c·t = CONST, THAT CLICKS EPISODE 41 / Part V — where the tool breaks

The monstrous number was a familiar one

Time runs from where
the tool reaches Measured in bits, Penrose's \(10^{10^{123}}\) is the same number that has appeared three times already.
And the direction of time can be restated as the reach of a conformal transformation.

What you need: Episode 19's scale, Episode 24, Episode 31's CCC, Episode 33's three-step test, Episode 40\(3.27\times10^{122}\) bits — the same number, a fourth time

Last time we found that the early universe's gravitational entropy is exactly zero (FLRW has Weyl \(=0\)). This time we look at a proposal that turns that into a law — the Weyl curvature hypothesis. We count out what the famous number \(10^{10^{123}}\) actually is, and restate which way the arrow of time points in this series' own language.

01What the Weyl curvature hypothesis says

in
At initial singularities, the Weyl curvature \(C=0\)Penrose's (1979) demand
out
At final singularities (black holes), no restrictionWeyl may be as large as it likes
A law that makes beginning and end asymmetricand that asymmetry, the claim goes, is the origin of the arrow of time
In four dimensions the Riemann curvature has 20 independent components $$\underbrace{10}_{\text{Ricci: fixed by matter}}\;+\;\underbrace{10}_{\text{Weyl: free. Gravitational waves, tides}}$$

Conclusion of §01

The hypothesis switches off exactly half the curvature's degrees of freedom on the initial surface.

02Why the demand is needed — turning the headroom into a probability

Statistical mechanics: the phase-space volume ratio is \(e^{\Delta S/k_B}\) $$\exp\!\left(2.265\times10^{122}\right)=10^{\,9.84\times10^{121}}$$

so the initial state was a one-in-\(10^{10^{122}}\) draw

Penrose writes \(10^{10^{123}}\) — his estimate of \(S_{\max}\) is slightly larger; it is the same number and the same argument.

◇ ◇ ◇

03The core — measuring that monstrous number in bits

By Episode 19's practice $$\text{surprise}=\frac{S_{\max}}{\ln 2}=\mathbf{3.27\times10^{122}\ \text{bits}}$$
The \(10^{122}\)s so farbits
Ep. 24: information the universe has processed, \(N=C\cdot t_0\)\(3.11\times10^{122}\)
Ep. 40: the holographic bound\(3.27\times10^{122}\)
Ep. 40: \(S\) with all the mass as one black hole\(3.27\times10^{122}\)
Ep. 41: how special the initial state was\(3.27\times10^{122}\)

The main point of this episode

All four are the same number.
Penrose's \(10^{10^{123}}\) is, in this series' currency, \(3.3\times10^{122}\) bitsa familiar number.
── After Episodes 26 and 40, the compression works a third time.

04The ledger — is this the series' best deal?

TheoryPays [bits]BuysAmount bought
Inflation (Ep. 27)\(10.73\)horizon, flatness, \(n_s\)assessed at \(-6.5\)
Cosmon (Ep. 32)\(10.73\)the size of \(\rho_\Lambda\)up to \(408\)
MOND (Ep. 29)\(21.46\)rotation curves\(+1971\) (a loss)
Weyl curvature hypothesis\(5.37\)how special the initial state was\(3.27\times10^{122}\)

On the ledger it is better by many orders of magnitude than anything else in this series (a ratio of \(6\times10^{121}\)). But — the price "one law = one parameter = 5.37 bits" is a convenience from Episode 5, and there is no guarantee that a boundary condition can be bought at a parameter's price. That is this episode's weakest point (honest line 2).

05The arrow of time runs from where the tool reaches to where it does not

EpochWeylStep (Ep. 33)Conformal transformation
Initial singularity (the hypothesis)\(C=0\)Step 2: conformally flatreaches
The universe now (globally FLRW)\(C\approx0\)Step 2nearly reaches
The end state (black holes, Ep. 39)\(C\ne0\)Step 3does not reach

Conclusion of §05

The arrow of time runs from where conformal transformations reach to where they do not.
── The "the tool breaks / the tool cannot reach" of Part V turns out to be the direction of time itself.

