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On 10 July 1908, Heike Kamerlingh Onnes's Leiden laboratory circulated purified helium through liquid air and evaporatively cooled liquid-hydrogen precooling, a regenerative heat exchanger, and an expansion valve. Returning helium cooled incoming helium; liquid collected in a transparent vacuum vessel. Reflected light revealed its liquid meniscus, crossed by thermocouple wires. Subsequent mass density and spectroscopy checks tested contamination.^[ref]

A constant-volume helium gas thermometer below the valve supplied the temperature indication. An oxygen boiling reference check disagreed with its expected reading; the real gas correction to thermodynamic temperature remained provisional. A temperature plateau suggested immersion before the liquid meniscus was recognized.^[ref]

The measurement principles separate two questions. A visible liquid meniscus establishes an interface; a gas thermometer estimates temperature through an equation of state. A stationary temperature trace can also result from a steady state without phase coexistence, so it cannot replace the interface observation. Conversely, seeing a liquid does not calibrate a thermometer. Thermometer calibration, the gas thermometer dead-space correction, departures from the ideal gas law and incomplete thermal equilibrium all matter when assigning an absolute temperature. A check far from the intended measurement range tests the instrument but does not by itself bound every low-temperature error.

Latent heat is the heat absorbed or released in a first-order phase transition through its entropy jump. For an equilibrium conversion from thermodynamic phase $\alpha$ to thermodynamic phase $\beta$ at fixed pressure, evaluate both thermodynamic phases at the same temperature and pressure on their phase coexistence curve:Use one normalization throughout: $L$, $S$ and $H$ may all be per mole, giving $L$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or all per unit mass, giving $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Here $S$ is entropy and $H$ is enthalpy on that basis. Equal Gibbs free energies of the coexisting thermodynamic phases give $\Delta H=T_{\mathrm{tr}}\Delta S$. Identifying this enthalpy difference with supplied heat assumes a closed system and an equilibrium conversion with only pressure-volume work. ^[ref]

For a pure substance with positive latent heat per unit mass $L$, let $x$ be the phase fraction of the higher-enthalpy thermodynamic phase, measured by mass. While both thermodynamic phases coexist at the fixed pressure,Adding heat changes $x$ rather than the temperature; this creates the horizontal portion of a heating curve. Once $x=1$, additional heat raises the temperature according to specific heat capacity at constant pressure. Reversing the conversion releases heat. The plateau represents changing enthalpy, rather than a finite single-thermodynamic phase heat capacity satisfying $Q=C_p\Delta T$. ^[ref]

For a constructed fixed-pressure model, take $m=1\mathrm{kg}$, initial temperature $280\mathrm{K}$, transition temperature $T_{\mathrm{tr}}=300\mathrm{K}$, solid specific heat capacity at constant pressure $c_s=2\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, latent heat $L=100\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$, and liquid specific heat capacity at constant pressure $c_l=3\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. These are stipulated model parameters, not empirical properties of a named material. Heating the solid to $300\mathrm{K}$ takes $40\mathrm{kJ}$ of sensible heat; converting it fully to liquid takes another $100\mathrm{kJ}$ of heat; heating the liquid to $320\mathrm{K}$ takes $60\mathrm{kJ}$. The total is $200\mathrm{kJ}$. After a total heat input of $90\mathrm{kJ}$, $x=(90-40)/100=1/2$ and the temperature is still $300\mathrm{K}$. The corresponding per-unit-mass entropy jump is $L/T_{\mathrm{tr}}=\frac{1}{3}\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$.

A continuous phase transition has no latent heat. Its heat capacity can have a jump or a singularity without a finite interval in which added heat converts coexisting thermodynamic phases at a constant temperature. A heating plateau therefore requires the stated phase coexistence conditions; it is not a feature of every phase transition. ^[ref]

An ice bath can absorb heat while its temperature barely changes: the heat converts ice into water. The latent heat of fusion quantifies this melting budget. For a pure substance at its equilibrium melting temperature $T_m(p)$ and specified pressure $p$, define it here per unit mass:Here $h_l$ and $h_s$ are the specific enthalpies of the coexisting bulk liquid and solid, and $H$ is enthalpy. For ordinary melting, $L>0$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$; an amount $m$ has a total fusion enthalpy $mL$ in joules. A molar fusion enthalpy instead has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and equals $ML$, where $M=m/n$ is mass divided by amount of substance in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Always check the normalization before inserting a tabulated value into an energy balance.^{[ref] [ref]}

