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In the nondegenerate regime, $\mu$ lies well inside the gap, with $\mu \gg kT$ and $E_g-\mu \gg kT$. Then conduction occupations are approximately $e^{(\mu -E)/(kT)}$, while the hole probability in a valence state is approximately $e^{(E-\mu )/(kT)}$. Using the near-edge density of states and the supplied gamma integral givesThereforeThis assumes the edge approximations are valid over the thermally populated ranges. For an intrinsic semiconductor $n=p$, so each carrier density is proportional to $T^{3/2}{e}^{-E_g/(2kT)}$. Both mobile conduction electrons and holes contribute to conductivity, $\sigma =|e|(n{b}_e+p{b}_h)$ with mobilities $b_e,b_h$. The exponentially varying carrier densities explain strong temperature dependence even when the mobilities vary only algebraically.

Donors supply electrons and acceptors remove electrons to create holes. With complete ionization and negligible minority carriers, the majority densities are $n\simeq N_d$ on the n side and $p\simeq N_a$ on the p side. The corresponding isolated chemical potentials areCarrier diffusion at contact leaves positively charged donors and negatively charged acceptors in a depletion layer. At equilibrium the electrochemical potential is constant; the electron energy shift is $-|e|\varphi$. Hence the n-minus-p electric potential isAll $N$ quantities must have a consistent normalization: if $N_c,N_v$ are state densities per volume, $N_d,N_a$ are dopant densities rather than bare atom counts. The formula neglects incomplete ionization, degeneracy and changes in band structure.

At a common temperature the two equilibrium junction/contact effects do not drive a perpetual current. At different junction temperatures, electrons and holes transport entropy differently, giving different Seebeck coefficients on the two sides. The loop has a thermoelectric electromotive force, oriented as ${\int }_{T_1}^{T_2}(S_n-S_p)dT$, and therefore generally carries current when closed. This is the Seebeck effect: energy is supplied by heat flow from hot to cold. The temperature dependence of the junction barrier is useful intuition, but an exact terminal voltage is not obtained simply by subtracting two equilibrium built-in potentials; it also depends on bulk transport and contact conditions. The given equilibrium density-of-states data do not determine the numerical thermopowers.

Chemical equilibrium for $p+e^{-}\leftrightarrow H+\gamma$ implies ${\mu }_H={\mu }_p+{\mu }_e$, because equilibrium photons have zero chemical potential. Dividing the three nonrelativistic Maxwell-Boltzmann number densities therefore givesThe Hydrogen binding energy is $I=m_p+m_e-m_H=13.6\mathrm{eV}$ in natural units. For ground-state hydrogen, with hyperfine splittings unresolved, $g_H=4$ and $g_p=g_e=2$, so the degeneracy ratio is one. Equivalently one can omit nuclear-spin degeneracy consistently from both the proton and atom. Excited hydrogen levels are exponentially suppressed at recombination temperatures. Also $m_H/m_p=1+O(m_e/m_p)$, so its effect in the prefactor is negligible. ThusThe sign of the exponential follows from the smaller rest mass of the bound atom: cooling favours neutral hydrogen once translational entropy no longer compensates the binding energy.

The chemical potential is $\mu =(\partial E/\partial N{)}_{S,V}$, equivalently $(\partial F/\partial N{)}_{T,V}$: it is the free-energy cost of adding a particle at fixed temperature and volume. With units $k_B=1$, a large reservoir has entropy $S_R(E_{\mathrm{tot}}-E,N_{\mathrm{tot}}-N)=S_R^0-E/T+\mu N/T$ to first order. Since the number of compatible reservoir states is proportional to its exponential, the grand canonical ensemble assignsFor independent fermions, a one-particle state of energy $\epsilon$ has occupation zero or one. Its partition factor is $1+e^{-(\epsilon -\mu )/T}$, so the mean occupation is the Fermi-Dirac distribution $[e^{(\epsilon -\mu )/T}+1{]}^{-1}$. Multiplication by the given density of states yields $dN=C{\epsilon }^{1/2}d\epsilon /[e^{(\epsilon -\mu )/T}+1]$.

