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Pressure and its physical terms. Solving the molar Van der Waals equation for pressure givesThe excluded-volume parameter $b$ reduces the volume available for translational motion, raising the kinetic pressure relative to $RT/V$. The attraction parameter $a$ lowers the pressure because attractive interactions reduce the momentum transferred to the boundary. This mean-field model treats $a,b$ as temperature-independent material constants. For one mole, $V$ is the sample volume with the molar amount already fixed; $b$ has units of that volume and $a$ of pressure times volume squared.

volume dependence of entropy. The Helmholtz free energy has differential $dF=-SdT-PdV$. Equality of mixed derivatives yields the Maxwell relationFor $a=0$ this derivative is $R/(V-b)$, givingEquivalently, $S=R\ln (V-b)+f(T)$ with a common reference volume understood inside the logarithm. The equation of state fixes this volume dependence, not the otherwise unspecified function of temperature.

Excess entropy at fixed volume and temperature. The ideal gas has $(\partial S_{\mathrm{id}}/\partial V{)}_T=R/V$. ThereforeIntegrate from infinite volume, where the entropy difference is specified to vanish:It is negative because excluding configurations reduces the accessible state count. It tends to zero as $V\to \infty$ and has leading dilute limit $-Rb/V$. This is the entropy excess of a van der Waals gas.

Effect of attraction. Restoring a temperature-independent $a$ changes the pressure by $-a/V^2$, but not its derivative with respect to temperature. With the same infinite-volume entropy reference, it therefore does not change $\Delta S^{\mathrm{ex}}$ at fixed $V,T$. This can also be seen fromThe attraction affects internal energy and the equilibrium volume, so it can affect entropy indirectly when different state points are compared. If $a$ were temperature dependent, its derivative would contribute directly to the entropy; that is outside the stipulated constant-parameter model.

Coexistence isotherm. Below the Van der Waals critical point, the formal homogeneous isotherm has a local minimum and maximum. Its middle branch has $\partial P/\partial V>0$, hence negative isothermal compressibility, and is mechanically unstable. Portions of the negative-slope branches can be metastable rather than globally stable. The equilibrium isotherm replaces the loop between liquid volume $V_l$ and vapour volume $V_g$ by a constant saturation pressure $P_s$. The two volume are fixed byThe Maxwell equal-area construction expresses equality of the two phases' chemical potentials. Intermediate total volume are mixtures: changing the vapour fraction changes total volume at the same pressure, rather than following the unstable homogeneous curve. The spinodals are the turning points of the formal isotherm; they are distinct from the coexistence endpoints.

Dense-phase volume at low temperature. At a general subcritical temperature, the two coexistence equations above must be solved together. For this equation of state the area condition is explicitlyThis supplies an exact implicit determination of the liquid volume even when a simple closed form is unavailable.

For the asymptotically low-temperature limit, put $\tau =RTb/a\ll 1$. The liquid is dense, $V_l\simeq b$, while the vapour is dilute, $P_s{V}_g\simeq RT$ and $V_g\gg V_l$. The area condition then givesThus the vapour volume grows exponentially in $1/\tau$, and the saturation pressure is exponentially small. It is consequently consistent to neglect $P_s$ in the liquid equation to algebraic order in $\tau$:Choose the root near $b$, not the large-volume zero-pressure root:Expanding gives the leading requested dense-phase behavior,This is the low-temperature liquid volume of a van der Waals gas. Setting $P_s=0$ is an asymptotic approximation, not an exact coexistence condition at every temperature below $T_c$. The figure illustrates generic subcritical coexistence; its finite-temperature liquid volume is determined by the full equal-area equations. The van der Waals liquid also ceases to be a realistic description at sufficiently low physical temperature, where other phases or quantum effects can matter.

Configurational entropy counts the arrangements compatible with a macroscopic configuration: $S=k_B\ln \Omega$. Let $x=L/(Na)$, $N_{\downarrow }=N(1+x)/2$ and $N_{\uparrow }=N(1-x)/2$. The multiplicity is $N!/(N_{\downarrow }!N_{\uparrow }!)$. Stirling's approximation therefore givesWith downward $L$ positive, the gravitational potential energy is $-mgL$. Minimizing the Helmholtz free energy $F=-mgL-TS$ gives $\mathrm{artanh}x=mga/(k_{B}T)$, henceRaising temperature reduces the extension: the entropic preference for many up/down arrangements opposes alignment by the load. For small loads, $L\simeq N{a}^{2}mg/(k_{B}T)$.