For a closed system converting completely from solid to liquid at fixed pressure, with only pressure-volume work and negligible changes in bulk kinetic energy and potential energy, the first law of thermodynamics gives $Q_p=\Delta H=mL$. Both endpoints are at $T_m(p)$: warming initially colder ice or subsequently warming the resulting water requires additional sensible heat. Since $H=U+pV$, the corresponding specific internal energy jump is $\Delta u=L-p\Delta v$, where $v=V/m$; $L$ includes the work associated with the change in volume. Heat is transferred energy, whereas enthalpy is a state function. ^[ref]

Let $x=m_l/m$ be the liquid phase fraction by mass, rather than by volume. Neglecting interfacial contributions to enthalpy, equilibrium phase coexistence at fixed pressure givesIn the model above, reaching $T_m$ takes $2.10\mathrm{kJ}$ of sensible heat; the plateau then absorbs $33.40\mathrm{kJ}$, ending at a cumulative $35.50\mathrm{kJ}$. At $18.80\mathrm{kJ}$, half the mass is liquid and the temperature is still $T_m$. The final warming takes another $4.18\mathrm{kJ}$. The rounded $L$ and $c_l$ are inspired by the introductory ice-melting calorimetry values in Case Western's demonstration; constant properties and exact equilibrium throughout are stipulated model assumptions. ^[ref]

Within one thermodynamic phase, specific heat capacity at constant pressure is $c_p=(\partial h/\partial T{)}_p$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; a sample's heat capacity at constant pressure is $C_p=m{c}_p$, in $\mathrm{J}{\mathrm{K}}^{-1}$. The plateau is a finite enthalpy change at zero temperature change through varying $x$, so neither a finite single-phase $c_p$ nor $Q=m{c}_p\Delta T$ represents it. The heating curve needs separate sensible heat intervals and a fusion contribution. ^[ref]

The plateau assumes a pure substance, a stable bulk solid-liquid phase coexistence pair, fixed pressure, and sufficiently slow, uniform heating to remain near thermodynamic equilibrium. The total mass and composition must remain fixed, with negligible evaporation and no other phase transition. A fast heater can produce temperature gradients, so a local thermometer need not read the equilibrium $T_m$. Neither $T_m$ nor $L$ is universal across pressures or compositions: common-pressure melting-point slope describes the pressure dependence of $T_m$, while mixtures can melt over a temperature range as their coexisting compositions change. Their liquidus must be consulted instead of assuming a pure-substance plateau. ^[ref]

This makes ice useful as a temperature buffer: under these conditions it can absorb $m_{s}L$ before the remaining solid mass $m_s$ disappears, after which the water warms. If a stipulated constant net input power $P$ goes entirely into melting, that buffer lasts $m_{s}L/P$. Reversing the equilibrium conversion releases the same fusion enthalpy during freezing. The beautiful thermodynamic connection is $L=T_m(s_l-s_s)$, with $s$ the specific entropy and $T_m$ absolute temperature: equal Gibbs free energies at phase coexistence allow different enthalpies and entropies. The nearly stationary thermometer hides a finite, measurable change of state. ^[ref]

A lattice gas describes particle positions by occupations of fixed cells. In the single-occupancy model, $n_{\alpha }\in \{0,1\}$ and $N={\sum }_{\alpha }{n}_{\alpha }$. Its grand canonical partition function sums configurations with weight $\exp [-\beta (U-\tilde{\mu }N)]$. Discretizing a classical position integral into cells of volume $b$ gives $\tilde{\mu }=\mu +k_{B}T\ln (b/{\lambda }^3)$, with thermal de Broglie wavelength $\lambda$.

For energy $U=-4J{\sum }_{⟨\alpha {\alpha }^{\prime }⟩}{n}_{\alpha }{n}_{{\alpha }^{\prime }}$ on a regular lattice of coordination $q$, the substitution $n_{\alpha }=(1+{\sigma }_{\alpha })/2$ maps the model to the Ising model with coupling $J$ and effective field $h=qJ+\tilde{\mu }/2$. The mean Ising spin is a density imbalance, $⟨\sigma ⟩=2bn-1$. For $J>0$ below the critical temperature, the field-driven Ising transition becomes gas-liquid phase coexistence.

For one mole, liquid–vapour phase coexistence requires equal pressure and chemical potential. At $\tau =RTb/a\ll 1$, the saturation pressure is exponentially small. Neglecting it in the dense-phase equation of state gives $RT{V}_l^2-a{V}_l+ab=0$. The root near the excluded volume isThis is an asymptotic low-temperature result, not an exact coexistence volume at every subcritical temperature. Away from that limit, solve equal pressure together with the Maxwell equal-area construction.

