Worked examples — Statistical mechanics — microstate, macrostate

See the parent topic for the setup, and Binomial and Gaussian distributions for the maths of the peak.

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The scenario matrix

Every counting problem in this topic is one of these cells. We will hit each one.

Cell	What makes it special	Example that covers it
A — Edge macrostate ($n=0$ or $n=N$)	Only $\Omega =1$ microstate; a "corner" of the picture	Ex 1
B — Peak macrostate ($n=N/2$)	Largest $\Omega$; this is equilibrium	Ex 2
C — Off-peak, generic $n$	Ordinary combination; ratios to the peak	Ex 3
D — Degenerate input ($N=0$, $N=1$, empty)	Formula must not break; check $0!=1$	Ex 4
E — Large-$N$ limiting behaviour	Peak sharpens $\sim 1/\sqrt{N}$; Gaussian shape	Ex 5
F — Biased two-state (uneven weights)	Microstates not equally likely a priori — subtle!	Ex 6
G — Continuous system (gas: $V,E$ scaling)	Count volumes in phase space, not lists	Ex 7
H — Real-world word problem	Translate everyday shaking into $\Omega$	Ex 8
I — Exam twist (composite / multi-state)	Combine two subsystems; multiplicities multiply	Ex 9

Before any of it, one figure that fixes the whole geography of the coin/spin problem.

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Ex 1 — Cell A: the corner macrostate

Forecast: Guess before reading — is it $6$? $36$? Something bigger?

1. Write the multiplicity. The macrostate "$n$ up out of $N$" has $\Omega (n)=\left(\genfrac{}{}{0ex}{}{N}{n}\right)$. Why this step? This is the counting rule the parent derived: pick which $n$ of $N$ slots are up.
2. Plug $n=N=6$. $\Omega (6)=\left(\genfrac{}{}{0ex}{}{6}{6}\right)=\frac{6!}{6!0!}=\frac{720}{720\cdot 1}=1.$ Why this step? Every slot is forced up — there is literally one sequence $\uparrow \uparrow \uparrow \uparrow \uparrow \uparrow$.
3. Plug $n=0$. $\Omega (0)=\left(\genfrac{}{}{0ex}{}{6}{0}\right)=\frac{6!}{0!6!}=1.$ Why this step? "All down" is also a single sequence. The formula gives $1$ only because $0!=1$ — that convention is what makes the edges behave.

Verify: $\left(\genfrac{}{}{0ex}{}{6}{6}\right)=\left(\genfrac{}{}{0ex}{}{6}{0}\right)=1$, and both are the smallest entries of row $N=6$ (see the pyramid in the figure). ✓

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Ex 2 — Cell B: the peak (equilibrium)

Forecast: The peak sits at $n=N/2=3$. Guess $\Omega (3)$ before computing.

1. Evaluate the middle. $\Omega (3)=\left(\genfrac{}{}{0ex}{}{6}{3}\right)=\frac{720}{6\cdot 6}=\frac{720}{36}=20.$ Why this step? $n=3$ splits the $6$ slots evenly; $3!$ orders the ups and $3!$ orders the downs, both divided out.
2. Compare to the corner. $\Omega (3)/\Omega (6)=20/1=20$. Why this step? Even for a tiny system, "half up" already has 20× more ways than "all up". This ratio is the whole reason equilibrium exists.
3. Probability of the peak. $P(3)=\Omega (3)/2^N=20/2^6=20/64=\mathrm{0.3125.}$ Why this step? By the Fundamental postulate of statistical mechanics, each of the $2^6=64$ microstates is equally likely, so a macrostate's probability is its share $\Omega /2^N$.

Verify: row-6 of Pascal's triangle is $1,6,15,20,15,6,1$; sum $=64=2^6$, and $20$ is indeed the max. ✓

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Ex 3 — Cell C: an off-peak macrostate and its odds

Forecast: Bigger or smaller than the peak $\Omega (5)$? By how much?

1. Compute $\Omega (2)$. $\left(\genfrac{}{}{0ex}{}{10}{2}\right)=\frac{10!}{2!8!}=\frac{10\cdot 9}{2}=45.$ Why this step? Choose which 2 of 10 slots are heads; the $10\cdot 9$ ordered picks are halved because the 2 chosen slots are interchangeable.
2. Compute the peak $\Omega (5)$. $\left(\genfrac{}{}{0ex}{}{10}{5}\right)=\frac{10!}{5!5!}=252.$ Why this step? We need something to compare against — the equilibrium value.
3. Probability. $P(2)=45/2^{10}=45/1024\approx \mathrm{0.0439.}$ Why this step? Same $\Omega /2^N$ rule. Off-peak macrostates are rare but not impossible.
4. Ratio to peak. $45/252\approx 0.179$ — already almost 6× less likely than half-heads, for only $N=10$. Why this step? Shows the peak dominating even at modest $N$; foreshadows Ex 5.