At zero temperature the occupation is a step function, filled below $\mu$ and empty above it, so $\mu ={\epsilon }_F$ and $N=(2C/3){\epsilon }_F^{3/2}$. At fixed particle number, the low-temperature shift of $\mu$ is $O(T^2/{\epsilon }_F)$ and does not affect the leading estimate. In the narrow interval of width $T$ above ${\epsilon }_F$, replace the density of states by $C\sqrt{{\epsilon }_F}$ and put $u=(\epsilon -{\epsilon }_F)/T$. ThenEquivalently, the fraction above the Fermi energy is asymptotic to $(3\log 2/2)T/{\epsilon }_F$.

The degeneracy factor $g_A$ counts internal states with the same energy, including the allowed spin states. The chemical potential ${\mu }_A$ measures the free-energy cost of changing the number of particles of species $A$; in this convention the rest energy remains in the exponential of the nonrelativistic Maxwell--Boltzmann number density.

The chemical potential is fixed by the densities of conserved charges and by chemical equilibrium between reactions. For a reaction ${\sum }_A{\nu }_{A}A=0$, chemical equilibrium requires ${\sum }_A{\nu }_A{\mu }_A=0$. Equivalently, when these are the only constraints, ${\mu }_A$ is a linear combination of the chemical potentials of the conserved charges carried by $A$. An unconstrained species has zero chemical potential. In particular, the photon chemical potential vanishes in thermal equilibrium when photon-number-changing reactions are effective.

For $p+e\rightleftharpoons H+\gamma$, chemical equilibrium and the vanishing photon chemical potential imply ${\mu }_H={\mu }_p+{\mu }_e$. Taking the ratio of the three nonrelativistic Maxwell--Boltzmann number densities cancels these chemical potentials:The ground-state hydrogen atom has four combined electron and proton spin states if the small hyperfine splitting is ignored, whereas $g_p=g_e=2$. Hence $g_H/(g_p{g}_e)=1$. With Hydrogen binding energy $B=M_p+M_e-M_H$ and $M_H\simeq M_p$, the Saha ionization equation givesThe approximation in the prefactor neglects relative corrections of order $(M_e+B)/M_p$; the Hydrogen binding energy must be retained in the exponential.

For a nonzero uniform wavefunction, the Laplacian vanishes. Dividing the stationary cubic-quintic Gross–Pitaevskii equation by ${\psi }_0$ yieldsEquivalently, the interaction energy density is $e(n)=V_0{n}^2/2+W_0{n}^3/3$ and its derivative $e^{\prime }(n_0)$ is the chemical potential. Positive local compressibility requires ${\mu }^{\prime }(n_0)=V_0+2W_0{n}_0>0$. This stability condition is distinct from the sign of $\mu$ itself.

At fixed particle number $N_p=\int |\psi {|}^{2}d\mathbf{x}$, an equilibrium wavefunction is a constrained critical point of the energy functional. Introduce the chemical potential $\mu$ as the Lagrange multiplier for $N_p$, and impose $\delta (E-\mu N_p)=0$. The resulting stationary cubic-quintic Gross–Pitaevskii equation isThe time-dependent equilibrium has $\psi (\mathbf{x},t)=e^{-i\mu t/\hslash }{\psi }_0(\mathbf{x})$, so its number density is stationary even though its quantum phase changes uniformly. The specified particle number fixes $\mu$ together with the boundary conditions. In an infinite uniform bulk, particle number is understood through a finite volume or a prescribed bulk density.

For the displayed scaling, assume $\mu >0$ and $n_0>0$. SetSubstituting in the stationary cubic-quintic Gross–Pitaevskii equation and dividing by $\mu \sqrt{n_0}$ givesThe uniform-state relation proves $\alpha +\beta =1$, so the real bulk amplitude is one; its constant quantum phase can be chosen to make $\tilde{\psi }\to 1$ in the zero-winding sector. The unit of distance $\ell$ is the kinetic-to-chemical-potential scale, reducing to the stated healing length convention for purely repulsive two-body coupling.

The positive-$\mu$ assumption is necessary for the printed sign convention. If $\mu <0$, the real length is $\hslash /\sqrt{2m|\mu |}$, and the dimensionless linear term is $-1$, with interaction coefficients divided by $|\mu |$; the displayed equation with linear term $+1$ does not follow. At $\mu =0$ this scaling is undefined. These are limitations of this particular nondimensionalization, not a claim that all cubic-quintic bulk states have positive chemical potential.

