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). $\endgroup$ – Thomas Fritsch Aug 27 '19 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? $\endgroup$ – Mauro Giliberti Aug 27 '19 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). $\endgroup$ – probably_someone Aug 27 '19 at 17:11
  • $\begingroup$ @probably_someone yes, that's correct, I meant the standards Schwartzschild coordinates en.m.wikipedia.org/wiki/Schwarzschild_coordinates $\endgroup$ – Mauro Giliberti Aug 28 '19 at 16:44

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.


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 '19 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 '19 at 16:01
  • $\begingroup$ Shouldn't the stress-energy tensor T be an energy density, as @G._Smith said? $\endgroup$ – Mauro Giliberti Aug 29 '19 at 8:01
  • $\begingroup$ @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 '19 at 12:35
  • $\begingroup$ Ok, that's what I thought, I was looking for S. I. units but now I got it, thanks $\endgroup$ – Mauro Giliberti Aug 29 '19 at 12:39

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