c·t = CONST, THAT CLICKS EPISODE 2 ／ Last time the memory, this time the clock
The universe has two clocks, and they run at different rates
Two clocks that
do not mesh
140 steps of logarithmic cosmic time, 73 steps of logarithmic energy.
Divide them and the expansion law comes out — and \(c\cdot t=\text{const}\) demands they mesh 1:1.
Last time we divided the memory by the operations. This time it is the clock. The universe carries two rulers — the logarithm of time and the logarithm of energy — and both can be counted in steps. The first is the axis of cosmology, the second the axis of the renormalisation group. Divide one by the other and, once again, the expansion law comes out. Only this time the verdict on \(c\cdot t=\text{const}\) arrives with no dynamics at all — from two observed ratios and nothing else.
01The logarithm may be the real clock
Last time we counted \(8.08\times10^{60}\) ticks for the universe. But the number that matters, for a machine, is this one.
and memory per step (since \(N\propto t^2\))
$$e^{2}=7.389\ \text{per step}\qquad\Longrightarrow\qquad e^{2\times140.24}=10^{121.8}\ \checkmark$$140 steps, memory up by a factor of 7.4 at each one — that alone reproduces \(10^{122}\) bits. The universe is a machine that has made only 140 moves. The raw \(10^{60}\) is less faithful to its history than this 140, because the history is not "\(10^{60}\) repetitions of one operation" but 140 doublings.
02There is a second ruler
In Episode 8 of the previous series we saw that quantum field theory needs a "standard of fineness" \(\mu\). Coupling constants run along that axis (\(1/137\to1/128\)), and that running is the breaking of conformal symmetry. This axis, too, can be counted logarithmically.
In cosmology, \(\mu\) is temperature. How many steps from the Planck temperature down to today's CMB?
73 steps — about half the clock of time. The same cosmic history, counted on two rulers, gives two different numbers of steps.
03The core — divide
The result of this episode
$$\bar p=\frac{\mathcal{N}_T}{\mathcal{N}_t}=\frac{73.03}{140.24}=0.5207$$The ratio of the two clocks is the logarithmically averaged expansion index of cosmic history. With \(T\propto1/a\) and \(a\propto t^{p}\) we have \(\ln(T_P/T_0)=p\,\ln(t_0/t_P)\) — that is the whole derivation.
Photon number is conserved as the universe expands, so temperature falls as \(1/a\). Each time a species annihilates, though, its energy is handed to the photons and the fall slows a little — the effect is exactly the change in \(g_{*s}\).
so
$$\bar p=\frac{73.03-1.10}{140.24}=0.5129$$| Universe | \(p\) | length of the energy axis | verdict |
|---|---|---|---|
| observed (from \(T_0\) and \(t_0\)) | 0.513 | 71.9 steps | ── |
| radiation-dominated | 0.500 | 70.1 | essentially matches |
| matter-dominated | 0.667 | 93.5 | does not match |
| \(c\cdot t=\)const | 1.000 | 140.2 | off by a factor 1.92 |
The observed 0.513 sits just above radiation. The matter era occupies most of the universe's age, yet contributes almost nothing logarithmically — the handover happens around step 132, leaving only 8 steps. Counted in logarithms, cosmic history is almost entirely radiation-dominated.
04The ninth characterisation
that is
$$\text{one step of cosmic time}\ \longleftrightarrow\ p\ \text{steps of the renormalisation group}$$This is the ninth entry, after ⑧ (operations per bit) from Episode 1. And it is a relative of ⑥ "one cell per tick" — that one synchronises the horizon with light, this one synchronises the universe with the renormalisation group. Every property \(c\cdot t=\text{const}\) gathers has the same shape: two clocks falling into step.
05The verdict, from two numbers
What \(c\cdot t=\)const is demanding
140.24 ＝ 73.03
No dynamics were used. Only today's temperature, 2.7255 K, and today's age, 13.8 billion years.
