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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}]

enter image description here

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?

enter image description here

mayi
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    Ummm... in which direction do you think the red ball would rotate? (Rotate, not Move.) – David G. Stork Nov 06 '21 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. – Hans Lundmark Nov 06 '21 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 07 '21 at 06:04
  • Can you imagine the ball changing directions without rotating? Supposedly that's what a zero-curl field would make it do. – Karl Nov 07 '21 at 06:11
  • @Karl Sorry, I can't imagine. I think the direction changed can only be because this little ball rotated. – mayi Nov 07 '21 at 06:38
  • You'll need to take into account the speed of the flow. If the flow on one side of the ball is faster than on the other side, it will cause the ball to rotate, even if the velocity vectors are parallel. Have a look at this 3b1b video: https://www.youtube.com/watch?v=rB83DpBJQsE (the part about the curl starts at 4:32). – Hans Lundmark Nov 07 '21 at 07:54
  • @HansLundmark Note in 3:52s said, we usually use color present the length so that the line segments don't overlap, but not missing information as you said above. – mayi Nov 07 '21 at 08:43
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    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. – Hans Lundmark Nov 07 '21 at 11:49
  • By the way, that's not what's said at 3:52. Did you mean to write some other timestamp? – Hans Lundmark Nov 07 '21 at 11:54
  • @HansLundmark Oh, god. I first time to know the video in youtube is a omit version. I'm not sure if you can see the Chinese version of this video. It is in the 3:52s in the Chinese full version – mayi Nov 07 '21 at 12:06
  • Ah, OK, then it makes more sense! – Hans Lundmark Nov 07 '21 at 12:40
  • The fact that the flow goes from the east, then from the south is a separate issue from whether there is curl (and attendant rotation). I see no differential in flow that would spin a ball in the red position. – David G. Stork Nov 07 '21 at 16:47
  • @HansLundmark Note VectorPlot[{1 + y^2, 0}, {x, -3, 3}, {y, -3, 3}, VectorScaling -> Automatic] is different with VectorPlot[{1, 0}, {x, -3, 3}, {y, -3, 3}, VectorScaling -> Automatic] – mayi Nov 08 '21 at 10:01
  • Yes, of course. It's the results from StreamPlot that look the same (except for the colouring). – Hans Lundmark Nov 08 '21 at 12:39

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