c·t = CONST, THAT CLICKS BONUS ⑤ ／ dug after the main series closed

The shape of beta alone hits all three problems, with zero false positives

The prior was never
ours to choose The renormalisation group hands out the measure — and for a flow it is unique.
And I ended up closing two escape routes I had built myself.

What you need: Episode 3's procedure, Episode 35 on schemes, Episode 37 on running, Episode 48's priors, bonuses ②③④Twenty cases, three hits, no false positives

Bonus ④ collapsed surprise, naturalness and numerical coincidence onto one question: is there a canonical measure? Bonus ③ answered "compact gives Haar, non-compact gives nothing" — but that was too coarse. Something besides group symmetry hands out measures. And following it, the three great fine-tuning problems come out with zero false positives.

01The renormalisation group hands out the measure

For a one-dimensional flow $dg/dt=\beta(g)$, the measure invariant under the flow $$\frac{d}{dt}\int\rho\,dg=0 \iff \frac{d(\rho\beta)}{dg}=0 \iff \rho\beta=\text{const} \iff \boxed{\rho\propto\frac1\beta}$$

unique up to normalisation — and $\int dg/\beta$ is RG time itself

Conclusion of §01

"The prior probability = the RG time the theory spends near that value."
── This is not a choice. The invariant measure of a flow is unique.

02Why must the prior be RG-invariant?

A parameter's value depends on the scale you quote it atEpisode 37 — $\alpha$ is 128 at $M_Z$ and 137 at low energy

If the verdict changed with the scale, it would depend on a conventionEpisode 3: an answer that changes with a convention is not an answer

◎

So the prior must be RG-invariant — and by §01 that fixes it uniquelythe reading was derived, not assumed

Scale	$\alpha_s$	"unnaturalness" under a uniform prior
around 1 GeV	$0.500$	$1.00$ bit
around $m_b$	$0.214$	$2.22$ bit
$M_Z$	$0.118$	$3.08$ bit
$M_{\rm Planck}$ (one-loop)	$0.0191$	$5.71$ bit

The same coupling of the same theory moves by 4.71 bits just by changing the scale. A uniform prior is not RG-invariant — so it cannot serve as a verdict.

03The shape of the measure is fixed by the shape of $\beta$

Shape of $\beta$	$dg/\beta$	Induced measure	Price
$\beta=0$	degenerate	fall back to the group (Haar)	depends
$\beta\propto g$ (multiplicative)	$dg/g$	log-uniform	cheap
$\beta\propto g^2$	$dg/g^2$	$1/g^2$ weighting	cheap
$\beta=$const (additive)	$dg$	linear	expensive

Conclusion of §03

And $\beta\propto g$ happens exactly when $g=0$ has enhanced symmetry —
with a symmetry at $g=0$, $g$ can only be generated in proportion to itself; without one, other masses generate it additively.
── This is 't Hooft's naturalness criterion itself.

◇　◇　◇

04The core — scoring all 20 Standard Model parameters by the shape of $\beta$

Parameter	Count	Shape of $\beta$	Protecting symmetry	Price	How it is treated
Gauge couplings $g_1,g_2,g_3$	3	$\propto g^3$ (multiplicative)	gauge symmetry	cheap	not a problem
Yukawa couplings	9	$\propto y$ (multiplicative)	chiral symmetry	cheap	't Hooft's own example
CKM: 3 angles + 1 phase	4	multiplicative, compact	compactness	cheap	only $\theta_{13}$ at 8.8 bits
Higgs quartic $\lambda$	1	$\supset-6y_t^4$ (additive)	none	flagged	→ vacuum metastability
Higgs mass$^2$ $m^2$	1	$\supset M^2$ (if new physics)	none	expensive	the hierarchy problem
$\theta_{\rm QCD}$	1	$\beta=0$	compactness only	expensive	the strong CP problem
Cosmological constant $\Lambda$	1	$\supset m^4$ (additive)	none	expensive	the CC problem

The main point of this episode

Scoring by the shape of $\beta$ alone, exactly 3 come out "expensive".
They are the hierarchy, strong CP and the cosmological constant — the three known fine-tuning problems.
── Twenty cases, three hits, zero false positives and zero false negatives.
And $\lambda$ being "flagged" is not a miss — it correctly catches vacuum metastability, a problem of a different kind.

