CONFORMAL TRANSFORMATIONS THAT CLICKEPISODE 5 ／ The high point — a cosmology table becomes a particle-physics table

Put matter in and the dilaton can no longer stay gauge: it starts rolling down a slope

The equation of state
was a potential In Episode 4 gravity became a single scalar field. Now we add matter.
And the first table in every cosmology textbook gets translated, wholesale, into another language.

What you need: one derivative, exponentials and logarithms, and "acceleration = minus the slope of the potential" \(U(\phi)\propto-\phi^{\,1-3w}\)

Episode 4 concluded like this: conformally transform gravity and the scale factor becomes a dilaton field \(\phi\), while "where the expansion is carried" is nothing but a choice of gauge. No physical degrees of freedom were added. But that was because we were looking at gravity alone. The universe contains matter. The moment matter is added, \(\phi\) can no longer be a freely choosable gauge: it takes on a potential and starts to move. And the shape of that potential is fixed completely by a single property of the matter.

01What drives the dilaton is the trace of the stress tensor

First, let us guess which property of matter matters. The symmetry we acquired in Episode 4 was \(\tilde g\to\Omega^2\tilde g,\ \phi\to\phi/\Omega\). As long as matter does not break it, \(\phi\) can remain gauge.

The quantity that measures whether matter breaks the conformal transformation is the trace of the stress–energy tensor, \(T^\mu{}_\mu\). In cosmology matter is described by a density \(\rho\), a pressure \(p\) and an equation of state \(p=w\rho c^2\), so

How badly conformal symmetry is broken
$$T^\mu{}_\mu = -\rho c^2+3p = (3w-1)\,\rho c^2$$

This vanishes only for \(w=1/3\). In other words only radiation (light) leaves the conformal transformation alone. Light has no mass and imports no standard of size — swap the ruler and light cannot tell.

Conversely, any matter with \(w\ne1/3\) breaks conformal symmetry, and the amount of breaking, \((3w-1)\rho\), becomes exactly the force that pushes the dilaton. The rest of this episode extracts that force as an equation.

02Reading the Friedmann equation as a problem in mechanics

We translate the basic equation of an expanding universe (the Friedmann equation) into the language of Episode 4's stage — conformal time \(\eta\) and the field \(\phi=a\). Below, \(' = d/d\eta\).

The calculation — three lines

① Write the Friedmann equation in conformal time

In cosmic time \(H=\dot a/a\), and since \(dt=a\,d\eta\) we have \(H=a'/a^2\). Substituting into \(H^2=\tfrac{8\pi G}{3}\rho\) and rearranging,

$$(\phi')^2=\frac{8\pi G}{3}\,\rho\,\phi^4$$

② Insert how matter dilutes, \(\rho\propto\phi^{-3(1+w)}\)

$$(\phi')^2=K\,\phi^{\,1-3w},\qquad K\equiv\frac{8\pi G\rho_0}{3}$$

③ Rearrange into "kinetic + potential"

$$\tfrac12(\phi')^2\;+\;\underbrace{\Big(-\tfrac{K}{2}\phi^{\,1-3w}\Big)}_{U(\phi)}\;=\;0$$

The left-hand side is exactly the mechanical energy of a unit-mass particle running along the \(\phi\) axis in a potential \(U\). The Friedmann equation says that energy is exactly zero.

Savour this. The equation of an expanding universe has become a ball rolling in one dimension. And with total energy exactly zero — the familiar slogan "the total energy of the universe is zero" is, precisely, this equation.

Differentiate the energy equation with respect to \(\eta\) (\(\phi'\phi''+U'\phi'=0\)) and a familiar form appears.

The conclusion of this episode
$$\phi''=-\,U'(\phi),\qquad\qquad \boxed{\;U(\phi)=-\frac{K}{2}\,\phi^{\,1-3w}\;}$$

Acceleration = minus the slope. The matter's equation of state \(w\) was the exponent of the dilaton's potential.

Look at the exponent — \(1-3w\). It is the same number, up to sign, as the trace \((3w-1)\rho\) from section 01. That is no coincidence: the amount by which conformal symmetry is broken becomes, directly, the shape of the potential.

◇ ◇ ◇

03Building the table — four cosmological characters, four shapes

All that remains is to substitute values of \(w\). Here are the four familiar characters of cosmology.

