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For two subsystems with independently specified macrostates, suppose every allowed microstate of $A$ can be paired with every allowed microstate of $B$. Their allowed set is a Cartesian product. With equal probability for each pair, Boltzmann's entropy formula givesFor example, take two independent copies of the toy model, each with two labels choosing between its two boxes. Each copy has four configurations and Boltzmann entropy $k_B\ln 4$. Combining their independent labels gives sixteen configurations and Boltzmann entropy $k_B\ln 16=2k_B\ln 4$. This explains why entropy adds even though microscopic counts multiply. ^[ref]

The product assumption matters. A constraint shared between the subsystems can forbid some pairs; for example, fixing only their combined energy generally requires summing over its possible divisions. See entropy maximization under thermal contact for how thermal equilibrium selects the dominant division. Statistical independence cannot be inferred merely from drawing a boundary between two parts.

The Boltzmann distribution assigns probabilities to individual microstates in a canonical ensemble, a statistical ensemble for thermal contact. The system exchanges energy with a thermal reservoir at fixed positive absolute temperature $T$, while its particle number $N$ and external parameters, such as volume, stay fixed. The thermal reservoir is large enough that these exchanges barely change its temperature; the coupling permits thermal equilibrium but contributes negligible interaction energy. All the microstates considered have the same $N$. No chemical potential term is needed in their weights; fixed $N$ does not imply that the chemical potential is zero.

For a finite discrete set of microstates with energies $E_i$, writeHere $k_B$ is the Boltzmann constant, $\beta$ is the inverse temperature, $e^{-\beta E_i}$ is a Boltzmann factor, and $Z$ is the partition function. The Boltzmann factor is a weight; dividing by the partition function makes it a probability. ^[ref]

The useful surprise is that a fixed temperature still permits thermal fluctuations in the system's energy. The Boltzmann distribution predicts their probabilities and expected values without tracking every microscopic exchange. Its exponential function also makes rare excited microstates sensitive to temperature: an energy gap much larger than $k_{B}T$ gives strong Boltzmann suppression, while warming increases their relative populations. A population ratio can therefore reveal a gap or serve as a thermometer, provided thermal equilibrium and the relevant state degeneracies are known.

Historically, Josiah Willard Gibbs's Elementary Principles in Statistical Mechanics (1902) presents the canonical ensemble in Chapter IV. His equations 90–92 give an exponential function of energy and its normalization, in classical phase space. This classical formulation exhibits the same weighting principle. ^{[ref] [ref]}

Group microstates by their energy $E$. The state degeneracy $g(E)$ counts how many microstates have that energy; their aggregate probability isHere $P(E)$ is the probability of the macrostate retaining only energy, while $p_i$ is the probability of one individual microstate. For two such macrostates,Counting more microstates can compensate for a smaller Boltzmann factor. Omitting the state degeneracy confuses a sum over microstates with a sum over distinct energies. ^[ref]

Consider two distinguishable, noninteracting two-level systems, each with energies $0$ and $\Delta >0$. Their four microstates $(0,0),(0,1),(1,0),(1,1)$ have total energies $0,\Delta ,\Delta ,2\Delta$. Thus the three energy macrostates have state degeneracies $1,2,1$. This is a stipulated finite model, not a measured material.

Set $q=e^{-\beta \Delta }$. The four individual Boltzmann factors are $1,q,q,q^2$, so the partition function and aggregate probabilities areEach of the two middle microstates has probability $q/Z$, while their shared energy macrostate has probability $2q/Z$.

At the temperature satisfying $\Delta =k_{B}T\ln 2$, the Boltzmann factor is $q=1/2$ and the partition function is $Z=9/4$. ThereforeThe middle energy is as likely as the ground state energy because its state degeneracy is twice as large. Yet each individual middle microstate is only half as likely as the ground state microstate. The diagram places the four energies beside their individual probabilities, then shows the sums within each macrostate.

The expected value of energy follows directly by weighting the three possible energies:At $q=1/2$ this is $2\Delta /3$. An individual energy measurement still returns $0$, $\Delta$ or $2\Delta$; the expected value describes the ensemble average, rather than a fourth possible measured energy.

For two individual microstates, the partition function cancels:At positive temperature, a higher-energy individual microstate is less probable, and equal-energy individual microstates are equally probable. For example, a gap $E_i-E_j=k_{B}T\ln 2$, using the natural logarithm, gives $p_i/p_j=1/2$. This comparison concerns individual microstates; a whole macrostate can contain several of them.

