Chapter 2: The Second Law

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.

Suppose you flip $20$ fair coins.

(a) How many possible outcomes (microstates) are there?

(b) What is the probability of getting the sequence HTHHTTTHTHHHTHHHHTHT (in exactly that order)?

(c) What is the probability of getting $12$ heads and $8$ tails (in any order)?

Problem $2.20$. Suppose you were to shrink Figure$2.7$until the entire horizontal scale fits on the page. How wide would the peak be?

Use a computer to plot formula $2.22$directly, as follows. Define $z$=$q\_\left\{A\right\}/q$, so that $(1-z)$=$q\_\left\{B\right\}/q$. Then, aside from an overall constant that we'll ignore, the multiplicity function is $\left[4z\right(1-z){]}^N$, where $z$ranges from $0$to$1$and the factor of $4$ensures that the height of the peak is equal to $1$for any $N$. Plot this function for$N$=$1,10,100,1000$, and $10,000$. Observe how the width of the peak decreases as$N$increases.

This problem gives an alternative approach to estimating the width of the peak of the multiplicity function for a system of two large Einstein solids.

(a) Consider two identical Einstein solids, each with $\mathrm{N}$oscillators, in thermal contact with each other. Suppose that the total number of energy units in the combined system is exactly $2\mathrm{N}$. How many different macrostates (that is, possible values for the total energy in the first solid) are there for this combined system?

(b) Use the result of Problem$2.18$to find an approximate expression for the total number of microstates for the combined system. (Hint: Treat the combined system as a single Einstein solid. Do not throw away factors of "large" numbers, since you will eventually be dividing two "very large" numbers that are nearly equal. Answer: $2^{4\mathrm{N}}/\sqrt{8\mathrm{\pi N}}$.)

(c) The most likely macrostate for this system is (of course) the one in which the energy is shared equally between the two solids. Use the result of Problem $2.18$to find an approximate expression for the multiplicity of this macrostate. (Answer:$2^{4\mathrm{N}}/\left(4\mathrm{\pi N}\right)$ .)

(d) You can get a rough idea of the "sharpness" of the multiplicity function by comparing your answers to parts (b) and (c). Part (c) tells you the height of the peak, while part (b) tells you the total area under the entire graph. As a very crude approximation, pretend that the peak's shape is rectangular. In this case, how wide would it be? Out of all the macrostates, what fraction have reasonably large probabilities? Evaluate this fraction numerically for the case $\mathrm{N}={10}^{23}$.

Consider a two-state paramagnet with ${10}^{23}$elementary dipoles, with the total energy fixed at zero so that exactly half the dipoles point up and half point down.

(a) How many microstates are "accessible" to this system?

(b) Suppose that the microstate of this system changes a billion times per second. How many microstates will it explore in ten billion years (the age of the universe)?

(c) Is it correct to say that, if you wait long enough, a system will eventually be found in every "accessible" microstate? Explain your answer, and discuss the meaning of the word "accessible."

For a single large two-state paramagnet, the multiplicity function is very sharply peaked about $N_{\uparrow }=N/2$.

(a) Use Stirling's approximation to estimate the height of the peak in the multiplicity function.

(b) Use the methods of this section to derive a formula for the multiplicity function in the vicinity of the peak, in terms of $x\equiv N_{\uparrow }-(N/2)$. Check that your formula agrees with your answer to part (a) when $x=0$.

(c) How wide is the peak in the multiplicity function?

(d) Suppose you flip $1,000,000$coins. Would you be surprised to obtain heads and 499,000 tails? Would you be surprised to obtain 510,000 heads and 490,000 tails? Explain.

The mathematics of the previous problem can also be applied to a one-dimensional random walk: a journey consisting of $N$steps, all the same sic, cache chosen randomly to be cither forward or backward. (The usual mental image is that of a drunk stumbling along an alley.)

(a) Where are you most likely to find yourself, after the end of a long random walk?

(b) Suppose you take a random walk of $10,000$steps (say each a yard long). About how far from your starting point would you expect to be at the end?

(c) A good example of a random walk in nature is the diffusion of a molecule through a gas; the average step length is then the mean free path, as computed in Section $1.7.$Using this model, and neglecting any small numerical factors that might arise from the varying step size and the multidimensional nature of the path, estimate the expected net displacement of an air molecule (or perhaps a carbon monoxide molecule traveling through air) in one second, at room temperature and atmospheric pressure. Discuss how your estimate would differ if the clasped time or the temperature were different. Check that your estimate is consistent with the treatment of diffusion in Section$1.7.$

Consider an ideal monatomic gas that lives in a two-dimensional universe ("flatland"), occupying an area $A$instead of a volume $V$. By following the same logic as above, find a formula for the multiplicity of this gas, analogous to equation $2.40$.

Rather than insisting that all the molecules be in the left half of a container, suppose we only require that they be in the leftmost $99\%$(leaving the remaining $1\%$completely empty). What is the probability of finding such an arrangement if there are $100$molecules in the container? What if there are $10,000$molecules? What if there are ${\mathbf{10}}^{\mathbf{23}}$?