c·t = CONST, THAT CLICKS EPISODE 6 ／ Three episodes of dividing — now open the contents
Being empty turned out to be why this tool works
Only \(10^{-18}\) of the
memory is in use
A machine that performs one operation per 28.5 bits holds that memory almost entirely empty.
Count the emptiness and out falls the reason this series exists at all.
Episode 1 divided memory by operations, Episode 2 divided two clocks, Episode 5 divided fit by parameters. This time we do not divide — we open the memory and look inside. Capacity: \(10^{122}\) bits. In use: \(1.5\times10^{-18}\) of it. Essentially empty. And chasing that emptiness lands us on why the conformal transformation works on the universe at all — the foundation this whole series stands on.
01Capacity, and how much is used
Capacity (the holographic bound on the horizon)
$$\frac{S_{\max}}{k_B}=\frac{A_H}{4\ell_P^2}=2.05\times10^{122}\qquad(=2.96\times10^{122}\ \text{bit})$$In use (the census of Egan & Lineweaver 2010)
$$\frac{S_{\rm obs}}{k_B}=3.1\times10^{104}$$Divide
$$\text{occupancy}=1.51\times10^{-18}$$Written out in zeros, 99.9999999999999998% of it is free. The previous series quoted this number once; here we dig past it.
02The occupancy was a ratio of areas all along
Today's entropy is almost entirely supermassive black holes. And a black hole's entropy and the cosmic horizon's entropy are written by exactly the same formula — \(S=k_BA/4\ell_P^2\). So the division cancels \(\ell_P\) and leaves a bare ratio of areas.
Conclusion of §02
$$\text{occupancy}=\frac{\sum A_{\rm BH}}{A_H}=1.51\times10^{-18}$$Add up the horizons of every black hole in the universe and divide by the area of the cosmic horizon — that is what "memory occupancy" is.
as a radius
$$r=\sqrt{\frac{\sum A_{\rm BH}}{4\pi}}=1.61\times10^{17}\ \mathrm{m}=\mathbf{17\ \text{light years}}$$Sew together the horizons of every black hole in the observable universe and you get a sphere of radius 17 light years — 34 across, about far enough from the Sun to just include Vega and Arcturus. Nearly all of the universe's information is written on that.
03Which side is in use?
Extra 2 of the previous series had a table splitting the gravitational field in two — the conformal-factor side (gauge, bookkeeping, no arrow of time) and the Weyl-tensor side (physics, gravitational entropy, arrow of time). That was structure only. Now we can put numbers in.
| Contents | \(S/k_B\) | Share | Which side |
|---|---|---|---|
| Supermassive black holes | \(3.1\times10^{104}\) | ≈ 1 | Weyl side |
| CMB photons | \(2.03\times10^{88}\) | \(6.5\times10^{-17}\) | matter/radiation |
| Neutrinos | \(1.93\times10^{88}\) | \(6.2\times10^{-17}\) | matter/radiation |
Conclusion of §03
99.999999999999986% of today's cosmic entropy sits on the gravitational side.
The table from Extra 2 has acquired numbers. The memory in use is almost entirely the Weyl side. The conformal-factor side — the \(\phi\) this series has been moving all along, the expansion, the scale factor — carries no entropy whatsoever.
04Following the history — full at the start, then empty
Entropy density \(s\propto a^{-3}\), bound \(\propto H\propto1/t\)
$$\frac{s}{3H/4\ell_P^2}\ \propto\ t^{\,1-3p}$$\(t^{-1/2}\) in radiation (\(p=1/2\)), \(t^{-1}\) in matter (\(p=2/3\)); stacking from the Planck era
$$\underbrace{1.84\times10^{-28}}_{\text{Planck}\to\text{equality}}\times\underbrace{3.68\times10^{-6}}_{\text{equality}\to\text{today}}=6.7\times10^{-34}$$Entropy itself really has grown — from \(2\times10^{88}\) around recombination to \(3.1\times10^{104}\) today, 16.2 orders. But almost all of that increase is on the gravitational side.
The thing this episode most wants to say
The universe began thermally full and gravitationally empty.
Expansion widened the capacity and the thermal occupancy kept falling.
And only black holes are filling it back in — that is what the arrow of time is.