Changing the reference for energy leaves every probability unchanged. If all energies become $E_i^{\prime }=E_i+C$ for the same constant $C$, thenThe partition function itself changes, but its common factor cancels in the Boltzmann distribution. Thus choosing a ground state energy of zero simplifies calculations without changing populations. The expected value of energy shifts by $C$, as it must when its reference changes.

Keep the gap $\Delta$ fixed and vary positive temperature. With $x=k_{B}T/\Delta$, the Boltzmann factor is $q=e^{-1/x}$. In the four-microstate example, as $T\to 0^{+}$, $q\to 0$: the unique ground state has probability tending to one, and the other microstates have probabilities tending to zero. If several microstates instead shared the lowest energy, their state degeneracy would split that limiting population equally.

As $T\to \infty$ in this fixed finite model, all four individual Boltzmann factors tend to one, so all four individual microstates have probability tending to $1/4$. The three aggregate probabilities instead approachThe equal-population limit applies to this finite set of microstates. An infinite collection of microstates cannot all have the same positive normalized probability, so the statement cannot be extended to an unbounded energy spectrum. Real systems may also acquire additional accessible microstates as temperature rises.

The plotted crossing occurs at $x=1/\ln 2$, where $P(0)=P(\Delta )=4/9$. For larger positive temperatures, the middle macrostate is more probable than the lowest-energy macrostate, although the lowest-energy individual microstate remains more probable than either individual middle microstate. The expected value of energy rises from zero toward $\Delta$ in this finite model.

Take $N=20$ distinguishable, labeled particles and two boxes of equal volume. Retain only whether each particle is in the left or right box: use a bit $0$ for left and $1$ for right. Each length-20 string is one coarse microstate of this toy model. Assume the labels make independent side choices with equal probability, so the $2^{20}=1048576$ strings are equiprobable. Equal volumes motivate these weights for noninteracting particles with no preference for either box; equal volumes alone would not establish them for an arbitrary interacting gas.

Now retain only the number $k$ in the left box as the macrostate. Choosing which $k$ labels are left gives a binomial coefficient, soThe factorials count assignments to labels; the probability of a macrostate sums the probabilities of its constituent microstates. This is the binomial distribution with parameters $20$ and $1/2$. The same binomial coefficient appears in the counting of distinguishable two-level systems, although here the macrostate records occupancy rather than energy. ^[ref]

The largest binomial coefficient occurs at $k=10$, the balanced macrostate. All-left means $k=20$ and supplies just one string. ThusThere are $184756$ balanced strings for every all-left string, even though each individual string has the same probability. Exact balance has about $17.6\%$ of the total probability; equilibrium reasoning concerns the many macrostates near balance too, rather than certainty of exactly ten particles on each side. The diagram compares the counts with their natural logarithms.

Four fair, independent coins give a smaller worked example with the same binomial distribution. If $k$ is the number of heads, the five macrostates $k=0,1,2,3,4$ contain respectively $1,4,6,4,1$ microstates. Their probabilities are respectively $1/16,4/16,6/16,4/16,1/16$. In particular, two heads is one macrostate containing six ordered coin sequences, with Boltzmann entropy $k_B\ln 6$ in this counting analogy. A prescribed ordered sequence has one realization. The next diagram enumerates the same sixteen strings as left/right assignments.

In Boltzmann entropy of a two-box occupancy, impose an initial restriction that all labels are left. There is one allowed coarse microstate. Remove that restriction, allowing all occupancy values: the allowed set is now the union of every occupancy macrostate, containing $2^N$ strings. If the final description assigns equal probability to this whole set, its Boltzmann entropy change isThe counting identity is the binomial theorem. For twenty labels, $\Delta S=20k_B\ln 2\approx 13.862944k_B$. This final description fixes neither box occupancy; it is coarser than the particular macrostate $k=10$, whose Boltzmann entropy is $k_B\ln 184756$. The two counts answer different questions, so substituting the balanced macrostate for the full allowed set would change the calculation.

For four labels, the initial all-left string is 0000; removing the restriction makes all sixteen strings available. The $k=2$ column has only six strings. Accordingly, the whole allowed set has Boltzmann entropy $k_B\ln 16=4k_B\ln 2$, while the balanced macrostate has Boltzmann entropy $k_B\ln 6$.

