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Finite gravitational phase and entropy reference

1 · Prerequisites

2 · Summary

This page develops finite classical gravitational phase models with explicit SI phase references, softened canonical preparations, variational entropy, collapse and nonextensivity bounds, and a regulated relative Kepler shell. It separates fixed-Hamiltonian canonical response from Gibbs-volume response, compares finite-system entropy conventions, and proves the prescribed static-background classical gas identities. The finite-mode Fermi occupation result is conditional on the current batch-45 draft trace lemma. Its statement matches the planned interface, and its authored proof, precheck, and ready authoring record are present; the consuming result makes no claim of independent supplier approval. The local Gaussian and annular integration results and the shell-measure normalization are explicit page-local foundations. No item asserts empirical verification, gravitational backreaction, or a general relativistic quantum spectrum.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

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Finite gravitational phase and entropy reference

Definition

Use SI dimensions for the numerical model parameters: Newton constant G has units m3 kg−1 s−2, each mass mi has units kg, each position qi has units m, each momentum pi has units kg m s−1, energy H has units J, Boltzmann constant kB has units J K−1, and inverse energy β has units J−1. Fix an action reference h∗>0 in J s; it is a normalization convention and is not identified with Planck's constant.

For an integer N≥1, any Lebesgue measurable position region B⊆R3, and equal particle mass m>0, define the ordered phase space ΓN=BN×R3N with coordinates (q1,…,qN,p1,…,pN). This phase space and measure are allowed even when B is unbounded or has infinite volume; finite canonical preparations separately require a positive finite partition integral. The dimensionless reference measure for identical classical particles is dνN=d3Nq d3NpN!h∗3N. For a measurable real Hamiltonian H:ΓN→R and β>0 with 0<ZN(β):=∫ΓNe−βH dνN<∞, define the canonical density pβ=ZN(β)−1e−βH relative to νN. A probability density relative to νN is a measurable function ρ:ΓN→[0,∞) with ∫ΓNρ dνN=1. Set 0log⁡0:=0. For such a density with ∫ΓN∣ρlog⁡ρ∣ dνN<∞, define its reference-relative entropy by S(ρ)=−kB∫ΓNρlog⁡ρ dνN.

When H is fixed and the Gibbs-volume function Ω(E):=νN{H≤E} satisfies 0<Ω(E)<∞, define the Gibbs-volume entropy by SG(E)=kBlog⁡Ω(E). Where ω(E):=Ω′(E)>0, define the distinct surface-entropy convention, for a fixed energy width ΔE>0 measured in the same units as E, by SB(E)=kBlog⁡(ω(E)ΔE). Since νN and Ω are dimensionless, ω has units inverse to energy and ωΔE is dimensionless. For either α∈{G,B}, where Sα is differentiable and Sα′(E)>0, define Tα(E):=[Sα′(E)]−1; where also Tα′(E)≠0, define Cα(E):=dE/dTα=1/Tα′(E). These are finite-system conventions, and their temperatures and capacities need not agree.

The measure is dimensionless because d3Nq d3Np and h∗3N have the same units. The definition supplies no physical assertion that the chosen Hamiltonian describes nature and no mechanism of dynamical equilibration.

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Finite-dimensional Gaussian normalization and moment envelope

Statement

Let d≥1 be an integer, m>0, and 0<β0≤β1<∞. For every β∈[β0,β1], ∫Rde−β∣p∣2/(2m) dp=(2πmβ)d/2. For every integer s≥0, the polynomially weighted Gaussian has a uniform finite bound sup⁡β∈[β0,β1]∫Rd(1+∣p∣2s)e−β∣p∣2/(2m) dp<∞.

Facts & Assumptions

Given: An integer d≥1, mass parameter m>0, and compact positive inverse-temperature interval [β0,β1].

[F1]

The one-dimensional Gaussian integral is ∫Re−x2 dx=π. (The Gaussian integral ∫−∞∞e−x2 dx=π)

[F2]

For sigma-finite measure spaces and a nonnegative product-measurable function, Tonelli's theorem identifies the product integral with either iterated integral. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

Proof

1.1F1F2algebra

In one dimension set y=β/(2m) x; the substitution and [F1] give ∫Re−βx2/(2m)dx=2πm/β. Taking the product of d such factors and applying [F2] repeatedly gives the displayed exact d-dimensional integral.

1.2F1algebra

Put α=β0/(2m)>0. For any integer s≥1, the function x↦∣x∣2se−αx2/2 is bounded: it is continuous on [−1,1] and tends to zero at both infinite ends. Hence ∣x∣2se−αx2≤Cs,αe−αx2/2; on [−1,1] the same bound holds after increasing Cs,α. By the scaled one-dimensional Gaussian integral, both ∫e−αx2dx and ∫∣x∣2se−αx2dx are finite.

2.1F2step 1.2algebra∎

For β≥β0, e−β∣p∣2/(2m)≤e−α∣p∣2. If s≥1, (∑i=1dpi2)s≤ds−1∑i=1d∣pi∣2s; for s=0 the weight is one. Factor the remaining coordinate Gaussians and apply [F2] and the one-dimensional bounds from step 1.2. The resulting finite upper bound depends only on d,s,m,β0, not on β in the stated interval.

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Radial integration over a bounded Euclidean shell

Statement

Let 0≤a<b<∞ and let F:[a,b]→R be continuous. The closed radial shell Aa,b={x∈R3:a≤∣x∣≤b} is Jordan measurable, and ∫Aa,bF(∣x∣) dx=4π∫abF(r)r2 dr. The angular area measure used below is the pushforward of sin⁡θ dθ dϕ on [0,π]×[0,2π] under (θ,ϕ)↦(sin⁡θcos⁡ϕ,sin⁡θsin⁡ϕ,cos⁡θ); its total mass is 4π.

Facts & Assumptions

Given: A compact radial interval [a,b]⊂[0,∞) and a continuous real function F.

[F1]

A bounded subset of Euclidean space is Jordan measurable exactly when its boundary has content zero. (A bounded set in Rm is Jordan measurable iff its boundary is null, equivalently of content zero)

[F2]

Continuous real functions on compact Jordan sets are Riemann integrable. (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set)

[F3]

For an injective C1 map with invertible derivative on an open neighborhood of a compact Jordan set, change of variables holds with the absolute Jacobian determinant. (Change of variables for an injective C1 map on a compact Jordan set)

[F4]

Such a map sends a compact Jordan set to a compact Jordan image. (An injective C1 map with invertible derivative sends compact Jordan sets to compact Jordan sets)

Proof

1.1F1algebra

Fix a sphere of radius r>0. At least one coordinate of every point on it has absolute value at least r/3. On each of the six patches where a specified coordinate is largest in absolute value and has a specified sign, write that coordinate as ±r2−u2−v2 over a bounded disk with r2−u2−v2≥r2/3. The graph gradient is uniformly bounded. A grid of squares of side δ covers the disk with O(δ−2) squares; the graph over each square fits in a rectangular box of base area δ2 and height O(δ), so the total covering volume is O(δ). Thus the sphere has content zero. The sphere at radius zero is a point and is covered by one cube of arbitrarily small volume. Since ∂Aa,b is contained in the union of the spheres of radii a and b, [F1] makes the bounded shell Jordan measurable.

