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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    $\begingroup$ 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. $\endgroup$ Aug 26, 2012 at 21:08
  • 4
    $\begingroup$ I have a long, detailed, careful treatment of this topic in section 5.11 of my GR book, lightandmatter.com/genrel . It gets rather complicated. A given tensor can have different units in different coordinate systems, different components of the same tensor can have different units, and there are multiple conventions to be found in the literature that result in different units being assigned to different quantities. As an example of how conventions can vary, see Dicke, Phys Rev 125 (1962) 2163. He lets the metric have units of distance. $\endgroup$
    – user4552
    Feb 25, 2018 at 16:19

4 Answers 4


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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    $\begingroup$ These statements about the units of tensors could be made more precise by noting that they apply to the components of the tensors, and they hold only in a coordinate system in which the coordinates have units of distance (which they often do not, e.g., Schwarzschild coordinates). $\endgroup$
    – user4552
    Feb 25, 2018 at 16:11
  • $[T_{\mu\nu}]$ is $J/m^3$

  • $[g_{\mu\nu}]$ is $1$

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

  • $\begingroup$ 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? $\endgroup$
    – user17589
    Jan 7, 2013 at 23:49
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    $\begingroup$ 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. $\endgroup$
    – Michael
    Jan 8, 2013 at 1:25
  • $\begingroup$ Regarding this point: Why are the the elements of the stress-energy tensor not all symmetrical, and the units are not all the same? $\endgroup$
    – Quillo
    May 10, 2023 at 8:00

The answers above are correct, but the notation is bad. When they write "m" they mean "meters" and the correct unit for dimensional analysis is "length = L.

$[R_{\mu\nu}]$, $[Λ]$, and $[R]$ have units of $\rm\frac{1}{L^2}$

$[T_{\mu\nu}]$ has units of energy/volume = pressure = force/area = $\rm\frac{mass}{[L*t^2]}$

Einstein's constant $k$ converts these. It is $k = \frac{8\pi G}{c^4}$ and has units of $\rm\frac{t^2}{L*mass}$, so it converts the stress-energy $[T_{\mu\nu}]$ to the units of the other side of the field equation, which has all the metric-related parts, each term of which is dimensionally $\rm\frac{1}{L^2}$.


I recommend this review for a detailed analysis of dimensional analysis in relativity, its connection with the operational meanings of the tensors, and a review of the literature:

Here is a summary from it.

For dimensional analysis I use ISO 80000 conventions and notation. I sometimes use notation such as $\pmb{T}{}_{\bullet}{}^{\bullet}$ to indicate that the tensor $\pmb{T}$ is covariant in its first slot and contravariant in its second; I call this a "co-contra-variant tensor".


First of all, a coordinate is just a function that associates a physical quantity with every event on the (spacetime) manifold, or on a region thereof. Together with the other coordinates, such function allows us to uniquely identify the event within that region. A coordinate can thus potentially have any dimensional units. It could be the distance from something, and so have dimensions $\mathrm{L}$; or the time elapsed since something, and so $\mathrm{T}$; or an angle, $1$; or even a temperature, dimensions $\Theta$.

The dimensions of the coordinates don't matter, as we'll now see.


Consider a system of coordinates $(x^i)$ with dimensions $(\mathrm{X}_i)$.

