In the geometry of GR, the metric tensor $g$ can determine the Riemannian connection and curvature tensor by combining the spatial derivatives (w.r.t. the 4d coordinate system) properly.

I am curious about the dimensional analysis of the metric tensor.

According to the geometric picture of GR, the connection as the potential is related with energy and the curvature tensor is related with force strength (with the mass to connect the connection/curvature with energy/force).

Then what's the dimension of the metric tensor? Intuitively it should be dimensionless, but how its first/secondary spatial derivative is related with energy/mass and force/mass=acceleration respectively?

Another observation is from the representation of Lorentz group. Where the rotation/boost is related with $SU(2)$ and $SL(2)$ transformations. If we take the $SU(2)$ or $SL(2)$ as transformations $U$ on quantum states, then they are dimensionless. So the acceleration (boost/time) can be regarded as $$dU/dt=H/\hbar=1/t$$ so we get boost is dimensionless, so time=length and energy=mass (these are normal conclusions since we usually take c=1). The reason that I check the Lorentz group representation is that the general metric tensor is generated from the Minkowski metric by dimensionless operation $GL(4)$, so this seems to confirm that the metric tensor should be dimensionless.

But if we go back to the former analysis, where the spatial derivative of the dimensionless metric tensor gives energy/mass, which is then also dimensionless since energy=mass. So we get the spatial derivative of a dimensionless value is still dimensionless.

There must be something wrong with my deduction. Can anybody help to clarify this?

  • $\begingroup$ Evidently $[g] =L^2$ since coordinates in RG have no dimension and $[ds^2]=L^2$ $\endgroup$ – Valter Moretti Jun 19 '17 at 19:25
  • $\begingroup$ @Valter Moretti Then what's the dimension of the potential and the curvature? How they correspond to the field potential and field strength? Thanks. $\endgroup$ – XXDD Jun 20 '17 at 3:10
  • $\begingroup$ Do you mean that then the spatial derivative will not change the dimension so both the potential and field strength have the same dimension of $L^2$? $\endgroup$ – XXDD Jun 20 '17 at 3:42
  • $\begingroup$ Yes I do, coordinate derivatives do not change physical dimensions. $\endgroup$ – Valter Moretti Jun 20 '17 at 3:46
  • $\begingroup$ @X.Dong Note that Valter Moretti is using convention #2 in the list I give in my answer, which, while perfectly acceptable, is not the most common choice. It's more common to give coordinates in GR the dimensions of $L^1$. $\endgroup$ – tparker Jun 20 '17 at 3:49

The line element $ds^2 = g_{\mu \nu} dx^\mu dx^\nu$ has the dimensions of length$^2$. But there are several different conventions for how to distribute those dimensions across the factors:

  1. Some people like to have the metric dimensionless and have the coordinate $dx^\mu$ have the dimension of $L^1$. This is my personal favorite, because then you can figure out the dimension of the various curvature tensors by just counting how many spacetime derivatives they're made up from (one factor of $L^{-1}$ for each derivative).

  2. Some people like to have the coordinates $dx^\mu$ dimensionless, in which case the metric and all the curvature tensors have the dimensions of $L^2$.

  3. Some people like to have different coordinates and different components of the metric have different dimensions - e.g. for the Euclidean metric $ds^2 = dr^2 + r^2 d\theta^2$, $[r] = L^1$, $[\theta] = L^0$, $[g_{rr}] = L^0$, $[g_{r\theta} = L^1]$, and $[g_{\theta \theta}] = [L^2]$. In this case, the different components of the various curvature tensors have different dimensions as well.

No matter which convention you use, the dimensions always work out correctly at the end of the day, when all indices have been contracted down to physically observable Lorentz scalars.

  • $\begingroup$ Thanks for the answer. I prefer to follow your first idea since it's conventional and helps to clarify the physical picture. I do not understand if the metric tensor is dimensionless, how we can derive the dimension of the potential(connection) and the dimension of the field strength (curvature). Could you please help to clarify this? Why the $L^{-1}$ connection works as a potential and the $L^{-2}$ curvature works as a field strength? Thanks. $\endgroup$ – XXDD Jun 20 '17 at 3:01
  • $\begingroup$ Or I am trying to understand if the metric tensor corresponds to a dimensionless $GL(4)$ transformation, and if we take this transformation as a general operation on a quantum state (as an extension of unitary operations), then what's the correspondence of the potential and the curvature? Which one of them correspond to the Hamiltonian operator? $\endgroup$ – XXDD Jun 20 '17 at 3:08
  • $\begingroup$ If the $GL(4)$ is taken as the dimensionless boost ($L/t$), then the Hamiltonian operator $H/\hbar$ has a dimension of $t^{-1}=L^{-1}$ and it should correspond to the acceleration (or force or the field strength divided by mass). But the curvature (field strength) should have a dimension of $L^{-2}$ but not $L^{-1}$. So I am confused. $\endgroup$ – XXDD Jun 20 '17 at 3:21
  • $\begingroup$ @X.Dong I'm not an expert of the formulation of classical E&M as a U(1) fiber bundle, but I think the confusion is that you are identifying the EM gauge connection $A_\mu$ with the Christoffel connection $\Gamma^\mu_{\nu \rho}$. This is certainly understandable because of the similarities of the names, but I think it's actually more natural to identify the EM gauge connection $A_\mu$ with the GR metric $g_{\mu \nu}$ itself. (Note that both are dimensionless in units where $\hbar = c = 1$, and both contain a gauge freedom.) The curvature of the EM gauge connection - the field strength ... $\endgroup$ – tparker Jun 20 '17 at 3:59
  • $\begingroup$ $F_{\mu \nu}$ - corresponds to the GR Cristoffel connection $\Gamma^\mu_{\nu \rho}$. Both have dimensions $L^{-1}$, are gauge-invariant, and give the curvature of the gauge connection or the spacetime respectively. The EM matter source field $J^\mu$ corresponds to the GR matter source field $T_{\mu \nu}$ - both have dimensions $L^{-2}$. So everything works out - the metric is like a potential and has units of energy/mass, the Christoffel connection is like the electromagnetic field strength and has units of force/mass, and the Einstein tensor/stress-energy tensor is like the source current. $\endgroup$ – tparker Jun 20 '17 at 4:06

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