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Consider $N$ noninteracting bosons in one internal quantum state, at thermal equilibrium in the three-dimensional harmonic oscillator potentialHere $m$ is the mass, the positive ${\omega }_i$ are angular frequencies, and $\bar{\omega }$ is their geometric mean. Measure the one-particle energy $ϵ$ and chemical potential $\mu$ from the ground-state energy. The quantum harmonic oscillator levels then have $ϵ=\hslash ({\omega }_x{n}_x+{\omega }_y{n}_y+{\omega }_z{n}_z)$ with nonnegative integer $n_i$ and reduced Planck constant $\hslash =h/(2\pi )$.

When $ϵ\gg \hslash {\omega }_i$, counting lattice points can be approximated by the volume of a tetrahedron in the positive octant of $(n_x,n_y,n_z)$:This is a density of states, with units of inverse energy. The factor $1/6$ is the unit tetrahedron volume; differentiating produces the factor $1/2$. Treat the single ground state separately: its macroscopic population would be lost by replacing every level with a smooth density of states.

For $k_{B}T\gg \hslash {\omega }_i$, integrate the Bose-Einstein distribution over excited energy eigenstates. As $\mu$ approaches zero from below, their maximum population isHere $k_B$ is the Boltzmann constant and $\zeta$ the Riemann zeta function. A short proof uses $1/(e^x-1)={\sum }_{\ell \ge 1}{e}^{-\ell x}$ and ${\int }_0^{\infty }{x}^2{e}^{-\ell x}dx=2/{\ell }^3$. The nonnegative terms permit exchanging summation and integration by Tonelli's theorem, giving ${\int }_0^{\infty }{x}^2/(e^x-1)dx=2\zeta (3)$. Near zero the integrand is proportional to $x$, so its integral converges. That finite excited-state capacity is the mechanism of Bose-Einstein condensation.

Set $N_{\mathrm{ex}}^{\mathrm{sat}}(T_c)=N$. Particle conservation gives, to this continuum accuracy,Above $T_c$, the macroscopic condensate fraction is zero and the chemical potential adjusts below zero. This cubic power belongs to a three-dimensional harmonic trap; it is not the homogeneous three-dimensional ideal gas exponent. For $N=10^5$, $\bar{\omega }/(2\pi )=100\mathrm{Hz}$ and $\zeta (3)\simeq 1.202056903$, the model gives $T_c\simeq 209.5\mathrm{nK}$. At $T=0.8T_c\simeq 167.6\mathrm{nK}$, $N_0/N=0.488$, or approximately $48800$ condensed bosons. These are chosen parameters and calculated values.

Finite $N$ gives a smooth crossover rather than a sharp phase transition. The exact ideal gas sum over excited quantum harmonic oscillator levels differs from the continuum integral, especially when $k_{B}T$ is not large relative to every $\hslash {\omega }_i$. A sharp trapped-gas limit takes $N\to \infty$, $\bar{\omega }\to 0$ with $N{\bar{\omega }}^3$ fixed and fixed frequency ratios. Interactions also alter the critical temperature, condensate fraction and release profile; a dilute gas can have appreciable interaction energy compared with the trap level spacing. The cubic law is a useful baseline, not a fit guaranteed for either 1995 experiment.

For molar equation $p=RT/(v-b)-a/v^2$, the critical values are $v_c=3b$ and $T_c=8a/(27Rb)$. At $v=v_c$, $(\partial p/\partial v{)}_T=-R(T-T_c)/(4b^2)$. Thus ${\kappa }_T=-v_c^{-1}(\partial v/\partial p{)}_T=4b/[3R(T-T_c)]$ on the stable branch above $T_c$. The mean-field compressibility exponent is one; the continuation below the critical temperature is unstable rather than an equilibrium negative compressibility.

The critical-isotherm exponent is defined at the critical temperature by $m\sim |B{|}^{1/\delta }$ as the conjugate field $B$ tends to zero.

The critical temperature is the temperature at which the excited states of a Bose system are just saturated as the chemical potential approaches the lowest one-particle energy. Cooling below it puts the excess particles into the condensate.

In a dilute gas, Bose-Einstein condensation makes a single one-particle state hold a substantial fraction of the bosons. This is a population effect: cooling reduces the number that the excited energy eigenstates can accommodate at thermal equilibrium. The surplus accumulates in a Bose-Einstein condensate. For an ideal gas the occupied mode is the trap ground state; interactions can deform that mode, as described by the Gross–Pitaevskii equation. Dilute does not mean classical: a small number density can still give overlapping thermal de Broglie wavelengths at sufficiently low temperature.