Starting with an example, take a contra-contra-co-variant tensor $\pmb{A}$, with components $(A{}^{ij}{}_k)$ in some coordinate system. Then the component $A{}^{ij}{}_k$ must have the following dimensions: $$ \dim(A{}^{ij}{}_k) = \mathrm{D}\, \mathrm{X}_i \, \mathrm{X}_j \, {\mathrm{X}_k}^{-1},\tag{1}\label{1} $$ where $\mathrm{D}$ is the same in all components. The reason for this is simple. Written in invariant form, the tensor is $$\begin{aligned} \pmb{A} &= A{}^{ij}{}_k\;\partial_{x^i}\otimes\partial_{x^j}\otimes\mathrm{d}x^k \\ &\equiv A{}^{00}{}_0\;\partial_{x^0}\otimes\partial_{x^0}\otimes\mathrm{d}x^0 +A{}^{00}{}_1\;\partial_{x^0}\otimes\partial_{x^0}\otimes\mathrm{d}x^1 +\dotsb \end{aligned}$$ and all terms must have the same dimensions. This is only possible if the components have dimensions as in $\eqref{1}$. This also means that $\dim\pmb{A} = \mathrm{D}$ independently of any coordinates. For the present discussion we may call these the "absolute" dimensions of the tensor. I believe that this is the point of view and terminology of Schouten (1989), chap. VI.

What we have just seen is obviously consistent under coordinate changes. For example, transforming components to a primed system, $$ A'{}^{ij}{}_{k} = A{}^{lm}{}_{n}\; \frac{\partial x'{}^i}{\partial {x}^{l}}\, \frac{\partial x'{}^j}{\partial {x}^{m}}\, \frac{\partial x^n}{\partial x'{}^{k}} $$ and the transformation coefficients take care of the dimensional changes.

This example generalizes to tensors of any type in an obvious way.

Tensor operations

Applying the kind of reasoning just discussed we can find the dimensional effect of the main operations on tensors:

  • tensor multiplication $\otimes$ multiplies the dimensions: $\dim(\pmb{A}\otimes\pmb{B}) = \dim(\pmb{A})\dim(\pmb{B})$;
  • same for the exterior product $\land$;
  • same for contraction (but without raising or lowering indices! see below);
  • pull-back and push-forward don't change the dimensions of the tensor they map;
  • the Lie derivative with respect to a vector field $\pmb{v}$ multiplies by the absolute dimensions of this vector: $\dim(\mathrm{L}_{\pmb{v}}\pmb{A}) =\dim(\pmb{v})\dim(\pmb{A})$;
  • same for the interior product $\mathrm{i}_{\pmb{v}}$;
  • the exterior derivative $\mathrm{d}$ doesn't alter the dimensions of the form on which it operates: $\dim(\mathrm{d}\pmb{\omega}) = \dim(\pmb{\omega})$ (we could use the Cartan identity to check this);
  • same for the integration of a form over a submanifold;
  • the covariant derivative operator $\nabla$ doesn't alter the dimensions either: $\dim(\nabla\pmb{A}) = \dim(\pmb{A})$. But note that $\dim(\nabla_{\pmb{v}}\pmb{A}) = \dim(\pmb{v})\dim(\pmb{A})$.

The dimensional effect of the covariant derivative operator can be quickly checked by noting that the expression of $\nabla\pmb{A}$ contains the following term: $$ \nabla\pmb{A} = \dotsb + \partial_{x^l}A{}^{ij}{}_{k}\; \partial_{x^i}\otimes\partial_{x^j}\otimes\mathrm{d}x^k\otimes\mathrm{d}x^l +\dotsb. $$ From the same expression we also find that

  • the Christoffel symbol $\varGamma{}^i{}_{jk}$ has dimensions $$\dim(\varGamma{}^i{}_{jk}) = \mathrm{X}_i\,{\mathrm{X}_j}^{-1}\,{\mathrm{X}_k}^{-1}.$$


Consider a curve to the manifold, $c\colon s \mapsto P$, where the parameter $s$ has dimension $\mathrm{S}$. If we consider the manifold as "dimensionless" (if this makes sense), then the dimensions of the tangent vector $\dot{c}$ to the curve are $\dim(\dot{c}) = \mathrm{S}^{-1}$. This follows either from $\dot{c} := \partial x^i[c(s)]/\partial s\; \partial_{x^i}$, or considering that $\dot{c}$ can be interpreted as the push-forward of $\partial_s$, that is, $c_*(\partial_s)$.

