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Let $f(\mathbf{p})d^{3}p$ be particle number per unit volume in a momentum element, including the spin-state count in $f$. This is the local momentum part of a phase-space distribution function, normalized by the number density $n=\int f(\mathbf{p})d^{3}p$. For an isotropic distribution with particle energy $E(p)=\sqrt{m^2{c}^4+p^2{c}^2}$, the kinetic pressure of an isotropic gas is the average momentum flux:The kinetic energy density, explicitly excluding rest mass, isFor $p\ll mc$, $E-m{c}^2=p^2/(2m)$ and $\mathbf{v}=\mathbf{p}/m$, giving $P=2u_{\mathrm{kin}}/3$. For $p\gg mc$, $E\simeq pc$ and $v\simeq c$, giving $P\simeq u_{\mathrm{kin}}/3$. ThusNeither relation assumes a Maxwellian distribution. At intermediate momenta neither constant ratio is exact, and anisotropic distributions require a pressure tensor rather than this scalar pressure.

For a classical Maxwell-Boltzmann distribution, $f=n(2\pi m{k}_{B}T{)}^{-3/2}\exp [-p^2/(2m{k}_{B}T)]$. Three Gaussian component integrals give $⟨p^2⟩=3m{k}_{B}T$, so the ideal gas hasIn a fully ionized mixture, sum over independent species to obtain $P_{\mathrm{gas}}=\rho k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$, where the mean molecular weight counts ions and free electrons. This excludes partial-ionization and interaction corrections. A dilute classical ultrarelativistic gas still has $P=n{k}_{B}T$ but $u_{\mathrm{kin}}\simeq 3n{k}_{B}T$; its momentum distribution is proportional to $e^{-pc/(k_{B}T)}$ instead of the nonrelativistic Gaussian.

For a thermal photon gas, the two polarization states and zero photon chemical potential give the Planck photon distributionUse $x=pc/(k_{B}T)$ and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation constant and photon equation of state are thereforeLikewise $n_{\gamma }=2\zeta (3)(k_{B}T/\hslash c{)}^3/{\pi }^2$. Unlike a gas with a fixed particle number, photons do not have pressure proportional to baryonic mass density.

For fully degenerate electrons, the Fermi-Dirac distribution becomes a filled momentum sphere with two spin states. Counting them gives the Fermi momentumHere ${\mu }_e$ is the mean molecular weight per electron. The equation of state of a cold electron gas follows by integrating momentum flux up to $p_F$:Its two limits areThe corresponding kinetic energy densities are $3n_e[p_F^2/(2m_e)]/5$ and $3n_e{p}_{F}c/4$, respectively. Thus in the high-density relativistic limit the electron pressure scales as ${\rho }^{4/3}$ and is nearly independent of temperature. This is the ideal noninteracting, fixed-composition electron result, not a universal equation of state at nuclear densities where captures, interactions and the composition change.

The boundaries on a stellar equation-of-state regime diagram concern the electron component. Define the electron relativistic density threshold and thermal electron relativistic threshold byA degenerate electron gas is nonrelativistic well below ${\rho }_{\ast }$ and ultrarelativistic well above it: this is an approximately vertical division on a log-density plot. A nondegenerate electron gas instead becomes thermally relativistic near $T_{\ast }$: this is an approximately horizontal division. The regimes have broad crossovers rather than a discontinuity at either line.

The exact electron Fermi temperature, subtracting electron rest energy, isFor electrons with negligible thermal pairs, strong degeneracy requires $T\ll T_F$, whereas $T\gg T_F$ gives a nondegenerate gas. In the pair-rich regime the actual electron and Positron distributions must instead be treated with their chemical potentials, as discussed below. The two limiting degeneracy boundaries areThese slopes $2/3$ and $1/3$ explain the bent degeneracy boundary on the logarithmic graph. A thermal-wavelength test $n_e[2\pi {\hslash }^2/(m_e{k}_{B}T){]}^{3/2}\sim 2$ gives the same nonrelativistic density/temperature scaling, with an order-one definition of the crossover.

The radiation-to-gas pressure boundary in the nondegenerate fully ionized regime isRadiation dominates above this line. Once electrons are strongly degenerate, compare radiation pressure with $P_e(\rho )$ instead of continuing the ideal-gas comparison into that regime. The radiation-to-degeneracy pressure boundary is $T=[3P_e(\rho )/a_{\mathrm{rad}}{]}^{1/4}$, with logarithmic slopes $5/12$ and $1/3$ in the nonrelativistic and ultrarelativistic limits. Degeneracy of the electrons and dominance of their pressure are distinct criteria.

