CONFORMAL TRANSFORMATIONS THAT CLICKEPISODE 4 ／ From the metric to the action — why was that rewriting allowed?

The scale factor was not a property of spacetime but a field living on it

Gravity becomes
a single scalar In Episode 3 we rewrote the metric. But why such a rewriting is permitted
cannot be seen by staring at the metric. The reason lies one level deeper — in the action.

What you need: integration by parts, exponentials and logarithms, and the idea of an "action" (supplied below) \(\phi^2\tilde R + 6(\partial\phi)^2\)

In Episode 3 we found that the \(c\cdot t=\text{const}\) universe becomes Minkowski under a conformal transformation, and that reading that same metric in cosmic time gives "the speed of light decreases". Three pictures, one physics. But why is such a convenient rewriting allowed at all? Looking only at the metric will not tell you. Apply the conformal transformation to the laws themselves — to the action — and the reason becomes plain. And along the way something slightly startling happens: gravity turns into a single scalar field.

01First: what is an "action"?

There are two ways to write physical law. One is an equation of motion — "move according to this formula". The other is the action, which is a scoring table. You assign a score in advance to every conceivable path, and nature takes the path whose score is stationary.

The scoring table is the more convenient of the two because symmetries are visible at a glance. Rather than glaring at ten equations of motion, you look at one table and say "this operation does not change the score". That is exactly what we want to do here.

Gravity's scoring table is called the Einstein–Hilbert action.

Gravity's scoring table
$$S = \frac{c^4}{16\pi G}\int\! \sqrt{-g}\;R\;d^4x$$

This much is all you need in order to read it. \(R\) is how curved spacetime is (the scalar curvature). \(\sqrt{-g}\) is how volume is measured, the factor that corrects for the metric stretching. Together: "the curvature of spacetime, summed over all of spacetime". Demanding that this score be stationary gives Einstein's equations.

Notation (skippable) Signature \((-,+,+,+)\). \((\partial\phi)^2\) abbreviates \(g^{\mu\nu}\partial_\mu\phi\,\partial_\nu\phi\), and \(\Box\) is \(g^{\mu\nu}\nabla_\mu\nabla_\nu\). Tilded quantities (\(\tilde g,\tilde R,\tilde\Box\)) are the ones after the conformal transformation.

02Putting the conformal transformation into the scoring table

We make the same substitution as in Episode 3 — but this time into the action rather than the metric.

$$g_{\mu\nu} = \phi^2\,\tilde g_{\mu\nu}$$

Here \(\phi\) plays the role that the scale factor \(a\) played in Episode 3 (\(\phi=a\)). But from now on we treat \(\phi\) as a "field" carrying a value at each point of spacetime. That single step is what this episode is about.

Two formulas are needed. The volume one is straightforward — four dimensions, so four factors of \(\phi\). The curvature one is a little more involved, but it is the standard textbook result.

The calculation — it finishes in three lines

① The two formulas (writing \(\omega=\ln\phi\))

$$\sqrt{-g}=\phi^4\sqrt{-\tilde g},\qquad R=\phi^{-2}\Big[\tilde R-6\tilde\Box\omega-6(\tilde\partial\omega)^2\Big]$$

② Multiply and convert \(\omega\) back to \(\phi\)

$$\sqrt{-g}\,R=\sqrt{-\tilde g}\;\phi^2\Big[\tilde R-6\tilde\Box\omega-6(\tilde\partial\omega)^2\Big] =\sqrt{-\tilde g}\Big[\phi^2\tilde R-6\,\phi\,\tilde\Box\phi\Big]$$

(using \(\Box\ln\phi=\Box\phi/\phi-(\partial\phi)^2/\phi^2\), the \((\partial\phi)^2\) terms cancel neatly)

③ Integrate by parts and the sign flips as it survives

$$\int\!\sqrt{-\tilde g}\,\big(-6\phi\tilde\Box\phi\big)=+6\int\!\sqrt{-\tilde g}\,(\tilde\partial\phi)^2$$

A derivative moves from one factor to the other, one minus sign appears, and two minus signs cancel. Exactly the integration by parts you learned in school.