Figure: the 20 parameters sorted by the shape of $\beta$. Left is multiplicative (log measure, cheap); right is additive or zero (linear or Haar, expensive). Move the slider for where you put new physics — only the Higgs mass jumps from left to right the moment you place it.

scale of new physics, $\log_{10}(M/\text{GeV})$ 19

multiplicative (log measure) — cheap additive or $\beta=0$ — expensive

05The remaining freedom turns into a question of physics

$\int dg/\beta$ diverges at a fixed point$\beta\to0$, so a range must still be cut

◎

But that cut is not arbitraryit is "how far does the theory remain valid?" — where new physics enters

→

And that is exactly what decides whether the hierarchy problem existsthe $m^2$ row of §04 — the only freedom left in the prior coincided with a question of physics

06The seventh and eighth compressions

Seventh $$\underbrace{\text{Ep.48 "is there a reason for the prior?"}}_{\text{criterion}} =\underbrace{\text{bonus ③ "is it compact?"}}_{\text{theorem}} =\underbrace{\text{"is }\beta\text{ multiplicative?"}=\text{"is }g=0\text{ a symmetry point?"}}_{\textbf{mechanism}}$$

The three faces of $\theta_{\rm QCD}$	Consequence
$\beta=0$, so it does not run	the RG cannot hand out a measure
Not running, it is RG-invariant	the only constant among independent inputs (bonus ③)
Being an angle, it is compact	only the Haar measure is available

Conclusion of §06 — the eighth compression

The three are not independent facts but three faces of $\beta=0$.
── And its being "the one uncontested fine-tuning problem" follows from the same single reason.

07Closing two of my own escape routes

③

Correcting bonus ③: non-compact spaces do get a measure, from the RGa measure is missing only when $\beta=0$ and non-compact, and the Standard Model has no such case — naturalness is more well-posed than I said

Correcting Episode 48: $\rho_\Lambda$'s $\beta$ has $m^4$ additivelyso the canonical measure is linear, not log-uniform — 408 bits is the right answer, and the cosmological constant problem has no escape

"It moves 400 bits with the prior" came from not knowing the canonical measure── I closed an escape route I had built myself

Is bonus ② safe? — a robustness check The value of $\log_2B$: 8.67 under a uniform measure, 8.66 under the RG measure — a difference of 0.01 bits.
(Only log-uniform differs, at 11.46 — still the same order.) The compression law does not depend on the choice of measure.

08What can now be said that could not before

Model	What to look at	Consequence
Adding a new scalar	does its mass $\beta$ gain an additive term?	if so, it creates a new hierarchy problem
Supersymmetry	boson–fermion cancellation removes the additive term	which is why it solves the hierarchy problem
The axion	gives $\theta$ a $\beta$ and makes it move	solves it by breaking the $\beta=0$ degeneracy
Any CC mechanism	how to remove the additive $m^4$	nobody has managed it

Conclusion of §08

Supersymmetry "solving" the hierarchy problem means removing the additive term in $\beta$ and turning a linear measure back into a logarithmic one (408 → about 7 bits).
── Write down a model's $\beta$ and you know on the spot whether it has a fine-tuning problem.

The honest line — and the hole that is not filled

(1) §01's uniqueness holds only for a one-dimensional flow. In several dimensions the invariant measure is not unique (many $\rho$ satisfy $\nabla\!\cdot\!(\rho\beta)=0$) — gauge couplings run independently at one loop so the 1-D argument applies to them, but not to all parameters. This is the most technical weakness here.

(2) $\beta$ is scheme-dependent (the same weakness as Episode 35's account of asymptotic safety). Whether it is multiplicative or additive is scheme-independent; the coefficients are not.

(3) §07's "linear is canonical for $\rho_\Lambda$" is not settled physics. In dimensional regularisation the $m^4$ terms appear differently, and it depends on the renormalisation conditions — doubt this and you return to Episode 48's "it cannot be decided".

(4) §04's table restates 't Hooft naturalness in the language of measures. The physics is known; what is new is only the reading that it closes as a question about priors — and twenty cases is a small sample, with "the three great problems" itself a convention of the literature.

(5) And the biggest hole. §02 establishes that the prior must be RG-invariant. But — having a measure and that measure being a probability are two different things. Reading "values where more RG time is spent are more likely" is natural but not proved. That hole is not filled. The same requirement does yield the Jeffreys prior in statistics, so it is not an isolated position — but it is still a position.

