Bonus ② concluded that it was constraining the wrong thing. So what is the right way to say it?
In bonus ② we found that \(c\cdot t=\text{const}\) is the unique expansion law with a constant comoving Hubble radius — the condition that a computer's address space does not move. This time we push that down into an even more primitive language: nothing but cells and ticks. It becomes one line — the horizon advances one Planck length per Planck time. And the real subject of this episode is what comes next. Why "one"? The answer sits at the equality case of the most basic inequality in general relativity.
Compute the growth rate of the Hubble radius \(R_H=c/H\). For \(a\propto t^p\) we have \(H=p/t\), hence \(R_H=ct/p\), and so
$$\frac{dR_H}{dt}=\frac{c}{p}$$Measure that in Planck units (length \(\ell_P\), time \(t_P\), with \(\ell_P/t_P=c\)) and every unit cancels, leaving a bare number.
| \(p\) | cells/tick | What happens |
|---|---|---|
| 1/2 (radiation) | 2.000 | The horizon outruns light → regions never in contact come in |
| 2/3 (matter) | 1.500 | Likewise |
| 1 (\(c\cdot t=\)const) | 1.000 | Lockstep with light. Nothing enters, nothing leaves |
| 2 | 0.500 | Light is faster → regions already in contact leave |
The horizon advances one Planck length per Planck time. The lockstep of a cellular automaton, exactly.
Integrating \(dR_H/dt=c\) gives \(R_H=ct\). Measure both sides in Planck units —
Cells across the horizon
$$\frac{R_H}{\ell_P}=\frac{1.306\times10^{26}\,\mathrm{m}}{1.616\times10^{-35}\,\mathrm{m}}=8.0776\times10^{60}$$Ticks elapsed
$$\frac{t_0}{t_P}=\frac{4.355\times10^{17}\,\mathrm{s}}{5.391\times10^{-44}\,\mathrm{s}}=8.0776\times10^{60}$$Ratio
$$\frac{R_H/\ell_P}{t_0/t_P}=1.000000$$The universe has run \(8\times10^{60}\) ticks, and its radius is \(8\times10^{60}\) cells. One cell per tick, exactly.
The memory, though, does not follow. The bit count is an area — proportional to the square of the radius.
$$N=\frac{\pi(R_H/\ell_P)^2}{\ln2}=2.96\times10^{122}\ \text{bit},\qquad \frac{dN}{d(\text{tick})}=7.3\times10^{61}\ \text{bit/tick}$$The radius gains one cell per tick while the memory gains \(10^{61}\) bits per tick. That is the origin of bonus ②'s "rich in memory, poor in clock cycles" — the area goes as the square of the radius.
This is the heart of the episode. The "one" is not a postulate; it appears as the equality case of a more basic inequality.
What governs how a bundle of light rays or observers converges is the Raychaudhuri equation. For a congruence of comoving observers (no shear, no vorticity), the expansion \(\theta=3H\) obeys
The Raychaudhuri equation
$$\frac{d\theta}{d\tau}=-\frac{\theta^2}{3}-4\pi G(\rho+3p)$$Insert FLRW (\(\theta=3H\)) and a familiar equation appears
$$\frac{\ddot a}{a}=-\frac{4\pi G}{3}(\rho+3p)$$The \(\rho+3p\) on the right is the active gravitational mass. Positive means pulling together (deceleration); negative means pushing apart (acceleration).
The condition \(\rho+3p\ge0\) has a name — the strong energy condition (SEC), the requirement that ordinary matter acts to converge light rays. And
\(c\cdot t=\text{const}\) is the exact saturation of the strong energy condition — it sits on the equality of the inequality.
At saturation the second term of the Raychaudhuri equation vanishes entirely.
$$\frac{d\theta}{d\tau}=-\frac{\theta^2}{3}\qquad\Longrightarrow\qquad \theta=\frac{3}{t}$$The expansion decays only because of itself. Matter contributes nothing at all to the focusing of light rays. Melia himself calls this "zero active gravitational mass".
If matter contributes nothing to the focusing, the causal structure ought to be the same as with no matter at all. And it is.
| scale factor | spatial curvature | density | expansion \(\theta\) | |
|---|---|---|---|---|
| Milne universe (empty) | \(a\propto t\) | \(k=-1\) | \(\rho=0\) | \(3/t\) |
| \(R_h=ct\) (with matter) | \(a\propto t\) | \(k=0\) | \(\rho\ne0\) | \(3/t\) |
The expansion history of the congruence — how light rays converge — agrees exactly. Only the spatial curvature differs. And the amount of matter whose net gravitational effect is zero, in today's universe, is
$$M=\frac{c^2R_H}{2G}=8.8\times10^{52}\ \mathrm{kg}=4.4\times10^{22}\ M_\odot$$Four times \(10^{22}\) suns' worth of matter, with exactly zero net gravitational effect. Push and pull balance precisely.
