The Anchor's First Test And the Constant It Would Have Eliminated:

Scholl, Martin

The Anchor's First Test And the Constant It Would Have Eliminated

Eliminating the Hubble Rate Between a₀ and the Geometric Mean — an Over-Constrained Test, Currently Failed, and One Robustness Result That Survives It

Martin Scholl — Independent Researcher · It Is All One — Cosmology Notes · July 2026 (working draft)

Abstract

The geometric-mean anchor, √(T_Planck·T_horizon) ≈ 19.4 K against a measured 2.725 K, has sat in this corpus's quarantine with a value and no explanation — in particular with an unexplained factor of 7.12 between the two. This note reports that the factor has a candidate identity, that the identity yields a closed-form prediction for the one constant the corpus calls a free input, and that the prediction then fails an over-constrained test against a₀. The identity: the curvature of metric D's radial-transverse slice changes sign at exactly two e-foldings, where the temperature is T₀·e² = 20.14 K, within 3.8% of the geometric mean. Taking that seriously turns the anchor into T₀ = √(T_Planck·T_horizon)/e², which eliminates T₀ as a free parameter — the anchor problem's own constant, gone. The price is that the Hubble rate becomes over-determined, and it does not survive: eliminating H between this relation and a₀ = cH/2e gives a₀ = π·c·k_B·T₀²·e³/(ħ·T_P), predicting 1.2985×10⁻¹⁰ against an observed 1.2×10⁻¹⁰, high by 8.2%. Read as three routes to H: a₀ gives 67.15 km/s/Mpc, direct measurement 67.40, the anchor 72.66. Two agree to 0.4% and the anchor is the outlier, so the suspect is the factor e², which is to say the areal radius. A natural one-parameter family of seals is then tested and cannot rescue it: the anchor requires p = 1.102 while a regular centre requires p ≥ 2, and the two are incompatible. What survives, and is worth more than the coincidence, is a robustness result discovered in the same family: the audited angular turnaround sits at exactly one curvature radius for every member, so z = e−1 is not a fitted number and cannot become one. Every flag flown.

1The Unexplained Factor, and a Candidate for It

The anchor has always been quoted as a near-miss of the wrong size: √(T_Planck·T_horizon) = 19.396 K against T₀ = 2.72548 K — a ratio of 7.1166 that nobody could name. The Core of Metric D note supplies a candidate. The Gaussian curvature of the radial-transverse slice, K_G = −R″/R, is proportional to (kr − 2) and therefore changes sign at exactly two e-foldings, where the temperature along the sight-line is

T = T₀ · e² = 20.139 K , against √(T_Planck·T_horizon) = 19.396 K (+3.83%)

Or, in distance rather than temperature: the geometric mean sits at 1.9624 e-foldings and the curvature flip at 2.0000, a discrepancy of 1.9%. The unexplained factor of seven has a candidate identity: two e-foldings, from the observer to the surface where the geometry stops being sphere-like.

2The Constant That Would Be Eliminated

Taken as exact, √(T_P·T_hor) = T₀·e² solves for T₀ — the number the Tolman note proved is otherwise structurally free, because the equilibrium's global period is a free parameter and metric D has no Killing horizon at finite distance to fix it:

T₀ = √(T_Planck · T_horizon)/e² = ħ^(3/4)·c^(5/4)·H^(1/2) / ( e²·√(2π)·G^(1/4)·k_B )

This is the anchor problem's target: the CMB temperature in closed form from ħ, c, G, k_B and H alone. Note the scaling, T₀ ∝ √H — a weak lever, which matters below.

3The Test: Eliminate H

Two relations of this corpus now contain H. Set them against each other and the Hubble rate cancels:

a₀ = π·c·k_B·T₀²·e³ / (ħ·T_P)

a relation between the rotation-curve scale and the background temperature with no cosmological parameter in it at all. It predicts a₀ = 1.2985×10⁻¹⁰ m/s² against an observed 1.2×10⁻¹⁰: high by 8.21%. Equivalently, as three independent routes to the Hubble rate: from a₀ = cH/2e, 67.15 km/s/Mpc; from direct measurement, 67.40; from the anchor, 72.66. Two agree with each other to 0.4% and the third is 8% away. The anchor is the outlier, so the suspect is not a₀ and not the Hubble rate but the factor e² — which is to say the areal radius R(r). It is worth naming the tempting move and refusing it. The anchor becomes exact at H = 72.66 km/s/Mpc, which is close to the local distance-ladder value, and there is even a respectable argument for preferring it, since a static cosmology has no business inheriting a Hubble rate obtained by fitting an expanding model to the microwave sky. But H is not free: raising it by 7.8% to rescue the anchor pushes a₀ off by the same 7.8%, and a₀ currently sits at +0.37%, one of the corpus's better lines. One knob, two duties, pulling opposite ways.

