c·t = CONST, THAT CLICKS EPISODE 12 ／ Closing Part II's cosmology run
In this picture the cosmological constant grows fastest of all
Substituting
into the vacuum
\(\tilde\rho_\Lambda\propto t^4\), \(\tilde\rho_m\propto t\), \(\tilde\rho_r=\)const.
The ranking reverses completely, and the word "constant" turns out to be bookkeeping.
The last episode of Part II's cosmology run. Substituting into the vacuum energy produces something rather amusing — in this picture the cosmological constant grows fastest of all (\(\propto t^4\)). And the thing that is genuinely constant is radiation, which diluted fastest in the standard picture. The ranking flips completely. Which is to say: the name "cosmological constant" was itself picture-dependent.
01Transforming all three components at once
Energy density has weight \(-4\), so here \(\tilde\rho=a^4\rho\). Just multiply by the standard dilution.
| Component | Standard picture | This picture | Rank |
|---|---|---|---|
| Radiation | \(\propto a^{-4}\) (dilutes fastest) | constant | 1st ↔ 3rd |
| Matter | \(\propto a^{-3}\) | \(\propto t\) | 2nd ↔ 2nd |
| Cosmological constant \(\Lambda\) | constant (hence "constant") | \(\propto t^4\) (grows fastest) | 3rd ↔ 1st |
Conclusion of §01
The name "cosmological constant" depends on the picture.
What is genuinely constant here is radiation, the component that diluted fastest in the standard picture.
── Episode 3's surgery on the series title reaches even this far.
Episode 11 counted "the photon gas is completely at rest". The first row here is the energy-density version of that: \(\rho_r=7.05\times10^{-14}\ \mathrm{J/m^3}\), the same value throughout cosmic history.
02And the cosmological constant problem does not move a millimetre
The cosmological constant problem is the \(10^{120}\) gap between the naive field-theory estimate and the observed value. If \(\rho_\Lambda\) grows as \(t^4\) here, does the problem move?
to the fourth power, the famous number
$$\frac{\rho_\Lambda}{M_{\rm Pl}^4}=1.13\times10^{-123}$$\(\rho_\Lambda\) grows as \(t^4\) — and so does \(M_{\rm Pl}^4\), since \(M_{\rm Pl}\) is a mass of weight \(-1\). They cancel exactly in the ratio.
Conclusion of §02
The cosmological constant problem (\(10^{-123}\)) does not move a millimetre when the picture changes.
The ratio is dimensionless — exactly the structure of Episode 10's Landauer cost.
03The "why now?" problem does not vanish either
Standard picture
$$\frac{\rho_\Lambda}{\rho_m}=\frac{1}{a^{-3}}\propto a^3$$This picture
$$\frac{\tilde\rho_\Lambda}{\tilde\rho_m}=\frac{a^4}{a}=a^3\qquad\text{── the same}$$| Value | In this picture | |
|---|---|---|
| Today's \(\rho_\Lambda/\rho_m\) | 2.175 | invariant |
| When \(\Lambda\) equals matter | \(a=0.772\) (\(z=0.30\)) | invariant |
| When radiation equals matter | \(a=2.9\times10^{-4}\) (\(z=3400\)) | invariant |
Not only the ratio but its time dependence is the same. So the degree of strangeness in "why exactly now?" is untouched. Change the picture and the puzzle remains a puzzle.
Figure: the three energy densities. Switching the way of speaking rotates the three lines and reverses their ranking (at the right, radiation is flat and \(\Lambda\) is steepest). And yet the two crossings — the equality epochs — do not move at all.
At the left (standard) the red plunges and the green is flat — the familiar textbook figure. Drag right and the three rotate until, at the far right, red is flat, dark green is \(+1\), light green is \(+4\). It looks like a different universe, and yet not one crossing has moved.
