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:

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

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?


1 Answer 1


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 $M_{\rm Pl}$.

  • $\begingroup$ 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. $\endgroup$
    – MadMax
    Commented May 22 at 17:23
  • $\begingroup$ @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. $\endgroup$
    – Andrew
    Commented May 22 at 18:26
  • $\begingroup$ 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 $\endgroup$
    – MadMax
    Commented May 22 at 18:44
  • $\begingroup$ Not commenting on that specific paper, but I totally agree there is a lot of misunderstanding of effective field theory in the literature, and people computing effects outside the regime of validity of a given theory. $\endgroup$
    – Andrew
    Commented May 22 at 18:56
  • 1
    $\begingroup$ @Caspar201 I would call it a "best guess" based on extrapolating our current best theories of gravity and quantum mechanics. Assuming our current framework works, something special needs to happen at or before the Plank scale. Or, our rules totally break down. But often we make progress in physics by extrapolating the current laws beyond where they've been strictly tested, often we get correct predictions, and sometimes we find the extrapolation leads to a problem which requires a modification to the original laws (which is the most exciting situation!) $\endgroup$
    – Andrew
    Commented May 23 at 16:30

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