c·t = CONST, THAT CLICKS EPISODE 1 ／ Starting from the specification sheet

Divide the memory by the operations and out comes the equation of state

The universe has computed
0.035 operations per bit "The universe is a computer with finite resources" — we take that view and write it out
to the end, not as a metaphor but as a specification sheet. The very first sheet already goes strange.

What you need: division, Planck units, logarithms\(\Omega/N=\dfrac{\ln 2}{3\pi^2(1+w)}\)

The previous series, "Conformal Transformations That Click", closed by ruling that \(c\cdot t=\text{const}\) was flawless as a way of speaking, disqualified as a model. This new series digs into the one thing left over there — that \(c\cdot t=\text{const}\) is what physics looks like seen through information theory. We begin by taking the specifications of the machine: memory, clock, operation count. Line all three up, divide them, and the most basic quantity in cosmology comes back wearing a completely different face.

01Memory — how many bits fit on the horizon

The first specification of any computer is its memory. For the universe, the holographic principle already has the answer: information is written not on the volume but on the area. Divide the horizon area \(A\) by the Planck area, then by \(4\ln 2\), and you have a bit count.

Calculation — memory

Bits on the horizon

$$N=\frac{A}{4\ell_P^2\ln 2}=\frac{\pi}{\ln 2}\left(\frac{R_H}{\ell_P}\right)^2$$

Numbers (\(R_H=c\,t_0=1.305\times10^{26}\) m, \(\ell_P=1.616\times10^{-35}\) m)

$$\frac{R_H}{\ell_P}=8.075\times10^{60}\qquad\Longrightarrow\qquad N=2.956\times10^{122}\ \text{bit}$$

That is the entire memory of the universe. All the storage on Earth comes to something like \(10^{22}\) bits, so this is \(10^{100}\) times more.

02Clock — how many ticks have gone by

The second specification is the clock. Count one Planck time as one tick.

Calculation — clock $$\frac{t_0}{t_P}=\frac{4.354\times10^{17}\,\mathrm{s}}{5.391\times10^{-44}\,\mathrm{s}}=8.075\times10^{60}\ \text{ticks}$$

Counted logarithmically

$$\ln\frac{t_0}{t_P}=140.2$$

You will have noticed that \(R_H/\ell_P\) for the memory and \(t_0/t_P\) for the clock came out as the same \(8.075\times10^{60}\). That is "one cell per tick" from bonus episode ③ of the previous series — \(dR_H/dt=c\), which is to say \(c\cdot t=\text{const}\) itself. We leave that alone here and press on.

The logarithm may be the real clock The number 8×10⁶⁰ is less meaningful, as a machine specification, than \(\ln(t_0/t_P)=140\). The history of the universe is not "10⁶⁰ repetitions of the same operation" but 140 doublings. Indeed, memory grows by a factor of \(e^2=7.39\) per logarithmic tick (since \(N\propto t^2\)); raise that to the 140th and you get \(e^{280}\approx10^{122}\) — the memory figure comes back exactly. The universe is a machine that has run only 140 steps.

03Operations — how many times has the state changed

The third specification is the operation count. The physical floor on "one operation" is the Margolus–Levitin limit: a system of energy \(E\) needs at least \(\pi\hbar/2E\) to move to an orthogonal state. So the rate of operations is bounded by \(2E/\pi\hbar\).

The total energy inside the horizon is, in any flat FLRW universe, \(E=Mc^2=c^4R_H/2G\) — the identity from §06 of bonus episode ③ (the one that turned out to be the real content of Dirac's large-number hypothesis). Writing the expansion law as \(a\propto t^{p}\) gives \(R_H=ct/p\), so the rate grows in proportion to time.

Calculation — operations, in three lines

① The rate

$$\frac{d\Omega}{dt}=\frac{2E}{\pi\hbar}=\frac{c^4R_H}{\pi\hbar G}=\frac{c^5\,t}{\pi\hbar G\,p}$$

② Integrate

$$\Omega=\frac{c^5t^2}{2\pi\hbar G\,p}$$

③ Use \(\ell_P^2=\hbar G/c^3\) and every unit cancels

$$\Omega=\frac{1}{2\pi p}\left(\frac{ct}{\ell_P}\right)^2$$

Putting in numbers for \(p=1\) (that is, \(c\cdot t=\text{const}\)) gives \(\Omega=1.038\times10^{121}\) operations — the same order as Lloyd's well-known estimate (2002) that the universe has performed \(10^{120}\) operations.