Maxwell relations turn pressure, volume and temperature measurements into information about entropy. For example, the pressure rise when heating a rigid sealed vessel determines the entropy change when expanding the same sample isothermally:The quantities on the two sides describe different experiments, yet both are mixed partial derivatives of one Helmholtz free energy. This is the useful surprise: a mechanical measurement can reveal a thermal property. Similarly, the coefficient of thermal expansion gives $(\partial S/\partial P{)}_{T,N}=-\alpha V$. ^[ref]

For a simple compressible system, hold particle number $N$ and composition fixed, and allow only pressure-volume work. Write $U$ for internal energy, $H=U+PV$ for enthalpy, $F=U-TS$ for Helmholtz free energy, and $G=H-TS$ for Gibbs free energy. Their natural-variable differentials areThus the first partial derivatives, with the subscript recording the variables kept fixed, areThe symmetry of second derivatives produces all four standard Maxwell relations, including their signs:For a brief derivation of the $F$ relation, differentiate $F_T=-S$ with respect to $V$ and $F_V=-P$ with respect to $T$: $-S_V=F_{TV}=F_{VT}=-P_T$. Each other row follows in the same way. Assume continuous second partial derivatives of the relevant thermodynamic potential in the region considered. At a phase transition, the equilibrium potential may lose this smoothness; apply the identities within each smooth thermodynamic phase, and handle latent heat and phase coexistence separately. If particle numbers vary, the potential differentials acquire chemical potential terms, so the fixed-$N$ constraints must remain explicit. ^[ref]

For an ideal gas with fixed $N$, the ideal gas law gives $P=N{k}_{B}T/V$. At fixed volume, heating changes the pressure at rate $N{k}_B/V$. Consequently an isothermal expansion at positive temperature hasFor one mole with $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $V_2=2V_1$, this is $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. At $300\mathrm{K}$, a reversible isothermal expansion absorbs $Q=T\Delta S\simeq 1728.85\mathrm{J}$ and performs equal work, since its internal energy is unchanged. The entropy difference is a state function difference: the same equilibrium endpoints give the same $\Delta S$ in an insulated free expansion, although there $Q=0$ and the process is irreversible.

The same Maxwell relations let thermal measurements determine the difference between constant-pressure and constant-volume heat capacities. In a smooth single thermodynamic phase,using heat capacity at constant volume, heat capacity at constant pressure, coefficient of thermal expansion and isothermal compressibility. Along fixed pressure, $dV=\alpha VdT$, soThus measured expansion and compression responses determine a heat capacity difference without measuring both heat capacities separately. They do not determine the absolute heat capacities or absolute entropy by themselves; an independent thermal measurement supplies that missing information. For an ideal gas, $\alpha =1/T$ and ${\kappa }_T=1/P$ reduce the result to $C_P-C_V=PV/T=N{k}_B$, which is Mayer's relation.

When the necessary higher partial derivatives exist and commute, differentiating the Maxwell relations once more also givesIntegrating these slopes determines how a heat capacity changes with volume or pressure at fixed temperature. For an ideal gas, both right-hand sides vanish because the ideal gas law is linear in temperature at either fixed volume or fixed pressure. Consequently $C_V$ and $C_P$ depend only on temperature at fixed $N$, although they need not be constant in temperature. ^[ref]

Historically, James Clerk Maxwell's Theory of Heat, first published in 1871, presents these connections using the geometry of isothermal processes and reversible adiabatic processes. His fourth edition, page 167, explicitly relates expansion on heating at fixed pressure to heat emitted during an isothermal compression: the modern $G$ relation gives $\delta Q_{\mathrm{rev}}=TdS=-T\alpha VdP$. This explains why a substance with positive coefficient of thermal expansion emits heat when compressed isothermally, whereas one with negative coefficient absorbs it. ^{[ref] [ref]}

Choose a hypothetical fluid with molar constants $a_m=0.360\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$ and $b_m=4.00\times 10^{-5}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, using $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These stipulated constants are not a fit to carbon dioxide. The molar critical formulas giveAt $T=360\mathrm{K}$ and $v_m=240\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, first convert $v_m$ to $2.40\times 10^{-4}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. The repulsive-volume term and attractive correction areso $p=8.716033\mathrm{MPa}$. The ideal gas law would instead give $RT/v_m=12.471694\mathrm{MPa}$. The excluded volume raises pressure relative to that ideal value, but attraction more than offsets the increase here. Neither correction alone describes this example.