The mismatch is \(e^{67.2}=1.55\times10^{29}\). Here is what that means, put two ways.
holding \(T_0\) fixed and running back with \(T\propto1/t\)
$$T(t_P)=T_0\cdot\frac{t_0}{t_P}=2.7255\times8.075\times10^{60}=2.20\times10^{61}\ \mathrm{K}=1.55\times10^{29}\,T_P$$equivalently, this universe reaches the Planck temperature only at
$$t=8.4\times10^{-15}\ \mathrm{s}\qquad(\text{step }67.2)$$the time at which the temperature falls to 0.8 MeV (\(9.28\times10^9\) K)
$$t=t_0\frac{T_0}{T}=4.354\times10^{17}\times\frac{2.7255}{9.28\times10^{9}}=1.28\times10^{8}\ \mathrm{s}=\mathbf{4.05\ \text{years}}$$in standard cosmology
$$t=1.2\ \text{seconds}$$The main point of this episode
In a \(c\cdot t=\text{const}\) universe, the temperature at which neutrons freeze out is reached after 4 years.
A free neutron lives 880 seconds. By then there would be none left.
Bonus episode ① of the previous series reached this through a race of reaction rates (\(\Gamma\propto T^5\) against \(H\)). Here we get the same conclusion without touching the race at all — the temperature–time relation alone shifts the moment by a factor of \(6\times10^8\). This is the shortest possible statement of that verdict.
(A) Ways of speaking: one and the same spacetime can be written as "space stretches" or "light slows" or "masses grow" — exactly equivalent under a conformal transformation, and observationally indistinguishable.
(B) The expansion law: that this \(a(t)\) goes as \(\propto t\) — not a rewriting but a claim about which spacetime you are in. It moves dimensionless quantities (the time dependence of \(Q/k_BT\), of \(\Gamma/H\), and today's \(\mathcal{N}_T/\mathcal{N}_t\)), so observation can decide it.
All three pictures assume the same \(a\propto t\), so all three receive the same verdict. "Equivalent" refers to the three pictures among themselves; it was never a claim that \(a\propto t\) and \(\Lambda\)CDM are equivalent.
Figure: horizontal axis, the logarithmic age of the universe in steps; vertical axis, \(\ln(T_P/T)\), the step count of the renormalisation group. The line always passes through today's point (140.2, 73.0) and its slope is \(p\). Move the knob and watch where the early universe goes
At \(p=0.513\) the line very nearly passes through the origin — precisely, through \(T_P/3.0\), and that factor of 3 is exactly the \(g_{*s}\) correction of 1.10 nats from §03. The satisfying picture of "Planck temperature at the Planck time" holds because the observed \(T_0\) and \(t_0\) say so, not because we assumed it.
Push the knob right to \(p=1\) and the line shoots far below the origin, sinking the early universe into the grey band (hotter than the Planck temperature). At the same time the crossing with the vermilion line (neutron freeze-out) slides far to the right — that is what "the cooling is too slow" actually looks like. Step 97.5 (0.12 s) becomes step 118.3 (4.07 years).
06Putting the staircase of forgetting on this axis
Now that the two axes are joined, we can place the a-theorem from bonus episode ⑦ of the previous series on them. That result was: the renormalisation group has a direction, \(a_{\rm UV}>a_{\rm IR}\). Since \(a\) is the coefficient of the entanglement entropy across a sphere, coarse-graining loses information. In the Standard Model \(a\) falls by a factor of 16.06 from the Planck scale to today.
| what drops out | temperature | time | step | \(a\) | drop |
|---|---|---|---|---|---|
| (Planck) | \(T_P\) | \(t_P\) | 0.0 | 995.5 | ── |
| top, Higgs, W, Z | 173 GeV | 7.8 ps | 74.1 | 772.5 | 0.254 nat |
| bottom | 4.2 GeV | 15 ns | 81.6 | 739.5 | 0.044 |
| tau, charm | 1.3 GeV | 0.16 μs | 84.0 | 695.5 | 0.061 |
| QCD confinement | 0.2 GeV | 7.7 μs | 87.9 | 89.5 | 2.050 |
| electron | 0.511 MeV | 2.8 s | 100.7 | 78.5 | 0.131 |
| neutrinos | 0.05 eV | 4.4 Myr | 132.2 | 62.0 | 0.236 |
Put on the logarithmic clock of the universe, the a-theorem says this: of 140 moves, the universe forgot nothing until move 74, then lost three quarters of everything at move 88.
