For various reasons, some people seem to think that the following modification to Einstein Gravity
(1)$S= \int \tfrac{1}{2} d\phi\wedge *d\phi + \tfrac{\kappa^2}{2} *\mathcal{R} + {\color{red} \tfrac{3 \phi}{192\pi^2 f}Tr(R\wedge R)}$
is
interesting to consider. In some toy world, it might be
^{1}. But in the real world, there are nearly massless neutrinos. In the Standard Model,
$U(1)_{BL}$ has a gravitational
ABJ anomaly (where, in the real world, the number of generations
$N_f=3$)
(2)$d * J_{BL} = \frac{N_f}{192\pi^2} Tr(R\wedge R)$
which, by a
$U(1)_{BL}$ rotation, would allow us to
entirely remove^{2} the coupling marked in red in (
1).
In the real world, the neutrinos are not massless; there’s the Weinberg term
(3)$\frac{1}{M}\left(y^{i j} (H L_i)(H L_j) + \text{h.c.}\right)$
which explicitly breaks
$U(1)_{BL}$. When the Higgs gets a VEV, this term gives a mass
$m^{i j} = \frac{\langle H\rangle^2 y^{i j}}{M}$
to the neutrinos, So, rather than completely decoupling,
$\phi$ reappears as a (dynamical) contribution to the
phase of the neutrino mass matrix
(4)$m^{i j} \to m^{i j}e^{2i\phi/f}$
Of course there
is a CPviolating 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 largescale structure — are …
bizarre.
Update:
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,
$\langle H \rangle=0$ and the effect of a
$\phi$dependent phase in the Weinberg term (
3) is
even more suppressed.
^{1} An analogy with Yang Mills might be helpful. In pure YangMills, the $\theta$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 $\theta$ 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,
$\mathcal{A}$, for
$U(1)_{BL}$ and modify (
1) to
(5)$S= \int \tfrac{1}{2} (d\phif\mathcal{A})\wedge *(d\phif\mathcal{A}) + \tfrac{\kappa^2}{2} *\mathcal{R} + {\color{red} \tfrac{3 \phi}{24\pi^2 f}\left[\tfrac{1}{8}Tr(R\wedge R)+d\mathcal{A}\wedge d\mathcal{A}\right]}$
Turning off the Weinberg term, the theory is invariant under
$U(1)_{BL}$ gauge transformations
$\begin{split}
\mathcal{A}&\to \mathcal{A}+d\chi\\
\phi&\to \phi+ f \chi\\
Q_i&\to e^{i\chi/3}Q_i\\
\overline{u}_i&\to e^{i\chi/3}\overline{u}_i\\
\overline{d}_i&\to e^{i\chi/3}\overline{d}_i\\
L_i&\to e^{i\chi}L_i\\
\overline{e}_i&\to e^{i\chi}\overline{e}_i\\
\end{split}$
where the anomalous variation of the fermions cancels the variation of the term in red. Note that the first term in (
5) is a gaugeinvariant mass term for
$\mathcal{A}$ (or would be if we promoted
$\mathcal{A}$ to a dynamical gauge field). Choosing
$\chi = \phi/f$ eliminates the term in red. Turning back on the Weinberg term (which explicitly breaks
$U(1)_{BL}$) puts the coupling to
$\phi$ into the neutrino mass matrix (where it belongs).
Posted by distler at 1:34 PM 
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