Exercises — Statistical mechanics — microstate, macrostate

A microstate is the complete ordered pattern (which specific coin shows what). A macrostate is a bulk count you could read off without knowing individual coins.

• (a) "HTH" — the exact face of each of the 3 coins → microstate.
• (b) "two heads" — only the count $n=2$ → macrostate.
• (c) a named face for every coin → microstate.
• (d) "$n=1$" — a count → macrostate.

Most: the count nearest the middle, $n=1$ or $n=2$ (both tie at $\left(\genfrac{}{}{0ex}{}{3}{1}\right)=\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$). Fewest: the extremes $n=0$ and $n=3$, each with exactly $\left(\genfrac{}{}{0ex}{}{3}{0}\right)=\left(\genfrac{}{}{0ex}{}{3}{3}\right)=1$ microstate. The bar chart is symmetric and peaks in the middle — that is all you need to recognise the answer.

(a) Total $=2^6=64$. Why: each of 6 objects doubles the count. (b) $\Omega (2)=\left(\genfrac{}{}{0ex}{}{6}{2}\right)=\frac{6!}{2!4!}=\frac{720}{2\times 24}=\frac{720}{48}=15.$ So 15 of the 64 microstates show exactly two ups.

$\left(\genfrac{}{}{0ex}{}{6}{0}\right),\left(\genfrac{}{}{0ex}{}{6}{1}\right),\dots ,\left(\genfrac{}{}{0ex}{}{6}{6}\right)=1,6,15,20,15,6,1.$ Sum $=1+6+15+20+15+6+1=64=2^6.✓$ Why the check works: summing multiplicities over all macrostates must rebuild every microstate exactly once — this is the binomial theorem with $1+1=2$.

The fundamental postulate says every accessible microstate is equally likely, so $P(n)=\frac{\Omega (n)}{2^N}=\frac{\left(\genfrac{}{}{0ex}{}{6}{3}\right)}{2^6}=\frac{20}{64}=\mathrm{0.3125.}$ The peak macrostate $n=3$ carries $31.25\%$ of the probability — more than any other single macrostate.

$\Omega (5)=\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$ and $\Omega (0)=\left(\genfrac{}{}{0ex}{}{10}{0}\right)=1$, so the ratio is $\boxed{252}$. What it measures: the "balanced" macrostate ($n=5$) is realised in 252 times as many ways as the "all down" extreme. Under the equal-microstate postulate, the balanced macrostate is therefore 252× more likely. This ratio of multiplicities is exactly what makes equilibrium overwhelming — and it is the raw material of entropy differences, since $S=k\ln\Omega$ turns a ratio of $\Omega$ into a difference of entropies.

Relative width $\frac{\sigma }{\mu }=\frac{\frac{1}{2}\sqrt{N}}{N/2}=\frac{1}{\sqrt{N}}$.

• $N=100$: $1/\sqrt{100}=0.1$ → the up-count typically lands within $\pm 10\%$ of half.
• $N=10^4$: $1/\sqrt{10^4}=0.01$ → within $\pm 1\%$. Implication: as $N$ grows, the fractional wobble shrinks like $1/\sqrt{N}$. For lab-sized $N\sim 10^{23}$ the fluctuations are $\sim 10^{-11.5}$ — utterly unmeasurable. You always see the peak macrostate: that is equilibrium. (This is the Gaussian limit of the binomial.)

The four microstates with exactly one up (call up $\uparrow$, down $\downarrow$): $\uparrow \downarrow \downarrow \downarrow ,\downarrow \uparrow \downarrow \downarrow ,\downarrow \downarrow \uparrow \downarrow ,\downarrow \downarrow \downarrow \uparrow .$ That is $\left(\genfrac{}{}{0ex}{}{4}{1}\right)=4$. Why order is invisible to the macrostate: the macrostate only records the number $n=1$; it never asks which slot. The four patterns differ microscopically but are identical macroscopically — a perfect miniature of "many micros, one macro."

Joint multiplicity $=\left(\genfrac{}{}{0ex}{}{4}{n_A}\right)\left(\genfrac{}{}{0ex}{}{4}{4-n_A}\right)$:

Most probable split: $n_A=2$ (product $=36$) — energy shared equally, exactly what "thermal equilibrium" should mean. Check: $1+16+36+16+1=70=\left(\genfrac{}{}{0ex}{}{8}{4}\right)$. $✓$ Synthesis point: the joint system, treated as one pool of 8 spins with 4 ups, has $\left(\genfrac{}{}{0ex}{}{8}{4}\right)=70$ microstates; splitting by "how many are in A" just partitions those 70. The equilibrium split maximises ${\Omega }_A{\Omega }_B$, which by $S=k\ln \Omega$ maximises $S_A+S_B$ — the second law emerging from pure counting. This partitioned counting is exactly the microcanonical logic.

First, the constant we need. Boltzmann's constant $k$ (sometimes written $k_B$) is the fixed conversion factor that turns "a count of microstates" into thermodynamic entropy in $S=k\ln\Omega$. Its value is $k=1.381\times 10^{-23}\text{J/K}.$ It is tiny because a single microstate carries only a minuscule amount of entropy; you need Avogadro-many of them to reach everyday J/K.

Now the counting. $\Omega \propto V^N$, so doubling $V$ multiplies $\Omega$ by $2^N$ (each of the $N$ particles independently gains twice the room, and choices multiply): $\frac{{\Omega }_2}{{\Omega }_1}={\left(\frac{2V}{V}\right)}^N=2^N.$ Entropy change: $\Delta S=k\ln \frac{{\Omega }_2}{{\Omega }_1}=k\ln 2^N=Nk\ln 2.$ Numerically $\Delta S=(6.022\times 10^{23})(1.381\times 10^{-23})(0.6931)\approx 5.76\text{J/K}.$ This is the textbook entropy of free expansion — and it dropped straight out of counting position cells $V^N$. Counting predicts thermodynamics.