Figure: the life of the universe seen through Weyl curvature. At the left (the beginning) Weyl is 0 and we are in Step 2; further right Weyl grows and we cross into Step 3. Move the slider to see where the tool stops reaching — that boundary is the arrow of time.

84.1
Step 2 (conformally flat, the tool reaches) Step 3 (Weyl not 0, it does not)

06The link to Episode 31 — the same demand, seen twice

31
CCC: to glue aeons there must be no ruler at the boundary= only the conformal structure survives there = the demand that Weyl \(=0\)
41
The hypothesis: Weyl \(=0\) at the initial singularity= the explanation of why the beginning was special
The same demand put to two different usesCCC for "there is a continuation", the hypothesis for "the beginning was special" — Penrose's two proposals share one root

07Can it be observed?

Weyl curvature is the gravitational-wave degree of freedom $$\text{Weyl}=0\ \text{initially}\ \Longrightarrow\ \text{no primordial gravitational waves}$$

the current bound is \(r<0.036\) (BICEP/Keck 2021, 95% CL) → 4.8 bits by Episode 19's practice

So far no primordial gravitational waves have been found — a tailwind for the hypothesis. But the logic is not airtight: the hypothesis is a demand on the singularity itself, while inflation concerns the stage after it. A non-zero \(r\) would not immediately refute the hypothesis.

08The objection — inflation is not a substitute

The debate Objection: "Is the special beginning not already explained by inflation?"
Penrose's reply: starting inflation itself requires a low-entropy initial state. It presupposes the very thing to be explained.
── The two are not substitutes, on Penrose's view. This debate is unsettled, and this document endorses neither side.
The honest line

(1) The numbers in §02 and §03 inherit Episode 40's estimates directly. \(S_{\max}\) is the holographic bound counted on the Hubble sphere, and it moves by an order of magnitude depending on which horizon is used (Episode 40, caveat 2). The difference between \(10^{10^{122}}\) and \(10^{10^{123}}\) is of that size.

(2) §04's ledger is this episode's weakest point. "One law = 5.37 bits" was built in Episode 5 as the price of a parameter, and there is no guarantee that a boundary condition can be bought at that price. Strictly, specifying an initial condition should be paid for by the description length needed to write that condition down, which could be far larger than 5.37 bits — §04 is an observation that it looks like a good deal, not a proof.

(3) "Phase-space volume ratio \(=e^{\Delta S}\)" may be too naive once gravity is involved. Whether black-hole entropy really is the logarithm of a phase-space volume is an open question in quantum gravity, and §02's probability reading rests on that assumption.

(4) §07's observational link is not airtight. The hypothesis is a demand at the singularity and does not directly say that no primordial gravitational waves are generated during the evolution afterwards — read "the bound on \(r\) is a tailwind" as circumstantial. The 4.8 bits also assumes a uniform prior over \(r\in[0,1]\).

(5) The Weyl curvature hypothesis is a hypothesis, not an established law. Even its formulation is debated (how to state "Weyl \(=0\)" rigorously at a singularity is not obvious), and this document endorses neither it nor inflation. The academic standard remains the \(\Lambda\)CDM model including inflation.