For the positive-feedback entropic chain with replacement force $f(L)=f_{0}L/a$, $f_0>0$, the conservative potential is $U(L)=-{\int }_0^{L}f(s)ds=-f_0{L}^2/(2a)$. Using $-f(L)L$ would overcount it by a factor two. The stationary equation becomesExpansion about $x=0$ givesFor $T>T_c$ the stable extension is zero. Below $T_c$, nonzero minima appear, with $x\simeq \pm \sqrt{3(T_c-T)/T}$ close to the transition and $|x|\to 1$ as $T\to 0$. Restricting to downward extensions selects the positive branch. The onset is continuous in this model, with a nonanalytic mean-field change at $T_c$; a finite jump in $L$ does not follow from the printed force law, despite the question's discontinuity wording. A finite chain has rounded fluctuations rather than a literal thermodynamic singularity.

This is the same self-consistency equation as the zero-field mean-field treatment of the Ising model: the segment orientation corresponds to a spin, extension to magnetization, and the extension-dependent force to the self-consistent molecular field. The competition between alignment energy and binary configurational entropy explains the shared temperature dependence.

Write $b=\pi k_B^2/(3\hslash c)$, so $U=bL{T}^2$. At fixed length, $dS=C_{L}dT/T$, giving $S=2bLT$ with the entropy constant fixed by the unique photon vacuum. The thermal Helmholtz free energy and generalized pressure follow:The entropy constant can also be checked from the exact discrete photon partition function; the continuum approximation should not be extrapolated blindly below its level-spacing regime. In one dimension $P=U/L$, whereas the isotropic three-dimensional relation is $p=U/(3V)$. Both follow from the number of spatial directions carrying momentum flux.

Entropy is a thermodynamic state function. For a reversible heat transfer, $dS=\delta Q_{\mathrm{rev}}/T$; the Clausius inequality generalizes this to irreversible processes. The combined entropy of a system and its surroundings does not decrease. A subsystem can lose entropy by exporting heat or matter, so the second-law statement must specify the whole isolated combination. In a simple compressible system,which identifies temperature, pressure and chemical potential as derivatives of the energy representation of the state function. Thermodynamic equilibrium maximizes entropy under fixed total energy, volume and particle number. A heat reservoir instead leads to minimizing the Helmholtz free energy at fixed $T,V,N$, or the Gibbs free energy at fixed $T,P,N$.

Statistical mechanics explains the state count. In a microcanonical ensemble, equally probable compatible microstates give Boltzmann's entropy formula, $S=k_B\ln \Omega$. With general discrete probabilities, the Gibbs entropy isThe microcanonical formula follows by setting every nonzero probability to $1/\Omega$. For a canonical ensemble, $p_j=e^{-\beta E_j}/Z$, soThis turns microscopic energy levels and their populations into a measurable thermodynamic quantity. Classical phase-space counting needs a consistent microscopic normalization and indistinguishability convention; quantum state counting supplies that normalization. Independent systems have multiplicative state counts and therefore additive entropies.

For an ideal gas freely expanding from volume $V$ to $2V$ in an insulated enclosure, energy remains fixed while each particle gains twice the accessible position volume. The state count increases by $2^N$, giving $\boxed{\Delta S=N{k}_B\ln 2}$. The actual irreversible process transfers no heat. A reversible isothermal comparison path transfers heat $N{k}_{B}T\ln 2$ and computes the same entropy change because the initial and final thermodynamic equilibrium states are the same.

Calorimetry supplies another connection to experiment: at fixed pressure and within one phase, $S(T)-S(0)={\int }_0^T{C}_P(T^{\prime })d{T}^{\prime }/T^{\prime }$, with latent-heat contributions $\mathcal{L}/T$ at first-order transitions. The third law of thermodynamics fixes $S(0)=0$ for an ideal perfect crystal with a unique ground state. Degeneracy or frozen disorder can produce residual entropy. Entropy thus links heat-capacity measurements, phase thermodynamic equilibrium and the number of accessible microscopic configurations.

