c·t = CONST, THAT CLICKS EPISODE 46 ／ Part VI — examining the procedure

We found where the plausibility came from

Twelve statements,
three inputs Collect every way of saying $a\propto t$ and count the independent ones.
And you find out why $c\cdot t=$const looked so plausible.

What you need: Episode 19's scale, Episode 25's ledger, Episode 26's compression, Episode 33's three-step testOnly $H_0t_0$ misses by as little as 4.9 per cent

Part VI puts the procedure itself under examination. First, we look head-on at the $a\propto t$ this series has spent 45 episodes on — how many different ways are there to say it? We collect them all, count them, and count how many are independent. And what emerges is why $c\cdot t=$const looked so plausible in the first place.

01Collecting every way of saying $a\propto t$

No.	Statement	What it needs	Kind
A1	$a(t)\propto t$	the definition itself	kinematic
A2	$\ddot a=0$	differentiate twice	kinematic
A3	deceleration parameter $q=-\ddot aa/\dot a^2=0$	a rewriting of A2	kinematic
A4	$H=\dot a/a=1/t$	differentiate once	kinematic
A5	$H\cdot t=1$	a rewriting of A4 (dimensionless)	kinematic
A6	comoving Hubble radius $(aH)^{-1}$ constant	because $aH=1$	kinematic
A7	conformal time $\eta=\int dt/a=\ln t$	just integrate	kinematic
A8	Hubble radius $c/H=c\,t$ exactly	a rewriting of A4	kinematic
A9	the particle horizon diverges	$\int dt/t$ diverges	kinematic
B1	equation of state $w=-1/3$	needs the Friedmann equations	dynamical
C1	$R=6(1+k)/t^2$ (Episode 33)	needs $k$ as well	geometric
C2	$k=-1$ gives $R=0$ (Milne)	a special case of C1	geometric

Conclusion of §01

Twelve in all. Of those, nine (A1–A9) are equivalent by kinematics alone — mere rewritings by differentiation and integration.
── On Episode 19's scale, A1 to A9 are 0 bits apart (identities).

02How many independent inputs?

The expansion law $a\propto t$this alone yields all of A1–A9

The Einstein equationsyields B1 ($w=-1/3$) — dynamics are needed

The spatial curvature $k$yields C1 and C2 — geometry as well

Compression $$\frac{12\ \text{statements}}{3\ \text{inputs}}=\mathbf{4.0\times}$$

After Episode 26 (24 → 12), Episode 40 (three $10^{122}$s → one) and Episode 41 (four → one), this is the fourth compression — and this time it is applied to the series' own subject.

03The two $a\propto t$s are different things

Which $a\propto t$	$R$	Riemann	In Episode 33
Milne ($k=-1$, $\rho=0$)	$0$	zero	Step 1: Minkowski in disguise
$k=0$ with $w=-1/3$ matter	$6/t^2\ne0$	$\ne0$	Step 2: genuinely curved

"$a\propto t$" is one condition, but the spacetimes satisfying it are not one thing. Empty, it is flat; with matter in it, it is curved — the same expansion law, different spacetimes.

◇　◇　◇

04The core — which can be tested, and by how much do they miss?

Characterisation	Prediction	Observed	$\sigma$	bits
$q=0$ (A3)	$0$	$-0.53\pm0.04$	$13.2$	$130$
$w=-1/3$ (B1)	$-0.333$	$-1.03\pm0.03$	$23.2$	$395$
$H_0t_0=1$ (A5, Planck)	$1$	$0.951\pm0.007$	$6.8$	$36$
$H_0t_0=1$ (A5, local)	$1$	$1.030\pm0.014$	$2.1$	$4.9$

The main point of this episode

The Hubble tension straddles $H_0t_0=1$ exactly.
Planck's value gives 0.951 (below); the local measurement gives 1.030 (above).
── Meanwhile $q=0$ misses by 13$\sigma$ and $w=-1/3$ by 23$\sigma$ — by orders of magnitude.