Gravity's scoring table, conformally transformed
$$S = \frac{c^4}{16\pi G}\int\!\sqrt{-\tilde g}\;\Big[\;\phi^2\tilde R\;+\;6\,(\tilde\partial\phi)^2\;\Big]d^4x$$

The original table contained only \(R\). Now there are two terms: the curvature one, and one that penalises \(\phi\) for moving. Gravity has split into geometry and a field.

◇ ◇ ◇

03What has appeared — the "1/6" is no accident

This form is in fact in the particle-physics textbooks. Pull out a factor of 6.

$$S = \frac{6c^4}{16\pi G}\int\!\sqrt{-\tilde g}\;\Big[(\tilde\partial\phi)^2+\tfrac{1}{6}\,\phi^2\tilde R\Big]d^4x$$

What is inside the brackets is exactly a conformally coupled scalar field. When you place a scalar field in curved spacetime, the strength \(\xi\) of its coupling to curvature (the term \(\xi\phi^2R\)) is in principle free to choose. But

The conformal coupling (in \(D\) dimensions)
$$\xi=\frac{D-2}{4(D-1)}\;\xrightarrow{\;D=4\;}\;\frac{2}{12}=\frac{1}{6}$$

Only at this special value is the massless scalar theory invariant under conformal transformations. And when we rewrote gravity, that special value came out by itself. It is no accident — \(\phi\) was introduced as the conformal factor in the first place, so it was bound to couple in the way conformal transformations like. But it is satisfying to see the inevitable appear on the page.

04The scale factor was a field

Here is the biggest statement of the episode.

In the original picture, \(a(t)\) was a property of spacetime itself — a fact about the container: "space is stretching". But the \(\phi\) after the rewriting is a field that rides on spacetime and carries a value. Container \(\tilde g\) and content \(\phi\) have separated. Expansion has emigrated from geometry into a field.

This \(\phi\) has a name: the dilaton (the field of stretching and shrinking). And recalling from Episode 3 that mass becomes \(\phi\) times its old value, the role of this field is clear — the dilaton is the field that sets the scale of every mass.

't Hooft's proposal In 2014–15 Gerard 't Hooft made a proposal starting from precisely this rewriting: "spontaneously breaking conformal symmetry happens automatically in the Einstein–Hilbert action by multiplying the metric tensor with the square of a scalar dilaton field, which takes over the role of the conformal factor." On that basis he argued that local conformal symmetry should be regarded not as an approximate and badly broken symmetry but as an exact symmetry that is spontaneously broken — the same structure as the Higgs mechanism. He goes so far as to suggest it may be as fundamental as Lorentz invariance.

05Why the rewriting was allowed: there is one more symmetry

At last we can answer the opening question. Look at the new scoring table. \(\tilde g\) and \(\phi\) appear only ever in the combination \(g=\phi^2\tilde g\). Which means —

The symmetry we have just acquired
$$\tilde g_{\mu\nu}\to\Omega^2\tilde g_{\mu\nu},\qquad \phi\to\phi/\Omega \qquad\Longrightarrow\qquad S\ \text{is unchanged}$$

Multiply \(\tilde g\) by \(\Omega^2\) and divide \(\phi\) by \(\Omega\), and the product \(\phi^2\tilde g\) does not move. So the score does not change by a single point. The conformal transformation has become a symmetry of this theory.

And if there is a symmetry, then how you divide things up is your choice. Whether the one fact of expansion is carried by \(\tilde g\) or by \(\phi\) is ours to decide — and that is what the three pictures of Episode 3 really were. The difference between the pictures is not a difference of physics but a choice of gauge.