Verify: $\left(\genfrac{}{}{0ex}{}{10}{2}\right)=45$, $\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$, $45/1024\approx 0.04395$. ✓

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Ex 4 — Cell D: degenerate tiny systems

Forecast: A one-coin system — how many macrostates, how many total microstates?

1. $N=1$: list macrostates. $n=0$ (down) and $n=1$ (up). Why this step? With one object there are only two "how many up" values.
2. Their multiplicities. $\left(\genfrac{}{}{0ex}{}{1}{0}\right)=1,\left(\genfrac{}{}{0ex}{}{1}{1}\right)=1$; total $=1+1=2=2^1$. ✓ Why this step? Confirms ${\sum }_n\left(\genfrac{}{}{0ex}{}{N}{n}\right)=2^N$ still holds at the smallest nontrivial size.
3. $N=0$: the empty system. There is one macrostate ($n=0$) with $\left(\genfrac{}{}{0ex}{}{0}{0}\right)=\frac{0!}{0!0!}=1$, and $2^0=1$ total microstate. Why this step? The "empty configuration" is a legitimate single microstate. Multiplicity is never $0$ here — the machine degrades gracefully.

Verify: $\left(\genfrac{}{}{0ex}{}{1}{0}\right)+\left(\genfrac{}{}{0ex}{}{1}{1}\right)=2=2^1$ and $\left(\genfrac{}{}{0ex}{}{0}{0}\right)=1=2^0$. ✓

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Ex 5 — Cell E: the large-$N$ limit and the sharp peak

Forecast: Will $P(50)$ be close to $1$, or small? (Trap: it is actually small — read on.)

1. Peak probability. $P(50)=\left(\genfrac{}{}{0ex}{}{100}{50}\right)/2^{100}\approx \mathrm{0.0796.}$ Why this step? Even the single most-likely macrostate is under 8%. The peak is broad in $n$ but narrow relative to $N$ — those are different statements.
2. Relative width. For a fair two-state system the standard deviation of $n$ is $\sigma =\frac{1}{2}\sqrt{N}$. Here $\sigma =\frac{1}{2}\sqrt{100}=5$. Why this step? The binomial spread grows like $\sqrt{N}$ while the centre $n=N/2$ grows like $N$.
3. Fractional spread. $\frac{\sigma }{N/2}=\frac{5}{50}=0.10$, i.e. $10\%$. Why this step? This is $1/\sqrt{N}=1/10$. For $N=10^{23}$ it becomes $\sim 3\times 10^{-12}$ — you only ever measure the peak. That is equilibrium.

Verify: $\sigma =\frac{1}{2}\sqrt{100}=5$, relative spread $=1/\sqrt{100}=0.1$, and $\left(\genfrac{}{}{0ex}{}{100}{50}\right)/2^{100}\approx 0.0796$. ✓

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Ex 6 — Cell F: biased objects (microstates NOT equally likely)

Forecast: Does $\Omega (3)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ still appear? Yes — but it is no longer the whole story.

1. Count the sequences. There are still $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ ordered patterns with 3 heads. Why this step? The combinatorics of arrangement is unchanged — bias does not change how many patterns exist.
2. Weight each pattern. Each specific "3H,2T" sequence now has probability $p^3{q}^2=0.7^3\cdot 0.3^2=0.343\cdot 0.09=0.03087$. Why this step? When outcomes are biased, microstates are not equally likely, so we cannot just divide by $2^N$. We multiply the count by the per-sequence weight.
3. Combine. $P(3\text{ H})=\left(\genfrac{}{}{0ex}{}{5}{3}\right)p^3{q}^2=10\times 0.03087=\mathrm{0.3087.}$ Why this step? This is the general binomial distribution; the fair coin is the special case $p=q=\frac{1}{2}$, where $p^n{q}^{N-n}=2^{-N}$ for every sequence and we recover $\Omega /2^N$.

Verify: $10\cdot 0.7^3\cdot 0.3^2=0.3087$. ✓

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Ex 7 — Cell G: a continuous system (ideal gas)

Forecast: Multiplicity grows by $2^{\text{(something)}}$ — is the exponent $N$, $3N/2$, or $2N$?

1. Isolate the $V$-dependence. Only $V^N$ changes. $\frac{{\Omega }_{2V}}{{\Omega }_V}=\frac{(2V{)}^N}{V^N}=2^N.$ Why this step? Each of the $N$ particles independently gains access to a doubled position space — factors multiply, giving one factor of $2$ per particle.
2. Plug $N=100$. Factor $=2^{100}\approx 1.27\times 10^{30}$. Why this step? Concrete size: even 100 particles gain $10^{30}$ more microstates by doubling the box. This is why a gas rushes to fill available volume.
3. Take the log (entropy language). $\ln (2^{100})=100\ln 2\approx \mathrm{69.3.}$ Why this step? Entropy is $S=k\ln \Omega$ (Entropy and Boltzmann's relation S = k ln Ω), so $\Delta S=k\cdot 100\ln 2$ — a positive jump, matching the second law.

