11.01.03 · stat-mech-physics / thermodynamics

Third law of thermodynamics: Nernst heat theorem, absolute entropy, and zero-point entropy

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Anchor (Master): Landau & Lifshitz, Statistical Physics, Part 1, 3rd ed. (1980), §20–23; Pathria & Beale, Statistical Mechanics, 3rd ed. (2011), Ch. 3

Intuition Beginner

The first law says energy is conserved. The second law says entropy tends to increase. The third law says something about what happens at the coldest possible temperature.

As you cool a system toward absolute zero — temperature $T$ approaching zero — the entropy $S$ approaches a minimum value. Not just "a" minimum, but the minimum: a universal floor that you can set to zero by convention. The third law states that the entropy change in any process between equilibrium states vanishes as the temperature approaches zero.

In practical terms, you can never actually reach absolute zero. You can get closer and closer — current experiments reach below one nanokelvin — but every step toward zero temperature becomes harder. The third law guarantees this: cooling a system by any finite process requires an entropy change, and as $T$ approaches zero, no entropy change is available to drive the cooling.

Three consequences follow immediately. Heat capacities must go to zero at $T = 0$ (otherwise you could extract entropy at arbitrarily low temperature). Thermal expansion coefficients must go to zero (otherwise you could use mechanical work to create an entropy change). And the Gibbs and Helmholtz free energies become indistinguishable from the internal energy as $T \to 0$ , because the $T S$ term vanishes.

Visual Beginner

Picture a plot of entropy $S$ versus temperature $T$ for a pure substance. The curve starts at the origin $(T = 0, S = 0)$ — this is the third law's content — and rises monotonically. Near $T = 0$ the curve is very flat: the entropy barely increases because the heat capacity is vanishingly small. At higher temperatures the curve climbs more steeply.

Now overlay a second curve: the entropy of the same substance in a different phase or under a different pressure. Both curves start at $S = 0$ when $T = 0$ . Their difference — the entropy change for the transition between the two states — squeezes to zero at the origin. The two curves may diverge at higher temperatures, but they converge at $T = 0$ .

Worked example Beginner

Take a solid at very low temperature. Experimentally, the heat capacity at constant volume $C_{V}$ follows the Debye $T^{3}$ law:

C_{V} = \frac{12 π ^{4}}{5} N k_{B} (\frac{T}{Θ _{D}})^{3},

where $Θ_{D}$ is the Debye temperature (a material constant). The entropy is the accumulated sum of $C_{V} / T$ over all temperatures from zero to $T$ . Substituting the Debye form and carrying out the sum:

S (T) = \frac{4 π ^{4}}{5} N k_{B} (\frac{T}{Θ _{D}})^{3} .

As $T \to 0$ , the entropy $S \to 0$ — exactly as the third law requires. The heat capacity also vanishes ( $C_{V} \sim T^{3} \to 0$ ), consistent with the consequence that no finite heat capacity can persist to absolute zero.

Check your understanding Beginner

Formal definition Intermediate+

Let $S (T, X)$ denote the entropy of a thermodynamic system at temperature $T$ in equilibrium state $X$ , where $X$ denotes the remaining thermodynamic variables (volume, pressure, magnetic field, etc.).

Nernst heat theorem (1906). For any two equilibrium states $X_{1}$ and $X_{2}$ of a system in internal equilibrium,

T \to 0^{+} lim [S (T, X_{2}) - S (T, X_{1})] = 0.

Planck's stronger form (1911). The entropy of every pure substance in internal equilibrium approaches zero as $T \to 0$ :

T \to 0^{+} lim S (T, X) = 0.

Planck's form sets a universal entropy scale. The absolute entropy of a substance at temperature $T$ is then

S (T) = \int_{0}^{T} \frac{C ( T ^{'} )}{T ^{'}} d T^{'},

where $C$ is the appropriate heat capacity and the integral is evaluated along a reversible path from $T = 0$ to $T$ . The convergence of this integral requires $C (T) \to 0$ as $T \to 0$ — itself a consequence of the third law.