2.1F2F3F4algebra

If a=b, then Aa,b is a sphere of Jordan content zero by step 1.1, so both integrals are zero. Now assume 0<a<b. For 0<ε<min⁡{π/4,(b−a)/2} put Kε=[a,b]×[ε,π−ε]×[ε,2π−ε] and Ψ(r,θ,ϕ)=(rsin⁡θcos⁡ϕ,rsin⁡θsin⁡ϕ,rcos⁡θ). On the open domain with r>0, 0<θ<π and 0<ϕ<2π, Ψ is injective and C1 with determinant r2sin⁡θ>0; its inverse is given by radius, polar angle and the unique azimuth in (0,2π). The compact set Kε is Jordan, and [F2] makes x↦F(∣x∣) integrable on Ψ(Kε). Applying [F3] gives ∫Ψ(Kε)F(∣x∣) dx=(∫abF(r)r2 dr)(∫επ−εsin⁡θ dθ)(2π−2ε).

3.1F1F2F3F4F5step 2.1algebra∎

The missing part of Aa,b outside Ψ(Kε) is contained in the union of the polar-cap box [−bsin⁡ε,bsin⁡ε]2×[−b,b] and the azimuth-seam box [−b,b]×[−bsin⁡ε,bsin⁡ε]×[−b,b]. Their total volumes are at most 8b3(ε2+ε), which tends to zero. Since F is bounded, the difference between its integral on the whole shell and on Ψ(Kε) tends to zero. Letting ε↓0 in step 2.1 gives the formula and the angular factor 2⋅2π=4π. If a=0, first apply the result to [η,b] and then let η↓0; the omitted ball lies in [−η,η]3 and contributes at most 8η3sup⁡[0,b]∣F∣, while the absolute value of the omitted radial integral is at most sup⁡[0,b]∣F∣η3/3. Both tend to zero.

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Finite softened gravitational Hamiltonian and domain

Definition

Use the phase space and dimensionless reference measure of Finite gravitational phase and entropy reference. Fix an integer N≥1, a bounded Lebesgue measurable B⊆R3 with positive finite volume, a common mass m>0, a gravitational coupling G>0, and a softening length a>0. On ΓN=BN×R3N define HN,a(q,p)=∑i=1N∣pi∣22m+Ua(q),Ua(q)=−∑1≤i<j≤NGm2∣qi−qj∣2+a2. The positions and momenta are mathematical phase variables; the softening is a fixed regulator parameter. The Hamiltonian is measurable and its potential is bounded because each pair denominator is at least a. A point-particle model is a different Hamiltonian and is not included at a=0.

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Regulated relative Kepler phase volume and shell law

Definition

Use SI units, positive masses m1,m2 in kg, reduced mass μ=m1m2/(m1+m2), gravitational parameter A=Gm1m2 in J m, relative position q∈R3 in m, relative momentum p∈R3 in kg m s−1, and a fixed action reference h∗>0 in J s. Assume r0>0 and R>16r0. The relative phase domain is Γrel={(q,p):r0≤∣q∣≤R, p∈R3},dνrel=d3q d3ph∗3, with no center-of-mass variables or factor 1/2!. The relative Hamiltonian is h(q,p)=∣p∣22μ−A∣q∣.

For E=−e<0 with A/R<e≤A/(16r0), put δ=er0/A∈(0,1/16] and J(δ)=∫δ1s1/2(1−s)3/2 ds,Ω(E)=νrel{(q,p)∈Γrel:h(q,p)≤E}. The radial integration formula of Radial integration over a bounded Euclidean shell gives the explicit phase volume Ω(−e)=(4π)(4π/3)(2μ)3/2A3h∗3 e−3/2J(δ). This is finite and positive on the stated branch.

Define S2={u∈R3:∣u∣=1} and p∗(r,E)=2μ(E+A/r) for r0≤r≤A/e. Let dσ be the angular area measure of total mass 4π from Radial integration over a bounded Euclidean shell. Push the finite measure (μp∗/h∗3)d3q dσ(u) on {r0≤∣q∣≤A/e}×S2 forward by (q,u)↦(q,p∗(∣q∣,E)u); call the resulting measure ΣE. Its candidate normalized shell law is PE=ΣE/Ω′(E). The separate result Finite regulated Kepler shell measure is normalized ↗ proves Ω′(E)>0, finite total mass, and PE({h=E})=1. Thus the energy-shell preparation is explicitly defined; it is not inferred from phase-volume counting alone.

The entropy convention attached to this model is Gibbs volume, SG(E)=kBlog⁡Ω(E), with 1/TG=dSG/dE and CG=dE/dTG where defined. Other finite-system entropy conventions are distinct.

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Static relativistic finite gas normalization and hydrostatics

Statement

Let Ix0⊆R be an open interval and B⊂R3 an open bounded coordinate box with compact closure and positive volume. On a neighborhood of Ix0×B‾, with x0=ct assigned metres and t in seconds, prescribe the smooth static metric g=−N(x)2(dx0)2+hij(x)dxidxj, where hij is positive definite with eigenvalues bounded above and away from zero, and the lapse satisfies 0<N0≤N(x)≤N1<∞. The static unit observer is u=N−1∂x0, so g(u,u)=−1; this is a prescribed background, not a solution sourced by the gas.

Use SI units: particle mass m>0 is in kg, c>0 is in m s−1, orthonormal momentum p^ is in kg m s−1, and local energy E is in J. In any local orthonormal coframe, pi=eiap^a and hij=∑aeiaeja, so dVh d3p^=d3x d3pi; orthonormal-frame changes are rotations and preserve d3p^. This defines a fibre measure without assuming a global frame. Use the fixed action reference h∗>0 in J s and βK>0 in J−1: E(p^)=m2c4+c2∣p^∣2,HK(x,p^)=N(x)E(p^),Z1(βK)=∫B∫R3e−βKHKdVh d3p^h∗3. Then 0<Z1<∞, with all polynomial momentum moments and compact-positive-βK derivatives finite. For each fixed particle count M≥1, the classical ideal-gas preparation with Gibbs factor 1/M! has ZM=Z1M/M!; a grand fugacity preparation with dimensionless z>0 has Ξ=exp⁡(zZ1).

