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If the metric tensor is unitless, why do its perturbation terms pick up units of Newton's constant?

In the following expansion, metric perturbations pick up a factor of $\kappa\propto\sqrt{G}$ \begin{equation} g_{\mu\nu}=\eta_{\mu\nu}+\kappa h_{\mu\nu}+ \kappa^{2}h_{\mu\lambda}h^{\lambda}_{\nu}+\cdots \end{equation} For example, in this paper on pg.5. Also in this paper by 't Hooft, for the expansion on pg.2 and defined on pg.3.

What is the explicit origin of $\kappa$ and why does it have dimensions while the metric $g_{\mu\nu}$ doesn't?

Qmechanic
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ConfusedStudent1234
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2 Answers2

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You are right that $g_{\mu \nu}$ is dimensionless. But this is not true for $h_{\mu \nu}$: it has dimension $1$ because it was canonically normalized so that the kinetic term in the perturbative expansion of the Einstein-Hilbert action has no coupling (just what we expect in a standard QFT). And since $[\kappa] = -1$, each term in the expansion of $g_{\mu \nu}$ is dimensionless. See my answer here for a brief explanation of how this works: https://physics.stackexchange.com/a/467869/133418.

Canonical normalization is a standard redefinition that one usually makes in a QFT when, for some reason, the Lagrangian does not have the kinetic and interaction terms in their usual forms. The redefinition then puts them in the usual form we know in which the perturbative methods for computing scattering amplitudes have been developed. In principle, the physics is the same whether you perform a redefinition or not, but it's just more convenient and practical to use the existing methods and apply them on similar-looking Lagrangians.

Avantgarde
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Let's for simplicity work in units where $\hbar=1=c$. Recall that the Einstein-Hilbert (EH) action in $d$ spacetime dimensions is $$ S~=~ \frac{1}{16\pi G} \int \! d^dx \sqrt{-g} (R-2\Lambda).\tag{1} $$ Recall that the coupling constant $G$ has dimension $$ [G]~=~L^{d-2},\tag{2}$$ cf. e.g. this Phys.SE post. The perturbative ansatz $$ g_{\mu\nu}~=~\eta_{\mu\nu}+2\sqrt{8\pi G} h_{\mu\nu} +{\cal O}(G)\tag{3} $$ is to ensure that the free part of EH Lagrangian density ${\cal L}$ is canonically normalized in the field theory, i.e. schematically of the form $${\cal L}~=~ \frac{1}{2}(\partial h)^2 + {\cal O}(\sqrt{G}).\tag{4} $$ Explicitly the dimensions of the components of the metric tensors are $$ [g_{\mu\nu}]~=~L^0, \qquad [h_{\mu\nu}]~=~L^{\frac{2-d}{2}}. \tag{5} $$ For more information, see e.g. my related Phys.SE answers here and here.

Qmechanic
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