A 1% fluctuation means $n-\mu =0.01\mu =0.01\cdot \frac{N}{2}=0.005N$. Measure it in units of $\sigma =\frac{1}{2}\sqrt{N}$: the number of standard deviations is $z=\frac{n-\mu }{\sigma }=\frac{0.005N}{\frac{1}{2}\sqrt{N}}=0.01\sqrt{N}.$ For $N=10^{20}$: $z=0.01\times 10^{10}=10^8$. The exponent is $-z^2/2=-\frac{1}{2}(10^8{)}^2=-5\times 10^{15}$. So $P\sim e^{-5\times 10^{15}}$ — a probability so small it has no physical meaning. A 1% fluctuation in a macroscopic sample never happens in the age of the universe. This is the statistical bedrock of irreversibility.

Step 1 — write the log of the central binomial. Since $\left(\genfrac{}{}{0ex}{}{N}{N/2}\right)=\frac{N!}{[(N/2)!{]}^2}$, $\ln \left(\genfrac{}{}{0ex}{}{N}{N/2}\right)=\ln N!-2\ln \left(\frac{N}{2}\right)!.$ Step 2 — insert Stirling with its $\frac{1}{2}\ln (2\pi N)$ tail (we keep this tail because it is exactly what produces the $\pi N$ factor at the end). For the numerator, $\ln N!\approx N\ln N-N+\frac{1}{2}\ln (2\pi N).$ For each factorial in the denominator, replace $N\to N/2$: $\ln \left(\frac{N}{2}\right)!\approx \frac{N}{2}\ln \frac{N}{2}-\frac{N}{2}+\frac{1}{2}\ln \left(2\pi \frac{N}{2}\right)=\frac{N}{2}\ln \frac{N}{2}-\frac{N}{2}+\frac{1}{2}\ln (\pi N).$ Step 3 — subtract. Multiply the denominator line by 2 and subtract from the numerator line: $\ln \left(\genfrac{}{}{0ex}{}{N}{N/2}\right)\approx [N\ln N-N+\frac{1}{2}\ln (2\pi N)]-[N\ln \frac{N}{2}-N+\ln (\pi N)].$ The $-N$ and $+N$ cancel. Group the "big" logs: $N\ln N-N\ln \frac{N}{2}=N\ln \frac{N}{N/2}=N\ln 2$. That is where the $N\ln 2$ (the same scale as $2^N$) comes from. Step 4 — collect the leftover $\frac{1}{2}\ln$ terms, which produce the $\pi N/2$: $\frac{1}{2}\ln (2\pi N)-\ln (\pi N)=\frac{1}{2}\ln (2\pi N)-\frac{1}{2}\ln ((\pi N{)}^2)=\frac{1}{2}\ln \frac{2\pi N}{(\pi N{)}^2}=\frac{1}{2}\ln \frac{2}{\pi N}=-\frac{1}{2}\ln \frac{\pi N}{2}.$ Putting Steps 3–4 together: $\boxed{\ln \left(\genfrac{}{}{0ex}{}{N}{N/2}\right)\approx N\ln 2-\frac{1}{2}\ln \left(\frac{\pi N}{2}\right).}$ Step 5 — divide by $2^N$ (subtract $N\ln 2$) to get the peak probability: $\ln P(N/2)\approx -\frac{1}{2}\ln \left(\frac{\pi N}{2}\right)⟹P(N/2)\approx \sqrt{\frac{2}{\pi N}}.$ Step 6 — evaluate. For $N=100$: $\sqrt{2/(100\pi )}=\sqrt{0.006366}=0.07979\approx \mathrm{0.0798.}$ Exact: $\left(\genfrac{}{}{0ex}{}{100}{50}\right)/2^{100}=0.07959$. Agreement to $\sim 0.2\%$ — Stirling nails it. Mastery insight: the peak probability falls like $1/\sqrt{N}$, yet the peak still dominates because every other macrostate falls even faster — the distribution as a whole collapses to a spike of relative width $1/\sqrt{N}$ (Exercise 3.2). This is the binomial → Gaussian crossover in action.

The microstates keep changing constantly (the microscopic pattern of ups is never still). The only postulate is that in equilibrium each accessible microstate is equally likely — no pattern is favoured. Given that even-handedness, macrostates are not equal: a macrostate's chance is just its share of microstates, $\Omega (n)/2^N$. Because $\Omega$ is astronomically peaked at the balanced count (the $\left(\genfrac{}{}{0ex}{}{N}{n}\right)$ tower), almost every microstate belongs to that one macrostate. So you see it not by decree but by sheer weight of numbers — the relative width shrinks like $1/\sqrt{N}$ until any deviation is impossible to observe. "Equilibrium" is simply "wherever the microstates overwhelmingly are." The lone assumption is equal microstate probability; every other conclusion — the peak, its sharpness, irreversibility — is pure arithmetic riding on that one even-handed rule.

• L1: microstate = ordered pattern; macrostate = a count. Peak in the middle, valleys at the extremes.
• L2: total $=2^N$; $\Omega (n)=\left(\genfrac{}{}{0ex}{}{N}{n}\right)$; $P=\Omega /2^N$.
• L3: ratios of $\Omega$ drive probability; relative width $=1/\sqrt{N}$ shrinks.
• L4: independent systems → multiply $\Omega$, add $S=k\ln \Omega$; gas $\Omega \propto V^N{E}^{3N/2}$ gives $\Delta S=Nk\ln 2$ on doubling $V$.
• L5: Gaussian/Stirling quantify the spike; the only postulate is equal microstate probability.