Question bank — Statistical mechanics — microstate, macrostate

True — this is the Fundamental postulate of statistical mechanics; every accessible microstate carries equal weight, and everything else (which macrostate you see) follows from just counting them.

False — a macrostate's probability is $\Omega /{\Omega }_{\text{total}}$ (its own count over the total $2^N$), so macrostates with more microstates behind them are hugely more likely; the flat "equal" rule applies to microstates only.

False — it has multiplicity $\Omega =1$, so it is allowed; it is merely fantastically improbable ($1/2^N$), which is a different statement from impossible.

False — a microstate fixes every coordinate, hence it fixes every bulk average too, so it belongs to exactly one macrostate; the many-to-one map runs the other way ("many micros, one macro").

True for discrete counting like $\left(\genfrac{}{}{0ex}{}{N}{n}\right)$ — you are counting distinct configurations; for continuous phase space $\Omega$ becomes a volume and must be divided by the cell size $h^{3N}$ (with $h$ Planck's constant) to become a pure number.

False — the microstate churns wildly and never repeats; only the macrostate is frozen because almost every microstate the system wanders into shares that same most-probable macrostate.

True as microstates (each specific ordered sequence has probability $1/2^N$); the confusion is that "about half heads" is a macrostate bundling many such sequences, so it wins.

False — since $\Omega \propto V^N$ (each of $N$ particles independently gets more room $V$), doubling $V$ multiplies $\Omega$ by $2^N$, an astronomically larger factor; this violent growth is exactly why gas expands to fill its container (Second law and irreversibility).

True in absolute terms (width $\sim \sqrt{N}$), but the relative width $\sqrt{N}/N=1/\sqrt{N}$ shrinks — so for large $N$ the distribution is a razor-sharp spike (Binomial and Gaussian distributions).

Reversed — the microstate is the single complete arrangement; the macrostate is the bulk description that many microstates share.

The choices multiply, not add: $N$ objects with 2 options each give $2\times 2\times \cdots =2^N$, by the fundamental counting principle, not $2N$.

Wrong count — "all up" is one specific microstate ($\Omega =1$), so $P=1/2^N$; the $N$-fold reasoning describes "exactly one up", which is $\left(\genfrac{}{}{0ex}{}{N}{1}\right)=N$.

The sum over the macrostate label $n$ is $2^N$ (binomial theorem with $1+1$); it recovers every microstate once, and $2^N\ne N!$ except at tiny $N$.

It scales as $E^{3N/2}$ — energy fixes the radius $\sqrt{2mE}$ of a $3N$-dimensional momentum sphere, and its surface scales like $E^{(3N-2)/2}\approx E^{3N/2}$, not linearly per particle.

$\ln 1=0$, so that macrostate has zero entropy — the perfectly ordered "all up" state is the low-entropy extreme, not the high one (Entropy and Boltzmann's relation S = k ln Ω).

Listing $\sim 2^{10^{23}}$ items is impossible, but combinatorics ($\left(\genfrac{}{}{0ex}{}{N}{n}\right)$, phase-space volume) gives $\Omega$ in closed form — the whole power of the method is a count without enumeration.

Its multiplicity dwarfs all others and the relative fluctuation shrinks as $1/\sqrt{N}$; at $N\sim 10^{23}$ that spread is around $10^{-11}$, so measured bulk variables look perfectly steady.

Only if every microstate is equally weighted can $P(\text{macrostate})=\Omega /{\Omega }_{\text{total}}$; without it we'd have no principled way to assign probabilities and "counting" would be meaningless.

The "spread out" macrostate has overwhelmingly more microstates than the "clustered in one corner" one, so the system almost surely evolves toward it — irreversibility is just a multiplicity landslide.

$N!$ orders all objects, but re-ordering the $n$ ups among themselves or the $N-n$ downs among themselves gives the same macrostate, so those internal permutations must be divided out.

More energy opens a bigger momentum sphere, so $\Omega$ grows ($\propto E^{3N/2}$); since $S=k\ln \Omega$ increases with $E$, this is precisely what makes $\partial S/\partial E$ positive, and by definition $1/T=\partial S/\partial E$ (Temperature as ∂S/∂E).

$\left(\genfrac{}{}{0ex}{}{N}{0}\right)=1$ — there is exactly one way to have everybody down, the same lonely count as "all up", so both extremes are equally (un)likely.

Two microstates (up, down) and two macrostates ($n=0,n=1$) — a degenerate case where the map is one-to-one and the "many micros per macro" picture hasn't switched on yet.

It goes to zero like $1/\sqrt{N}$, so the macrostate becomes effectively deterministic — this limit is why bulk thermodynamics is sharp and reproducible.

Only weakly: $\Omega (1)=2$ versus $\Omega (0)=\Omega (2)=1$, so $P(n=1)=1/2$ — the "sharp peak" story is a large-$N$ phenomenon and simply doesn't exist for tiny systems.

Yes if it violates a constraint (e.g. $n=N+1$ ups is impossible), giving $\Omega =0$ and $P=0$ — such macrostates are simply not accessible and drop out of the counting.

Fixing $E$ pins the radius of the $3N$-dimensional momentum sphere, so allowed microstates lie on a thin shell (a surface), not a single point — this shell is the microcanonical counting region (Microcanonical ensemble).