Question about both sides of the Einstein field equations

Question

If both sides of the Einstein field equations need to have the same measurement units, how can the coefficient of the energy momentum tensor convert all of the units of the energy momentum tensor (energy density, stress, and momentum density) into the units of the Einstein tensor?

Answer 1

This is a question that appears much simpler than the answer turns out to be. It has relatively little to do with the EFE and more to do with how to handle units in general with tensors. There are at least 4 approaches that I am aware of:

1. Perhaps the most common initial approach is to consider the coordinates to have units of length. This is often the beginning approach and $ct$ is introduced as the time coordinate. So the fact that $ds^2=g_{\mu\nu}dx^\mu dx^\nu$ has units of length squared implies that $g_{\mu\nu}$ is dimensionless.

2. However, things like Schwarzschild coordinates call that approach into question, with its two angular coordinates. So frequently the second concept is to consider each coordinate to have its own unit. In this example $t$ has units of time, $r$ has units of length, and $\theta$ and $\phi$ are dimensionless. This implies that $g_{\mu\nu}$ also has different units for each coordinate such that $g_{tt}$ has units of $L^2 T^{-2}$, $g_{rr}$ is dimensionless, $g_{\phi\phi}$ has units of $L^2$, and so forth.

3. This quickly becomes a nightmare to keep track of. So, the typical next approach is to say that coordinates are unitless. Coordinates are simply numerical labels on events. So $ds^2=g_{\mu\nu}dx^\mu dx^\nu$ implies that the metric has units of length squared. In other words, even if the $z$ coordinates are spaced $1 \mathrm{\ m}$ apart, that information is contained in the metric, not in the coordinates.

4. The final approach that I am aware of is to recognize that $dx^\mu$ is shorthand for the vector $dx^\mu \hat e_{(\mu)}$. In this approach the coordinates are still considered unitless, but the basis vectors have units. This means that even in coordinates like Schwarzschild while $dx^\theta$ is dimensionless $dx^\theta \hat e_{(\theta)}$ has units of length. Now we have come full circle and $ds^2=g_{\mu\nu}dx^\mu dx^\nu$ again implies that $g_{\mu\nu}$ is dimensionless, but the dimensionality is tied to the basis vectors rather than the coordinates. So that dimensionality holds even with dimensionless coordinates.

That last approach may be the best one. Certainly, the other approaches are either exhausting or confusing to me. So with that approach the answer for the question regarding the EFE is clear. $$G_{\mu\nu}+\Lambda g_{\mu\nu} = \kappa T_{\mu\nu}$$ $g_{\mu\nu}$ is unitless and $G_{\mu\nu}$ has units of $L^{-2}$, so $\Lambda$ must also have units of $L^{-2}$. Then on the right hand side $T_{\mu\nu}$ has units of pressure, which is force per area. So to match the left hand side $\kappa$ must have units of inverse force.

The current estimate of $\Lambda$ is $1.47 \ 10^{-52} \mathrm{\ m^{-2}}$, and $\kappa$ is $2.08 \ 10^{-43} \mathrm{\ N^{-1}}$.

Answer 2

Special and general relativity are typically done in units in which the speed of light is set to be $c=1$; this means that distance and time are measured in the same units, or equivalently, that velocity is dimensionless. While energy density, momentum density, and stress have different units classically, those units always differ by factors of the velocity, so in units in which $c=1$, they have the same dimensions.

This may be more familiar with the components of the energy-momentum four-vector, rather than the stress-energy four-tensor, but the idea is the same. It is conventional to write the energy-momentum for a particle as $p^{\mu}=(E,\vec{p}\,)=(E,p_{x},p_{y},p_{z})$ and state that is has mass squared $m^{2}=p^{\mu}p_{\mu}=E^{2}-\vec{p}\,^{2}$. Classically, energy, momentum, and mass all have different units, but $E$, $pc$, and $mc^{2}$ all have the same units (of energy). This is why particle masses are typically quoted in energy units, MeV or GeV; what is strictly meant is that the mass has the given value in MeV$/c^{2}$ or GeV$/c^{2}$.

