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I was trying to figure out Curl in other coordinates. The Curl documentation (https://reference.wolfram.com/language/ref/Curl.html ) didn't really say how Mathematica interpreted the input.

Anyway, I tried the following just to figure it out. {0,x} should have curl of 1 everywhere, where {,} is a Cartesian coordinate.

Then, I converted to polar (thinking the answer should still be 1). I found the magnitude r = x and theta = Pi/2 because my vector points in the y direction. Anyway, I ran that and got something funky instead of 1 (see below).

Curl[{0,x},{x,y}, "Cartesian"] (*Out[1]= 1*)
Curl[{r Cos[theta],Pi/2},{r,theta},"Polar"] (*Out[2]= -((-(\[Pi]/2)-r Sin[theta])/r)) *)

So the question is how does Mathematica interpret curl inputs, why I didn't get 1 as an output, and how to fix that?

I am a beginner so please keep it simple. Links to suitable resources for this are also helpful.

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    $\begingroup$ Have you seen this? $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Mar 28, 2019 at 5:05
  • $\begingroup$ Sorry, I don't really understand too much about tensors. $\endgroup$
    – ions me
    Commented Mar 28, 2019 at 5:06
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    $\begingroup$ Oh wait, it looks like it takes local coordinates as input. $\endgroup$
    – ions me
    Commented Mar 28, 2019 at 5:37
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    $\begingroup$ Try this TransformedField["Cartesian"->"Polar",{0,x},{x,y}->{r,θ}]//Curl[#,{r,θ},"Polar"]&. $\endgroup$
    – Silvia
    Commented Mar 28, 2019 at 5:58
  • $\begingroup$ Ok, I get it now. Should I delete the post? $\endgroup$
    – ions me
    Commented Mar 28, 2019 at 18:44

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