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04The core — divide the memory by the operations

Three specifications in hand. When you judge a computer, this is the number you most want to see: how many operations per bit? Is it a machine that is all memory and no work, or the other way round?

Try it, and something surprising happens.

Calculation — just divide $$\frac{\Omega}{N}=\frac{1}{2\pi p}\left(\frac{ct}{\ell_P}\right)^{2}\ \Bigg/\ \frac{\pi}{\ln 2}\left(\frac{ct}{p\,\ell_P}\right)^{2} =\frac{p\,\ln 2}{2\pi^{2}}$$

The \(t\) has cancelled cleanly. Memory and operation count both grow as \(t^2\), so the ratio does not depend on the age of the universe. One second after the beginning, today, or \(10^{100}\) years from now, the number of operations per bit is exactly the same.

And what is left, \(p\), is the expansion law itself. Substituting the relation to the equation of state, \(p=2/3(1+w)\) —

The result of this episode

$$\boxed{\ \frac{\Omega}{N}=\frac{\ln 2}{3\pi^{2}\,(1+w)}\ }$$

The number of operations per bit is a pure number fixed by the equation of state alone. It does not depend on the age of the universe, nor its size, nor any detail of its contents.

The most basic quantity in cosmology (the equation of state) and the most basic ratio in computer science (operations divided by memory) are one and the same number.

Run it through the decision procedure Following the method of the previous series' finale, sort this quantity into one of two columns. Both \(\Omega\) and \(N\) are plain counts; they have no dimensions. So a conformal transformation — a change of ruler — leaves them alone: right-hand column, physics. And not merely unmoved: it reads off \(w\) directly. Which means "measuring the expansion law of the universe" and "measuring its computational efficiency" are the same job.

05Four universes, side by side

Universe\(w\)\(p\)\(\Omega/N\)one operation per…
stiff (kination)\(+1\)1/30.0117185.4 bits
radiation\(+1/3\)1/20.0175657.0 bits
matter\(0\)2/30.0234142.7 bits
\(c\cdot t=\)const\(-1/3\)10.0351228.5 bits
de Sitter\(-1\)∞divergesgrows with time

Not one of them comes close to 1. In 13.8 billion years the universe has never once performed one operation per bit. Nor is that because it is young: the ratio does not depend on time, so it has always been like this.

06The eighth characterisation

Let us restate the \(c\cdot t=\text{const}\) row a little more precisely. Since \(p=ct/R_H\), the boxed result can be written like this:

The same formula, as a ratio $$\frac{\Omega}{N}=\frac{\ln 2}{2\pi^{2}}\cdot\frac{ct}{R_H}$$

and therefore

$$R_h=ct\qquad\Longleftrightarrow\qquad \frac{\Omega}{N}=\frac{\ln 2}{2\pi^{2}}=0.0351152\cdots$$

Bonus episode ③ of the previous series listed seven characterisations of \(a\propto t\) and remarked that "having this many properties meet at a single point is itself remarkable". We can add an eighth.

①
The potential is a pure quadratic\(U=-\phi^2/2t_0^2\) (Conformal Transformations, Ep.5)
②
Conformal time spans all of \((-\infty,\infty)\)it covers the whole of Minkowski space (Ep.6)
③
Zero acceleration\(\ddot a=0\); the watershed for the horizon problem (Ep.5)
④
The comoving Hubble radius is constantthe address space does not move (bonus ②)
⑤
Both the particle horizon and the event horizon are infiniteread and write access to the whole memory
⑥
\(dR_H/dt=c\), cell count = tick countone cell per tick (bonus ③)
⑦
Saturation of the strong energy condition\(\rho+3p=0\); the boundary of the singularity theorems (bonus ③)
⑧
Operations per bit equal \(\ln 2/2\pi^2\)the ratio of operations to memory matches the ratio of distance-light-has-travelled to horizon radius (this episode)

⑥ and ⑧ are two faces of the same fact. What ⑥ says in "lengths and times", ⑧ says in "operations and bits" — the dimensionful phrasing and the dimensionless one. By the decision procedure of the previous series, ⑧ sits in the right-hand column.

