Third law of thermodynamics — S → 0 as T → 0: Visual walkthrough

Step 1 — What is a "microstate"? Draw the box first

WHAT. Before any formula, we need a picture of the thing we are counting. Imagine a tiny crystal as a grid of atoms. Each atom can be arranged in some way — here, each atom carries a little arrow (call it its orientation or its wiggle direction). One complete choice of every arrow is one microstate: one specific way the whole crystal can look.

WHY. Entropy is not "heat" and not "temperature" — it is a count of possibilities. To count, we must first know what we are counting. That "what" is the microstate. Nothing else on this page makes sense until this box exists.

PICTURE. Look at the figure: the same crystal (same energy, same temperature — the macrostate) drawn three different ways. All three are allowed; each is one microstate. The number of such distinct pictures is what we will call $W$ .

Figure — Third law of thermodynamics — S → 0 as T → 0

Step 2 — Count the pictures: introduce $W$

WHAT. Give the count a name. Let $W = distinct internal pictures number of microstates consistent with the macrostate.$ Here $W$ is a plain counting number: $W = 1, 2, 3, \dots$ . If the crystal has $N$ atoms and each can point 2 ways, then $W = 2^{N}$ (multiply the choices).

WHY. We want a single number that measures "how many ways can this look". A big $W$ means lots of freedom = lots of disorder. A small $W$ means the system is pinned down. This is the raw material the Third Law will squeeze to $1$ .

PICTURE. The figure shows the choices multiplying: 1 atom → 2 pictures, 2 atoms → 4 pictures, 3 atoms → 8 pictures. Each new atom doubles the count, which is why powers appear.

Step 3 — Why a logarithm? Turn multiplying into adding

WHAT. Boltzmann defines entropy as $S = J/K, sets the units k_{B} natural log ln count W .$ The $ln$ is the natural logarithm: $ln W$ asks "e to what power gives $W$ ?"

WHY this tool and not $W$ itself? Two reasons, both visible in the picture:

Additivity. Put two independent crystals side by side. Their combined count multiplies: $W_{tot} = W_{1} W_{2}$ . But we want entropy to add (double the system → double the entropy). The logarithm is the one function that turns multiplication into addition: $ln (W_{1} W_{2}) = ln W_{1} + ln W_{2}$ . No other simple tool does this.
Taming the explosion. $W = 2^{N}$ is astronomically large. $ln 2^{N} = N ln 2$ is a sane, size-proportional number.

PICTURE. The figure shows two boxes fusing: the counts multiply below, but the $ln$ values stack (add). That stacking is the whole reason $ln$ is chosen.

Recall Why not use

$W$ directly as entropy? Because entropy of two independent systems must add, while their microstate counts multiply. ::: The logarithm converts $W_{1} W_{2}$ into $ln W_{1} + ln W_{2}$ , making entropy additive.

Step 4 — Temperature as "how many rungs can I climb?"

WHAT. Energy in a crystal comes in a ladder of levels: a lowest rung (the ground state) and higher rungs (excited states). Temperature $T$ sets how much thermal energy, roughly $k_{B} T$ , is on hand to hop up the ladder.

WHY. The number of accessible microstates depends on how many rungs the system can reach. Hot ( $k_{B} T$ large) → many rungs reachable → many microstates → big $W$ → big $S$ . This links the abstract count $W$ to the physical dial $T$ we actually turn.

PICTURE. The energy ladder: at high $T$ the amber glow reaches many rungs (many arrangements lit up); as $T$ drops, the glow shrinks toward the bottom rung only.

Step 5 — Cool to the bottom: $T \to 0$ forces the ground state

WHAT. Take $T \to 0$ . Then $k_{B} T \to 0$ : there is essentially no thermal energy to climb any rung. The system is forced onto the single lowest rung, the ground state.

WHY. Any higher rung sits an energy gap $Δ E$ above the bottom. The chance of occupying it fades like $e^{- Δ E / (k_{B} T)}$ ; as $T \to 0$ that exponential crashes to $0$ . So every excited possibility switches off and only the ground rung survives.

PICTURE. Same ladder as Step 4, but now the amber glow has collapsed entirely onto the bottom rung. All upper rungs are dark.

Step 6 — For a PERFECT crystal the bottom rung is unique: $W = 1$

WHAT. For a perfect crystal — every atom locked in its unique lattice site, no defects, no orientation ambiguity — the ground state is a single arrangement. Counting it: $W = 1$ .