The appeal is that changing the trap changes a calculable Quantum Hamiltonian, while absorption imaging makes its population distribution visible. The Bose condensation in a three-dimensional harmonic trap below turns density of states into a prediction for a critical temperature and condensate fraction. These are separate questions from measuring matter-wave coherence or superfluidity.

Eric A. Cornell and Carl E. Wieman's joint lecture emphasizes the competition between useful elastic collisions, which restore thermal equilibrium during evaporative cooling, and collisions that remove or heat the trapped sample. A long trap lifetime alone is insufficient: reequilibration must keep up with evaporation.^[ref]

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.

Freeze the magnitude of an order parameter and retain just one common orientation $\theta$ for $N$ rigidly aligned cells. Give this collective coordinate energy $E(\theta )=-Nh\cos \theta$ at positive temperature $T$. The Boltzmann distribution is the displayed normalized probability density function, where $k_B$ is the Boltzmann constant. Here $h$ is energy per cell, a different normalization from the source density in continuous vacuum selection by an external source. This is a deliberately restricted model of selection among already ordered orientations: it does not derive alignment or an ordering critical temperature from a spatial interaction.

Its mean orientation is an elementary ratio of integrals:At $h=0$ every angle has equal probability, and the numerator vanishes exactly. For small $x$, expansion of the exponential gives $m_N(h)=x/2+O(x^3)$ because the angular mean of ${\cos }^2\theta$ is $1/2$. For $x=1$ and $5$ the means are approximately $0.446390$ and $0.893383$. At fixed $h>0$, increasing $N$ concentrates the probability density function near $\theta =0$; expanding $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$ gives angular variance approximately $1/x$. ConsequentlyThe uniform angular average remains a symmetric mixture if one sets $h=0$ first. The other order selects a definite direction. This is the continuous-orientation version of the symmetry-breaking order of limits; it illustrates selection without replacing the separate problem of stability against spatial fluctuations.

For a uniform three-dimensional ideal gas of spinless bosons with energy $p^2/(2m)$ and ground energy set to zero, the excited number density is $n_{\mathrm{ex}}={\lambda }_T^{-3}{\mathrm{Li}}_{3/2}(z)$, with quantum gas fugacity $z=e^{\mu /(k_{B}T)}\le 1$. Its finite maximum $\zeta (3/2)/{\lambda }_T^3$ gives the displayed critical temperature. Below it, particle conservation gives condensate fraction $N_0/N=1-(T/T_c{)}^{3/2}$. The excited-state internal energy is $U=(3/2)V{k}_{B}T{\lambda }_T^{-3}{\mathrm{Li}}_{5/2}(z)$, with $z\to 1$ below the transition. The exponent differs from that of a harmonic trap. Condensation is not by itself a demonstration of superfluidity; the ideal quadratic dispersion has vanishing Landau critical speed.

Universality means that systems with different microscopic interactions can have the same asymptotic critical behaviour. Under renormalization-group flow, irrelevant microscopic differences shrink, and systems whose critical flows approach the same renormalization-group fixed point belong to one universality class. Important determinants are spatial dimension, order-parameter symmetry and number of components, the range of interactions and any additional relevant constraints. Microscopic Bravais lattice shape alone does not normally change the short-range scalar universality class.

Critical exponents, properly normalized scaling functions and certain dimensionless amplitude ratios are universal within a universality class. For example, the universal specific-heat amplitude ratio compares the two singular amplitudes above and below the transition, canceling their common normalization factors. Scaling functions require choices of thermal, field and length metric factors before comparisons between microscopic systems make sense. The critical temperature, critical field, Bravais lattice spacing, individual correlation-length and magnetic susceptibility amplitudes, and analytic free-energy backgrounds are not generally universal. Thus two models can share critical exponents without sharing their transition temperature or the absolute size of their response.

Use the usual homogeneous, local, analytic Landau-Ginzburg theory; rotational symmetry and Z2 symmetry alone do not specify locality, analyticity, or the signs of the coefficients. The local derivative expansion contains every rotational scalar with an even total number of fields:The omitted terms include all allowed higher field powers and contracted spatial derivatives. Terms differing by a total derivative are equivalent for bulk behavior with suitable boundary conditions. The displayed gradient energy assumes spatial homogeneity. If only rotations about an origin are imposed, rotationally invariant position-dependent coefficients are also possible.

