The Muon g-2 Discrepancy Now Turns on One Number: Hadronic Vacuum Polarisation — Epoche C2
The muon's anomalous magnetic moment is now measured to 0.19 parts per million, and whether that measurement disagrees with the Standard Model depends entirely on a single term in the prediction — the hadronic vacuum polarisation — for which two independent methods of evaluation give answers differing by more than either quotes as its own error. This note records where that situation actually stands, because the headline version, a five-sigma departure from theory and hence new particles, omits the step on which everything currently depends. Natural units are used, $\hbar = c = 1$, with energies in MeV and GeV. Following the convention of the field, the anomaly and its contributions are quoted in units of $10^{-11}$. 1. The quantity, and why the muon is the right probe Before the dispute can be stated, the quantity in dispute has to be pinned down, together with the reason anyone expects it to be sensitive to physics beyond the Standard Model. The magnetic moment of a spin-half particle is $\boldsymbol{\mu} = g\,(e/2m)\,\mathbf{s}$; the Dirac equation gives $g = 2$ exactly, and the anomaly is defined as the fractional excess, $a_\mu = (g-2)/2$. Schwinger's 1948 calculation of the leading radiative correction gave $a = \alpha/2\pi$ for any charged lepton. With $\alpha^{-1} = 137.036$ this evaluates to $1/(137.036 \times 6.28319) = 1.1614097\times10^{-3}$, that is $116\,140\,973\times10^{-11}$. The full electrodynamic contribution is a series in $\alpha/\pi = 2.3228\times10^{-3}$, with coefficients $C_n$ that depend on the lepton mass ratios and have been computed through $n = 5$: order contribution, units of $10^{-11}$ $(\alpha/\pi)^1$ 116 140 973.3 $(\alpha/\pi)^2$ 413 217.6 $(\alpha/\pi)^3$ 30 141.9 $(\alpha/\pi)^4$ 381.0 $(\alpha/\pi)^5$ 5.1 total 116 584 718.9 The five terms sum to $116\,584\,718.9$, and Schwinger's single term supplies $116\,140\,973/116\,584\,719 = 99.62\%$ of it. The successive ratios — $413\,218/116\,140\,973 = 3.6\times10^{-3}$, then $30\,142/413\,218 = 7.3\times10^{-2}$, then $381/30\,142 = 1.3\times10^{-2}$ — show a series whose coefficients grow but not fast enough to defeat the expansion parameter, which is why the quoted uncertainty on the total is $0.1\times10^{-11}$, a part in $10^{9}$ of the whole. Everything interesting therefore lives in the remaining hundred parts per million of $a_\mu$. Sensitivity to a heavy new particle follows from dimensional analysis of the one-loop diagram in which the particle, of mass $M$, is exchanged on the muon line. The vertex correction is a loop, giving $1/16\pi^2$; it must vanish as $M \to \infty$ and it flips the muon's chirality, which costs one factor of $m_\mu$, while the operator itself carries a second, so the amplitude scales as $m_\mu^2/M^2$. For a coupling of order unity, $\delta a_\mu \sim (1/16\pi^2)(m_\mu/M)^2$. The $m_\mu^2$ is why the muon and not the electron is the probe: the ratio of sensitivities is $(m_\mu/m_e)^2 = (105.658/0.511)^2 = 4.3\times10^{4}$, and no electron measurement, however precise, recovers that factor. The same relation, run backwards, says what mass a generic explanation would need. Inverting for the size of the discrepancy as it stood in 2021, $\delta a_\mu = 249\times10^{-11} = 2.49\times10^{-9}$, gives $(m_\mu/M)^2 = 16\pi^2 \times 2.49\times10^{-9} = 157.9 \times 2.49\times10^{-9} = 3.93\times10^{-7}$, hence $m_\mu/M = 6.27\times10^{-4}$ and $M = 105.7\ \mathrm{MeV}/(6.27\times10^{-4}) = 1.7\times10^{5}\ \mathrm{MeV} = 1.7\times10^{2}$ GeV. A generic explanation therefore sits at the electroweak scale, which is precisely where the LHC has looked hardest and found nothing. That is one reason to look very carefully at the theory side before concluding anything. The electroweak contribution of the Standard Model itself illustrates the same scaling and can be checked in a line. The one-loop $W$ and $Z$ exchanges give approximately $a_\mu^{\mathrm{EW},1} \approx 5G_F m_\mu^2/(24\sqrt{2}\pi^2)$; with $G_F = 1.1664\times10^{-5}\ \mathrm{GeV}^{-2}$ and $m_\mu^2 = 0.011164\ \mathrm{GeV}^2$, the numerator is $5\times1.302\times10^{-7} = 6.51\times10^{-7}$ and the denominator is $24\times1.4142\times9.8696 = 335.0$, giving $1.94\times10^{-9}$, or $194\times10^{-11}$. Two-loop corrections, enhanced by logarithms of $M_Z/m_\mu$, subtract about $41\times10^{-11}$, leaving the value used below. Since $G_F \propto 1/M_W^2$, this is the $(m_\mu/M)^2$ law with $M = M_W$, and it lands where the estimate says it should. 2. The measurement, and why it is not the contested part Having fixed what is being predicted, the next question is how well it is measured, and the answer is: well enough that the measurement is not where the argument is. A muon storage ring measures not $g$ but the difference between the spin precession frequency and the cyclotron frequency, $\omega_a = \omega_s - \omega_c = (e/m)\,a_\mu B$, which is proportional to the anomaly itself rather than to $g$ — so a one per mille knowledge of $B$ buys a one per mille knowledge of a quantity that is already a part in a thousand of the whole. Electric quadrupoles are needed for vertical focusing, and they contaminate $\omega_a$ with a term proportional to $a_\mu - 1/(\gamma^2-1)$. That term vanishes at the "magic" Lorentz factor $\gamma^2 = 1 + 1/a_\mu = 1 + 857.7 = 858.7$, so $\gamma = 29.30$ and the stored momentum is $p = \gamma\beta m_\mu = 29.30 \times 0.99942 \times 105.658\ \mathrm{MeV} = 3.094$ GeV. Both experiments discussed here ran at that momentum. The anomaly is then extracted from the ratio of $\omega_a$ to the free-proton Larmor frequency measured in the same field, so the absolute magnetometry cancels into externally known constants rather than entering directly. Brookhaven's E821 reached 0.54 ppm and published its final report in 2006. Fermilab's Run-1 result in 2021 reproduced it at 0.46 ppm, and the 2023 publication of Runs 1 to 3 gave a world average $a_\mu(\mathrm{exp}) = 116\,592\,059\,(22)\times10^{-11}$, a fractiona