Thinking in Waves Episode 5 ／ Part II · Constants as waves

Change the axis and the answer flips

The running coupling ──
here the constant really was an integral Last time, differentiating with respect to time gave zero. We had the wrong axis.
Differentiate with respect to the logarithm of energy and $\alpha$ does move ── $1/137.036\to1/128.940$.
And that change is precisely the integral of the beta function. A one-loop calculation reproduces the published value.

Prerequisites: Episode 4Computation: kenshou/hashiru.js

"Are all physical constants integrals of something?" ── there is exactly one place where the answer is literally yes.
The coupling constants.

01Change the axis

Couplings run with energy $$\frac{d\alpha}{d\ln\mu}=\beta(\alpha)$$ $$\alpha(\mu)=\alpha(\mu_0)+\int\beta(\alpha)\,d\ln\mu\qquad\textbf{★ constant ＝ integral}$$

This is not a hypothesis but a measured fact. Let us compute it.

02Why does it run? ── the vacuum polarises

Around a charge the vacuum polarises into electron–positron pairs. Seen from afar it is screened and weak; get closer and you see more of the bare charge, so it grows.

One-loop vacuum polarisation (leptons only) $$\Delta\alpha_{\text{lep}}(\mu)=\frac{\alpha}{3\pi}\sum_\ell\left[\ln\frac{\mu^2}{m_\ell^2}-\frac53\right],\qquad \alpha(\mu)=\frac{\alpha(0)}{1-\Delta\alpha}$$

Particle	Mass [MeV]	$\ln(m_Z^2/m_\ell^2)$	incl. $-5/3$	Contribution $\Delta\alpha$
electron	0.5110	24.184	22.517	$1.743\times10^{-2}$
muon	105.658	13.521	11.854	$9.178\times10^{-3}$
tau	1776.860	7.876	6.209	$4.808\times10^{-3}$
total			40.581	$0.031421$

Cross-check for §02

Computed $\Delta\alpha_{\text{lep}}(m_Z)=\mathbf{0.031421}$; published 0.031498.
One loop gets you that far.

03The hadronic contribution ── this part cannot be computed

The quark contribution cannot be obtained perturbatively, because the strong interaction takes over at low energy. It is extracted from experiment ($e^+e^-\to$hadrons).

Putting it together $$\Delta\alpha_{\text{had}}^{(5)}(m_Z)=0.02766\quad(\text{measured; published value})$$ $$\Delta\alpha=0.031421+0.02766=0.059081$$ $$\alpha(m_Z)=\frac{7.297353\times10^{-3}}{1-0.059081}=7.755558\times10^{-3}$$

	Our calculation	Published
$1/\alpha(m_Z)$	128.940	$\approx$ 128.95

04This is the answer ── the constant is moving

Energy	$1/\alpha$	$\alpha$
0 (infinite distance)	137.036	$7.297\times10^{-3}$
$m_Z=91.2$ GeV	128.940	$7.756\times10^{-3}$

The heart of §04

$1/\alpha$ moves from 137.0 to 128.9 ── a 5.9 % change.
"The fine-structure constant, 1/137" is only its low-energy limit.

And that change is exactly an integral:

$$\Delta\!\left(\frac1\alpha\right)=-\frac{1}{3\pi}\sum_\ell\left[\ln\frac{\mu^2}{m_\ell^2}-\frac53\right]-(\text{hadrons})$$

★ "Constants are integrals" ── here it is literally true. But the integration variable is not time; it is $\ln(\text{energy})$.

05Keep running it and it diverges

Take the one-loop formula and keep pushing the energy up. (Leptons only in this section — hadrons and the W are omitted, so the value at $m_Z$ differs from the previous section. This is about the shape.)

Energy $\mu$	$1/\alpha(\mu)$	Note
0.511 MeV	137.213	electron mass
1.777 GeV	135.237	tau mass
91.2 GeV	132.730	Z boson mass
1 TeV	131.206
$1.22\times10^{19}$ GeV	107.625	Planck mass
$10^{93}$ GeV	0	★ Landau pole. $\alpha$ diverges

This extrapolation is formal

In reality other interactions take over long before, and $10^{93}$ GeV is $10^{74}$ times the Planck mass ── not a regime where a gravity-free calculation means anything.
Still, the structure is interesting ── keep integrating and it diverges at a finite energy. The "constant" turns out to be a quantity running toward a divergence.