Write $g=V_0$ for the effective coupling and keep $N$ for the number obtained by the stated two-dimensional normalization. A repulsive interaction, $g>0$, permits the Thomas–Fermi approximation for a condensate: density-gradient kinetic energy is small compared with the trapping and interaction terms over most of the cloud. The stationary Gross–Pitaevskii equation then givesand $n_{\mathrm{TF}}=0$ outside the cloud, up to the narrow edge region. HereNormalize the parabolic number density directly:Consequently the two-dimensional Thomas–Fermi condensate parameters areWith $a_{\mathrm{ho}}=\sqrt{\hslash /(m\omega )}$ and central healing length $\xi =\hslash /\sqrt{2m\mu }$, the bulk approximation requiresIndeed $R/\xi =2\mu /(\hslash \omega )\gg 1$. In the coupling convention printed in the paper, formal substitution of $g=4\pi {\hslash }^{2}a/m$ givesThis is the requested relationship in that convention; $\omega$ sets the physical size and chemical potential, while the interaction-strength ratio to $\hslash \omega$ is independent of $\omega$.

There is a dimensional convention to make explicit. A physical three-dimensional scattering length $a$ makes $4\pi {\hslash }^{2}a/m$ a three-dimensional contact coupling, whereas a genuinely two-dimensional particle-number normalization requires coupling units of energy times area. For an axial factor $\chi (z)$ normalized to one, the effective two-dimensional contact coupling is $g_{2D}=(4\pi {\hslash }^{2}a/m)\int |\chi {|}^{4}dz$. For a Gaussian of width $a_z$ this is $4\pi {\hslash }^{2}a/(m\sqrt{2\pi }{a}_z)$, and the physical TF condition becomes $Na/a_z\gg 1$ up to numerical constants. Alternatively the printed coupling can be retained if $N$ is a line density. The paper supplies no axial scale, so its $Na$ expressions require that effective-unit or line-density convention; the formulas in terms of $g$ above are unambiguous. The approximation fails inside vortex cores and at the cloud edge, and it is not a repulsive TF ground-state construction for $g\le 0$.

For a positive uniform number density $n_{\ast }$ and constant superfluid velocity $v\hat{\mathbf{x}}$, current divergence vanishes. The stationary density equation therefore imposes $(\gamma -\Gamma n_{\ast })n_{\ast }=0$, giving $n_{\ast }=\gamma /\Gamma$. The quantum potential vanishes, so the phase equation sets $\mu =m{v}^2/2+V_0\gamma /\Gamma$. Thus the uniform-current pumped polariton condensate isDirect substitution verifies the equation: the kinetic term gives $m{v}^2/2$, the interaction gives $V_0\gamma /\Gamma$, and the gain-loss term is zero. A constant overall phase is arbitrary.

With the stated $\gamma ,\Gamma >0$, this nonzero plane wave exists for any real $v$ and real interaction coefficient $V_0$ when $\mu$ is determined by the solution. If a particular real chemical potential is prescribed instead, the existence condition isEquality gives the uniform condensate at rest. A nonzero density generally requires $\gamma /\Gamma >0$; a zero nonlinear-loss coefficient with positive pumping would not allow this nonzero stationary balance. The vacuum is a separate zero solution, unstable to small perturbations for positive pump.

Existence should not be confused with stability. For completeness, linearize in the comoving frame with $\psi ={\psi }_{\ast }[1+u+iw]$, and let ${ϵ}_k={\hslash }^2{k}^2/(2m)$. The linear real-amplitude and phase equations givesoFor $V_0\ge 0$ both growth rates have nonpositive real parts, including the neutral uniform-phase mode. For $V_0<0$, sufficiently small nonzero $k$ makes the constant term negative, so one growth rate is positive: the plane wave still exists but is unstable in an infinite system. Uniform drift adds only a Doppler phase to these growth rates in this model.

Here equilibrium means a time-independent number density and current, with a common temporal quantum phase. Write $\psi (\mathbf{x},t)=e^{-i\mu t/\hslash }\Psi (\mathbf{x})$ with real $\mu$, since an imaginary temporal frequency would change the density. The stationary pumped Gross–Pitaevskii equation isThe chemical potential is the common phase-rotation frequency times $\hslash$; this open system has no conserved particle-number constraint fixing it as in the conservative problem. Local gain and loss need not cancel separately in an inhomogeneous equilibrium, because a stationary condensate number current can transport particles between gain and loss regions.

