CONFORMAL TRANSFORMATIONS THAT CLICKBONUS ② (full length) / Putting the series' own motivation on trial

\(c\cdot t=\text{const}\) came out of the idea that the universe is a computer with finite resources. Was that idea right?

Is the universe a computer
with finite resources? In bonus episode ① nucleosynthesis judged the expansion law \(a\propto t\) folded into \(c\cdot t=\text{const}\).
What goes on trial now is one step earlier — the motivation itself.

What you need: logarithms, division, and counting in bits comoving Hubble radius \(=c/\dot a\)

The auxiliary line \(c\cdot t=\text{const}\) originally came from an information-theoretic intuition: the universe is a computer with finite resources, so the bigger it gets, the slower the computation per unit place must run. Let us now turn the decision procedure built in Episode 10 on that motivation itself. The conclusion, up front — the motivation is remarkably accurate and the implementation misses its target. Exactly the shape of the diagnosis passed on VSL in bonus episode 3 of the previous series. And chasing the miss leads to a place where gravity, quantum theory, information and time meet at a single point.

01What it is, information-theoretically: the address space does not move

Seen as a computer, the most basic resource of the universe is "how far can I get a signal to". The quantity for that is the comoving Hubble radius.

The computer's address space
$$r_H=\frac{c}{aH}=\frac{c}{\dot a}$$

It is "the range you can reach causally right now", measured in comoving coordinates. Look for the expansion law that holds it fixed and the answer is unique.

The calculation — one line $$r_H=\frac{c}{\dot a}=\text{const}\quad\Longleftrightarrow\quad \dot a=\text{const}\quad\Longleftrightarrow\quad a\propto t$$

For \(a\propto t^p\), \(r_H\propto t^{1-p}\). It comes to a dead stop only at \(p=1\).

Expansion law\(r_H\)What happens to the computer
\(p<1\) (radiation, matter)growsNew comoving regions come into view — memory appears that was never in contact (the horizon problem)
\(p=1\) (\(c\cdot t=\)const)constantNothing enters, nothing leaves. The address space does not change
\(p>1\) (Λ, inflation)shrinksRegions already in contact become unreachable — information is lost

So \(c\cdot t=\text{const}\) is the unique expansion law for which a finite-resource computer's address space does not move. The motivation was accurate. It is the fourth face of Episode 5's "triple boundary at \(w=-1/3\)".

Which resource you hold fixed decides the expansion law Hold the number of bits fixed (horizon entropy \(\propto (c/H)^2\)) — you get \(a\propto e^{Ht}\), de Sitter.
Hold the address space fixed (comoving Hubble radius) — you get \(a\propto t\), \(c\cdot t=\)const.
The choice of resource is what fixes the shape of the universe.

02Measuring this computer

Today's universe: the spec sheet

Memory (bits on the horizon)

$$N=\frac{\pi R_H^2}{\ell_P^2\ln 2}=2.96\times10^{122}\ \text{bit}$$

Total operations (integrating the Margolus–Levitin bound \(2E/\pi\hbar\))

$$\Omega=2.1\times10^{121}\ \text{ops}\qquad(\text{agreeing with Lloyd's }10^{120}\text{, 2002})$$

Clock (times light has crossed the horizon)

$$\ln\frac{t_0}{t_P}=\mathbf{140}\ \text{times}$$

Against \(10^{122}\) bits of memory, the clock has ticked 140 times. A machine wildly rich in memory and poor in cycles.

And this computer runs exactly at the thermodynamic limit.

Energy per bit = the Landauer limit
$$\frac{E}{N}=2.672\times10^{-53}\,\mathrm{J},\qquad k_BT_H\ln2=2.671\times10^{-53}\,\mathrm{J}$$

The ratio is 1.0000. This is the identity \(E=T_HS\) (it holds in any FLRW), but it reads as: the universe runs the Landauer limit with zero margin. Episode 10 of the previous series — "\(k_BT\ln2\) to erase one bit" — balances on a cosmic scale.

03So far so good — let us actually build the model

If the motivation is accurate, it should turn into a model. Take Episode 3's "masses grow" not as a gauge statement but as physics.