For $n_{\mathrm{mol}}=0.500\mathrm{mol}$, the same state occupies $V=n_{\mathrm{mol}}{v}_m=120\mathrm{c}{\mathrm{m}}^3$. The total-volume formula gives $n_{\mathrm{mol}}{b}_m=20\mathrm{c}{\mathrm{m}}^3$ and the same $8.716033\mathrm{MPa}$. Halving the sample has halved both its physical volume and its excluded volume, while retaining its pressure. Conversely, if the two critical measurements $T_c,p_c$ are used to determine model constants, inversion givesAn independent critical molar volume then tests the prediction $v_{m,c}=3R{T}_c/(8p_c)$; agreement with two fitted numbers does not ensure agreement with the third.

At $t=0.9$, the Van der Waals liquid-gas coexistence example below becomes $T\simeq 288.6537\mathrm{K}$, $p_{\ast }\simeq 5.391653\mathrm{MPa}$, $v_{m,l}\simeq 72.40823\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and $v_{m,g}\simeq 281.8611\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. For a mean molar volume of $120\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, the lever rule gives gas mole fraction $f_g\simeq 0.227219$. Its volume fraction is instead $f_g{v}_{m,g}/v_m\simeq 0.533703$: less than one quarter of the molecules occupy more than half the volume. The flat equilibrium pressure corresponds to changing these proportions. It does not describe a homogeneous sample sliding along the unstable loop.

A first-order phase transition has a discontinuity in a first derivative of the equilibrium free energy, such as entropy or the order parameter; a thermal first-order transition normally has latent heat. At a continuous phase transition the order parameter goes continuously to zero and there is no latent heat, although response functions and higher derivatives of the free energy can be singular.

For the symmetric scalar Landau free energy at zero conjugate field, minimizeIf $u>0$, the sixth-order term is unnecessary near $r=0$. For $r>0$ the minimum is $M=0$, whereas for $r<0$ there are two symmetry-related minima with $M^2=-r/u$ to leading order. The order parameter vanishes continuously as $r\uparrow 0$. The equilibrium free-energy density changes from zero to $-r^2/(4u)$, so its first temperature derivative is continuous and its second derivative has a jump. This is the ordinary mean-field continuous phase transition.

If $u<0$, keep $v>0$ for stability. A nonzero stationary order parameter, with $s=M^2$, obeys $r+us+v{s}^2=0$. For phase coexistence with $M=0$ its free energy must also vanish. Substituting $r=-us-v{s}^2$ into $V$ givesThe nonzero solution of $V=0$ is thereforeThe competing minima exchange stability at positive $r_t$, with a finite jump of the order parameter. This is a first-order phase transition. The metastability region is also visible: nonzero stationary points first appear at the spinodal $r=u^2/(4v)$, and the disordered local minimum loses stability at $r=0$.

For example, holding $u,v$ fixed and taking $r=a(T-T_0)$, the ordered phase has entropy lower than the disordered phase by $a{M}_t^2/2$ at coexistence. Heating through the transition absorbs latent heat per volume $T_{t}a{M}_t^2/2$. More generally the temperature dependence of all coefficients contributes to this entropy difference. With no inversion symmetry, a cubic term can also produce a first-order phase transition, as comparison of the competing minima of the Landau free energy shows.

The uniform stationarity equation and local stability condition areFor $u>0$, $h=0$, and small negative $r$, the stable nonzero solution satisfies $M^2=-r/u+O(r^2)$. Thus the order parameter vanishes continuously as $r\uparrow 0$. The equilibrium Landau free energy has leading singular contribution $f_{\min }=-r^2/(4u)$ below the transition and zero above it. Its first thermal derivative is continuous, while its second derivative has a finite jump: the mean-field approximation predicts a continuous phase transition with no latent heat. The magnetic susceptibility follows by differentiating the stationarity equation:It diverges on approaching the ordinary thermodynamic critical point, and the quadratic Landau scalar correlation length behaves as $\xi \sim \sqrt{\kappa /|r|}$, with a different amplitude on the two sides.

For $u<0$, a nonzero stationary point at $h=0$ has $r+u{M}^2+v{M}^4=0$. Substitution into $V(M)-V(0)$ gives $-u{M}^4/4-v{M}^6/3$. Setting this to zero yieldsThe finite jump of $M$ makes this a first-order phase transition. For a thermal path $r=r(T)$ with $r^{\prime }(T)\ne 0$ and the other coefficients fixed, $\partial f_{\min }/\partial T=r^{\prime }(T)M^2/2$ jumps, giving an entropy discontinuity and nonzero latent heat. The Landau free energy itself remains continuous. The disordered minimum loses stability at $r=0$; the two nonzero stationary solutions merge at $r=u^2/(4v)$. These spinodal points bound metastability, but neither replaces the equilibrium phase coexistence condition.