Figure: the gauge knob. However you split the expansion between "the share carried by the metric" and "the share carried by the field", the observed ratio does not move
s = 0.00 (the standard picture)　at t = 0.5t₀:　space ã = 0.500　field φ = 1.000　→　ratio ã·φ = 0.500
Carried by the metric, ã (stretching of space) Carried by the field, φ (= mass m̃/m) The observed ratio ã·φ (does not move)

Blue and green swap places violently as you drag the knob, while the navy line does not twitch. What the navy line represents is "distance between galaxies ÷ radius of an atom" — a dimensionless ratio you can actually measure with a telescope and a ruler. The atomic radius goes as \(1/\tilde m\) and \(\tilde m=\phi m\), so the ratio is proportional to \(\tilde a\cdot\phi\). And \(\tilde a\cdot\phi = a\) does not depend on the choice of gauge.

Shaking hands with bonus episode 8 (previous series) In "Not fixing it was the strongest move", we wrote that Yang–Mills won because it chose a description that keeps the symmetry manifest. What we did here is exactly the same trick. By bringing in one field \(\phi\), the fixed fact of "expansion" was turned into a freedom you may re-choose, and made manifest as a symmetry. The moment it became manifest, the three pictures listed in Episode 3 line up as three gauges of one theory.

06Counting — but nothing was added

We should stop here for a moment. We added a field: did the physics grow? Let us count.

+1
One field addedThe scalar field \(\phi\) (the dilaton)
−1
One symmetry addedThe conformal transformation \(\tilde g\to\Omega^2\tilde g,\ \phi\to\phi/\Omega\). A symmetry eats one degree of freedom
= 0
Net physical degrees of freedom: zeroThat is why the rewriting is safe — and, conversely, why nothing new has happened

This is an operation that rewrites the bookkeeping, not one that adds physics. The backbone of the previous series has surfaced again. But the rewriting bought us one thing: the freedom of choice became visible — the single, and greatest, profit of the exercise.

07The price: one sign points the wrong way

Finally, we leave behind something uncomfortable. Look at the resulting table once more.

$$S \supset \frac{c^4}{16\pi G}\cdot 6\int\!\sqrt{-\tilde g}\,(\tilde\partial\phi)^2$$

For a healthy scalar field this term should begin with a minus, as \(-\tfrac12(\partial\phi)^2\). Here it is \(+6\). The sign points the wrong way. Such a field is called a ghost — the more it moves, the lower its energy, so left alone it runs away without limit.

For now this causes no trouble. As we saw in section 05, \(\phi\) is entirely gauge: a single conformal transformation fixes it to any value you like. Something with no freedom to move has no way to run away.

The trouble comes when you quantise. A path integral sums over every value of every field, so "it is gauge, so do not look" no longer works. In 1978, Gibbons, Hawking and Perry pointed out that because of this sign the gravitational action is unbounded below — the conformal factor problem. It is one of the oldest sore spots in quantum gravity and is still not fully resolved. Episode 9 takes it head on.

Being straight with you

What we did here is a Stückelberg-type rewriting — we did not construct a new conformally invariant theory of gravity. Its physical content is identical to ordinary Einstein gravity; the counting in section 06 is the proof. Conformal symmetry can be manufactured after the fact, in unlimited quantities, this way — do not forget that.

't Hooft's claim is a hypothesis beyond this point: the proposal that we treat it not as a symmetry that can be bolted on afterwards but as one nature actually has, spontaneously broken. It is attractive, but it is not yet a claim with evidence behind it. All today's calculation guarantees is that the rewriting can be done correctly.