In the mean-field approximation, take a spatially uniform order parameter and minimize its Landau free energy. For $\gamma >0$, $u>0$ and $r=a(T-T_c)$ with $a>0$, keeping the quadratic and quartic terms givesThe single minimum splits into two minima as the temperature passes below the critical temperature. Choosing one minimum gives spontaneous symmetry breaking of the Z2 symmetry and a continuous phase transition. The order parameter is the expectation in a selected ordered state; averaging equally over both states gives zero. A Z2 symmetry does not guarantee a continuous phase transition: a negative quartic coefficient stabilized by a positive sextic coefficient can instead give a first-order phase transition.

Relaxing the Z2 symmetry permits odd terms. A nonzero linear source generally rounds this ordinary continuous phase transition. A cubic term can instead give a first-order phase transition: for $V(\varphi )=r{\varphi }^2/2+w{\varphi }^3/3+u{\varphi }^4/4$, $u>0$ and $w\ne 0$, equal minima occur atThe order parameter jumps from zero to ${\varphi }_{\ast }$. A continuous phase transition without an exact Z2 symmetry is still possible with additional tuning; odd terms do not logically exclude it.

For the multicritical even Landau potential, set $r={\mu }^2=a(T-T_c)$ and assume an integer $n\ge 2$, ${\lambda }_n>0$ and $\gamma >0$. For $n>2$, all lower even interactions must be absent or tuned to zero to obtain these mean-field critical exponents. The stationary equation isIts nonzero solutions are minima because $V^{\prime \prime }({\varphi }_0)=-(2n-2)r>0$. Thus the order-parameter critical exponent isSubstituting into the minimum free energy density givesThe singular heat capacity per unit volume is $c_s=-T{\partial }_T^2{f}_{\min }$. Smooth variation of the other coefficients does not change the leading power. ConsequentlyFor $n=2$, this means a finite heat capacity jump, with ${\alpha }_{\mathrm{MF}}=0$; for $n=3$, the tricritical point has ${\beta }_{\mathrm{MF}}=1/4$ and ${\alpha }_{\mathrm{MF}}=1/2$. The case $n=1$ would contain only a quadratic potential and would not stabilize an ordered phase.

The Gaussian fluctuation correction near a critical point is more singular than the mean-field approximation contribution whenThis is the upper critical dimension of an even scalar interaction. Equivalently, its coupling has engineering dimension $2n-(n-1)d$, positive below $d_c$. Dominant fluctuations signal breakdown of the mean-field approximation; the Gaussian expression is not then a calculation of the exact interacting heat-capacity critical exponent. At $d=d_c$, equality of the powers allows logarithmic corrections.

At the critical temperature $T_B$, the excited states are just saturated and the condensate is about to form, soTherefore

Renormalization compares the same long-distance physics at different observation scales. The starting object is a microscopic or cutoff statistical Hamiltonian ${\mathcal{H}}_g$ with every symmetry-allowed coupling $g$. Because a critical correlation length is much larger than a lattice spacing, repeatedly eliminating short-distance detail can reveal scale-invariant behavior.

In a real-space transformation choose a normalized blocking kernel $B(S|s)$ for retained variables. DefineSumming over $S$ proves exact preservation of $Z$ before truncation. The transformed Hamiltonian generally gains longer-range and many-body interactions and a field-independent constant. Observables must be blocked consistently. In momentum-shell language split $\varphi ={\varphi }_{<}+{\varphi }_{>}$, integrate $\Lambda /b<|q|<\Lambda$, then use $x^{\prime }=x/b$ and a field rescaling to restore cutoff and kinetic normalization. Coarse-graining, rescaling and tracking generated interactions are distinct steps.

Repeated transformations generate a flow $g^{\prime }=R_b(g)$ with composition $R_b{R}_c=R_{bc}$ when the same exact coarse-graining family is used. Losing degrees of freedom generally makes this a semigroup, despite its conventional group name. A renormalization-group fixed point is invariant after rescaling. At a continuous transition the large-distance theory approaches a critical renormalization-group fixed point; finite correlation length instead eventually flows into an ordinary thermodynamic phase regime.