For equal-size molecules, let $f$ be the mixing Helmholtz free energy per lattice site. Flory–Huggins solution theory reduces to the regular solution model:The first two terms are minus the combinatorial entropy of mixing; the last is the excess cost of unlike contacts. A positive Flory interaction parameter penalizes such contacts. Additive terms linear in composition change the chosen pure-component reference but do not change stability.

A small composition fluctuation at fixed mean composition costs energy proportional to $f^{\prime \prime }$. Thus local stability requires $f^{\prime \prime }>0$, and the spinodal isInside this curve, $f^{\prime \prime }<0$ and arbitrarily small composition fluctuations grow. The minimum of ${\chi }_s$ is at $\varphi =1/2$, giving the critical value ${\chi }_c=2$. At the thermodynamic critical point, $f^{(2)}=f^{(3)}=0$ while $f^{(4)}>0$.

For $\chi =0$, $f$ is strictly convex with its minimum at $1/2$. At $\chi =2$, the central quadratic term vanishes but a positive quartic term remains. At $\chi =2.5$, there are two minima and a central maximum; the spinodal compositions are $(1\pm \sqrt{1-2/\chi })/2\simeq 0.2764,0.7236$. The common-tangent construction gives coexisting compositions ${\varphi }_b\simeq 0.1448$ and $1-{\varphi }_b$, satisfying $\ln [{\varphi }_b/(1-{\varphi }_b)]+\chi (1-2{\varphi }_b)=0$. Between a binodal endpoint and the spinodal, the homogeneous state is locally stable but metastable; global stability requires the convex envelope of $f$.

The Flory interaction parameter measures the energetic preference for unlike rather than like contacts in Flory–Huggins solution theory. For polymer volume fraction $\varphi$ and $N$ lattice sites per chain, the dimensionless mixing Helmholtz free energy per site isThe logarithmic terms represent combinatorial entropy; polymer connectivity reduces the chain's translational mixing entropy by the factor $N$. In a simple contact model $\chi =z[{ϵ}_{ps}-({ϵ}_{pp}+{ϵ}_{ss})/2]/(k_{B}T)$, where $z$ is the coordination number. Positive $\chi$ penalizes polymer-solvent contacts. Increasing it can overcome mixing entropy and cause phase separation; a negative second derivative of the mixing free energy marks the spinodal region. For a long dilute chain, $\chi <1/2$ corresponds to a good solvent, $\chi =1/2$ to the idealized theta condition, and $\chi >1/2$ to a poor solvent. Real effective parameters can depend on temperature and concentration, rather than being universal chemical constants.

For a freely jointed Gaussian chain, independent isotropic segment vectors have zero mean and coordinate variance ${\ell }^2/3$. The central limit theorem gives the end-to-end vector densityThe number of conformations in a fixed small vector-volume element is proportional to this density. Taking its logarithm gives the configurational entropyThis is the vector-constrained entropy appropriate to an anchored strand. A density for the scalar length alone also has a $4\pi R^2$ shell factor and is a different constraint. The approximation requires many segments and extensions well below $N\ell$.

If the internal energy has negligible extension dependence, the Helmholtz free energy is $F=F_0+3k_{B}T{R}^2/(2N{\ell }^2)$. Differentiation gives an entropic force with effective spring constant $3k_{B}T/(N{\ell }^2)$. In an affine isotropic network, averaging over initially oriented strands gives rubber elasticityfor an incompressible material; ${\nu }_s$ is the number of elastically active strands per volume. For example a volume-preserving stretch with principal factors $\lambda ,{\lambda }^{-1/2},{\lambda }^{-1/2}$ has strain-energy density $W=G({\lambda }^2+2/\lambda -3)/2$. Its nominal tensile stress is $G(\lambda -{\lambda }^{-2})=3G(\lambda -1)+O((\lambda -1{)}^2)$, proving the stated small-strain Young's modulus.

Chemically, increasing the effective cross-link density, for example by changing vulcanization, generally increases the modulus by shortening strands between cross-links. Dangling ends, loops and changes of network structure mean the chemical junction count need not equal ${\nu }_s$.

