Question

Use Stirling's approximation to find an approximate formula for the multiplicity of a two-state paramagnet. Simplify this formula in the limit $N_{\downarrow }\ll N$to obtain $\Omega \approx {\left(Ne/N_{\downarrow }\right)}^{N_{\downarrow }}$. 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.

Short answer

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

$\Omega \approx {\left[\frac{Ne}{N_{\downarrow }}\right]}^{N_{\downarrow }}$

Step-by-step solution

Two state of paramagnet 

Consider two-state paramagnet with $N$magnetic dipoles and $N_{\downarrow }$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:

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

Writing out the binomial coefficient:

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

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

$\Omega \left(N,N_{\downarrow }\right)=\frac{\left(N_{\downarrow }+N\right)!}{N_{\downarrow }!N!}$

take the natural logarithms for both sides:

$\ln \left(\Omega \right)=\ln \left(\frac{\left(N_{\downarrow }+N\right)!}{N_{\downarrow }!N!}\right)$

but, we have these logarithmic relations:

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

Logarithm

$\ln \left(\Omega \right)=\ln \left(\frac{\left(N_{\downarrow }+N\right)!}{N_{\downarrow }!N!}\right)=\ln \left(N_{\downarrow }+N\right)!-\ln (N!)-\ln \left(N_{\downarrow }!\right)$

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

$\ln (n!)\approx n\ln \left(n\right)-n$

so,we get:

$\ln \left(\Omega \right)\approx \left(N_{\downarrow }+N\right)\ln \left(N_{\downarrow }+N\right)-\left(N_{\downarrow }+N\right)-N\ln \left(N\right)+N-N_{\downarrow }\ln \left(N_{\downarrow }\right)+N_{\downarrow }$

$\ln \left(\Omega \right)\approx \left(N_{\downarrow }+N\right)\ln \left(N_{\downarrow }+N\right)-N\ln \left(N\right)-N_{\downarrow }\ln \left(N_{\downarrow }\right)$

If we now use the assumption that $N>>N_{\downarrow }$, we get:

$\ln \left(\Omega \right)\approx \left(N_{\downarrow }+N\right)\ln \left[N\left(\frac{N_{\downarrow }}{N}+1\right)\right]-N\ln \left(N\right)-N_{\downarrow }\ln \left(N_{\downarrow }\right)$

we factored out $N$from the first logarithm:

$\ln \left(\Omega \right)\approx \left(N_{\downarrow }+N\right)\left[\ln \left(N\right)+\ln \left(\frac{N_{\downarrow }}{N}+1\right)\right]-N\ln \left(N\right)-N_{\downarrow }\ln \left(N_{\downarrow }\right)$

but, we have the logarithmic relation:

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

for $b>>a$, so:

To neglect second term 

$\ln \left(\Omega \right)\approx N_{\downarrow }\ln \left(N\right)+N_{\downarrow }\frac{N_{\downarrow }}{N}+N\ln \left(N\right)+N\frac{N_{\downarrow }}{N}-N\ln \left(N\right)-N_{\downarrow }\ln \left(N_{\downarrow }\right)$$\ln \left(\Omega \right)\approx N_{\downarrow }\ln \left(N\right)+\frac{{\left(N_{\downarrow }\right)}^2}{N}+N_{\downarrow }-N_{\downarrow }\ln \left(N_{\downarrow }\right)$

$\ln \left(\Omega \right)\approx \left(N_{\downarrow }+N\right)\left[\ln \left(N\right)+\frac{N_{\downarrow }}{N}\right]-N\ln \left(N\right)-N_{\downarrow }\ln \left(N_{\downarrow }\right)$

$\ln \left(\Omega \right)\approx N_{\downarrow }\left[\ln \left(N\right)-\ln \left(N_{\downarrow }\right)\right]+\frac{{\left(N_{\downarrow }\right)}^2}{N}+N_{\downarrow }$

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

$\ln \left(\Omega \right)\approx N_{\downarrow }\ln \left[\frac{N}{N_{\downarrow }}\right]+\frac{{\left(N_{\downarrow }\right)}^2}{N}+N_{\downarrow }$

use the assumption that$N>>N_{\downarrow }$, to neglect the second term:

$\ln \left(\Omega \right)\approx N_{\downarrow }\ln \left[\frac{N}{N_{\downarrow }}\right]+N_{\downarrow }$

Exponentiating both sides 

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

$$\begin{gathered} e^{\left(\ln \right(\Omega \left)\right)}\approx e^{\left(N_{\downarrow }\ln \left[\frac{N}{N_{\downarrow }}\right]+N_{\downarrow }\right)} \\ \Omega \approx e^{\left(\ln \left[\frac{N}{N_{\downarrow }}\right]\right)}{e}^{\left(N_{\downarrow }\right)} \\ \Omega \approx \left[e^{{\left.\left(\ln \left[\frac{N}{N_{\downarrow }}\right]\right)\right]}^{N_{\downarrow }}}{e}^{\left(N_{\downarrow }\right)}\right.\backslash \end{gathered}$$

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

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

$\to \Omega \approx {\left[\frac{N_e}{N_{\downarrow }}\right]}^{N_{\downarrow }}$

The corresponding result in $N<<N\downarrow$case is:

$\Omega \approx {\left[\frac{Ne}{N_{\downarrow }}\right]}^{N_{\downarrow }}$

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

$\Omega \left(N,N_{\downarrow }\right)=\frac{\left(N_{\downarrow }+N\right)!}{N_{\downarrow }!N!}$