Linearize in scaling coordinates $u_i$:For increasing observation length, $y_i>0$ is relevant and must be tuned, $y_i<0$ is irrelevant, and $y_i=0$ is marginal and needs nonlinear analysis. Discrete multipliers $b^{y_i}$ must be compared with one, not with zero. The stable manifold forms the critical surface after relevant directions have been set to zero.

After subtracting regular additive contributions, ordinary hyperscaling givesChoosing $b=|t{|}^{-1/y_t}$ or choosing the field as scale yieldsFor a critical two-point function $G(r)\sim r^{-(d-2+\eta )}$, the order parameter scaling dimension is $x_m=(d-2+\eta )/2$, and its conjugate field has $y_h=d-x_m=(d+2-\eta )/2$. This recovers the Q1 identities while distinguishing the gap exponent $\Delta$ from the field's scaling dimension $y_h$.

As a concrete dimensional check, start from a scalar gradient action. Restoring its coefficient uses ${\varphi }^{\prime }(x^{\prime })=b^{(d-2)/2}{\varphi }_{<}(b{x}^{\prime })$, so at the Gaussian fixed pointQuartic interactions are relevant below four dimensions. They can produce an interacting Wilson-Fisher fixed point, whose exponents differ from naive Gaussian counting. Above four, the quartic coupling is irrelevant at the Gaussian fixed point but dangerous for ordered observables: ${\varphi }_0^2=-r/u$ is singular when $u$ is dropped. Thus the simple two-variable hyperscaling formula is not universal above the upper critical dimension.

Different microscopic systems can flow to the same long-distance renormalization-group fixed point, explaining universality of exponents and normalized scaling forms. Critical temperatures, metric factors and regular free energies remain nonuniversal. Dimension, order parameter symmetry, interaction range and additional soft variables constrain the universality class. Truncations, perturbative expansions and numerics approximate the flow; they must not be confused with the exact elimination definition.^[ref]

Entropy at absolute zero. The third law of thermodynamics concerns the low-temperature limit of equilibrium states. In the zero-entropy formulation, the entropy of a perfect crystal with a nondegenerate ground state tends to zero as $T\to 0$. The statistical interpretation is $S=k_B\ln \Omega$: a unique accessible ground state has $\Omega =1$. The entropy-difference formulation of the third law of thermodynamics states that entropy differences associated with regular changes of mechanical parameters tend to zero as absolute zero is approached. These statements require equilibrium and appropriate ground-state assumptions; a frozen disordered material or a degenerate ground state can carry residual entropy.

For a phase of fixed composition and pressure,A finite limiting entropy constrains this integral. Ordinary equilibrium heat capacities therefore vanish as temperature approaches zero, with familiar examples $C\propto T^3$ for three-dimensional acoustic phonons and $C\propto T$ for a normal degenerate Fermi liquid. A constant positive heat capacity extrapolated to zero would give a logarithmically divergent entropy difference and is impossible in this regime. Discontinuous transitions and anomalous limits need separate treatment rather than extending a high-temperature approximation indefinitely.

Gibbs free energy. For fixed particle numbers,Thus, when $S\to 0$, the Gibbs free energy approaches its zero-temperature value with a horizontal tangent:The equality $G(0,p)=H(0,p)$ uses the enthalpy $H=U+pV$; neither limiting value needs to be zero. The low-temperature Gibbs free-energy expansion makes the approach quantitative. For $C_p=a{T}^r$ with $r>0$ and $S(0)=0$,The second derivative is ${\partial }_T^{2}G=-C_p/T\le 0$, although it need not tend to zero. For a reaction between such equilibrium phases, ${\partial }_T\Delta G=-\Delta S\to 0$, and $\Delta G$ and $\Delta H$ approach one another. If a genuine residual entropy $S_0$ is retained instead, $G=G(0,p)-S_{0}T+\cdots$ has limiting slope $-S_0$. A horizontal tangent cannot be asserted without the zero-entropy condition.

Why helium remains a liquid. Helium atoms attract each other weakly and are exceptionally light. Their zero-point energy is large compared with the binding that would localize them in a solid. Both helium-4 and helium-3 can therefore remain liquid down to the lowest temperatures at low pressure; substantial applied pressure is needed to solidify them. This is not a violation of the third law. A quantum liquid can have a unique low-entropy ground state, and zero-point motion does not itself imply thermal entropy. The third law does not require every substance to crystallize at atmospheric pressure.

