c·t = CONST, THAT CLICKS EPISODE 27 ／ Part IV — the same surgery, applied to other theories
The most famous motivation turned out to be the weakest argument
Inflation, on the
same operating table
"It solves the horizon problem" holds two different things under one name.
Cut them apart and what survives is \(n_s\).
Remember the surgery from Episode 3 — check whether one name contains two different things. The watchword held both "an equivalent rewriting" and "a claim observation can judge", under a single label. In Part IV we apply that surgery to other theories. The first patient is inflation.
01Splitting "inflation solves the horizon problem"
Exactly the structure of Episode 3. So how "cheap" is (A)? That is where we start counting.
02(A) costs zero e-folds if \(a\propto t\)
for \(a\propto t^p\)
$$d_p=a(t)\int_0^t\frac{c\,dt'}{a(t')}=\frac{ct}{1-p}$$at \(p=1\), \(\int dt'/t'\) diverges logarithmically, so
$$d_p=\infty$$| Expansion law | \(d_p\) | Horizon problem? |
|---|---|---|
| \(p=1/2\) (radiation) | \(2ct\) | yes |
| \(p=2/3\) (matter) | \(3ct\) | yes |
| \(p=0.99\) | \(100\,ct\) | yes |
| \(p=1\) (c·t=const) | \(\infty\) | no |
Conclusion of §02
\(c\cdot t=\)const removes the horizon problem with zero e-folds and zero parameters.
So (A) is not an achievement peculiar to inflation — the horizon problem was never "the universe's problem" but "the decelerating universe's problem".
Episode 17 counted the information this problem needs — about \(10^4\) causally disconnected patches at recombination, 20 KB to be agreed on. At \(p=1\) there is one patch, so the information needed is 0 bits.
03What (A) demands of inflation in e-folds
today's comoving Hubble radius must lie inside the one at the start of inflation
$$e^N\ \ge\ \frac{a_e}{a_0}\cdot\frac{H_{\rm inf}}{H_0}$$| \(V^{1/4}\) | \(H_{\rm inf}\) [1/s] | \(T_{\rm reh}\) [GeV] | \(N_{\min}\) |
|---|---|---|---|
| \(10^{16}\) GeV (GUT) | \(3.6\times10^{37}\) | \(4.1\times10^{15}\) | 62.1 |
| \(10^{13}\) GeV | \(3.6\times10^{31}\) | \(4.1\times10^{12}\) | 55.2 |
| \(10^{10}\) GeV | \(3.6\times10^{25}\) | \(4.1\times10^{9}\) | 48.3 |
| \(10^{6}\) GeV | \(3.6\times10^{17}\) | \(4.1\times10^{5}\) | 39.1 |
The commonly quoted "\(N\approx60\)" is the GUT-scale value. Here is the crux of the surgery: \(N\) is not a free parameter — it is fixed by requirement (A).
04And the same \(N\) predicts \(n_s\)
put in the \(N=62.1\) fixed by (A)
$$n_s=0.9678$$observed (Planck 2018)
$$n_s=0.9649\pm0.0042\qquad\Longrightarrow\qquad N=57.0\pm6.8$$discrepancy
$$0.75\sigma$$The thing this episode most wants to say
Two \(N\) values fixed by entirely different requirements agree within \(1\sigma\).
One from "today's universe fitting inside one causal patch", the other from the tilt of the CMB fluctuations. Why those two should be connected emerges only once inflation is assumed.
Figure: e-folds \(N\) across. Blue is the \(N_{\min}\) demanded by (A) (which moves with the reheating scale); orange is the \(N\) implied by (B)'s \(n_s\). Move the slider through reheating scales to see where they overlap.
05Measuring the surprise with Episode 19's procedure
| Method | Content | Surprise |
|---|---|---|
| A: relative width from \(n_s\) | \(\sigma_N/N=0.120\), prior \(\ln\) range 4.61 | 5.3 bit |
| B: probability of the band | probability of \(N\in[55,70]\) is 0.052 | 4.3 bit |
About 4–5 bits. On Episode 19's scale that falls in the coincidence band (4.7–7.4 bits), but this one has an explanation — the structure whereby a single \(N\) fixes both. In Episode 19's classification, an explained agreement moves to physics. That is the difference from Episode 18's 1.96 fm (7.4 bits, unexplained).
06The reveal — same surgery, different outcome
| Pays | Buys | Net | |
|---|---|---|---|
| Inflation | \(N\) + the shape of \(V\) ≈ 2 parameters (\(-10.7\) bit) | predicts \(n_s\) at 4.3 bit | \(-6.5\) bit (an underestimate) |
| c·t=const | one fewer parameter (\(+5.4\) bit) | the horizon problem disappears | \(-148.3\) bit (Episode 25) |
Inflation's \(-6.5\) is an underestimate — only \(n_s\) is credited here, while the same two parameters also buy the bound on \(r\), adiabaticity, gaussianity, the super-horizon TE anticorrelation, flatness and the monopole problem. Whereas \(c\cdot t=\)const's \(-148.3\) is the loss on fit itself, and is not an underestimate.
Conclusion of §06
The same surgery, and what survives is different.
Discard (A) and inflation still has (B).
\(c\cdot t=\)const solves (A) for free but has nothing corresponding to (B).
① \(N=60\) depends strongly on the reheating scale. As the table shows, \(V^{1/4}=10^6\) GeV gives \(N_{\min}=39\) and the GUT scale gives 62 — a spread of 23. The agreement with \(n_s\) assumes the GUT scale, and does not hold for low-scale models.
