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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?

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    $\begingroup$ Evidently $[g] =L^2$ since coordinates in RG have no dimension and $[ds^2]=L^2$ $\endgroup$ Commented Jun 19, 2017 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
    Commented Jun 20, 2017 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
    Commented Jun 20, 2017 at 3:42
  • $\begingroup$ Yes I do, coordinate derivatives do not change physical dimensions. $\endgroup$ Commented Jun 20, 2017 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
    Commented Jun 20, 2017 at 3:49

2 Answers 2

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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.

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  • $\begingroup$ 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
    Commented Jun 20, 2017 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
    Commented Jun 20, 2017 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
    Commented Jun 20, 2017 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
    Commented Jun 20, 2017 at 3:59
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    $\begingroup$ @Dale It depends on what you mean by "work", but on second thought I'm inclined to agree with you that convention 1 is awkward for general coordinate systems. I like convention 3 now, because it seems natural (to me) to keep e.g. Cartesian coordinates on Euclidean space dimensionful. Really a matter of personal preference. $\endgroup$
    – tparker
    Commented Dec 23, 2019 at 4:21
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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.
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