The reference area is $A_0=10^{-5}{\mathrm{m}}^2$. Neglect self-weight, thermal expansion and energetic elasticity, keep the same permanent network, and heat quasistatically in the rubbery regime. Then $G$ increases from $20$ to $22\mathrm{MPa}$. With a constant suspended load, the exact affine-network balance isIt gives ${\lambda }_{300}=1.01662$ and ${\lambda }_{330}=1.01509$. ThusLinear elasticity instead gives extensions $16.35$ and $14.86\mathrm{mm}$, a $1.49\mathrm{mm}$ rise, consistent at this small strain. Incompressibility makes the area $A_0/\lambda$, already included by using nominal stress. Heating increases entropic stiffness, so the loaded rubber contracts and performs work on the mass. Strong thermal expansion, changes in cross-linking or leaving the rubbery regime would require additional material data.

Expand the Grassmann fields in fermionic Matsubara frequencies ${\omega }_l=(2l+1)\pi /\beta$. For $U=0$, the Grassmann Gaussian integral factorizes into thermal fermionic one-level thermal determinants:Each factor counts the empty and occupied states of one spin orbital. Equivalently, the regularized Matsubara sum gives the logarithm of that factor; the raw infinite determinant requires thermal normalization of a fermion determinant. ThereforeThe printed $F$ is the grand potential $\Omega$, because the trace contains $H-\mu \hat{N}$. The Helmholtz free energy at fixed mean particle number is $F_{\mathrm{H}}=\Omega +\mu ⟨\hat{N}⟩$.

Entropic elasticity. A rubber contains flexible polymer strands tied into a network by cross-links. Stretching biases the strands toward relatively extended configurations, reducing their configurational entropy. The restoring entropic force follows from the increase in Helmholtz free energy $F=U-TS$. For a Gaussian strand with $N_s$ statistical steps of length $b$, its deformation-dependent free energy is approximately $3k_{B}T{R}^2/(2N_s{b}^2)$. Its stiffness therefore increases with absolute temperature, even if the underlying bond stiffness changes little.

In the ideal affine-network model of rubber elasticity, the shear modulus iswhere ${\nu }_s$ is the number density of elastically active network strands. Thus heating increases the modulus approximately in proportion to $T$, provided the network density and rubbery state remain approximately unchanged. Thermal expansion, changing relaxation times or damage can modify that simple trend.

Examples are lightly crosslinked natural rubber, based on cis-rich polyisoprene, and crosslinked polydimethylsiloxane. The crosslinks prevent the long-time macroscopic flow of an unconnected chain solution.

Density estimate and the unspecified modulus convention. With room temperature $T\simeq 298\mathrm{K}$, the Boltzmann constant gives $k_{B}T\simeq 4.11\times 10^{-21}\mathrm{J}$. If the stated modulus is $G=2\times 10^5\mathrm{Pa}$, thenThe molar value counts moles of active strands. If “modulus” instead means Young's modulus $E$, a nearly incompressible isotropic rubber has $E\simeq 3G$, and the inferred value isThus the robust order-of-magnitude estimate is $10^{25}{\mathrm{m}}^{-3}$. The cross-link density inferred from rubber elasticity is usually an effective strand density; the actual junction density is ${\nu }_j=2{\nu }_s/f$ for junction functionality $f$ in an ideal network without inactive loops or dangling ends. For four-functional junctions this would be half the strand density. The paper supplies neither modulus type nor junction functionality, so a unique chemical junction count cannot be extracted.

Here $T$ is the absolute temperature and $k_B$ the Boltzmann constant. An adsorption blob contains $g\sim (\xi /b{)}^2$ monomers, so there are $N/g\sim N(b/\xi {)}^2$ such blobs. Confining a blob removes an amount of conformational entropy of order $k_B$; thus the positive confinement contribution to the Helmholtz free energy isThe statement about one blob's entropy is used as an entropy loss relative to the free chain, rather than as an absolute entropy.

In the assumed slab model, contacts occur within a monomer-scale distance $b$ of the substrate. The contact fraction is approximately $b/\xi$, giving $N_c\simeq Nb/\xi$. Multiplying this by the negative contact energy yieldsThe two contributions express the competition between loss of conformational entropy and gain of adsorption energy. The numerical coefficients are the ones stipulated by this blob approximation.