The strong energy condition carries one more heavy role — it is the hypothesis of the Hawking–Penrose singularity theorems.
| \(w\) | SEC | The singularity theorems |
|---|---|---|
| \(w>-1/3\) | holds | The theorems apply → the singularity is unavoidable (radiation and matter live here) |
| \(w=-1/3\) | saturated | sitting exactly on the boundary |
| \(w<-1/3\) | violated | The hypothesis fails (inflation lives here) |
In Episode 6 we wrote that after the conformal transformation the singularity appeared to vanish, and that building a dimensionless ratio showed it barely surviving, logarithmically. That was no accident. \(c\cdot t=\text{const}\) sits exactly on the boundary of the singularity theorems, so the singularity does not disappear, but survives in the weakest possible form.
Indeed the proper time of a comoving observer is \(\tau=t\), so \(t=0\) is reached in finite time — geodesically incomplete, a singularity. But every divergence is logarithmic (Episode 6's \(\eta\to-\infty\), bonus ②'s sound horizon). The weakest singularity there is.
While we are here, a famous "coincidence" gets disposed of. In flat FLRW,
In Planck units
$$\frac{M}{m_P}=\frac{R_H}{2\ell_P}$$This is an identity for any \(p\). The Hubble radius is always exactly the Schwarzschild radius of its contents.
So Dirac's "coincidence of large numbers", \(M/m_P\sim R/\ell_P\), is — a restatement of the Friedmann equation. Not a coincidence but a relation that follows from the definitions. Check: \(R/\ell_P=8.078\times10^{60}\), \(M/m_P=4.039\times10^{60}\), a ratio of exactly 1/2.
Put into the language of information it is more compact still.
Here are all the characterisations of \(a\propto t\) the series has produced.
All seven are restatements of \(a\propto t\). What Episode 5 called a "triple boundary" has become sevenfold. That so many properties collect at a single point is itself remarkable — which is, I think, why this auxiliary line has held people's attention for so long.
Beauty and correctness are different things — the consistent position of this series. Let us confirm the verdict one last time.
| Verdict | Content | Source |
|---|---|---|
| ✗ | Nucleosynthesis makes only \(10^{-3}\) of the observed helium | bonus ① |
| ✗ | Radiation cannot couple, so it breaks for \(z>103\); the horizon problem returns | bonus ② |
| ✗ | \(w_{\rm DE}=-0.476\) (18σ from the observed \(-1.03\pm0.03\)); \(\beta=0.954\) (280× Cassini) | bonus ② |
| ✗ | Violates Bousso's covariant entropy bound for \(t<5\) s (25× at \(t=1\) s; \(10^{88}\) at the Planck era) | this episode |
The last one is this episode's calculation, and it is worth doing carefully — we apply Bousso's covariant entropy bound to this universe head on.
① Pick the surface
Bousso's bound says: the entropy crossing a converging (\(\theta\le0\)) null hypersurface — a light sheet — emanating from a two-dimensional surface of area \(A\) is at most \(A/4\ell_P^2\). In flat FLRW the natural surface is the apparent horizon \(R_A=1/H\) (the Hubble radius). Light rays leaving it inward and toward the past see both \(a\) and \(r\) decrease, so the area decreases — a legitimate light sheet.
② The entropy crossing the light sheet
Every piece of matter inside the comoving ball of radius \(r_A\) crosses this light sheet exactly once. So
$$S=s(t)\cdot\frac{4\pi}{3}R_A^3,\qquad A=4\pi R_A^2$$③ Rearranging \(S\le A/4\ell_P^2\)
$$\boxed{\ \ s\ \le\ \frac{3H}{4\ell_P^2}\ \ }$$The ceiling on entropy density is set by the Hubble rate.
Now the time dependence. With \(s\propto a^{-3}\propto t^{-3p}\) and \(H\propto1/t\),
$$\frac{s}{3H/4\ell_P^2}\ \propto\ t^{\,1-3p}$$| \(p\) | \(1-3p\) | Going back in time |
|---|---|---|
| 1/3 (stiff, \(w=1\)) | 0 | The ratio is constant — saturate once and you stay saturated |
| 1/2 (radiation) | −0.5 | Approaches slowly |
| 2/3 (matter) | −1 | Approaches |
| 1 (\(c\cdot t=\)const) | −2 | Violates fastest of all |
Put numbers in and the gap against standard cosmology becomes bizarre.
