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.
This post imported from StackExchange Physics at 2025-01-23 14:51 (UTC), posted by SE-user pglpm
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