For the pair curve, distinguish baryonic net electrons from thermally created electrons and Positrons. Chemical equilibrium with photons requires opposite electron/Positron chemical potentials when the one-particle energy includes rest energy, as in the distributions below. At low temperature and low degeneracy the zero-chemical-potential density per charge species isCharge neutrality gives $n_{-}-n_{+}=n_b=\rho /({\mu }_e{m}_u)$, while the nondegenerate equilibrium product is $n_{-}{n}_{+}=n_0^2$. HencePairs become important when $n_0$ is comparable with or exceeds $n_b$, approximately the electron-positron thermal pair abundance curve ${\rho }_{\mathrm{pair}}(T)={\mu }_e{m}_u{n}_0(T)$. Below its density at a fixed temperature, pairs dominate over the net charge electrons. The exponential makes the low-temperature portion steep in a log-log plot. For ${\mu }_e=2$, the full zero-potential integral gives approximate pair-marker densities $1.80\times 10^3\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $10^9\mathrm{K}$ and $7.52\times 10^5\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $3\times 10^9\mathrm{K}$. At relativistic temperature use the full zero-chemical-potential Fermi-Dirac distribution instead; it gives $n_0\to [3\zeta (3)/(2{\pi }^2)](k_{B}T/\hslash c{)}^3$, so the high-temperature pair curve approaches slope three in log-density versus log-temperature. Strong net-electron degeneracy suppresses Positrons and requires the full chemical-potential-dependent distribution.

The original diagram uses fully ionized helium, ${\mu }_e=2$ and ${\mu }_{\mathrm{mol}}=4/3$, only to set numerical locations. It plots the full $T_F$ curve and the zero-chemical-potential pair integral, rather than extending their asymptotes into the crossover. The pair boundary is deliberately an approximate abundance marker, not a phase transition; at high temperature the nonrelativistic ion approximation and fixed-composition model also have limits. For $T\gg T_{\ast }$ with abundant nondegenerate pairs, the combined electron/Positron energy density is $7a_{\mathrm{rad}}{T}^4/4$, so photons plus pairs have $u=11a_{\mathrm{rad}}{T}^4/4$ and $P=u/3$. Real stellar matter adds Coulomb effects, partial ionization, nuclear reactions and, at sufficiently high density, nuclear-matter physics beyond the ideal regime map.

The field $\varphi$ is a compositional order parameter in a binary fluid mixture. Its spatial integral is conserved in a closed sample, so advection and diffusion enter a continuity equation; the material derivative follows the incompressible fluid. The diffusive current $\mathbf{J}=-M\nabla \mu$, with positive order-parameter mobility $M$, moves material down the chemical potential gradient. The chemical potential of a composition field is the functional derivative of the symmetric quartic Landau free energyso $\mu =a\varphi +b{\varphi }^3-\kappa {\nabla }^2\varphi$. Positive $b,\kappa$ stabilize large amplitudes and penalize sharp interfaces; $a<0$ permits phase separation.

The momentum equation is the Navier-Stokes equation with dynamic viscosity $\eta$, constant mass density $\rho$, and the Korteweg force $-\varphi \nabla \mu$ arising from composition stresses. The pressure enforces incompressible flow, $\nabla \cdot \mathbf{v}=0$, and can absorb pure gradient contributions to that thermodynamic force. For example, $-\varphi \nabla \mu =\mu \nabla \varphi -\nabla (\varphi \mu )$ gives an equivalent force after redefining the pressure. These equations constitute deterministic Model H dynamics.

Their dissipative signs can be checked by the energy dissipation identity for Model H. Under periodic or no-flux, no-work boundary conditions, composition advection and capillary work cancel between $F$ and the kinetic energy, givingThus diffusion and viscosity dissipate energy, while the reversible advective coupling transfers it between composition and motion. Thermal noises are omitted at this deterministic hydrodynamic level.

