CONFORMAL TRANSFORMATIONS THAT CLICKBONUS / Judging the expansion law, not the way of speaking

Ten episodes of saying "these are equivalent". So what happens if you run it at face value?

What happens if you run it
at face value The conformal transformation gave \(c\cdot t=\text{const}\) three ways of speaking.
But having three ways of speaking and having a correct model are entirely different things.

What you need: the exponential function, and building ratios \(n/p=e^{-Q/k_BT_f}\)

This series has said, consistently: "space stretches", "light slows down" and "atoms shrink" are three ways of telling the same physics. A correct claim. But as the honest-line box of Episode 5 said, whether the expansion law \(a\propto t\) is itself correct is an entirely separate question. Today we go to receive the verdict on that. The courtroom is Big Bang nucleosynthesis. And to say the outcome first — at face value, it loses. Moreover, defend it in any of the three pictures and the verdict is identical.

01Why is nucleosynthesis the courtroom?

Within the first few minutes of the universe, protons and neutrons bound together to make helium. The amount is well measured: a mass fraction of \(Y_p\simeq0.245\) — about a quarter of the matter in the universe is helium. Standard cosmology accounts for this beautifully.

And nucleosynthesis is extraordinarily sensitive to how the universe expands, because it comes down to a race.

The race that decides nucleosynthesis

The weak interaction (shuttling neutrons ⇄ protons, at rate \(\Gamma\))
          vs
the expansion of the universe (pulling things apart and stopping the reactions, at rate \(H\))

While \(\Gamma\) is winning, the ratio of neutrons to protons takes its equilibrium value, set by the temperature. The instant expansion wins (\(\Gamma=H\)) the ratio freezes out, and every remaining neutron ends up in helium — the "bits freezing" of Episode 10 of the previous series.

02With \(a\propto t\), the cooling is too slow

When freeze-out happens is decided by a contest between the temperature dependences of \(\Gamma\) and \(H\).

Calculation — estimating the freeze-out temperature

The weak interaction rate (the same in either universe)

$$\Gamma\;\propto\;T^5$$

Standard cosmology (radiation dominated, \(a\propto t^{1/2}\))

\(T\propto t^{-1/2}\) and \(H\propto1/t\), so \(H\propto T^2\). The freeze-out condition gives

$$T^5\sim T^2\;\Longrightarrow\;T_f\sim0.8\ \mathrm{MeV}$$

\(c\cdot t=\text{const}\) (\(a\propto t\))

\(T\propto1/a\propto1/t\) and \(H=1/t\), so \(H\propto T\). The freeze-out condition becomes

$$T^5\sim T\;\Longrightarrow\;\text{freeze-out does not happen until far lower temperature}$$

Drop the exponent of the temperature from 2 to 1 and the contest reverses. Expansion is too feeble and the weak interaction keeps winning — estimates in the literature put the freeze-out temperature roughly two orders of magnitude lower than standard.

03Delay the freeze-out and the neutrons disappear

This is the fatal blow. The equilibrium neutron-to-proton ratio, written with the mass difference \(Q=1.293\) MeV, is

$$\frac{n}{p}=e^{-Q/k_BT}$$

An exponential. The lower the temperature, the more exponentially the neutrons are depleted. Delay the freeze-out and you are made to sit through all of that decline.

Freeze-out \(T_f\)\(n/p\)Helium \(Y_p\)Verdict
0.8 MeV (standard)0.1990.33 → 0.25*agrees with the observed 0.245
0.5 MeV0.0750.14not enough
0.3 MeV0.0130.027nowhere near enough
0.1 MeV\(2.4\times10^{-6}\)\(4.8\times10^{-6}\)no helium can be made
0.05 MeV\(5.9\times10^{-12}\)\(1.2\times10^{-11}\)annihilated

* In the standard case some neutrons β-decay after freeze-out, taking \(n/p\) from 1/6 to 1/7, giving \(Y_p=2(1/7)/(1+1/7)=0.25\).

Lower the freeze-out temperature by a mere factor of 8 and helium drops by five orders of magnitude. That is what happens when \(c\cdot t=\text{const}\) is run at face value. Lewis, Barnes and Kaushik (2016) did the actual computation and concluded that the helium mass fraction comes out of order \(10^{-3}\) against the observed 0.245. The title of their paper is "Primordial nucleosynthesis in the \(R_h=ct\) cosmology: pouring cold water on the simmering Universe".

Figure: lower the freeze-out temperature \(T_f\) and the helium abundance dies exponentially. The green line is the observed value 0.245
T_f = 0.80 MeV n/p = 0.199 → Y_p = 0.331 (the standard freeze-out temperature)
Helium mass fraction Y_p Observed 0.245 Standard freeze-out 0.8 MeV

Drag the slider left. Going from 0.8 to 0.3 alone sends the curve off a cliff; at 0.1 it is pinned to the floor of the plot. Exponentials show no mercy.