Minimize the Helmholtz free energy at fixed $N,T$. Its derivative isThe physical stationary point for $ϵ>0$ isTo check the minimum directly, put $x=b/\xi$. Then $F/(N{k}_{B}T)=(x-ϵ/2{)}^2-{ϵ}^2/4$, which is globally minimized at $x=ϵ/2$. The binding free energy gain per monomer is consequently $\mu =k_{B}T{ϵ}^2/4>0$.

The weak-adsorption blob picture requires $b\ll {\xi }_{\ast }\ll b\sqrt{N}$, or $2/\sqrt{N}\ll ϵ\ll 2$ in this model. If $ϵ$ is too large, the layer saturates at a monomer-scale thickness; if too small, a many-blob adsorbed layer has not formed.

The stretched segment is a Gaussian chain of $n$ Kuhn length steps, with unperturbed root-mean-square end-to-end distance $l=b\sqrt{n}$. Its end-to-end probability density is proportional to $\exp [-3z^2/(2n{b}^2)]$, so its leading elastic Helmholtz free energy is $3k_{B}T{z}^2/(2n{b}^2)$. Adding the remaining segment's adsorption free energy givesAn additive constant independent of $n,z$ is irrelevant. At the optimal adsorption height this becomesThe original PDF's hint labels the positive quantity proportional to $k_{B}T{R}^2/l^2$ an entropy. It has units of energy and is the stretching free energy. The corresponding change of conformational entropy is $\Delta S=-3k_B{R}^2/(2l^2)$. This dimensional and sign correction is essential to the minimization.

This is the leading extensive Gaussian chain approximation: endpoint normalization and junction corrections give subleading terms. It requires many stretched monomers and $z\ll nb$, rather than extension near the full contour length.

During slow peeling in atomic force microscopy, $n$ adjusts to minimize the Helmholtz free energy at each imposed height. Define the positive binding gain $\mu =k_{B}T[ϵb/\xi -(b/\xi {)}^2]$ for the adsorbed segment. The stationary point satisfiesSince ${\partial }_n^{2}F=3k_{B}T{z}^2/(b^2{n}^3)>0$, this is a minimum while $0<n_{\ast }<N$. The upward pulling force is the height derivative of the minimized free energy. The implicit $n_{\ast }(z)$ term vanishes because ${\partial }_{n}F=0$, givingUsing $\xi =2b/ϵ$,It is independent of $z$: extra pulling height releases more monomers while leaving the fractional extension of the stretched part unchanged. Equivalently $F_{\min }=-N\mu +f_{\mathrm{peel}}z$. This is the equilibrium peeling force of an adsorbed Gaussian chain.

The plateau ends when $n_{\ast }=N$, atThereafter all monomers are detached and the Gaussian chain force is $f=3k_{B}Tz/(N{b}^2)$, agreeing continuously with the plateau at $z_{\det }$. Very near the start and final release, finite-chain and junction effects lie outside the blob approximation.

The square lattice has coordination number of a lattice $z=4$. The mean-field decoupling isAccounting for the factor $1/2$ in the directed-neighbour sum gives the Hamiltonian in the mean-field approximation:The constant term is necessary to correct the double counting of the interaction. Let $h=4JS+B$. The one-site partition function isso $Z_{\mathrm{MF}}=e^{-2\beta JN{S}^2}{Z}_1^N$. The Helmholtz free energy is thereforeThe equilibrium mean-field approximation is obtained by minimizing this expression over the allowed $S$. The $s=0$ state supplies the extra $1$ in $Z_1$; omitting it would instead describe the two-state Ising model. This is the Blume-Capel mean-field transition at zero single-ion anisotropy.

The self-consistency equation requires that the mean spin computed using the single-site Boltzmann distribution equals the spin $S$ used to form the effective field. With $h=4JS+B$,ThusEquivalently, differentiating part (b) gives $N^{-1}\partial F_{\mathrm{MF}}/\partial S=4J[S-⟨s{⟩}_h]$, so stationarity of the Helmholtz free energy in the mean-field approximation is exactly this condition. Among multiple stationary points one must select the stable minimum.

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.

