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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?

Qmechanic
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Caspar201
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1 Answers1

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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 $M_{\rm Pl}$.

Andrew
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