Microscopic interpretation

The Boltzmann entropy $S = k_{B} ln Ω$ connects the third law to the ground-state degeneracy $Ω_{0}$ of a quantum system. The third law requires $Ω_{0}$ to be small — specifically, $ln Ω_{0} / N \to 0$ as $N \to \infty$ for a system of $N$ particles. A non-degenerate ground state ( $Ω_{0} = 1$ , $S = 0$ ) satisfies this exactly. A ground state with a degeneracy that grows sub-exponentially in $N$ also satisfies it.

Systems that violate the strong form — so-called frustrated systems — have a residual entropy $S_{0} > 0$ at $T = 0$ . The classic example is ice, where proton disorder on the hydrogen-bond network produces $Ω_{0} \sim (3/2)^{N}$ , giving $S_{0} \approx N k_{B} ln (3/2)$ . Such systems do not violate the Nernst form (entropy differences still vanish at $T = 0$ for processes between states of the same material), but they violate Planck's stronger statement.

Key derivation: Consequences of the third law Intermediate+

Proposition. The third law implies the following thermodynamic consequences:

Vanishing heat capacities. $C_{V} \to 0$ and $C_{P} \to 0$ as $T \to 0$ .
Vanishing thermal expansion. $α_{V} := V^{- 1} (\partial V / \partial T)_{P} \to 0$ as $T \to 0$ .
Equivalence of free energies. $G - H \to 0$ as $T \to 0$ (where $G$ is the Gibbs free energy and $H$ is the enthalpy).

Proof of (1). The entropy at constant volume is $S (T, V) = \int_{0}^{T} C_{V} (T^{'}, V) / T^{'} d T^{'}$ . By the third law this integral converges. For the integral $\int_{0}^{T} (C_{V} / T^{'}) d T^{'}$ to converge at its lower limit, $C_{V} / T^{'}$ must be integrable near $T = 0$ . If $C_{V}$ approached a non-zero constant $c$ as $T \to 0$ , the integrand would behave as $c / T^{'}$ near $T^{'} = 0$ , giving a logarithmic divergence. Hence $C_{V} \to 0$ .

More precisely, the Debye model and experiment show $C_{V} \sim T^{3}$ for insulators and $C_{V} \sim T$ for metals at low $T$ . Both vanish at $T = 0$ .

Proof of (2). The Maxwell relation $(\partial S / \partial P)_{T} = - (\partial V / \partial T)_{P}$ gives $(\partial S / \partial P)_{T} = - V α_{V}$ . By the third law, $(\partial S / \partial P)_{T} \to 0$ as $T \to 0$ (since $S$ itself approaches a pressure-independent constant). Therefore $V α_{V} \to 0$ , and since $V$ remains finite, $α_{V} \to 0$ .

Proof of (3). The Gibbs free energy is $G = H - T S$ . As $T \to 0$ , the product $T S \to 0$ provided $S$ is bounded. Since $S \to 0$ by the third law and $S$ is continuous for systems in internal equilibrium, $T S \to 0$ and $G \to H$ .

Bridge. The third law builds toward the low-temperature behaviour of quantum statistical systems — Bose-Einstein condensation 11.05.04 and the Fermi gas 11.05.05 — where the heat capacity vanishes as $T \to 0$ in precise quantitative agreement with these thermodynamic consequences. The key insight is that the third law is not merely a convention about entropy zero-points but a substantive physical constraint that appears again in the microscopic theory: the ground state of a quantum system has a degeneracy that does not grow exponentially with $N$ , and this is dual to the thermodynamic statement that $C_{V} \to 0$ . The foundational reason is that quantum mechanics quantises the energy levels, creating a gap between the ground state and the first excited state that suppresses entropy at low temperature.

Exercises Intermediate+

Exercise 6 (hard, symbolic).

Prove the unattainability principle: show that no finite number of adiabatic demagnetisation steps can cool a paramagnetic salt to $T = 0$ . Assume the entropy at applied field $B$ and temperature $T$ satisfies $S (B, T) = S (0, T^{'})$ for an adiabatic process connecting $(B, T)$ to $(0, T^{'})$ , and use the third law.

Hint

An adiabatic process conserves entropy. Sketch the $S (T)$ curves for $B = 0$ and $B > 0$ and show they both approach $S = 0$ as $T \to 0$ , so no finite step can reach $T = 0$ .