07The reveal — why there are two
| clock of time \(\mathcal{N}_t\) | clock of energy \(\mathcal{N}_T\) | |
|---|---|---|
| what it counts | steps the horizon has grown | steps the standard of fineness has fallen |
| origin | geometry (\(R_H=c/H\)) | matter (photon conservation, \(T\propto1/a\)) |
| where it lived before | the conformal-factor side (bookkeeping) | the anomaly side (physics) |
| what it measures | memory, operations (Ep.1) | \(\beta\) functions, \(a\), forgetting |
The table from bonus episode ② of the previous series — where the gravitational field itself splits into conformal factor (bookkeeping, no arrow of time) and Weyl tensor (physics, arrow of time) — is at work here too. The clock of time is kept by geometry; the clock of energy by matter. It is precisely because the two are independent that dividing them yields something meaningful. Had they been two counts of the same thing, the ratio would be 1 by construction and would measure nothing.
And \(c\cdot t=\text{const}\) is exactly the model that demands the ratio be 1 — forcing two independent clocks into step. A beautiful demand; the universe answers 0.51.
\(T_P=m_Pc^2/k_B\) is a convention. It gives the Planck temperature no physical significance; it is simply where we put the origin of the energy axis. Move the origin and \(\mathcal{N}_T\) changes — but \(\bar p\) is a ratio between two points, so as long as both ends move together the conclusion does not.
"The universe began at \(t_P\)" is also an assumption. Inflation would rewrite the early logarithmic time, and entropy release at reheating would break \(T\propto1/a\). The \(g_{*s}\) correction of 1.10 nats counts only standard particle content and nothing else.
The "4.05 years" and "1.2 seconds" of §05 come respectively from \(T\propto1/t\) and from the standard \(t=2.42g_*^{-1/2}(\mathrm{MeV}/T)^2\,\mathrm{s}\). The figure draws a single power law, so even at \(p=0.513\) it puts 0.8 MeV at about 0.12 s (the careful standard calculation gives 1.2 s) — that is the price of applying a log-averaged \(p\) to all epochs. The claim is the eight-order gap between four years and one second, not the coefficients.
The staircase in §06 inherits every caveat from bonus episode ⑦ of the previous series — \(a\) is properly defined only at conformally invariant fixed points. Speaking of "the current \(a\)" at intermediate energies is a free-field cartoon and is not a test of the a-theorem; massive vectors (W, Z) and confinement are handled crudely. The neutrino step (4.4 Myr) uses a matter-era approximation. And \(\bar p=0.513\) is a logarithmic average, not an instantaneous index — the real universe runs radiation (0.5) → matter (0.667) → \(\Lambda\) (accelerating), and radiation occupies most of the logarithmic history, which is why the average lands near 0.5.
Exercises (everything you need is above)
- How many steps long is the energy axis in a universe that stays matter-dominated forever?
Show the answer
\(p=2/3\), so \(\mathcal{N}_T=0.667\times140.24=93.5\) steps — more than 20 above the observed 71.9. Today's CMB would then be \(e^{21.6}=2.4\times10^{9}\) times too cold. Matter domination alone does not work either. - Why is \(\bar p=0.513\) so close to radiation's 0.5, when the matter era lasts longer?
Show the answer
Because we are counting logarithms. The radiation-to-matter handover is at \(z\simeq3400\), around step 132, leaving only 8 steps to today. Logarithmically, 94% of cosmic history is radiation-dominated, so the average sits near the radiation value. - In a \(c\cdot t=\text{const}\) universe, when does the temperature fall to 0.1 MeV (where deuterium forms)?
Show the answer
\(t=t_0T_0/T\) with \(T=1.16\times10^{9}\) K gives \(t=1.02\times10^{9}\) s \(=\) 32 years. In standard cosmology it is 132 seconds. Since neutrons are gone after 880 seconds, there is nothing left to make deuterium with. - What fraction of the universe's forgetting is due to QCD confinement alone?
Show the answer
\(\ln(695.5/89.5)/\ln(995.5/62.0)=2.050/2.776=\) 73.9%. The eight gluons alone contribute \(8\times62=496\), roughly half of the Standard Model's \(a\). Three quarters of everything the universe forgot happened in the single instant of step 87.9. - (Harder) How is ⑨ "the two clocks mesh 1:1" related to ⑥ "one cell per tick"?