Removing a barrier changes the allowed set immediately, but it does not instantaneously randomize the actual probability distribution. Motion and equilibration must make those configurations accessible. The counting result illustrates the configurational contribution to free expansion; it does not describe the relaxation time or every contribution to the entropy of a real gas.

At inverse temperature $\beta =1/(k_{B}T)$, the Boltzmann factor $e^{-\beta E}$ is the relative canonical ensemble weight of a microstate with energy $E$.

Why does a gas spread through an available container? One part of the answer is counting: a nearly uniform macrostate can be realized by vastly more microstates than a macrostate with every particle crowded into one corner. Boltzmann's entropy formula turns that multiplicity into entropy. The useful insight is that a large entropy measures how many microscopic realizations fit a specified macroscopic description, rather than how untidy a picture looks.

For a specified macrostate $M$ containing $\Omega (M)$ accessible, equally likely microstates, writeHere $k_B$ is the Boltzmann constant, and $\ln$ is the natural logarithm, also written $\log$ in this article's title. The count $\Omega (M)$ is dimensionless; entropy has units of energy divided by temperature. Always specify the macrostate, the constraints and the model's microscopic resolution before counting. A microcanonical ensemble provides a familiar equilibrium setting for the equal-probability assumption. ^[ref]

The natural logarithm compresses an enormous count into a manageable number and makes products of counts into sums of entropies; see additivity of Boltzmann entropy. With this counting convention, one allowed microstate has zero Boltzmann entropy. This zero refers to the chosen model and resolution; it does not by itself assign an absolute entropy to a real material.

The partition function $Z(\beta )={\sum }_s{e}^{-\beta E_s}$ is the sum of the Boltzmann weights over individual microstates. Dividing by it makes their probabilities sum to one. When summing distinct energies instead, include the state degeneracy: $Z={\sum }_{E}g(E)e^{-\beta E}$. For two independent distinguishable two-level systems,The middle coefficient counts two microstates, exactly as in the preceding figure. For $N$ such units, $Z_N=Z_1^N$; this factorization requires additive energy and independence, and generally fails with interactions.

The usefulness goes beyond normalization. For fixed energies, derivatives of the natural logarithm of the partition function yield measurable quantities:Here $U$ is internal energy, the second expression is the energy variance, and $S$ is Gibbs entropy. For one two-level system at fixed $\Delta$, its heat capacity $C=\partial U/\partial T$ and entropy areAt $\beta \Delta =\ln 3$, these give $U/\Delta =0.25$, $C/k_B\simeq 0.2263$ and $S/k_B\simeq 0.5623$. For independent units, all three quantities add.

The figure shows why heat capacity peaks. At low temperature, the excited energy is rarely occupied; at high temperature, the probabilities are almost equal and further warming barely changes internal energy. The largest response lies between these limits, at $k_{B}T/\Delta \simeq 0.4168$, where $C/k_B\simeq 0.4392$. Meanwhile entropy rises from zero toward $k_B\ln 2$, reflecting two equally likely alternatives. This high-temperature limit belongs to the stipulated two-level system; extra accessible energy eigenstates would change it.

Entropy is an extensive quantity and a state function of an equilibrium system, measured in $\mathrm{J}{\mathrm{K}}^{-1}$. Differences between equilibrium states are calculated along a reversible thermodynamic process:Here heat entering the system is positive and $T$ is its absolute thermodynamic temperature. The reversible path used for calculation need not be the path actually taken. Dividing the actual heat by a single temperature generally fails for an irreversible process.

For example, put two rigid bodies, each with constant heat capacity $C=100\mathrm{J}{\mathrm{K}}^{-1}$, at $600\mathrm{K}$ and $300\mathrm{K}$ in an otherwise isolated system. Assume each body has a uniform internal temperature, no phase transition occurs, and no work is extracted. Their final thermal equilibrium is at $450\mathrm{K}$, because the internal energy lost by one equals that gained by the other. Using entropy from heat capacity,so $\Delta S_{\mathrm{tot}}=C\log (9/8)=11.78\mathrm{J}{\mathrm{K}}^{-1}$. One body's entropy decreases while the total increases. ^[ref]

In statistical mechanics, Boltzmann's entropy formula counts equally likely microstates compatible with a macrostate; Gibbs entropy handles a specified probability distribution over microstates. If the number of equally likely microstates doubles, entropy increases by $k_B\log 2$. The logarithm makes multiplicative counts additive for independent systems. These are precise statements about states and probabilities, not a universal measure of visible messiness. Calling entropy merely "disorder" obscures the thermal calculation above and hides which microstates are being counted. ^[ref]

The usefulness of entropy is that measurable heat capacities determine its changes, and the Second law of thermodynamics turns those changes into constraints on processes. It also connects macroscopic measurements to microscopic descriptions without requiring us to track every particle.