One hypothesis, and w = −1/3 falls out by itself

The hypothesis (this alone)

$$\frac{m}{M_{\rm Pl}}\propto a$$

Then

$$\rho_m=n\,m\propto a^{-3}\cdot a=a^{-2}$$

Which is exactly the \(\rho\propto a^{-2}\) that \(w_{\rm tot}=-1/3\) demands. "Masses grow in proportion to \(a\)" alone gives \(w=-1/3\) with no tuning. And the problem from bonus ① — a negative dark-energy density once matter is included — disappears at the same time.

This is coupled quintessence, and the parameters are not free.

Fixed by \(\Omega_m\) alone (zero freedom)
$$\lambda^2=\frac{4}{2-3\Omega_m},\qquad \beta=\frac{\lambda}{2} \qquad\xrightarrow{\ \Omega_m=0.30\ }\qquad \lambda=1.907,\ \ \beta=0.954$$

Integrate numerically and this solution turns out to be an attractor — perturb \(\rho_m\) or \(\dot\sigma\) by \(\pm50\%\) and it comes back to \(Ht\to1\), \(\Omega_m\to0.30\). For the first time, "why is \(w\) exactly \(-1/3\)?" has an answer: because the solution attracts. Up to here it is good news for the model.

◇ ◇ ◇

04And then it breaks — the dilaton cannot rescue light

There is exactly one thing this picture cannot save: radiation.

The strength with which the dilaton exchanges information with matter is proportional to the trace of the stress–energy tensor, \(T^\mu{}_\mu\) (Episode 5). But —

Episode 7's one line, doing its work here
$$\text{radiation is conformally invariant}\ \Longrightarrow\ T^\mu{}_\mu=0\ \Longrightarrow\ \text{zero coupling}\ \Longrightarrow\ \rho_r\propto a^{-4}\ \text{unchanged}$$

The dilaton can rescue mass, but it cannot rescue light.

\(\rho_r\propto a^{-4}\) grows faster going back than \(\rho_{\rm tot}\propto a^{-2}\), so radiation must eventually take over the total. The moment is

$$\sqrt{\Omega_r}=9.6\times10^{-3}\qquad\Longrightarrow\qquad \text{radiation dominates for }z>103$$

Recombination (\(z=1100\)) lies entirely inside that. \(c\cdot t=\text{const}\) cannot hold before recombination. That leaves two branches.

A
Keep \(a\propto t\) at all epochsThe dark-energy density has to be negative — \(-110\) times at recombination, \(-1.5\times10^{13}\) times at nucleosynthesis. And bonus ①'s nucleosynthesis verdict still stands.
B
Let radiation give \(a\propto t^{1/2}\)\(z>103\) becomes standard cosmology, so nucleosynthesis and the acoustic peaks both work — they work because it became standard. And the horizon problem returns (figure below).
Figure: the history of the address space (comoving Hubble radius). Turn the radiation knob from 0 up to its real value and \(c\cdot t=\text{const}\)'s only selling point evaporates
Ω_r = 0 the address space is exactly constant → growth from the Planck era to today = 1 (and 1 by volume)
Comoving Hubble radius r_H Where radiation starts to dominate Today's value (reference)

Only at knob position 0 is the line perfectly horizontal — the address space frozen. Add even a trace of radiation and the line bends on the left, squeezing the flat part into the right-hand edge. At the real value:

 \(c\cdot t=\)const (no radiation)with radiation (branch B)ΛCDM
Growth of address space (volume)1\(1.2\times10^{89}\)\(\sim10^{88}\)
Causally disconnected patches1\(1.2\times10^{4}\)\(9.6\times10^{3}\)
Range where \(a\propto t\) holdsall of it6.4% of logarithmic history──

The horizon problem returns, at essentially ΛCDM's severity. The universe as a computer cannot manage its own radiation — and radiation is exactly what dominates the early universe. A design that freezes the address space fails to operate in the era where it is most needed.

05The memory was almost empty

Let us measure the universe-as-computer once more.

 Entropy \(S/k_B\)Fraction of capacity
Horizon capacity (maximum)\(2.05\times10^{122}\)1
Total entropy today\(3.1\times10^{104}\)\(1.5\times10^{-18}\)
CMB photons (the early state)\(2.0\times10^{88}\)\(1.0\times10^{-34}\)

Today's total entropy is dominated by supermassive black holes (Egan & Lineweaver 2010). Even so, only \(10^{-18}\) of the memory is filled.

