Metric tensor: Why relate it to Cartesian/Minkowski coordinates?

Question

Why does the metric tensor always relate to cartesian coordinates?

Let's take the simple case for the metric tensor in 3D-space without a time dimension,

$g_{ij}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2\; \sin^2(\theta) \end{bmatrix}$

here, the $\sin^2(\theta)$ stems from the fact that we originally derived the distances in cartesian coordinates as $\rm ds=\sqrt{dx^2+dy^2+dz^2}$ and then know the transformation between cartesian and polar. So the exact form of $g_{ij}$ as function of it's target coordinates, is always derived from the original coordinates, which are the cartesian ones.

But why don't we describe the metric tensor based on some other original coordinates, like hyperbolic and transform then to spherical ones (apart from the fact that it would be ugly business)?

So cartesian coordinates seem in some way special, my first idea was that maybe because they're an inertial frame of reference they would provide a natural basis for GR. But this can't be the case, as differential geometry comes from pure math, which doesn't care about inertial/noninertial statements.

So what is going on, is it the fact that we simply 'discovered' math in euclidean space first and later learned how to relate different coordinate systems to the euclidean one?

Same question naturally extends to relativity and minkowski coordinates.

Answer 1

Unless I misunderstood your question and example...

Here's my 2 cents.

In your example you are using spherical coordinates to express the location of points in an otherwise Euclidean space. The distance between points does not change and the topology of the underlying point set is not changed. By one interpretation you can indeed use intersecting Hyperbolas, etc to create hyperbolic coordinates.

But I think you may be asking, why assume space has an underlying (global if not local) structure of E3? I agree, why? The answer would be experience. That geometry describes our 3dim space in which we observe things. At least it did good enough until Einstein. Now we know that Lorentz invariance governs space-time intervals. We still need to describe isometry in 3d as it's a part of many physical theories. Keep in mind that all of differential geometry grew from an abstraction or generalization of Euclidian geometry so it was natural to say that the local structure of the metric (measure of line segments) would be diag(1, 1, 1). The local structure of space-time if diag(-1, 1, 1, 1) or diag(1, -1, -1, -1).

If I misunderstood please comment and I'll try to explain more.

Answer 2

I think it's only for habit. There is not, in my point of view, a practical reason. Perhaps getting the equations of motion is easier to obtain if you consider Cartesian coordinates. But if you note, the same equations of motion can similarly be obtained in any metric. For example, if you consider the following integral action$$S\left[\phi\right]=\int_\Omega\mathcal{L}\left(x,\phi,\partial_{\mu}\phi\right)\sqrt{-g}~d^4x,$$so, the correspondent Euler-Lagrange's equations is$$\frac{1}{\sqrt{-g}}\partial_{\mu}\left(\sqrt{-g}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\nu}\phi\right)}\right)-\frac{\partial\mathcal{L}}{\partial\phi}=0.$$

Thus, if you assume a Minkowski metric in spherical coordinates, $$g^{\mu\nu}=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -\dfrac{1}{r} & 0 \\ 0 & 0 & 0 & -\dfrac{1}{r\sin\theta} \end{pmatrix},$$ whose $$g=\det\left(g_{\mu\nu}\right)=-r^2\sin^2\theta,$$the motion equation obtained will already come out directly in spherical coordinates, without the need to use any other transformation relations. Nothing prevents you from doing this. However, you will have a little more work.