Chemical equilibrium gives ${\mu }_{\mathrm{He}}=2{\mu }_p+2{\mu }_n$, since the photon chemical potential is zero. Set $A=(2\pi m_{p}kT/h^2{)}^{3/2}$. The degeneracies and approximate masses give prefactors $2A,2A,8A$ for protons, neutrons and helium. ThusHere the binding energy uses $2m_p+2m_n-m_{\mathrm{He}}$; the PDF prints $2n_n$ in that mass relation, a dimensional typo.

Baryon conservation gives $n_B\propto a^{-3}$. In adiabatic expansion without photon-producing entropy release, $T\propto a^{-1}$ and $n_{\gamma }\propto T^3\propto a^{-3}$, so $\eta =n_B/n_{\gamma }$ stays constant. This statement is for that epoch; entropy transfer such as electron-positron annihilation modifies the photon normalization.

Since the density ratio equals $n_B^3{X}_p^2{X}_n^2/X_{\mathrm{He}}$, substitution of the given photon density yieldsAt $m_p{c}^2\simeq 938$ MeV, $B_{\mathrm{He}}=30$ MeV and $kT=0.3$ MeV, the right side is about $1.22\times 10^{-29}{\eta }^{-3}$. Unity therefore corresponds to $\eta \simeq 2.3\times 10^{-10}$, illustrating the stated very rough temperature estimate for a cosmological baryon-to-photon ratio of order $10^{-10}$ to $10^{-9}$. A numerical value of $\eta$ is necessary; it is not specified in the question. The exponential makes the temperature estimate far less sensitive than the ratio itself to order-one prefactors.

Actual helium synthesis must proceed through light intermediates. The deuterium bottleneck persists because deuterium's much smaller binding energy allows the abundant high-energy photon tail to destroy it. Only at a lower temperature do enough deuterons survive to initiate efficient reactions leading to helium. Direct four-nucleon assembly at the equilibrium temperature is ineffective.

Periodic boundary conditions give wavevectors $\mathbf{k}=(2\pi /L)(n_1,n_2)$. Each state occupies area $(2\pi /L{)}^2$ in wavevector space. A ring of radius $k$ contains $L^2(2\pi kdk)/(2\pi {)}^2$ states. With $ϵ={\hslash }^2{k}^2/(2m)$, $kdk=mdϵ/{\hslash }^2$, givingThere is no spin degeneracy. At temperature $T>0$, set $z=e^{\mu /(k_{B}T)}<1$. The Bose occupation integral givesThe integral follows by expanding the denominator as ${\sum }_{\ell \ge 1}{z}^{\ell }{e}^{-\ell ϵ/(k_{B}T)}$ and integrating each positive term.

As $\mu \uparrow 0$, this density diverges. Thus the excited states never have a finite maximum capacity at a positive temperature: every finite prescribed density can be accommodated with a strictly negative chemical potential. The ground-state occupation then stays finite as area tends to infinity and its fraction vanishes. Hence there is no Bose-Einstein condensation at any positive temperature in this homogeneous ideal two-dimensional gas. At exactly zero temperature the ground state is occupied; the printed phrase about any temperature is understood as absence of a nonzero critical temperature, not exclusion of that zero-temperature limit.

The nucleons are nonrelativistic, so the Maxwell-Boltzmann distribution givesBoth spin degeneracies are two. Taking the ratio and using the specified negligible chemical potentials yields the neutron-to-proton ratioThe mass prefactor differs from unity only at the per-mille level. The dependence on $T_F$ is much stronger: $d\ln r_F/d\ln T_F=Q/T_F>0$. Thus earlier, hotter weak freeze-out leaves more neutrons relative to protons. For illustration, $T_F=1\mathrm{MeV}$ gives $r_F\simeq 0.28$; $T_F=0.8\mathrm{MeV}$ gives $r_F\simeq 0.20$. Free-neutron decay after freeze-out lowers the ratio further before most Big Bang nucleosynthesis takes place.

A periodic lattice Hamiltonian commutes with lattice translations. The Bloch theorem therefore classifies its states by characters of the translation group:where $u$ is lattice periodic. For each crystal momentum $\mathbf{k}$, solving on a primitive cell with the corresponding quasiperiodic boundary condition gives discrete eigenvalues indexed by $n$. Their continuous dependence on $\mathbf{k}$ produces the energy bands $E_n(\mathbf{k})$ as $\mathbf{k}$ ranges through the Brillouin zone. Bragg mixing opens gaps at suitable degeneracies, although bands can overlap in three dimensions and gaps are not guaranteed between every pair.