Define f(x,p^)=zh∗−3e−βKHK and the proper local densities n(x)=∫f d3p^,ε(x)=∫Ef d3p^,P(x)=13∫p^⋅∇p^E f d3p^. Their units are m−3, J m−3, and J m−3, respectively. With βloc=βKN(x) and TK=(kBβK)−1, the exact model identities are Tloc(x)=1kBβloc=TKN(x),P=nkBTloc,∂iP=−(ε+P)∂ilog⁡N. If z=eβKμK with μK in joules, the local chemical parameter is μloc=μK/N. No Einstein equation, self-consistent gas backreaction, interacting quantum trace, or arbitrary curved-space mode spectrum is asserted.

Facts & Assumptions

Given: A bounded positive-volume region, smooth prescribed static metric data satisfying the stated uniform bounds, m,c,h∗,kB,βK,z>0, and the classical ideal-gas preparation.

[F1]

The phase-reference item fixes the action unit h∗ and the convention that a phase integral is dimensionless after division by its corresponding power. (Finite gravitational phase and entropy reference)

[F2]

If an integrable section is differentiable in a parameter and its derivative has an integrable uniform majorant on a neighborhood, differentiation passes through the integral. (Differentiation under the integral sign)

[F3]

Tonelli factors nonnegative measurable integrals on products of sigma-finite measure spaces. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

Proof

1.1F1algebra

In a local orthonormal coframe, pi=eiap^a and hij=∑aeiaeja, so ∣det⁡(eia)∣=det⁡h. Hence d3pi=det⁡h d3p^ and d3x d3pi=dVh d3p^. The transition matrix between orthonormal coframes is orthogonal and has absolute determinant one, so this fibre measure is independent of the chosen local coframe.

1.2F1F2F3givenalgebra

In the orthonormal momentum variables, E(p^)≥c∣p^∣, so e−βKNE≤e−βKN0c∣p^∣. The right side is integrable on R3: ∣p^∣≥(∣p^1∣+∣p^2∣+∣p^3∣)/3 reduces its integral to a product of three finite one-dimensional exponential integrals. The same comparison with polynomial factors gives every energy moment and a common majorant for βK in compact positive intervals. Each βK derivative of the integrand is (−HK)je−βKHK; [F2] therefore gives Z1(j)=∫(−HK)je−βKHKdVh d3p^/h∗3 for every finite order. The proper volume of B‾ is finite by compactness and the metric eigenvalue bounds, and positive because B has positive volume and det⁡h>0. Thus Z1 is finite and strictly positive. For M independent labelled particles, Tonelli [F3] factors the product integral; the fixed classical indistinguishable-particle counting convention divides by M!, giving ZM=Z1M/M!. Summing grand weights yields ∑M≥0(zZ1)M/M!=ezZ1.

2.1givenstep 1.2algebra

At a fixed x, put βloc=βKN(x). On a momentum cube [−R,R]3, integrate ∑i=13∂p^i(p^ie−βlocE)=3e−βlocE−βloc(p^⋅∇p^E)e−βlocE. The constant factor zh∗−3 in f cancels from the moments. Here ∇p^E=c2p^/E and p^⋅∇p^E≤c∣p^∣, so the divergence terms are integrable under the exponential bound of step 1.2. On a face with ∣p^i∣=R, ∣p^∣≥R and ∣p^∣≥(∣p^j∣+∣p^k∣)/2, hence ∣p^∣≥R/2+(∣p^j∣+∣p^k∣)/(22). Each face integral is therefore bounded by a constant times Re−βloccR/2∫R2e−βlocc(∣y∣+∣z∣)/(22)dy dz, which tends to zero. The integral of the divergence over R3 is zero, and division by 3βloc gives P=n/βloc=nkBTloc.

3.1F1F2step 1.2step 2.1algebra∎

For each spatial coordinate xi, ∂if=−βK(∂iN)Ef. The derivatives of the lapse are bounded on the compact closure, N≥N0, and Ef has the integrable majorant from step 1.2. Hence [F2] gives ∂in=−βK(∂iN)ε. Differentiating P=n/(βKN) and substituting this identity yields ∂iP=−εN∂iN−nβKN2∂iN=−(ε+P)∂ilog⁡N. Finally, rewriting βK(NE−μK) as βloc(E−μloc) gives μloc=μK/N.

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Finite gravitational Gibbs normalization and differentiable moments

Statement

Use the box, mass m, integer N≥1, regulator a>0, and Hamiltonian HN,a=KN+Ua of Finite softened gravitational Hamiltonian and domain, with KN=∑i∣pi∣2/(2m). Put CN=Gm2N(N−1)/(2a) and use the reference measure of Finite gravitational phase and entropy reference. For every β>0, 0<∣B∣NN!h∗3N(2πmβ)3N/2≤ZN(β)≤eβCN∣B∣NN!h∗3N(2πmβ)3N/2<∞. The density pβ=ZN−1e−βHN,a is a positive probability density. For each integer j≥0, ∫ΓN∣HN,a∣je−βHN,adνN is finite, and ZN is C∞ on (0,∞) with ZN(j)(β)=∫ΓN(−HN,a)je−βHN,a dνN. On every compact positive β interval, one integrable majorant works for each fixed derivative order.

Facts & Assumptions

Given: A bounded measurable box B⊆R3 with 0<∣B∣<∞, N≥1, m,G,a,β>0, and the finite phase model.

[F1]

The softened model has HN,a=KN+Ua on BN×R3N, where each pair term in Ua is −Gm2/∣qi−qj∣2+a2. (Finite softened gravitational Hamiltonian and domain)

[F2]

For d≥1, m>0, and a compact interval [β0,β1]⊂(0,∞), all Gaussian momentum moments ∫Rd(1+∣p∣2j)e−β∣p∣2/(2m)dp have a common finite bound. (Finite-dimensional Gaussian normalization and moment envelope)

[F3]

If an integrable section is differentiable in a parameter and its derivative is dominated by one integrable majorant on an open interval, the derivative passes through the integral. (Differentiation under the integral sign)

[F4]

Tonelli factors nonnegative measurable integrals on finite products of Euclidean Lebesgue spaces. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

Proof

1.1givenF1F2F4

Each pair denominator is at least a, so −CN≤Ua≤0 and KN≥0. Consequently e−βKN≤e−βHN,a≤eβCNe−βKN. Tonelli [F4] and [F2] evaluate the momentum integral as (2πm/β)3N/2; the position integral is ∣B∣N, so the stated positive lower and finite upper bounds follow after division by N!h∗3N.