Figure: age of the universe (in orders of magnitude above the Planck time) across, memory occupancy up. Blue-grey is the thermal side (falling from full), old-gold is the gravitational side (refilling once structure forms). The slider picks an epoch and reports the breakdown.
The blue-grey line touches the ceiling at the far left (the Planck era) and falls from there without stopping. The thermal history of the universe is a history of memory emptying out. The old-gold line rises only near the right edge and overtakes the blue-grey by 16 orders — the universe only started remembering anything very recently.
05The reveal — the tool works because that side is empty
Here is the point of the episode: why does the conformal transformation work so well on the universe?
An FLRW universe is exactly conformally flat — its Weyl tensor vanishes identically. That is what let Episode 3 of the previous series transform a \(c\cdot t=\text{const}\) universe into Minkowski. But the Weyl tensor is gravitational entropy, so \(C=0\) means "the gravitational memory is empty".
| Conformal-factor side | Weyl-tensor side | |
|---|---|---|
| Under a conformal transformation | moves (this is the gauge) | does not move (\(C\) is conformally invariant) |
| Entropy | none | yes (= gravitational entropy) |
| In use today | \(0\) | \(3.1\times10^{104}\) (essentially all of it) |
| This series' tool | moves only this | cannot touch it |
Conclusion of §05
A conformal transformation can only move the unused side of the universe's memory.
The occupancy \(1.5\times10^{-18}\) is, directly, a measurement of this tool's range.
So when Episode 4 deleted everything deletable and was left with one mass, everything that vanished lay on the empty side — expansion, curvature, temperature. Conversely, everything that could not be deleted — the conformal factor problem of Episode 9 of the previous series, the \(N=mc^2t/\hbar\) of Episode 6, black hole entropy — sits on the side that does not move.
① The usage figure is the census of Egan & Lineweaver (2010) (\(S_{\rm obs}=3.1\times10^{104}k_B\), dominated by supermassive black holes). It depends strongly on the SMBH mass function, and the authors themselves acknowledge order-of-magnitude uncertainty.
② "Capacity" means \(A/4\ell_P^2\) on the horizon, not what can actually be stored. The holographic bound is an inequality saying "no more than this fits", not a guarantee that this much is usable. Free space does not mean usable space.
③ "Conformally flat" means the Weyl tensor vanishes, not that curvature vanishes. FLRW has \(C=0\) but \(R\ne0\) (Episode 6 of the previous series).
④ The Weyl curvature hypothesis is a proposal, not a theorem. It is Penrose's conjecture and is neither proved nor disproved.
⑤ Gravitational entropy has no established definition — there is still no standard prescription for how to count \(C\). What the text calls "entropy on the Weyl side" is in fact the sum of black hole entropies, not a quantity computed from \(C\). In addition, the history in §04 assumes comoving conservation of entropy (no reheating production), and the gold curve in the figure is schematic — Egan & Lineweaver give only today's value.
Exercises (solvable with this episode's formulas alone)
- Why does "memory occupancy" reduce to a ratio of areas?
Show the answer
Because a black hole and the cosmic horizon carry entropy by the same formula \(S=k_BA/4\ell_P^2\). Dividing cancels \(4\ell_P^2\) and leaves \(\sum A_{\rm BH}/A_H\). Even the Planck length disappears — occupancy is a pure geometric ratio. - Glue every black hole horizon into one sphere: what radius, in light years?
Show the answer
\(\sum A_{\rm BH}=1.51\times10^{-18}\times2.14\times10^{53}=3.24\times10^{35}\ \mathrm{m^2}\), so \(r=\sqrt{A/4\pi}=1.61\times10^{17}\) m = 17 light years. Nearly all of the universe's information is written on a surface 34 light years across. - Why does occupancy fall with time even though entropy is rising?
Show the answer
Because the denominator rises faster. Capacity grows as \(A_H\propto R_H^2\propto t^2\), while thermal entropy is comoving-conserved (roughly constant). So the ratio falls. Rising entropy and falling occupancy are perfectly compatible — the second law is about the numerator, expansion is about the denominator. - Which side of the memory can a conformal transformation move?
