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In regular three dimensional space we always limit ourselves to Cartesian (i. e. orthonormal) frames. This has lots of advantages: dot products are easy, no need to distinguish between vectors and covectors, finding components of vectors is simple, etc. Even when using curvilinear coordinates, our basis vectors are orthonormal.

Of course linear algebra tells us that we can use whatever basis we want. So I ask: is there any situation in things are easier in a non Cartesian frame?

Edit: I'm not talking about coordinates. As mentioned in the comments, lots of different coordinate systems (spherical, cylindrical, etc) are used when there is a special geometry that makes things easier. But at least in spherical and cylindrical coordinates, the basis vectors at each point are orthonormal. I want to know whether there are reasons for using basis vectors that are not orthonormal.

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closed as too broad by Kyle Kanos, Carl Witthoft, JamalS, Jim, ACuriousMind Jan 8 '15 at 19:09

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ How about spherically symmetric potentials? $\endgroup$ – glS Jan 8 '15 at 13:16
  • $\begingroup$ Something like this? en.wikipedia.org/wiki/Skew_coordinates They give one (rather lame) example of solving Laplace's equation in a parallelogram. I'm sure there's better ones though $\endgroup$ – tpg2114 Jan 8 '15 at 13:22
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    $\begingroup$ I downvoted because (a) it's a bad question (b) shows no prior investigation into finding such an answer and (c) appears to be a list-based question (which is actually off-topic here). $\endgroup$ – Kyle Kanos Jan 8 '15 at 14:11
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    $\begingroup$ @Hindsight: (a) yes, downvotes are up to personal judgments, (b) OP has a fair few questions in E&M where non-Cartesian frames are very common, (c) list-based questions are ones in which answers are basically lists $\endgroup$ – Kyle Kanos Jan 8 '15 at 14:19
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    $\begingroup$ @Hindsight: I see now, definitely not clear with the original question. I've retracted my downvote, but I still think this is too broad because it's still asking for a list of situations. $\endgroup$ – Kyle Kanos Jan 8 '15 at 14:57
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Yes! You ask for an example. String theory is much, much more simple in the so-called light cone gauge. It is simply a choice of coordinates in Minkowski spacetime in which the coordinate axes are null geodesics:

$$ ds^2 = c^2 dt^2 - dx^2 - ... = du \cdot dv - ... $$

Here $(u,v,...)$ are no orthogonal coordinates.

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    $\begingroup$ UPD: the term "Cartesian" is not the same as "orthogonal" (which I think you meant). This is probably the reason for misunderstandings and downvotes. $\endgroup$ – Prof. Legolasov Jan 8 '15 at 14:41
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One example of useful non-orthogonal coordinates sometimes used in quantum mechanics is the coordinates

$$f_1=\vec x\cdot\vec x,\ f_2=\vec y\cdot\vec y,\ f_3=\vec x\cdot\vec y,$$

where $\vec x$ and $\vec y$ are two Jacobi coordinates (those which are not position of barycenter) in a three-body problem. The above mentioned coordinates have rotational invariance and thus are useful to reformulate quantum three body problem in terms of only internal motion of the system, separating global rotation away. These coordinates are used in e.g. this paper (see neighborhood of equation $(30)$ there).

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In a stationary axisymmetric space-time one has time-like and axial Killing fields $\xi^{\mu},\psi^{\mu}$ that one can use to construct a coordinate system $(t,x^1,x^2,\phi)$ adapted to these symmetries i.e. in which $\xi^{\mu} = \delta^{\mu}_t, \psi^{\mu} = \delta^{\mu}_{\phi}$ and $\partial_t g_{\mu\nu} = \partial_{\phi}g_{\mu\nu} = 0$. The conceptual and calculational advantages of these coordinates is self-evident. In general, unless $\xi_{[\mu}\nabla_{\nu}\xi_{\gamma]} = 0$ thus yielding a static space-time, one cannot make the $dtd\phi$ (that is, $g_{t\phi}$) term vanish in the metric.

This is of course due to the presence of frame-dragging in a non-static space-time. One can instead choose to use a coordinate system in which the frame-dragging effect is eliminated, which would be the comoving coordinates of the family of zero angular momentum observers, but this would make the metric more complicated in the sense of leading to conceptual and calculation difficulties not the least of which is because the metric will be time-dependent and the coordinate system itself will be non-rigid.

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