Turn that around — the universe is still 99.9999999999999998% conformally flat. Which is why FLRW works, and why this series' tool works.

06The gravitational field itself splits into bookkeeping and physics

This is the deepest point of the episode. In Episode 7 we counted "10 components of a four-dimensional metric = 9 light cones + 1 ruler marking". That "9" has a name — the Weyl tensor \(C_{\mu\nu\alpha\beta}\), a quantity that a conformal transformation does not move at all.

The conformal-factor side (1 component)The Weyl-tensor side (9 components)
moves under a conformal transformationdoes not move (\(C\) is conformally invariant)
gauge, bookkeeping (Ep.4: net zero)physics
ghost = wrong-sign kinetic term (Ep.9)the propagating gravitational-wave degrees of freedom
carries no gravitational entropyis gravitational entropy (Penrose)
carries \(a(t)\)carries structure, clumping, black holes
initial condition is free\(C=0\) initially (Weyl curvature hypothesis)
carries no arrow of time★ the arrow of time lives only here

The previous series' motto — units are bookkeeping, dimensionless is physics — is realised as the internal structure of the gravitational field itself. And the arrow of time lives only on the physics side.

How broken the tool is, is the arrow of time Penrose's Weyl curvature hypothesis says the universe began with \(C=0\). And \(C=0\) is precisely the condition (conformal flatness) that let Episode 3 transform to Minkowski. This series' tool works perfectly where gravitational entropy is zero, and fails in proportion to how much of it there is. The degree to which the tool is broken is itself the arrow of time — and the \(10^{-18}\) of section 05 is the number saying how broken it is right now.

07So it was constraining the wrong thing

At last we can say why no prediction came out.

The conclusion of this bonus episode

\(c\cdot t=\text{const}\) is a condition on \(a(t)\) = a condition on the conformal factor = a condition on the bookkeeping.
Constrain the bookkeeping and no prediction follows.
To have predictions you must constrain the Weyl side — causal structure — light sheets.

And a "finite resource" principle of exactly that shape already exists.

Bousso's covariant entropy bound
$$S\le \frac{A}{4\ell_P^2}\qquad(\text{on a light sheet})$$

It is fully conformally invariant — light sheets (null surfaces) do not move under a conformal transformation (Episode 7), and \(A/\ell_P^2\) is dimensionless (even in the Weyl frame, \(A\to A/a^2\) and \(\ell_P^2\to\ell_P^2/a^2\) leave it alone). A gauge-independent finite-resource condition — and therefore one shaped so that it can predict.

The motivation was right. Writing the universe as a computer with finite resources is a legitimate idea. It is just that — what should have been constrained was not \(a(t)\), but the information on light sheets.

Being straight with you

The numbers for branch B in section 04 (patch count \(1.2\times10^4\), acoustic-peak position) are estimates from the simplified hybrid model \(H^2=H_0^2[\Omega_ra^{-4}+(1-\Omega_r)a^{-2}]\), not a Boltzmann-code calculation. Read them as order-of-magnitude. While writing this article the author estimated the acoustic peak with radiation left out and got the order of magnitude wrong — the very theme of this episode, that the component you drop is the one that matters most, tripped up the writer.

In the table of section 06, "ghost = gravitational entropy upside down" is continuous with the Gross–Perry–Yaffe negative mode of Euclidean Schwarzschild (the negative specific heat of black holes), but recent work shows that mixing between the transverse-traceless and trace modes makes a naive identification delicate. No claim beyond "they share the same sign" is made here. The definition of gravitational entropy itself (how to count \(C\)) is also not settled.

