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Selected stationary thermal equilibrium — Examples

1 · Prerequisites

2 · Summary

These examples compute the exact lapse, local temperature, and observer acceleration in the Schwarzschild exterior, the Rindler right wedge, and a rigidly rotating Minkowski domain. Each uses the selected Killing flow and stays inside its timelike domain. The rotating quantum trace check is restricted to the cited one-mode model; it is not a continuum-field theorem. No horizon, Hawking, Unruh, or reported experimental temperature is inferred.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: Literature-sourcedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Schwarzschild supported thermal bath

Physical scope

Static test bath on the Schwarzschild exterior r>r_s, with asymptotic Killing-time normalization, exact lapse and acceleration, and a bounded weak-field expansion; no horizon or quantum-temperature claim.

Example

Let ms>0 be a source mass in kilograms, distinct from any kinetic particle mass, let GN>0 be Newton's constant in m3 kg−1 s−2, and let c>0 be the speed of light in m s−1. Set

rs:=2GNmsc2,f(r):=1−rsr,r>rs.

In physical length coordinates x0=ct, use the Schwarzschild exterior metric

ds2=−f(r)(dx0)2+f(r)−1dr2+r2(dθ2+sin⁡2θ dϕ2),

and select ξ=∂x0 with the asymptotic normalization Nξ→1 as r→∞. The time orientation is chosen so ξ is future-directed. Under the selected stationary-bath assumption, the local proper temperature satisfies

T(r)=T∞1−rs/r,T∞:=TK>0.

The static observer congruence has outward covariant acceleration component ar=rs/(2r2f), contravariant component ar=rs/(2r2), and physical proper-acceleration magnitude

c2gabaaab=GNmsr2f(r).

For a fixed z0∈(0,1) and 0<rs/r≤z0, the weak-field expansion has the explicit bound

∣T(r)T∞−1−GNmsc2r∣≤38(1−z0)−5/2(rsr)2.

This is a conditional calculation in the specified exterior model. At r=rs, the chosen Killing field is null and the normalized stationary observers do not exist. The limiting divergence as r↓rs is not a constructed horizon state, a Hawking temperature, or a claim that every observer measures an infinite temperature.

Facts & Assumptions

Given: The Schwarzschild exterior metric and constants above, the chosen future Killing field ξ, and the selected stationary thermal-bath assumption on r>rs.

[F1]

Conditional Tolman–Ehrenfest relation: for one selected connected stationary bath with smooth future timelike Killing field ξ, positive proper temperature T, and Killing β=u/(kBT), the constant TK>0 satisfies NξT=TK and a=∇log⁡Nξ. Its draft proof is conditional and its current Step 1 record is authoring readiness only.

[A1]

ms>0, GN>0, and c>0 have the stated SI units; the metric is taken as the model geometry on r>rs, and the time orientation makes ∂x0 future-directed there.

[A2]

This example adopts the selected bath only on the connected exterior, with the chosen asymptotic normalization. No matter equation of state, support apparatus, experiment, or state construction is assumed.

Verification

technique · Compute the norm of the chosen Killing field, apply the conditional Tolman theorem, and differentiate the explicit lapse
1.1givenA1algebra

The metric coefficients are independent of x0, so ξ=∂x0 is Killing. Its norm is g(ξ,ξ)=−f(r), and f(r)>0 exactly on r>rs. Therefore it is future timelike on this domain and Nξ=f(r).

2.1givenA2F1step 1.1

The connected exterior, smooth positive lapse, selected stationary bath, and Killing inverse-temperature field meet [F1]. Hence NξT=TK throughout the exterior, so T=TK/f. Since Nξ→1 as r→∞, write this reference constant as T∞=TK.

2.2F1step 1.1algebra

From a=∇log⁡Nξ in [F1], ar=∂r(12log⁡f)=rs/(2r2f). Raising the index with grr=f gives ar=rs/(2r2), and then gabaaab=f−1(ar)2, so c2∣a∣=c2rs/(2r2f)=GNms/(r2f).

3.1givenalgebra∎

Put z=rs/r. For 0<z≤z0<1, Taylor's theorem at z=0 for φ(z)=(1−z)−1/2 gives φ(z)=1+z/2+φ′′(θz)z2/2 for some θ∈(0,1), and φ′′(y)=34(1−y)−5/2. Thus the remainder is at most 38(1−z0)−5/2z2. Since z/2=GNms/(c2r), this proves the displayed bound.

ExampleConstruction: Literature-sourcedVerification: AI-generatedOpen item page →

Rindler and rigid rotating bath domains

Physical scope

Rindler right wedge and rigidly rotating Minkowski domain inside its light cylinder; arbitrary positive classical reference temperature; differential rotation and one-mode trace restrictions kept distinct.

Example

Rindler right wedge. Let a0>0 be an acceleration in m s−2, use coordinates (x0,x,y,z) with x0=ct, and take the open domain x>0 with metric

ds2=−(a0xc2)2(dx0)2+dx2+dy2+dz2.

The selected Killing field ξR=∂x0 has lapse NR=a0x/c2 and observer acceleration ax=1/x, so the physical proper-acceleration magnitude is c2/x. The selected stationary-bath model gives

TR(x)=TKc2a0x.

At the reference point xref=c2/a0, NR=1 and TR=TK. Any TK>0 is allowed by this classical geometric model; the metric and temperature relation do not select the Unruh value.