4Can the Seal Absorb It? No

The data demands the flip at 1.9624 e-foldings rather than 2 — a −1.9% correction to the seal. Test the natural one-parameter family, leaving the lapse and hence the redshift untouched:

R(r) = r · exp( −(kr)^p / p )

Its properties come out in three lines. The turnaround, R′ = 0, occurs where x^p = 1, hence at x = 1 for every p. The flip, R″ = 0, occurs where x^p = p + 1, hence at x = (p+1)^(1/p). And the near-origin source behaves as −2(p + 2 + 1/p)·x^(p−2), which reproduces the known −8k/r at p = 1 and is finite only for p ≥ 2. The anchor requires the flip at 1.9624, hence p = 1.1017. A regular centre — caveat (i-a), closed only this morning — requires p ≥ 2, which puts the flip at √3 = 1.732 and the temperature there at 15.4 K. The two duties are incompatible in this family. Buying the anchor means reinstating the singularity that was just removed.

5What Survives: the Turnaround Cannot Be Fitted

The first line of §4 is a gift, and it is independent of everything that failed. The turnaround sits at x = 1 for every member of the family — for every exponent p, without adjustment. The audited angular turnaround at z = e − 1 = 1.71828 is therefore robust against this entire class of corrections to the seal. It was not fitted, and it cannot be made a fitted number by any deformation of this kind. That is worth recording in the ledger on its own.

6Verdict, for the Quarantine Card

The geometric-mean anchor was given its first quantitative test and did not pass. It is 8% adrift of two constants that agree with each other to 0.4%; the seal correction it demands is excluded by the regularity condition; and its whole chain is conditional on an areal radius that is not derived. It moves from quarantined-and-neutral to failing-its-first-test. It is not dead — kill it properly only when R(r) is known — but it should no longer be listed as though it were merely awaiting attention.

openWhat is owed if anyone wants to revive it: a derived R(r) whose curvature flip lands at 1.9624 e-foldings and whose centre is regular. Those two conditions are not satisfied together by any member of the family tested here, and any candidate must satisfy both.

7The Sentence

The anchor's unexplained factor of seven turned out to have a name — two e-foldings, to the surface where the geometry stops being sphere-like — and the name was good enough to eliminate the one constant this cosmology calls free, until the elimination was carried through to a₀ and came out eight percent high, which is how an over-constrained programme is supposed to behave when a coincidence is only a coincidence.

References

The papers and notes of this series (the Allgemeine Feldtheorie — the quarantine list and caveat (iii); The Core of Metric D — the regularity condition and the curvature flip; Tolman from Staticity — why T₀ is otherwise a free input; The Quaternion Screw — a₀ = cH/2e; The Family Law's Cosmic Rung). Verification script: Cosmology/anchor_elimination_check.py. (Citations from memory; the literature-verification pass applies.) Acknowledgment: the observation that the over-constraint would eliminate a constant is the author's. Computation and drafting by machine (Claude, Anthropic).

8Verification

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.

anchor_elimination_check.py — anchor_elimination_check

runs in your browser

==========================================================================
1. THE ELIMINATION — H cancels, and a0 meets T0 directly
==========================================================================
   a0 = cH/(2e)                     ->  H = 2e a0 / c
   T0 = sqrt(T_P T_hor)/e^2,  T_hor = hbar H/(2 pi kB)
                                    ->  H = 2 pi kB T0^2 e^4/(hbar T_P)
   equate:
        a0  =  pi c kB T0^2 e^3 / (hbar T_P)      <-- NO H

   a0 predicted = 1.2985e-10 m/s^2
   a0 observed  = 1.2000e-10 m/s^2      deviation +8.21%

==========================================================================
2. WHICH CONSTANT IS ELIMINATED
==========================================================================
   Before: T0 is a free input (the anchor problem) AND a0's denominator is fitted.
   After : T0 is PREDICTED from H. The anchor's constant is gone -- that is the
           one constant eliminated. The price is that H becomes over-determined.

   three independent routes to H:
     from a0 = cH/2e (obs a0)       H =   67.15 km/s/Mpc
     direct measurement (corpus)    H =   67.40 km/s/Mpc
     from the anchor T0=.../e^2     H =   72.66 km/s/Mpc

   a0 and the direct measurement agree to 0.4%. The ANCHOR is the odd one out,
   by +8%. So the suspect is the anchor's factor e^2 -- i.e. R(r).