Matter\(w\)\(\rho\propto\)Potential \(U(\phi)\)\(\phi(\eta)\)\(a(t)\)
Radiation\(1/3\)\(a^{-4}\)flat (zero force) \(U=-K/2\)\(\propto\eta\)\(t^{1/2}\)
Matter\(0\)\(a^{-3}\)a constant-slope ramp \(U\propto-\phi\)\(\propto\eta^2\)\(t^{2/3}\)
\(c\cdot t=\)const\(-1/3\)\(a^{-2}\)a mass term \(U=-\dfrac{\phi^2}{2t_0^{2}}\)\(e^{\eta/t_0}\)\(t\)
Cosmological constant Λ\(-1\)constantself-coupling \(U\propto-\phi^4\)\(-1/(H_0\eta)\)\(e^{H_0t}\)

The left half is the first page of a cosmology textbook; the right half is the language of particle physics. Radiation is a free field with no potential, matter is a field pushed by a constant force, the cosmological constant is a \(\phi^4\) self-coupling — all faces you meet daily in field theory. And the red row in the middle is this series' protagonist.

Figure: turn the \(w\) knob and the shape of the potential on the left and the expansion history on the right change together. The curve on the right is the result of rolling down the hill on the left
w = −0.333 (c·t=const)　U ∝ −φ² (a mass term)　→　a ∝ t^1.00
Potential U(φ) Present position (φ = 1) Expansion a(t)

Push the knob to the right (radiation) and the hill goes perfectly flat; move it left and the downhill gets steeper and steeper. The ball always sets off from \(\phi=0\) (the Big Bang) and rolls right — with total energy zero, its speed at the summit is necessarily zero.

04\(c\cdot t=\text{const}\) was special in three ways at once

Look closely at the red row. \(w=-1/3\) stands on a boundary in three apparently unrelated senses — and the three are one and the same fact.

①
The potential is exactly quadratic — that is, a mass termOf the four, only here is \(U\propto\phi^2\). Quadratic makes the equation of motion \(\phi''=K\phi\) linear, so the solution is uniquely an exponential. It is the simplest possible field theory.
②
Exactly the boundary at which conformal time becomes infiniteThe time needed to get back to the summit \(\phi=0\) is \(\displaystyle\int_0\frac{d\phi}{\phi'}=\frac{1}{\sqrt K}\int_0\phi^{-\frac{1-3w}{2}}d\phi\). This diverges when the exponent is at least 1, i.e. \(w\le-1/3\). \(w=-1/3\) is exactly where it turns into a logarithmic divergence, and from there the Big Bang recedes to infinity.
③
The acceleration is exactly zeroSince \(\ddot a\propto-(1+3w)\), the universe accelerates when \(w<-1/3\). At \(w=-1/3\) it neither accelerates nor decelerates — linear expansion. It is the watershed for whether the horizon problem can be solved.

In Episode 3 we wrote that with \(a\propto t\), conformal time \(\eta\) runs all the way to \(-\infty\) and covers the whole of Minkowski space. The reason is now clear — because the potential is quadratic. The summit of a quadratic hill can only be reached in infinite time; it is the same feeling as the vertex of a parabola, where the speed goes to zero.

The only scale that appears in this theory At \(w=-1/3\) we get \(K=1/t_0^2\) (insert \(\rho_0=3/8\pi Gt_0^2\)), so the potential is \(U=-\phi^2/2t_0^2\). The one number that fixes its curvature is \(1/t_0=H_0\). Converted to an energy, $$\hbar H_0 = 6.58\times10^{-16}\,\mathrm{eV\cdot s}\times2.30\times10^{-18}\,\mathrm{s^{-1}}\approx 1.5\times10^{-33}\ \mathrm{eV}$$ So the conformally transformed \(c\cdot t=\text{const}\) universe is a theory consisting of flat spacetime plus one field of mass \(10^{-33}\) eV, and nothing else.

05Why is every potential a hill?

Looking at the table again, something should bother you. Every \(U\) carries a minus sign in front. Not valleys but hills. The ball rolls down and never comes back.

This is Episode 4's homework, collected. There we saw that the dilaton's kinetic term has the wrong sign (a ghost): where a healthy field would have \(-\tfrac12(\partial\phi)^2\), we had \(+6(\partial\phi)^2\). If the sign of the kinetic term is reversed, so is the way the potential acts. Where a healthy field settles into a valley, this field runs down a hill.

Episode 4's ghost shows its face here
$$\underbrace{+6(\partial\phi)^2}_{\text{wrong sign}}\qquad\Longrightarrow\qquad \underbrace{U\propto-\phi^{\,1-3w}}_{\text{hills everywhere}}$$

And rolling forever down a hill means it never stops. You may read this as: expansion does not end because of that sign. Of course that is the language of the rewritten side; in the original picture one simply says "the universe keeps expanding". The same fact wears a different face in a different language.

Being straight with you

The \(U(\phi)\) of this episode is an effective potential, extracted by rewriting the Friedmann equation. It does not say "a field with this potential exists in the universe" — it is a dictionary that shows how the matter's equation of state looks when translated into the dilaton's language. Episode 4's counting (one field, minus one symmetry, net zero) still applies.