Exercises

Why must the prior be RG-invariant?
Show the answer
Otherwise the verdict changes with the scale. Under a uniform prior, $\alpha_s$ reads 1.00 bits at 1 GeV and 5.71 at $M_{\rm P}$ — the same coupling moving by 4.71 bits. Episode 3: an answer that changes with a convention is not an answer.
How many RG-invariant measures are there?
Show the answer
For a one-dimensional flow, exactly one: $\rho\beta=$const, i.e. $\rho\propto1/\beta$, which is $\int dg/\beta=$ RG time. ── But per caveat (1), in several dimensions it is not unique.
How does the shape of $\beta$ map to the price?
Show the answer
$\beta\propto g$ (multiplicative) → log measure → cheap. $\beta=$const (additive) → linear measure → expensive. $\beta=0$ → back to the group. And $\beta\propto g$ happens exactly when $g=0$ has enhanced symmetry — 't Hooft's criterion.
Scoring the 20, how many come out "expensive"?
Show the answer
Three — $m^2$ (with new physics), $\theta_{\rm QCD}$ and $\Lambda$. They are the hierarchy, strong CP and the cosmological constant: the three known problems, with zero false positives and zero false negatives. $\lambda$'s "flagged" correctly catches vacuum metastability.
(Harder) What is the biggest unfilled hole here?
Show the answer
That having a measure and that measure being a probability are different things. RG-invariance fixes the measure uniquely, but the reading "values where more RG time is spent are more likely" is not proved. If that fails, everything from §04 onward fails with it.

Summary: the prior was never ours to choose

For a one-dimensional flow $dg/dt=\beta(g)$, the invariant measure is $\rho\propto1/\beta$, and it is unique — it is RG time. And why it must be RG-invariant can be said too: a verdict that changes with the scale depends on a convention, and is therefore not a verdict (Episode 3). Under a uniform prior, $\alpha_s$'s "unnaturalness" moves by 4.71 bits.

The shape of the measure follows from the shape of $\beta$ alone — multiplicative gives a log measure and is cheap; additive gives a linear one and is expensive. And $\beta\propto g$ happens exactly when $g=0$ carries a symmetry: 't Hooft's criterion itself.

Scoring all 20 Standard Model parameters by the shape of $\beta$, only three come out expensive — the hierarchy, strong CP and the cosmological constant, the three known problems, with zero false positives and zero false negatives. $\lambda$'s flag is not a miss either: it correctly catches vacuum metastability.

Even the remaining freedom — the range of integration — was not arbitrary. It is "how far does the theory remain valid?", i.e. where new physics enters, which is precisely what decides whether the hierarchy problem exists. The one freedom left in the prior coincided with a question of physics.

And two of my own escape routes closed. Bonus ③'s "non-compact means ill-posed" was too coarse — naturalness is more well-posed than I said. Episode 48's "it moves 400 bits with the prior" too — $\rho_\Lambda$'s $\beta$ is additive, so the canonical measure is linear, 408 bits is the right answer, and the cosmological constant problem has no escape.

But — having a measure and that measure being a probability are two different things. That hole is still open.

In closing — what the five bonus episodes found ①: whether the mass variation is universal can be measured ($\mu$, 23.3 bits) + a 10.2-bit degeneracy in the quark-mass direction.
②: a hierarchy shrinks to its own logarithm ($B\to\log_2B$; robustness confirmed in ⑤).
③: the zero column is not homogeneous, and almost no constants survive (only $\theta_{\rm QCD}$ remains).
④: a 4.1-bit prediction beats a 15.7-bit discovery — "theory first" means fixing the measure first.
⑤: and that measure was being handed out by the renormalisation group — the shape of $\beta$ alone hits the three great problems.
── All five stand on the single procedure of Episode 3. And in ③ and ⑤ that procedure deleted two of the criteria this series had built and closed two of its escape routes. The tool is still working.