05Why is $H_0t_0=1$ the only near miss?

Compute $H_0t_0$ in $\Lambda$CDM $$H_0t_0=\frac{2}{3\sqrt{\Omega_\Lambda}}\,\mathrm{asinh}\sqrt{\frac{\Omega_\Lambda}{\Omega_m}}=0.9510$$

$a\propto t$ predicts 1 — a miss of only 4.9 per cent

∫

$H_0t_0$ integrates the whole history of the universethe decelerating and accelerating eras cancel

now

$q$ and $w$ are quantities of this instantthere is nothing for them to cancel against

◎

Which is why $c\cdot t=$const looked plausiblenearly right on the integrated observable, out by orders on the instantaneous ones — and the previous series' verdict (it contradicts nucleosynthesis) was about that instantaneous side

Figure: $\Lambda$CDM's $a(t)$ and the straight line $a\propto t$. Match height and slope at today (right edge) and the middle still diverges — yet by area (that is, by integral) they nearly agree. Move the slider — $q$ is the curvature of an instant; $H_0t_0$ is the whole history.

time $t/t_0$ 1.00

$\Lambda$CDM $a(t)$ $a\propto t$ ($q=0$) today ($t=t_0$)

06The ledger

Direction	What	Amount
buys	the horizon problem disappears (A6 and A9)	──
buys	one fewer parameter	part of Episode 25's $-148.3$
pays	$q=0$ fails	$130$ bits
pays	$w=-1/3$ fails	$395$ bits

The paying side is larger by orders of magnitude — the same conclusion as Episode 25's ledger, reached by a different road.

The honest line

(1) §04's "1.030 with the local measurement" is not a calculation inside a consistent model. $t_0=13.797$ Gyr comes from Planck's $\Lambda$CDM, and multiplying a locally measured $H_0$ by that $t_0$ does not cohere (change $H_0$ and $t_0$ changes too) — read it as an indication that a larger $H_0$ can push $H_0t_0$ above 1.

(2) §04's $\sigma$s and bits simply combine the quoted errors. The $\pm0.04$ on $q_0$ and $\pm0.03$ on $w$ depend on the analysis and move with the treatment of systematics — the orders of magnitude (130, 395 bits) stand, but do not trust the significant figures. The value of $q_0$ itself is obtained within $\Lambda$CDM.

(3) §02's "4.0× compression" follows from having counted twelve statements. The number of statements can be inflated or deflated at will — the substance is the structure "nine are identities, three inputs are independent", and the number 4.0 itself means nothing (the same caution as Episode 26 §02).

(4) §05's "integral versus instant" explains why it looks plausible; it does not defend $a\propto t$. Explaining why something looks plausible is not evidence that it is right — if anything the opposite: knowing where the plausibility comes from makes the verdict clearer than before.

(5) This series' verdict has not changed since Episode 3. $c\cdot t=$const is a notation, not new physics, and extrapolated at face value to the early universe it contradicts nucleosynthesis (previous series). The academic standard is the $\Lambda$CDM model.