04The reveal — "constant" is a word with a hidden comparison
| Compare \(\rho_\Lambda\) with… | Result |
|---|---|
| energy per comoving volume (standard) | constant → "cosmological constant" |
| particle masses (this picture) | \(\propto t^4/t^4=\)constant |
| a fixed ruler's volume (this picture) | \(\propto t^4\) → not a constant |
| \(M_{\rm Pl}^4\) | \(1.13\times10^{-123}\), invariant ← this is the physics |
Only the last row is a genuinely invariant statement. That, and the equation of state \(w=-1\), which is also dimensionless and therefore the same in every picture. Those two are all that can be said physically about \(\Lambda\).
05An aside — the two smallest numbers in nature
ratio
$$\frac{m_\nu}{\rho_\Lambda^{1/4}}=22.3$$The two smallest scales known in nature are only 22 apart, when everything else is separated by factors of \(10^{25}\). This coincidence, pointed out in Extra 5 of the previous series, is of course dimensionless — so it does not move with the picture either. No theory currently explains it.
① \(\tilde\rho=a^4\rho\) follows from energy density having weight \(-4\) (energy \(-1\) plus volume \(+3\)). Standard counting.
② "The cosmological constant problem is \(10^{120}\)" is the ratio of a naive cutoff estimate (\(\rho_{\rm vac}\sim M_{\rm Pl}^4\)) to the observed value. Whether that estimate is legitimate is disputed, and supersymmetry or the treatment of renormalisation change the number substantially. The claim here is only that whatever the ratio is, swapping pictures does not move it.
③ The "why now" problem is quantified via the time dependence of \(\rho_\Lambda/\rho_m\). That is one formulation among several, and whether the coincidence problem is a problem at all is itself debated (anthropic explanations, dynamical dark energy, and so on).
④ \(\Lambda\) is treated as a perfect fluid with \(w=-1\). With dynamical dark energy (quintessence and the like) \(\rho_\Lambda\) varies in time and the third row of §01 changes — though the conclusion that it does not move with the picture is the same.
⑤ \(m_\nu=0.05\) eV is the indicative lower bound on the heaviest neutrino from oscillation experiments. Its factor-of-22 proximity to \(\rho_\Lambda^{1/4}\) is an unexplained numerical coincidence, with no known theoretical relation (Extra 5 of the previous series).
Exercises (solvable with this episode's formulas alone)
- How do the three energy densities behave here, and what happens to the ranking?
Show the answer
\(\tilde\rho=a^4\rho\), so radiation \(a^4a^{-4}=\)const, matter \(a^4a^{-3}=a\propto t\), \(\Lambda\) \(a^4\propto t^4\). The ranking reverses completely: the "constant" \(\Lambda\) grows fastest, and the fastest-diluting radiation becomes the genuine constant. - \(\rho_\Lambda\) grows as \(t^4\); why does the cosmological constant problem not move?
Show the answer
Because \(M_{\rm Pl}^4\), the thing it is compared with, grows as the same \(t^4\) (\(M_{\rm Pl}\) is a mass, weight \(-1\)). Dividing cancels it, leaving \(\rho_\Lambda/M_{\rm Pl}^4=1.13\times10^{-123}\) invariant. - Is the "why now" problem eased in this picture?
Show the answer
No. \(\rho_\Lambda/\rho_m\) goes as \(\propto a^3\) in the standard picture and as \(a^4/a=a^3\) here — even the time dependence is identical. Today's 2.175 and the equality at \(z=0.30\) do not move. Good puzzles are written dimensionlessly, so rewriting cannot reach them. - What is genuinely "constant" in this picture?
Show the answer
The radiation energy density (\(7.05\times10^{-14}\ \mathrm{J/m^3}\)) — the energy-density version of Episode 11's "the photon gas is completely at rest". The name "cosmological constant" was picture-dependent. - (Harder) List everything about \(\Lambda\) that can be said independently of the picture.
