Is length adimensional when space is not flat?

Question

Consider the two manifolds $\mathbb{R}^2$, equipped with the usual metric $g_{ij}=\delta_{ij}$, and $\mathbb{H}^2=\{(x, y)\,:\,y>0\}$, equipped with the hyperbolic metric $h_{ij}=\delta_{ij}/y^2$. Assume that the coordinates $x, y$ have the dimension of length. Then in the Euclidean case the length of a curve has the dimension of length (obviously): $$ ds=\sqrt{dx^2+dy^2}.$$ What about the hyperbolic case? Here it seems to me that the length of a curve is adimensional: $$ds =\frac{\sqrt{dx^2+dy^2}}{y}.$$

Questions. Am I right? Does all of this have a physical meaning of some sort, or it is just that I am not considering some dimensioned constant which has been set equal to $1$?

Answer 1

Assume that the coordinates $x,y$ have the dimension of length

The correct interpretation of the situation is that the above assumption is false.

There's no reason why the coordinate functions of a Riemannian manifold must have dimension of length. A simple example is the radial coordinate system $(r,\theta)$ for $\mathbb{R}^2 \setminus \{0\}$. The angular variable $\theta$ is usually measured in radians, which as a unit is given by length / length, and so actually has no dimensions.

Answer 2

Coordinates are diffeomorphisms $\varphi_i:U_i\to \mathbb R^m$ where $U_i$ is an open subset of the manifold. A smooth manifold is covered by many such coordinate charts. Indeed, if $f= (f_1, f_2): U\to\mathbb R^2$ is a chart, then so is $\tilde f = (f_1+(f_2)^3-e^{f_2},f_2)$. This should suggest that the coordinates $f_1,f_2$ are unitless.

Furthermore, the coordinate charts are defined on any abstract smooth manifold, which does not have any notion of length. We can talk about length after introducing a Riemannian metric $g_{ij}$ on the manifold. The components $g_{ij}$ can be understood as measured in units of length.

Answer 3

The relevant difference between the Euclidean and hyperbolic situations is that hyperbolic spaces come with a natural unit of length, because they have (constant) curvature. One takes advantage of that by normalizing the unit of length so that the curvature is $-1$. That normalization was already incorporated in the formula you quoted for the metric, and that makes the metric dimensionless.

Answer 4

I believe that (at least in physics) keeping track of the units is powerful. In mathematics one can avoid mistakes by assigning units to quantities (e.g. Riemannian geometry, differential equations), even though they are abstract and originally dimensionless. This is just a prejudgement, but it works for some people.

Anyway, what is the meaning of a quantity having units in mathematics? Hardly one one can give an answer withoug evoking physics. Thus, rather than using reductio ad absurdum to conclude that coordinates cannot have the dimension of length, I think that the units used to write the metric are "natural units" (in the sense of physics, e.g. $\hbar=1=c$) because of the following fact:

In physics, the canonical coordinates of $\mathbb{R}^m$ have units of length.

If you want to keep track of units the metric should be $h_{ij}=\kappa \delta_{ij}/y^2$, as one for instance writes metrics if one does not work in natural units:

$$ds^2=c^2 {d \tau}^{2} = \left(1 - \frac{r_S}{r} \right) c^2 dt^2 - \left(1-\frac{r_S}{r}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)$$ (the Schwarzschild metric). A second example why is not convenient to forget that coordinates have units is when one writes an action functional like $\int_M L\,d(vol) $. Here $d(vol)$ has dimension (length$)^n$ so $L$ must have dimension $(length)^{-n}$. How would you rule out so many possible actions if not by dimensional analysis and requiring that (some of) the coordinates have units? My point is that, even though you cannot assign length to all coordinates, once one can consistently impose that $ds^2$ has square length units: because of the way $g_{ij}(x)$ transforms, any change of coordinates will give you the right factors.

To addres the title question, flatness does not have to do with having dimensionless metrics. For instance, $(x^2+y^2)(dy^2+dx^2)$ is a metric which is not flat, and, if you assign length units to these coordinates, the metric won't be dimensionless.