Exercises

Of the twelve statements, how many are identities of one another?
Show the answer
Nine (A1–A9) — $\ddot a=0$, $q=0$, $H=1/t$, $Ht=1$, constant $(aH)^{-1}$, $\eta=\ln t$, $c/H=ct$, the diverging particle horizon. They are rewritings by differentiation and integration, so 0 bits apart on Episode 19's scale.
How many independent inputs are there?
Show the answer
Three: (i) the expansion law $a\propto t$ (giving A1–A9), (ii) the Einstein equations (giving B1), (iii) the spatial curvature $k$ (giving C1 and C2). Twelve statements came from three inputs — the fourth compression.
Are Milne and "$k=0$ with $w=-1/3$" the same spacetime?
Show the answer
No. Milne ($k=-1$, $\rho=0$) has $R=0$ and vanishing Riemann — Minkowski in disguise (Episode 33, Step 1). With $k=0$ and $w=-1/3$ matter, $R=6/t^2\ne0$ and it is genuinely curved (Step 2).
Of $q=0$, $w=-1/3$ and $H_0t_0=1$, which comes closest?
Show the answer
$H_0t_0=1$. Even with Planck's value it is 0.951 (a 4.9 per cent miss), and with the local measurement it becomes 1.030, straddling 1. Meanwhile $q=0$ misses by 13$\sigma$ and $w=-1/3$ by 23$\sigma$ — by orders of magnitude.
(Harder) Why is $H_0t_0$ the only near miss?
Show the answer
Because it is an integral. $H_0t_0$ integrates the whole history, so the decelerating and accelerating eras cancel. $q$ and $w$ are quantities of this instant, with nothing to cancel against — which is why $c\cdot t=$const looked plausible. Per caveat (4), this is not a defence of it.

Summary: we found where the plausibility came from

Collecting every way of saying $a\propto t$ gives twelve statements, of which nine (A1–A9) are identities of one another — $\ddot a=0$, $q=0$, $H=1/t$, $Ht=1$, constant comoving Hubble radius, $\eta=\ln t$, $c/H=ct$, the diverging particle horizon. All rewritings by differentiation and integration. The rest are dynamical ($w=-1/3$) and geometric ($R=6(1+k)/t^2$), leaving three independent inputs — the fourth compression, after Episodes 26, 40 and 41.

And the same $a\propto t$ covers two different spacetimes: Milne ($k=-1$, empty) is flat, Minkowski in disguise; with $k=0$ and matter it is genuinely curved — one condition does not fix one spacetime.

Only what sits in a dimensionless quantity can be tested. $q=0$ misses by 13$\sigma$, 130 bits; $w=-1/3$ by 23$\sigma$, 395 bits — by orders of magnitude. But $H_0t_0=1$ is different — 0.951 with Planck's value (a 4.9 per cent miss), 1.030 with the local measurement. The Hubble tension straddles 1 exactly.

Why is only one of them close? Because $H_0t_0$ is an integral — integrating the whole history, the decelerating and accelerating eras cancel. $q$ and $w$ are quantities of this instant, with nothing to cancel against.

Which is why $c\cdot t=$const looked plausible. Nearly right on the integrated observable, out by orders on the instantaneous ones — and the previous series' verdict (it contradicts nucleosynthesis) was about that instantaneous side. Knowing where the plausibility comes from makes the verdict clearer than before.

Next time — Episode 47 This time only the characterisations placed in dimensionless quantities could be tested ($q$, $w$, $H_0t_0$). Next time we draw the map of dimensionless quantities — every one that appears in physics, laid out on one page, with the border between what is physics and what is bookkeeping drawn across it. And we look at where the International System of Units drew that line in 2019 — the kilogram prototype was retired, and mass is now built from $\hbar$. The line this series drew by hand in Episode 3 turns out to be exactly where the world's metrology drew it too.

This document is Episode 46 of "c·t = const, That Clicks" (the first of Part VI), written for physics-minded high-school and university readers. The characterisations of $a\propto t$, the Milne universe and $H_0t_0$ in $\Lambda$CDM are all standard, and nothing here is a new claim — the numbers are computed in kenshou/calc50.py. §04's "1.030 with the local measurement" is not a calculation inside a consistent model: $t_0=13.797$ Gyr comes from Planck's $\Lambda$CDM, and multiplying a locally measured $H_0$ by that $t_0$ does not cohere (change $H_0$ and $t_0$ changes too) — read it as an indication that a larger $H_0$ can push $H_0t_0$ above 1. §04's significances simply combine the quoted errors; the $\pm0.04$ on $q_0$ and $\pm0.03$ on $w$ depend on the analysis, so the orders of magnitude stand but the significant figures should not be trusted (and $q_0$ itself is obtained within $\Lambda$CDM). §02's "4.0× compression" follows from having counted twelve statements, and the count can be inflated or deflated at will — the substance is the structure, not the number. §05's "integral versus instant" explains why it looks plausible and does not defend $a\propto t$ — knowing where plausibility comes from makes the verdict clearer, not weaker. This series' verdict has not changed since Episode 3: $c\cdot t=$const is a notation, not new physics, and extrapolated at face value to the early universe it contradicts nucleosynthesis. The academic standard is the $\Lambda$CDM model. ── To make a PDF, use your browser's Print dialogue (sliders freeze and answers are hidden in the print version).