This document is bonus episode ⑤ of "c·t = const, That Clicks", written after the main 50 episodes closed, for physics-minded high-school and university readers. The numbers are computed in kenshou/calc63.py and calc64.py. The renormalisation group, invariant measures of flows, 't Hooft naturalness and vacuum metastability are all standard material, and §04's table restates 't Hooft naturalness in the language of measures — the physics is known; what is new is only the reading that it closes as a question about priors. §01's uniqueness holds only for a one-dimensional flow; in several dimensions the invariant measure is not unique (gauge couplings run independently at one loop, so the argument applies to them, but not to all parameters) — this is the most technical weakness here. $\beta$ is scheme-dependent: whether it is multiplicative or additive is not, but the coefficients are. §07's "linear is canonical for $\rho_\Lambda$" is not settled physics — in dimensional regularisation the $m^4$ terms appear differently and it depends on the renormalisation conditions; doubt it and you return to Episode 48's "it cannot be decided". Twenty cases is a small sample and "the three great problems" is itself a convention of the literature. And the biggest hole: RG-invariance fixes the measure uniquely, but having a measure and that measure being a probability are two different things, and the reading "values where more RG time is spent are more likely" is not proved (the same requirement yields the Jeffreys prior in statistics, so it is not isolated — but it is still a position). ── 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 scale of new physics and watch the Higgs mass change sides. "Show the answer" opens each solution.

Scale	\(\alpha_s\)	"unnaturalness" under a uniform prior
around 1 GeV	\(0.500\)	\(1.00\) bit
around \(m_b\)	\(0.214\)	\(2.22\) bit
\(M_Z\)	\(0.118\)	\(3.08\) bit
\(M_{\rm Planck}\) (one-loop)	\(0.0191\)	\(5.71\) bit

The prior was never
ours to choose The renormalisation group hands out the measure — and for a flow it is unique.
And I ended up closing two escape routes I had built myself.

01The renormalisation group hands out the measure

02Why must the prior be RG-invariant?

03The shape of the measure is fixed by the shape of \(\beta\)

04The core — scoring all 20 Standard Model parameters by the shape of \(\beta\)

05The remaining freedom turns into a question of physics

06The seventh and eighth compressions

07Closing two of my own escape routes

08What can now be said that could not before

Summary: the prior was never ours to choose

Shape of \(\beta\)	\(dg/\beta\)	Induced measure	Price
\(\beta=0\)	degenerate	fall back to the group (Haar)	depends
\(\beta\propto g\) (multiplicative)	\(dg/g\)	log-uniform	cheap
\(\beta\propto g^2\)	\(dg/g^2\)	\(1/g^2\) weighting	cheap
\(\beta=\)const (additive)	\(dg\)	linear	expensive

Parameter	Count	Shape of \(\beta\)	Protecting symmetry	Price	How it is treated
Gauge couplings \(g_1,g_2,g_3\)	3	\(\propto g^3\) (multiplicative)	gauge symmetry	cheap	not a problem
Yukawa couplings	9	\(\propto y\) (multiplicative)	chiral symmetry	cheap	't Hooft's own example
CKM: 3 angles + 1 phase	4	multiplicative, compact	compactness	cheap	only \(\theta_{13}\) at 8.8 bits
Higgs quartic \(\lambda\)	1	\(\supset-6y_t^4\) (additive)	none	flagged	→ vacuum metastability
Higgs mass\(^2\) \(m^2\)	1	\(\supset M^2\) (if new physics)	none	expensive	the hierarchy problem
\(\theta_{\rm QCD}\)	1	\(\beta=0\)	compactness only	expensive	the strong CP problem
Cosmological constant \(\Lambda\)	1	\(\supset m^4\) (additive)	none	expensive	the CC problem

The three faces of \(\theta_{\rm QCD}\)	Consequence
\(\beta=0\), so it does not run	the RG cannot hand out a measure
Not running, it is RG-invariant	the only constant among independent inputs (bonus ③)
Being an angle, it is compact	only the Haar measure is available

Model	What to look at	Consequence
Adding a new scalar	does its mass \(\beta\) gain an additive term?	if so, it creates a new hierarchy problem
Supersymmetry	boson–fermion cancellation removes the additive term	which is why it solves the hierarchy problem
The axion	gives \(\theta\) a \(\beta\) and makes it move	solves it by breaking the \(\beta=0\) degeneracy
Any CC mechanism	how to remove the additive \(m^4\)	nobody has managed it

The prior was neverours to choose The renormalisation group hands out the measure — and for a flow it is unique.And I ended up closing two escape routes I had built myself.

01The renormalisation group hands out the measure

02Why must the prior be RG-invariant?

03The shape of the measure is fixed by the shape of \(\beta\)

04The core — scoring all 20 Standard Model parameters by the shape of \(\beta\)

05The remaining freedom turns into a question of physics

06The seventh and eighth compressions

07Closing two of my own escape routes

08What can now be said that could not before

Summary: the prior was never ours to choose

The prior was never
ours to choose The renormalisation group hands out the measure — and for a flow it is unique.
And I ended up closing two escape routes I had built myself.