Component chemical potentials. Define the per-spin quantum concentration $n_Q={\lambda }_T^{-3}=(m{k}_{B}T/(2\pi {\hslash }^2){)}^{3/2}$. For a classical component with one-particle potential energy $U_{\sigma }(\mathbf{r})$, the local Helmholtz free energy density is $k_{B}T{n}_{\sigma }[\ln (n_{\sigma }/n_Q)-1]+n_{\sigma }{U}_{\sigma }$. Its derivative with respect to number density gives the full local chemical potential, including the magnetic energy. With the up state defined as the lower level,The Bohr magneton ${\mu }_B$ is a magnetic moment; ${\mu }_{B}B$ is an energy. The factor two for the two spin states is not already included in $n_Q$.

Total equilibrium chemical potential. If spin conversion is allowed, a particle can be added in either spin state at the same equilibrium free-energy cost. Thusnot their sum or zero. The total particle number, or a reservoir, fixes this common value; the supplied information does not specify a numerical chemical potential. A useful local expression is $\mu =k_{B}T\ln [n/(2n_Q\cosh x)]$, with $x={\mu }_{B}B/(k_{B}T)$, whose density dependence is found below.

Why it is uniform and common. Moving a small number $dN$ of one species from ${\mathbf{r}}_1$ to ${\mathbf{r}}_2$ changes the free energy by $[{\mu }_{\sigma }({\mathbf{r}}_2)-{\mu }_{\sigma }({\mathbf{r}}_1)]dN$. If this difference were nonzero, a transfer could lower the free energy. Equilibrium under diffusion therefore requires each full chemical potential to be spatially constant. A spin flip transferring $dN$ from down to up similarly changes free energy by $({\mu }_{\uparrow }-{\mu }_{\downarrow })dN$, proving equality when spin relaxation is permitted. This is chemical equilibrium of a classical spin gas. Mere thermal contact at fixed separate spin populations would not enforce the latter equality; full spin equilibrium is the necessary assumption. A spatially varying magnetic field is balanced by a spatially varying density, not by a varying full chemical potential.

Magnetization. Invert the component equations at the common $\mu$:The two levels have moments $+{\mu }_B$ and $-{\mu }_B$ along the field. The magnetization, magnetic moment per unit volume, is thereforeIts SI units are $\mathrm{A}{\mathrm{m}}^{-1}$. This expression neglects interactions and magnetic feedback of the gas on the specified external field.

High-temperature response. For $|x|\ll 1$, $\sinh x\simeq x$ and $\cosh x\simeq 1$, yieldingAt fixed number density, this is the Curie law: the susceptibility with respect to $B$ is $n{\mu }_B^2/(k_{B}T)$. The conventional dimensionless susceptibility with respect to $H$ is approximately ${\chi }_v={\mu }_{0}n{\mu }_B^2/(k_{B}T)$ when $B\simeq {\mu }_{0}H$. At fixed $\mu$, however, $n_Q{e}^{\beta \mu }$ changes with temperature, so $M$ itself need not scale simply as $1/T$. The moment per particle has the Curie behaviour; the density constraint must be stated.

Total density. Adding the two populations givesFor a finite container and fixed $N$, normalization gives $e^{\beta \mu }=N/[2n_Q\int \cosh x(\mathbf{r})d^{3}r]$. In a weak field the leading density is uniform, with changes only of order $x^2$. Particles preferentially occupy regions where the magnitude of the field lowers one of the spin energies, even though the common chemical potential remains uniform.

One particle. Use the classical phase-space normalization of one quantum state per volume $h^3$, with no unspecified internal degeneracy. For $a>0$ and $\beta =1/(k_{B}T)$, the canonical partition function factorizes into momentum and position integrals:The Gaussian momentum integral is $(2\pi m{k}_{B}T{)}^{3/2}$. In spherical coordinates, put $t=\beta ar$:Therefore the classical ideal gas in a spherical linear trap hasHere $a$ is a force, in $\mathrm{J}{\mathrm{m}}^{-1}$, so $k_{B}T/a$ is a length and $Z_1$ is dimensionless. A nonconfining $a\le 0$ would not give this finite spatial integral. As required by the classical phase-space approximation, the temperature must exceed the trap's discrete level scale.