A helium-4 atom has integer spin zero and is a boson. Below the lambda point near $2.17\mathrm{K}$ at saturated vapor pressure, its liquid is superfluid. Its strong interactions mean that it is not quantitatively an ideal Bose gas, and the condensate fraction need not equal the superfluid fraction. The two-fluid model of superfluidity separates an entropy-carrying normal component of a superfluid from the superfluid component, whose entropy is zero. At sufficiently low temperature the dominant thermal excitations are acoustic phonons: their linear dispersion in three dimensions gives $U-U_0\propto V{T}^4$, $C_V\propto V{T}^3$ and $S\propto V{T}^3$. Higher-energy rotons are exponentially suppressed. Thus the normal component of a superfluid and entropy disappear as $T\to 0$, while superfluid flow and quantum motion can persist.

A helium-3 atom has spin $1/2$ and is a fermion. Well below its Fermi temperature, but above its superfluid transition, liquid helium-3 is a degenerate interacting Fermi liquid. Only quasiparticles close to the Fermi surface are thermally excited, giving $C_V\propto T$ and $S\propto T$. At temperatures of order a millikelvin, helium-3 also becomes superfluid through formation of Cooper pairs of fermions. This is not Bose-Einstein condensation of individual helium-3 atoms. Pairing reorganizes the low-energy excitation spectrum; in the fully gapped bulk B phase, the fermionic quasiparticle contribution to heat capacity is exponentially suppressed at temperatures well below the pairing gap. Gapless collective modes can still give thermal terms following a power law. The entropy nevertheless tends to zero. The normal Fermi-liquid law cannot be extrapolated through its pairing transition unchanged.^[ref]

Helium-3 also illustrates the importance of nuclear spin entropy. Over an intermediate low-temperature range, localized spins in the solid can retain approximately $k_B\ln 2$ of entropy per atom, while the degenerate liquid has much less. Reversible conversion of liquid to the higher-entropy solid then absorbs heat; adiabatic compression along the melting curve can cool the sample. This is Pomeranchuk cooling. At still lower temperature, interactions between the nuclear spins remove this nearly free-spin entropy, so a constant $k_B\ln 2$ cannot be extrapolated to absolute zero.^[ref]

Liquefaction and cooling methods. A gas must be below its liquid-gas critical point to form a distinct liquid phase by compression. Compression above the critical temperature alone cannot achieve ordinary liquid-vapour coexistence. Practical liquefaction removes the compression heat, cools the compressed gas in a heat exchanger and uses expansion to lower its temperature further.

A throttling valve operates approximately at constant specific enthalpy when kinetic-energy changes and heat leakage are negligible. The Joule-Thomson coefficient isusing consistent molar quantities, or consistent total quantities for a fixed amount. A pressure drop cools the gas if ${\mu }_{\mathrm{JT}}>0$. An ideal gas has zero Joule-Thomson coefficient; real intermolecular interactions are essential. In particular, room-temperature helium warms on throttling, so it must first be brought into a region of positive Joule-Thomson coefficient. Expansion through an engine or turbine can instead remove work and produce approximately isentropic cooling.

A regenerative liquefier sends the cold returning gas through a counterflow heat exchanger to precool the incoming compressed gas before its next expansion. Expansion engines and external precooling can supply the initial cooling; repeated operation then produces liquid. Modern helium liquefaction commonly combines work-producing expansion with a final throttle. ^[ref]

Once liquid exists, pumping away vapour lowers its saturation pressure and boiling temperature. The escaping vapour carries latent heat away, cooling the remaining liquid. At still lower temperatures, a dilution refrigerator exploits the different entropy carried by helium-3 in concentrated and dilute helium mixtures; adiabatic demagnetization and Pomeranchuk cooling provide other routes.

Approaching zero. The unattainability formulation of the third law states that regular cooling operations with finite resources cannot reach absolute zero in finitely many stages. As the entropy available to remove becomes small and thermal couplings weaken, each method loses its ability to take the final step. An isentropic cooling trajectory starting at nonzero entropy cannot reach a common zero-entropy endpoint through regular finite changes of parameters. A fully general finite-time claim also requires assumptions about available controls and heat-transfer rates. Quantum liquids and refrigeration methods can approach $0\mathrm{K}$ while remaining consistent with vanishing equilibrium entropy; they do not provide a finite-operation route to absolute zero.

