I have read quite a lot of times now that the only coordinate system whose coordinate basis is orthonormal is the cartesian coordinate system.

Even though this makes sense, more or less, I never succeed when I try to prove it for myself. Could anyone give me a hint, or a sketch of a part of the proof, so I can work from there?

EDIT: by "coordinate basis", I mean the basis you get by doing

$$ \vec{e}_\alpha = \frac{\partial\vec{x}}{\partial q^\alpha} $$

with $\vec{x}$ the position vector and $(q^1,...,q^n)$ the coordinates of a point in $\mathbb{R}^n$ according to your coordinate system.


It is not entirely clear to me what you are looking for.

There is a linear algebra question: Is the canonical ONB on $\Bbb R^n$ the only ONB? This is false as $\frac1{\sqrt{2}}(1,1)$ and $\frac1{\sqrt2}(1,-1)$ form an ONB of $\Bbb R^2$.

There is a differential geometry question: If we have a coordinate chart on $\Bbb R^n$ such that basis induced on tanget spaces is an ONB at every point, is this chart an affine linear transformation of the standard chart?

Here it is true, as we are demanding that the coordinate chart locally maps an ONB to another ONB, ie the map should be an isometry locally. Since the map is invertible it will be a global isometry. The isometry group of euclidean $\Bbb R^n$ is $O(n)\ltimes \Bbb R^n$, that is these are all affine linear maps. See here.

  • $\begingroup$ Sorry for not being clear enough, but yeah, my question was precisely the differential geometry one. I have been taught coordinate systems and tensor fields without being taught topology or differential geometry (thank you, physics carreer), so my terminology is quite poor. $\endgroup$ – TeicDaun Jun 17 '18 at 11:56
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    $\begingroup$ Don't fret, I was a physicist too, it can't stop you from learning math. (Although knowing math can stop you from understanding what physicists are saying.) $\endgroup$ – s.harp Jun 17 '18 at 16:01

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