2.1F1F2step 1.1algebra

Fix a compact interval [β0,β1]⊂(0,∞). Since ∣Ua∣≤CN and KN=∣p∣2/(2m) in d=3N momentum coordinates, for every fixed j there is a finite Aj such that ∣HN,a∣je−βHN,a≤Aj(1+∣p∣2j)e−β0∣p∣2/(2m) for every β∈[β0,β1] and every phase point. The right side is integrable by [F2] times the finite position volume ∣B∣N. This proves all absolute energy moments and supplies an integrable compact-parameter envelope.

3.1F2F3step 2.1algebra∎

For each j, the parameter derivative of (−HN,a)je−βHN,a is (−HN,a)j+1e−βHN,a; its measurability and an integrable uniform bound are provided by step 2.1 at order j+1. Apply [F3] successively on compact subintervals of (0,∞), starting with j=0, to differentiate under the integral any finite number of times. The formula for ZN(j) follows by induction, and the moment of order zero gives the positive normalized density.

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Point collapse and unscaled gravity nonextensivity

Statement

Use the dimensionless phase reference of Finite gravitational phase and entropy reference and the softened Hamiltonian of Finite softened gravitational Hamiltonian and domain.

  1. For two equal positive masses in a bounded three-dimensional box with nonempty interior, the distinct point Hamiltonian H2pt(q,p)=∣p1∣2+∣p2∣22m−Gm2∣q1−q2∣ makes the canonical partition integral infinite for every β>0.

  2. For fixed a>0, let BN be a sequence of bounded boxes containing one fixed cube C of volume v>0 and diameter D<∞, with constants b1,b2>0 such that b1N≤∣BN∣≤b2N. The softened canonical partition functions satisfy log⁡ZN(β)N≥βc(N−1)−log⁡N+Kβ,c=Gm22D2+a2>0, where Kβ is independent of N. Hence log⁡ZN/N→+∞ and the free energy per particle −β−1log⁡ZN/N tends to −∞.

  3. For the unscaled point attraction −Gm2/r restricted to the hard-core position domain DN={q∈BnN:∣qi−qj∣>a for i≠j} with fixed a>0, there is a fixed-density sequence of integers n≥2, N=n3, boxes Bn=(0,nd)3 of side O(n) for fixed d>a, and positive-volume admissible position sets on which the interaction energy is at most −c0N5/3 for some c0>0. Its canonical partition functions therefore also satisfy log⁡ZN/N→+∞.

For the sequences in clauses 2 and 3, write FN(β)=−β−1log⁡ZN(β). In clause 2, the free energy per particle FN/N tends to −∞. Since b1≤∣BN∣/N≤b2, the free energy per volume FN/∣BN∣ also tends to −∞. Clause 3 gives the same per-particle limit along its fixed-density sequence, and there ∣Bn∣/N=d3, so the per-volume free energy also tends to −∞. Thus finite boxes at each particle count, fixed softening in clause 2, and hard-core exclusion with unscaled attraction in clause 3 do not ensure a finite real thermodynamic-limit value for either normalized free energy in these sequences. These results do not address differently scaled interactions or every possible gravitational thermodynamic limit.

Facts & Assumptions

Given: The phase reference, the softened Hamiltonian, G,m,a,β>0, and the box or packing hypotheses in the Statement.

[F1]

The phase reference is dνN=d3Nq d3Np/(N!h∗3N). (Finite gravitational phase and entropy reference)

[F2]

The softened model has kinetic energy ∑i∣pi∣2/(2m) and pair attraction −Gm2/∣qi−qj∣2+a2. (Finite softened gravitational Hamiltonian and domain)

[F3]

For d=3N, the momentum Gaussian integral is (2πm/β)3N/2 and its polynomial moments are finite. (Finite-dimensional Gaussian normalization and moment envelope)

[F4]

Tonelli's theorem applies to the nonnegative Boltzmann integrand on product Lebesgue spaces. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

Proof

1.1givenF1F3F4algebra

Choose an interior point q0 of the box and a length ℓ>0 small enough that the cubes below lie in the box for all sufficiently large integers n. Let Q1,n=q0+(0,ℓ/n)3 and Q2,n=q0+(2ℓ/n,0,0)+(0,ℓ/n)3. They are disjoint, have volumes (ℓ/n)3, and every pair satisfies 0<∣q2−q1∣≤11ℓ/n. On their product the point-potential Boltzmann weight is at least exp⁡(βGm2n/(11ℓ)), so its positional integral is at least (ℓ/n)6exp⁡(βGm2n/(11ℓ)). This lower bound tends to infinity, proving divergence of the positional integral. The independent momentum Gaussian integral is positive and finite by [F3]; Tonelli [F4] therefore makes the full partition integral infinite.

1.2F1F2F3givenalgebra

In the softened model, when every qi∈C, all pair distances are at most D and hence Ua(q)≤−Gm2N(N−1)/(2D2+a2)=−cN(N−1). Restricting the nonnegative partition integral to CN×R3N and integrating momenta gives ZN(β)≥vNN!h∗3NeβcN(N−1)(2πmβ)3N/2. Use N!≤NN, divide logarithms by N, and gather all N-independent terms into Kβ to obtain the claimed bound. Since βcN−log⁡N→+∞, the free-energy conclusion follows.

2.1F1F3givenalgebra∎

For each fixed N=n3, the hard-core condition bounds each pair attraction below by −Gm2/a; the finite box and Gaussian momentum integral therefore make its partition function finite. Fix d>a+2δ with 0<δ<d. In Bn=(0,nd)3, place centers at d(k1+12,k2+12,k3+12) for ki∈{0,…,n−1} and choose the open cube of side δ centered at each center. These cubes lie in Bn and have positive volume. Distinct grid centers are at least d apart, and two points in their cubes differ from their centers by at most 3δ; hence every particle separation exceeds d−3δ>a, since 2>3. The product cubes are admissible. Each pair distance is at most 3(n−1)d+3δ≤crnd with cr=3. Since ∣Bn∣=n3d3=Nd3, this sequence has fixed number density d−3. There are N(N−1)/2 pairs; for N≥2, UN≤−Gm2crndN(N−1)2≤−c0N5/3,c0=Gm24crd>0, because n=N1/3. Restricting the finite partition integral to these product cubes, integrating momenta, and using N!≤NN gives log⁡ZN/N≥βc0N2/3−log⁡N+Kβ→+∞.