Measure single-particle energy from the quantum harmonic oscillator ground state energy $\hslash \omega$, and write $\tilde{\mu }={\mu }_{\mathrm{physical}}-\hslash \omega \le 0$. The Bose-Einstein distribution and the supplied density of states giveThis excited-state population increases with the chemical potential and has the finite saturation valueAny additional particles must occupy the ground state: this is two-dimensional harmonic-trap Bose condensation. The critical temperature and the condensate fraction are thereforeHere $N$ is the total particle number, the integral of the inhomogeneous number density over the trap; there is no fixed uniform area in which $N$ itself is a number density. Equivalently $N\kappa ={\pi }^{2}m(k_B{T}_c{)}^2/(6{\hslash }^2)$ relates the trap population to the critical temperature. Harmonic confinement changes the low-energy density of states and permits this saturation in two spatial dimensions.

These formulas are leading large-$N$ results with $k_B{T}_c\gg \hslash \omega$. A sharp transition belongs to the trapped-gas thermodynamic limit $N\to \infty$, $\omega \to 0$ at fixed $N{\omega }^2$; a finite trap has finite-size rounding. In particular, finite $N_0$ requires $\tilde{\mu }=-k_{B}T\log (1+1/N_0)<0$, approaching zero in the condensation limit.

At the thermodynamic critical point, the two spinodal branches merge: $f^{\prime \prime }=0$ and $f^{(3)}=0$. The third derivative isFor $0<\varphi <1$, its zero gives $1-{\varphi }_c=\sqrt{N}{\varphi }_c$. Substitution in the spinodal relation yieldsAn absolute critical temperature requires an interaction–temperature relation. Under the usual energetic approximation $\chi =C/T$, define the theta temperature $\Theta =2C$, at which an infinite chain has $\chi =1/2$. ThenThus $T_c$ tends to a nonzero constant, $T_c\sim N^0$, while $\Theta -T_c\sim N^{-1/2}$. It is the temperature deficit, not $T_c$ itself, that scales as $N^{-1/2}$.

More generally, if a smooth $\chi (T)$ obeys $\chi (\Theta )=1/2$ and ${\chi }^{\prime }(\Theta )\ne 0$, thenThis is the critical point of a Flory–Huggins polymer solution. The same leading chain-length exponent follows for a smooth temperature law, while its coefficient depends on the omitted solvent thermodynamics. The lattice interpretation connects the chain-length dependence of mixing entropy to the theta condition $\chi =1/2$.^[ref].

Equilibrium requires $h=am+b{m}^3$. On the critical isotherm $a=0$, soHere $\delta$ is defined by $h\propto \mathrm{sgn}(m)|m{|}^{\delta }$ at the critical temperature, or equivalently $|m|\propto |h{|}^{1/\delta }$. The positive quartic coefficient selects a stable equilibrium, including the flat but stable zero-field thermodynamic critical point.

Near zero the Langevin function has expansionFor positive $x$, $\mathcal{L}(x)$ rises toward one and its ratio $\mathcal{L}(x)/x$ decreases from $1/3$ to zero. Therefore a line $\tau x$ has only the zero intersection for $\tau \ge 1/3$, and also a nonzero positive intersection for $0<\tau <1/3$. In a fixed signed component there is a corresponding negative solution; in the isotropic vector problem the nonzero solutions have all spin-space orientations.

The mean-field transition temperature isAt the critical temperature the line is tangent at the origin. Just below it, $S\simeq \sqrt{(5/3)(1-T/T_c^{\mathrm{MF}})}$, the usual mean-field square-root onset. The ordered minimum is stable below this temperature while $S=0$ becomes unstable.

This is a mean-field prediction, not a genuine finite-temperature ordering transition for the ideal two-dimensional short-range isotropic model. Long-wavelength spin fluctuations invalidate the predicted nonzero thermodynamic magnetization at every $T>0$. The relevant obstruction is the Mermin-Wagner theorem for continuous spin symmetry; anisotropy, long-range interactions or a finite sample change the assumptions. Its original paper establishes the absence of finite-temperature magnetic ordering for one- and two-dimensional finite-range isotropic Heisenberg models.^[ref]

At zero field, write the right-hand side of the self-consistency equation as $G_T(S)=g(4\beta JS)$, with $g(x)=2\sinh x/[1+2\cosh x]$. The origin is always a solution, and the slope there isThe critical crossing occurs when this equals one, givingThis is a continuous phase transition in the mean-field approximation. For $x>0$,so the positive branch is a strictly concave function, while $g$ is odd and bounded by one. For $T\ge T_c$, the slope at zero is at most one and strict concavity puts $G_T(S)$ below $S$ for every $S>0$, leaving only the unmagnetized solution. For $T<T_c$, it initially lies above $S$ but has $G_T(1)<1$; hence there is one positive intersection and, by spin inversion symmetry, one negative intersection.