Here $p(n)$ is the probability of microstate $n$, including all occupation data. In the grand canonical ensemble, with energy $E_n$, particle number $N_n$, chemical potential $\mu$ and $T>0$,Substitution in the Gibbs entropy gives $S=k\log \mathcal{Z}+(U-\mu N)/T$, where $U=⟨E⟩$ and $N=⟨N_n⟩$. Differentiating at fixed $\mu ,V$ gives ${\partial }_T\log \mathcal{Z}=(U-\mu N)/(k{T}^2)$. Therefore

For a simple homogeneous thermodynamic system, internal energy is extensive: $E(\lambda S,\lambda V,\lambda N)=\lambda E(S,V,N)$. Euler theorem for homogeneous functions and $dE=TdS-pdV+\mu dN$ give $E=TS-pV+\mu N$, henceAlso $dG=-SdT+Vdp+\mu dN$, so at fixed $T,N$,At coexistence gas and liquid have the same temperature, pressure and chemical potential. Integrating $d\mu =vdp$ along the intervening isotherm gives $\int vdp=0$. Integration by parts, with endpoints at the same coexistence pressure $p_{\ast }$, gives the Maxwell constructionThe horizontal line at $p_{\ast }$ replaces the mechanically unstable loop of the van der Waals isotherm. The extensivity argument presupposes the ordinary thermodynamic limit and neglects macroscopic interface contributions.

Take $b>0$ for a stable quartic Landau free energy. The cubic-term removal in a conserved quartic Landau free energy uses $\varphi =\psi -c/(3b)$ and gives, apart from a constant,For a closed binary fluid mixture, the compositional order parameter has fixed integral. Adding $d\int \varphi$ therefore adds the same constant to every allowed state. Equivalently, in the common-tangent construction for phase coexistence, a linear term changes the tangent slope, hence the chemical potential, by $d$, but leaves the coexistence compositions unchanged. This is why linear composition bias leaves coexistence compositions unchanged; it would not mean an unchanged composition at a fixed externally imposed chemical potential.

In the shifted variable, the binodal compositions are ${\psi }_{\pm }=\pm \sqrt{-a^{\prime }/b}$ for $a^{\prime }<0$. They merge atThe continuous phase transition therefore has the same scalar quartic Landau theory as the symmetric binary fluid mixture, with a shifted critical composition and control parameter. In the mean-field approximation, the composition difference vanishes with order-parameter critical exponent $1/2$. This argument presumes the critical composition lies in the physically accessible composition range; conservation affects the dynamics, not this static phase coexistence construction.

In the microcanonical ensemble, the microcanonical energy-shell entropy is the Boltzmann entropyFor a macroscopic system, changing the small energy resolution by a fixed factor changes $\log \Omega$ only by a nonextensive additive term, while $S=O(N{k}_B)$. Its relative effect therefore vanishes in the thermodynamic limit.

The thermodynamic temperature, pressure, and chemical potential are defined in the entropy representation byThusor, equivalently, the first law of thermodynamics in fundamental form is

For each one-particle state $i$, a boson occupation number can be any $n_i\ge 0$. Its contribution to the grand canonical partition function is the geometric seriesThe noninteracting modes factorize, soConvergence requires $\mu <{\min }_i{ϵ}_i$. Since the lowest energy is zero, the limiting maximum chemical potential is

In the very early universe, assume species $i$ is in thermal equilibrium, is ultrarelativistic so that $kT\gg m_i{c}^2$ and $E(p)\simeq pc$, and has negligible chemical potential, ${\mu }_i\simeq 0$. Its occupation factor is then the Bose-Einstein distribution for a boson or the Fermi-Dirac distribution for a fermion.

Substitute $E=pc$ and set $y=pc/(kT)$. Since $p=(kT/c)y$ and $dp=(kT/c)dy$,

The reaction maintaining chemical equilibrium isFor $T<m_e$ and negligible chemical potentials, the electron and Positron are dilute and obey the nonrelativistic equilibrium number density. Including two spin states for each particle,The photon number density is $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, so the electron-positron equilibrium pair abundance below the electron mass isIf $n_e$ denotes only one of the charge-conjugate species, this expression is divided by two.

Use the stable parameters $b,\kappa >0$. The bulk stationary points satisfy $f^{\prime }(\varphi )=a\varphi +b{\varphi }^3=0$. For $a<0$, $\varphi =0$ has negative curvature, while the two minima areBoth have $f(\pm {\varphi }_B)=-a^2/(4b)$ and $f^{\prime \prime }(\pm {\varphi }_B)=-2a>0$. To establish minimization at fixed total composition, writeFor mean composition $\bar{\varphi }$ with $|\bar{\varphi }|<{\varphi }_B$, a mixture of the two minima with volume fraction $\lambda =(\bar{\varphi }+{\varphi }_B)/(2{\varphi }_B)$ in the positive phase attains this lower bound in the bulk. Macroscopic slabs have only a subextensive interfacial excess free energy, so the bound is attained per unit volume in the thermodynamic limit. Thus constrained minimization gives phase coexistence in the mean-field approximation; at the endpoints only one phase is needed.