Answer

In adiabatic demagnetisation, the system starts at $(B_{1}, T_{1})$ and is adiabatically (isentropically) reduced to $B_{2} = 0$ at temperature $T_{2}$ . Conservation of entropy gives $S (B_{1}, T_{1}) = S (0, T_{2})$ .

By the third law, $S (0, T) \to 0$ only as $T \to 0$ . For the final temperature to be $T_{2} = 0$ , we need $S (B_{1}, T_{1}) = S (0, 0) = 0$ . But $S (B_{1}, T_{1}) > 0$ for any $T_{1} > 0$ (since entropy is monotonically increasing in $T$ ). So $T_{2} > 0$ always.

Repeating the process: start at $(B_{2}, T_{2})$ with $T_{2} < T_{1}$ but $T_{2} > 0$ . Each step reduces $T$ but never to zero, because the entropy curves for $B > 0$ and $B = 0$ both approach zero only asymptotically. The temperature after $n$ steps satisfies $T_{n} > 0$ for all finite $n$ . Absolute zero is unattainable by any finite number of steps — this is the unattainability statement of the third law.

Exercise 7 (hard, symbolic).

Ice Ih has a residual entropy $S_{0} = N k_{B} ln (3/2)$ due to proton disorder (Pauling's estimate). Show that this violates Planck's form of the third law but not the Nernst heat theorem. Compute $S_{0}$ in J/mol/K for one mole of ice.

Hint

Pauling's argument: each water molecule has two protons close and two far. For each O-O bond, there are 2 allowed proton positions out of $2^{2} = 4$ total, giving a factor $3/2$ per molecule in the degeneracy estimate.

Answer

Pauling's estimate: each water molecule donates two protons to its four hydrogen bonds, with 2 of 4 orientations close. The number of allowed configurations per molecule is approximately $3/2$ , giving $Ω_{0} \approx (3/2)^{N}$ and $S_{0} = k_{B} ln (3/2)^{N} = N k_{B} ln (3/2)$ .

This violates Planck's form because $S (T = 0) = S_{0} > 0$ . The Nernst heat theorem is not violated because entropy differences between states of the same ice structure still vanish as $T \to 0$ — the residual entropy is the same for all such states.

Numerically: $S_{0} = N_{A} k_{B} ln (3/2) = R ln (3/2) = 8.314 \times 0.405 = 3.37$ J/(mol K), in reasonable agreement with the experimental value $3.41$ J/(mol K).

Exercise 8 (hard, symbolic).

A two-level system has energy levels $E_{0} = 0$ and $E_{1} = ϵ$ with degeneracies $g_{0}$ and $g_{1}$ respectively. Compute the entropy at temperature $T$ and find the condition on $g_{0}$ and $g_{1}$ for the third law to hold (in Planck's form).

Hint

Compute the partition function, then $S = k_{B} (ln Z + β ⟨ E ⟩)$ , and take $T \to 0$ .

Answer

The partition function is $Z = g_{0} + g_{1} e^{- β ϵ}$ . The average energy is $⟨ E ⟩ = g_{1} ϵ e^{- β ϵ} / Z$ . The entropy is

S = k_{B} [ln Z + \frac{β g _{1} ϵ e ^{- β ϵ}}{Z}] .

As $T \to 0$ ( $β \to \infty$ ): $Z \to g_{0}$ , $⟨ E ⟩ \to 0$ , and $S \to k_{B} ln g_{0}$ . Planck's form ( $S \to 0$ ) requires $g_{0} = 1$ : the ground state must be non-degenerate. If $g_{0} > 1$ , the system has residual entropy $S_{0} = k_{B} ln g_{0}$ at $T = 0$ — a violation of Planck's form but not necessarily of the Nernst theorem (since different macroscopic states may share the same ground-state degeneracy, making entropy differences vanish).

For the third law to hold in full generality for all substances, the overwhelming majority of pure crystalline solids have $g_{0} = 1$ . Frustrated systems with $g_{0} > 1$ are the exception.