Show the answer
Both have the shape "two clocks fall into step", but they synchronise different things. ⑥ pairs the horizon with light (\(dR_H/dt=c\), geometry with geometry); ⑨ pairs the universe with the renormalisation group (\(d\ln T/d\ln t=-1\), geometry with matter). Both are equivalent to \(a\propto t\), but ⑨ is the stronger statement — ⑥ closes within geometry, whereas ⑨ brings in a clock on the matter side (\(T\propto1/a\)) and can therefore be decided by observation. Indeed, this episode's verdict came from ⑨.
Summary Take the ratio, and out comes the expansion law again
The universe carries two logarithmic rulers: the clock of time, \(\mathcal{N}_t=\ln(t_0/t_P)=140.24\) steps (memory up \(e^2=7.39\) per step, \(10^{121.8}\) after 140), and the clock of energy, \(\mathcal{N}_T=\ln(T_P/T_0)=73.03\) steps. Geometry keeps the first, matter the second.
Their ratio is \(\bar p=0.5207\) (0.5129 after the \(g_{*s}\) correction) — the logarithmically averaged expansion index of cosmic history. Just above radiation's 0.5, which tells us 94% of the logarithmic history was radiation-dominated. In differential form, \(d\ln T/d\ln t=-p\): one step of cosmic time is \(p\) steps of the renormalisation group. Only \(p=1\) meshes them 1:1, and that is the ninth characterisation of \(a\propto t\).
Then the verdict. \(p=1\) demands \(140.24=73.03\) — a mismatch of \(1.55\times10^{29}\), obtained with no dynamics, from today's temperature and today's age alone. Concretely: the temperature at which neutrons freeze out, 0.8 MeV, is reached after 4.05 years (standard: 1.2 seconds). Five orders of magnitude too late for a neutron that lives 880 seconds. Bonus episode ① of the previous series showed this through a race of reaction rates; here it falls out of a ratio.
And placing the a-theorem staircase on this axis: nothing happens for the first 74 steps, 73.9% of all forgetting occurs in the single step at 87.9 (QCD confinement), and the last drop is at step 132.2 (neutrinos). Of 140 moves, the universe lost three quarters of everything at move 88.
This document is Episode 2 of "c·t = const, That Clicks", written for physics-minded high-school and university readers. The adiabatic relation \(T\propto g_{*s}^{-1/3}/a\), the radiation-era relation \(t=2.42\,g_*^{-1/2}(\mathrm{MeV}/T)^2\) s, and the change of \(g_{*s}\) from 106.75 to 3.909 in the Standard Model are all standard. We use \(T_P=m_Pc^2/k_B=1.4168\times10^{32}\) K, \(T_0=2.7255\) K, \(t_0=4.3536\times10^{17}\) s. The quantities \(\mathcal{N}_t=140.24\), \(\mathcal{N}_T=73.03\), \(\bar p=0.5207\) (0.5129 after the \(g_{*s}\) correction), and the observation that \(d\ln T/d\ln t=-p\) meshes 1:1 only at \(p=1\), are calculations and readings made here. So are the \(c\cdot t=\)const figures: 0.8 MeV at 4.05 years, 0.1 MeV at 32 years, the Planck temperature at \(8.4\times10^{-15}\) s (step 67.2). The free neutron lifetime is 879.4 s. That linear expansion applied to the early universe destroys nucleosynthesis is the numerical result of Lewis, Barnes & Kaushik (2016, MNRAS 460, 291); this episode reproduces that conclusion from the temperature–time relation alone. The a-theorem is due to Komargodski & Schwimmer (2011), but \(a\) is defined only at conformal fixed points, and the staircase in §06 is a free-field cartoon, not a test of the theorem (free scalar : Weyl : vector \(=1:11/2:62\)). The values of \(a\), the 73.9% drop at QCD confinement, and the step numbers are computed here; massive vectors and confinement are treated crudely. \(\bar p\) is a logarithmic average, not an instantaneous index. Taking \(T_P\) as the origin of the energy axis, and "the universe began at \(t_P\)", are conventions and assumptions respectively. Linear expansion (\(c\cdot t=\)const, \(R_h=ct\)) is a minority model still under test; Melia and collaborators argue it is favoured at low redshift, and the mismatch here is confined to extrapolation into the early universe. 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).