Indistinguishable dilute particles. Independent labeled particles would give $Z_1^N$. In the classical dilute regime, division by $N!$ removes the overcounting of permutations of identical particles:This is the Maxwell-Boltzmann limit, not an exact quantum canonical partition function at high occupation. Exchange effects cannot be captured merely by continuing this formula beyond its dilute regime.

Free energy, energy and entropy. The Helmholtz free energy isWith leading Stirling's approximation, $\ln N!\simeq N\ln N-N$, this becomesAt fixed trap strength, ${\partial }_T\ln Z_1=9/(2T)$. Differentiating $F$ gives the requested entropy,The $11/2$ is $1+9/2$: the first term comes from indistinguishability through Stirling's formula, and the second from the temperature dependence of $Z_1$. Without the large-$N$ approximation, $S=k_B[N\ln Z_1-\ln N!+(9/2)N]$. Independently,The kinetic contribution is $3N{k}_{B}T/2$ and the linear-trap potential contribution is $3N{k}_{B}T$, also obtained from generalized equipartition.

Entropy curves and reversible cooling. For a fixed $N$ and a reference pair $(T_{\ast },a_{\ast })$, the dependence can be written without logarithms of dimensional quantities:At each fixed $a$, the entropy increases logarithmically and is concave downward in $T$; decreasing $a$ shifts the curve upward. Its formal classical divergence as $T\to 0$ marks failure of classical statistics, not a physical negative-infinite entropy.

For a reversible adiabatic change, $S$ is constant at fixed $N$, soHence reducing $a$ lowers the temperature: this is isentropic decompression of a spherical linear trap. The gas does work on the control that weakens the trap. Equivalently $dU=⟨\partial H/\partial a⟩da=N⟨r⟩da$, with $⟨r⟩=3k_{B}T/a$; inserting $U=9N{k}_{B}T/2$ gives the same relation. Reversibility requires an equilibrium sequence without heat input or entropy production. Because $Z_1$ stays constant on this adiabat, $N/Z_1$ also stays constant: this mechanism cools the cloud but does not by itself increase quantum degeneracy.

Chemical potential and quantum threshold. From the extensive free energy,Thus the quantum gas fugacity is $z=e^{\mu /(k_{B}T)}\simeq N/Z_1$. Classical statistics requires $z\ll 1$; quantum effects become important as the chemical potential comes within order $k_{B}T$ of the lowest one-particle energy. With the semiclassical zero at the potential minimum, the quantum degeneracy threshold in a spherical linear trap is consequentlyThere are no numerical values of $m,T,a$ in the paper, so this parameter-dependent estimate is the requested atom number. It also follows from the peak density: $n(0)=N/[8\pi (k_{B}T/a{)}^3]$, and degeneracy starts when $n(0){\lambda }_T^3$ is order one. The criterion is a crossover estimate, not an assertion that the classical formula remains valid at $\mu =0$. For spinless bosons the continuum excited-state saturation is more specifically $N_c=\zeta (9/2)Z_1$, while for fermions the onset is occupation constrained by exclusion. At temperatures comparable with the quantum trap spacing, the actual ground energy must be included consistently rather than identified exactly with the classical minimum.

Availability and total entropy. The reservoir changes by $d{U}_R=-dU$ and $d{V}_R=-dV$. Its particle number and the system's particle number are fixed. The reservoir's first law of thermodynamics therefore givesAdd the system's entropy change. Because $T_R,p_R$ are constant,Thus define the availabilityAn additive constant in $A$ is immaterial; its value above the equilibrium reference is the exergy. The Second law of thermodynamics gives $\Delta S_{\mathrm{tot}}\ge 0$, hence $\Delta A\le 0$. A stable equilibrium maximizes total entropy over the allowed variations and therefore minimizes $A$. The environmental $T_R,p_R$ enter this definition even when the system has not yet reached their values. Because both thermal and mechanical exchanges are allowed, equilibrium here includes mechanical balance; equality of temperatures alone does not ensure the minimum.

Probability and curvature. For a fixed value of a macroscopic coordinate $x$, let $S(x)$ count the compatible system microstates. The product of the system and reservoir multiplicities isAll factors independent of $x$ enter a normalization constant, givingHere $A(x)$ is the constrained or effective availability for this coordinate, including the multiplicity of any eliminated variables. A nonuniform coordinate measure must likewise be retained or incorporated in that effective availability.