Show the answer
Two things only: ① the equation of state \(w=-1\) (dimensionless), and ② \(\rho_\Lambda/M_{\rm Pl}^4=1.13\times10^{-123}\) (dimensionless). The value and time dependence of \(\rho_\Lambda/\rho_m\) follow and are also invariant. That it "is constant" cannot be said independently of the picture — which is the most amusing part of this episode.
Summary — the name "constant" was bookkeeping
Energy density has weight \(-4\), so \(\tilde\rho=a^4\rho\). Applied to the three components: radiation constant, matter \(\propto t\), cosmological constant \(\propto t^4\). The ranking reverses completely. The fastest-diluting radiation becomes the genuine constant, and the thing called "constant" grows fastest. The name "cosmological constant" was itself picture-dependent.
And still the puzzles do not move. \(\rho_\Lambda\) grows as \(t^4\), \(M_{\rm Pl}^4\) grows by the same factor, so \(\rho_\Lambda/M_{\rm Pl}^4=1.13\times10^{-123}\) is invariant. The cosmological constant problem does not shift a millimetre. Nor does "why now": \(\rho_\Lambda/\rho_m\) goes as \(\propto a^3\) in both pictures, today's 2.175 and the equality at \(z=0.30\) unchanged. Good puzzles are written dimensionlessly from the start, so rewriting cannot reach them.
The reveal is Episode 3's again — "is constant" means nothing until you say constant relative to what. What can be said about \(\Lambda\) independently of the picture is exactly two things: \(w=-1\) and \(\rho_\Lambda/M_{\rm Pl}^4\). And as an aside, the two smallest scales in nature (\(\rho_\Lambda^{1/4}/M_{\rm Pl}=1.83\times10^{-31}\) and \(m_\nu/M_{\rm Pl}=4.10\times10^{-30}\)) sit only a factor 22 apart — a coincidence that, being dimensionless, likewise stays put.
This document is Episode 12 of "c·t = const, That Clicks", written for physics-minded high-school and university readers. That energy density has conformal weight \(-4\) (energy \(-1\) plus volume \(+3\)), and that standard cosmology has \(\rho_r\propto a^{-4}\), \(\rho_m\propto a^{-3}\), \(\rho_\Lambda=\)const, are standard. The results \(\tilde\rho_r=\)const, \(\tilde\rho_m\propto t\), \(\tilde\rho_\Lambda\propto t^4\) (the reversal of ranking) are this document's calculation from those two. \(\rho_\Lambda^{1/4}=2.240\) meV, \(\rho_\Lambda^{1/4}/M_{\rm Pl}=1.835\times10^{-31}\) and \(\rho_\Lambda/M_{\rm Pl}^4=1.13\times10^{-123}\) are computed here from \(h=0.674\), \(\Omega_\Lambda=0.685\). The statement "the cosmological constant problem is \(10^{120}\)" is the ratio of a naive cutoff estimate \(\rho_{\rm vac}\sim M_{\rm Pl}^4\) to the observed value, and the legitimacy of that estimate is disputed — the claim made here is only that whatever the ratio is, swapping pictures does not move it. Likewise, whether the coincidence problem ("why now") is a problem at all is debated. \(\Lambda\) is treated as a perfect fluid with \(w=-1\); with dynamical dark energy the third row of §01 changes, though the picture-independence conclusion does not. The equality epochs \(a=(\Omega_m/\Omega_\Lambda)^{1/3}=0.772\) (\(z=0.30\)) and \(a=\Omega_r/\Omega_m=2.9\times10^{-4}\) (\(z=3400\)) are computed here. \(m_\nu=0.05\) eV is the indicative lower bound on the heaviest neutrino from oscillation experiments, and its factor-of-22 proximity to \(\rho_\Lambda^{1/4}\) is an unexplained numerical coincidence (Extra 5 of the previous series). Linear expansion (\(c\cdot t=\)const, \(R_h=ct\)) is a minority model under examination and conflicts with nucleosynthesis when extrapolated into the early universe (Lewis, Barnes & Kaushik 2016). 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).