Dimensional analysis of metric tensor 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?
 A: 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:


*

*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).

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

*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.
A: There's always a bit of confusion regarding coordinates and their dimensions. A coordinate is, from a physical point of view, a quantity associated with every event in a region of spacetime (the domain of the chart), in such a way that the values of a set of such quantities uniquely identify the events in that region. Any quantity will do: the distance from something, the time elapsed since something, an angle – but also a temperature or the value of a field. So we could have a local coordinate system where the coordinates have dimensions of length, angle (that is, "1"), magnetic flux, and temperature.
As tparker points out, this implies that different components of the metric tensor will have different dimensions. But every tensor has an absolute dimension, as Schouten (1989) calls it. It's the dimension of the tensor as a geometric object, independently of any coordinate system. It's the dimension of the sum
$$g_{00}\;\mathrm{d}x^0 \otimes \mathrm{d}x^0 + 
g_{01}\;\mathrm{d}x^0 \otimes \mathrm{d}x^1 + \dotsb
\equiv \pmb{g}.$$
There are different choices for the absolute dimension of the metric tensor: $\text{length}^2$, $\text{time}^2$, and so on. My favourite is $\text{time}^2$, because if we transport a clock from an event $E_1$ to an event $E_2$ (timelike separated) along a timelike path $s \mapsto c(s)$, the clock will show an elapsed time (proper time)
$$\int_{c} \sqrt{\Bigl\lvert \pmb{g}[\dot{c}(s),\dot{c}(s)] \Bigr\rvert}\; \mathrm{d}s,$$
which is independent of the parametrization $s$. Assuming $c$ to be adimensional means that $\pmb{g}$ must have dimensions $\text{time}^2$. But some authors, eg Curtis & al (1985), define the elapsed time as $\frac{1}{c}$ times the integral above, so that $\pmb{g}$ has absolute dimension $\text{length}^2$ instead. Anyway, the point is that $\pmb{g}$, as an intrinsic geometric object, has a dimension that is independent of any coordinates.
Note that $\pmb{g}$'s absolute dimension causes differences in the absolute dimensions of tensors obtained from one another by raising or lowering indices.
Regarding a connection – independently of any metric – consider the action of its covariant derivative $\nabla$ on the coordinate vectors:
$$\nabla \frac{\partial}{\partial x^\lambda} =
\sum_{\mu\nu} \varGamma{}^{\nu}{}_{\mu\lambda}\;
\frac{\partial}{\partial x^\nu}\otimes\mathrm{d}x^{\mu}.$$
To ensure that the terms in the sum and the left side have the same dimension, the Christoffel symbol $\varGamma{}^{\nu}{}_{\mu\lambda}$ must have dimensions $\mathrm{K}\,\dim(x^{\nu})\,\dim(x^{\mu})^{-1}\,\dim(x^{\lambda})^{-1}$, where $\mathrm{K}$ is arbitrary. The effect of the covariant derivative is thus to multiply the dimension of its argument by $\mathrm{K}$. It seems very natural to take $\mathrm{K}=1$, otherwise we would have troubles with the definition of the Riemann tensor:
$$R(\pmb{u},\pmb{v})\pmb{w} = 
\nabla_{\pmb{u}}\nabla_{\pmb{v}}\pmb{w}
-\nabla_{\pmb{v}}\nabla_{\pmb{u}}\pmb{w}
-\nabla_{[\pmb{u},\pmb{v}]}\pmb{w},$$
where $\nabla$ appears twice in two summands and once 
in one summand.
From this it follows that the Riemann tensor $R{}^\bullet{}_{\bullet\bullet\bullet}$ and the Ricci tensor $R_{\bullet\bullet}$ are adimensional.
See this answer for a longer discussion.
References


*

*Curtis, Miller (1985): Differential Manifolds and Theoretical Physics (Academic Press); chap. 11, eqn (11.21).

*Schouten (1989): Tensor Analysis for Physicists (Dover, 2nd ed.); chap. VI.