② \(n_s\approx1-2/N\) is model dependent. It holds for \(R^2\) (Starobinsky) type and α-attractors, but not for every inflationary model. This is a prediction of a well-used family of models, not "the prediction of inflation".
③ Instantaneous reheating is assumed. Prolonged reheating moves \(N_{\min}\) by several to tens more.
④ §06's ledger is a rough calculation to put things in common units. The parameter price uses Episode 5's \(N_{\rm data}=1701\), and only \(n_s\) is credited on inflation's side. Read "\(-6.5\) versus \(-148\)" only as a comparison of orders.
⑤ That the horizon problem vanishes at \(p=1\) does not support that expansion law. Episode 3's judgement (helium abundance from nucleosynthesis) is unchanged. §02 says only that (A) is cheap.
⑥ This episode does not refute inflation. Quite the opposite — it confirms that (B) survives the surgery.
Exercises (solvable with this episode's formulas alone)
- Find the particle horizon for \(a\propto t^p\) and show it diverges at \(p=1\).
Show the answer
\(d_p=a\int_0^t c\,dt'/a=ct^p\int_0^t t'^{-p}dt'=ct^p\cdot t^{1-p}/(1-p)=ct/(1-p)\). As \(p\to1\), \(\int dt'/t'=[\ln t']_0^t\) diverges logarithmically at the lower limit. - From \(n_s=0.9649\), find \(N\) and its error.
Show the answer
\(N=2/(1-n_s)=2/0.0351=57.0\); the error is \(dN/dn_s=2/(1-n_s)^2=1623\) times \(0.0042\), i.e. \(\pm6.8\). \(N=57.0\pm6.8\). - How many \(\sigma\) is the agreement with the GUT-scale \(N_{\min}=62.1\)?
Show the answer
\((62.1-57.0)/6.8=0.75\sigma\) — within \(1\sigma\). - Convert the agreement into bits using Episode 19's procedure.
Show the answer
With a log-uniform prior \(N\in[10,1000]\), the \(\ln\) range is 4.61 and the probability of \(N\in[55,70]\) is \(\ln(70/55)/4.61=0.052\), so \(-\log_2 0.052=\) 4.3 bits; by relative width, 5.3. Having an explanation, it classifies as physics rather than coincidence. - (Harder) Why is "inflation solves the horizon problem" a weak argument?
Show the answer
Because \(a\propto t\) does the same with zero e-folds and zero parameters. So (A) does not distinguish inflation from the alternatives. What supports inflation is (B) — \(n_s\), adiabaticity, gaussianity, super-horizon correlations. The point of the surgery is to name which argument is actually paying.
Summary — (A) is cheap; what survives is (B)
We cut "inflation solves the horizon problem" in two with Episode 3's surgery — (A) establishing causal contact, (B) producing a fluctuation spectrum.
(A) proved cheap. The particle horizon of \(a\propto t^p\) is \(ct/(1-p)\), which at \(p=1\) diverges logarithmically to infinity. So \(c\cdot t=\)const removes the horizon problem with zero e-folds and zero parameters — and Episode 17's 20 KB becomes 0 bits when there is one patch. The horizon problem was never "the universe's problem" but "the decelerating universe's problem".
For inflation, (A) fixes \(N\) — \(N_{\min}=62.1\) at the GUT scale. And the same \(N\) predicts \(n_s\approx1-2/N=0.968\), against an observed \(0.9649\pm0.0042\), i.e. \(N=57.0\pm6.8\) — an agreement at 0.75σ. Measured by Episode 19's procedure that is 4–5 bits of surprise, classified as physics rather than coincidence because it has an explanation.
As a ledger, inflation comes to \(-6.5\) bits (an underestimate, crediting only \(n_s\)) against \(c\cdot t=\)const's \(-148.3\). The same surgery, and what survives is different — discard (A) and inflation still has (B), while \(c\cdot t=\)const solves (A) for free and has nothing corresponding to (B). And the most famous motivation of all, "it solves the horizon problem", turns out to be the weakest argument — that is the yield of the surgery.
This document is Episode 27 of "c·t = const, That Clicks", written for physics-minded high-school and university readers. The particle horizon \(d_p=ct/(1-p)\), the e-fold requirement \(e^N\ge(a_e/a_0)(H_{\rm inf}/H_0)\), the slow-roll result \(n_s\approx1-2/N\), and Planck 2018's \(n_s=0.9649\pm0.0042\) are all standard. Computed here are the \(N_{\min}\) per reheating scale (62.1 at the GUT scale, 39.1 at \(10^6\) GeV), the inversion \(N=57.0\pm6.8\) from \(n_s\), the 0.75σ discrepancy, and the 4.3–5.3 bits of surprise by Episode 19's procedure (kenshou/calc31.py). \(N=60\) depends strongly on the reheating scale (a spread of 23), and \(n_s\approx1-2/N\) is a result for a family of models (\(R^2\), α-attractors) rather than a prediction of every inflationary model. Instantaneous reheating is assumed; prolonged reheating moves \(N_{\min}\) further. §06's ledger is a rough calculation for common units, with the parameter price from Episode 5's \(N_{\rm data}=1701\) and only \(n_s\) credited on inflation's side — an underestimate, to be read only as a comparison of orders. That the horizon problem vanishes at \(p=1\) does not support that expansion law (Episode 3's judgement is unchanged). Linear expansion (\(c\cdot t=\)const, \(R_h=ct\)) is a minority model under examination, and 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).