Exercises
  1. Find the expansion law for which the comoving Hubble radius is constant.
    Show answer
    \(r_H=c/\dot a\), so \(\dot a=\)const, that is \(a\propto t\). It is the unique solution. For \(a\propto t^p\), \(r_H\propto t^{1-p}\), which is stationary only for \(p=1\).
  2. Why can radiation not couple to the dilaton?
    Show answer
    Because the coupling is proportional to \(T^\mu{}_\mu\), and radiation is conformally invariant so \(T^\mu{}_\mu=0\) (Episode 7). What has no scale cannot exchange information with the field that sets the scale.
  3. What fraction of the horizon's memory has today's universe used?
    Show answer
    \(3.1\times10^{104}/2.05\times10^{122}=1.5\times10^{-18}\). Nearly empty. Which means the universe is still extremely conformally flat — and that is why both FLRW and this series' tool work.
  4. (Harder) Why can \(c\cdot t=\text{const}\) not, in principle, carry a prediction? Use the counting of Episode 4.
    Show answer
    \(a(t)\) is carried by the conformal factor, which by Episode 4's counting (field +1, symmetry −1, net zero) is gauge = bookkeeping. A condition on the bookkeeping fixes no observable. Predictions live only on the Weyl side (the conformally invariant causal structure), and Bousso's bound is what constrains it. This holds only so long as no extra physical hypothesis such as \(m/M_{\rm Pl}\propto a\) is added — section 03 added exactly that, and obtained predictions (and refutations).

SUMMARYAccurate motivation, misdirected implementation

Seen information-theoretically, \(c\cdot t=\text{const}\) has a clear identity — the unique expansion law for which the comoving Hubble radius is constant. The computer's address space neither grows nor shrinks. It is exactly the watershed between the horizon problem (memory appears that was never contacted) and information loss (contacted memory becomes unreachable). Today's spec sheet: \(10^{122}\) bits of memory, \(10^{121}\) operations, and a clock that has ticked 140 times — with the energy per bit sitting exactly on the Landauer limit.

But build the model by taking "masses grow \(\propto a\)" as physics (\(\lambda\) and \(\beta\) fixed by \(\Omega_m\) alone, and the solution an attractor) and radiation alone cannot be saved — it is conformally invariant and cannot couple to the dilaton (Episode 7). Radiation dominates for \(z>103\), the one selling point evaporates, and the horizon problem returns at ΛCDM's level. The deeper reason is that the gravitational field itself splits into the conformal factor (bookkeeping, ghost, no arrow of time) and the Weyl tensor (physics, gravitational entropy, the arrow of time). \(c\cdot t=\text{const}\) was constraining the bookkeeping side. What should have been constrained is the information on light sheets.

This document is bonus episode ② of the series "Conformal Transformations That Click", written for physics-minded high-school and university students. That the expansion law with constant comoving Hubble radius \(c/\dot a\) is restricted to \(a\propto t\) is elementary. The horizon entropy \(S=k_BA/4\ell_P^2\), the Margolus–Levitin bound \(2E/\pi\hbar\), and the total operation count of the universe \(\sim10^{120}\) (Lloyd 2002) are standard. \(E=T_HS\) — hence the coincidence of the energy per bit with the Landauer limit — is an identity valid in any FLRW and is not specific to \(c\cdot t=\text{const}\). The identification with coupled quintessence in section 03 assumes universal coupling; \(\lambda^2=4/(2-3\Omega_m)\) and \(\beta=\lambda/2\) are derived and numerically checked in this article (the solution was also verified numerically to be an attractor). However \(\beta=0.954\) greatly exceeds Cassini's \(|\beta|<3.4\times10^{-3}\) and the CMB bound \(\beta\lesssim0.05\) for coupled dark matter. The branch-B numbers in section 04 are simple estimates from \(H^2=H_0^2[\Omega_ra^{-4}+(1-\Omega_r)a^{-2}]\), not Boltzmann-code calculations. The entropy budget follows Egan & Lineweaver (2010, ApJ 710, 1825) with \(S_{\rm obs}=3.1\times10^{104}k\). The Weyl curvature hypothesis is Penrose's proposal and the definition of gravitational entropy is unsettled. The relation between the conformal-factor negative mode and the Gross–Perry–Yaffe negative mode of Euclidean Schwarzschild has recently been shown to resist naive identification because of transverse-traceless/trace mixing. The covariant entropy bound is due to Bousso (1999). The academic standard is the ΛCDM model including inflation. — To print, use your browser's Print → Save as PDF (in the print version the slider is frozen and answers are hidden).

Print / PDF: ⌘+P (Ctrl+P on Windows). On screen the slider changes the amount of radiation and shows the constancy of the address space breaking down. "Show answer" opens the solutions.