# What are the units of the quantities in the Einstein field equation?

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 $T_{\mu\nu}$?

• 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
• 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. – Ben Crowell Feb 25 '18 at 16:19

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 $\mathrm N/\mathrm m^2$.

• 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). – Ben Crowell Feb 25 '18 at 16:11
• $[T_{\mu\nu}]$ is $J/m^3$

• $[g_{\mu\nu}]$ is $1$

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

• 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 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 $\rm\frac{1}{L^2}$.

Let's start from more general facts about dimensional analysis on differential manifolds.

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".

## Coordinates

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.

## Tensors

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 $$\dim(\varGamma{}^i{}_{jk}) = \mathrm{X}_i\,{\mathrm{X}_j}^{-1}\,{\mathrm{X}_k}^{-1}.$$

## Curves

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, $$c_*(\partial_s)$$.

This has an interesting, quirky implication. Given a vector field $$P\mapsto\pmb{V}(P)$$ we say that $$c$$ is an integral curve for it if $$\pmb{V}[c(s)] = \dot{c}(s).$$ But this equation is only valid if $$\pmb{V}$$ has dimensions $$\mathrm{S}^{-1}$$. For the general case a constant dimensional factor needs to be introduced in the equation above.

## 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 $$s$$.

If the path is timelike, this "length" can be measured by a clock having that path as worldline – it's its proper time. Thus, for me $$\mathrm{Z}^{1/2} = \mathrm{T}$$, a time, and therefore the absolute dimensions of the metric tensor are time squared: $$\dim(\pmb{g})=\mathrm{T}^2.$$

I believe that these dimensions also make sense for spacelike paths: in this case we would have to measure the "length" by dividing it in very small pieces and using radar coordinates on each piece. So we're measuring the "length" by checking clocks, to see how long it takes for the light to bounce back: time $$\mathrm{T}$$, again.

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 $$\mathrm{T}^{-2}$$. 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 $$\mathrm{T}^2$$, and "rising an index" multiplies them by $$\mathrm{T}^{-2}$$ (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. I'll try to sketch an informal argument that reflects my point of view. The argument can be made more rigorous but that would take too long to do here.

The dynamics equation $$\nabla\cdot\pmb{T}=0$$ holds in general-relativistic (thermo)mechanics, and also in Newtonian (thermo)mechanics when no body forces and no body heating are present. In Newtonian mechanics it's the formal combination of the balances of momentum density and energy density – which incidentally have the same dimensions $$\mathrm{M}\,\mathrm{L}^{-1}\,\mathrm{T}^{-3}$$, energy/(volume × time).

The divergence of the stress-energy-momentum gives us a 4-force density, just like the 3-divergence of the stress gives us a force density. Please check Misner & al (1973), chap. 14, for a very interesting discussion of these matters, and also Eckart (1940) and Burke (1980, 1987).

Further, the 4-force is an object that, integrated over a path, gives us an energy density (cf Milne 1951 chap. IV, and Burke again). The integral of a force in Newtonian mechanics is the work done by the force. In general-relativistic mechanics, the timelike component of the 4-force additionally gives us the increase in energy owing to heating (Eckart 1940).

So $$\nabla\cdot\pmb{T}\equiv T{}_i{}^j{}_{;\,j}\;\mathrm{d}x^i$$ has the dimensions of energy density, $$\mathrm{M}\,\mathrm{L}^{-1}\mathrm{T}^{-2}$$. The co-contra-variant stress-energy-momentum $$\pmb{T}{}_{\bullet}{}^{\bullet}$$ has therefore the same dimensions. But the co-co-variant tensor, obtained by contraction with the metric, $$\pmb{T}_{\bullet\bullet} \equiv \pmb{T}\cdot \pmb{g}$$, has dimensions of energy density times squared time: $$\mathrm{M}\,\mathrm{L}^{-1}$$, a mass over length.

Einstein's constant $$\kappa$$ therefore relates a dimensionless quantity and a mass over length: $$\pmb{G}_{\bullet\bullet} = \kappa \pmb{T}_{\bullet\bullet}\;.$$ Its dimension must be $$\mathrm{M}^{-1}\,\mathrm{L}$$, and it's easily seen that these are the dimensions of $$G/c^2$$. So I'm one of those people (like Fock 1964 p. 199) who define $$\kappa = 8\pi G/c^2.$$

## References

• Burke (1980): Spacetime, Geometry, Cosmology (University Science Books)
• Burke (1987): Applied Differential Geometry (Cambridge)
• Eckart (1940): The thermodynamics of irreversible processes. III. Relativistic theory of the simple fluid, Phys. Rev. 58, 919.
• Fock (1964): The Theory of Space, Time and Gravitation (Pergamon)
• Misner, Thorne, Wheeler (1973): Gravitation (Freeman)
• Schouten (1989): Tensor Analysis for Physicists (Dover, 2nd ed.)