Finite-size rounding replaces a limiting discontinuity or divergence by a finite crossover or peak. For a spatial system near a continuous phase transition, the bulk description ceases to apply unchanged when the correlation length becomes comparable with the sample size. How the crossover width and location depend on size also depends on the model, geometry and boundary conditions; there is no universal $1/N$ rule for every phase transition.

An exact finite example is the Curie–Weiss model with $s_i=\pm 1$, $J>0$, and a conjugate field $h$ in energy units. Set $K=J/(k_{B}T)$ and $u=h/(k_{B}T)$. Its statistical Hamiltonian and finite partition function areThe binomial coefficient counts the microstates with $k$ positive Ising spins. The diagonal terms in the square add a configuration-independent constant to the energy; retaining them fixes the normalization used here. Divide each summand by $Z_N$ to obtain the probability mass function of $m_k=(2k-N)/N$. In particular,The variance expression makes the finite response nonnegative, while spin inversion symmetry gives $⟨m⟩=0$ exactly at $u=0$ for every plotted $N$.

Every curve is computed from the finite partition function, without selecting one minimum of the Bragg-Williams free energy of the Ising model. The field panel holds $K=1.5$ fixed and shows increasingly rapid switching between opposite spin magnetizations. The temperature panel uses $u=0$: its nonzero $⟨|m|⟩$ measures the typical magnitude, while the mean spin magnetization stays zero. The reference $k_{B}T/J=1$ marks the limiting critical temperature, not a singular point of any finite curve.

A short limiting argument explains the field panel. Combining the binomial coefficient with its leading entropy gives a Helmholtz free energy per spin whose stationarity equation is $m=\tanh (Km+u)$. For $K>1$ and $u=0$ the two ordered minima have $m=\pm m_{\ast }$, where $m_{\ast }=\tanh (K{m}_{\ast })$. For large $N$ well below the critical temperature, their relative Boltzmann weight is approximately $e^{2Nu{m}_{\ast }}$, giving $⟨m⟩\approx m_{\ast }\tanh (Nu{m}_{\ast })$. Thus this two-peak approximation has a field width of order $1/(N{m}_{\ast })$ in $u$. It is not valid at the critical temperature, where the peaks merge; the figure uses the exact finite sum throughout.

For finitely many microstates with finite energies $E_{\alpha }(h)$ that are real analytic functions of a real control $h$, the canonical partition function at $T>0$ is a finite positive sum:It and the resulting Helmholtz free energy are locally real analytic functions of $T$ and $h$. Hence such a finite Ising model has no literal nonanalytic equilibrium phase transition. A limit of these real analytic functions can nevertheless fail to be a real analytic function; the narrow but smooth crossover can become a sharp phase transition.^[ref]

Finite particle number alone is not the hypothesis of a finite sum. Classical phase space is continuous, and a system of quantum states can have infinitely many energy eigenvalues: its partition function requires convergence and local bounds that permit differentiation or an analytic extension of the integral or thermal trace. A divergent partition function, a singularly changed statistical Hamiltonian, or the zero-temperature limit falls outside the finite-positive-sum argument. Observing a steep change in a finite sample therefore motivates a limiting model; it does not measure an actual infinity.

Take a cycle graph with four labeled graph vertices and four distinct edges $\{1,2\},\{2,3\},\{3,4\},\{4,1\}$. Set $h=0$, $J>0$, and impose periodic boundary conditions by writing ${\sigma }_5={\sigma }_1$. The square drawing represents a one-dimensional ring: it has no diagonal interactions and is not a two-dimensional lattice.

Call an edge between opposite Ising spins a domain wall. If $d$ is their number, the four bond products sum to $4-2d$, so the energy isEach domain wall costs $2J$ relative to an aligned bond. Returning around the ring must return to the starting Ising spin, so $d$ is even: equivalently, ${\prod }_{i=1}^4({\sigma }_i{\sigma }_{i+1})=1$. Choosing an even set of $d$ wall edges and then choosing ${\sigma }_1$ determines the entire pattern. The state degeneracy is therefore $g_d=2\left(\genfrac{}{}{0ex}{}{4}{d}\right)$, using a binomial coefficient.