07So is there a universe that reaches "1"?

Solve the boxed result backwards. Demanding \(\Omega/N=1\) gives —

Solving backwards $$1+w=\frac{\ln 2}{3\pi^{2}}=0.02341\qquad\Longrightarrow\qquad \boxed{\,w=-0.97659\,}$$

Almost exactly \(-1\): just short of a cosmological constant. And the equation of state observed for dark energy is \(w=-1.03\pm0.03\) — already on the far side of it.

The universe has just entered its dark-energy-dominated era. Under de Sitter-like expansion \(N\) saturates to a constant while the operation rate does not fall off, so the ratio grows without bound.

The ratio in de Sitter $$\frac{\Omega}{N}=\frac{\ln 2}{\pi^{2}}\,Ht\qquad\Longrightarrow\qquad \frac{\Omega}{N}=1\ \text{at}\ Ht=\frac{\pi^{2}}{\ln 2}=14.24$$

with \(H_\Lambda=H_0\sqrt{\Omega_\Lambda}=1.81\times10^{-18}\,\mathrm{s^{-1}}\)

$$t\approx2.5\times10^{11}\ \text{years}\qquad(\text{about 250 billion years from now})$$

The main point of this episode

In 13.8 billion years the universe has never once performed one operation per bit.
And it has only just now entered the expansion law that will carry it across that line.

Figure: the horizontal axis is the logarithmic age of the universe (\(\ln(t/t_P)\); today is 140.2). In a power-law universe the operations-per-bit line is perfectly flat, and turning the \(w\) knob only raises or lowers it. The copper dashed line marks one operation per bit — a power-law universe reaches it only once \(w<-0.977\), that is, only just short of a cosmological constant

w = −0.333
operations per bit (power law) one operation per bit the Λ-dominated future

Push the knob left (\(w\to-1\)) and the flat line rises slowly. Even so it only touches the copper dashes once \(w=-0.977\) is passed — only just short of a cosmological constant does the machine stop being a bad deal. Push it right (matter, radiation, stiff) and the line only sinks. The grey dashes are the \(\Lambda\)-dominated future, the one place that is not flat.

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08The reveal — why both were \(t^2\)

The ratio came out independent of time because both \(N\) and \(\Omega\) went as \(t^2\). That is no accident.

quantitywhy \(t^2\)
memory \(N\)because it is an area: \(N\propto R_H^2\) and \(R_H\propto t\). This is the holographic principle
operations \(\Omega\)because the rate goes as \(E\propto R_H\propto t\) and you integrate it. The identity \(E=(c^2R_H/2G)\,c^2\) is what is doing the work

That second identity is the one bonus episode ③ used to dispose of Dirac's large-number hypothesis as "just an identity". The Hubble radius always equals the Schwarzschild radius of what is inside it. So the energy of the universe is proportional to its radius, and so is its operation rate.

The chain $$\underbrace{N\propto R_H^{2}}_{\text{holography}}\qquad \underbrace{\frac{d\Omega}{dt}\propto E\propto R_H}_{\text{the large-number identity}}\qquad\Longrightarrow\qquad \frac{\Omega}{N}\propto\frac{\int R_H\,dt}{R_H^{2}}=\text{(a function of the expansion law alone)}$$

"Operations are proportional to memory" is the product of holography and the large-number identity. Neither is special to \(c\cdot t=\text{const}\); both hold in any flat FLRW universe — which is why the division leaves nothing behind but the expansion law.

09An aside — memory is installed faster than work is done

Today's values (\(p=1\)) $$\frac{dN}{dt}=\frac{2N}{t_0}=1.36\times10^{105}\ \mathrm{bit/s},\qquad \frac{d\Omega}{dt}=\frac{2\Omega}{t_0}=4.77\times10^{103}\ \mathrm{ops/s}$$ $$\frac{dN/dt}{d\Omega/dt}=\frac{2\pi^{2}}{p\ln 2}=28.5$$

In the time it takes the universe to perform one operation, it installs 28.5 bits of memory. As a computer design that is bizarre — memory keeps piling up while the speed of processing what is written on it never comes close to keeping pace. When bonus episode ② of the previous series called the universe extremely memory-rich and clock-poor, this ratio is what it meant. Today that "richness" has a value: \(2\pi^2/\ln2\).