Exercises

  1. What does the hypothesis switch off on the initial surface? Put it in numbers.
    Show the answer
    Of the 20 independent components of the Riemann curvature in four dimensions, the 10 Weyl componentsexactly half the curvature's degrees of freedom. The other 10 are Ricci, fixed by matter through the Einstein equations.
  2. How many bits is Penrose's \(10^{10^{123}}\)?
    Show the answer
    \(3.3\times10^{122}\) bits (\(S_{\max}/\ln2\)) — the same number as Episode 24's \(N=C\cdot t_0\) and Episode 40's holographic bound. Four headline numbers turned out to be one.
  3. Where is §04's ledger weak?
    Show the answer
    In assuming a boundary condition can be bought at "one law = 5.37 bits". Episode 5's 5.37 bits is the price of a parameter, and there is no guarantee it applies to specifying an initial condition — strictly one should pay the description length of writing that condition down, which may be far larger.
  4. State the arrow of time in terms of Episode 33's three-step test.
    Show the answer
    It runs from Step 2 (conformally flat, the tool reaches) to Step 3 (Weyl \(\ne0\), it does not). The initial singularity has Weyl \(=0\); the final black holes have Weyl \(\ne0\) — Part V's "the tool breaks / cannot reach" turns out to be the direction of time itself.
  5. (Harder) How is Episode 31's CCC related to this hypothesis?
    Show the answer
    They are the same demand put to different uses. CCC requires that there be no ruler at the gluing boundary (= only the conformal structure survives = Weyl \(=0\)); the hypothesis places the same condition to explain why the beginning was special — Penrose's two proposals share one root.

Summary: the monstrous number was a familiar one

The Weyl curvature hypothesis demands Weyl \(=0\) at initial singularities and nothing at final ones. Of the 20 Riemann components in four dimensions it switches off the 10 Weyl components — exactly half — on the initial surface.

Why is it needed? Turning Episode 40's headroom into a probability, the initial state was a one-in-\(10^{10^{122}}\) draw (Penrose writes \(10^{10^{123}}\)). But measured in bits that is \(3.27\times10^{122}\) bitsexactly the same number as Episode 24's \(N=C\cdot t_0\) and Episode 40's holographic bound. Four headline numbers were one. After Episodes 26 and 40, that is the third compression.

On the ledger: pay 5.37 bits, buy \(3.27\times10^{122}\) — it looks like a spectacular deal. But there is no guarantee that a boundary condition can be bought at a parameter's price, and that is this episode's weakest point.

And the most important thing this time. Lining up beginning and end on Episode 33's three-step test: the beginning has Weyl \(=0\) and is Step 2 (conformally flat, the tool reaches); the final black holes have Weyl \(\ne0\) and are Step 3 (it does not). In other words — the arrow of time runs from where conformal transformations reach to where they do not. Part V's "the tool breaks / cannot reach" turns out to be the direction of time itself.

Finally, it is the same demand as Episode 31's CCC — no ruler at the boundary, which is Weyl \(=0\). Penrose's two proposals share one root.

This document is Episode 41 of "c·t = const, That Clicks" (the fifth of Part V), written for physics-minded high-school and university readers. The Weyl curvature hypothesis is Penrose's (1979) proposal, and the \(10^{10^{123}}\) estimate is his — the numbers are computed in kenshou/calc45.py. §02 and §03 inherit Episode 40's estimates directly: \(S_{\max}\) is the holographic bound counted on the Hubble sphere and moves by an order of magnitude depending on which horizon is used — the difference between \(10^{10^{122}}\) and \(10^{10^{123}}\) is of that size. §04's ledger is this episode's weakest point: "one law = 5.37 bits" was built in Episode 5 as the price of a parameter, and there is no guarantee a boundary condition can be bought at that price — strictly one should pay the description length of the condition, possibly far larger, so §04 is an observation that it looks like a good deal, not a proof. "Phase-space volume ratio = \(e^{\Delta S}\)" may be too naive once gravity is involved; whether black-hole entropy really is the logarithm of a phase-space volume is an open question in quantum gravity. §07's observational link is not airtight — the hypothesis is a demand at the singularity and does not directly forbid primordial gravitational waves generated afterwards, so the bound on \(r\) is circumstantial (and the 4.8 bits assumes a uniform prior). The hypothesis is a hypothesis, not an established law, and even its formulation is debated — this document endorses neither it nor inflation; the academic standard remains the \(\Lambda\)CDM model including inflation. The curve in the figure is schematic, not a numerical prediction. ── 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 slider to see where the tool stops reaching. "Show the answer" opens each solution.