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Finite regulated Kepler shell measure is normalized

Statement

Under Regulated relative Kepler phase volume and shell law, for every E=−e with A/R<e≤A/(16r0), the phase volume is finite and positive and has an ordinary derivative on the interior of this energy interval. At the endpoint e=A/(16r0), write Ω′(E) for the right derivative in E from within the interval. In both cases the pushed-forward shell measure ΣE has total mass Ω′(E)>0, so PE=ΣE/Ω′(E) is a probability measure supported on the exact energy surface {(q,p):h(q,p)=E}.

Facts & Assumptions

Given: The relative Kepler phase domain, its Lebesgue phase measure, the Hamiltonian h(q,p)=∣p∣2/(2μ)−A/∣q∣, and the energy range from the Definition.

[F1]

The Definition sets Ω(E)=νrel{h≤E} and defines the shell pushforward using p∗(r,E)=2μ(E+A/r). (Regulated relative Kepler phase volume and shell law)

[F2]

On a three-dimensional radial shell, ∫a≤∣x∣≤bF(∣x∣) dx=4π∫abF(r)r2dr for continuous F, including a=0. (Radial integration over a bounded Euclidean shell)

Proof

1.1F1F2algebra

The inequalities A/R<e and e≤A/(16r0) imply r0<A/e<R. At fixed q with r=∣q∣, the momentum sublevel h(q,p)≤−e is the ball ∣p∣≤2μ(A/r−e), which is nonempty exactly for r0≤r≤A/e. By [F2] its momentum volume is 4π3[2μ(A/r−e)]3/2. Integrating the resulting radial function over q by [F2], then substituting s=er/A, gives Ω(−e)=(4π)(4π/3)(2μ)3/2A3h∗3e−3/2J(δ),δ=er0A. The integration interval is compact and has positive length, while the integrand is continuous and positive in its interior, so Ω is finite and strictly positive.

2.1F1step 1.1algebra

Put f(s)=s1/2(1−s)3/2 for 0<s<1. For h≠0 with δ+h∈(0,1), the oriented-integral identity gives (J(δ+h)−J(δ))/h=−(1/h)∫δδ+hf(s) ds; its difference from −f(δ) tends to zero by continuity. At δ=1/16, restrict to h<0, which corresponds to increasing E within the branch, to obtain the right derivative in E. Thus J′(δ)=−δ1/2(1−δ)3/2, with that one-sided interpretation at the endpoint. Differentiating the explicit phase-volume formula with E=−e now yields Ω′(E)=(4π)(4π/3)(2μ)3/2A3h∗3e−5/2[32J(δ)+δ3/2(1−δ)3/2]>0. The bracket is finite and positive, so the derivative is finite and strictly positive.

3.1F1F2step 2.1algebra

Integrating the defining pushforward density over its entire source gives ΣE({h=E})=16π2μh∗3∫r0A/er22μ(A/r−e) dr=16π2μ2μh∗3A3e−5/2∫δ1s3/2(1−s)1/2 ds, where [F2] supplies the spatial radial measure and σ(S2)=4π supplies the momentum-direction factor. Put g(s)=s3/2(1−s)3/2. Direct differentiation and integration on [δ,1] give 32J(δ)+δ3/2(1−δ)3/2=32∫δ1s3/2(1−s)1/2 ds. Since μ2μ=12(2μ)3/2, this identity makes the preceding shell mass exactly the Ω′(E) in step 2.1.

4.1F1step 3.1algebra∎

Every image point has ∣p∣=p∗(r,E), hence h=∣p∣2/(2μ)−A/r=E. The measure is finite by step 3.1 and has total mass Ω′(E)>0; division by this total therefore defines a probability supported on the energy surface.

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Exact relative-shell canonical normalization and response

Statement

Use the relative Kepler Hamiltonian and shell Q={q∈R3:r0≤∣q∣≤R} from Regulated relative Kepler phase volume and shell law, with μ,A,r0>0 and R>16r0. For every β>0, define Zrel(β)=1h∗3∫Q∫R3e−βh(q,p) d3p d3q. Then 0<∣Q∣eβA/Rh∗3(2πμβ)3/2≤Zrel(β)≤∣Q∣eβA/r0h∗3(2πμβ)3/2<∞. All energy moments are finite, and on every compact positive β interval one integrable majorant justifies all derivatives of Zrel. Put dνrel=h∗−3d3q d3p on Q×R3 and pβ=Zrel(β)−1e−βh. For a measurable observable A:Q×R3→R with finite second moment, define ⟨A⟩β:=∫Apβ dνrel and Var⁡β(A):=∫(A−⟨A⟩β)2pβ dνrel. The normalized canonical density has ⟨h⟩β=−∂βlog⁡Zrel,d⟨h⟩βdβ=−Var⁡β(h),CK:=d⟨h⟩βdTK=kBβ2Var⁡β(h)≥0,TK=(kBβ)−1. This is the exact six-dimensional relative canonical preparation; center-of-mass degrees are absent.

Facts & Assumptions

Given: The relative phase shell and Hamiltonian, β>0, μ,A,r0>0, and R>16r0.

[F1]

The relative shell is Q={q:r0≤∣q∣≤R} and h(q,p)=∣p∣2/(2μ)−A/∣q∣, with measure d3q d3p/h∗3. (Regulated relative Kepler phase volume and shell law)

[F2]

Radial integration gives ∣Q∣=4π∫r0Rr2dr=4π3(R3−r03). (Radial integration over a bounded Euclidean shell)

[F3]

Gaussian momentum moments are finite and uniformly bounded on compact positive inverse-temperature intervals. (Finite-dimensional Gaussian normalization and moment envelope)

[F4]

Differentiation under an integral is valid when the parameter derivative has an integrable uniform majorant. (Differentiation under the integral sign)

[F5]

Tonelli factors nonnegative measurable integrals on products of sigma-finite Lebesgue spaces. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

Proof

1.1F1F2F3F5algebra

On Q, −A/r0≤−A/∣q∣≤−A/R. Therefore eβA/Re−β∣p∣2/(2μ)≤e−βh(q,p)≤eβA/r0e−β∣p∣2/(2μ). The position volume is positive and finite by [F2]. Tonelli [F5] and the Gaussian formula [F3] integrate the momentum factor as (2πμ/β)3/2, proving both strict positivity and the stated finite bounds.

1.2F1F2F3F4algebra

Fix [β0,β1]⊂(0,∞). Since ∣−A/∣q∣∣≤A/r0, for every integer j≥0 a finite constant Bj gives ∣h(q,p)∣je−βh(q,p)≤Bj(1+∣p∣2j)e−β0∣p∣2/(2μ). This is integrable on Q×R3 by [F2] and [F3], independently of β in the compact interval. The integrand derivatives are (−h)je−βh, so [F4] applied successively proves all moment and derivative formulas.

