91影视

Consider two-state paramagnet with $N$magnetic dipoles and $N_{鈫�}$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:

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

Writing out the binomial coefficient:

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

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

$惟\left(N,N_{鈫�}\right)=\frac{\left(N_{鈫�}+N\right)!}{N_{鈫�}!N!}$

take the natural logarithms for both sides:

$\ln \left(惟\right)=\ln \left(\frac{\left(N_{鈫�}+N\right)!}{N_{鈫�}!N!}\right)$

but, we have these logarithmic relations:

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

$\ln \left(惟\right)=\ln \left(\frac{\left(N_{鈫�}+N\right)!}{N_{鈫�}!N!}\right)=\ln \left(N_{鈫�}+N\right)!-\ln (N!)-\ln \left(N_{鈫�}!\right)$

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

$\ln (n!)鈮�n\ln \left(n\right)-n$

so,we get:

$\ln \left(惟\right)鈮�\left(N_{鈫�}+N\right)\ln \left(N_{鈫�}+N\right)-\left(N_{鈫�}+N\right)-N\ln \left(N\right)+N-N_{鈫�}\ln \left(N_{鈫�}\right)+N_{鈫�}$

$\ln \left(惟\right)鈮�\left(N_{鈫�}+N\right)\ln \left(N_{鈫�}+N\right)-N\ln \left(N\right)-N_{鈫�}\ln \left(N_{鈫�}\right)$

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

$\ln \left(惟\right)鈮�\left(N_{鈫�}+N\right)\ln \left[N\left(\frac{N_{鈫�}}{N}+1\right)\right]-N\ln \left(N\right)-N_{鈫�}\ln \left(N_{鈫�}\right)$

we factored out $N$from the first logarithm:

$\ln \left(惟\right)鈮�\left(N_{鈫�}+N\right)\left[\ln \left(N\right)+\ln \left(\frac{N_{鈫�}}{N}+1\right)\right]-N\ln \left(N\right)-N_{鈫�}\ln \left(N_{鈫�}\right)$

but, we have the logarithmic relation:

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

for $b>>a$, so:

$\ln \left(惟\right)鈮�N_{鈫�}\ln \left(N\right)+N_{鈫�}\frac{N_{鈫�}}{N}+N\ln \left(N\right)+N\frac{N_{鈫�}}{N}-N\ln \left(N\right)-N_{鈫�}\ln \left(N_{鈫�}\right)$$\ln \left(惟\right)鈮�N_{鈫�}\ln \left(N\right)+\frac{{\left(N_{鈫�}\right)}^2}{N}+N_{鈫�}-N_{鈫�}\ln \left(N_{鈫�}\right)$

$\ln \left(惟\right)鈮�\left(N_{鈫�}+N\right)\left[\ln \left(N\right)+\frac{N_{鈫�}}{N}\right]-N\ln \left(N\right)-N_{鈫�}\ln \left(N_{鈫�}\right)$

$\ln \left(惟\right)鈮�N_{鈫�}\left[\ln \left(N\right)-\ln \left(N_{鈫�}\right)\right]+\frac{{\left(N_{鈫�}\right)}^2}{N}+N_{鈫�}$

but, =$\ln \left(惟\right)鈮�N_{鈫�}\ln \left[\frac{N}{N_{鈫�}}\right]+N_{鈫�}$, so:

$\ln \left(惟\right)鈮�N_{鈫�}\ln \left[\frac{N}{N_{鈫�}}\right]+\frac{{\left(N_{鈫�}\right)}^2}{N}+N_{鈫�}$

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

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

$$\begin{gathered} e^{\left(\ln \right(惟\left)\right)}鈮�e^{\left(N_{鈫�}\ln \left[\frac{N}{N_{鈫�}}\right]+N_{鈫�}\right)} \\ 惟鈮�e^{\left(\ln \left[\frac{N}{N_{鈫�}}\right]\right)}{e}^{\left(N_{鈫�}\right)} \\ 惟鈮�\left[e^{{\left.\left(\ln \left[\frac{N}{N_{鈫�}}\right]\right)\right]}^{N_{鈫�}}}{e}^{\left(N_{鈫�}\right)}\right.\backslash \end{gathered}$$

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

$惟鈮�{\left[\frac{N}{N_{鈫�}}\right]}^{N_{鈫�}}{e}^{\left(N_{鈫�}\right)}$

$鈫�惟鈮�{\left[\frac{N_e}{N_{鈫�}}\right]}^{N_{鈫�}}$

The corresponding result in $N<<N鈫�$case is:

$惟鈮�{\left[\frac{Ne}{N_{鈫�}}\right]}^{N_{鈫�}}$

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

$惟\left(N,N_{鈫�}\right)=\frac{\left(N_{鈫�}+N\right)!}{N_{鈫�}!N!}$