Lean formalization Intermediate+

The third law has no formalization in Mathlib. The core mathematical content that would need to be formalised is:

A type ThermodynamicState carrying continuous functions $S$ , $C_{V}$ , $T$ , $P$ , $V$ satisfying thermodynamic identities (Maxwell relations, Euler relations).
The limit statement: for any two states in internal equilibrium, $lim_{T \to 0^{+}} ∣ S (T, X_{1}) - S (T, X_{2}) ∣ = 0$ .
Derived consequences: $C_{V} (T) \to 0$ , $α_{V} (T) \to 0$ .
The unattainability principle as an impossibility result: no finite sequence of adiabatic processes reaches $T = 0$ .

The obstruction is not deep mathematics but the absence of a thermodynamic-state infrastructure in Mathlib. Once such a structure exists with the appropriate continuity and monotonicity assumptions, the proofs are elementary limit arguments.

Advanced results Master

Simon's formulation and the role of internal equilibrium

The third law applies only to systems in internal equilibrium. A glass — a supercooled liquid frozen in a non-equilibrium configuration — can have a residual entropy at $T = 0$ that depends on its preparation history. Simon (1937) clarified that the third law applies to systems that have reached their thermodynamic equilibrium state, not to metastable or frozen non-equilibrium configurations.

This distinction matters operationally. The third law is sometimes stated as: the entropy of a system in internal equilibrium is zero at absolute zero. A glass is not in internal equilibrium, so its residual entropy does not constitute a violation. The rigorous formulation restricts attention to states on the equilibrium manifold — the set of states accessible by reversible (quasi-static) processes.

Negative temperatures and the third law

Systems with bounded energy spectra (e.g., nuclear spins in a magnetic field) can realise negative absolute temperatures. At negative $T$ , the population of the higher-energy state exceeds that of the lower-energy state. Entropy at negative $T$ is higher than at any positive $T$ , and the third law still holds: as $T \to 0^{+}$ , $S \to 0$ . The passage from positive to negative $T$ goes through $T = \pm \infty$ (maximum entropy), not through $T = 0$ . Negative temperatures do not violate the third law because they are hotter than any positive temperature — the system has more energy, not less.

The third law in quantum statistics

For a quantum system with Hamiltonian $\hat{H}$ and energy eigenvalues $E_{0} < E_{1} \leq E_{2} \leq \dots$ , the canonical partition function at inverse temperature $β$ is

Z (β) = n \sum g_{n} e^{- β E_{n}},

where $g_{n}$ is the degeneracy of level $n$ . The entropy is $S = k_{B} (ln Z + β ⟨ E ⟩)$ . As $T \to 0$ , $⟨ E ⟩ \to E_{0}$ and $Z \to g_{0} e^{- β E_{0}}$ , so

S \to k_{B} ln g_{0} .

The third law (Planck form) holds if and only if $g_{0} = 1$ . For a system with a non-degenerate ground state and an energy gap $Δ = E_{1} - E_{0} > 0$ , the approach to zero entropy is exponentially fast: $S \sim g_{1} / g_{0} \cdot e^{- β Δ}$ for $k_{B} T ≪ Δ$ .

For a gapless system ( $Δ = 0$ , a continuum of excitations starting at $E_{0}$ ), the approach to zero entropy depends on the density of states. A phonon spectrum with $ρ (ϵ) \sim ϵ^{2}$ in three dimensions gives $S \sim T^{3}$ ; a photon spectrum with $ρ (ϵ) \sim ϵ^{2}$ gives $S \sim T^{3}$ as well. In both cases $S \to 0$ , but the rate is power-law rather than exponential.

Residual entropy and frustration

A frustrated system has competing interactions that cannot all be satisfied simultaneously. The classic examples are:

Spin ice (Dy $_{2}$ Ti $_{2}$ O $_{7}$ , Ho $_{2}$ Ti $_{2}$ O $_{7}$ ): magnetic moments on a pyrochlore lattice obey "two-in, two-out" ice rules, producing a residual entropy identical in form to that of water ice.
Geometrically frustrated antiferromagnets: triangular or kagome lattice antiferromagnets where no spin configuration satisfies all anti-alignment constraints.
Structural glasses: configurational disorder frozen in at the glass transition temperature $T_{g}$ , with the residual entropy depending on the cooling rate.

These systems illustrate the boundary of the third law's applicability: the Nernst form ( $Δ S \to 0$ ) typically still holds, but Planck's absolute zero entropy fails because the ground-state manifold has macroscopic degeneracy.

