Tolman from Staticity The Temperature History Is a Theorem of P3 — and What It Cannot Decide
Two Independent Derivations of T·V = constant, the Redshift–Temperature Identity, and the Honest Standing of the Series' Most-Cited Audit
Abstract
The temperature history T(z) =
1What Is Being Claimed, and What It Is Worth
The relation in question is one line:
T(z) =
It says the same bath, looked at further away, is hotter in proportion to its redshift. The series has audited it at z = 2.4 and it passes. The question this note asks is not whether it is true but what its truth buys — and the answer turns out to be both less and more than the ledger currently records. Less, because the relation is a consequence of staticity alone and is therefore blind to everything that distinguishes one static cosmology from another. More, because the cleanest derivation of it is not an import from 1930 but a two-line consequence of this series' own first postulate, which makes
2Route One: the W-Circle (this series' own machinery)
Postulate P1 carries time on the quaternion's real axis as an imaginary value, W = iτ. Section 4 of the Allgemeine Feldtheorie has already drawn the consequence: multiplying by i is a quarter-turn, so time behaves like an angle, and an angle can close. A system at temperature T repeats itself when carried once around a W-circle whose circumference, measured in that system's own proper time, is
β_proper =
This is the Kubo–Martin–Schwinger condition, and in this series it is not borrowed but structural: temperature IS the reciprocal circumference of the W-circle. Now add P3. A static spacetime has a global time symmetry — one timelike Killing vector, hence one Killing time coordinate t shared by every observer, however deep in the field they sit. Continue t to imaginary values and the equilibrium state is a state periodic in that imaginary Killing time. Here is the one physical premise, and it is worth stating alone because everything turns on it: the system is in a single equilibrium, so the W-circle closes once, with one global period in Killing time. Call it β_∞. Two regions with different Killing-time periods would not be one equilibrium; they would be two systems, and the Euclidean section would not close as a single manifold. Equilibrium is exactly the statement that there is one period. The rest is kinematics. Proper time and Killing time are not the same clock: at a point where the lapse is V = √(−g_tt)/
β_proper(x) = V(x) · β_∞
long in the proper time of an observer sitting at x. Combine with the KMS circumference and the lapse cancels out of nothing — it survives:
3Route Two: Hydrostatic Equilibrium (independent, and it needs no quantum mechanics)
The second route uses no algebra of this series and no KMS condition — only the conservation of energy–momentum, which any metric theory supplies. Take a general static, spherically symmetric metric and leave both of its free functions unspecified:
ds² = −
Φ(r) is the potential — the lapse is V = e^Φ — and R(r) is the areal radius, the number that says how big a sphere at coordinate r actually is. Fill this spacetime with a photon gas at local temperature T(r), so that ρc² = aT⁴ and p = aT⁴/3, and demand hydrostatic equilibrium, ∇_μ T^μν = 0. The radial component reads
dp/dr = −(ρc² + p) · dΦ/dr
which is the relativistic Euler equation: pressure gradients hold matter up against the potential, and the inertia being held up includes the pressure itself. Substitute the radiation equation of state. Both sides carry the same factor (4/3)aT⁴ and it cancels, leaving
dT/T = −dΦ ⟹ T·e^Φ = T·V = constant
The same relation, from thermodynamics rather than algebra. The companion script performs this calculation symbolically for the general metric — it solves the conservation equation for T(r) without being told the answer and returns T(r) = C·e^(−Φ(r)) — and then verifies all four components of ∇_μ T^μν vanish identically for Metric D with T(r) =
4Why Redshift and Temperature Carry the Same Factor
Now the step that decides what the audit is worth. In a static spacetime the gravitational redshift is not an independent piece of physics to be computed; it is the lapse, by definition. A photon of proper frequency ν_e emitted at r_e and received at r_o has
1 + z = ν_e / ν_o = V(r_o) / V(r_e)
because the Killing time between successive wave crests is conserved along the ray — that is what a time symmetry means — and each observer converts it to proper time with their own lapse. Put the observer at the origin, normalise V(0) = 1, and 1+z = 1/V(r_e).