06The other couplings run too ── and they cross

Interaction	Sign of $\beta$	At high energy	Reason
Electromagnetic (U(1))	$+$	grows	screened by vacuum polarisation; stronger up close
Weak (SU(2))	$-$	shrinks	gauge-boson self-interaction wins
Strong (SU(3))	$-$	shrinks	★ asymptotic freedom; weaker up close

Since some grow and some shrink, they may cross somewhere. In the Standard Model they "nearly" meet around $10^{13}\sim10^{17}$ GeV; adding supersymmetry they agree well at $2\times10^{16}$ GeV ── the case for grand unification.

The heart of §06

"Integrating the constants" is itself the central tool of particle physics.
The intuition has been standard in the field for fifty years ── only as an integral over scale, not over time.

07Put the two derivatives side by side

Differentiate with respect to	Symbol	Result	Episode
time	$d\alpha/dt$	measured zero (bound $10^{-17}$/yr)	Ep. 4
$\ln E$	$d\alpha/d\ln\mu$	★ non-zero. Measured	this episode

Conclusion of this episode

For the same $\alpha$, the answer flips depending on the axis you differentiate along. Constant in time, variable in energy.

In wave language ── $\alpha$ is DC in time and non-DC in log-energy.
So whether something is a constant depends on which axis's spectrum you look at.

⇒ Answer to "are all physical constants integrals of something?":
★ for the coupling constants, exactly so. The key is guessing what the integration variable is.

Exercises

Why is "1/137" the famous number?
Show answer
Because everyday energy scales (atoms, chemistry) are effectively the $\mu\to0$ limit.
An atom is $\sim10^{-10}$ m across, corresponding to energies of eV, far below the electron mass (0.511 MeV). So vacuum polarisation barely acts and $\alpha(0)=1/137.036$ is what you see.
Only at accelerator energies does $1/129$ become the number you actually use.
Why is the hadronic contribution taken from experiment alone?
Show answer
Because $\alpha_s$ grows at low energy and the perturbative expansion does not converge (the flip side of asymptotic freedom).
Instead one measures the $e^+e^-\to$hadrons cross-section and converts it via a dispersion relation (a consequence of causality — Episode 8) into $\Delta\alpha_{\text{had}}$.
In other words, "the incomputable part is filled in with measurement plus causality" ── which is itself a theme of this series.
Is "constant in time but variable in energy" a contradiction?
Show answer
No, because they are different quantities.
$\alpha(\mu)$ is the effective coupling strength measured at scale $\mu$, a function of $\mu$. Whether "$\alpha$ varies in time" asks whether the value measured at the same $\mu$ differs between epochs.
The former is a consequence of the renormalisation group (inside the theory); the latter would be new physics (outside it). Conflating them produces the false claim "$\alpha$ runs, therefore it varies".

Summary　It was the axis that was wrong

Last time, differentiating in time gave zero. But couplings run with energy.

We computed the one-loop vacuum polarisation ── $\Delta\alpha_{\text{lep}}(m_Z)=\mathbf{0.031421}$ (published 0.031498). Adding the hadronic 0.02766 gives $1/\alpha(m_Z)=128.940$ (published 128.95).

$1/137.036\to1/128.940$: a 5.9 % change. And the change is precisely the integral of the beta function ── "constants are integrals" is literally true here.

Keep running and it diverges at a finite energy ($10^{93}$ GeV, a formal extrapolation). The constant turns out to be a quantity racing toward a divergence. And because the three couplings run in different directions, they cross at high energy ── the case for grand unification.

Whether something is a constant depends on which axis's spectrum you look at.
$\alpha$ is DC in time and non-DC in log-energy.

Next time ── into Part III The constants are settled. From here we look at physical quantities themselves as waves.
First, mass. The proposal was "the integral of the universe's gravity", but ── seen as a wave, a completely different and completely successful answer appears.
Mass is the cutoff frequency of the vacuum, regarded as a waveguide.

This document is Episode 5 of the series "Thinking in Waves". The computation is in kenshou/hashiru.js. On the numbers: §02 uses the standard one-loop QED vacuum-polarisation formula; we merely put numbers into it. $\Delta\alpha_{\text{had}}^{(5)}=0.02766$ is a published value derived from experiment, not computed here. The $1/\alpha(m_Z)=128.940$ in §03 is the five-flavour QED value; the frequently quoted 127.95 is the $\overline{\text{MS}}$ value including the electroweak sector ── the value is scheme-dependent, so compare with care. §05 is a one-loop extrapolation with three leptons only, and the Landau pole at $10^{93}$ GeV is a formal number (gravity, hadrons and the W are all ignored, so it has no physical meaning). The unification scales in §06 summarise the literature. ── To print, use your browser's Print → Save as PDF.

Energy	\(1/\alpha\)	\(\alpha\)
0 (infinite distance)	137.036	\(7.297\times10^{-3}\)
\(m_Z=91.2\) GeV	128.940	\(7.756\times10^{-3}\)