Verify: $(2V{)}^{100}/V^{100}=2^{100}$ and $\ln 2^{100}=100\ln 2\approx 69.31$. ✓

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Ex 8 — Cell H: real-world word problem

Forecast: More or less than a coin-flip's 50%? Guess before summing.

1. Model it. Each marble is a fair two-state object (L/R). "Number on the left" $=n$ is the macrostate; total microstates $=2^{20}=1,048,576$. Why this step? Same coin machine — the word "marbles" is just costume.
2. Sum the favourable multiplicities. Balanced means $n\in \{8,9,10,11,12\}$: ${\Omega }_{\text{bal}}=\left(\genfrac{}{}{0ex}{}{20}{8}\right)+\left(\genfrac{}{}{0ex}{}{20}{9}\right)+\left(\genfrac{}{}{0ex}{}{20}{10}\right)+\left(\genfrac{}{}{0ex}{}{20}{11}\right)+\left(\genfrac{}{}{0ex}{}{20}{12}\right).$ Why this step? Macrostate probabilities add because the macrostates are mutually exclusive; each contributes its own count of microstates.
3. Numbers. $125970+167960+184756+167960+125970=772616.$ Why this step? Note the symmetry $\left(\genfrac{}{}{0ex}{}{20}{8}\right)=\left(\genfrac{}{}{0ex}{}{20}{12}\right)$ — the distribution is mirror-symmetric about $n=10$.
4. Probability. $P=772616/1048576\approx \mathrm{0.737.}$ Why this step? Nearly 74% — the balanced band captures most of the probability, exactly because macrostates near the peak dominate (Ex 5's lesson in disguise).

Verify: $\left(\genfrac{}{}{0ex}{}{20}{8}\right)=125970$, sum $=772616$, and $772616/2^{20}\approx 0.7368$. ✓

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Ex 9 — Cell I: exam twist (composite system)

Forecast: Do the two counts multiply, or add? And is the "joined" count bigger or smaller than the "separate" count?

1. Separate multiplicities. ${\Omega }_A=\left(\genfrac{}{}{0ex}{}{3}{1}\right)=3,{\Omega }_B=\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6.$ Why this step? Each subsystem is counted with the same rule, independently.
2. Combine — multiply. ${\Omega }_{A+B}^{\text{(fixed each)}}={\Omega }_A\cdot {\Omega }_B=3\times 6=18.$ Why this step? Independent choices multiply (fundamental counting principle): any of $A$'s 3 patterns pairs with any of $B$'s 6.
3. Now relax the constraint. If only the total $n=3$ out of $N=7$ is fixed, ${\Omega }_{\text{total}}=\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35.$ Why this step? Removing the internal partition opens new microstates (e.g. $n_A=0,n_B=3$) that the rigid split forbade.
4. Interpret. $35>18$: letting the systems share raises the microstate count. This is why heat flows — the combined system slides to the arrangement with the most microstates (see Temperature as ∂S/∂E and the Microcanonical ensemble). Why this step? It connects pure counting to the direction of spontaneous change.

Verify: $\left(\genfrac{}{}{0ex}{}{3}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{4}{2}\right)=3\cdot 6=18$; $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$; and the identity ${\sum }_k\left(\genfrac{}{}{0ex}{}{3}{k}\right)\left(\genfrac{}{}{0ex}{}{4}{3-k}\right)=\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$ (Vandermonde) confirms the relaxed count collects all internal splits. ✓

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Recall Which cell was which?

Edge $n=0,N$ ::: Ex 1 (Cell A) Peak / equilibrium ::: Ex 2 (Cell B) Generic off-peak ratio ::: Ex 3 (Cell C) Degenerate $N=0,1$ ::: Ex 4 (Cell D) Large-$N$ sharpening $1/\sqrt{N}$ ::: Ex 5 (Cell E) Biased (weighted) microstates ::: Ex 6 (Cell F) Continuous gas $V^N{E}^{3N/2}$ ::: Ex 7 (Cell G) Word problem ::: Ex 8 (Cell H) Composite / multiplied multiplicities ::: Ex 9 (Cell I)

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Connections

• Parent — microstate & macrostate
• Binomial and Gaussian distributions
• Fundamental postulate of statistical mechanics
• Entropy and Boltzmann's relation S = k ln Ω
• Second law and irreversibility
• Microcanonical ensemble
• Temperature as ∂S/∂E
• Phase space and Liouville's theorem