Statistical mechanics — microstate, macrostate

WHY does this distinction exist?

WHY: A litre of gas has $\sim 10^{22}$ molecules. You can never write down all positions and velocities, and you don't need to — thermometers and pressure gauges only see averages. So physics splits into two descriptions:

• Microscopic reality — every coordinate of every particle (huge information).
• Macroscopic measurement — a handful of bulk variables (tiny information).

The deep idea: the bridge between them is counting. Equilibrium is not magic — it's just the macrostate that the largest number of microstates corresponds to.

HOW to count: the coin / spin model (derive from scratch)

WHAT we model: $N$ distinguishable two-state objects (spins up/down, or coins H/T). This is the simplest non-trivial counting problem and all the logic of stat-mech is already here.

• Microstate: the full ordered list, e.g. for $N=4$: $\uparrow\downarrow\uparrow\uparrow$ — a specific sequence.
• Macrostate: just "how many are up", $n$. We don't care which ones.

Why total microstates $=2^N$? Each object independently has 2 choices, and choices multiply (fundamental counting principle): $\Omega_\text{total} = \underbrace{2\times 2 \times \cdots \times 2}_{N} = 2^N$

Why the macrostate $n$ has multiplicity $\binom{N}{n}$? We must count how many distinct sequences have exactly $n$ ups. Pick which $n$ of the $N$ slots are up — order of the slots themselves doesn't matter — so it's a combination: $\Omega(n) = \binom{N}{n} = \frac{N!}{n!\,(N-n)!}$

Why this step? $N!$ orders all objects; but permuting the $n$ ups among themselves ($n!$) and the $N-n$ downs ($\,(N-n)!$) gives the same macrostate, so we divide them out.

Sanity check (Forecast-then-Verify): sum over all macrostates must rebuild every microstate: $\sum_{n=0}^{N} \binom{N}{n} = 2^N \quad\checkmark \text{(binomial theorem with }1+1)$

Worked Example 1 — Four coins by hand

Macrostate = number of heads $n$.

Total $=1+4+6+4+1=16=2^4$. Why this step? Confirms we counted every microstate exactly once.

Most probable macrostate is $n=2$ with $P=6/16=0.375$. The "all heads" macrostate exists but has $P=1/16$.

Worked Example 2 — Probability of an extreme fluctuation

For $N=100$ spins, what's the probability all point up? $P = \frac{\Omega(100)}{2^{100}} = \frac{1}{2^{100}} \approx 8\times10^{-31}$ Why this step? Only one microstate ($\Omega=1$) realises this macrostate, divided by the total $2^{100}$. Even for a tiny system, the extreme macrostate is essentially impossible — this is the statistical origin of irreversibility.

Worked Example 3 — Ideal gas, why $\Omega$ grows with $V$ and $E$

For $N$ free particles in volume $V$ at energy $E$, count microstates as (position cells)×(momentum-shell states): $\Omega \propto V^N \times (\text{area of momentum sphere of radius }\sqrt{2mE})^{} \propto V^N E^{3N/2}$ Why this step? Each particle can be anywhere in $V$ (so $V^N$), and energy $E=\sum p_i^2/2m$ fixes the radius of a $3N$-dimensional momentum sphere; its surface "volume" scales like $E^{(3N-2)/2}\approx E^{3N/2}$. WHY it matters: larger $V$ or $E$ ⇒ vastly more microstates ⇒ higher entropy ⇒ gas spontaneously expands and shares energy. Counting predicts thermodynamics.

Flashcards

Connections

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

Concept Map

Hinglish (regional understanding)