Statistical mechanics — microstate, macrostate: Visual walkthrough

We assume you know nothing but counting. Every symbol is earned before it is used.

Step 1 — What is a "state"? Two words we must not mix

WHAT. Take $N$ coins in a row. Each coin is either H (heads) or T (tails). Two totally different questions:

"Give me the exact pattern" — like H T T H. That full ordered list is a microstate.
"Just tell me how many are heads" — one number $n$ . That is a macrostate.

Here $N$ = the number of coins, and $n$ = the number of heads. Nothing else yet.

WHY split them? Because in a real gas you can only measure the second kind (a couple of bulk numbers). You can never write down the first kind. So physics is forced to ask: how many microstates share the same macrostate?

PICTURE. The left column is one microstate (a full pattern). The right box groups many patterns under a single label "2 heads".

Step 2 — Count every microstate: why $2^{N}$

WHAT. How many complete patterns exist for $N$ coins? Coin 1 has 2 choices; for each of those, coin 2 has 2 choices; and so on.

WHY multiply? This is the fundamental counting principle: independent choices multiply, because every choice for coin 1 can be paired with every choice for coin 2. Two coins → $2 \times 2 = 4$ patterns; three coins → $2 \times 2 \times 2 = 8$ .

$Ω_{total} = N coins 2 \times 2 \times \dots \times 2 = 2^{N}$

$2$ — the number of options per coin (H or T).
$N$ — how many coins, i.e. how many times we multiply.
$Ω_{total}$ — the grand total of distinct microstates.

PICTURE. A branching tree: each level doubles the number of leaves. Count the leaves at the bottom — that is $2^{N}$ .

Recall Why not add the 2's?

Adding ( $2 + 2 + \dots$ ) would count choices for one coin at a time in isolation. But a pattern needs a choice for every coin simultaneously — the choices stack, and stacking independent choices means multiply. ::: multiply, because each coin's option pairs with every option of the others.

Step 3 — Count one macrostate: why $(n N)$

WHAT. Fix the macrostate to "exactly $n$ heads". How many patterns realise it? We must choose which $n$ of the $N$ slots are heads.

WHY a combination and not a permutation? Because the heads are identical — swapping two heads gives the same pattern. We only care which slots are heads, not in what order we picked them. "Choose $n$ slots out of $N$ , order irrelevant" is exactly the combination (see Binomial and Gaussian distributions):

$Ω (n) = (n N) = \frac{N !}{n ! ( N - n )!}$

Reading it term by term:

$N!$ — line up all $N$ coins in every order ( $N$ factorial $= N \times (N - 1) \times \dots \times 1$ ).
$n!$ — but the $n$ heads are interchangeable; their $n!$ reorderings give one pattern, so divide them out.
$(N - n)!$ — likewise the $N - n$ tails are interchangeable; divide out their reorderings too.
$(n N)$ — read " $N$ choose $n$ ": the number of distinct patterns with $n$ heads.

PICTURE. Below: the $N$ slots as boxes; we shade a chosen subset of $n$ . Different shadings that reuse the same slot-set are the same macrostate — that's why we divide.

Step 4 — The whole distribution: plot $Ω (n)$

WHAT. Compute $(n N)$ for every $n$ from $0$ to $N$ and plot it. For $N = 10$ the values are the row $1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1$ of Pascal's triangle.

WHY plot it? A picture reveals the punchline instantly: the counts are not flat. The extremes ( $n = 0$ or $n = N$ ) have $Ω = 1$ ; the middle ( $n = N /2$ ) towers above everything.

PICTURE. Bars of $Ω (n)$ vs $n$ . The green centre bar dwarfs the red edge bars. That height difference is why you never see all-heads.

Step 5 — The forecast that must come true: $\sum_{n} (n N) = 2^{N}$

WHAT. Add up the multiplicities of all macrostates. If our two counts (Step 2 and Step 3) are consistent, this sum must rebuild every microstate exactly once.

$\sum_{n = 0}^{N} (n N) = 2^{N}$

$n = 0 \sum N$ — sweep $n$ across every macrostate, $0, 1, \dots, N$ .
$(n N)$ — microstates in that one macrostate.
$2^{N}$ — the grand total from Step 2.

WHY it works. The binomial theorem says $(a + b)^{N} = \sum_{n} (n N) a^{n} b^{N - n}$ . Put $a = b = 1$ : every term becomes just $(n N)$ , and the left side is $(1 + 1)^{N} = 2^{N}$ . So partitioning microstates by macrostate loses none.

PICTURE. The bars from Step 4, stacked end-to-end, exactly fill the total $2^{N}$ — no overlap, no gaps.

Step 6 — From counts to probability, and the degenerate edges

WHAT. By the Fundamental postulate of statistical mechanics — every accessible microstate is equally likely — the probability of a macrostate is its share of the total:

$P (n) = \frac{Ω ( n )}{2 ^{N}} = \frac{1}{2 ^{N}} (n N)$

$Ω (n)$ — ways to get $n$ heads (favourable microstates).
$2^{N}$ — all microstates (equally likely, by the postulate).
$P (n)$ — the probability you measure macrostate $n$ .