The curvature of the Helmholtz free energy in the mean-field approximation is $4J[1-G_T^{\prime }(S)]$ per site. It is negative at $S=0$ below $T_c$ and positive at the nonzero intersections, identifying the ordered minima and the unstable zero solution.

The critical temperature and branches here belong to the mean-field approximation. In the exact finite system, spin inversion symmetry forces zero average magnetization at zero field; selecting a broken-symmetry phase requires the thermodynamic limit with a vanishing field of specified sign.

An equation of state relates equilibrium pressure, temperature and density. The ideal-gas form $p=n{k}_{B}T$ neglects intermolecular forces and excluded volume. A low-density real-gas expansion iswhere the virial coefficients encode interaction effects and generally depend on temperature.

A simple phenomenological example is the molar Van der Waals equationHere v is molar volume, b represents excluded volume and the attractive interaction lowers pressure through a. Its dilute expansion gives $pv/(RT)=1+[b-a/(RT)]/v+\cdots$. It recovers the ideal-gas limit but is not a quantitatively universal equation for all real gases.

Below its critical temperature, the raw isotherm contains a region with $(\partial p/\partial v{)}_T>0$, corresponding to negative isothermal compressibility and mechanical instability. Equilibrium phase separation replaces that segment by a coexistence pressure satisfying equal chemical potentials, equivalently the Maxwell equal-area construction. The critical point follows from $p_v=p_{vv}=0$:The distinction between an unstable homogeneous branch and a stable two-phase state is essential when interpreting a real-gas equation of state.

Use $v$ for molar volume to distinguish it from the fluctuating parcel volume below. The equation is $p=RT/(v-b)-a/v^2$. The parameter $b>0$ models excluded volume, while $a>0$ models the reduction in pressure from attractive interactions.

Below the critical temperature, a raw isotherm has a loop. The positive-slope segment is mechanically unstable and surrounding metastable portions do not give the stable equilibrium mixture. Replace the loop by a constant coexistence pressure $p_s$ satisfying the Maxwell equal-area condition,The endpoints are liquid and vapour molar volumes; intermediate mean volumes correspond to phase mixtures with changing proportions. Above the critical temperature, liquid-like and gas-like states connect continuously without a coexistence plateau.

In the pressure-temperature diagram, the liquid-vapour coexistence line ends at the critical point. Below it, the low-pressure side is vapour and the high-pressure side liquid. “Gas” conventionally denotes the region above the critical temperature, where compression alone cannot cross a liquid-vapour transition. The model has no solid phase.

The critical isotherm has an inflection: $p_v=p_{vv}=0$. Thuswhich giveFor fluctuations about a stable equilibrium value $x_0$, the system plus reservoir has probability proportional to $e^{-A(x)/(k_{B}T)}$. Expanding availability,gives a Gaussian distribution and, to this order, $⟨x⟩=x_0$ withThis requires positive curvature and sufficiently small fluctuations; it is not exact when higher-order terms matter.

The whole cylinder's volume is fixed. For the volume fluctuations of a small fluid parcel, interpret the small quantity as a material parcel containing $s$ moles, whose volume can fluctuate against the much larger remainder. Its mean volume at the critical density is $\bar{V}=s{v}_c=3sb$. At fixed temperature its availability is $A(V)=F_s(V,T)+p_{\mathrm{bath}}V$, soOn the critical isochore, approached from $T>T_c$,ConsequentlyEquivalently, $⟨\delta V^2⟩=k_{B}T{\kappa }_T\bar{V}$, with ${\kappa }_T=4b/[3R(T-T_c)]$. The mean-field variance diverges as $(T-T_c{)}^{-1}$, so its critical exponent is $1$; the root-mean-square fluctuation has exponent $1/2$. A fixed spatial observation box instead has number fluctuations, not fluctuating geometric volume. Finite size and non-Gaussian terms cut off the literal divergence; the approximation fails exactly at the critical point.