And almost entirely empty By the count in bonus episode ②, the entropy the universe actually uses is \(1.5\times10^{-18}\) of its capacity. Which is to say — a machine that performs one operation per 28.5 bits is using \(10^{-18}\) of that memory. Both writing and computing are nowhere near their limits. The universe is indeed a computer with finite resources, but it is not remotely using up the finite resources it has. What that emptiness means is the subject of Episode 6.
Being straight with you

"Operation" is defined by the Margolus–Levitin limit as an upper bound on transitions to orthogonal states. It is not a claim that meaningful computation is going on, nor that the universe actually runs at that bound — this is a specification sheet, not a benchmark.

The coefficients are sensitive to convention. Taking the rate as \(2E/\pi\hbar\) and \(E\) as the total energy inside the horizon \(Mc^2=c^4R_H/2G\) gives \(\Omega=1.04\times10^{121}\); the difference from Lloyd's (2002) \(\sim10^{120}\) is exactly this choice. Whether bits are divided by \(\ln2\) (bits) or not (nats) shifts \(\Omega/N\) by a further factor of 1.44. So the structure — that it does not depend on time and is fixed by \(w\) alone — is the substance here, not the digits.

\(w\) is the effective equation of state of the universe as a whole, not of a single component. The real universe hands off from radiation to matter to \(\Lambda\), so \(\Omega/N\) takes different values over different intervals and would need \(\int R_H dt/R_H^2\) followed numerically. The table idealises "if that \(w\) held forever".

The "about 250 billion years" in §07 approximates everything after today as pure de Sitter. And \(w=-1.03\pm0.03\) comes from a fit assuming constant \(w\); the phantom side (\(w<-1\)) is a region future measurements may still move. "Already on the far side" is a statement about the central value, not a settled fact.

Exercises (everything you need is above)

  1. How many operations per bit in a radiation-dominated universe (\(w=1/3\))?
    Show the answer
    \(\Omega/N=\ln2/(3\pi^2\cdot4/3)=\ln2/(4\pi^2)=0.01756\), i.e. one operation per 57 bits. Exactly half the \(c\cdot t=\text{const}\) value — radiation is an even less computational universe.
  2. Why does \(\Omega/N\) not depend on the age of the universe? One line.
    Show the answer
    Because \(N\propto R_H^2\propto t^2\) (holography) and \(\Omega\propto\int E\,dt\propto\int R_H\,dt\propto t^2\) (from the identity \(E=c^4R_H/2G\)). Both are the same \(t^2\), so all that survives the ratio is information about the expansion law.
  3. Is \(\Omega/N\) bookkeeping or physics?
    Show the answer
    Physics. \(\Omega\) and \(N\) are plain counts and therefore dimensionless, so a conformal transformation cannot move them. Moreover it determines \(w\) uniquely, so if it did move you would genuinely be in a different universe — right-hand column, by the decision procedure of the previous series' finale.
  4. Find the \(w\) that gives \(\Omega/N=1\) and say what it means.
    Show the answer
    \(1+w=\ln2/3\pi^2=0.0234\), so \(w=-0.9766\) — just short of a cosmological constant. For a power-law universe to reach one operation per bit it must already be almost fully accelerating; conversely, a universe dominated by matter or radiation is, in principle, a bad computational deal.
  5. (Harder) Show that ⑥ "one cell per tick" and ⑧ "\(\Omega/N=\ln2/2\pi^2\)" are the same fact.
    Show the answer
    Write \(\Omega/N=(\ln2/2\pi^2)(ct/R_H)\). Statement ⑥ is \(dR_H/dt=c\); integrating gives \(R_H=ct\), i.e. \(ct/R_H=1\). Substituting yields \(\Omega/N=\ln2/2\pi^2\), and the converse likewise. "The horizon advances at the speed of light" and "the ratio of operations to memory is \(\ln2/2\pi^2$" are two readings of the single equation \(ct/R_H=1\) — the first dimensionful, the second dimensionless.

