c·t = CONST, THAT CLICKS EPISODE 37 ／ Part V — where the tool breaks
"Nothing happens to light" was a classical statement
Quantum theory writes
into the zero column
Quantum theory cannot stay at \(D=4\) — Episode 34's \(\Omega^{D-4}\) becomes the breaking itself.
And we measure that breaking in bits.
Episode 11 counted "nothing happens to light". That was a classical statement. Quantised, a resolution scale \(\mu\) enters and conformal symmetry breaks — and the breaking is exactly the exponent \(\Omega^{D-4}\) from Episode 34. This time we measure the breaking in bits. And we find that its size is nothing but a count of the degrees of freedom.
01Quantum theory cannot stay at \(D=4\)
the exponent vanishes only at \(D=4\) — this is Episode 11's "nothing happens"
But a quantum calculation cannot stay at \(D=4\).
\(\alpha\) is dimensionless only at \(D=4\); at \(\varepsilon\ne0\) it carries a dimension
Once it carries a dimension, it needs exactly the treatment of Episode 35 — pair it with a scale to make it dimensionless. What asymptotic safety does by forming \(g=Gk^2\), QED does by writing \(\alpha(\mu)\).
Conclusion of §01
\(\alpha\), which sat in the "zero column" of Episode 16's weight map, acquires a \(\mu\) once quantised.
── Quantum theory writes a number into the column this series has been calling untouchable.
02How much does it move?
| \(1/\alpha\) | where it comes from | |
|---|---|---|
| Low energy (\(q\to0\)) | \(137.036\) | CODATA 2022 |
| Electron loop alone, up to \(M_Z\) | \(134.47\) | \(\ln(M_Z/m_e)=12.09\), \(\Delta=2.566\) |
| Measured \(1/\alpha(M_Z)\) | \(127.951\) | PDG (\(\overline{\text{MS}}\)) |
The electron loop alone accounts for 28 per cent of the total shift. The muon, tau and quarks fill in the rest — except that the hadronic part cannot be computed perturbatively and is put in from measured \(e^+e^-\to\)hadrons data. That piece is an experimental input, not a prediction.
Conclusion of §02
$$\frac{\alpha(M_Z)}{\alpha(0)}=1.0710\qquad\text{── }\textbf{7.1 per cent larger}$$
03Measuring the "effective weight"
| Charged fermions included | \(d\ln\alpha/d\ln\mu\) |
|---|---|
| The electron only | \(1.55\times10^{-3}\) |
| Three charged leptons | \(4.65\times10^{-3}\) |
| All Standard Model charged fermions | \(1.03\times10^{-2}\) |
The classical weight is 0. Quantised, an "effective weight" of order \(10^{-3}\) appears. Per e-fold it is small, but accumulated over 12.1 e-folds it becomes 7.1 per cent.
04The core — measuring the breaking in bits
the laboratory precision on \(\alpha\) (CODATA 2022) = the noise floor
$$1.6\times10^{-10}\;\to\;32.5\ \text{bits}$$measure the \(m_e\to M_Z\) running in units of that floor
$$\frac{7.10\times10^{-2}}{1.6\times10^{-10}}=4.44\times10^{8}\;\to\;\mathbf{28.7\ bits}$$| On Episode 19's scale | Surprise |
|---|---|
| An identity (Episode 24's \(C\cdot t=N\)) | \(0\) bit |
| The band of coincidences (Episode 36) | \(5.6\) bit |
| Koide's relation | \(15.7\) bit |
| The breaking of conformal symmetry in QED | \(28.7\) bit |
| The uniformity of the CMB (Episode 17) | \(1.6\times10^5\) bit |
The main point of this episode
The breaking sits 29 bits above the noise.
── Far above the band of coincidences (4 to 7), a measured effect that cannot possibly be chance.
Episode 11's "nothing happens" does not survive quantisation.
Figure: the running of \(1/\alpha\). Classically it would be a horizontal line (weight 0); quantised it acquires a slope. Move the slider to change how many charged fermions are counted — the slope is nothing but a count of the fields.
05Does this contradict Episode 30?
Conclusion of §05
Episode 3's procedure applies directly.
── "If you have not named what you are comparing to, you have not yet made a sentence." Here, whether the comparison is across epochs or across scales flips the answer to its opposite.