Metric tensor

From the above discussion we see that the component $g_{ij}$ of the metric $\pmb{g}$ has dimensions $\dim(g_{ij}) = \mathrm{Z}\,\mathrm{X}_i\,{\mathrm{X}_j}^{-1}\,{\mathrm{X}_k}^{-1}$, where $\mathrm{Z}$ are the absolute dimensions of the metric. What are these absolute dimensions?

The answer probably depends on how you see the operational meaning of the metric. Here I offer my personal point of view. We can use the metric to measure the "length" of (timelike or spacelike) paths in spacetime. The "length" of a path $c(s)$ with $s\in [a,b]$ is $$ \int_a^b\!\!\!\mathrm{d}s\; \sqrt{\Bigl\lvert g_{ij}[c(s)]\;\dot{c}^i(s)\,\dot{c}^j(s) \Bigr\rvert}. $$ We see that this "length" has dimensions $\mathrm{Z}^{1/2}$ (and not unexpectedly it doesn't depend on the dimensions of the curve parameter $s$). Therefore $$\dim(\pmb{g})=\mathrm{L}^2\ .$$

However, note that a few important relativity authors (see references in the review cited above) focus on timelike paths, for which the "length" is measured by a clock having that path as worldline – it's its proper time. Thus some authors instead define $$\dim(\pmb{g})=\mathrm{T}^2\ .$$

By our usual argument it's possible to see that the Riemann curvature tensor $\pmb{R}{}^{\bullet}{}_{\bullet\bullet\bullet}$, the Ricci tensor $\pmb{R}_{\bullet\bullet}$, and the Einstein tensor $\pmb{G}_{\bullet\bullet}$ are dimensionless – $1$ – and the scalar curvature has dimensions $\mathrm{L}^{-2}$. Note that the Riemann and Ricci tensors (with the contra/co-variant type specified above) do not require a metric for their definition, but an affine connection. They are dimensionless no matter what dimensions we give the metric. By construction the (fully co-variant) Einstein tensor is always dimensionless, too.

An important operation done with the metric:

  • "lowering an index" of a tensor multiplies its dimensions by $\mathrm{L}^2$, and "rising an index" multiplies them by $\mathrm{L}^{-2}$ (if you agree with my discussion above).

Stress-energy-momentum tensor

What are the absolute dimensions of the co-contra-variant stress-energy-momentum tensor $\pmb{T}{}_{\bullet}{}^{\bullet}$? We must look for an operational meaning here too. There's still ongoing research on this matter (see the review above). The main points are summarized in this answer. The literature offers three main conventions:

  1. $\operatorname{dim}(\pmb{T}) := \mathrm{E}\mathrm{L}^{-1} = \mathrm{M} \mathrm{L} \mathrm{T}^{-2}$

  2. $\operatorname{dim}(\pmb{T}) := \mathrm{M} \mathrm{L}^{-1}$

  3. $\operatorname{dim}(\pmb{T}) := \mathrm{M}\mathrm{L}^{-3} \mathrm{T}^{2}$

The first is by far the most common, the second is used by a few but important authors, the third by McVittie.

Einstein constant

Einstein's constant $\kappa$ therefore relates a dimensionless quantity and the dimension of the stress-energy-momentum tensor: $$\operatorname{dim}(\pmb{G}_{\bullet\bullet}) = \operatorname{dim}(\kappa) \times \operatorname{dim}(\pmb{T}_{\bullet\bullet})\ .$$

If we use convention 1. above, then it's easily seen that $\operatorname{dim}(\kappa) = \mathrm{E}^{-1}\,\mathrm{L}$, and these are the dimensions of $8\pi G/c^4$. This is most widely used convention.

If we use convention 2. above, then $\operatorname{dim}(\kappa) = \mathrm{M}^{-1}\,\mathrm{L}$, and these are the dimensions of $8 \pi G/c^2$. This value for Einstein's constant is indeed used by Fock (1964 p. 199) and a few other authors (eg Synge, Adler-Bazin-Schiffer, McVittie).

Additional references