Question

Use the methods of this section to derive a formula, similar to equation$2.21$, for the multiplicity of an Einstein solid in the "low-temperature" limit,$q\ll N$ .

Short answer

The formula of the Similar Equation

$\Omega ={\left(\frac{eN}{q}\right)}^q$

Step-by-step solution

Solve the problem of solution

The number of microstates in an Einstein solid with $N$oscillators and $q$energy quanta is:

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

For any macroscopic solid, both $\mathrm{q}$and$\mathrm{N}$ are large numbers (on the order of Avogadro's number, orlocalid="1650382250464" ${10}^{23}$) so the factorials in $\Omega$are very large numbers, not calculable on most computers. To get estimates of$\Omega$we can use Stirling's approximation for the factorials. The derivation of this approximation for the high temperature case$q>>N$(lots more energy quanta than oscillators) is given in Schroeder's book, so at low temperature case we have$q<<N$(lots more oscillators than energy quanta). Writing out the binomial coefficient:

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

We can now take logarithms for both sides, we get,

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

but, $\ln \left(\frac{x}{y}\right)$=$\ln \left(x\right)-\ln \left(y\right)$and $\ln \left(x\right)-\ln \left(y\right)$=$\ln \left(x\right)+\ln \left(y\right)$, so:

$\to \ln \left(\Omega \right)$=$\backslash \ln \left[\right(q+N)!]-\backslash \ln (q!)-\backslash \ln (N!)$

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

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

The assumption of Equation

so equation (l) will become:

$$\begin{gathered} \left(\Omega \right)=(q+N)\ln (q+N)-(q+N)-q\ln \left(q\right)+q-Nln\left(N\right)+N \\ \ln \left(\Omega \right)=(q+N)\ln (q+N)-q\ln \left(q\right)-Nln\left(N\right) \end{gathered}$$

If we now make the assumption that$q<<N$, we get:

$$\begin{gathered} \ln \left(\Omega \right)=(q+N)\ln \left[N\left(1+\frac{q}{N}\right)\right]-q\ln \left(q\right)-N\ln \left(N\right) \\ \ln \left(\Omega \right)=(q+N)\left(\ln \left(N\right)+\ln \left(1+\frac{q}{N}\right)\right)-q\ln \left(q\right)-N\ln \left(N\right) \end{gathered}$$

but for$N>>q$, we have,$\ln \left(1+\frac{q}{N}\right)$ =\frac$\left\{q\right\}\left\{N\right\}$, so:

$$\begin{gathered} \ln \left(\Omega \right)=(q+N)\left(\ln \left(N\right)+\frac{q}{N}\right)-q\ln \left(q\right)-Nln\left(N\right) \\ \ln \left(\Omega \right)=q\ln \left(N\right)+\frac{q^2}{N}+N\ln \left(N\right)+q-q\ln \left(q\right)-Nln\left(N\right) \\ \ln \left(\Omega \right)=\left(q\ln \right(N)-q\ln (q\left)\right)+\frac{q^2}{N}+q \end{gathered}$$

but,$q\backslash \ln \left(N\right)-q\backslash \ln \left(q\right)$$q\ln \left(\frac{N}{q}\right)$,= so:

$\ln \left(\Omega \right)=q\ln \left(\frac{N}{q}\right)+\frac{q^2}{N}+q$

Neglect the second term

but as$q<<N$, so we can neglect the second term$\frac{q^2}{N^{\text{'}}}$, so:

$\ln \left(\Omega \right)=q\ln \left(\frac{N}{q}\right)+q$

Exponentiating this equation gives the approximate value for$\Omega$:

$e^{\ln \left(\Omega \right)}=e^{\left[q\ln \left(\frac{N}{q}\right)+q\right]}$

$\Omega ={\left[e^{\ln \left(\frac{N}{q}\right)}\right]}^q\times e^q$

but,$e^{lnx}=x$, so:

$\Omega ={\left(\frac{N}{q}\right)}^q\times e^q$

$\to \Omega ={\left(\frac{eN}{q}\right)}^q$