Synthesis. The third law sits at the boundary between classical thermodynamics and quantum mechanics — it is the bridge from the macroscopic entropy concept to the microscopic ground-state structure. The central insight is that the third law encodes the quantum mechanical quantisation of energy: the gap between ground and first excited state is what forces entropy to zero at $T = 0$ . This is dual to the statement that heat capacities vanish, which appears again in the Debye model 11.05.06 and the Fermi gas 11.05.05. The foundational reason the third law holds for most (but not all) physical systems is that quantum mechanics provides a non-degenerate ground state separated by a finite gap from excited states; frustrated systems with degenerate ground states represent the exceptional case where the generalisation of the third law must be refined. Putting these together, the third law is the thermodynamic shadow of quantum mechanics: it is exactly the macroscopic manifestation of the quantum energy gap.

Full proof set Master

Proposition (Simon's formulation). For any system in internal equilibrium, the entropy difference $Δ S$ between any two equilibrium states at the same temperature vanishes as $T \to 0$ . Equivalently, the isotherms in the $S$ - $X$ plane (where $X$ denotes all other thermodynamic variables) converge as $T \to 0$ .

Proof. Let $S_{1} (T)$ and $S_{2} (T)$ be the entropies of two equilibrium states at temperature $T$ . By the second law, any reversible isothermal process connecting the two states at temperature $T$ has entropy change $Δ S (T) = S_{2} (T) - S_{1} (T)$ . The heat absorbed is $Q = T Δ S (T)$ .

By the third law (Nernst form), $lim_{T \to 0} Δ S (T) = 0$ . Hence $Q = T Δ S (T) \to 0$ as well (since $T \to 0$ and $Δ S$ is bounded). This means no isothermal process at $T = 0$ can transfer heat — the system is thermally inert at absolute zero.

For the stronger (Planck) form: if the ground state is unique ( $g_{0} = 1$ ), then $S (T) = \int_{0}^{T} C / T^{'} d T^{'}$ and the integral converges. Conversely, if the integral converges for every reversible path from $T = 0$ , then $S (0)$ is well-defined and can be set to zero by convention. The convergence of $\int_{0}^{T} C / T^{'} d T^{'}$ is equivalent to $C (T) / T$ being integrable near $T = 0$ , which requires $C (T) \to 0$ faster than $1/∣ ln T ∣$ (a very weak constraint). In practice, all known materials have $C \sim T^{n}$ for $n \geq 1$ at low $T$ , far stronger than the minimum requirement. $□$

Connections Master

11.05.05 The Fermi gas at low temperature has $C_{V} \sim T$ , satisfying the third law with the linear coefficient $γ$ providing direct information about the density of states at the Fermi level.
11.05.06 The Debye model gives $C_{V} \sim T^{3}$ at low $T$ , and the Debye temperature $Θ_{D}$ governs the rate at which entropy approaches zero — the third law constrains the phonon density of states at low energy.
11.06.01 The ferromagnetic Ising model has a non-degenerate ground state (all spins up or all spins down, related by symmetry) with $S (T = 0) = 0$ , satisfying the third law; frustrated spin models provide counterexamples.
11.01.02 Thermodynamic potentials and Legendre transforms relate the third law's consequences to the behaviour of $G$ , $F$ , and $H$ at low temperature — all coincide at $T = 0$ .
11.04.01 The canonical ensemble partition function $Z = \sum_{n} g_{n} e^{- β E_{n}}$ gives the microscopic mechanism: the ground-state degeneracy $g_{0}$ controls whether $S \to k_{B} ln g_{0}$ or $S \to 0$ .

Historical and philosophical context Master

Walther Nernst proposed the heat theorem in 1906 on the basis of extensive measurements of chemical equilibria at low temperatures. His original formulation was thermodynamic: the entropy change in any chemical reaction vanishes as $T \to 0$ . Planck (1911) strengthened this to the statement that every pure substance has zero entropy at $T = 0$ , giving an absolute entropy scale.

The third law occupies a curious position in thermodynamics. Unlike the first and second laws, which are empirical generalisations with no known exceptions, the third law has known exceptions (frustrated systems, glasses). Its validity is tied to quantum mechanics: a quantum system with a non-degenerate ground state automatically satisfies it, while a classical system — with a continuous phase space — has no mechanism to force $S \to 0$ .