Show the answer
The conformal-factor side — i.e. the unused side only. The Weyl tensor is conformally invariant and cannot be touched. Since today's \(3.1\times10^{104}\) is essentially all on the Weyl side, this series' tool cannot reach the memory that is actually in use. - (Harder) What does "how badly the tool is breaking is the arrow of time" mean?
Show the answer
The conformal transformation works perfectly when \(C=0\) (conformal flatness), and by the Weyl curvature hypothesis the universe started there. Growing gravitational entropy = growing \(C\) = a growing component that cannot be rewritten conformally. So "the direction in which the tool's range narrows" is itself the direction of time, and \(1.5\times10^{-18}\) is the current reading. Note, per caveat ⑤, that gravitational entropy has no settled definition, so this correspondence is not a rigorous theorem.
Summary — being empty is why the tool works
The universe's memory: capacity \(2.96\times10^{122}\) bits, usage \(3.1\times10^{104}\), occupancy \(1.51\times10^{-18}\) — essentially empty. And because black holes and horizons obey the same \(S=k_BA/4\ell_P^2\), that occupancy reduces to a bare area ratio \(\sum A_{\rm BH}/A_H\) — glue every black hole horizon together and you get a sphere of radius 17 light years, carrying nearly all the information in the universe.
Counting which side that "nearly all" is on gives 99.999999999999986% gravitational (Weyl). The table from Extra 2 of the previous series — the gravitational field splitting into conformal factor and Weyl tensor — now has numbers in it. The conformal-factor side, the \(\phi\) this series keeps moving, carries no entropy at all.
The history came in three stages: occupancy ≈ 1 in the Planck era (full); \(7\times10^{-34}\) today counting only matter and radiation (emptied out); \(1.5\times10^{-18}\) once black holes are included (gravity refilled 15.4 orders). Entropy itself rose 16.2 orders, almost all of it gravitational. The universe began thermally full and gravitationally empty.
And the reveal: a conformal transformation can only move the unused side of the memory. The Weyl tensor is invariant and untouchable. So everything Episode 4 could delete lay on the empty side, and everything it could not (the conformal factor problem, \(N=mc^2t/\hbar\), black hole entropy) lay on the used side. If the universe began at \(C=0\) as Penrose's hypothesis says, then this tool worked perfectly at the beginning and fails in proportion to the gravitational entropy since accumulated. The breakage of the tool is the arrow of time.
This document is Episode 6 of "c·t = const, That Clicks", written for physics-minded high-school and university readers. The horizon entropy \(S=k_BA/4\ell_P^2\), the identical form of the Bekenstein–Hawking entropy of a black hole, and the fact that FLRW spacetimes are conformally flat with vanishing Weyl tensor are all standard. The entropy budget of the observable universe — \(S_{\rm obs}=3.1\times10^{104}k_B\) (dominated by supermassive black holes), CMB photons \(2.03\times10^{88}k_B\), neutrinos \(1.93\times10^{88}k_B\) — is from Egan & Lineweaver (2010, ApJ 710, 1825). These values depend on the supermassive black hole mass function, and the authors themselves state an order-of-magnitude uncertainty. The occupancy \(1.51\times10^{-18}\), its reduction to the area ratio \(\sum A_{\rm BH}/A_H\), the equivalent radius of 17 light years, the gravitational share 99.999999999999986%, and the three-stage history (Planck era \(O(1)\), today's thermal side \(7\times10^{-34}\)) are all computed here. The history in §04 rests on \(s\le3H/4\ell_P^2\) from Extra 3 of the previous series (Bousso's 1999 covariant entropy bound applied to the apparent horizon) and assumes comoving conservation of entropy; the "\(O(1)\) at the Planck era" comes from the same one-loop estimate. The gravitational curve in the figure is schematic — Egan & Lineweaver give only today's value. The Weyl curvature hypothesis is Penrose's proposal, not a theorem, and gravitational entropy has no established definition — what the text calls "entropy on the Weyl side" is the sum of black hole entropies, not a quantity computed from \(C\). The holographic bound is an inequality on what fits, not a guarantee of storable capacity. The academic standard remains the \(\Lambda\)CDM model including inflation. ── To make a PDF, use your browser's Print dialogue (sliders freeze and answers are hidden in the print version).