\right)$

use Stirling's approximation for the logarithm of a factorial:

$\mathrm{ln}\left(n!\right)\approx n\mathrm{ln}\left(n\right)-n$

so,we get:

$\mathrm{ln}\left(\Omega \right)\approx \left({N}_{↓}+N\right)\mathrm{ln}\left({N}_{↓}+N\right)-\left({N}_{↓}+N\right)-N\mathrm{ln}\left(N\right)+N-{N}_{↓}\mathrm{ln}\left({N}_{↓}\right)+{N}_{↓}$

$\mathrm{ln}\left(\Omega \right)\approx \left({N}_{↓}+N\right)\mathrm{ln}\left({N}_{↓}+N\right)-N\mathrm{ln}\left(N\right)-{N}_{↓}\mathrm{ln}\left({N}_{↓}\right)$

If we now use the assumption that $N>>{N}_{↓}$, we get:

$\mathrm{ln}\left(\Omega \right)\approx \left({N}_{↓}+N\right)\mathrm{ln}\left[N\left(\frac{{N}_{↓}}{N}+1\right)\right]-N\mathrm{ln}\left(N\right)-{N}_{↓}\mathrm{ln}\left({N}_{↓}\right)$

we factored out $N$ from the first logarithm:

$\mathrm{ln}\left(\Omega \right)\approx \left({N}_{↓}+N\right)\left[\mathrm{ln}\left(N\right)+\mathrm{ln}\left(\frac{{N}_{↓}}{N}+1\right)\right]-N\mathrm{ln}\left(N\right)-{N}_{↓}\mathrm{ln}\left({N}_{↓}\right)$

but, we have the logarithmic relation:

$\mathrm{ln}\left(\frac{a}{b}+1\right)\approx \frac{a}{b}$

for $b>>a$, so:

## Step 3 To neglect second term

$\mathrm{ln}\left(\Omega \right)\approx {N}_{↓}\mathrm{ln}\left(N\right)+{N}_{↓}\frac{{N}_{↓}}{N}+N\mathrm{ln}\left(N\right)+N\frac{{N}_{↓}}{N}-N\mathrm{ln}\left(N\right)-{N}_{↓}\mathrm{ln}\left({N}_{↓}\right)$$\mathrm{ln}\left(\Omega \right)\approx {N}_{↓}\mathrm{ln}\left(N\right)+\frac{{\left({N}_{↓}\right)}^{2}}{N}+{N}_{↓}-{N}_{↓}\mathrm{ln}\left({N}_{↓}\right)$

$\mathrm{ln}\left(\Omega \right)\approx \left({N}_{↓}+N\right)\left[\mathrm{ln}\left(N\right)+\frac{{N}_{↓}}{N}\right]-N\mathrm{ln}\left(N\right)-{N}_{↓}\mathrm{ln}\left({N}_{↓}\right)$

$\mathrm{ln}\left(\Omega \right)\approx {N}_{↓}\left[\mathrm{ln}\left(N\right)-\mathrm{ln}\left({N}_{↓}\right)\right]+\frac{{\left({N}_{↓}\right)}^{2}}{N}+{N}_{↓}$

but, =$\mathrm{ln}\left(\Omega \right)\approx {N}_{↓}\mathrm{ln}\left[\frac{N}{{N}_{↓}}\right]+{N}_{↓}$, so:

$\mathrm{ln}\left(\Omega \right)\approx {N}_{↓}\mathrm{ln}\left[\frac{N}{{N}_{↓}}\right]+\frac{{\left({N}_{↓}\right)}^{2}}{N}+{N}_{↓}$

use the assumption that$N>>{N}_{↓}$, to neglect the second term:

$\mathrm{ln}\left(\Omega \right)\approx {N}_{↓}\mathrm{ln}\left[\frac{N}{{N}_{↓}}\right]+{N}_{↓}$

## Step 4 Exponentiating both sides

Exponentiating both sides this equation gives the approximate value for $\Omega$, so:

${e}^{\left(\mathrm{ln}\left(\Omega \right)\right)}\approx {e}^{\left({N}_{↓}\mathrm{ln}\left[\frac{N}{{N}_{↓}}\right]+{N}_{↓}\right)}\phantom{\rule{0ex}{0ex}}\Omega \approx {e}^{\left(\mathrm{ln}\left[\frac{N}{{N}_{↓}}\right]\right)}{e}^{\left({N}_{↓}\right)}\phantom{\rule{0ex}{0ex}}\Omega \approx \left[{e}^{{\left(\mathrm{ln}\left[\frac{N}{{N}_{↓}}\right]\right)]}^{{N}_{↓}}}{e}^{\left({N}_{↓}\right)}\right\$

but,$e^\left\{\\mathrm{ln}\left(x\right)\right\}$=$x$, SO:

$\Omega \approx {\left[\frac{N}{{N}_{↓}}\right]}^{{N}_{↓}}{e}^{\left({N}_{↓}\right)}$

$\to \Omega \approx {\left[\frac{{N}_{e}}{{N}_{↓}}\right]}^{{N}_{↓}}$

The corresponding result in $N< case is:

$\Omega \approx {\left[\frac{Ne}{{N}_{↓}}\right]}^{{N}_{↓}}$

which could have been predicted easily, since$N$and$N↓$appear symmetrically in the following approximation:

$\Omega \left(N,{N}_{↓}\right)=\frac{\left({N}_{↓}+N\right)!}{{N}_{↓}!N!}$ ### Want to see more solutions like these? 