The twelve middle microstates consist of eight patterns with one exceptional Ising spin and four patterns with two adjacent $+1$ Ising spins and two adjacent $-1$ Ising spins. The two alternating patterns have four domain walls. All $2+12+2=16$ microstates are shown below.

Writing $K=\beta J$, the partition function and the probability of each energy class areFor an original exact example, choose $K=(\ln 2)/4$. Then $e^{4K}=2$, $Z=17$, and $P(d=0)=4/17$, $P(d=2)=12/17$, $P(d=4)=1/17$. Each aligned microstate has probability $2/17$, each middle microstate $1/17$, and each alternating microstate $1/34$: multiply by state degeneracy before comparing whole classes. As $K\to 0$, all sixteen microstates become equally likely, but the class probabilities approach $1/8,3/4,1/8$. As $K\to \infty$, the two all-aligned microstates each have limiting probability $1/2$.

Take three Ising spins $s_i\in \{-1,1\}$, with an antiferromagnetic interaction of energy scale $J$ on each edge of a triangle. An individually satisfied bond has $s_i{s}_j=-1$. But multiplying all three bond products gives $(s_1{s}_2)(s_2{s}_3)(s_3{s}_1)=1$, whereas three satisfied bonds would give $-1$. At least one bond must therefore remain unsatisfied. This is an exactly finite example of geometrical frustration.

There are six ground states with one unsatisfied bond and ground-state energy $-J$. The two all-aligned microstates have energy $3J$. No interaction is switched off in a ground state: the system minimizes their sum by accepting one local disappointment. Allowing continuous three-component spins instead of Ising spins changes this optimization problem and permits the $120$-degree arrangement described under geometrical frustration.

Writing $\beta =1/(k_{B}T)$ for inverse temperature, the exact partition function and mean internal energy areAt $T>0$ these expressions are smooth: a finite triangle has no thermodynamic phase transition. In zero magnetic field, spin inversion symmetry also gives $⟨s_i⟩=0$ at every site, even as the Boltzmann distribution concentrates on the six frustrated ground states.

A heat bath, or thermal reservoir, is an idealized system so large that exchanging a finite amount of heat does not appreciably change its temperature.

The ideal heat bath supplies a fixed absolute temperature $T_b$ while its internal energy changes through heat transfer. A real body's suitability as a heat bath depends on its heat capacity relative to the exchange being studied, and on how quickly its internal temperature gradients disappear. Large physical size alone supplies neither condition. A finite heat bath makes the approximation quantitative; a regulated calibration bath instead uses feedback control to maintain its temperature by exchanging energy with external equipment.

For a weakly coupled system that exchanges energy but no matter with an ideal heat bath, the canonical ensemble assigns the Boltzmann distributionHere $E_n$ is a microstate's energy, $k_B$ the Boltzmann constant and $Z$ the partition function. The fixed temperature removes the need to track every exchange with the environment. A substantially changing bath temperature no longer gives this one fixed Boltzmann distribution.^[ref]

The Ising model is a spin model with an Ising spin ${\sigma }_v\in \{-1,1\}$ at every graph vertex $v$. For a finite graph with vertex set $V$ and edge set $E$, a uniform coupling $J$ and field parameter $h$ give the HamiltonianEach edge is counted once. Here $T>0$ is temperature, $k_B$ is the Boltzmann constant, and $\beta$ is inverse temperature; $J$ and $h$ have units of energy. For magnetic applications, $h$ includes the magnetic moment multiplying the applied magnetic field. The canonical ensemble assigns the Boltzmann distributionThe partition function $Z$ makes the probabilities sum to one. Positive $J$ rewards equal neighboring Ising spins, modeling ferromagnetism; negative $J$ rewards opposite neighbors, giving an antiferromagnetic Ising model. Positive $h$ rewards $+1$, while negative $h$ rewards $-1$.^[ref]

A single local rule can organize a whole graph: rewarding agreement on each edge makes all-equal Ising spin patterns favorable even though no edge directly couples distant vertices. The four-spin Ising ring makes this visible with only sixteen microstates. It also separates three different questions: whether neighbors agree, whether the mean spin magnetization is nonzero, and how large its thermal fluctuations are.

