I'm suddenly getting confused on what should be a very simple point. Recall that the $g$-factor of a particle is defined as $$\mu = \frac{ge}{2m} L$$ where $L$ is the spin angular momentum. For any classical system where the charge and mass distributions are identical, $g = 1$. However, one of the great successes of Dirac's theory of the electron was to show that $g = 2$ instead.

The Dirac equation describes spin $1/2$ particles, so it should also apply to the proton. But the proton instead has $g \approx 5.6$. The standard explanation for this is that the proton is a complicated bound state of quarks and gluons, so we shouldn't expect the Dirac equation to apply. But I don't understand why, in detail, it doesn't! More precisely, any argument I can think of that shows $g \neq 2$ for the proton also applies to the electron.

Naively, we can measure the $g$-factor by using light of very low frequencies, with wavelengths much larger than the compositeness scale of the proton. At these distance scales, the proton should look just like a point charge. The only accessible states at these energies are "spin up" and "spin down", just like an electron. From the perspective of early 20th century physics, the electron and proton looked equally pointlike.

So one would naively think that the electron and proton both obey the Dirac equation. But that's not right, because high energy physics can still affect low energy observables. For example, in QFT, we would compute the $g$-factor using the amplitude $\langle p' | j^\mu | p \rangle$ where the external states are proton or electron states. This can be expanded perturbatively, but for the proton there are large loop corrections because the strong coupling is strong at low energies. Meanwhile, the loop corrections are small for the electron because it couples predominantly electromagnetically, and $\alpha_e \approx 1/137$.

This is all standard. My confusion is two-fold. First, at the time of Dirac, we knew of three apparently fundamental particles, the electron, proton, and neutron. If $g \approx 2$ only worked for one of them, how was this a success of the Dirac equation? How was the failure for the other two explained? It couldn't have been that people said the proton and neutron were composite, because we didn't learn that until 50 years later.

Second off, it would seem that $g \approx 2$ for the electron places extremely strong constraints on new physics. For example, if the electron is composite due to a confining force at scales above what we've probed, then this new force almost certainly contributes significantly to $g$, just like the strong force does for the proton. It would seem that this rules out just about all electron compositeness models, but that also seems like too strong of a conclusion to be believed. Does $g \approx 2$ really show this?

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    $\begingroup$ what is wrong in accepting that the Dirac equation is strict for elementary particles only? en.wikipedia.org/wiki/Elementary_particle , and the complexity of gluons quarks antiquarks etc that enter into hadrons will need a much more complicated system ( lattice QCD). see a visualization of the proton profmattstrassler.com/articles-and-posts/largehadroncolliderfaq/… $\endgroup$ – anna v Oct 4 '19 at 5:42
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    $\begingroup$ @annav What puzzles me is then, why it took 50 years after the Dirac equation to discover the proton was composite? Couldn't we have immediately concluded that, from its failure to obey the Dirac equation? $\endgroup$ – knzhou Oct 4 '19 at 7:19
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    $\begingroup$ It took us many years to stop thinking of the proton and neutron as elementary particles and develop the standard model with its axiomatic elementary particles. Have a look here en.wikipedia.org/wiki/Neutron_magnetic_moment to get a feeling how the g of the neutron ( which should not exist if it were elemenary, has a valude because of the compositeness. $\endgroup$ – anna v Oct 4 '19 at 8:21
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    $\begingroup$ and this for the proton en.wikipedia.org/wiki/Proton_magnetic_moment $\endgroup$ – anna v Oct 4 '19 at 8:23
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    $\begingroup$ I'm not sure why you think "a decent argument that 𝑔≈2 for the electron but 𝑔≉2 for the proton, which works independent of physics we haven't observed yet" exists. You've argued very well that if all we know about electron and proton is their charge and their spin, then we'd expect them to have the same g-factor! That the proton has a different g-factor then necessarily means there have to be physics we don't know yet. $\endgroup$ – ACuriousMind Oct 4 '19 at 16:50

I suspect you are relying on the modern language, which is yet controverted by the effective theory community these days, if I am not too cut off from recent developments... I believe it is all hiding behind the receding obsession with renormalizability, and thus minimal coupling, obviated by the Wilson revolution.

The point is the minimal-coupling gauge-invariant, renormalizable, Dirac action was perfectly adequate to describe g=2 through the Gordon decomposition term of the current associated with the magnetic dipole density of the electron, $$ -j'_\mu A^\mu\sim -(e/2m) \left (\frac{1}{2} F^{\mu\nu} \bar \psi \sigma _{\mu\nu}\psi\right ). $$

A physicist of the late 30s (I'm cluelessly guessing here!), knowing the nucleon magnetic moments were not canonical, would augment his minimal coupling Dirac action for them with an extra, nonminimal-coupling (unrenormalizable, which he wouldn't know about) Pauli moment term, stuck in by hand, $$ -(e/2M) \left (\frac{1}{2} F^{\mu\nu} \bar \psi \sigma _{\mu\nu}\psi\right ), $$ perhaps to be added to the above Gordon current piece (which would vanish for the neutral neutron! whose magnetic moment was measured by Alvarez & Bloch, 1939), for a phenomenological parameter M. He would fit everything to determine M for experimentally determined nuclear magnetons; note is not the nucleon's mass, but merely of its rough order of magnitude; and hope for the future to clarify things. Not having a clue about the mysteries of nature, he'd leave it at that.

The late 40s revolution in renormalization allowed computation of corrections to the g of the electron; but, due to non-renormalizability, not for the nucleon, what with the above nasty dimension-5 spatchcocked Pauli term, with its mystery scale M. (As an aside, this term is dear to the heart of extended-supergravitists, M being the Planck scale.)

Then, in the mid-60s, during the triumphant march of quark composition, the said Pauli moment terms were further calculated from a loosely-bound constituent-quark wavefunction. I would not be surprised if today's lattice geeks can specify the exact parameters in the effective Dirac action-cum-Pauli moment involved.

In the subsequent SM years, launched by 't Hooft's proof of SSB YM renormalizability, giddy quasi-religious attachment to renormalizability was lavished on these systems—until Ken Wilson restored humility by reminding us we all live in a resolutely effective action world. But, "elementary" was a virtual shorthand for a field described by a renormalizable action.

So, in the early 80s, hyperambitious model builders were ready to contemplate compositeness even for pure Dirac action particles like the leptons, included in the sample papers of my comment above, and Harari 1982. Now they had the reverse problem: how to constrain the scales of compositeness, so in effect, how to make the M of an extraneous Pauli moment be enormous. I hope you are not asking about that, since these guys went pretty deeply pretty fast. And then they appeared to mumble, shrug, and walk away.

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    $\begingroup$ I'm confused - you say that your first expression is minimally coupled and renormalizable but your second expression is non-minimally-coupled and non-renormalizable. But the two expressions are identical except for the the capitalization of the letter "m". How can this change whether or not the term is renormalizable? $\endgroup$ – tparker Jan 24 '20 at 4:29
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    $\begingroup$ m is the particle mass of the renormalizable Dirac Lagrangian, in the Gordon decomposition, where the e.o.m. were applied. It is a rewriting of a renormalizable theory and m is matched to its further occurrences in the delicate dance of systematic cancelations of renormalizability. M is an arbitrary mass parameter pointedly different from the particle mass, and, prima facie non-renormaliable -- unless you conjured up something freaky lying unobserved for a century. Note that for the neutron there is no electric current to Gordon-decompose! $\endgroup$ – Cosmas Zachos Jan 24 '20 at 11:13

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