2.1F3F4step 1.1step 1.2algebra∎

The first two derivatives are Zrel′=−∫he−βhdν and Zrel′′=∫h2e−βhdν. Dividing by Zrel>0 and differentiating the quotient gives d⟨h⟩β/dβ=−⟨h2⟩β+⟨h⟩β2=−Var⁡β(h). Since dTK/dβ=−1/(kBβ2), the chain rule yields the displayed CK≥0.

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Negative Gibbs-volume capacity in a finite regulated shell

Statement

Under Regulated relative Kepler phase volume and shell law, let E=−e, δ=er0/A∈(0,1/16], and z(δ)=δ3/2(1−δ)3/2J(δ),B(δ)=32+z(δ). For the Gibbs-volume entropy SG(E)=kBlog⁡Ω(E) and temperature convention 1/TG=dSG/dE, TG(E)=ekBB(δ),CG(E):=dEdTG=−kBB(δ)2B(δ)−δB′(δ)<0. The denominator is at least 1. Derivatives are ordinary for 0<δ<1/16; at δ=1/16, compute dSG/dE and dTG/dE as right derivatives from within the branch and define CG=1/(dTG/dE). This is a finite regulated mathematical branch with the stated entropy convention; it does not identify a universal operational temperature or dynamical stability.

Facts & Assumptions

Given: Positive μ,A,r0, R>16r0, energy E=−e with A/R<e≤A/(16r0), and the Gibbs-volume convention.

[F1]

The relative Hamiltonian, phase volume, and entropy convention are defined by the regulated shell item, with Ω(−e)=CA3e−3/2J(er0/A) for a positive constant C. (Regulated relative Kepler phase volume and shell law)

Proof

1.1F1algebra

Write e=−E and δ=er0/A. For f(s)=s1/2(1−s)3/2 and h≠0 with δ+h∈(0,1), the oriented-integral identity gives (J(δ+h)−J(δ))/h=−(1/h)∫δδ+hf(s) ds; continuity makes this tend to −f(δ). At δ=1/16, restrict to h<0, which corresponds to right increments in E within the branch. Hence J′(δ)=−δ1/2(1−δ)3/2, with the one-sided interpretation at the endpoint. Differentiating log⁡Ω=C1−32log⁡e+log⁡J(δ) with respect to E=−e gives dSG/dE=kB(32+z)/e=kBB/e. Its reciprocal is the stated positive TG.

1.2F1algebra

On [1/4,1/2]⊆[δ,1], s1/2≥1/2 and (1−s)3/2≥1/4, so J(δ)≥1/32. Also δ3/2≤1/64 and (1−δ)3/2≤1, hence 0<z≤1/2.

2.1step 1.1algebra

Differentiating TG=e/[kBB(δ)] with respect to e, using dδ/de=δ/e, yields dTGde=B−δB′kBB2. Since E=−e, inversion gives the displayed formula for CG, provided B−δB′>0.

3.1step 2.1step 1.2algebra∎

Logarithmic differentiation of z gives δz′=32z1−2δ1−δ+z2≤32z+z2, because 0<δ≤1/16 makes (1−2δ)/(1−δ)≤1. Therefore B−δB′=32+z−δz′≥32−z2−z2≥1 when 0<z≤1/2. Step 2.1 now gives CG<0 throughout the branch.

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Nonnegative fixed-Hamiltonian canonical heat capacity

Statement

Under the hypotheses of Finite gravitational Gibbs normalization and differentiable moments, keep the box, particle number, mass, regulator, reference measure, and Hamiltonian fixed while varying β>0, and set pβ=ZN(β)−1e−βHN,a. For a measurable observable A:ΓN→R with finite second moment, define ⟨A⟩β:=∫ΓNApβ dνN and Var⁡β(A):=∫ΓN(A−⟨A⟩β)2pβ dνN. Define TK=(kBβ)−1 and CK=d⟨HN,a⟩β/dTK. Then ⟨HN,a⟩β=−∂βlog⁡ZN,∂β⟨HN,a⟩β=−Var⁡β(HN,a),CK=kBβ2Var⁡β(HN,a)≥0. This is a response identity for a fixed Hamiltonian and fixed regulator; it does not assert stability when the Hamiltonian or background varies with temperature.

Facts & Assumptions

Given: The finite softened Gibbs model, β>0, kB>0, and its temperature convention TK=(kBβ)−1.

[F1]

The finite Gibbs density is normalized, and the partition function is twice differentiable with ZN(j)=∫(−HN,a)je−βHN,adνN for j=0,1,2. (Finite gravitational Gibbs normalization and differentiable moments)

Proof

1.1F1algebra

Differentiating log⁡ZN and using ZN′=−∫HN,ae−βHN,adνN gives −∂βlog⁡ZN=ZN−1∫HN,ae−βHN,adνN=⟨HN,a⟩β.

2.1F1step 1.1algebra

The quotient rule and ZN′′=∫HN,a2e−βHN,adνN give ∂β⟨HN,a⟩β=−ZN′′ZN+(ZN′ZN)2=−(⟨HN,a2⟩β−⟨HN,a⟩β2)=−Var⁡β(HN,a). All terms are finite by [F1].

3.1F1step 2.1algebra∎

Since dTK/dβ=−1/(kBβ2), the chain rule yields CK=(d⟨H⟩/dβ)/(dTK/dβ)=kBβ2Var⁡β(HN,a). The variance is the expectation of the nonnegative square (HN,a−⟨HN,a⟩)2, so CK≥0.

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Unique finite gravitational Gibbs variational optimum

Statement

Under the hypotheses of Finite gravitational Gibbs normalization and differentiable moments, let ρ:ΓN→[0,∞) be a measurable probability density relative to νN, so ∫ΓNρ dνN=1, satisfying ∫ΓN∣ρlog⁡ρ∣ dνN<∞,∫ΓN∣HN,a∣ρ dνN<∞, with 0log⁡0=0. For any probability density σ with integrable energy, define its mean energy by ⟨HN,a⟩σ:=∫ΓNHN,aσ dνN. Define S(ρ)=−kB∫ΓNρlog⁡ρ dνN and D(ρ∥pβ):=∫ΓNρlog⁡(ρ/pβ) dνN, with 0log⁡(0/pβ):=0. Then S(ρ)−kBβ⟨HN,a⟩ρ≤kBlog⁡ZN, with equality exactly when ρ=pβ almost everywhere. Among densities with the same mean energy as pβ, this makes pβ the unique entropy maximizer.

