My question is pretty straight-forward: what are the units of the tensors in General Relativity? This should sound easy, but I always studied those in natural units ($c=1$) so I can't figure it out. In particular, what are the units of

  • $G_{\mu\nu}$
  • $g_{\mu\nu}$
  • $R^\rho_{\mu\sigma\nu}$
  • $R_{\mu\nu}$
  • $R$
  • $T_{\mu\nu}$
  • $\Gamma^\lambda_{\mu\nu}$


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    $\begingroup$ It depends on the units of your coordinates ($x^0, x^1, x^2, x^3$) which can be anything (length, angle, time, etc). $\endgroup$ Aug 27, 2019 at 15:04
  • $\begingroup$ @ThomasFritsch you are right: this makes my question even more mysterious for me: if I have an action (whose units I know well) expressed as a function of say $g$ and $R$, can I say something about their units? Also: let's say my coordinates are spherical as in Schwarzschild's metric $(ct, r, \theta, \varphi)$, can I then say something about the tensors I specified in the question? $\endgroup$ Aug 27, 2019 at 15:15
  • $\begingroup$ @MauroGiliberti Your spherical coordinates don't work, as their units are mixed (two are distance, two are ratios of distances and are unitless). $\endgroup$ Aug 27, 2019 at 17:11
  • $\begingroup$ @probably_someone yes, that's correct, I meant the standards Schwartzschild coordinates en.m.wikipedia.org/wiki/Schwarzschild_coordinates $\endgroup$ Aug 28, 2019 at 16:44

2 Answers 2


Here's the general rule I use in my own calculations.

In cartesian like coordinates $x^{\mu} \sim \mathrm{L}^1$ : \begin{align} g_{\mu \nu} &\sim \mathrm{L}^0, \\[12pt] \Gamma_{\mu \nu}^{\lambda} &\sim \mathrm{L}^{-1}, \\[12pt] R^{\lambda}_{\; \kappa \mu \nu} &\sim \mathrm{L}^{-2}, \\[12pt] R_{\mu \nu} &\sim \mathrm{L}^{-2}, \\[12pt] G_{\mu \nu} &\sim \mathrm{L}^{-2}, \\[12pt] R &\sim \mathrm{L}^{-2}, \\[12pt] T_{\mu \nu} &\sim \mathrm{L}^{-4}, \\[12pt] \kappa \equiv 8 \pi G &\sim \mathrm{L}^{2}, \\[12pt] \end{align}

Take note that the dimensions of these quantities heavily depend on the dimensions of your coordinates, which are totally arbitrary. However, whatever the coordinates $x^{\mu}$, you have \begin{equation} ds^2 = g_{\mu \nu} \, dx^{\mu} \, dx^{\nu} \sim \mathrm{L}^2. \end{equation} Also, tensor invariants do not depend on the coordinates and have the same dimensions in all coordinates systems. For example: \begin{align} R \equiv g^{\mu \nu} \, R_{\mu \nu} &\sim \mathrm{L}^{-2}, \\[12pt] R^{\mu \nu \lambda \kappa} \, R_{\mu \nu \lambda \kappa} &\sim \mathrm{L}^{-4}. \end{align}

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    $\begingroup$ There is a caveat. Often you have coordinates that have various dimensionalities, and then you generally apply $g_{\mu\nu} x^\mu x^\nu \sim L^2$. The most common example are polar units where you usually have $g_{\phi \phi} \sim L^2, g^{\phi\phi} \sim L^{-2}$ and similar things happen to all coordinate components of tensors. $\endgroup$
    – Void
    Aug 27, 2019 at 15:46
  • $\begingroup$ @Void, yes I agree, but in all cases, you still have $ds^2 \sim \mathrm{L}^2$. $\endgroup$
    – Cham
    Aug 27, 2019 at 16:01
  • $\begingroup$ Shouldn't the stress-energy tensor T be an energy density, as @G._Smith said? $\endgroup$ Aug 29, 2019 at 8:01
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    $\begingroup$ @MauroGiliberti, in geometrical units, everything is expressed as a length $\mathrm{L}$ with some exponent. Energy density is then $\mathrm{L}^{-4}$. For the Einstein equation, this is important since $G_{\mu \nu} \sim \mathrm{L}^{-2}$ (in cartesian likes coordinates) and $8 \pi G \sim \mathrm{L}^2$. $\endgroup$
    – Cham
    Aug 29, 2019 at 12:35
  • $\begingroup$ Ok, that's what I thought, I was looking for S. I. units but now I got it, thanks $\endgroup$ Aug 29, 2019 at 12:39

If you choose coordinates with the units of length, such as $(ct, x, y, z)$, then the metric tensor and its inverse are dimensionless, the Christoffel symbols have the dimensions of inverse length, and the curvature tensors are inverse length squared. In these coordinates the energy-momentum tensor has the dimensions of energy density.


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