Today
$$s=2.9\times10^{9}\ \mathrm{m^{-3}},\qquad \frac{3H_0}{4\ell_P^2}=2.2\times10^{43}\ \mathrm{m^{-3}}\qquad\Longrightarrow\qquad \text{usage}=1.3\times10^{-34}$$\(a\propto t\) (ratio \(\propto t^{-2}\))
$$\text{ratio}=1\ \text{at}\ t=5.0\ \text{s},\qquad 25\ \text{at}\ t=1\ \text{s},\qquad 8.6\times10^{87}\ \text{at}\ t_P$$Standard cosmology (radiation dominated, ratio \(\propto T\))
$$s=\frac{2\pi^2}{45}g_*T^3,\quad H=1.66\sqrt{g_*}\,\frac{T^2}{M_{\rm Pl}}\quad\Longrightarrow\quad T\le\frac{2.84\,M_{\rm Pl}}{\sqrt{g_*}}\ \ (0.28\,M_{\rm Pl}\ \text{for}\ g_*=100)$$In other words standard cosmology sits on the bound at \(O(1)\) in the Planck era and satisfies it ever after, whereas \(a\propto t\) violates it by \(10^{88}\) in the Planck era.
That the violation begins at \(t=5\) seconds matters too — the neutron freeze-out of nucleosynthesis (\(t\sim1\) s) falls entirely inside it. Bonus ①'s failure of nucleosynthesis and this overrun of the information budget point at the same epoch.
One last question, asked backwards. What universe do you get if you demand \(s=3H/4\ell_P^2\) identically?
That is \(w=1\) (stiff matter / kination), matching the \(1-3p=0\) entry in the table above ✓
Hold the bit count (horizon entropy) fixed → \(a\propto e^{Ht}\) de Sitter
Hold the address space (comoving Hubble radius) fixed → \(a\propto t\) \(c\cdot t=\)const
Hold the memory usage (holographic saturation) fixed → \(a\propto t^{1/3}\) \(w=1\)
So — \(c\cdot t=\text{const}\) optimises the addressing, not the memory usage. As a finite-resource computer, which resource you choose to hold fixed splits the answer three ways. This is the most concrete form of bonus ②'s conclusion that it was constraining the wrong thing: constrain the information on light sheets and out comes \(a\propto t^{1/3}\), not \(a\propto t\).
Section 01's "the horizon outruns light" (\(dR_H/dt>c\) for \(p<1\)) is not a violation of causality. The Hubble radius is not a causal boundary, just the length \(c/H\); nothing is moving faster than light. "Cells per tick" is a metaphor from computing, not a claim that spacetime is really a lattice.
Section 08's \(s\le3H/4\ell_P^2\) is Bousso's covariant entropy bound applied to one particular surface, the apparent horizon. Bousso's bound makes a claim about every surface, so other surfaces could give stronger conditions — what is shown here is that it fails on at least this one. The calculation also assumes that entropy is comovingly conserved; the real universe has entropy production (reheating and so on), so \(s\) further back may have been smaller. But the violation for \(a\propto t\) is \(10^{88}\), which is a great deal to make up that way. The \(T\le0.28M_{ m Pl}\) on the standard-cosmology side is likewise a one-loop-level estimate whose coefficient depends on the choice of \(g_*\).
Drop \(c\cdot t=\text{const}\) into the language of computers and it becomes one line — \(dR_H/dt=c\): the horizon advances one Planck length per Planck time. Integrate and \(R_H/\ell_P=t/t_P=8.08\times10^{60}\): the cell count and the tick count agree exactly. The name Melia gave it, \(R_h=ct\), was "one cell per tick" all along.
And the answer to "why one?" is in the Raychaudhuri equation. The right-hand side of \(\ddot a/a=-(4\pi G/3)(\rho+3p)\) vanishes — the exact saturation of the strong energy condition. Matter contributes nothing to the focusing of light rays, and the causal structure is that of an empty universe (memory and addressing separated). It also sits on the boundary of the singularity theorems, so the singularity does not vanish but survives only in the weakest form. As a bonus, Dirac's large-number hypothesis turned out to be a mere identity. Seven characterisations converging on one point — that is why the auxiliary line is beautiful. And still, nucleosynthesis, radiation, observation and the holographic bound each say no, independently.
The last of those came out of this episode, by applying Bousso's covariant entropy bound to the apparent horizon — \(s\le3H/4\ell_P^2\). Standard cosmology sits on that bound at \(O(1)\) in the Planck era and satisfies it ever after; \(a\propto t\) begins violating it going back from \(t=5\) seconds and is \(10^{88}\) over by the Planck era. And working backwards from the bound, the expansion law that saturates it identically is \(a\propto t^{1/3}\) (\(w=1\)), not \(a\propto t\). What \(c\cdot t= ext{const}\) optimises is the addressing, not the memory usage.
Print / PDF: ⌘+P (Ctrl+P on Windows). On screen the slider changes the exponent of the expansion law and shows where the horizon's staircase coincides with light. "Show answer" opens the solutions.