Since energy and momentum have units that differ by an (irrelevant in relativity) factor of $c$, so do the components of the stress-energy tensor that represent energy and momentum densities. They just differ from the above by dividing by a factor of volume. Stress—force per unit area (pressure being the simplest kind of a stress)—also has units of energy density; both are force times distance divided by volume.

Answer 3

That's a good question, actually, because a careful analysis of this reveals several wide-spread discrepancies and errors propagated on the net and even in the peer-reviewed literature.

First, as the names of the components of the stress, energy and momentum tensor themselves makes clear (e.g. "energy density", "momentum density"), the actual geometric object is not a tensor at all, but a tensor density; specifically, a rank $(1,1)$ tensor density: $^ρ_ν$.

(I have more to say on that matter here: What kind of geometric object is the stress tensor density?.)

In contrast, what is normally called the "stress energy tensor" is converted to an ordinary tensor by dividing out by $\sqrt{|g|}$, where $g$ is the determinant of the underlying metric $g_{μν}$: $T^ρ_ν = ^ρ_ν/\sqrt{|g|}$. The form that people are more used to seeing is with both indices raised lowered, $T_{μν} = g_{μρ} T^ρ_ν$ or $T^{ρσ} = g^{νσ} T^ρ_ν$, where $g^{μν}$ are the components of the inverse of the metric. I'll have more to say on that, at the end.

In the following, I'll use the term "stress tensor [density]" for short.

The continuity equation that goes with the stress tensor density is: $$∂_ρ \left(^ρ_μ X^μ\right) = 0,$$ which is valid if $X = X^μ ∂_μ$ is a Killing vector field. It may be thought of as the continuity equation for the $p_X$ component of the momentum, where the direction of the vector $X$ is variable.

In flat space-time, the vectors $X = ∂_ν$, for each $ν$, are Killing vectors, so one obtains a continuity equation for the momentum component $p_ν$: $∂_μ ^μ_ν = 0$.

More generally, for curvilinear coordinates or in curved space-time we have the covariant continuity equation $$∂_μ ^μ_ν = Γ^ρ_{μν} ^μ_ρ,$$ which may be thought of as a continuum version of the equation for the momentum component $p_ν$ of a test body of mass $m$: $$\frac{dp_ν}{ds} = Γ^ρ_{μν} p_ρ \frac{dx^μ}{ds},\quad p_ν = m g_{μν} \frac{dx^μ}{ds},$$ that exists on a time-like world-line that's parametrized as $x^μ = x^μ(s)$ with $s$ being the proper time for the world-line. The body, itself, has a singular stress-energy-momentum tensor density given by: $$^ρ_ν(x) = \int p_ν(s) \frac{dx^ρ}{ds} δ^4(x - x(s)) ds.$$ The continuity equation for the stress tensor density generalizes this case.

(An example where this is used: Test Bodies: Auto-Parallel Or Geodesic).

The dimensional analysis comes straight out of the definition of the canonical momentum or action - and also out of the definition for the Einstein-Hilbert stress tensor density.

First, consider the canonical stress tensor density derived from a field, whose dynamics are given by an action integral of the form $S = \int (q,v) d^4 x$, with the Lagrangian density $(q,v)$, dependent on field components $q^a(x)$ and their gradients $v^a_μ(x) = ∂_μ q^a(x)$. Associated with this is the canonical stress tensor density, which I will denote using $$: $$^ρ_ν = \frac{∂}{∂v^a_ρ} v^a_ν - δ^ρ_ν .$$

Now, let's do some dimensional analysis, where we'll use $[\_]$ to denote the dimension of a quantity, with $L$, $T$ and $M$, respectively, for length, duration and mass. Denote the following: $$ [μ] = \left[dx^μ\right],\quad Ω = \left[d^4 x\right] = [0][1][2][3],\quad \left[g_{μν} dx^μ dx^ν\right] = A = L^2,\quad [S] = H = \frac{ML^2}{T}. $$ For the field components, let $$[a] = \left[q^a\right].$$

Then, we have the following: $$ \left[g_{μν}\right] = \frac{A}{[μ][ν]},\quad \left[g^{μν}\right] = \frac{[μ][ν]}{A},\quad [\sqrt{|g|}] = \frac{A^2}{Ω},\\ \left[∂_μ\right] = \frac1{[μ]},\quad [v^a_μ] = \frac{[a]}{[μ]},\quad [] = \frac{H}{Ω},\quad \left[^ρ_ν\right] = \frac{H}{Ω}\frac{[ρ]}{[ν]}. $$