◇ ◇ ◇

04Defend it in any of the three pictures — the same verdict

Here the series' work pays off. The verdict does not change with the picture you speak in.

PictureWhat it says is happening
① Space stretchesThe temperature falls as \(1/t\). It falls too slowly for freeze-out to arrive in time.
② Light slows downThe same (only the time coordinate has changed — Episode 3).
③ Masses growThe temperature is constant. The mass difference \(Q\) grows as \(\propto t\), too slowly, so by the time \(Q/k_BT\) exceeds 1 the weak interaction has used the neutrons up.

Version ③ is especially satisfying. As Episode 3 showed, in the conformally transformed picture the universe does not cool; masses grow instead. What decides whether helium is made is, in either picture, the dimensionless ratio \(Q/k_BT\). Whether the numerator grows or the denominator falls is all that differs; the value of the ratio is the same — so the verdict is the same too.

Use the finale's decision procedure, unchanged

\(Q/k_BT\) is dimensionless → it does not move with the gauge → physics
Hence no picture can secure an acquittal.

05Another, harsher criticism

Separate from nucleosynthesis, there is a more fundamental objection. Mitra (2014) argued that the \(R_h=ct\) universe can be written in static form by a coordinate transformation, which in the flat case implies zero density — a vacuum — making it, like the Milne universe, "a vacuum in disguise".

The difference from the Milne universe (this matters) The Milne universe (\(a\propto t\), curvature \(k=-1\), no matter) becomes Minkowski spacetime by a pure coordinate transformation — no conformal transformation is needed, because it is empty. To make \(R_h=ct\) (\(a\propto t\), \(k=0\), with matter) Minkowski, on the other hand, required the conformal transformation we performed in this series (Episode 3). Whether a coordinate change suffices, or a conformal transformation is required — that difference is precisely the point at issue in whether the two are the same thing. Mitra's claim has been disputed and is not settled.

06The verdict

As a way of speaking: acquitted — indeed, excellentThe conformal transformation is exact, and the three pictures are observationally indistinguishable (Episodes 3–7). The auxiliary line "light slows down" was a legitimate gauge choice.
The expansion law \(a\propto t\): rejected in the early universeWhat is rejected is not "writing \(c\cdot t=\text{const}\)" but the expansion law folded into it. Applying \(a\propto t\) at face value into the nucleosynthesis era leaves helium short by more than two orders of magnitude (Lewis et al. 2016). Without substantial new physics it does not stand.
?
Whether it is a "vacuum in disguise": under disputeMitra's (2014) claim and rebuttals from Melia and others remain opposed. At minimum it is distinguished from the Milne universe by whether a conformal transformation is required.
As a tool: it survivesThe structure whereby \(w=-1/3\) is a triple boundary — the potential becoming quadratic, conformal time becoming infinite, acceleration vanishing (Episode 5) — is correct regardless of whether the model is right. A point on a boundary is always mathematically special.

The verdict in Episode 7 of the previous series read "rejected at face value; acquitted in an \(\alpha\)-invariant gauge but unobservable". Restated in the language of conformal transformations, it becomes: the equivalence is a consequence of relativity (\(c_B\cdot a=\text{const}\)); what fails is the expansion law substituted into it (\(a\propto t\)). And what today makes plain is that this series built the tool for not confusing those two.

Let us be precise about the wording

The equivalence derived from relativity is, as bonus episode ④ of "Cosmology That Clicks" showed, \(c_B\cdot a=\text{const}\) (together with \(\alpha\) invariance). Constant when multiplied by \(a\) — not by \(t\). That relation holds whatever the form of \(a(t)\), and applies unchanged to a \(\Lambda\)CDM universe. The equivalence itself is therefore not the kind of claim observation can reject.

So where does \(c\cdot t=\text{const}\) come from? It is \(c_B\cdot a=\text{const}\) with \(a\propto t\) substituted in. Indeed \(c_B\cdot t=c_0\,t/a\), which is constant only when \(a\propto t\). In the actual universe (log-averaged \(a\propto t^{0.51}\)) one gets \(c_B\cdot t\propto t^{0.49}\), growing by a factor of \(10^{29.7}\) between the Planck time and today.

So two different things were travelling under one name — the equivalence (\(c_B\cdot a=\text{const}\); a consequence of relativity; always true) and the expansion law (\(a\propto t\); decided by observation). This bonus episode rejects only the latter. The way of speaking — "light slows down" — is left entirely untouched.

Being straight with you

The estimate of \(T_f\) in section 02 is an order-of-magnitude argument. Comparing \(\Gamma\propto T^5\) with \(H\) is the standard way to estimate a freeze-out temperature, but obtaining an accurate \(Y_p\) requires solving the whole reaction network numerically (which is what Lewis et al. 2016 did). The table in section 03 is likewise a guide, computed from the equilibrium value at freeze-out, and does not include the details of subsequent neutron decay and nuclear reactions.

