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The Einstein field equations (EFE) may be written in the form:

$$R_{\mu\nu}-\frac {1}{2}g_{\mu\nu}R+g_{\mu\nu}\Lambda=\frac {8\pi G}{c^4}T_{\mu\nu}$$ where the units of the gravitational constant $G$ are $\mathrm{\frac{N\,m^2}{kg^2}}$ and the units of the speed of light are $\mathrm{\frac{m}{s}}$.

What are the units of the Ricci curvature tensor $R_{\mu\nu}$, the scalar curvature $R$, the metric tensor $g_{\mu\nu}$, the cosmological constant $\Lambda$ and the stress-energy tensor $T_{\mu\nu}$?

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Of course, almost everyone who does this picks the unit of time so that $c = 1$, and the unit of mass so that $G=1$, so that everything is measured in inverse units of length squared. – Jerry Schirmer Aug 26 '12 at 21:08

2 Answers 2

  • $[T_{\mu\nu}]$ is $J/m^3$

  • $[g_{\mu\nu}]$ is $1$

  • $[R_{\mu\nu}]$, $[\Lambda]$, and $[R]$ is $1/m^2$

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But in spherical coordinates the angular components of the metric tensor is proportional to $r^2$. So the dimension of this matrix component is $L^2$! Isn't it? – user17589 Jan 7 '13 at 23:49
It's sheer laziness. To be absolutely explicit and consistent people should right $\left( r/l \right)^2$, where $l$ is some length unit, and define the angular variable $ \theta l $. The $l$ cancels out of the line element $\mathrm{d}s^2$ which seems to be why nobody bothers to write it in the metric. But you can't have one component of a tensor have different units than another component because under Lorentz transformations they get mixed together. – Michael Brown Jan 8 '13 at 1:25

The metric tensor is unitless. That can be seen from the fact that $g_{\mu\nu}v^\mu v^\nu$ gives the square of the four-vector length of $v$, and thus has the unit of $v^2$.

The scalar curvature is a contraction of the Ricci tensor. A contraction doesn't change the units. Also the Ricci tensor is a contraction of the Riemann tensor.

The Riemann tensor is made of coordinate derivatives of the connection coefficients, which are made of coordinate derivatives of the metric. Since each coordinate derivative adds a unit $m^{-1}$, the Ricci tensor and curvature scalar have both unit $\mathrm{m}^{-2}$.

The cosmological constant then of course also has to have the unit $\mathrm m^{-2}$, so that the units match.

$T$, the stress-energy tensor, has the unit of energy density, or pressure (both are actually the same unit, if you look closer), that is, $\mathrm J/\mathrm m^3$ or $\mathrm N/\mathrm m^2$.

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protected by Qmechanic Jan 8 '13 at 12:00

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