It has recently come to my attention that the operations carried out with a Del operator do not match their notation.

That is, the divergence or curl are written as

$\nabla \cdot \vec{v}$ and $\nabla \times \vec{v}$.

The Del operator is a vector operator, and has been expressed as above given that it outputs divergence and curl in Cartesian coordinates; however, this notation is also used in curvilinear coordinates. This troubles me because that is not really the case for curvilinear coordinates (if i carry out the dot product or the cross product literally I will get the wrong answer), yet we still use vector products notation and we are not carrying out those operations at all but rather carrying out integrals.

Can I carry out a dot-product in cylindrical coordinates and arrive to the correct form of the divergence? Or the vector notation does not represent the actual divergence and curl, and it is just customary.

Thank you for your time and answers.


No, it is still the dot product, even in other coordinate systems. The inner product is actually a $(0,2)$ tensor, and thus is coordinate invariant. Just because $(a^1,a^2)\boldsymbol{\cdot}(b^1,b^2)=a^1b^1+a^2b^2$ in the standard basis does not mean that this is how we compute the dot product in other bases. E.g, for two vectors with components $(u^1,u^2)~;~(v^1,v^2)$ with respect to the polar coordinate system, the dot product is $$(u^1,u^2)\boldsymbol{\cdot}(v^1,v^2)=u^1 v^1 \cos(u^2-v^2)$$ There are similar constructions for the cross product.

Unfortunately, there is no easy formula to take the dot product between vectors expressed via some arbitrary coordinate system without first having to convert back to the standard basis.

  • $\begingroup$ How does one derive what the actual inner product is? Shouldn't it be able to come out using vector notation? I have only seen it derived through integration. $\endgroup$
    – RSM
    Jan 25 at 18:47
  • 1
    $\begingroup$ @RSM You convert the vectors back to their representations in the standard basis and take the usual inner product. There is no better way to do it than that, unfortunately. $\endgroup$
    – K.defaoite
    Jan 25 at 20:21

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