For every bounded measurable real u with ∫pβu dνN=0, ∫HN,apβu dνN=0, and u≠0 in L2(pβdνN), the perturbation ρt=pβ(1+tu) is a probability density of the same mean energy for ∣t∣∥u∥∞≤1/2. For sufficiently small nonzero t, S(ρt)<S(pβ),D(ρt∥pβ)=t22∫pβu2 dνN+Rt,∣Rt∣≤23∣t∣3∥u∥∞∫pβu2 dνN. This is a variational claim in the stated density class, not a dynamical equilibration theorem.

Facts & Assumptions

Given: The positive normalized density pβ, its finite energy moments, and the reference-relative entropy convention.

[F1]

For the softened finite model, pβ=ZN−1e−βHN,a is positive and normalized, and its energy moments are finite. (Finite gravitational Gibbs normalization and differentiable moments)

[F2]

The phase-reference item defines S(ρ)=−kB∫ρlog⁡ρ dνN for probability densities relative to the dimensionless measure νN. (Finite gravitational phase and entropy reference)

[F3]

A nonnegative measurable function with integral zero vanishes almost everywhere. (A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere)

Proof

1.1algebra

For s>0, ϕ(s)=slog⁡s−s+1 has derivative log⁡s, is decreasing for 0<s<1, and increasing for s>1. Its continuous value at s=0 is 1, so ϕ(s)≥0 with equality exactly at s=1.

2.1F1step 1.1algebra

Since pβ>0, set r=ρ/pβ almost everywhere. The assumptions and log⁡pβ=−βHN,a−log⁡ZN make ρlog⁡(ρ/pβ) integrable. Expanding ϕ and using ∫ρ=∫pβ=1 gives D(ρ∥pβ)=∫pβϕ(r) dνN=∫ρlog⁡ρ dνN+β⟨HN,a⟩ρ+log⁡ZN≥0.

3.1F1F2F3step 2.1algebra

Rearranging step 2.1 and using the entropy convention [F2] gives the stated inequality. Equality forces ϕ(r)=0 almost everywhere for the probability measure pβdνN, so r=1 almost everywhere by [F3], and ρ=pβ. Conversely, if ρ=pβ almost everywhere then D(ρ∥pβ)=0 in step 2.1 and the inequality is an equality. At equal mean energy this is the unique entropy-maximizing density.

4.1F1F2step 2.1algebra∎

Put ψ(x)=(1+x)log⁡(1+x)−x. For ∣x∣≤1/2, ψ(0)=ψ′(0)=0 and ψ′′(x)=1/(1+x), hence ∣ψ′′(x)−1∣≤2∣x∣. Integrating this second-derivative bound twice from zero gives ∣ψ(x)−x2/2∣≤∣x∣3/3≤2∣x∣3/3. The two centering conditions make ρt normalized and energy preserving, while ∣tu∣≤1/2 makes it nonnegative. Moreover [F1] gives ∫∣HN,a∣pβ<∞ and ∫pβ∣log⁡pβ∣<∞ by log⁡pβ=−βHN,a−log⁡ZN; since 1+tu and log⁡(1+tu) are bounded, these imply ∫∣HN,a∣ρt<∞ and ∫∣ρtlog⁡ρt∣<∞. Thus ρt belongs to the stated density class. Therefore D(ρt∥pβ)=∫pβψ(tu) dνN=t22∫pβu2 dνN+Rt with the stated bound. Since u is nonzero in L2(pβdνN), the quadratic term dominates the cubic bound for sufficiently small nonzero t. The equal-energy identity in step 2.1 then gives S(ρt)<S(pβ).

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Supplied static finite-mode Fermi occupation identity

Statement

This is a finite-mode conditional construction. Assume the Axiom of Countable Choice and the exact current interface for the batch-45 draft Explicit diagonal Fock domains and occupation trace, whose Statement matches its planned manifest entry on qsm-bose-fermi-gases. Its current carrier has a complete authored proof and precheck, plus a Step-1 ready authoring record; those records do not assert independent review approval. Let J={1,…,M} be a nonempty finite set of supplied modes, let m,c>0, let Nj>0 be supplied dimensionless lapse values, and let p^j∈R3 be supplied mode momenta. Put ϵj=m2c4+c2∣p^j∣2,Ej=Njϵj,HF=C2M, with occupation basis ∣n1,…,nM⟩, nj∈{0,1}, and diagonal operators H∣n⟩=(∑jEjnj)∣n⟩, N∣n⟩=(∑jnj)∣n⟩. For βK>0 and finite μK∈R, conditional on that exact supplier result, ΞF=Tr⁡e−βK(H−μKN)=∏j=1M(1+e−βK(Ej−μK))<∞, and the normalized diagonal state has occupation mean E[nj]=1eβK(Ej−μK)+1. With βj=βKNj and μj=μK/Nj, this is exactly E[nj]=1eβj(ϵj−μj)+1. No arbitrary curved-space Dirac spectrum or self-consistent quantum-gravity state is asserted.

Facts & Assumptions

Given: A finite supplied mode set, positive finite mode energies, βK>0, μK∈R, positive lapse values, and the named conditional supplier interface.

[A1]

The current batch-45 draft supplier Explicit diagonal Fock domains and occupation trace assumes ACω; for fermionic modes it assumes qj>0, ∑jqj<∞, and only finitely many gaps ϵj−μ are negative, and concludes ΞF=Tr⁡Gβ,μ=∏j(1+qj)<∞ with normalized diagonal state whose finite-mode occupation probabilities are P(nj=1)=qj/(1+qj). Its exact current Statement matches the planned batch-45 manifest; its authored proof and precheck are present and its Step-1 record says ready, but this is not independent review approval. For our finite set the summability and negative-gap hypotheses hold automatically. We apply the supplier with its mode energy set equal to {Ej} and its chemical potential equal to μK; its interface also allows countably many modes under the stated conditions. (Explicit diagonal Fock domains and occupation trace)

[A2]

The Axiom of Countable Choice, ACω, is the stated assumption used by the current batch-45 QSM draft supplier. (The Axiom of Countable Choice (ACω))

Proof

1.1givenA1A2

For each finite mode, Ej−μK is real and finite, so qj=e−βK(Ej−μK)>0. A finite family of positive numbers has finite sum, and it has only finitely many negative energy gaps. Therefore the finite-mode hypotheses of [A1] hold under [A2]; conditional on the supplier's stated result, which is not independently approved here, it gives the exact trace product and normalized diagonal state.