Set that beside the result of Sections 2 and 3, T(r)·V(r) =
T(r) =
The relation is exact, and it is exact for the same reason a tautology is: the redshift and the temperature ratio are the same function V, met twice. Nothing was fitted, nothing was measured, no property of the source entered, and no feature of the geometry beyond staticity was used. This is the honest content of the audit. T(z) =
5What the Test Cannot Decide — Stated Plainly
Three consequences follow, and the first two subtract from the ledger. It cannot distinguish Metric D from any other static metric. Section 3 showed that R(r) cancels. Metric D's whole remaining content, once the potential is fixed, is its areal radius R(r) = r·e^(−kr) — the function that produces the angular-diameter turnaround at z = e−1 and everything geometric that follows. Tolman is blind to it. Two static cosmologies with the same potential and wildly different R(r) return exactly the same temperature history. So this audit constrains one of Metric D's two functions and says nothing whatever about the other. It cannot distinguish this cosmology from an expanding one. In a Friedmann universe a blackbody bath cools adiabatically as T ∝ 1/a, and since 1+z = 1/a there also, the prediction is T(z) =
6What the Theorem Does Not Fix: the Normalization
One more limitation, and it points at a wound the series already knows by another name. The constant in T·V = constant is not determined by the theorem. Route one makes the reason transparent: the global period β_∞ is a free parameter of the equilibrium. In one familiar situation it is not free — a spacetime with a Killing horizon fixes it, because the Euclidean section must be smooth where the horizon closes, and that single regularity condition is what produces the Hawking and Gibbons–Hawking temperatures. Metric D has no such closure available at finite proper distance: its lapse e^(−kr) sinks toward zero only as r → ∞, and the structure at r =
7Status of This Note
What is established: that T·V = constant follows from P3 plus equilibrium, by two independent routes, one of them native to P1; that T(z) =
On authorship, since this series' first open item is precisely an independent re-derivation by the author: this note was drafted and its algebra verified by machine. It is a walkthrough to be checked, not a discharge of that item. The item stands until M.S. has walked both routes himself — which, for Route One, is three lines and an afternoon.
8The Sentence
The temperature history is not a measurement the cosmology passed but a shape the postulate could not have avoided: one equilibrium closes the W-circle once, the lapse makes it shorter for whoever stands deeper, and the same lapse that reddens the light warms its source — so T(z) =
References
R. C. Tolman, Phys. Rev. 35, 904 (1930); R. C. Tolman and P. Ehrenfest, Phys. Rev. 36, 1791 (1930); R. Kubo, J. Phys. Soc. Jpn. 12, 570 (1957); P. C. Martin and J. Schwinger, Phys. Rev. 115, 1342 (1959); W. G. Unruh, Phys. Rev. D 14, 870 (1976); G. W. Gibbons and S. W. Hawking, Phys. Rev. D 15, 2738 (1977); and the papers and notes of this series (the Allgemeine Feldtheorie — the theorem tree and
9Verification
The companion scripts, with their recorded output. Each script's docstring states what it establishes and what it does not; the Source tab shows the file itself, unedited.
tolman_check.py — tolman_check
======================================================================
ROUTE 2 (hydrostatic): photon gas at T(r) in a GENERAL static metric
======================================================================
conservation r-component = 4*a*(T(r)*Derivative(Phi(r), r) + Derivative(T(r), r))*T(r)**3/3
solving for T(r): Eq(T(r), C1*exp(-Phi(r)))
=> T * exp(Phi) = const <-- Tolman-Ehrenfest, ANY static metric, any R(r)
(note: R(r) never enters -- the areal radius is irrelevant to Tolman)
======================================================================
METRIC D: the specific case, exp(Phi) = exp(-k r)
======================================================================
with T(r) = T_0 e^{kr}:
(div T)_t = 0
(div T)_r = 0
(div T)_theta = 0
(div T)_phi = 0
lapse V = sqrt(-g_tt)/c = exp(-k*r)
T * V = T_0 <-- constant, independent of r
redshift: 1+z = V(0)/V(r) = exp(k*r)
therefore T(z)/T_0 = exp(k*r) == 1+z IDENTICALLY
The SAME lapse V carries both the redshift and the temperature.
