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The $\nabla$-operator is simple in cartesian coordinates, $[\partial_x,\partial_y,\partial_z]$, but in spherical coordinates, it becomes $[\partial_r, \frac{1}{r}\partial_\theta, \frac{1}{r\sin\theta}\partial_\varphi]$ and in cylindrical coordinates $[\partial_\rho, \frac{1}{\rho}\partial_\varphi, \partial_z]$; is there a general formula for converting into a different coordinate system, perhaps in terms of a Jacobian?

(Sub-question: Is there any reason why similar operators like $\hat{\nabla} = [\partial_r, \partial_\theta, \partial_\varphi]$ aren't in use?)

Frank Vel
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  • To the subquestion: we know that $\nabla$ transforms as a vector under rotations. How does $\hat{\nabla}$ transform?. – NickD May 08 '17 at 14:22
  • https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates – Kenny Wong May 08 '17 at 14:24
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    This is because vector calculus notation is full of old fashioned notions. If you want to understand what is going on with the $\nabla$ operator in vector calculus, you should really use the modern language of differential forms. You can see that all the operators $\text{div}$, $\text{curl}$, $\text{grad}$, $\text{lapl}$ are just different combinations of something called the exterior derivative $\text{d}$ and the Hodge dual $\star$, and all integral theorems (Gauss, Stokes, fundamental theorem of calculus) are just different expressions of a single one (Kelvin-Stokes). – Jackozee Hakkiuz May 18 '18 at 16:35

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There is a large body of literature on this subject. I found a section in an old calculus book on Orthogonal Curvilinear Coordinates in the chapter on Vector Analysis. It's too complex to present here. (The book is Advanced Calculus for Applications by Hildebrand.) However, you can get started here: Curvilinear coordinates.

Cye Waldman
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