WHY. "Perfect" means there is literally only one way to place all the atoms in the lowest energy pattern. No choices remain, so the count of ground-state pictures is exactly $1$ .

PICTURE. One clean lattice, every arrow identical and aligned. There is no alternative picture to draw — the point of the figure is its uniqueness.

Now plug the count into Boltzmann: $S = k_{B} ln W = k_{B} ln 1 = k_{B} \cdot 0 = 0.$ Term by term: $W = 1$ (one ground picture) → $ln 1 = 0$ (log of one is zero) → multiplied by $k_{B}$ still $0$ . Zero possibilities-beyond-one means zero entropy.

Step 7 — The degenerate case: when the bottom rung has $g$ floors

WHAT. Suppose the ground state is not unique — it is $g$ -fold degenerate (there are $g$ equally-low arrangements the system can freeze into). Then even at $T = 0$ , $W = g$ , and $S (0) = k_{B} ln g .$

WHY. Real solids like CO can freeze with each molecule pointing CO or OC. Cooling removes the thermal wiggling but cannot decide which frozen orientation each molecule took — that choice gets trapped. With $g > 1$ , $ln g > 0$ , so entropy does not reach zero. This leftover is residual entropy.

PICTURE. A CO crystal at $T = 0$ : molecules randomly locked as CO/OC. Count $g = 2^{N}$ for $N$ molecules — the figure highlights that each site keeps a stuck binary choice.

See also Residual entropy of ice and CO.

Step 8 — The payoff: climbing back up gives ABSOLUTE entropy

WHAT. Because the Third Law nails $S (0) = 0$ (for a perfect crystal), we can add up entropy from the very bottom: $S (T) = = 0 S (0) + \int_{0}^{T} \frac{C _{p} ( T ^{'} )}{T ^{'}} d T^{'} .$ Term by term: $S (0) = 0$ is the anchor the Third Law gives us; $C_{p} (T^{'})$ is the heat capacity (heat needed per degree); dividing by $T^{'}$ comes from $d S = δ Q_{r e v} / T$ .

WHY does the integral not blow up at the bottom? Near $T = 0$ solids obey the Debye law $C_{p} = a T^{3}$ (see Heat capacity and Debye T-cubed law). Then the integrand is $a T^{'3} / T^{'} = a T^{'2} \to 0$ : it vanishes smoothly at the origin, so the area is finite. The Third Law and the vanishing of $C_{p}$ are self-consistent.

PICTURE. The area under the $C_{p} / T$ curve from $0$ to $T$ — shaded amber — starts flat at the origin (no divergence) and accumulates into $S (T)$ .

The one-picture summary

The whole journey on a single $S$ -vs- $T$ chart: the perfect crystal (cyan) glides down to exactly zero at $T = 0$ ; the imperfect crystal (amber) levels off at the residual plateau $k_{B} ln g > 0$ ; both curves flatten (slope $\to 0$ ) because $C_{p} \to 0$ . That flattening-together is also why you can never reach $T = 0$ (unattainability — see Adiabatic demagnetization and reaching low temperatures).

Recall Feynman: the whole walkthrough in plain words

We drew a box of atoms and said: one exact arrangement is a microstate, and $W$ counts them. Independent atoms multiply their choices, so $W$ explodes as powers — that's why we take a logarithm, which turns "multiply" into "add" and keeps entropy sane and additive. Temperature is just the energy budget for climbing an energy ladder: hot means many rungs reachable, many arrangements, big entropy. Cool it and the budget dies — the probability of any upper rung fades like $e^{- Δ E / k_{B} T}$ — so everyone falls to the bottom rung. If the crystal is perfect, there's exactly one bottom arrangement, $W = 1$ , and $ln 1 = 0$ , so entropy hits zero. If instead the bottom has several stuck configurations (CO pointing this way or that), a bit of mess freezes in: residual entropy $k_{B} ln g$ . Finally, because zero is pinned down, we can add up heat capacity from the bottom to get absolute entropy — and the sum behaves, because $C_{p}$ dies as $T^{3}$ near zero.

Connections

Boltzmann entropy S = k ln W — the counting engine used in Steps 1–3.
Heat capacity and Debye T-cubed law — why the Step 8 integral converges.
Absolute entropy and standard molar entropy — the practical payoff of $S (0) = 0$ .
Residual entropy of ice and CO — the $g > 1$ case of Step 7.
Adiabatic demagnetization and reaching low temperatures — unattainability in the summary figure.
Second law of thermodynamics — supplies $d S = δ Q_{r e v} / T$ .