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For a nonideal chemical solution, write its thermodynamic activities on a mole fraction basis as $a_A=x{\gamma }_A$ and $a_B=(1-x){\gamma }_B$. Use composition-independent pure-component reference chemical potentials at the same temperature and pressure:The activity coefficients describe departure from the ideal solution under that reference convention.^[ref]

At fixed temperature and pressure, insert these expressions into the Gibbs-Duhem equation. The ideal solution terms cancel because $xd\ln x+(1-x)d\ln (1-x)=dx-dx=0$. What remains isThus two independent fits to measured activity coefficients can each look smooth while failing to describe one possible Gibbs free energy. This is a necessary consistency condition; satisfying it alone does not establish experimental accuracy or thermodynamic stability.

For an original algebraic example, suppose at one fixed temperature and pressure that $\ln {\gamma }_A=\alpha (1-x{)}^2$, with dimensionless constant $\alpha$. The constraint requires $d\ln {\gamma }_B/dx=2\alpha x$. With the pure-B normalization ${\gamma }_B(0)=1$, integration givesFor $\alpha =1/2$ and $x=1/4$, the derivatives of the two logarithms are $-3/4$ and $1/4$; their weighted sum is $-3/16+3/16=0$. Choosing $\ln {\gamma }_B=\alpha x$ instead would give a weighted sum $3/16$ at that composition, contradicting the Gibbs-Duhem equation. These are invented model functions, with no claim to fit a particular mixture.

The fixed-temperature, fixed-pressure restriction matters. Along a path with changing temperature or pressure, the general equation contains $SdT-Vdp$, and the reference chemical potentials and $RT$ also change. Across phase coexistence, apply the relation separately within each smooth thermodynamic phase; do not differentiate a discontinuous switch between phases as if it were a single composition curve.

For a single-component equilibrium system with extensive internal energy $E(S,V,N)$, scaling its entropy, volume and particle number together gives $E(\lambda S,\lambda V,\lambda N)=\lambda E(S,V,N)$. The Euler theorem for homogeneous functions therefore gives $E=S{\partial }_{S}E+V{\partial }_{V}E+N{\partial }_{N}E$. The first law of thermodynamics identifies these derivatives as $T,-P,\mu$, respectively, proving $E=TS-PV+\mu N$. Dividing by volume yields the entropy density formula $s=(\rho +P-\mu n)/T$. This step fixes the entropy normalization rather than merely identifying its differential. Differentiating the Euler relation and subtracting $dE=TdS-PdV+\mu dN$ gives the Gibbs-Duhem equation $SdT-VdP+Nd\mu =0$. The argument requires extensivity; it is not an assumption about arbitrary nonadditive systems.

Let $f$ be a differentiable function on an open domain stable under positive rescaling, with $f(\lambda x)={\lambda }^{k}f(x)$ for $\lambda >0$. The chain rule differentiates this homogeneous function identity at $\lambda =1$ to giveIf $f$ is twice differentiable, differentiating once more gives ${\sum }_i{x}_i{\partial }_i{\partial }_{j}f=(k-1){\partial }_{j}f$. In degree one its Hessian matrix therefore annihilates the radial vector $x$. This is the identity used in the no-scale identity from degree-one homogeneity. For an extensive quantity, the degree-one formula also supplies the thermodynamic identity behind the Gibbs-Duhem equation.

The Euler theorem for homogeneous functions applied to an extensive quantity, together with $dE=TdS-pdV+\mu dN$ give

The Gibbs-Duhem equation constrains changes of the intensive quantities describing one uniform thermodynamic phase. For a mixture with $k$ independently specified component amounts, its molar form isHere $S$ is entropy, $V$ is volume, $T$ is temperature, $p$ is pressure, $n_i$ is the amount of substance of component $i$, and ${\mu }_i$ is its molar chemical potential. This assumes a simple compressible system whose bulk internal energy is an extensive quantity; additional work coordinates require additional terms.^[ref]

The proof exposes the source of the constraint. For internal energy $E(S,V,n_1,\dots ,n_k)$, replication of the whole thermodynamic phase by a factor $\lambda$ gives $E(\lambda S,\lambda V,\lambda n_1,\dots ,\lambda n_k)=\lambda E(S,V,n_1,\dots ,n_k)$. The Euler theorem for homogeneous functions and the fundamental differential of internal energy therefore giveTaking the differential of the first equation produces all the terms in the second, plus $SdT-Vdp+{\sum }_i{n}_{i}d{\mu }_i$. Subtraction proves the Gibbs-Duhem equation. Thus the constraint comes from two descriptions of the same state function, rather than an assumed microscopic interaction model.

At fixed temperature and pressure, divide by $n={\sum }_i{n}_i>0$ and introduce the mole fractions $x_i=n_i/n$:The chemical potentials cannot be varied independently. In a two-component mixture, put $x_A=x$ and $x_B=1-x$. Within a smooth single-thermodynamic phase region with $0<x<1$,One chemical potential fixes the other's change, but determining its absolute value still requires a reference value. This does not say ${\mu }_A={\mu }_B$: equality at phase coexistence concerns the same chemical species in different thermodynamic phases. Nor does the constraint say $dG=0$ whenever temperature and pressure are fixed; changing composition still contributes ${\sum }_i{\mu }_{i}d{n}_i$ to the Gibbs free energy.

