# How do you understand the nature of curl?

My teacher tells me the curl describes the component of rotation at a point in a vector field. When a ball is placed in a vector field with a non-zero curl, it tends to rotate.

Let's consider a field like $$\{2 x y-\sin (x),x^2+e^{3 y}\}$$.We can easily calculate the curl of this vector field is $$0$$. We can visualize this vector field by Wolfram Mathematica like follow:

StreamPlot[{2 x y - Sin[x], x^2 + E^(3 y)}, {x, -30, 30}, {y, -30, 30}]


But when I saw this vector field that I had visualized, I began to doubt my own understanding. Could it be that if I place a small ball on this red circle, the ball really won't spin? Or did my teacher lie to me?

• Ummm... in which direction do you think the red ball would rotate? (Rotate, not Move.) Nov 6, 2021 at 21:26
• With StreamPlot you lose information about the length of the vectors, so it's useless for thinking about the curl. For example, the plots of the vector fields $(1,0)$ and $(1+y^2,0)$ look identical (except for some length-based colouring with the new defaults in Mathematica 12), but they don't have the same curl; the first one is irrotational, the second one isn't. Nov 6, 2021 at 22:43
• @DavidG.Stork Notice that those arrows flow from the east to the north. Then I think there must be a clockwise turning point on the inside of the bend and a counterclockwise turning point on the outside of the bend. Do you think it is reasonable for me to think so?
– mayi
Nov 7, 2021 at 6:04
• Can you imagine the ball changing directions without rotating? Supposedly that's what a zero-curl field would make it do.
– Karl
Nov 7, 2021 at 6:11
• In the video, all the vectors are drawn with (more or less) the same length, but the point is that the actual flow is animated so that you can see how fast it goes. Even if you colour the streamlines or the vector field, it's very hard to get a feeling for how little balls would rotate from a non-moving picture. Nov 7, 2021 at 11:49