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

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 $$\kappa\propto\sqrt{G}$$ $$$$g_{\mu\nu}=\eta_{\mu\nu}+\kappa h_{\mu\nu}+ \kappa^{2}h_{\mu\lambda}h^{\lambda}_{\nu}+\cdots$$$$ 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?

• The expansion with $\kappa$ made explicit is arbitrary. You could absorb the $\kappa$ constant into the definition of $h_{\mu \nu}$ so it becomes unitless (most authors do this).
– Cham
Jul 3, 2023 at 20:32
• I am really confused. Why does everyone seem to think that the metric tensor is dimensionless? I always thought that it is/has the dimension of an area. Jul 3, 2023 at 20:55
• Ah, I am talking about $g$, whereas you are talking about $g(e_i,e_j)$. I guess that both $g$ and $e$ are not dimensionless such that $g_{ij}$ is dimensionless. Right? Jul 3, 2023 at 20:59
• @Filippo There is more than one convention for the units of the metric tensor. Take a look at the line element $ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}$, the line element has units of $[L]=2$. We may choose to have either the metric components to be unitless or the coordinates $dx^{\mu}$. Jul 3, 2023 at 21:01
• @ConfusedStudent1234 Since$$g=g_{\mu\nu}dx^{\mu}\otimes dx^{\nu}$$you just confirmed my point, namely that $g$ transforms like an area. Jul 3, 2023 at 21:13

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.
• By the kinetic term not having a coupling, do you mean that it shouldn't have a coupling constant at all? Or do you mean that the product of the kinetic term and coupling should be unitless? In your other answer, the first order term after redefinition is $S_{EH} = \int d^4 x \ \left(h \partial^2 h \right)$ making the first term unitless in the Lagrangian density. Where $h$ has units $[L]=1$ and the $\partial$ having $[L]=-1$. Jul 3, 2023 at 10:30
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 $$g_{\mu\nu}~=~\eta_{\mu\nu}+\sqrt{8\pi G} h_{\mu\nu} +{\cal O}(G)\tag{3}$$ 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.