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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:

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