# Where does Planck's constant come from in non-renormalizability of quantum gravity?

I am trying to understand the idea that gravity breaks down at the Planck scale, but I am confused by the use of natural units ($$c = \hbar = 1$$). The Einstein-Hilbert action in natural units is:

$$$$S_{EH} = \frac{1}{16 \pi G}\int d^4 x \sqrt{-g} R.$$$$

I have seen different arguments, but the simplest one just notes that a perturbative expansion of $$S_{EH}$$ leads to $$E^2/G$$ terms so the expansion blows up when $$E > \sqrt{G}$$. See for example p. 172 of Zee.

In natural units, the Planck length $$l_P$$ is equal to $$\sqrt{G}$$. Therefore, it seems that beyond the Planck scale perturbation theory breaks down.

Now I would like to put back all of the constants. From the action itself, I can see that $$c$$ will be there somewhere. But where does $$\hbar$$ come from? Please don't tell me it's just dimensional analysis, because all dimensional analysis tells me is that we need some constant with the right dimensions not necessarily equal to $$\hbar$$. Is there a specific physical reason it is $$\hbar$$?

EDIT: Of course we expect $$\hbar$$ because we are interested in quantum gravity. But I want to know where specifically in this sort of calculation $$\hbar$$ appears - which equations are appealed to?

• Commented May 22 at 16:43

One way to think about it is in terms of the path integral. For perturbative quantum gravity around flat space, we expand the metric as $$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$$ where $$h_{\mu\nu}$$ is a perturbation. (Actually at this point you could think of this as a field redefinition, but to calculate anything you need $$h$$ to be small).
Then the path integral (ignoring gauge fixing terms, Fadeev Popov ghosts, etc) is $$Z = \int D h_{\mu\nu} \ e^{i\frac{S_{EH}}{\hbar}}$$ where $$S_{EH} = \frac{c^4}{16\pi G} \int d^4 x \sqrt{-g} R$$ The reason $$\hbar$$ shows up in this place in the path integral is a postulate, it's equivalent to having $$\hbar$$ appear in the canonical commutation relations. We do it because it works in other theories.
Now if we expand the exponent of the path integral's integrand in powers of $$h$$ we get, schematically, $$\frac{S_{EH}}{\hbar} = \frac{c^4}{16 \pi G \hbar} \left(\partial^2 h^2 + \partial^2 h^3 + \cdots + \partial^2 h^n + \cdots \right)$$ We then canonically normalize the field $$h$$ -- in other words, we choose a standard normalization for $$h$$ so that we can apply standard power counting theorems in quantum field theory. We do this by defining $$\tilde{h} = \frac{h}{M_{\rm Pl} c^2}$$ where the reduced Planck mass is given by $$M_{\rm Pl} \equiv \sqrt{\frac{\hbar c}{8\pi G}}$$ Then $$S_{EH}/\hbar$$ has a canonical kinetic term, with a series of non-renormalizable interactions (we know they are non-renormalizable by power-counting, which we can do easily because of the canonical normalization) $$S_{EH}/\hbar \sim -\frac{1}{2}(\partial h)^2 + \frac{\partial^2 h^3}{M_{\rm Pl}} + \cdots + \frac{\partial^2 h^n}{M_{\rm Pl}^{n-2}} + \cdots$$ Then, by standard effective field theory arguments, we expect to be able to treat this as an effective field theory that breaks down at energy scales of order $$M_{\rm Pl}$$.
• Very informative answer. The central point relevant to the original question is that we can't talk about $\hbar$ or $G$ in isolation. Quantum gravity effect manifests itself at the scale of $M_{\rm Pl} \equiv \sqrt{\frac{\hbar c}{8\pi G}}$. Only concentrating on $\hbar$ or $G$ individually is pointless when we talk about quantum gravity. It's is ratio of $\frac{G}{\hbar}$ that really counts. Commented May 22 at 17:23
• @MadMax Right, I agree. You can always choose units where $\hbar=1$. The physics is that there is an energy scale at which quantum corrections to the classical answer become of order 1. It's also important that $\hbar$ is involved because classically, pure GR with no matter is scale invariant. The scale $M_{\rm Pl}$ only arises when you account for quantum corrections. Commented May 22 at 18:26
• As you emphasized, from an effective field theory point of view, GR breaks down at energy scales of order $M_{\rm Pl}$. Therefore, any meaningful calculation involving Planck scale effects should take into account infinite numbers of high-energy terms NOT present in GR. The numerous HEP papers of Planck scale effects of Black Hole or Big Bang using GR ONLY are just, you know, Bull*t. For example, the very existence of Black Hole event horizon is highly questionable, see a new paper published today here: arxiv.org/abs/2405.12685 Commented May 22 at 18:44