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The definition of Planck units is such that $$G \hbar = c^3 l_P^2$$ so it makes sense that quantum gravity practitioners, when not using natural "equal one" units, can choose between $\hbar$ and $l_P^2·$ It is specially noticeable in the path integral, because the Einstein-Hilbert action is $$S=\frac{1}{16\pi} \frac{c^3}{G} \int R \sqrt{-g} \, d^4x$$ and so the weight $\exp(iS/\hbar)$ is $$ e^{\frac{i}{\ell_p^2} \int R \sqrt{-g} \, d^4x } $$

So I am wondering, is there some school or family of authors that explicitly prefers to use $l_p^2$? It sounds me as a possible schism as the one in the metric sign, but in this case it seems most authors avoid to choose, invoking natural units.

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    $\begingroup$ Would it matter if they did? $\endgroup$
    – Andrew
    Commented Oct 26 at 19:50
  • $\begingroup$ Why do you call this a choice between $\hbar$ and $l_P^2$? It’s a choice between writing the prefactor as $c^3/G\hbar$ or $1/l_P^2$. Given that the integral is a length-squared, the latter seems much clearer. $\endgroup$
    – Ghoster
    Commented Oct 26 at 20:28
  • $\begingroup$ @Andrew I guess not, pretty as it happens with the +--- vs -+++ debate. So I framed the question in this context, schools. And of course, QFT practitioners would hate exp(i$\frac{G}{c^3l_P^2}S_{YM}$) $\endgroup$
    – arivero
    Commented Oct 26 at 21:43
  • $\begingroup$ It's irrelevant. Use whatever you wish to use. Why do you care if other people use it or not? $\endgroup$
    – Prahar
    Commented Oct 26 at 21:55
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    $\begingroup$ Some equations need to change signs if you use one metric signature convention vs another, so it's not trivial to switch between different conventions. In pure gravity, one could just replace $\hbar$ with $\ell_{\rm Pl}^2$, so that's an easier conversion. Most people tend to use $\hbar\rightarrow 0$ rather than $\ell_{\rm Pl}\rightarrow 0$ for the classical limit in the technical literature. In popular books, speaking of the Planck length being small is probably more common than saying $\hbar$ is small. However, it's also common to use Planck units in quantum gravity, where $\hbar=c=G=1$. $\endgroup$
    – Andrew
    Commented Oct 26 at 23:07

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As far as I know, no textbook author uses $l_P$ as the normalisation constant for the path integral of quantum gravity.

This is because while $\hbar$ is a constant, $l_P$ is a unit. They are in different epistemic categories.

More specifically, we use mass, length and time as fundamental magnitudes because of their apparent linearity: the mass of two objects is the sum of their masses, the length of two objects is the length of their concatenation, the duration of two intervals is the duration of their sequence. We choose units to measure such magnitudes: MKS, cgs, or even Imperial.

One specific selection of units is $(l_P,m_P,t_P)$. In these units, the values of $\hbar$, of $G$ and of $c$ are equal to one. Technically we should write "$c=$ 1 $l_p/t_P$" as we write "$c=$ 299792458 $m/s$". We omit the mention to the units and we say "working in units where $\hbar=G=c=1$" as an abuse of language. We also say "units where $\hbar=c$ to refer to an undetermined system of units where $G$ must not need to be one.

With this view, it stills make sense to say $\hbar \to 0$, but it does not make sense $l_P \to 0$. Of course there is a quantity that could be a constant of Nature, call it lowercase $l_p$, whose value is $l_p = 1 l_P$ measured in the natural system of units, and we could consider $l_p\to 0$.

Instead of going into such subtleties, people prefers to keep $\hbar$ in all the weights, at least until $l_p$ were not proved to be as fundamental as $c$ or $\hbar$. The situation could be different if it was definitely proven that $l_p$ is the smallest possible length, or $l_p^2$ the smallest possible area, or $cl_p^2$ the smallest possible areal speed. And note that it the last two cases we would surely choose a new name for the new constant of Nature.

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