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.

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    $\begingroup$ The coordinates are chosen based on symmetry of the problem. Exploiting the symmetry simplifies the calculation. If the problem doesn't have spherical symmetry - say it has the symmetry of a flat disk, then you shouldn't use spherical coordinates - you would want to use Polar coordinates instead. And Spherical coordinates are not Cartesian coordinates - but you can transform back to Cartesian coordinates from either Spherical or Polar coordinates at any time. $\endgroup$ May 15, 2019 at 19:25
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    $\begingroup$ You don't need to start with Cartesian coordinates, you can get the spherical metric components directly from Euclidean geometry. $\endgroup$
    – jacob1729
    May 15, 2019 at 19:25
  • $\begingroup$ @jacob1729: Yes, that's the point of my question: Why do we always use euclidean geometry as basis? The functional form of $g_{ij}$ in any literature derives from its euclidean derivation. This makes the euclidean system special, and I'm asking why it's special. $\endgroup$ May 15, 2019 at 19:39
  • $\begingroup$ Note $s=\sqrt{dx^2+dy^2+dz^2}$ is incorrect - it should read $ds=\sqrt{dx^2+dy^2+dz^2}$ for Cartesian coordinates. It's metric tensor would be $g_{ij}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\\ \end{bmatrix}$ The metric tensor in your post is for Spherical coordinates. You should be able to write down the arch length - $ds$ for spherical coordinates by inspection using the metric tensor. $\endgroup$ May 15, 2019 at 19:41
  • $\begingroup$ The $3$ dimensional Euclidean system is special because it's the space you live in - and it's where physical measurements are intərprəted $\endgroup$ May 15, 2019 at 19:51

2 Answers 2


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.

  • $\begingroup$ I was aiming more for the following: The exact functional form of $g_{ij}$ on the coordinates used, is always derived based on cartesian coordinates. Why cartesian? Not sure this answers it, I'll update the question accordingly tho. $\endgroup$ May 15, 2019 at 18:59
  • $\begingroup$ I am not sure your assumption is true. How is it based on Cartesian? $\endgroup$
    – user196418
    May 15, 2019 at 19:00
  • $\begingroup$ I think I'm thinking of those 'proof-by-picture' sessions when you start out as freshman and derive e.e. volume elements by sketching 3D volumes with their sides being of arc length $dx_1 = r sin(\theta)$ etc. I'm not sure how you would derive arc lengths (and thus the metric tensor..) in GR without pictures and more formally. $\endgroup$ May 15, 2019 at 19:03
  • $\begingroup$ For one thing those pics are meant to illustrate how an infinitesimal arc, area, or volume is expresses in curved coordinates that tell us how to find points in otherwise flat space. For another, differential geometry assumes that local neighborhoods of any point look "flat". This allows us to carry over all we learned. The only potential issue is dealing with closed manifolds that require more than one patch to cover. $\endgroup$
    – user196418
    May 15, 2019 at 19:07
  • $\begingroup$ Then your answer would essentially be: "Because euclidean came first, and we constructed everything from it." So I would be wondering about the Aliens that would discover spherical coordinates first, and then construct euclidean geometry from that. How would $g_{ij}$ look? Would it just be our $g^{ij}$? $\endgroup$ May 15, 2019 at 19:41

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.

  • $\begingroup$ Fair point, but then how do we know the form of $g^{\mu\nu}$, is this not in the end derived from some basis transformation that goes from cartesian to spherical? $\endgroup$ May 16, 2019 at 3:43
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    $\begingroup$ form of minkowski metric in spherical coordinates is fixed by spherical symmetry, meaning of the coordinates and the fact that the curvature is zero. You don't need cartesian coordinates anywhere in the derivation. Try googling something like the most general metric in spherical coordinates and you should be able to find how it is done. $\endgroup$
    – Umaxo
    May 16, 2019 at 6:41
  • $\begingroup$ @AtmosphericPrisonEscape : Well, integral action is built in a covariant form is not toa. It is done so that in the end the form of the field equations does not change, regardless of the particular choice of coordinates. So, you must first decide which scenario you want to describe your theory. Once this is done, you build your metric based on this choice. For example, the metric is constructed by the relation $g_{\mu\nu} =\vec{{e}}_{\mu}\cdot\vec{{e}}_{\nu}$. So, as you may notice, you must first know a relationship of transformation between the base vectors and from there build your metric. $\endgroup$
    – lucenalex
    May 16, 2019 at 9:00
  • $\begingroup$ @lucenalex: Thanks for your comment. Yes, but which transformation relation would that be? You always transfrom from a-coordinates to b-coordinates. b being spherical in this case, and a is usually cartesian. $\endgroup$ May 16, 2019 at 14:22
  • $\begingroup$ @AtmosphericPrisonEscape: I don't know if I understand well your question. But, let's go there! I would say that any relation of transformation between the coordinates of the same point of the Minkowski spacetime: from cartesian to spherical, from spherical to cylindrical, etc. But there is something that needs to be made clear, the transformations between two points distinct of Minkowski's spacetime which must be taken into account are the Lorentz transformations, which is different of transformation in the same point of Minkowski spacetime. $\endgroup$
    – lucenalex
    May 16, 2019 at 15:58

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