Exercises (solvable with this episode's equations alone)
  1. Fix \(\phi\) to a constant (\(\phi=1\)). What does the new scoring table become?
    Show answer
    \(\partial\phi=0\), so the second term vanishes and \(S=(c^4/16\pi G)\int\sqrt{-\tilde g}\,\tilde R\) — back to the original Einstein–Hilbert action. This is "the standard picture" (the gauge in which the metric carries the expansion).
  2. Conversely, fix \(\tilde g\) to Minkowski (\(\tilde R=0\)). What is left in the table, and what theory is it?
    Show answer
    Only \(S=(6c^4/16\pi G)\int(\tilde\partial\phi)^2\). A theory of one scalar field moving on flat spacetime — gravity has turned entirely into the dilaton. This is the action-level face of Episode 3's "masses grow" picture.
  3. What is the conformal coupling \(\xi\) in \(D=2\)? What happens to the \(\phi^2R\) term there?
    Show answer
    \(\xi=(D-2)/4(D-1)=0\). In two dimensions the coupling to curvature disappears and a massless scalar field is conformally invariant as it stands. The special power of two-dimensional conformal field theory is continuous with this fact.
  4. (Harder) In the figure, why does the navy line not move however far blue and green trade places? Explain with exponents.
    Show answer
    Taking the gauge \(s\) as \(\tilde a=a^{1-s},\ \phi=a^{s}\), we always have \(\tilde a\cdot\phi=a^{1-s}\cdot a^{s}=a\). The atomic radius goes as \(1/\tilde m=1/(\phi m)\) and the proper distance as \(\tilde a\), so the ratio is \(\tilde a\phi\propto a\), independent of \(s\). The exponents adding to one is exactly what gauge invariance is here.

SUMMARYExpansion emigrated from geometry into a field

Put \(g=\phi^2\tilde g\) into gravity's scoring table \(S=(c^4/16\pi G)\int\sqrt{-g}R\), integrate by parts, and out comes \(\;\phi^2\tilde R+6(\tilde\partial\phi)^2\;\). Pull a 6 out of the bracket and the coefficient is \(1/6\) — precisely the conformal coupling in four dimensions. Gravity has turned into a single conformally coupled scalar field. That \(\phi\) is the dilaton, the field that sets the scale of every mass ('t Hooft).

And this table is invariant under \(\tilde g\to\Omega^2\tilde g,\ \phi\to\phi/\Omega\) — the conformal transformation has become a symmetry. The three pictures of Episode 3 are three gauge choices of that symmetry. But since we added one field and one symmetry, the net physical degrees of freedom are zero: a rewriting of the bookkeeping, not new physics. The price is that \(\phi\)'s kinetic term has the wrong sign (a ghost) — harmless classically, awkward quantum mechanically.

This document is Episode 4 of the series "Conformal Transformations That Click", written for physics-minded high-school and university students. That under \(g_{\mu\nu}=\phi^2\tilde g_{\mu\nu}\) one has \(\sqrt{-g}=\phi^4\sqrt{-\tilde g}\) and \(R=\phi^{-2}[\tilde R-6\tilde\Box\ln\phi-6(\tilde\partial\ln\phi)^2]\), and that integration by parts brings the Einstein–Hilbert action to the form \(\phi^2\tilde R+6(\tilde\partial\phi)^2\), are standard results (surface terms are dropped). The conformal coupling is \(\xi=1/6\) in four dimensions and \(\xi=(D-2)/4(D-1)\) in general. This rewriting introduces a compensator (Stückelberg field) and adds no physical degrees of freedom — so it does not amount to constructing a new theory of gravity with conformal invariance. The proposal to regard local conformal symmetry as an exact, spontaneously broken symmetry is due to 't Hooft (arXiv:1410.6675 and others). That the kinetic term of the conformal factor has the wrong sign, and that this makes the Euclidean gravitational action unbounded below, has been known since Gibbons, Hawking & Perry (1978, Nucl. Phys. B138, 141). The figure visualises the split under the gauge \(\tilde a=a^{1-s},\ \phi=a^{s}\) with \(a(t)=t/t_0\). — 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 continuously how the expansion is shared between metric and field. "Show answer" opens the solutions.