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In elementary vector calculus we use different coordinate systems to tackle problems with different symmetry. The cartesian coordinate system has certain very nice properties, including that the differential does not depend on the coordinate value itself, simplifying integration, grads, curls etc. Meanwhile in curvilinear coordinates we do have such a dependence, for obvious geometrical reasons.

Are these coordinate systems then fundamentally distinct mathematical constructs? How can we describe the differences? Or is the cartesian case just a particularly well behaved special (possibly unique?) case of a general coordinate system? Is the linearity of the unit vectors key?

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I recommend looking into tensor calculus as applied to Riemannian geometries. This discipline answers all of your questions and more by allowing one to do things like vector calculus without specifying a coordinate system. Think of it as a way to do all the work up front and add a coordinate system later (when it becomes convenient). Actual tensor calculus is of course much deeper than this but this is a nice benefit that I think you would like.

The differences between coordinate systems are mostly baked into what we call the metric tensor. We can write operations like gradients, curls, integrals, divergence, and more in terms of a scalar/vector field and this metric tensor. Cartesian style coordinates are very nice. In many cases however, they are not significantly nicer than any coordinate system that has a constant, diagonal metric tensor (constant, orthogonal basis vectors). This boils down to any coordinate system that could be referred to as "stretched" cartesian coordinates.

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  • $\begingroup$ So the "niceness" of Cartesian coordinates is based on the diagonality of the metric tensor? $\endgroup$ Commented Oct 14, 2020 at 12:41
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    $\begingroup$ @JoshD I would say that there are three good things about Cartesian coordinates that contribute to it's "niceness". In my opinion and experience if I were to rank these properties from "nicest" to least "nice", I would say the following. (1): the metric tensor is constant thus the basis vectors are locally constant. (2): the metric tensor is diagonal thus the basis vectors are orthogonal at any point $p$ in the space. (3): the metric tensor has ones on the diagonal thus the basis vectors are unit length. $\endgroup$ Commented Oct 14, 2020 at 12:50
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    $\begingroup$ @JoshD The justification for my top choice is that in many applications we have to deal with derivatives/integrals of the metric tensor so it's nice when all the components are constant. $\endgroup$ Commented Oct 14, 2020 at 12:52
  • $\begingroup$ This seems like the key point thanks! So as the basis vectors are constant in space, this means that the coordinate axes are simply straight lines. Therefore all coordinate systems which are nice in this first way will have parallelograms (or higher dimensional equivalents) as volume elements. If you want a more 'interesting' volume element with curvature you'll have to introduce a non-constant basis vector. Thanks this has clarified the key points considerably. $\endgroup$ Commented Oct 14, 2020 at 13:26

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