This problem concerns a collection of N identical harmonic oscillators (perhaps an Einstein solid or the internal vibrations of gas molecules) at temperature T. As in Section 2.2, the allowed energies of each oscillator are 0, hf, 2hf, and so on. (

a) Prove by long division that

$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$

For what values of x does this series have a finite sum?

(b) Evaluate the partition function for a single harmonic oscillator. Use the result of part (a) to simplify your answer as much as possible.

(c) Use formula 6.25 to find an expression for the average energy of a single oscillator at temperature T. Simplify your answer as much as possible.

(d) What is the total energy of the system of N oscillators at temperature T? Your result should agree with what you found in Problem 3.25.

(e) If you haven't already done so in Problem 3.25, compute the heat capacity of this system and check t hat it has the expected limits as $T\to 0$ and $T\to \infty$.

This problem concerns a collection of N identical harmonic oscillators (perhaps an Einstein solid or the internal vibrations of gas molecules) at temperature T. As in Section 2.2, the allowed energies of each oscillator are 0, hf, 2hf, and so on.

(a) Divide 1 by 1 - x using long division, yielding:

(b) Assume that the oscillator's ground state energy is zero; the energies are provided by:

$E_n=nhfn=0,1,2,\dots$

The partition function is:

$Z=\sum _n{e}^{-\beta E_n}$

Substitute with n:

$$\begin{gathered} Z=\sum _n{e}^{-n\beta hf} \\ Z=\sum _n{\left(e^{-\beta hf}\right)}^n \end{gathered}$$

From part (a),

$\frac{1}{1-x}=1+x+x^2+x^3+\dots =\sum _n{x}^n$

Using the above expression, the partition function is:

$$\begin{gathered} Z=\sum _n{\left(e^{-\beta hf}\right)}^n=\frac{1}{1-e^{-\beta hf}} \\ Z=\frac{1}{1-e^{-\beta hf}}\left(1\right) \end{gathered}$$

(c)The average energy is given by:

$\overline{E}=-\frac{1}{Z}\frac{\partial Z}{\partial \beta }$

Substitute from part (b)

$$\begin{gathered} \overline{E}=-\left(1-e^{-\beta hf}\right)\frac{\partial }{\partial \beta }\left(\frac{1}{1-e^{-\beta hf}}\right) \\ \overline{E}=\left(1-e^{-\beta hf}\right)\frac{hf{e}^{-\beta hf}}{{\left(1-e^{-\beta hf}\right)}^2} \\ \overline{E}=\frac{hf{e}^{-\beta hf}}{1-e^{-\beta hf}} \\ \overline{E}=\frac{hf}{e^{\beta hf}-1} \end{gathered}$$

(d) The average energy multiplied by the number of oscillators equals the total energy of the system, so:

$U=N\overline{E}$

Substitute from part (c):

$U=\frac{Nhf}{e^{\beta hf}-1}$

(e) The heat capacity is equal to the partial derivative of total energy in terms of temperature, i.e.

$$\begin{gathered} C=\frac{\partial U}{\partial T}=\frac{\partial \beta }{\partial T}\frac{\partial U}{\partial \beta } \\ C=\frac{\partial \beta }{\partial T}\frac{\partial }{\partial \beta }\left(\frac{Nhf}{e^{\beta hf}-1}\right) \\ C={\left(\frac{\partial T}{\partial \beta }\right)}^{-1}\left(-\frac{N(hf{)}^2{e}^{\beta hf}}{{\left(e^{\beta hf}-1\right)}^2}\right) \end{gathered}$$

But, $\beta =1/kT\to T=1/k\beta$,So:

${\left(\frac{\partial T}{\partial \beta }\right)}^{-1}=-k{\beta }^2$

Therefore,

$C=\frac{Nk(\beta hf{)}^2{e}^{\beta hf}}{{\left(e^{\beta hf}-1\right)}^2}$

We can express the exponential as: at a high temperature $\beta \to 0$, we can write it as:

$e^{\beta hf}=1+\beta hf$

Therefore,

$$\begin{gathered} C=\frac{Nk\left(\beta hf{)}^2\right(1+\beta hf)}{(\beta hf{)}^2} \\ C=Nk(1+\beta hf) \end{gathered}$$

Since $\beta$is very small, we can approximate:

$C(\beta \to 0)\approx Nk$

As the temperature goes to zero, the Boltzmann factor therefore$\beta \to \infty ,\text{so}{e}^{\beta hf}\gg 1$hence we can approximate:

$$\begin{gathered} C\approx \frac{Nk(\beta hf{)}^2{e}^{\beta hf}}{e^{2\beta hf}} \\ C(\beta \to \infty )\approx Nk(\beta hf{)}^2{e}^{-\beta hf} \end{gathered}$$