There are also counter-arguments on the \(R_h=ct\) side. Melia and collaborators maintain that model-independent distance measurements favour \(R_h=ct\). The "rejection" here refers to extrapolating the model at face value into the early universe (the nucleosynthesis era) and is a separate matter from the debate over fits to low-redshift observations. Do not close any of these disputes after hearing only one side.

Exercises
  1. Compute \(n/p\) for \(T_f=0.3\) MeV. Use \(Q=1.293\) MeV.
    Show answer
    \(n/p=e^{-1.293/0.3}=e^{-4.31}=0.0134\), about one fifteenth of the standard value 0.199. Helium comes out at \(Y_p=2(0.0134)/(1.0134)=0.026\), a tenth of the observed 0.245.
  2. Why is the expansion "too feeble" when \(a\propto t\)? Explain with the temperature exponents.
    Show answer
    In the standard (radiation-dominated) case \(H\propto T^2\), while for \(a\propto t\), \(H\propto T\). Solving the freeze-out condition \(\Gamma\propto T^5=H\), the solution \(T_f\) is smaller for exponent 1 than for exponent 2 — that is, the weak interaction keeps winning down to much lower temperature.
  3. Can the nucleosynthesis problem be dodged by telling the story in the "growing mass" picture?
    Show answer
    No. Helium production is governed by the dimensionless ratio \(Q/k_BT\), which does not move with the gauge (the finale's decision procedure). Whether the temperature falls or the mass difference grows is a difference of wording; the value of the ratio and its evolution in time are the same.
  4. (Harder) What is the decisive difference between the Milne universe and \(R_h=ct\)?
    Show answer
    Milne (\(k=-1\), vacuum) becomes Minkowski by a coordinate transformation alone. \(R_h=ct\) (\(k=0\), with matter) needs more than that — it requires a conformal transformation (a swap of rulers), as in Episode 3. The former is the same spacetime written differently; the latter is a different spacetime mapped so that dimensionless quantities agree. In the language of Episode 7's "10 metric components = 9 light cones + 1 marking", the latter moves that one component.

SUMMARYThe equivalence stands; the substituted expansion law falls

Nucleosynthesis is extraordinarily sensitive to the expansion law. The neutron-to-proton ratio freezes out from the race between the weak interaction \(\Gamma\propto T^5\) and expansion \(H\), and that sets the helium abundance. In the standard case (\(H\propto T^2\)) freeze-out occurs at \(T_f\simeq0.8\) MeV and \(Y_p=0.25\). But with \(a\propto t\) one has \(H\propto T\), delaying freeze-out by about two orders of magnitude in temperature. Since \(n/p=e^{-Q/k_BT_f}\) is exponential, the neutrons are used up in the meantime and almost no helium is produced (Lewis et al. 2016).

And crucially, the verdict is the same however you defend it. What decides matters is the dimensionless ratio \(Q/k_BT\), which does not move with the gauge — the finale's decision procedure applies unchanged. The conformal transformation gave \(c\cdot t=\text{const}\) three ways of speaking, but adding ways of speaking does not earn a model a single extra point. Building the tool that keeps those two apart was the greatest achievement of these ten episodes.

This document is the bonus episode of the series "Conformal Transformations That Click", written for physics-minded high-school and university students. That the neutron-to-proton ratio freezes out when \(\Gamma_{\rm weak}\sim H\), that \(\Gamma\propto T^5\), that \(H\propto T^2\) in the standard radiation-dominated era giving \(T_f\simeq0.8\) MeV, that the equilibrium ratio is \(n/p=e^{-Q/k_BT}\) with \(Q=1.293\) MeV, that \(Y_p=2(n/p)/(1+n/p)\), and that the observed value is \(Y_p\simeq0.245\), are all standard results. That linear expansion (\(a\propto t\), \(R_h=ct\)) gives \(T\propto1/t\) and \(H=1/t\), hence \(H\propto T\), delaying freeze-out substantially and producing catastrophically little helium, is the conclusion of the numerical calculation of Lewis, Barnes & Kaushik (2016, MNRAS 460, 291), which reports a helium mass fraction of order \(10^{-3}\). Sections 02–03 and the figure are order-of-magnitude estimates, not the result of solving the reaction network. The claim that \(R_h=ct\) is a vacuum in disguise is due to Mitra (2014, MNRAS 442, 382); rebuttals by Melia and others exist and the matter is unsettled. That the Milne universe (\(k=-1\), vacuum) becomes Minkowski spacetime by a coordinate transformation alone, whereas \(R_h=ct\) (\(k=0\), with matter) requires a conformal transformation, was shown in Episode 3 of this series. Melia and collaborators also argue in favour of \(R_h=ct\) on low-redshift observations, and the "rejection" here is limited to extrapolation into the early universe. The academic standard is the ΛCDM model including inflation. — To print, use your browser's Print → Save as PDF (in the print version the slider is frozen and answers are hidden).

Print / PDF: ⌘+P (Ctrl+P on Windows). On screen the slider changes the freeze-out temperature and shows the helium vanishing. "Show answer" opens the solutions.