A motivating experiment is the University of Colorado Boulder's Curie-temperature demonstration. An iron rod suspended from a twisting wire on a stand is initially held near a permanent magnet by ferromagnetism. A gas burner supplied from a propane cylinder heats the rod; the readout is its motion away from the magnet, followed by return as it cools. Repeated cooling supplies a reversible comparison on the same sample. The source explicitly calls for heating the rod while keeping the magnet cool, since heating the magnet can reduce its strength. The inference is that warming the sample weakens its magnetic attraction across the Curie temperature. This is a qualitative force readout, with no calibrated temperature or magnetization trace supplied: it motivates competition between alignment and thermal fluctuations, but does not identify $J$, measure individual Ising spins, or realize a four-spin Ising ring.^[ref]

Ernst Ising's 1925 paper examined a linear chain of two-orientation magnetic elements with neighboring interactions. In its calculation for a large chain, the zero-field mean magnetization vanishes and the model has no spontaneous magnetization. The inspected English translation states this immediately after equation (8) in §2. This historical result concerns the chain; it does not establish the behavior of a two-dimensional Ising model.^[ref]

The Ising ring bond correlation is the neighboring two-point correlation function $c_1=⟨{\sigma }_i{\sigma }_{i+1}⟩$ in the four-spin Ising ring. All four edges are equivalent. At zero field each mean Ising spin is zero by spin inversion symmetry, so $c_1$ also equals the connected correlation function. Agreement gives bond product $+1$ and disagreement $-1$; thus $P({\sigma }_i={\sigma }_{i+1})=(1+c_1)/2$.

Differentiating the finite partition function with respect to $K=\beta J$ inserts the sum of four bond products. ThereforeHere $U$ is internal energy. At $K=(\ln 2)/4$, the class-weight calculation gives $c_1=(4-1)/17=3/17$ and $U=-12J/17$. The agreement probability for a selected bond is $10/17$, despite the exactly zero mean spin magnetization.

A second derivative gives the energy variance, and hence the heat capacity $C_H=dU/dT$ at fixed $J$ and $h=0$:To see the variance identity directly, a derivative of an expected value under weights $e^{KS}$ inserts $S-⟨S⟩$, where $S={\sum }_i{\sigma }_i{\sigma }_{i+1}=-H/J$. The factor $K^2$ follows from $dK/dT=-K/T$. At the same exact example, $\mathrm{Var}(H)/J^2=1216/289$. The heat capacity vanishes both as $K\to 0$ and as $K\to \infty$: at high temperature its dimensionless factor $K^2$ vanishes, while at low temperature excited microstates become exponentially rare.

Define the instantaneous spin magnetization $m=\frac{1}{4}{\sum }_i{\sigma }_i$. Global reversal sends $m$ to $-m$ without changing the zero-field energy. Pairing every microstate with its reversed partner proves $⟨m⟩=0$ at every positive temperature, even when almost all probability lies on the two fully aligned patterns. A vanishing mean does not imply weak alignment: the limiting probability distribution is half at $m=+1$ and half at $m=-1$.

The Ising ring magnetization fluctuations measure this distinction through variance. In the $d=0$ class, both patterns have $m^2=1$. Among the $d=2$ patterns, eight have $m^2=1/4$ and four have $m^2=0$; the two $d=4$ patterns also have $m^2=0$. ConsequentlyIt equals $6/17$ at $K=(\ln 2)/4$, approaches $1/4$ for four independent fair Ising spins at $K\to 0$, and approaches $1$ as $K\to \infty$. Unlike the energy variance, the spin magnetization variance remains large in the low-temperature symmetric mixture.

The same thermal fluctuation controls the field response. Allow $h$ to vary before setting it to zero; differentiating the Boltzmann distribution inserts $\beta {\sum }_i{\sigma }_i=4\beta m$, givingThis is a spin-normalized magnetic susceptibility with respect to the energy-valued parameter $h$; conversion to physical magnetization and magnetic field requires the magnetic moment and sample volume factors.

For finite $N$, $Z$ is a finite positive sum of exponential functions. Its free energy $-k_{B}T\log Z$ is a real analytic function for $T>0$, so the smooth heat capacity peak here is not a critical phase transition. Spontaneous magnetization concerns the order of limits: first a thermodynamic limit, then removal of a symmetry-breaking field. The four-spin Ising ring alone establishes neither that limit nor any two-dimensional Ising model result.^[ref]