Print / PDF: ⌘+P (Ctrl+P on Windows). On screen, move the time and see the middle diverge while the integral nearly agrees. "Show the answer" opens each solution.

Which \(a\propto t\)	\(R\)	Riemann	In Episode 33
Milne (\(k=-1\), \(\rho=0\))	\(0\)	zero	Step 1: Minkowski in disguise
\(k=0\) with \(w=-1/3\) matter	\(6/t^2\ne0\)	\(\ne0\)	Step 2: genuinely curved

Characterisation	Prediction	Observed	\(\sigma\)	bits
\(q=0\) (A3)	\(0\)	\(-0.53\pm0.04\)	\(13.2\)	\(130\)
\(w=-1/3\) (B1)	\(-0.333\)	\(-1.03\pm0.03\)	\(23.2\)	\(395\)
\(H_0t_0=1\) (A5, Planck)	\(1\)	\(0.951\pm0.007\)	\(6.8\)	\(36\)
\(H_0t_0=1\) (A5, local)	\(1\)	\(1.030\pm0.014\)	\(2.1\)	\(4.9\)

Twelve statements,
three inputs Collect every way of saying \(a\propto t\) and count the independent ones.
And you find out why \(c\cdot t=\)const looked so plausible.

01Collecting every way of saying \(a\propto t\)

02How many independent inputs?

03The two \(a\propto t\)s are different things

04The core — which can be tested, and by how much do they miss?

05Why is \(H_0t_0=1\) the only near miss?

06The ledger

Summary: we found where the plausibility came from

No.	Statement	What it needs	Kind
A1	\(a(t)\propto t\)	the definition itself	kinematic
A2	\(\ddot a=0\)	differentiate twice	kinematic
A3	deceleration parameter \(q=-\ddot aa/\dot a^2=0\)	a rewriting of A2	kinematic
A4	\(H=\dot a/a=1/t\)	differentiate once	kinematic
A5	\(H\cdot t=1\)	a rewriting of A4 (dimensionless)	kinematic
A6	comoving Hubble radius \((aH)^{-1}\) constant	because \(aH=1\)	kinematic
A7	conformal time \(\eta=\int dt/a=\ln t\)	just integrate	kinematic
A8	Hubble radius \(c/H=c\,t\) exactly	a rewriting of A4	kinematic
A9	the particle horizon diverges	\(\int dt/t\) diverges	kinematic
B1	equation of state \(w=-1/3\)	needs the Friedmann equations	dynamical
C1	\(R=6(1+k)/t^2\) (Episode 33)	needs \(k\) as well	geometric
C2	\(k=-1\) gives \(R=0\) (Milne)	a special case of C1	geometric

Twelve statements,three inputs Collect every way of saying \(a\propto t\) and count the independent ones.And you find out why \(c\cdot t=\)const looked so plausible.

01Collecting every way of saying \(a\propto t\)

02How many independent inputs?

03The two \(a\propto t\)s are different things

04The core — which can be tested, and by how much do they miss?

05Why is \(H_0t_0=1\) the only near miss?

06The ledger

Summary: we found where the plausibility came from

Twelve statements,
three inputs Collect every way of saying \(a\propto t\) and count the independent ones.
And you find out why \(c\cdot t=\)const looked so plausible.