Dimension of operators in modified gravity I was reading a paper arXiv:1903.06784 where the following action is considered:
\begin{equation}
S\sim \int d^4x \sqrt{-g}\bigg[M^2_{Pl}R-\frac{1}{2}(\nabla\phi)^2 -\frac{1}{2}\mu^2\phi^2-\frac{1}{2}\lambda\phi^4+\eta\, \phi^2\mathcal{G}\bigg]
\end{equation}
where $\mathcal{G}$ is of order $\mathcal{O}(R^2)$. They mention that the following term
$$\zeta\,\phi^4\mathcal{G}$$
would be higher-order and they say that figuring out the complete set of order-6 operators is left for future work.
My question is, how do you count the order of the operators in the action above? What would be an example of an order-6 operator?
 A: Note: the paper says dimension 6 operators, not order 6 operators. However, I should warn you that they are using a somewhat lazy definition of dimension here.
Consider a term in the Lagrangian (in $D$ spacetime dimensions) of the form
\begin{equation}
\mathcal{L} \sim \frac{1}{M^{N-D}} \mathcal{O}_N
\end{equation}
where $\mathcal{O}_N$ is an expression with mass dimension $N$, and $M$ is a dimensionful scale with mass dimension $1$. We say this term is an operator with dimension $N$.
To compute the mass dimension of a term $\mathcal{O}_N$, you add the mass dimensions of each factor. For example, the operator $\partial_\mu$ has units of 1/length, or (in units where $\hbar=c=1$) units of mass, and so has mass dimension 1.
Ordinarily, the mass dimension of a bosonic field $\phi$ is chosen so that the standard kinetic term $\sim (\partial \phi)^2$ has mass dimension $D$ in $D$ spacetime dimensions, so that the action $S\sim \int {\rm d}^D x (\partial \phi)^2 + ...$ is dimensionless (since the action has units of $\hbar$ and we have set $\hbar=1$). In $D=4$ spacetime dimensions, this means that a bosonic field $\phi$ has dimensions of mass. The reason for this normalization, is that under a change in energy scale, if every field, coordinate, and derivative is scaled according to its mass dimension, the action will tell you which terms become more important at larger and smaller energies.
For scalar fields this is straightforward. For the metric, things are a little trickier. To be consistent, we should write the metric perturbation as
\begin{equation}
g_{\mu\nu} = \eta_{\mu\nu} + \frac{h_{\mu\nu}}{M_{\rm pl}}
\end{equation}
where $h_{\mu\nu}$ has units of mass, but the metric itself $g_{\mu\nu}$ is dimensionless.
When you do this, you will find that the Einstein-Hilbert term (in $4$ spacetime dimensions) is actually made of operators with every integer dimension from $4$ to infinity:
\begin{equation}
\frac{M_{\rm pl}^2}{2} R \sim (\partial h)^2 + \frac{1}{M_{\rm pl}} h (\partial h)^2 + ... + \frac{1}{M_{\rm pl}^{n}} h^n (\partial h)^2 + ...
\end{equation}
There is physics in this statement: at low energies, the higher order interactions are not very important. This implies the existence of a low curvature regime where one can work to linear order in the metric perturbation $h_{\mu\nu}$. At high energies, one cannot ignore the interactions of $h$.
The authors of this paper are using a different set of rules for counting the scaling dimensions of the operators, which provides a language for describing the operators, but obscures this physics.
To spell it out: the authors start from saying the metric as dimensionless, and so treat the curvature $R$ an expression with mass dimension 2 (since the curvature is two derivatives of the metric), and $\mathcal{G}$ is an expression with mass dimension 4 (four derivatives). They treat $\phi$ as having mass dimension 1. So, $\frac{1}{M^2} \phi^2 \mathcal{G}$ is (in their language) a dimension 6 operator (the combined term has mass dimension 4, since it is a Lagrangian density, but $\phi^2 \mathcal{G}$ has mass dimension 6). Note that the dimensionful coupling constant that appears $\sim 1/M^2$ has mass dimension $2=6-4$, so this formally follows the pattern of counting dimensions I mentioned, where a $N$ dimensional operator has a dimensionful coupling constant $\sim 1/M^{N-D}$ in $D$ spacetime dimensions. This is why they call it a dimension six operator, but I would argue this is not really good terminology.
Let me relate this to relate to the way I would argue you should count dimensions. First, let me define $\zeta = \frac{\hat{\zeta}}{M_{\rm pl}^2}$, where $\hat{\zeta}$ is dimensionless. Then one can expand $\zeta \phi^2 \mathcal{G}$ in terms of the metric perturbation as
\begin{equation}
\frac{\hat{\zeta}}{M_{\rm pl}^2} \phi^2 \mathcal{G} \sim \frac{\hat{\zeta}}{M_{\rm pl}^6} \phi^2 (\partial h)^4 + \frac{\hat{\zeta}}{M_{\rm pl}^7} \phi^2 h (\partial h)^4 + ... +  \frac{\hat{\zeta}}{M_{\rm pl}^{6+n}} \phi^2 h^n (\partial h)^4 + ...
\end{equation}
In my opinion, this expansion contains more information, since it tells you how the operators scale with energy. This tells you that, for example, at leading order in an expansion in energy, $\phi^2 \mathcal{G}$ is comparable to the dimension 10 operator $\sim \frac{1}{M_{\rm pl}^6} h^6 (\partial h)^2$ in the Einstein-Hilbert action, assuming $\hat{\zeta}\sim 1$ and assuming all fields and derivatives are of the same order of magnitude. The definition the authors use, is just a fancy way of talking about the mass dimension of a given term in the Lagrangian.