Writing $r=at$ with $a>0$ near an ordinary thermodynamic critical point, the zero-field equation gives $M\sim (-t{)}^{1/2}$. At $r=0$, small $h$ obeys $h=u{M}^3+O(M^5)$, hence $M\sim \mathrm{sgn}(h)|h{|}^{1/3}$. The ordinary mean-field critical exponents are therefore $\beta =1/2$ and $\delta =3$. These are Landau theory predictions, not dimension-independent values for an interacting theory below its upper critical dimension.

A phase diagram divides a space of control parameters into equilibrium thermodynamic phases, marking phase coexistence boundaries and thermodynamic critical points. Its dimension counts independently variable controls, not spatial dimensions. An explicit three-dimensional example uses the controls $(r,u,h)$ of a sextic even Landau potential:Here $M$ is a scalar order parameter, $h$ its conjugate field, and $r$ a thermal control. The equilibrium magnetization minimizes $V$ globally. At $h=0$, $u>0$, the boundary $r=0$ is a continuous phase transition. At $h=0$, $u<0$, the boundary $r=3u^2/(16v)$ is a first-order phase transition between $M=0$ and $M=\pm \sqrt{-3u/(4v)}$. The two boundaries meet at the tricritical point $r=u=h=0$. A fixed strictly positive quartic coupling alone cannot produce this tricritical point; the stabilizing sextic term is essential when the quartic coefficient is tuned through zero.

The full phase diagram also contains the $h=0$ phase coexistence sheet between positive and negative ordered phases. Crossing it changes the sign of $M$ discontinuously. For $u<0$, the three-phase line $r=3u^2/(16v)$ is the junction of this sheet and two tricritical wings, on which two unequal same-sign equilibrium magnetizations coexist at nonzero $h$. The tricritical wing critical edges are lines of ordinary thermodynamic critical points; they terminate the first-order sheets and meet at the tricritical point. Away from these boundaries, varying $h$ produces a smooth crossover.

For an explicit construction of the tricritical wings, write their two coexisting order parameters as $M_1=s-d$ and $M_2=s+d$, with $s>0$ and $0<d\le s$. Equality of their Landau free energies and first derivatives givesIndeed, for these controls the tricritical wing coexistence factorization isThe last factor is $(M+2s{)}^2+s^2-d^2\ge 0$, proving that both specified minima are global. The reflected wing has negative $h$ and negative $M$. Letting $d\to 0$ gives the tricritical wing critical edge $u=-10v{s}^2/3$, $r=5v{s}^4$, $h=8v{s}^5/3$; there $V^{\prime \prime }(s)=V^{\prime \prime \prime }(s)=0$ and $V^{\prime \prime \prime \prime }(s)=40v{s}^2>0$. Letting $d=s$ recovers the three-phase line. This supplies the first-order sheets and their continuous boundary curves, rather than only a zero-field section.

A phase diagram partitions a space of thermodynamic controls into regions with distinct equilibrium phases. Its boundaries indicate phase coexistence or continuous phase transitions; a critical point is a termination of coexistence where the distinction between phases disappears. A concrete three-dimensional control space is $(T,\Delta ,h)$ for the Blume–Capel model: a spin can be $0$ or $\pm 1$, the exchange favors aligned nonzero spins, and $\Delta$ controls the energetic cost of nonzero spins. Varying $\Delta$ changes the balance between a continuous ordering transition and a first-order phase transition. Their meeting is a tricritical point.

Locally, the same structure is displayed by the Landau free energywith three controls $(r,u,h)$. For $u>0$, the line $r=h=0$ consists of ordinary critical points. For $u<0$, equal free energies of $M=0$ and a nonzero stationary point requireEliminating $r$ gives the tricritical three-phase lineHere $M=0$ and both signs of $M$ coexist, with a discontinuous order parameter. Below this line, or below $r=0$ when $u>0$, the plane $h=0$ is an ordered phase coexistence sheet: crossing it reverses the sign of $M$. Two tricritical wings also extend to nonzero $h$, separating weakly and strongly magnetized states of the same sign. Their ordinary critical points satisfy $f^{\prime }=f^{\prime \prime }=f^{\prime \prime \prime }=0$. Since $f^{\prime \prime \prime }=6uM+20v{M}^3$, their tricritical wing critical edges areAt these edges $f^{\prime \prime \prime \prime }=-12u>0$, giving an ordinary quartic critical expansion around $M_c$. All these structures end at the tricritical point $r=u=h=0$. The diagrams show both the zero-field section and the genuinely three-dimensional coexistence structure.