The quantum numbers are band index, wavevector modulo reciprocal lattice vectors, and spin where spin is conserved. In a finite crystal the allowed wavevectors are discrete; in the thermodynamic limit they become dense and the band description emerges. Spin-orbit coupling can replace a separately conserved spin label by an appropriate internal band label.

Dilute randomly placed impurities break exact translation symmetry, so crystal momentum is no longer an exact quantum number. Most extended states retain approximately the host band structure but acquire impurity scattering and disorder broadening. An impurity potential can also produce impurity levels in a crystal, often in a band gap: levels associated with a donor impurity lie near the conduction-band edge and can supply electrons, while those associated with an acceptor impurity lie near the valence-band edge and can create holes. The impurity's species and potential determine whether such levels exist and how deep they are; not every impurity necessarily makes a gap state. At low density these levels are nearly isolated, while overlap at increasing density can broaden them into an impurity band. Added or removed carriers shift the chemical potential as well.

At zero temperature, fill all momentum states inside the Fermi sphere, with two spin states per momentum. Phase-space counting givesFor dispersion $E=cp$, the ratio of energy to particle-number integrals isLet $\epsilon$ be the common chemical potential in the weak field, and let $+$ denote the lower Zeeman-energy branch $cp-\mu B$. Its populations areKeeping $N_{+}+N_{-}=N$ changes $\epsilon$ only at order $B^2$, so to first order it equals ${\epsilon }_F=c{p}_F$. Subtracting the two populations givesEach excess lower-energy spin contributes moment $\mu$ along the field. Hence the Pauli moment of a massless electron gas isThis includes the spin contribution specified in the problem. The field is weak enough that both spin branches remain occupied, and orbital magnetic effects are not part of the assumed energy model.

Write $ϵ=1-{\rho }_s/\rho >0$ and $\Delta v=1/{\rho }_s-1/\rho =ϵ/{\rho }_s$. When the ice sphere grows by volume $4\pi a^2\dot{a}dt$, it consumes liquid volume $({\rho }_s/\rho )4\pi a^2\dot{a}dt$. The difference is expelled from the fixed cavity. Thus mass conservation fixes the outward Darcy flux in the surrounding porous medium:The Darcy law $u_r=-(\Pi /\mu ){\partial }_{r}p$ and $p(\infty )=p_m$ implyMatching at the cavity wall, whose internal pressure is uniform, givesThe solid and liquid pressures are equal when the Gibbs--Thomson relation is neglected; nevertheless the coexistence temperature changes with their common pressure because their specific volumes differ. Along coexistence, equality of chemical potentials givesHence the common-pressure melting-point slope is $d{T}_m/dp=-T_m\Delta v/L$. To first order in the pressure change,The $T_m$ multiplying this coefficient is absolute temperature, approximately $273.15\mathrm{K}$ for water; a temperature difference may equally be measured in kelvin or degrees Celsius. Substituting the numerical value $0$ from the Celsius scale into a thermodynamic denominator would be incorrect.

In the quasi-stationary thermal conduction approximation, neglect advective heat transport and use the stated effective conductivity $k$ outside the sphere. The radial Laplace equation gives the bounded solid temperature and the liquid/porous-region temperature:This treats the exterior as a uniform conducting medium, as required by the printed reduction; distinct cavity and porous-medium conductivities would give an additional thermal resistance. There is no solid-side heat flux in the bounded quasi-stationary solution. The Stefan condition therefore becomesEliminate $T_i$ and $p$ to obtainFor $x=a/R$, $t=t_0\tau$ and $t_0=R^2{\rho }_{s}L/(k\Delta T)$, this isThe pressure follows either from the Darcy flux or the Stefan condition:In particular $T_i-T_{\infty }=\Delta TK/(x+K)$: rising pressure progressively reduces the available undercooling. This is spherical freezing inside a porous cavity, whose fixed hydraulic outlet radius $R$ distinguishes it from freezing directly through a porous matrix.

For the idealized zero-radius start, integration givesA finite seed replaces the right side by its difference from the initial value. The reduction ends when the sphere fills the cavity.

For $K\gg 1$, hydraulic drainage is easy and the pressure feedback is small throughout the cavity. The leading expressions areThis is ordinary conduction-controlled spherical solidification, with a completion time asymptotic to $t_0/2$.