T(z) = T_0 (1+z) is therefore a tautology of staticity, not a fit.
# -*- coding: utf-8 -*-
"""Independent check of the two routes to Tolman equilibrium, T*V = const."""
import sympy as sp
t,r,th,ph = sp.symbols('t r theta phi', positive=True)
c,k,a,T0 = sp.symbols('c k a T_0', positive=True)
X=(t,r,th,ph); n=4
def christoffel(g):
gi=g.inv()
return [[[sp.cancel(sum(gi[A,d]*(sp.diff(g[d,B],X[C])+sp.diff(g[d,C],X[B])-sp.diff(g[B,C],X[d]))/2
for d in range(n))) for C in range(n)] for B in range(n)] for A in range(n)]
def div_mixed(Tmix, Gam, g):
"""(nabla_mu T^mu_nu) for a mixed tensor."""
out=[]
for nu in range(n):
e=sum(sp.diff(Tmix[mu,nu],X[mu]) for mu in range(n))
e+=sum(Gam[mu][mu][lam]*Tmix[lam,nu] for mu in range(n) for lam in range(n))
e-=sum(Gam[lam][mu][nu]*Tmix[mu,lam] for mu in range(n) for lam in range(n))
out.append(sp.simplify(e))
return out
print("="*70)
print("ROUTE 2 (hydrostatic): photon gas at T(r) in a GENERAL static metric")
print("="*70)
Phi=sp.Function('Phi')(r); R=sp.Function('R')(r); T=sp.Function('T')(r)
g=sp.diag(-c**2*sp.exp(2*Phi), 1, R**2, R**2*sp.sin(th)**2)
Gam=christoffel(g)
rho=a*T**4; p=a*T**4/3 # radiation: rho c^2 = aT^4, p = aT^4/3
Tmix=sp.diag(-rho, p, p, p)
d=div_mixed(Tmix,Gam,g)
eq=sp.simplify(d[1])
print(" conservation r-component =", eq)
sol=sp.dsolve(sp.Eq(eq,0), T)
print(" solving for T(r):", sol)
print(" => T * exp(Phi) = const <-- Tolman-Ehrenfest, ANY static metric, any R(r)")
print(" (note: R(r) never enters -- the areal radius is irrelevant to Tolman)")
print()
print("="*70)
print("METRIC D: the specific case, exp(Phi) = exp(-k r)")
print("="*70)
gD=sp.diag(-c**2*sp.exp(-2*k*r), 1, sp.exp(-2*k*r)*r**2, sp.exp(-2*k*r)*r**2*sp.sin(th)**2)
GamD=christoffel(gD)
TD=T0*sp.exp(k*r) # the claimed history
rhoD=a*TD**4; pD=a*TD**4/3
TmixD=sp.diag(-rhoD,pD,pD,pD)
dD=div_mixed(TmixD,GamD,gD)
print(" with T(r) = T_0 e^{kr}:")
for i,nm in enumerate(['t','r','theta','phi']):
print(f" (div T)_{nm} = {sp.simplify(dD[i])}")
V=sp.sqrt(-gD[0,0])/c
print(f"\n lapse V = sqrt(-g_tt)/c = {sp.simplify(V)}")
print(f" T * V = {sp.simplify(TD*V)} <-- constant, independent of r")
print(f"\n redshift: 1+z = V(0)/V(r) = {sp.simplify(V.subs(r,0)/V)}")
print(f" therefore T(z)/T_0 = {sp.simplify(TD/T0)} == 1+z IDENTICALLY")
print("\n The SAME lapse V carries both the redshift and the temperature.")
print(" T(z) = T_0 (1+z) is therefore a tautology of staticity, not a fit.")