So I wonder why the gradient in coordniates other than cartesian ones bears coefficients. Let's take spherical coordinates for example. We have(adsbygoogle = window.adsbygoogle || []).push({});

(Source) - Sorry if image doesn't work - too lazy to get the TeX right.

From what I know, I don't see anything that raises cartesian coordinates above any other C/S' (apart from marginal properties such as whether the mapping is reversable). I hence would expect the gradient in spherical coordinates to be

[tex]\vec\nabla = \vec{e}_r\frac{d}{dr} + \vec{e}_\phi\frac{d}{d\phi} + \vec{e}_\theta\frac{d}{d\theta}[/tex],

where [tex]\vec{e}_\dots[/tex] is the unit vector in [tex]\vec{g}_{\dots} = \frac{\partial \vec{r}}{\partial \dots}[/tex] direction (where |g| is not normalized).

After all, the gradient in coordinate q descibes the change of intensity I in q direction. And when I change [tex]\vec{r}[/tex] by a little [tex]d\theta[/tex], the change in intensity is naturally [tex]\frac{dI}{d\phi}[/tex] andnot[tex]\frac{1}{r\sin(\phi)}\frac{\partial I}{\partial \theta}[/tex] (why would it be partial anyway?).

So why is transformed from cartesian coordinates? What validates cartesian coordinates as the origin of all transforms, so to say? I could understand the application of the Leibniz'en chain rule if [tex]\nabla[/tex] to a scalar potential in general coordinates were sought with respects to cartesian coordinates. But we are looking for the gradient in terms of spherical coordinates - so what do cartesian coordinates want to tamper with?

best regards

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# Cartesian coordinates vs. The rest of the world?

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