The edge / degenerate cases — none are skipped:

$n = 0$ (all tails) or $n = N$ (all heads): $Ω = 1$ , so $P = 1/ 2^{N}$ . The rarest allowed macrostates — they exist but are almost never seen.
$n = N /2$ (only when $N$ even): the largest $Ω$ , the most probable macrostate.
$N$ odd: no exact middle; the two central values $n = \frac{N - 1}{2}, \frac{N + 1}{2}$ tie for the peak.
$N = 1$ : only two macrostates, each $Ω = 1$ , each $P = 1/2$ — the split is invisible (need many coins for a sharp peak).

PICTURE. Probability curve $P (n)$ for a large $N$ , with the tiny edge bars marked in red and the towering peak in green — plus an arrow showing the peak width shrinking relative to $N$ .

Step 7 — Where does the width come from? Deriving $N$

Before we can talk about "the width shrinking", we must say what "width" means and why it equals $N$ — not just cite it.

WHAT & WHY (the trick that makes $N$ appear). Write the head-count as a sum of one little counter per coin. For coin $i$ let $x_{i} = 1$ if it's heads and $x_{i} = 0$ if tails. Then

$n = x_{1} + x_{2} + \dots + x_{N}$

$x_{i}$ — the $i$ -th coin's contribution: $1$ or $0$ .
$n$ — total heads, just the sum of those.

For one fair coin the mean is $\overset{x}{ˉ} = \frac{1}{2}$ and its variance is

$Var (x_{i}) = p (1 - p) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$

$p = \frac{1}{2}$ — probability that coin shows heads.
$p (1 - p)$ — variance of a single 0/1 coin (largest when $p = \frac{1}{2}$ , zero when the coin is rigged to always land one way).

The key rule (why variances, not $σ$ 's, add). For independent pieces, variances add — the squared strays accumulate without cross-terms because one coin's luck tells you nothing about another's:

$σ^{2} = Var (n) = coins N \times each = 1/4 p (1 - p) = \frac{N}{4}$

Take the square root to get the width itself:

$σ = \frac{N}{4} = \frac{1}{2} N \Rightarrow σ \sim N$

$N$ — number of independent coins; it enters once, linearly, inside the variance.
$\cdot$ — turning variance back into a width halves the exponent, so $N \to N$ .

PICTURE. Left: variance of one coin, $p (1 - p)$ , peaking at $p = \frac{1}{2}$ . Right: $N$ independent strays add as squares (Pythagoras-style), so the total width is $N$ longer, not $N$ longer.

Step 8 — Why $1 0^{23}$ coins means one answer forever

WHAT. The peak of $P (n)$ sits at the centre $\overset{n}{ˉ} = N /2$ . From Step 7 its width (standard deviation) is $σ = \frac{1}{2} N$ , while its position is $N /2$ . So the relative spread — width measured as a fraction of the whole — is

$\frac{width}{position} = \frac{σ}{n ˉ} = \frac{\frac{1}{2} N}{N /2} = \frac{1}{N} N \to \infty 0$

$σ = \frac{1}{2} N$ — the width just derived in Step 7.
$\overset{n}{ˉ} = N /2$ — where the peak sits.
$1/ N$ — the fractional fluctuation, vanishing for large $N$ .

WHY it matters. For $N \sim 1 0^{23}$ , $1/ N \sim 1 0^{- 11}$ : the fraction of heads is pinned to $50%$ to about $1$ part in $1 0^{11}$ . The measurable macrostate simply does not move. That frozen, most-probable macrostate is equilibrium — and its logarithm is entropy.

PICTURE. Three curves $N = 10, 100, 1000$ rescaled onto fraction-of-heads $n / N$ : as $N$ grows the bell collapses to a spike at $0.5$ .

The one-picture summary

Everything at once: the branching tree (all $2^{N}$ microstates) feeds into the bar chart ( $Ω (n)$ per macrostate), whose total re-sums to $2^{N}$ , whose normalised version is $P (n)$ , whose width $σ = \frac{1}{2} N$ sharpens relative to $N$ into equilibrium.

Recall Feynman retelling — the whole walkthrough in plain words

Line up $N$ coins. Each coin flips two ways, so there are $2^{N}$ complete patterns — that's every microstate. But you, the experimenter, only see one number: how many are heads, the macrostate. To find how many patterns give a certain head-count $n$ , you choose which $n$ slots are heads and ignore the order — that's $(n N)$ . Plot those counts and a mountain appears: one lonely pattern at "all heads", a towering pile at "half heads". Add all the piles back up and you recover $2^{N}$ exactly (nothing lost). Since every pattern is equally likely, the probability of a macrostate is just its pile-height over $2^{N}$ . Now the magic of big numbers: give each coin a little $1$ -or- $0$ counter; each wobbles by $\frac{1}{4}$ in variance, and because independent wobbles add as squares, the total wobble is $σ = \frac{1}{2} N$ — a random-walk $N$ , not a straight-line $N$ . So the mountain's width grows like $N$ but its centre grows like $N$ , making it razor-thin relative to its size. At $1 0^{23}$ coins it's a needle at exactly half heads. That immovable, overwhelmingly-favoured macrostate is what we call equilibrium — and heat, temperature, and irreversibility are all just this counting story wearing different clothes.

Connections

Parent: 2.4.08 Statistical mechanics — microstate, macrostate (Hinglish)
Entropy and Boltzmann's relation S = k ln Ω
Fundamental postulate of statistical mechanics
Binomial and Gaussian distributions
Microcanonical ensemble
Second law and irreversibility
Phase space and Liouville's theorem
Temperature as ∂S/∂E