And who supplies \(w=-1/3\) is a question this episode does not answer at all. Radiation, matter and dark energy are none of them at \(w=-1/3\), so the total for the whole universe has to come out at exactly \(-1/3\) (which is precisely what the \(R_h=ct\) universe demands). Whether that happens naturally is unresolved — and running it at face value back into the early universe wrecks nucleosynthesis. The bonus episode takes that head on. A beautiful potential and a correct model are two different things.

Exercises (solvable with this episode's equations alone)
  1. What shape does \(U\) have for radiation (\(w=1/3\))? How does the ball move? Derive \(a(t)\) from that.
    Show answer
    \(1-3w=0\), so \(U=-K/2\) is a constant — perfectly flat. Zero slope means zero force, so \(\phi'\) is constant and \(\phi\propto\eta\). Cosmic time is \(t=\int a\,d\eta\propto\eta^2\), hence \(\eta\propto t^{1/2}\) and \(a\propto t^{1/2}\). That is the radiation-dominated era.
  2. Why do the \(w\) at which the trace \(T^\mu{}_\mu=(3w-1)\rho c^2\) vanishes and the \(w\) at which the slope of \(U\) vanishes coincide?
    Show answer
    Because that trace is precisely the source that pushes the dilaton. Zero trace = matter does not break the conformal transformation = nothing is transmitted to the dilaton = zero force. They are two ways of saying one thing.
  3. For \(c\cdot t=\text{const}\) (\(U=-\phi^2/2t_0^2\)), how much conformal time does the ball need to return to the summit \(\phi=0\)?
    Show answer
    \(\phi'=\phi/t_0\), so \(d\eta=t_0\,d\phi/\phi\). Since \(\int_0 d\phi/\phi\) diverges logarithmically, the answer is infinite. Episode 3's "the Big Bang sits at \(\eta\to-\infty\)" is the same fact as this one line.
  4. (Harder) The horizon problem is solved for \(w<-1/3\). Explain, in the three ways of section 04, why \(c\cdot t=\text{const}\) stands exactly on the boundary.
    Show answer
    ① The exponent \(1-3w\) of the potential becomes 2 (quadratic). ② The exponent in the backward conformal-time integral \(\int_0\phi^{-(1-3w)/2}d\phi\) becomes 1, switching from convergence to logarithmic divergence. ③ \(\ddot a\propto-(1+3w)\) becomes zero. All three happen simultaneously at \(w=-1/3\) — they are not separate phenomena but three faces of one fact.

SUMMARYA cosmology table became a particle-physics table

Add matter and the dilaton can no longer be gauge. What pushes it is the trace of the stress–energy tensor, \((3w-1)\rho c^2\), and only radiation — for which it vanishes — leaves the conformal transformation alone. Rewriting the Friedmann equation in conformal time gives \(\tfrac12(\phi')^2+U(\phi)=0\), a rolling ball with exactly zero total energy, with potential \(U(\phi)=-\tfrac{K}{2}\phi^{1-3w}\) — the equation of state becomes the exponent of the potential.

Radiation is flat (a free field), matter is a constant ramp, the cosmological constant is \(\phi^4\), and \(c\cdot t=\text{const}\) is exactly a mass term, \(U=-\phi^2/2t_0^2\). That \(w=-1/3\) is the watershed where three things happen at once — ① the potential becomes quadratic, ② conformal time becomes infinite, ③ the acceleration vanishes — and the three are different faces of one fact. Every potential is a "hill" because Episode 4's ghost sign is still at work.

This document is Episode 5 of the series "Conformal Transformations That Click", written for physics-minded high-school and university students. The Friedmann equation in conformal time \((a')^2=\frac{8\pi G}{3}\rho a^4\), the perfect-fluid scaling \(\rho\propto a^{-3(1+w)}\), and the resulting \(a\propto t^{2/3(1+w)}\) are standard results. The \(U(\phi)=-\frac{K}{2}\phi^{1-3w}\) here is an effective potential read off by rewriting the Friedmann equation (the Hamiltonian constraint) as \(\frac12(\phi')^2+U=0\); it assumes no new physics. Its sign (\(U\le0\)) comes from the wrong-sign kinetic term of the conformal factor seen in Episode 4. That the trace \(T^\mu{}_\mu=(3w-1)\rho c^2\) vanishes at \(w=1/3\), and that the acceleration condition is \(w<-1/3\), are also standard. \(w=-1/3\) (linear expansion, \(R_h=ct\)) is a requirement on the effective equation of state of the universe as a whole and is not realised by a simple sum of the known components. On the conflict with nucleosynthesis when linear expansion is applied to the early universe, see Lewis, Barnes & Kaushik (2016, MNRAS 460, 291). The academic standard is the ΛCDM model including inflation. The figure is schematic with \(K=1\), and the right-hand curve plots \(a\propto t^{2/3(1+w)}\). — 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 varies the equation of state w and shows the potential's shape and the expansion history changing together. "Show answer" opens the solutions.