==========================================================================
3. WHAT THE DATA DEMANDS OF R(r)
==========================================================================
   flip must sit at x = ln(sqrt(T_P T_hor)/T0) = 1.9624, not 2.0000
   a -1.88% correction to the seal.

==========================================================================
4. A ONE-PARAMETER FAMILY THAT CAN CARRY IT
==========================================================================
   try  R = r exp(-(kr)^p / p)   [the lapse, hence the redshift, is untouched]
     R'  = (1 - x**p)*exp(-x**p/p)
     turnaround R'=0  ->  x^p = 1  ->  x = 1  FOR EVERY p.
        => the audited z = e-1 is automatic, not fitted. 
     R'' = -x**(p - 1)*(p - x**p + 1)*exp(-x**p/p)
     flip R''=0  ->  x^p = p+1  ->  x_flip = (p+1)^(1/p)
        p = 1.0   x_flip = 2.0000   T_flip =  20.139 K
        p = 1.1   x_flip = 1.9630   T_flip =  19.408 K
        p = 1.5   x_flip = 1.8420   T_flip =  17.196 K
        p = 2.0   x_flip = 1.7321   T_flip =  15.405 K
        p = 3.0   x_flip = 1.5874   T_flip =  13.330 K

   p that puts the flip on the anchor: p = 1.1017

==========================================================================
5. THE SECOND DUTY: does the same p fix the near field?
==========================================================================
   near x->0,  G^t_t ~ x^(p-2)  -> finite ONLY for p >= 2.
     p = 1.0000:  G^t_t(x=1e-3) =     -7999.00   diverging
     p = 1.1017:  G^t_t(x=1e-3) =     -3971.67   diverging
     p = 1.5000:  G^t_t(x=1e-3) =      -263.52   FINITE
     p = 2.0000:  G^t_t(x=1e-3) =        -9.00   FINITE

   p = 2 gives a regular centre, but puts the flip at (3)^(1/2) = 1.7321
   the anchor needs p = 1.102.  THE TWO DUTIES ARE INCOMPATIBLE in this family.

# -*- coding: utf-8 -*-
"""The anchor's joint test with a0 -- and the constant it would have eliminated.

Two relations of this corpus both contain H:
      a0  = c H / (2e)                        (the rotation-curve scale)
      T0  = sqrt(T_Planck x T_horizon) / e^2  (the geometric-mean anchor, IF the
                                               curvature sign flip at two
                                               e-foldings is what fixes it)
Eliminating H between them gives a relation with no Hubble rate in it at all:
      a0 = pi c k_B T0^2 e^3 / (hbar T_P)
and predicts a0 = 1.2985e-10 against an observed 1.2e-10: +8.21%.

WHAT WOULD HAVE BEEN ELIMINATED: T0 itself -- the anchor problem's own constant,
which the Tolman note proved is otherwise a free input. The price is that H
becomes over-determined, and the three routes do not agree:
      from a0 = cH/2e            67.15 km/s/Mpc
      direct measurement         67.40
      from the anchor            72.66
Two agree to 0.4%; the anchor is 8% out. The suspect is therefore the factor
e^2 -- which is to say R(r).

THE FAMILY, AND WHY IT FAILS. Try R = r exp(-(kr)^p/p), leaving the lapse and
hence the redshift untouched. Then:
      turnaround  R' = 0  <=>  x^p = 1  <=>  x = 1   FOR EVERY p
      flip        R''= 0  <=>  x^p = p+1 <=> x = (p+1)^(1/p)
      near-origin source  G^t_t ~ -2(p + 2 + 1/p) x^(p-2)
The first line is a gift and survives everything below: the audited turnaround
at z = e-1 sits at x = 1 for every member of the family, so that number is not
fitted and cannot be. The rest is a failure: the anchor needs p = 1.1017, a
regular centre needs p >= 2 (strictly), and the two are incompatible.

VERDICT ON THE GEOMETRIC-MEAN ANCHOR: it was given a chance to earn its keep and
did not. It is 8% adrift of two constants that agree with each other to 0.4%,
and it demands a seal that reinstates the very singularity caveat (i-a) closed.
Moved from quarantined to failing-its-first-test. Not dead -- the whole chain is
conditional on R(r), which is not derived -- but no longer neutral.
"""
import numpy as np, sympy as sp
hbar=1.054571817e-34; c=2.99792458e8; kB=1.380649e-23; G=6.67430e-11
Mpc=3.0856775814913673e22; T0=2.72548; a0_obs=1.2e-10; H_corpus=2.1843e-18
T_P=np.sqrt(hbar*c**5/(G*kB**2))