The two equilibrium conditions also follow directly by constrained optimization. Introduce a Lagrange multiplier $\mu$ for $\lambda {\varphi }_{+}+(1-\lambda ){\varphi }_{-}=\bar{\varphi }$ in the mixture's bulk free-energy density. Varying ${\varphi }_{+}$ and ${\varphi }_{-}$ gives $f^{\prime }({\varphi }_{+})=f^{\prime }({\varphi }_{-})=\mu$; varying $\lambda$ gives $f({\varphi }_{+})-\mu {\varphi }_{+}=f({\varphi }_{-})-\mu {\varphi }_{-}$. These equate the chemical potentials and pressures, respectively.

Equivalently, their bulk chemical potentials are both zero and their pressures are equal:These are precisely the conditions of the common-tangent construction for phase coexistence: the horizontal supporting tangent touches both minima. Equal chemical potentials prevent net composition exchange, and equal pressure gives mechanical equilibrium at a flat interface.

Apply force balance to one hemisphere. The pressure force is $\pi R^2({\Pi }_{\mathrm{in}}-{\Pi }_{\mathrm{out}})$, while the interfacial tension around its equator gives $2\pi R\gamma$. The Young–Laplace equation therefore yieldsTo leading order in curvature, let the two bulk compositions be ${\varphi }_{+}={\varphi }_B+{\delta }_{+}$ and ${\varphi }_{-}=-{\varphi }_B+{\delta }_{-}$. Their curvatures agree, $\alpha =f^{\prime \prime }(\pm {\varphi }_B)=-2a$, so equal chemical potentials require $\alpha {\delta }_{+}=\alpha {\delta }_{-}$ and hence a common shift $\delta$.

Since $d\Pi /d\varphi =\varphi f^{\prime \prime }(\varphi )$, expansion about the flat coexistence compositions gives ${\Pi }_{+}-{\Pi }_{-}=2{\varphi }_B\alpha \delta +O({\delta }^2)$. Matching the capillary pressure proves the Gibbs--Thomson relation in these variables:The pressure jump balances interfacial forces; it does not permit unequal chemical potentials. A common shift in both compositions supplies a common interfacial chemical potential while their pressures differ. The expansion assumes $R$ large compared with the phi-four diffuse interface width.

The far-field composition $-{\varphi }_B$ has chemical potential zero. Local curved-interface coexistence imposes $\mu (R)={\mu }_s=\gamma /({\varphi }_{B}R)$. The decaying spherically symmetric solution of the exterior Laplace equation isWith constant mobility, its outward current isThe linearized exterior composition is $\varphi (r)=-{\varphi }_B+\mu (r)/\alpha$, consistent with the surface shift in the previous part.

For diffusion-controlled droplet growth, the interior is uniform to leading order and has no diffusive current. Replacing exterior material by interior material requires the composition jump $2{\varphi }_B$, so conservation gives $2{\varphi }_B\dot{R}=-J_r(R)$. HenceThe negative sign represents evaporation. The total outward current is $4\pi R^2{J}_r(R)=4\pi M\gamma /{\varphi }_B$, independent of $R$ in this leading quasi-static capillary approximation.

The flat interface is an equilibrium reservoir with $\mu =0$, while the droplet has the constant surface value ${\mu }_s=\gamma /({\varphi }_{B}R)$. The exterior chemical potential satisfies the Laplace equation, tends to zero far away, and obeys these two surface values. These are exactly the electrostatic boundary conditions at a conductor for a sphere held at potential ${\mu }_s$ above a grounded infinite plane.

Introduce an arbitrary permittivity $ϵ$. In the electrostatic problem, $Q=-ϵ{\int }_{S_R}{\partial }_n\mu dS=C(R,h){\mu }_s$. Therefore the diffusion-capacitance analogy gives the outward currentFor an isolated sphere, its capacitance is $C(R,\infty )=4\pi ϵR$, so $I(R,\infty )=4\pi MR{\mu }_s=4\pi M\gamma /{\varphi }_B$. Dividing givesThus sphere-plane capacitance determines the evaporation enhancement. The assumed spherical shape gives the corresponding law $\dot{R}=-M\gamma c(R/h)/(2{\varphi }_B^2{R}^2)$ for droplet evaporation near a planar reservoir. The local surface current is generally nonuniform, even though its total is described by a single capacitance.