Einstein (1914) noted that the third law is essentially a quantum-mechanical statement about the behaviour of matter at low temperatures, not a purely thermodynamic law. This perspective was formalised by Simon (1930s), who distinguished between the unrestricted third law (which fails for glasses and frustrated systems) and the restricted third law (which applies only to systems in internal equilibrium and holds universally).

The unattainability principle — that $T = 0$ cannot be reached by any finite process — was shown by Belgiorno (2003) to be logically independent of the Nernst heat theorem under certain conditions, though in standard thermodynamics they are generally treated as equivalent.

Bibliography Master

@article{nernst1906,
  author = {Nernst, Walther},
  title = {Ueber die Berechnung des Sauerstoffes aus dem Wirmestrom},
  journal = {Nachr. Ges. Wiss. G\"ott.},
  year = {1906},
  pages = {1--38}
}

@book{planck1911,
  author = {Planck, Max},
  title = {Thermodynamik},
  publisher = {Veit \& Co.},
  address = {Leipzig},
  edition = {3rd},
  year = {1911}
}

@book{callen1985,
  author = {Callen, Herbert B.},
  title = {Thermodynamics and an Introduction to Thermostatistics},
  edition = {2nd},
  publisher = {Wiley},
  year = {1985}
}

@book{pathria-beale2011,
  author = {Pathria, R. K. and Beale, Paul D.},
  title = {Statistical Mechanics},
  edition = {3rd},
  publisher = {Academic Press},
  year = {2011}
}

@article{simon1930,
  author = {Simon, Franz},
  title = {Fifth International Congress of Refrigeration, Berlin},
  journal = {Z. f. K\"altetechnik},
  year = {1930}
}

Prerequisites

11.01.01
11.01.02

Tier anchors

beginner: Schroeder, An Introduction to Thermal Physics (2000), Ch. 3 (Entropy) and Ch. 4 (The Third Law)
intermediate: Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed. (1985), Ch. 11; Reif, Fundamentals of Statistical and Thermal Physics (1965), Ch. 5
master: Landau & Lifshitz, Statistical Physics, Part 1, 3rd ed. (1980), §20–23; Pathria & Beale, Statistical Mechanics, 3rd ed. (2011), Ch. 3

References

Nernst 1906 — Ueber die Berechnung des Sauerstoffes aus dem Wirmestrom · Nachr. Ges. Wiss. Gott. 1906, 1–38. Original statement: as T -> 0 the entropy change in any isothermal process approaches zero. This is the Nernst heat theorem, the earliest form of the third law.
Planck 1911 — Thermodynamik · Veit & Co. (Leipzig), 3rd ed. Planck strengthened Nernst by postulating S -> 0 as T -> 0 for every pure substance in internal equilibrium, giving an absolute entropy scale S(T) = integral from 0 to T of C/T dT.
Callen — Thermodynamics and an Introduction to Thermostatistics, 2nd ed. (Wiley, 1985) · Ch. 11 The Nernst Postulate. The third law as the statement that T = 0 is unreachable by finite processes. Consequences: heat capacities vanish at T = 0, thermal expansion coefficients vanish, the Gibbs free energy G and enthalpy H become indistinguishable.
Reif — Fundamentals of Statistical and Thermal Physics (McGraw-Hill, 1965) · Ch. 5 Statistical thermodynamics. The third law and its microscopic interpretation: S = k_B ln Omega, and the ground state is non-degenerate (or has a small degeneracy that does not grow with N).
Pathria & Beale — Statistical Mechanics, 3rd ed. (Academic Press, 2011) · Ch. 3 The canonical ensemble: low-temperature behaviour of the partition function Z = sum_n g_n exp(-beta E_n) and its connection to absolute entropy.
Landau & Lifshitz — Statistical Physics, Part 1, 3rd ed. (Pergamon, 1980) · Sec. 20 The third law of thermodynamics; Sec. 23 The sign of the heat capacity. Consequences for low-T thermodynamic behaviour.

Estimated time

beginner: 15 min
intermediate: 35 min
master: 60 min