print("="*74); print("1. THE ELIMINATION — H cancels, and a0 meets T0 directly"); print("="*74)
print("   a0 = cH/(2e)                     ->  H = 2e a0 / c")
print("   T0 = sqrt(T_P T_hor)/e^2,  T_hor = hbar H/(2 pi kB)")
print("                                    ->  H = 2 pi kB T0^2 e^4/(hbar T_P)")
print("   equate:")
print("        a0  =  pi c kB T0^2 e^3 / (hbar T_P)      <-- NO H")
a0_pred=np.pi*c*kB*T0**2*np.e**3/(hbar*T_P)
print(f"\n   a0 predicted = {a0_pred:.4e} m/s^2")
print(f"   a0 observed  = {a0_obs:.4e} m/s^2      deviation {100*(a0_pred/a0_obs-1):+.2f}%")

print("\n"+"="*74); print("2. WHICH CONSTANT IS ELIMINATED"); print("="*74)
print("   Before: T0 is a free input (the anchor problem) AND a0's denominator is fitted.")
print("   After : T0 is PREDICTED from H. The anchor's constant is gone -- that is the")
print("           one constant eliminated. The price is that H becomes over-determined.")
print("\n   three independent routes to H:")
for nm,H in [("from a0 = cH/2e (obs a0)", 2*np.e*a0_obs/c),
             ("direct measurement (corpus)", H_corpus),
             ("from the anchor T0=.../e^2", 2*np.pi*kB*T0**2*np.e**4/(hbar*T_P))]:
    print(f"     {nm:<30} H = {H*Mpc/1e3:>7.2f} km/s/Mpc")
print("\n   a0 and the direct measurement agree to 0.4%. The ANCHOR is the odd one out,")
print("   by +8%. So the suspect is the anchor's factor e^2 -- i.e. R(r).")

print("\n"+"="*74); print("3. WHAT THE DATA DEMANDS OF R(r)"); print("="*74)
B=np.sqrt(T_P*hbar*H_corpus/(2*np.pi*kB))/T0
print(f"   flip must sit at x = ln(sqrt(T_P T_hor)/T0) = {np.log(B):.4f}, not 2.0000")
print(f"   a {100*(np.log(B)/2-1):+.2f}% correction to the seal.")

print("\n"+"="*74); print("4. A ONE-PARAMETER FAMILY THAT CAN CARRY IT"); print("="*74)
x,p=sp.symbols('x p',positive=True)
g=x**p/p; R=x*sp.exp(-g)
Rp=sp.simplify(sp.diff(R,x)); Rpp=sp.simplify(sp.diff(R,x,2))
print(f"   try  R = r exp(-(kr)^p / p)   [the lapse, hence the redshift, is untouched]")
print(f"     R'  = {sp.simplify(Rp)}")
print(f"     turnaround R'=0  ->  x^p = 1  ->  x = 1  FOR EVERY p.")
print(f"        => the audited z = e-1 is automatic, not fitted. ")
print(f"     R'' = {sp.factor(sp.simplify(Rpp))}")
print(f"     flip R''=0  ->  x^p = p+1  ->  x_flip = (p+1)^(1/p)")
for pv in (1.0,1.1,1.5,2.0,3.0):
    print(f"        p = {pv:<4}  x_flip = {(pv+1)**(1/pv):.4f}   T_flip = {T0*np.exp((pv+1)**(1/pv)):7.3f} K")
f=lambda pv:(pv+1)**(1/pv)-np.log(B)
from scipy.optimize import brentq
p_anchor=brentq(f,1.001,3.0)
print(f"\n   p that puts the flip on the anchor: p = {p_anchor:.4f}")

print("\n"+"="*74); print("5. THE SECOND DUTY: does the same p fix the near field?"); print("="*74)
print("   near x->0,  G^t_t ~ x^(p-2)  -> finite ONLY for p >= 2.")
for pv in (1.0,p_anchor,1.5,2.0):
    Rf=lambda xx: xx*np.exp(-xx**pv/pv)
    h=1e-6; xs=1e-3
    d1=(Rf(xs+h)-Rf(xs-h))/(2*h); d2=(Rf(xs+h)-2*Rf(xs)+Rf(xs-h))/h**2
    Gt=(2*Rf(xs)*d2+d1**2-1)/Rf(xs)**2
    print(f"     p = {pv:.4f}:  G^t_t(x=1e-3) = {Gt:>12.2f}   {'FINITE' if abs(Gt)<1e3 else 'diverging'}")
print(f"\n   p = 2 gives a regular centre, but puts the flip at (3)^(1/2) = {3**0.5:.4f}")
print(f"   the anchor needs p = {p_anchor:.3f}.  THE TWO DUTIES ARE INCOMPATIBLE in this family.")