Let $\phi$ be a scalar field and $g_{\mu \nu} = \eta_{\mu \nu}+h_{\mu \nu}/M_p$ where $M_p$ is the Planck mass (so we assume we deal with perturbations). Let $\Lambda_2,\Lambda_3$ be energy scales such that $\Lambda_2 \gg \Lambda_3$. These are defined by $\Lambda^2_2 = M_p H_0$ and $\Lambda_3^3 = M_p H_0^2$. The Horndeski action is:
$$S = \int d^4 x \sqrt{-g} \sum^5_{i=2} \mathcal{L}_i,$$
where
\begin{align}
\mathcal{L_2}&=\Lambda_2^4 G_2,\nonumber \\
\mathcal{L_3}&=\Lambda_2^4 G_3 [\Phi], \nonumber \\
\mathcal{L_4}&= M_p^2 G_4 R + \Lambda_2^4 G_{4,X}([\Phi]^2 – [\Phi^2]),\nonumber \\
\mathcal{L_5}&= M_p^2 G_5 G_{\mu \nu}\Phi^{\mu \nu} – \frac{1}{6}\Lambda_2^4 G_{5,X}([\Phi]^3 – 3[\Phi][\Phi^2] + 2[\Phi^3]), \nonumber
\end{align}
where $G_2,G_3,G_4,G_5$ are functions of $\phi$ and $X = -\frac{1}{2}\nabla^\mu \phi \nabla_\mu \phi /\Lambda_2^4$, $\Phi^{\mu}_{ \ \nu}:= \nabla^\mu \nabla_\nu \phi/\Lambda_3^3$ and square brackets indicate the trace, e.g. $[\Phi^2] = \nabla^\mu \nabla_\nu \phi \nabla^\nu \nabla_\mu \phi/\Lambda_3^6$ and $,$ denote partial derivatives. We assume that $\langle \phi \rangle = 0$ on the background, bars will indicate evaluation at the background.
I am trying to derive the expression for the graviton propagator in this theory but it does not work out well. My idea was to identify the Lagrangian for $2$ gravitons (where $h = h^\mu_\mu$):
\begin{align}
\mathcal{L}_{hh} &= \Lambda^4_2 \bar{G}_2\sqrt{-g} + \bar{G}_4 M_{\mathrm{pl}}^2 R \sqrt{-g}\\
&\approx \bar{G}_2 H_0^2 \Big(1+\frac{1}{2}h + \frac{1}{8}h^2 – \frac{1}{4}h_{\mu \nu}h^{\mu \nu}\Big) + \bar{G}_4 \Big(-\frac{1}{2}(\partial_\sigma h)(\partial_\mu h^{\mu \sigma}) + \frac{1}{2}(\partial_\nu h)(\partial^\nu h) + \frac{1}{2}(\partial_\nu h^{\mu \nu})(\partial^\sigma h_{\sigma \mu})+ \frac{1}{2}(\partial^\sigma h_{\sigma \nu})(\partial_\mu h^{\mu \nu}) – \frac{1}{2}(\partial_\nu h)(\partial_\mu h^{\mu \nu}) – \frac{1}{2}(\partial_\beta h^{\mu \nu})(\partial^\beta h_{\mu \nu})\Big).
\end{align}
In ordinary GR one would add the gauge fixing term $-\bar{G}_4 (\partial_\nu h^{\mu \nu} – \frac{1}{2}\partial^\mu h)^2$ so that the last term becomes $\bar{G}_4(\frac{1}{4} (\partial_\nu h)^2 – \frac{1}{2} (\partial_\beta h^{\mu \nu})^2)$. However, in this case in the paper arXiv:1904.05874 they mention that the graviton propagator $\mathcal{P}^{\nu \beta}_{\mu \alpha}$ is found from ($\delta$ is the generalised Kronecker delta):
$$\frac{\bar{G}_4}{2}\mathcal{P}^{\nu \beta}_{\mu \alpha}(p) \delta^{\alpha \rho \mu^\prime}_{\beta \sigma \nu^\prime} p_\rho p^\sigma = -i \delta^{\mu^\prime}_\mu \delta^{\nu^\prime}_\nu.$$
My questions are: How do they get rid of the $\bar{G}_2$ term and what gauge fixing term has taken to arrive at this result for the definition of the propagator?
Best Answer
The propagator relation can be obtained by choosing the gauge term to be $-\bar{G}_2 H_0^2((1/8)h^2 - (1/4)h_{\mu \nu}^2)$ (the reason that this can be done comes from considering the first order equation of motion for the free graviton) since then the Lagrangian reads (the constant term and term proportional to $h$ in $\mathcal{L}_{hh}$ vanish by the equation of motion for $h_{\mu \nu}$):
$$\mathcal{L}_{hh} = -\frac{\bar{G}_4}{4} \delta^{\alpha \rho \mu}_{\beta \sigma \nu} (\partial_\rho h^\beta_\alpha)(\partial^\sigma h^\nu_\mu).$$