For $K\ll 1$, distinguish the initial crossover from the pressure-dominated regime. At $x\ll K$, equivalently $\tau \ll K^2$, the same square-root growth applies and $p-p_m\simeq p^{\ast }x/K$. At $K\ll x\le 1$, equivalently $K^2\ll \tau \lesssim 1/K$,The interface approaches $T_{\infty }$, the pressure approaches the finite coexistence value $p_m+p^{\ast }$, and growth becomes hydraulically limited. The leading completion time is $t_0/(3K)$. The cube-root approximation is not valid at the instant of nucleation, where the initial square-root regime must be retained.

For a nonrelativistic particle, $E(p)=m{c}^2+p^2/(2m)$. The Gaussian integral yieldsIndeed ${\int }_0^{\infty }{p}^2{e}^{-p^2/(2m{k}_{B}T)}dp=\sqrt{\pi }(2m{k}_{B}T{)}^{3/2}/4$, which combines with $4\pi /h^3$ to give the prefactor. For the two species this givesChemical equilibrium imposes ${\mu }_a+{\mu }_{\alpha }={\mu }_b+{\mu }_{\beta }$. If the two massless bath species have zero chemical potentials, then ${\mu }_a={\mu }_b$; their masses alone do not imply this chemical-potential condition.

Weak reactions such as $n+{\nu }_e\leftrightarrow p+e^{-}$, with leptons treated as relativistic, interconvert neutrons and protons. Negligible lepton chemical potentials give approximately $n_n/n_p=\exp [-(m_n-m_p)c^2/(k_{B}T)]$, since their degeneracies and masses are nearly equal. Equilibrium favors protons increasingly as temperature falls. But the weak reaction rate decreases faster than the expansion rate and eventually falls below it: cosmological weak freeze-out prevents the ratio from following its equilibrium exponential to zero. Free neutrons subsequently decay, reducing the frozen ratio further.

In Big Bang nucleosynthesis, cooling eventually makes deuterium survive photodissociation. Nuclear reactions then build helium through deuterium and the light intermediate nuclei, locking almost all surviving neutrons into stable helium-four. Neglecting binding-energy corrections and taking $r=n_n/n_p\le 1$, each helium nucleus uses two neutrons and two protons, so $n_{\mathrm{He}}=n_n/2$ andFor example $r=1/7$ gives a helium mass fraction $1/4$, with the remaining baryons predominantly hydrogen.

At a fixed subcritical temperature, coexistence requires equal pressure and equal chemical potential of the liquid and gas. The Maxwell equal-area construction chooses $p_0$ and the two outer intersection volumes $V_l,V_g$ so thatIndeed $d\mu =(V/N)dp$, and integration by parts gives $N({\mu }_g-{\mu }_l)=p_0(V_g-V_l)-{\int }_{V_l}^{V_g}p(V)dV$, so the equal-area condition is exactly equality of chemical potentials.

Let $V_{s,l}<V_{s,g}$ be the two spinodal volumes where $p^{\prime }(V)=0$. A homogeneous phase is mechanically unstable on $V_{s,l}<V<V_{s,g}$, since $p^{\prime }>0$ gives negative compressibility. The negative-slope branches $V_l<V<V_{s,l}$ and $V_{s,g}<V<V_g$ are metastable homogeneous liquid and gas respectively: they resist small perturbations but a two-phase state has lower free energy. The single-phase stable branches are $V<V_l$ and $V>V_g$. Throughout the coexistence interval $[V_l,V_g]$, the equilibrium state is a mixture at pressure $p_0$ with its phase fractions chosen to produce the total volume. The preceding figure labels both the formal homogeneous branches and the equilibrium coexistence segment.

The Gibbs free energy is $G=U-TS+pV$. For an extensive thermodynamic system, scaling the amount of matter at fixed pressure, temperature and composition scales $G$ by the same factor, hence $G(T,p,N)=N\mu (T,p)$. Euler's relation $U=TS-pV+\mu N$ gives the same conclusion. This uses the ordinary thermodynamic-limit extensivity assumption, rather than a universal claim about finite-size or long-range-interacting systems.

The fundamental thermodynamic identity gives $dG=-SdT+Vdp+\mu dN$. Comparing with $d(N\mu )$ at fixed $N$ yieldsThus the chemical potential changes with pressure at a rate equal to the volume per particle.