For the final critical exponent calculation, take $r=at$ with $a>0$, $t=(T-T_c)/T_c$, and hold the other controls fixed. At an ordinary critical point, the quartic term dominates the sextic term. For $h=0$ and $r<0$,whereas on the disordered side the singular contribution is zero. Two temperature derivatives give a finite jump in heat capacity, so $\alpha =0$. At $r=0$, the equation of state is $h=u{M}^3+O(M^5)$, so $\delta =3$.

Along the tricritical trajectory $u=0$, the nonzero minimum obeys $M^4=-r/v$ andThus two temperature derivatives produce $|t{|}^{-1/2}$. At $r=u=0$, $h=v{M}^5$. Therefore the mean-field critical exponents areWith $A$ the thermodynamic free energy, the physical sign is $C=-T{\partial }_T^{2}A$, rather than the positive sign printed in the paper. This correction affects the sign of the heat capacity, not its critical exponent. The tricritical result assumes that the independent quartic control is tuned to zero; a generic trajectory through nearby ordinary critical points need not have these exponents.

At a continuous phase transition the equilibrium order parameter tends continuously to its critical value, there is no latent heat, and the correlation length diverges. For $u>0$ at zero field, minimizing $V(M)=r{M}^2/2+u{M}^4/4$ gives $M=0$ when $r\ge 0$ and $M=\pm \sqrt{-r/u}$ when $r<0$. These minima join continuously at $r=0$. The minimized singular free-energy density is zero above and $-r^2/(4u)$ below, so its first thermal derivative is continuous, while its second derivative has a finite jump in mean-field approximation.

A first-order phase transition instead has competing minima whose free energies cross while their order parameters remain distinct. The first derivatives of the equilibrium free energy generally jump; along a temperature-driven crossing this gives latent heat. For the symmetric sextic potential with $u<0,v>0$, let $y=M^2>0$. At phase coexistence, the stationary points and equality of the two free energies requireEliminating $r$ gives $-u{y}^2/4-v{y}^3/3=0$, henceAt coexistence $V(M)=v{M}^2(M^2+3u/(4v){)}^2/6$, proving that all three degenerate minima are global. The equilibrium order parameter jumps from zero to one of the two nonzero values even though $r_{\mathrm{coex}}>0$. The disordered minimum loses local stability at $r=0$, and the nonzero stationary points first appear at $r=u^2/(4v)$. These spinodal points delimit metastability and are not the equilibrium transition line. Cubic terms, when symmetry permits them, provide another common route to a first-order transition.

A phase diagram locates stable pure thermodynamic phases and their phase coexistence boundaries in a space of control parameters. For a concrete three-dimensional example, take the temperature, crystal-field parameter, and magnetic field of the Blume–Capel model. Its spins take values $0,\pm 1$; changing the cost of the zero-spin state can turn a continuous ordering transition into a discontinuous one. A local scalar description near the resulting tricritical point is the Landau-Ginzburg theoryThe three local control coordinates are $(r,u,h)$, with $r$ varying linearly with the thermal scaling field. For $u>0$, $h=0$, the transition at $r=0$ is continuous. For $u<0$, zero-field coexistence is at $r=3u^2/(16v)$, and the magnetization jumps between zero and $\pm \sqrt{-3u/(4v)}$. The two loci meet at $r=u=h=0$, the tricritical point.

The full three-dimensional phase diagram has an ordered coexistence sheet at $h=0$, where the two opposite magnetizations exchange stability when $h$ changes sign. On its three-phase line it meets two nonzero-field tricritical wings. These wings are first-order phase transition surfaces between minima of the same sign but different magnitude, terminating at ordinary thermodynamic critical points. Their tricritical wing critical edges are found by $V^{\prime }=V^{\prime \prime }=V^{\prime \prime \prime }=0$:Indeed $V^{\prime \prime \prime }=6uM+20v{M}^3$ fixes $M_c^2$, and substituting into $V^{\prime \prime }=0$ and $V^{\prime }=0$ gives $r_c,h_c$; $V^{\prime \prime \prime \prime }=-12u>0$ ensures a stable critical quartic minimum. The two signs give the two wings. All transition surfaces and critical edges end together at the tricritical point. The following plot uses $v=1$; its wing surfaces come from the exact tricritical wing coexistence factorization, rather than schematic unconstrained surfaces.