If using coordinates indexed as $x^0 = t$, with $\left(x^1,x^2,x^3\right) = (x, y, z)$ being (locally and approximately) Cartesian, then $[0] = T$, while $[1] = [2] = [3] = L$, and we have $[Ω] = T L^3$ and the following dimensions: $$ \left[^0_0\right] = \frac{H}{Ω} = \frac{ML^2}{T}\frac1{T L^3} = \frac{M}{LT^2} = \frac1{L^3} \frac{ML^2}{T^2},\\ \left[^i_0\right] = \frac{H}{Ω} \frac{L}{T} = \frac{M}{T^3} = \frac1{L^3} \frac{ML^2}{T^2} \frac{L}{T},\\ \left[^0_j\right] = \frac{H}{Ω} \frac{T}{L} = \frac{M}{L^2T} = \frac1{L^3} \frac{ML}{T},\\ \left[^i_j\right] = \frac{H}{Ω} = \frac{M}{LT^2} = \frac1{L^3} \frac{ML}{T} \frac{L}{T} = \frac1{L^2} \frac{ML}{T^2}, $$ where $i, j = 1, 2, 3$.

Those are the dimensions, respectively, for energy density, energy flux, momentum density and momentum flux / pressure / stress.

The stress tensor density for a field theory is obtained from the canonical stress tensor density by a "correction", such as the Belinfante correction $^ρ_ν = ^ρ_ν + ∂_μ ^{ρμ}_ν$, where $^{ρμ}_ν = -^{μρ}_ν$. (Symmetrizing The Canonical Energy-Momentum Tensor.) Correspondingly, it has the same dimension: $$\left[^ρ_ν\right] = \left[^ρ_ν\right].$$

If using coordinates that are all set to length dimension, $[μ] = L$ (e.g. with $x^0 = c t$ where $c$ is light speed), then $Ω = L^4$ and the dimensions of all the components are the same: $H/Ω = M/{L^2T}$: momentum density.

For the connection coefficients, $Γ^ρ_{μν}$, assuming they are metrical, i.e. $∂_ρ g_{μν} = Γ^σ_{ρμ} g_{σν} + Γ^σ_{ρν} g_{μσ}$, then we have $$\left[Γ^ρ_{μν}\right] = \frac{[ρ]}{[μ][ν]}.$$ The curvature tensor, Ricci tensor and scalar and Einstein tensor are given, respectively, by $$ R^ρ_{σμν} = ∂_μ Γ^ρ_{νσ} - ∂_ν Γ^ρ_{μσ} + Γ^ρ_{μτ} Γ^τ_{νσ} - Γ^ρ_{ντ} Γ^τ_{μσ},\\ R_{μν} = R^ρ_{μρν},\quad R = g^{μν} R_{μν},\quad G_{μν} = R_{μν} - \frac12 g_{μν} R. $$ Correspondingly, their dimensions are $$ \left[R^ρ_{σμν}\right] = \frac{[ρ]}{[σ][μ][ν]},\quad \left[R_{μν}\right] = \frac1{[μ][ν]},\quad \left[R\right] = \frac1A = \frac1{L^2},\quad \left[G_{μν}\right] = \frac1{[μ][ν]}. $$

For the rank $(1,1)$ stress tensor, after dividing out by $\sqrt{|g|}$, and for its rank $(2,0)$ and rank $(0,2)$ variants, the dimensions are: $$ \left[T^ρ_ν\right] = \frac{H}{Ω}\frac{[ρ]}{[ν]}\frac{Ω}{A^2} = \frac{H}{A^2}\frac{[ρ]}{[ν]} = \frac{M}{L^2T}\frac{[ρ]}{[ν]},\\ \left[T_{μν}\right] = \frac{H}{A}\frac1{[μ][ν]} = \frac{M}{T}\frac1{[μ][ν]},\quad \left[T^{ρσ}\right] = \frac{H[ρ][σ]}{A^3} = \frac{M[ρ][σ]}{L^4T}. $$

A wide-spread error is attributing the dimensions of the stress tensor density $^ρ_ν$ to the stress tensor $T_{μν}$ or even to $T^{ρσ}$. This only works if all the coordinates are set to dimensions of length, in which case, all of these components have the dimension $M/(L^2T)$ of momentum density.