06The size of the breaking counts the degrees of freedom
\(C^2\) = the square of the Weyl curvature, \(E_4\) = the Gauss–Bonnet term
| Field content | \(a\) | \(c\) | \(c/a\) |
|---|---|---|---|
| One real scalar | \(0.0028\) | \(0.0083\) | \(3.000\) |
| One Weyl fermion | \(0.0306\) | \(0.0500\) | \(1.636\) |
| One vector field (the photon) | \(0.1722\) | \(0.1000\) | \(0.581\) |
| The Standard Model (\(N_0=4,\ N_{1/2}=45,\ N_1=12\)) | \(3.4528\) | \(3.4833\) | \(1.009\) |
The coefficients are fixed by the field content alone — \(a=(N_0+11N_{1/2}+62N_1)/360\), \(c=(N_0+6N_{1/2}+12N_1)/120\). The anomaly coefficient is just a count of the fields.
Conclusion of §06
The size of the breaking = the number of degrees of freedom. The same currency Episode 24 counted information in.
And the \(a\)-theorem (Komargodski–Schwimmer 2011) — \(a\) only decreases along the renormalisation-group flow (\(a_{\rm UV}>a_{\rm IR}\)).
── What Episode 4 called "coarse-graining is irreversible" has a field-theory counterpart, and it is a theorem.
07Was Episode 11 wrong?
(1) §02's \(1/\alpha(M_Z)=127.951\) is a measured value, not a result computed here. Only the electron loop's \(134.47\) comes from the one-loop formula; of the remaining contributions the hadronic part cannot be computed perturbatively and is taken from \(e^+e^-\to\)hadrons cross-section data. That is an experimental input, not a prediction of the theory.
(2) The value of \(\alpha(\mu)\) is scheme-dependent. \(127.951\) is the \(\overline{\text{MS}}\) value and other schemes give other numbers — so "\(\alpha\) moves by 7.1 per cent" is itself not strictly a sentence until the scheme is named (the same structure as §05). Physical scattering amplitudes are scheme-independent, and that the running shows up in experiment is well established.
(3) §04's 28.7 bits depends on choosing the laboratory precision on \(\alpha\) as the noise floor. Taking the floor to be the precision of \(\alpha(M_Z)\) (about \(10^{-4}\)) gives roughly 9.5 bits instead — the point is the placement "far above the band of coincidences", not the digits.
(4) Normalisations of \(a\) and \(c\) differ between references. Here we take \(a=(N_0+11N_{1/2}+62N_1)/360\) and \(c=(N_0+6N_{1/2}+12N_1)/120\). The ratio \(c/a\) and the structure "it counts fields" are normalisation-independent; the absolute values belong to this choice. \(N_{1/2}=45\) counts Weyl fermions without right-handed neutrinos. Also, this form of \(a,c\) is for conformally invariant field theories, so the value 3.45 obtained by applying it to the massive Standard Model should be read as an indication of size.
(5) The \(a\)-theorem is proved in four dimensions, but there is no corresponding theorem for \(c\) (this is distinct from the two-dimensional \(c\)-theorem). §06's link to "coarse-graining is irreversible" is this series' reading, not a claim of Komargodski and Schwimmer.
Exercises
- Why does quantisation break conformal symmetry? Put it in Episode 34's language.
Show the answer
As Episode 34 counted, the Maxwell action goes as \(S\to\Omega^{D-4}S\), and the exponent vanishes only at \(D=4\). But a quantum calculation uses \(D=4-\varepsilon\) (dimensional regularisation) or brings in a \(\mu\) (a cutoff), so it cannot stay at \(D=4\). The breaking is \(\Omega^{-\varepsilon}\ne1\) itself. - What is \(1/\alpha(M_Z)\) from the electron loop alone, and where does the rest come from?
Show the answer
\(137.036-\frac{2}{3\pi}\ln(M_Z/m_e)=137.036-2.566=\mathbf{134.47}\). The measured value is \(127.951\), so the electron supplies only 28 per cent of the shift. The rest is the muon, tau and quarks — but the hadronic part cannot be computed perturbatively and comes from \(e^+e^-\to\)hadrons data. - Do "\(\alpha\) is constant" (Episode 30) and "\(\alpha\) runs" (this episode) contradict each other?
Show the answer
No, because they are different questions. Episode 30 states \(\partial\alpha/\partial t=0\) (same \(\mu\), change the epoch); this episode states \(\partial\alpha/\partial\mu\ne0\) (same epoch, change \(\mu\)). "Constant" is not yet a sentence until you say constant with respect to what (Episode 3). - What does the trace-anomaly coefficient \(a\) count?
Show the answer
The number of fields — \(a=(N_0+11N_{1/2}+62N_1)/360\). The size of the breaking is the number of degrees of freedom itself, the same currency Episode 24 counted information in. And by the \(a\)-theorem, \(a\) only decreases along the RG flow, which corresponds to Episode 4's "coarse-graining is irreversible". - (Harder) How long is the footnote Episode 11 needs?
