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September 2, 2023


For various reasons, some people seem to think that the following modification to Einstein Gravity
(1)S=12dϕ*dϕ+κ 22*+3ϕ192π 2fTr(RR)S= \int \tfrac{1}{2} d\phi\wedge *d\phi + \tfrac{\kappa^2}{2} *\mathcal{R} + {\color{red} \tfrac{3 \phi}{192\pi^2 f}Tr(R\wedge R)}
is interesting to consider. In some toy world, it might be1. But in the real world, there are nearly massless neutrinos. In the Standard Model, U(1) BLU(1)_{B-L} has a gravitational ABJ anomaly (where, in the real world, the number of generations N f=3N_f=3)
(2)d*J BL=N f192π 2Tr(RR) d * J_{B-L} = \frac{N_f}{192\pi^2} Tr(R\wedge R)
which, by a U(1) BLU(1)_{B-L} rotation, would allow us to entirely remove2 the coupling marked in red in (1). In the real world, the neutrinos are not massless; there’s the Weinberg term
(3)1M(y ij(HL i)(HL j)+h.c.)\frac{1}{M}\left(y^{i j} (H L_i)(H L_j) + \text{h.c.}\right)
which explicitly breaks U(1) BLU(1)_{B-L}. When the Higgs gets a VEV, this term gives a mass m ij=H 2y ijM m^{i j} = \frac{\langle H\rangle^2 y^{i j}}{M} to the neutrinos, So, rather than completely decoupling, ϕ\phi reappears as a (dynamical) contribution to the phase of the neutrino mass matrix
(4)m ijm ije 2iϕ/fm^{i j} \to m^{i j}e^{2i\phi/f}
Of course there is a CP-violating phase in the neutrino mass matrix. But its effects are so tiny that its (presumably nonzero) value is still unknown. Since (4) is rigourously equivalent to (1), the effects of the term in red in (1) are similarly unobservably small. Assertions that it could have dramatic consequences — whether for LIGO or large-scale structure — are … bizarre.


The claim that (1) has some observable effect is even more bizarre if you are seeking to find one (say) during inflation. Before the electroweak phase transition, H=0\langle H \rangle=0 and the effect of a ϕ\phi-dependent phase in the Weinberg term (3) is even more suppressed.

1 An analogy with Yang Mills might be helpful. In pure Yang-Mills, the θ\theta-parameter is physical; observable quantities depend on it. But, if you introduce a massless quark, it becomes unphysical and all dependence on it drops out. For massive quarks, only the sum of θ\theta and phase of the determinant of the quark mass matrix is physical.
2 The easiest way to see this is to introduce a background gauge field, 𝒜\mathcal{A}, for U(1) BLU(1)_{B-L} and modify (1) to
(5)S=12(dϕf𝒜)*(dϕf𝒜)+κ 22*+3ϕ24π 2f[18Tr(RR)+d𝒜d𝒜]S= \int \tfrac{1}{2} (d\phi-f\mathcal{A})\wedge *(d\phi-f\mathcal{A}) + \tfrac{\kappa^2}{2} *\mathcal{R} + {\color{red} \tfrac{3 \phi}{24\pi^2 f}\left[\tfrac{1}{8}Tr(R\wedge R)+d\mathcal{A}\wedge d\mathcal{A}\right]}
Turning off the Weinberg term, the theory is invariant under U(1) BLU(1)_{B-L} gauge transformations 𝒜 𝒜+dχ ϕ ϕ+fχ Q i e iχ/3Q i u¯ i e iχ/3u¯ i d¯ i e iχ/3d¯ i L i e iχL i e¯ i e iχe¯ i \begin{split} \mathcal{A}&\to \mathcal{A}+d\chi\\ \phi&\to \phi+ f \chi\\ Q_i&\to e^{i\chi/3}Q_i\\ \overline{u}_i&\to e^{-i\chi/3}\overline{u}_i\\ \overline{d}_i&\to e^{-i\chi/3}\overline{d}_i\\ L_i&\to e^{-i\chi}L_i\\ \overline{e}_i&\to e^{i\chi}\overline{e}_i\\ \end{split} where the anomalous variation of the fermions cancels the variation of the term in red. Note that the first term in (5) is a gauge-invariant mass term for 𝒜\mathcal{A} (or would be if we promoted 𝒜\mathcal{A} to a dynamical gauge field). Choosing χ=ϕ/f\chi = -\phi/f eliminates the term in red. Turning back on the Weinberg term (which explicitly breaks U(1) BLU(1)_{B-L}) puts the coupling to ϕ\phi into the neutrino mass matrix (where it belongs).
Posted by distler at 1:34 PM | Permalink | Followups (3)