Summary　Divide, and out comes the equation of state

Written out as a specification sheet, the universe has memory \(N=\pi(R_H/\ell_P)^2/\ln2=2.96\times10^{122}\) bits, a clock of \(8.08\times10^{60}\) ticks (140 in the logarithm), and \(\Omega=(ct/\ell_P)^2/2\pi p=1.04\times10^{121}\) operations. So far this only re-confirms known estimates. What was interesting was dividing the memory by the operations.

Both grow as \(t^2\), so time cancels and only the expansion law is left: \(\Omega/N=\ln2/3\pi^2(1+w)\). The number of operations per bit is the equation of state. 1/57 for radiation, 1/42.7 for matter, 1/28.5 for \(c\cdot t=\text{const}\). In 13.8 billion years the universe has never once reached one operation per bit — and since the ratio does not depend on time, it never has at any epoch.

Writing it as \(\Omega/N=(\ln2/2\pi^2)(ct/R_H)\) shows that \(R_h=ct\) is exactly the statement that this ratio equals \(\ln2/2\pi^2\) — an eighth entry for the seven characterisations of bonus episode ③, the dimensionless counterpart of ⑥ "one cell per tick". Reaching "1" would require \(w=-0.977\), and the dark energy we observe is on the far side of that. Under a pure de Sitter approximation the universe first exceeds one operation per bit in about 250 billion years.

The reveal was two identities — memory goes as \(t^2\) because of holography, operations go as \(t^2\) because the Hubble radius equals the Schwarzschild radius of its contents (Dirac's large numbers). Both hold in any flat FLRW universe, so the division leaves nothing but the expansion law. And as an aside: memory is installed \(2\pi^2/\ln2=28.5\) times faster than operations are performed, on memory that is only \(10^{-18}\) used.

This document is Episode 1 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 Margolus–Levitin limit (an upper bound \(2E/\pi\hbar\) on the rate of transitions to orthogonal states), and the estimate that the universe has performed \(\sim10^{120}\) operations are standard since Lloyd (2002, PRL 88, 237901). \(M=c^2R_H/2G\) is an identity in flat FLRW, and that Dirac's large-number hypothesis is isomorphic to it was shown in bonus episode ③ of the previous series. The results \(\Omega=(ct/\ell_P)^2/2\pi p\), \(\Omega/N=p\ln2/2\pi^2=\ln2/3\pi^2(1+w)\), the value \(w=-0.97659\) giving \(\Omega/N=1\), and the de Sitter form \(\Omega/N=(\ln2/\pi^2)Ht\) (reaching 1 at \(Ht=\pi^2/\ln2=14.24\)) are all derived and computed here. Numbers use \(t_0=4.3536\times10^{17}\) s, \(\ell_P=1.6163\times10^{-35}\) m, \(t_P=5.3912\times10^{-44}\) s. "Operations" means the upper bound on transitions permitted by the energy, not meaningful computation. Conventions (whether the rate is \(2E/\pi\hbar\) or \(4E/h\), how \(E\) is taken, bits versus nats) shift \(\Omega/N\) by around a factor of 1.5 — the claim here is the structure (independent of time, fixed by \(w\)), not the digits. \(w\) is the effective equation of state of the universe as a whole, and the table idealises "if that \(w\) held for all time". The 250-billion-year figure assumes pure de Sitter after today (\(H_\Lambda=H_0\sqrt{\Omega_\Lambda}\), \(H_0=67.66\) km/s/Mpc, \(\Omega_\Lambda=0.685\)). The dark-energy value \(w=-1.03\pm0.03\) comes from a constant-\(w\) fit. The entropy occupancy \(1.5\times10^{-18}\) uses \(S_{\rm obs}=3.1\times10^{104}k_B\) from Egan & Lineweaver (2010, ApJ 710, 1825). Linear expansion (\(c\cdot t=\)const, \(R_h=ct\)) is a minority model still under test; extrapolated into the early universe it contradicts nucleosynthesis (Lewis, Barnes & Kaushik 2016, MNRAS 460, 291). 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).

Print / PDF: ⌘+P (Ctrl+P on Windows). On screen, the slider changes the equation of state and moves the operations-per-bit line. "Show the answer" opens each solution.