Show the answer
One line — "classically, that is". The photon's own weight is unchanged by quantisation and Episode 16's table stands. What broke was not the field but the coupling: \(\alpha\) acquired a \(\mu\). Episode 11's claim does not have to be withdrawn.
Summary: quantum theory writes a number into the zero column
The exponent counted in Episode 34, \(S\to\Omega^{D-4}S\), vanishes only at \(D=4\). And quantum theory cannot stay at \(D=4\) — \(D=4-\varepsilon\) with dimensional regularisation, a \(\mu\) with a cutoff. Either way \(\Omega^{-\varepsilon}\ne1\), and the exponent from Episode 34 becomes the breaking itself. Counting dimensions, \([e^2]=\)mass\(^{\,\varepsilon}\): \(\alpha\) is dimensionless only at \(D=4\). So it needs exactly Episode 35's treatment — pair it with a scale and write \(\alpha(\mu)\).
How much does it move? The electron loop alone takes \(1/\alpha\) from \(137.04\) to \(134.47\); the measured value at \(M_Z\) is \(127.951\). \(\alpha(M_Z)/\alpha(0)=1.0710\), 7.1 per cent larger. The "effective weight" per e-fold is only of order \(10^{-3}\), but twelve e-folds accumulate to that — quantum theory has written a small number into the "zero column" where \(\alpha\) sat on Episode 16's map.
Measure the breaking in bits. Taking the laboratory precision \(1.6\times10^{-10}\) (32.5 bits) as the noise floor, the 7.1 per cent running sits 28.7 bits above the noise — far above the band of coincidences (4 to 7 bits) found in Episode 36, and a measured effect that cannot possibly be chance.
Does that contradict Episode 30's "\(\alpha\) is constant to 26 bits"? No — they are different questions. Episode 30 holds \(\mu\) fixed and changes the epoch; this episode holds the epoch fixed and changes \(\mu\). "Constant" is not yet a sentence until you say constant with respect to what — Episode 3's procedure applies directly.
And what does the size of the breaking measure? The trace-anomaly coefficients \(a\) and \(c\) turn out to be nothing but counts of the fields (3.45 for the Standard Model). The size of the breaking = the number of degrees of freedom — the same currency Episode 24 counted information in. Further, by the \(a\)-theorem, \(a\) only decreases along the RG flow. What Episode 4 called "coarse-graining is irreversible" has a field-theory counterpart, and it is a theorem.
Was Episode 11 wrong? No — the footnote it needs is one line, "classically, that is". What broke was not the field but the coupling.
This document is Episode 37 of "c·t = const, That Clicks" (the first of Part V), written for physics-minded high-school and university readers. The trace (conformal) anomaly, the running of the QED coupling, and the \(a\)-theorem are all standard, established material and nothing here is a new claim — the numbers are computed in kenshou/calc41.py. \(1/\alpha(M_Z)=127.951\) is a measured value (PDG, \(\overline{\text{MS}}\)), not a result computed here — only the electron loop's 134.47 comes from the one-loop formula, and of the remaining contributions the hadronic part cannot be computed perturbatively and is taken from \(e^+e^-\to\)hadrons cross-section data. The value of \(\alpha(\mu)\) is scheme-dependent, so "it moves by 7.1 per cent" is not strictly a sentence until the scheme is named (physical amplitudes are scheme-independent, and the running is well established experimentally). The figure of 28.7 bits depends on choosing the laboratory precision on \(\alpha\) as the noise floor; taking the precision of \(\alpha(M_Z)\) (\(\sim10^{-4}\)) instead gives roughly 9.5 bits — the placement "far above the band of coincidences" is the point, not the digits. Normalisations of \(a\) and \(c\) differ between references; here \(a=(N_0+11N_{1/2}+62N_1)/360\) and \(c=(N_0+6N_{1/2}+12N_1)/120\) — this form is for conformally invariant field theories, so the value 3.45 obtained by applying it to the massive Standard Model is an indication of size (\(N_{1/2}=45\) counts Weyl fermions without right-handed neutrinos). The \(a\)-theorem (Komargodski & Schwimmer 2011) is proved in four dimensions but there is no corresponding theorem for \(c\), and the link drawn to Episode 4's "coarse-graining is irreversible" is this series' reading, not a claim of the original paper. ── To make a PDF, use your browser's Print dialogue (sliders freeze and answers are hidden in the print version).