Chapter 9: Statistical Mechanics

What information would you need to specify the macro-state of the air in a room? What information would you need to specify the microstate?

Obtain equation (9- 15) from (9-14). Make use or the following sums, correct when $\mathbf{\left|}\mathbf{x}\mathbf{\right|}\mathbf{<}\mathbf{1}$ :

$\begin{array}{l}{\sum }_{n=0}^{\infty }{x}^n=\frac{1}{1-x}\\ {\sum }_{n=0}^{\infty }n{x}^n=\frac{x}{{(1-x)}^2}\end{array}$

Show that equation (9- 16) follows from (9-15) and (9- 10).

Using the relationship between temperature and $\mathit{M}$and$\mathit{N}$ given in (9-16) and that between$\mathit{E}$ and$\mathit{n}$ in (9-6), obtain equation (9-17) from (9- 12). The first sum given In Exercise 30 will be useful.

Show that in the Iimit of large numbers, the exact probability of equation (9-9) becomes the Boltzmann probability of (9-17). Use the fact that $\frac{\mathbf{K}\mathbf{!}}{\left(K-k\right)\mathbf{!}}\mathbf{\equiv }{\mathit{K}}^{\mathbf{k}}$, which holds when $\mathit{k}\mathbf{<}\mathbf{<}\mathit{K}$.

The exact probabilities of equation (9-9) rest on the claim that the number of ways of adding$\mathit{N}$distinct non-negative integer to give a total of$\mathit{M}$ is $\frac{\mathbf{(}\mathbf{M}\mathbf{+}\mathbf{N}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{!}}{\mathbf{[}\mathbf{M}\mathbf{!}\mathbf{(}\mathbf{N}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{!}\mathbf{]}}$ . One way to prove it involves the following trick. It represents two ways that $\mathit{N}$distinct integers can add to$\mathit{M}\mathbf{-}\mathbf{9}$ and$\mathbf{5}$, respectively. In this special case.

The X's represent the total of the integers, $\mathit{M}$-each row has $\mathbf{5}$. The $\mathbf{1}\mathbf{\text{'}}\mathit{s}$represent "dividers" between the distinct integers of which there will of course be $\mathit{N}\mathbf{-}\mathit{I}$each row has$\mathbf{8}$ . The first row says that${\mathit{n}}_{\mathbf{1}}$ is$\mathbf{3}$ (three $\mathit{X}\mathbf{\text{'}}\mathit{s}$before the divider between it and${\mathit{n}}_{\mathbf{2}}$ ), ${\mathit{n}}_{\mathbf{2}}$is$\mathbf{0}$ (no$\mathit{X}\mathbf{\text{'}}\mathit{s}$ between its left divider with${\mathit{n}}_{\mathbf{1}}$ and its right divider with${\mathit{n}}_{\mathbf{3}}$ ),${\mathit{n}}_{\mathbf{3}}$ ) is$\mathbf{1}$ . ${\mathit{n}}_{\mathbf{4}}$through${\mathit{n}}_{\mathbf{6}}$ are$0$ , ${\mathit{n}}_{\mathbf{7}}$is$\mathbf{1}$ , and ${\mathit{n}}_{\mathbf{8}}$and ${\mathit{n}}_{\mathbf{9}}$are$\mathbf{0}$ . The second row says that ${\mathit{n}}_{\mathbf{2}}$is $\mathbf{2}$. ${\mathit{n}}_{\mathbf{6}}$is $\mathbf{1}$, ${\mathit{n}}_{\mathbf{9}}$is$\mathbf{2}$ , and all other$\mathit{n}$ are$\mathbf{0}$ . Further rows could account for all possible ways that the integers can add to$\mathit{M}$ . Argue that properly applied, the binomial coefficient (discussed in Appendix$\mathit{J}$ ) can be invoked to give the correct total number of ways for any$\mathit{N}$ and$\mathit{M}$ .

Consider a simple thermodynamic system in which particles can occupy only two states: a lower state, whose energy we define as 0 , and an upper state, energy$E_{u\circ }$

(a) Cany out the sum (with only two states, integration is certainly not valid) giving the average particle energy E. and plot your result as a function of temperature.

(b) Explain qualitatively why it should behave as it does,

(c) This system can be used as a model of paramagnetic, where individual atoms' magnetic moments can either be aligned or anti aligned with an external magnetic field, giving a low or high energy, respectively. Describe how the average alignment or antialignment depends on temperature. Does it make sense'?

Example 9.2 obtains a ratio of the number of particles expected in the n = 2state lo that in the ground state. Rather than the n = 2state, consider arbitrary n.

(a) Show that the ratio is $\frac{\mathbf{n}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{e}\mathbf{n}\mathbf{e}\mathbf{r}\mathbf{g}\mathbf{y}\mathbf{}{\mathbf{E}}_{\mathbf{n}}}{\mathbf{n}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{e}\mathbf{n}\mathbf{e}\mathbf{r}\mathbf{g}\mathbf{y}\mathbf{}{\mathbf{E}}_{\mathbf{1}}}\mathbf{=}{\mathit{n}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{13}\mathbf{.}\mathbf{6}\mathbf{c}\mathbf{V}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{n}}^{\mathbf{-}\mathbf{2}}\mathbf{)}\mathbf{/}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}$

Note that hydrogen atom energies are ${\mathit{E}}_{\mathbf{n}}\mathbf{=}\mathbf{-}\mathbf{13}\mathbf{.}\mathbf{6}\mathit{e}\mathit{V}\mathbf{/}\mathit{s}{\mathit{t}}^{\mathbf{2}}$.

(b) What is the limit of this ratio as n becomes very large? Can it exceed 1? If so, under what condition(s)?

(c) In Example 9.2. we found that even at the temperature of the Sun's surface$\left(~6000K\right)$, the ratio for n = 2 is only 10^-8 . For what value of nwould the ratio be 0.01?

(d) Is it realistic that the number of atoms with high n could be greater than the number with low n ?

Consider a system of one-dimensional spinless particles in a box (see Section 5.5) somehow exchanging energy. Through steps similar to those giving equation (9-27). show that

$\mathit{D}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{m}}^{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{L}}{\mathbf{\hslash }\mathbf{\pi }\sqrt{\mathbf{2}}}\frac{\mathbf{1}}{{\mathbf{E}}^{\mathbf{1}\mathbf{/}\mathbf{2}}}$

By carrying out the integration suggested just before equation (9-28), show that the average energy of a one-dimensional oscillator in the limit ${\mathbf{k}}_{\mathbf{\text{B}}}\mathbf{T}\mathbf{\gg }{\mathbf{h\omega }}_{\mathbf{0}}$is${\mathbf{k}}_{\mathbf{\text{B}}}\mathbf{T}$.