1.2A1algebra

The supplier's finite fermion marginal is E[nj]=qj/(1+qj)=1/(eβK(Ej−μK)+1) by substituting qj. Since Ej=Njϵj, this exponent is βK(Njϵj−μK).

2.1givenstep 1.2algebra∎

The definitions βj=βKNj and μj=μK/Nj give βj(ϵj−μj)=βK(Njϵj−μK). Substitution proves the displayed local-parameter form exactly for each supplied mode. The statement uses only this finite regulator and makes no claim about an unprovided continuum mode spectrum.

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Finite-bath entropy curvature and specified exchange stability

Statement

Let I1,I2⊆R be open energy intervals (energies in joules), and let Si:Ii→R be C2 entropies measured in joules per kelvin. Fix a total energy Etot in joules and define the open exchange domain I:={x∈I1:Etot−x∈I2}. For x∈I, set E1=x, E2=Etot−x, and Ti(Ei):=1/Si′(Ei) wherever Si′(Ei)>0. At a point x∈I with T1(E1)=T2(E2)=T>0 and Ti′(Ei)≠0, define the finite nonzero heat capacities Ci(Ei):=1/Ti′(Ei)=dEi/dTi. For Stot(x)=S1(x)+S2(Etot−x), one has Stot′′(x)=−1T2(1C1+1C2). In particular, if C1<0<C2, the critical point is a strict local entropy maximum in this energy-exchange direction whenever 0<C2<∣C1∣.

This condition is realized by taking S1=SG from Negative Gibbs-volume capacity in a finite regulated shell at an interior branch energy E∗ (for example, r0/R<δ<1/16), where T∗=TG(E∗)>0 and C1=CG(E∗)<0. The explicit phase-volume formula makes S1 C2 on an open energy interval around E∗. Choose a finite bath entropy Sb(Eb)=Cblog⁡(Eb−E0CbTref)+S0,Eb>E0, with Cb>0 in J K−1, E0 in J, Tref>0 in K, and S0 in J K−1. It has Tb=(Eb−E0)/Cb and heat capacity Cb; choose Eb=E0+CbT∗ to match the subsystem temperature T∗ and choose Cb<∣C1∣.

Separately adopt, on the open domain I and with time t in seconds, the effective scalar exchange equation x˙=LStot′(x),L>0 J K s−1. For each strict local maximum x∗∈I with Stot′′(x∗)<0, there are a,r>0 such that I0=[x∗−r,x∗+r]⊂I, Stot′′≤−a on I0, and every solution with x(0)∈I0 exists for all t≥0, remains in I0, and satisfies ∣x(t)−x∗∣≤e−Lat∣x(0)−x∗∣. Thus the equilibrium solution x(t)≡x∗ attracts those initial states exponentially. Along every solution, dStot/dt=L(Stot′)2≥0. This is stability of the specified one-dimensional exchange law, not of a general relativistic star or a causal transport equation.

Facts & Assumptions

Given: The entropy functions, their positive temperatures and finite nonzero heat capacities, a common-temperature critical point, and the separately adopted scalar exchange equation.

[F1]

At an interior regulated Kepler branch energy E∗, the Gibbs-volume response formulas supply T∗=TG(E∗)>0 and finite C1=CG(E∗)<0. (Negative Gibbs-volume capacity in a finite regulated shell)

[F2]

A continuous vector field locally Lipschitz in its state variable has a unique local solution through every point of its open ODE domain. (Picard-Lindelöf local existence and uniqueness for first-order systems)

[F3]

A Picard-Lindelof solution whose graph approaches a compact subset of its open ODE domain at a finite endpoint extends past that endpoint. (A solution whose graph approaches a compact interior region at a finite endpoint extends past that endpoint)

[F4]

For positive μ,A,r0,h∗ and R>16r0, the branch A/R<e<A/(16r0) has δ=er0/A and Ω(−e)=Ke−3/2J(δ), where K=(4π)(4π/3)(2μ)3/2A3/h∗3>0, J(δ)=∫δ1s1/2(1−s)3/2 ds, and SG=kBlog⁡Ω. (Regulated relative Kepler phase volume and shell law)

Proof

1.1givenalgebra

Since Si′=1/Ti, differentiation with respect to energy gives Si′′=−Ti−2(dTi/dEi)=−1/(Ti2Ci).

2.1givenstep 1.1algebra

The first derivative of Stot is S1′(x)−S2′(Etot−x), so it vanishes exactly at common temperature. The second derivative is the sum S1′′(x)+S2′′(Etot−x), which by step 1.1 is the stated expression. If C1<0<C2, then Stot′′<0 exactly when 1/C2>1/∣C1∣, equivalently C2<∣C1∣; continuity and the second-derivative test give a strict local maximum.

3.1F1F4step 2.1algebra

On 0<δ<1, the function f(δ)=δ1/2(1−δ)3/2 has continuous derivative. The identity J(δ+h)−J(δ)=−∫δδ+hf(s) ds and continuity of f give J′=−f, so J′′=−f′ is continuous; also J>0. Thus [F4] makes Ω positive and C2, and SG=kBlog⁡Ω is C2 on an open interval about the interior energy E∗. For the displayed bath entropy, Sb′=Cb/(Eb−E0), hence Tb=(Eb−E0)/Cb and dEb/dTb=Cb. Choose 0<Cb<∣CG(E∗)∣, Eb∗=E0+CbT∗, and Etot=E∗+Eb∗. Both entropies are C2 near these energies, their positive temperatures agree, and step 2.1 gives strictly negative total entropy curvature.

4.1F2F3givenstep 3.1algebra∎

At the stipulated maximum x∗ with Stot′′(x∗)<0, continuity gives a compact interval I0=[x∗−r,x∗+r]⊂I and a>0 with Stot′′≤−a there. Integrating from x∗, where Stot′=0, gives (x−x∗)Stot′(x)≤−a(x−x∗)2. The vector field LStot′ on R×I is C1 and hence locally Lipschitz, so [F2] gives a unique local solution. At the two endpoints of I0 the field points strictly inward, preventing a first exit, including for endpoint initial data. The chain rule gives dStot/dt=L(Stot′)2≥0 along every solution and, for those in I0, ddt∣x−x∗∣2≤−2La∣x−x∗∣2. Multiplying by e2Lat and differentiating yields ∣x(t)−x∗∣≤e−Lat∣x(0)−x∗∣. If its maximal forward endpoint b were finite, its graph for 0≤t<b would lie in the compact set [0,b]×I0⊂R×I, so [F3] would extend the solution, a contradiction. The local basin therefore exists for all forward times and converges exponentially.

5 · Examples, experiments and counterexamples

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