Rigid rotation in Minkowski spacetime. In cylindrical coordinates let

ds2=−(dx0)2+dr2+r2dϕ2+dz2,ξΩ=∂x0+Ωc∂ϕ,

where Ω is a constant angular velocity in s−1. For Ω≠0, use the annular domain 0<r<c/∣Ω∣; for Ω=0, ξΩ=∂x0 is timelike throughout Minkowski space. On the rotating domain,

NΩ=1−Ω2r2c2,TΩ(r)=TKNΩ.

The radial acceleration component is ar=−Ω2r/[c2(1−Ω2r2/c2)], directed inward, with physical magnitude γ2Ω2r, where γ=(1−Ω2r2/c2)−1/2. At the light cylinder r=c/∣Ω∣, ξΩ becomes null and the normalized stationary observer does not exist. A differential-rotation field ∂x0+[Ω(r)/c]∂ϕ is not Killing when Ω′(r)≠0: its metric Lie derivative has (Lξg)rϕ=r2Ω′(r)/c.

One-mode trace check. If an additional model specifies a single bosonic mode with lab Hamiltonian H=ϵN and angular momentum J=ℏjN, then its rotating generator is H−ΩJ. Its Gibbs trace is finite exactly when ϵ−ℏjΩ>0. This is the separate one-mode spectral condition of Rotating mode Gibbs obstruction; it does not prove a continuum-field trace obstruction or describe every rotating equilibrium state.

Domain sketches

The left panel uses dimensionless Minkowski coordinates (X/ℓ,X0/ℓ) for a fixed length ℓ>0 and the exact Rindler embedding

X=xcosh⁡(a0x0/c2),X0=xsinh⁡(a0x0/c2).

It maps x>0 onto the right wedge X>∣X0∣; the red curve is the constant-Rindler-position orbit x=ℓ, which obeys X2−(X0)2=ℓ2. The right panel is a fixed-z cross-section of the Ω≠0 rotating timelike domain. Its dashed circle is the excluded null boundary r=c/∣Ω∣, and the shaded open disk is where g(ξΩ,ξΩ)<0. The sketches show domains only; every lapse, acceleration, and temperature claim is established by the coordinate calculations in Verification.

RindlerrightwedgeX=`X0=`rightwedgex=`X=X0X=¡X0Rigidrotationat¯xedzXYc=j¬jtimelikenullboundary

Facts & Assumptions

Given: The displayed Rindler and Minkowski metrics, the constant parameters a0>0 and Ω, and a selected stationary bath only where its stated Killing field is future timelike.

[F1]

Conditional Tolman–Ehrenfest relation: on a selected connected stationary bath, NξT=TK and a=∇log⁡Nξ for the chosen future timelike Killing field. Its physical conclusion is conditional on the postulated bath, not a universal thermometer law.

[F2]

Rotating mode Gibbs obstruction: for a single bosonic occupation mode with ϵ>0, integer mode number, Ω in s−1, and positive inverse temperature, the diagonal partition sum is finite exactly when ϵ−ℏjΩ>0. The proposition explicitly constructs no continuum field state.

[A1]

The Rindler metric and the rotating Minkowski metric are the exact model geometries used in this example; ξR and ξΩ are the explicitly selected Killing generators on their respective timelike domains.

[A2]

The Rindler and rotating temperature fields are conditional classical bath assumptions. No equilibrium state is constructed, and no Unruh or Hawking temperature is inferred.

[A3]

The one-mode trace paragraph applies only when the additional Hamiltonian and angular-momentum assignments displayed in the Example are adopted.

Verification

technique · Compute the lapse and acceleration in each exact geometry, apply the conditional Tolman relation, and keep the single-mode trace premise separate
1.1givenA1algebra

The Rindler metric is independent of x0, so ξR is Killing. Its norm is g(ξR,ξR)=−(a0x/c2)2<0 on x>0, hence NR=a0x/c2. Differentiating log⁡NR gives ax=1/x, whose metric norm is 1/x.

1.2givenA1algebra

In the rotating Minkowski metric, ∂x0 and ∂ϕ are Killing fields and Ω is constant, so their displayed linear combination ξΩ is Killing. Its squared norm is −1+Ω2r2/c2, which is negative exactly for r<c/∣Ω∣ when Ω≠0; therefore NΩ=1−Ω2r2/c2.

1.3givenalgebra

For a nonconstant angular velocity, the coordinate formula for the Lie derivative gives (Lξg)rϕ=gϕϕ∂r(Ω/c)=r2Ω′(r)/c. Thus if Ω′(r)≠0, this field is not Killing and the selected stationary theorem does not apply.

2.1F1A2step 1.1algebra

By [F1] and [A2], the selected Rindler bath has TR=TK/NR=TKc2/(a0x) and physical proper acceleration c2∣a∣=c2/x. The point xref=c2/a0 has NR=1, so it provides the stated reference normalization.

2.2F1step 1.2algebra

Applying [F1] on that selected timelike domain yields TΩ=TK/NΩ. Differentiating log⁡NΩ gives ar=−Ω2r/[c2(1−Ω2r2/c2)]. Since grr=1, its physical magnitude is c2∣ar∣=γ2Ω2r. At the light cylinder the norm tends to zero and the timelike-field hypothesis of [F1] fails.

3.1F2A3step 1.2∎

Under [A3], the mode generator has coefficient a=ϵ−ℏjΩ in the exact occupation model of [F2]. Applying [F2] gives a finite trace exactly for a>0. This condition is logically separate from the geometric timelike-domain condition in step 1.2 and supplies no continuum-field conclusion.

Sources