# Power counting and (superficial) non-renormalizability

Comment: This stuff is new to me so it doesn't entirely make sense (yet).

Question: As I understand from Peskin and Schroeder chap 10 if you have a theory with interaction terms $$\lambda \phi^n$$ in the Lagrangian (density) and the dimension of the coupling constant ($$\lambda$$) is given by $$d-n(2-d)/2$$. Superficially, if $$d\geq0$$ the theory is renormalizable.

Secondly, I am aware that the Einstein Hilbert action $$S\rightarrow C*\int d^4x \sqrt{-g}R$$ is superficially non-renormalizable. (The constant $$C$$ in the above is of course proportional $$G^{-1}$$.) The constant $$C$$ ,itself, has mass dimension 2. Why do these dimensional criteria appear to be reversed for this form of action compared to the $$\phi^n$$ type terms? To rephrase another way: I understand that the dimension for $$G$$ itself is negative, but why is it important to look for the dimensionality of $$G$$ here instead of the dimensionality of what happens to actually be constant in front of the Lagrangian term as was the case in the $$\lambda \phi^n$$ example?

The key thing is that you need to be working with canonically normalized fields in order to use the power counting arguments.

Let's expand GR around flat space $$g_{\mu\nu} = \eta_{\mu\nu} + \tilde{h}_{\mu\nu}$$ The reason for the tilde will become clear in a second. So long as $\tilde{h}$ is "small" (or more precisely so long as the curvature $R\sim (\partial^2 \tilde{h})$ is "small"), we can view GR as an effective field theory of a massless spin two particle living on flat Minkowski space.

Then the Einstein Hilbert action takes the schematic form $$S_{EH}=\frac{M_{pl}^2}{2}\int d^4x \sqrt{-g} R = \frac{M_{pl}^2}{2} \int d^4x \ (\partial \tilde{h})^2 + (\partial \tilde{h})^2\tilde{h}+\cdots$$ where $M_{pl}\sim 1/\sqrt{G}$ in units with $\hbar=c=1$. $M_{pl}$ has units of mass. In this form you might thing that the interaction $(\partial \tilde{h})^2 \tilde{h}$ comes with a scale $M^2_{pl}$ with a positive power. However this is too fast--all the QFT arguments you have seen have assumed that the kinetic term had a coefficient of -1/2, not $M_{pl}^2$. Relatedly, given that $M_{pl}$ has units of mass and the action has units of $(mass)^4$, the field $\tilde{h}$ is dimensionless, so it is clearly not normalized the same way as the standard field used in QFT textbooks.

Now classically, the action is only defined up to an overall constant, so we are free to think of $M_{pl}^2$ as being an arbitrary constant. However, in QFT, the action appears in the path integral $Z=\int D\tilde{h}e^{iS[\tilde{h}]/\hbar}$ (note the notational distinction between $\tilde{h}$ and $\hbar$). Thus the overall constant of the action is not a free parameter in QFT, it is fixed and has physical meaning. Alternatively, you have to remember that the Einstein Hilbert action will ultimately be coupled to matter; when we do that, the scale $M_{pl}$ sitting in front of $S_{EH}$ will not multiply the matter action, and so $M_{pl}$ sets the relative scale between the gravitational action and the matter action.

The punchline is that we can't simply ignore the overall scale $M_{pl}^2$, it has physical meaning (ie, we can't absorb $M_{pl}$ into an overall coefficient multiplying the action). On the other hand, we want to put the action into a "standard" form where the overall scale isn't there, so we can apply the normal intuition about power counting. The solution is to work with a "canononically normalized field" $h$, related to $\tilde{h}$ by

$$\tilde{h}_{\mu\nu} = \frac{h_{\mu\nu}}{M_{pl}}$$

Then the Einstein Hilbert action takes the form

$$S_{EH} = \int d^4 x \ (\partial h)^2 + \frac{1}{M_{pl}} (\partial h)^2 h + \cdots$$

In this form it is clear that the interactions of the form $(\partial h)^2 h$ have a "coupling constant" $1/M_{pl}$ with dimensions 1/mass, which is non-renormalizable by power counting in the usual way.