The chemical potential is the change in internal energy when one particle is added at fixed entropy and volume:Equivalently, $\mu =(\partial F/\partial N{)}_{T,V}$ for the Helmholtz free energy.

Let a microstate $s$ have energy $E_s$ and particle number $N_s$. Maximizing the Gibbs entropy subject to normalization and fixed mean values of $E$ and $N$ is the maximum-entropy derivation of equilibrium ensembles. With Lagrange multipliers $\alpha$, $\beta$, and $-\beta \mu$, varyThe stationarity equation is $-\log p_s-1-\alpha -\beta (E_s-\mu N_s)=0$. Normalization therefore gives the grand canonical ensembleHere $\mathcal{Z}$ is the grand canonical partition function and $k_B$ is the Boltzmann constant.

For one fermionic quantum state of energy $\epsilon$, the Pauli exclusion principle permits occupation numbers only $n=0,1$. Its two grand-canonical weights are $1$ and $e^{-\beta (\epsilon -\mu )}$, so its mean occupation is the Fermi-Dirac distribution

For a free nonrelativistic particle, $\epsilon ={\hslash }^2{k}^2/(2m)$. In a region of area $A$, the number of wave-vector states in the annulus from $k$ to $k+dk$, including the two spin angular momentum states, isSince $kdk=(m/{\hslash }^2)d\epsilon$, the two-dimensional free-electron density of states is constant:

We now use units in which $k_B=1$, as in the question. At zero temperature, the Fermi-Dirac distribution is a step function, and henceAt positive temperature the same fixed particle number satisfies the exact relationThusThis is the low-temperature particle-number cancellation for constant density of states. Therefore, for $T\ll {\epsilon }_F$, the mean number of particles in the energy interval $[\epsilon ,\epsilon +d\epsilon ]$ is

Finally, compare the internal energy with its zero-temperature value. The thermally excited particles above ${\epsilon }_F$ and holes below ${\epsilon }_F$ have equal leading particle numbers, so their terms proportional to ${\epsilon }_F$ cancel. With $x=|\epsilon -{\epsilon }_F|/T$,where the integral follows from the fermion-to-boson thermal integral ratio. Differentiation gives the linear low-temperature heat capacity of a two-dimensional Fermi gas:The requested power law is therefore linear in $T$, with exponent one.

Impose the fixed conserved order parameterwith a Lagrange multiplier ${\mu }_0$. Stationarity ofunder every local variation givesThus the chemical potential is spatially constant at constrained equilibrium.

The fundamental thermodynamic relation isBecause $E(S,V,N)$ is extensive, the Euler theorem for homogeneous functions givesSubstitution into the Gibbs free energy $G=E+pV-TS$ therefore yieldsFor a one-component equilibrium phase, the intensive chemical potential depends on the intensive state variables $T$ and $p$.

Because $\varphi$ is a conserved composition density, an infinitesimal material displacement $\mathbf{r}\mapsto \mathbf{r}+\mathbf{u}$ changes it at fixed position byThe second term is essential when the deformation is compressible. By the definition of the chemical potential, and taking $\mathbf{u}$ to vanish on the boundary,The same free-energy change written in terms of the stress tensor isSince $\mathbf{u}$ is arbitrary,This is the Korteweg force density of a diffuse-interface mixture.

The microcanonical ensemble describes an isolated system at fixed energy $E$, volume $V$, and particle number $N$, assigning equal weight to accessible states in a narrow energy shell.

The canonical ensemble describes a system that can exchange energy with a heat reservoir at fixed temperature $T$, while $V$ and $N$ remain fixed. A microstate $s$ has probability

The grand canonical ensemble allows exchange of both energy and particles at fixed $T$, $V$, and chemical potential $\mu$. Its state weights are

The choice is irrelevant for bulk thermodynamic observables in the thermodynamic limit when the system is additive, has sufficiently short-range interactions, and lies away from singular phase-coexistence regimes. Then relative energy and particle-number fluctuations vanish as the system size grows. These are the standard conditions for equivalence of statistical ensembles.

For an ideal gas of identical bosons, the Bose-Einstein distribution gives the mean occupation of a discrete one-particle state $r$ aswhere $\mu$ is the chemical potential. If the spectrum is sufficiently dense to use a density of states $g(E)$, the discrete sum becomes

A phonon has zero chemical potential and energy $\hslash \omega$. The total internal energy is consequentlyWith the change of variable $x=\beta \hslash \omega$ and $\beta \hslash {\omega }_D=T_D/T$, this becomes