However, it's not natural to set all the coordinates to length units. In particular, besides the time $t$ coordinate, you may also have coordinates that are angular, such as for solutions like Schwarzschild, Reissner-Nordström, Kerr and Kerr-Newman (the fourth being the general case that includes the previous three).

Now, we'll match this to the Einstein Field Equations (with a cosmological coefficient $Λ$): $$G_{μν} + Λ g_{μν} = κ T_{μν}.$$ Matching dimensions, we get $$[Λ] = \frac1A = \frac1{L^2},\quad [κ] = \frac{A}{H} = \frac{T}{M}.$$ The value of $κ$ is a multiple of Newton's coefficient $G$, multiplied by some power of light-speed $c$, which have the dimensions $$[G] = \frac{L^3}{MT^2},\quad [c] = \frac{L}{T}.$$ In $3+1$ dimensions, the multiple is $8π$, but generally it depends on the number of space-time dimensions. In the literature, the $8π$ factor is glibly carried over to higher-dimensional gravity theories - which is wrong.

The literature is also confused in another way. You'll see any of the following in it: $G = 8πG/c^n$, for $n = 2, 3, 4$. Only one is correct. Throughout the net, and even on the wall over at Leiden, you'll see $n = 4$. They all wrong. The correct one is seen here: $$\left[\frac{G}{c^3}\right] = \frac{L^3}{MT^2}\frac{T^3}{L^3} = \frac{T}{M} = [κ].$$ Thus, $$κ = \frac{8πG}{c^3}.$$ The source of the wide-spread error is the confusion over the dimensional analysis of the stress tensor and the wide-spread confusion of stress tensor for stress tensor density. A lot of people have fallen to the Picard-Cardassian-Lights miscount spell: there are three powers of $c$ in the denominator, not four. (For Picard, it was four lights versus five.)

We can also see this, more directly, from the action integral for the Einstein Field Equations: $$S = \int d^n x,\quad = \sqrt{|g|} \frac{R - Λ}{2κ} + _M,$$ where $_M$ is the Lagrangian density (not the Lagrangian $L_M$, but the Lagrangian density $_M = \sqrt{|g|} L_M$) that yields the stress tensor density.

Again, for the dimensional analysis, we need $$\left[\sqrt{|g|} \frac{R - 2Λ}{2κ}\right] = [] = \frac{H}{Ω},$$ which leads to: $$[κ] = \frac{[\sqrt{|g|}][R]}{[]} = \frac{A^2}{Ω}\frac1A\frac{Ω}{H} = \frac{A}{H} = \frac{T}{M},\quad [Λ] = [R] = \frac1A = \frac1{L^2}.$$

The Einstein-Hilbert stress tensor density is derived from the Lagrangian density as the response field to the metric $g_{μν}$ - or (equivalently) - the (negative of) the response field to the inverse metric $g^{μν}$: $$_{μν} = -2\frac{∂_M}{∂g^{μν}}\quad ^{ρσ} = +2\frac{∂_M}{∂g_{ρσ}}.$$ The corresponding dimensions are: $$ \left[_{μν}\right] = \frac{\left[_M\right]}{\left[g^{μν}\right]} = \frac{H}{Ω}\frac{A}{[μ][ν]},\\ \left[^{ρσ}\right] = \frac{\left[_M\right]}{\left[g_{ρσ}\right]} = \frac{H}{Ω}\frac{[ρ][σ]}{A}. $$ From either case follows $$\left[^ρ_ν\right] = \frac{H}{Ω}\frac{[ρ]}{[ν]},$$ as before.

Here, I do a more comprehensive dimensional analysis that includes the electromagnetic and gauge fields, where you can also find the multiple for the $n$ dimensional coupling coefficient $κ_n$ versus $G_n/c^3$ where $G_n$ is the $n$-dimensional version of Newton's coefficient. Also: An earlier version of the analysis above.