The unheaded requests for critical exponents follow from the uniform equation of stateFor an ordinary thermodynamic critical point, $u>0$ is fixed and $r=At$, $A>0$. At $h=0$, $t<0$, the small ordered minimum obeys $M^2\sim -r/u$, so the order-parameter critical exponent is $\beta =1/2$. Above the transition $\chi =(\partial M/\partial h{)}_{h=0}=1/r$, giving the magnetic-susceptibility critical exponent $\gamma =1$. At $r=0$, $h\sim u{M}^3$, so the critical-isotherm exponent is $\delta =3$. At the tuned tricritical point, $u=0$, instead $M^4=-r/v$ below the transition and $h=v{M}^5$ on its critical isotherm, while the disordered susceptibility remains $1/r$. ThereforeBelow the transition the inverse susceptibility is $r+3u{M}^2+5v{M}^4$, giving $2|r|$ in the ordinary limit and $4|r|$ in the tricritical limit. Thus both sides give the same susceptibility power, with different amplitudes.

To derive the general Ginzburg criterion, tune away all lower even nonlinearities and let the leading stabilizer be $v_n{M}^{2n}/(2n)$, $n\ge 2$. The equation of state gives $M_0^2\asymp |r{|}^{1/(n-1)}$, and its curvature at the ordered saddle is $2(n-1)|r|$, so $\xi \asymp |r{|}^{-1/2}$. The fluctuations of the order parameter averaged over a correlation volume are estimated by its long-wavelength quadratic covariance:The integral has a fixed dimensionless upper limit after $q=p/\xi$; this is a coarse-volume fluctuation estimate, not the ultraviolet-dominated point variance. Consequently the multicritical Ginzburg ratio isThe Landau approximation is asymptotically controlled by this test only when the ratio tends to zero. This gives the upper critical dimensionThe same threshold follows by giving $M$ its Gaussian engineering dimension $(D-2)/2$: the coupling to $M^{2n}$ has scaling exponent $D-n(D-2)=2n-(n-1)D$, which becomes relevant below $D_c$.

For $D<D_c$, the fluctuation ratio diverges and the saddle calculation cannot predict the generic interacting critical exponents. At $D=D_c$ the test is marginal, not a proof of changed power exponents: the ratio has zero thermal power and marginal interactions can produce logarithmic corrections. Thus the PDF's inclusion of equality in a blanket claim about wrong power exponents needs this qualification. The Landau-Ginzburg theory as a fluctuating field theory remains useful; what loses validity is its fluctuation-free Landau approximation.

Set $h=0$ in the even sextic Landau free energy. For $u>0$, near $r=0$ the nonzero stationary branch satisfies $M^2=-r/u+O(r^2)$ for $r<0$. The minimum value is $V_{\min }=-r^2/(4u)+O(r^3)$ below the transition and zero above. The order parameter is continuous, and if $r$ crosses zero linearly with temperature, the first temperature derivative of the minimized free-energy density is continuous too. There is no latent heat; a second derivative has a finite jump at this level.

For $u<0$, solve the two conditions $V^{\prime }(M)=0$ and $V(M)=V(0)$ with $M\ne 0$. Setting $q=M^2$ gives $r+uq+v{q}^2=0$ and $r/2+uq/4+v{q}^2/6=0$. Eliminating $r$ givesBoth the disordered and ordered minima have positive curvature there, but their values coincide. Crossing this phase coexistence boundary changes the global minimum discontinuously. The order parameter jumps; for $r=r(T)$ at fixed $u,v$, ${\partial }_T{V}_{\min }=r^{\prime }(T)M^2/2$ jumps as well, generically yielding a nonzero latent heat. Local extrema can persist past coexistence as metastable states, which explains hysteresis in a dynamically delayed transition. The spinodal points, where local curvature vanishes, do not locate the equilibrium transition.

A first-order phase transition has a discontinuity in a first derivative of the equilibrium free energy, such as the entropy or order parameter. It can have latent heat when the entropy jumps. A continuous phase transition has a continuously vanishing order parameter and no latent heat, with singular higher derivatives and a diverging correlation length. The LG theory compares global minima, not merely the points where a local minimum loses stability.

For the uniform quartic Landau free energythe equation of state is $rm+u{m}^3=h$. At zero field the stable minimum is $m=0$ for $r>0$, and $m=\pm \sqrt{-r/u}$ for $r<0$. Thus tuning $r$ through zero gives a continuous transition. For a fixed $r<0$, varying $h$ through zero instead switches between the two ordered minima and makes $m$ jump, giving a first-order phase transition in the conjugate field.

