Curl in Polar Coordinates

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.

• Have you seen this? – J. M.'s technical difficulties Mar 28 '19 at 5:05
• Sorry, I don't really understand too much about tensors. – Ion Sme Mar 28 '19 at 5:06
• Oh wait, it looks like it takes local coordinates as input. – Ion Sme Mar 28 '19 at 5:37
• Try this TransformedField["Cartesian"->"Polar",{0,x},{x,y}->{r,θ}]//Curl[#,{r,θ},"Polar"]&. – Silvia Mar 28 '19 at 5:58
• Ok, I get it now. Should I delete the post? – Ion Sme Mar 28 '19 at 18:44