For a homogeneous extensive single-component thermodynamic phase, the Gibbs-Duhem equation at fixed temperature gives $dp=nd\mu$. With number density $n=N/V$, fixed $N$ gives $dn=-(n/V)dV$. The chain rule therefore givesThe derivatives are evaluated within the same smooth equation of state; an interface contribution or phase coexistence can invalidate this simple homogeneous derivation.

Let ${\psi }_0$ be the normalized stationary profile. Multiplying its stationary Gross–Pitaevskii equation by ${\psi }_0^{\ast }$, integrating and integrating the kinetic term by parts yieldsThis is not simply $H/N$: the quartic contribution is twice its contribution to that energy per particle. For an infinite periodic system one uses one cell and the corresponding cell particle number, or equivalently average densities.

At fixed lattice spacing, mean density $n$, coupling and temperature, the commonly used condensate compressibility in an optical lattice isIf compressibility means the thermodynamic volume response instead, the Gibbs-Duhem equation $dp=nd\mu$ at zero temperature gives ${\kappa }_T=[n^2(\partial \mu /\partial n{)}_s{]}^{-1}=\kappa /n$. Stating the convention removes this factor-of-density ambiguity; the lattice-depth derivative has the same sign at fixed $n$.

For the stable repulsive ground-band Bose-Einstein condensate, the intended trend is $\partial \kappa /\partial s<0$ for increasing nonzero lattice depth, with mean density held fixed. Deeper wells localize the field, enhancing the repulsive interaction response. To quantify this, in the weak-interaction band approximation write ${\psi }_0=\sqrt{n}{f}_s$, with the cell average of $|f_s{|}^2$ equal to one. Perturbation in $Un$ givesIn deep wells the localized orbital is approximately a normalized Gaussian of width $a_{\mathrm{ho}}=d/(\pi s^{1/4})$. Neglecting interwell overlap, $I\simeq d/(\sqrt{2\pi }{a}_{\mathrm{ho}})=\sqrt{\pi /2}{s}^{1/4}$, hence $\partial \kappa /\partial s\simeq -\kappa /(4s)<0$. This explicitly establishes the sign in that controlled regime; the repulsive ground-band calculation describes the broader trend.

The sign statement needs the repulsive, stable ground-band interpretation, not merely the literal assumption $U\ne 0$. It also need not be a strictly nonzero derivative at $s=0$: at weak interaction $f_s=1+(s/8)\cos (2\pi x/d)+O(s^2)$, with normalization corrected at second order, gives $I=1+s^2/32+O(s^3)$ and a zero initial slope. A leading density-dominated Thomas–Fermi approximation for a condensate with density positive everywhere gives $\mu =Un+s{E}_R/2$ and no depth dependence of this response at that approximation's order. These limits do not justify a strict universal sign for every unspecified interaction and parameter regime.

Let $V$ be the physical volume of a region with fixed comoving coordinates, so $V\propto a^3$, and write $U=\rho V$. The first law of thermodynamics is $dU=TdS-PdV+\mu dN$. Use extensive quantities: under replication, $U(\lambda S,\lambda V,\lambda N)=\lambda U(S,V,N)$. The Euler theorem for homogeneous functions givesSetting $\mu =0$ yieldsThis is the entropy form of the Gibbs-Duhem equation; it fixes the normalization by extensivity, rather than adding an arbitrary volume-independent entropy constant.

For equilibrium expansion of the same region, the first law and the cosmological perfect-fluid continuity equation implyTherefore $\boxed{\dot{S}=0}$, establishing cosmological entropy conservation for adiabatic equilibrium evolution. Here “comoving volume” means a patch moving with the cosmological fluid; its physical volume $V$ changes. The corresponding coordinate volume is constant. The local TeX's $\dot{V}/P$ in the continuity-equation hint is a transcription error: the PDF correctly has $\dot{V}/V$.

Replicating a macroscopic equilibrium system scales $S,V,N$, and $E$ by the same factor, so $E(S,V,N)$ is homogeneous of degree one. Euler's theorem and the first lawgiveDifferentiating this identity and comparing with the first law cancels the $TdS$, $-pdV$, and $\mu dN$ terms, leaving the Gibbs-Duhem equationTherefore

Differentiate the Euler relation for a single-component simple compressible system:The thermodynamic fundamental relation for the same system is $dU=TdS-pdV+\mu dN$. Subtraction givesDividing by $N$, with $s=S/N$ and $v=V/N$, proves the Gibbs-Duhem equationThe underlying Euler relation assumes extensive bulk thermodynamics and no additional work coordinates. The result says that the intensive quantities are constrained; the chemical potential cannot be varied independently of temperature and pressure within one homogeneous phase.

Use the Gibbs-Duhem equation for a homogeneous, extensive, single-component thermodynamic phase:At fixed temperature, this becomes $dp=nd\mu$, with number density $n=N/V$. Separately, at fixed particle number,The chain rule along the same isothermal process in the equation of state now givesConsequentlyThis is isothermal compressibility from chemical-potential response. Since chemical potential has units of energy per particle, the right-hand side has units of volume per energy, or $\mathrm{P}{\mathrm{a}}^{-1}$, as required. The derivation assumes a smooth bulk equation of state; changing the fractions of coexisting thermodynamic phases at fixed pressure needs separate treatment.