A temperature-like first-order transition can occur at zero field when $u<0$ and a positive sextic term $v{m}^6/6$ stabilizes the potential. Write $q=m^2$. Stationarity of a nonzero phase gives $r+uq+v{q}^2=0$, while equality with $V(0)=0$ gives $rq/2+u{q}^2/4+v{q}^3/6=0$. Solving these two conditions yieldsThe order parameter jumps from zero to $\pm \sqrt{q_{\mathrm{coex}}}$. At this pointso the competing stationary points are genuinely global minima. This phase coexistence condition differs from the spinodal points $r=0$ and $r=u^2/(4v)$, which mark loss or creation of local stability, not equilibrium coexistence.

If the symmetry permits a cubic term $w{m}^3/3$, a positive quartic coefficient does not preclude a first-order transition. For $V=r{m}^2/2+w{m}^3/3+u{m}^4/4$ with $w\ne 0,u>0$, stationarity and coexistence give $m=-2w/(3u)$ and $r=2w^2/(9u)$. This is the first-order transition in a cubic-quartic Landau potential; the symmetry restriction on the expansion is therefore part of the prediction.

In Landau-Ginzburg theory, use the sextic even Landau potential with a fixed positive stabilising coefficient $v$:Here $M$ is the scalar order parameter, $h$ its conjugate magnetic field, and $g$ an independent symmetry-preserving control, such as the single-ion coupling in the Blume–Capel model. Thus $g$ need not break $M\mapsto -M$, whereas $h$ does. The Blume–Capel mean-field tricritical parameters are one realization; generally, a tricritical point occurs whenTwo controls tune away the quadratic and quartic terms. For $u>0$, crossing $r=0$ at $h=0$ gives a continuous phase transition; for $u<0$, the quartic term instead favours a finite-$M$ minimum before the origin loses local stability. The sextic term bounds the free energy below. The tricritical point joins the continuous and discontinuous transition loci.

For the symmetric phase diagram, write $s=M^2$. A nonzero stationary point obeys $r+us+v{s}^2=0$. Its free-energy density relative to the origin is $rs/2+u{s}^2/4+v{s}^3/6$. Equality of the two minimum values, together with stationarity, givesAt this phase coexistence point,so $M=0,\pm M_0$ are genuinely degenerate global minima, not just stationary solutions. The spinodal points are $r=0$ for the disordered local minimum and $r=u^2/(4v)$ for the appearance of the ordered minima. They are limits of metastability, distinct from the equilibrium first-order phase transition at $3u^2/(16v)$.

Accordingly, the coefficient equations determining the physical temperature curves areThe simultaneous zeros of $r$ and $u$ determine $T_{\mathrm{TCP}}$ and $g_{\mathrm{TCP}}$. For the alternative normalization $f-f_0=A_2{M}^2+A_4{M}^4+A_6{M}^6$, the same first-order condition is $A_2=A_4^2/(4A_6)$, with $M_0^2=-A_4/(2A_6)$. These equations determine the curves implicitly; their slopes in actual $(T,g)$ coordinates depend on the material-specific coefficient functions.

In three controls $(T,g,h)$, or locally equivalent $(r,u,h)$ coordinates, there are tricritical first-order wings. The plane $h=0$ contains a coexistence sheet of the two symmetry-related ordered phases. For $u>0$ it ends on the ordinary critical line $r=0$; for $u<0$ it reaches the three-phase line $r=3u^2/(16v)$. Two further first-order surfaces extend from that line into $h>0$ and $h<0$. Across a wing, two minima of the same sign but different magnitudes exchange global stability; the field favours that sign. Each wing ends on an ordinary critical edge where those minima and their intervening maximum merge. The three surfaces and their critical boundaries meet at the tricritical point.

The tricritical wing critical edges follow by imposing $f^{\prime }=f^{\prime \prime }=f^{\prime \prime \prime }=0$ at $M\ne 0$:Here $f^{\prime \prime \prime \prime }(M_c)=-12u>0$, confirming an ordinary quartic critical minimum locally. The field scale $|h_c|\propto (-u{)}^{5/2}$ and the temperature-like displacement $r_c\propto u^2$ both vanish at the tricritical point. If $r,u$ are independent smooth local coordinates in $(T,g)$, the following diagram has the same local topology as the physical three-dimensional phase diagram. Its surfaces use exact equal-minimum conditions, as described by tricritical wing coexistence factorization.