I Intuitively understand the curl formula?

1. Aug 19, 2016

yosimba2000

Ok, so I know the curl represents how much something rotates about an axis.

Let's assume we have a vector field F = Fx + Fy + Fz, where x y and z are direction vectors.

So the rotation about the Z axis is made possible by a change in the Y direction and a change in the X direction.

But the formula for the curl around the Z-axis is given by: dFy/dx - dFx/dy

I pulled that from Equation 3 here: http://www.maxwells-equations.com/curl/curl.php

How does this represent the rotation about the X axis? I'm reading dFy/dx - dFx/dy as:
The change in the Y vector along the X direction, and The change in the X vector along the Y direction.

How does this represent curl around the Z axis? Also, why is it dFy/dx - dFx/dy?

Shouldn't dFy/dx still be a vector pointing in the Y direction, and dFx/dy a vector pointing in the X direction? So you can't just subtract them because they are perpendicular to each other, right? It's like saying I have Velocity in the Y direction and Velocity in the X direction, so the net velocity is the Y direction minus the X direction, which is incorrect. You have to use Pythagorean's Theorem to find the net velocity, so shouldn't you do the same here? As in the net curl is going to be sqrt[(dFy/dx)^2 + (dFx/dy)^2]

2. Aug 19, 2016

Staff: Mentor

3. Aug 19, 2016

yosimba2000

4. Aug 19, 2016

Staff: Mentor

5. Aug 19, 2016

Staff: Mentor

I found this other video which shows real world examples of curl via fluid flow:

6. Aug 19, 2016

yosimba2000

http://mathinsight.org/curl_components

Ok, so the curl, being here the counterclockwise rotation around the Z axis , is made possible if the upward Y force on the right is greater than the upward Y force on the left, and also made possible if the rightward X force on the bottom is greater than the rightward X force on the top.

And so the curl is defined as the change in force in Y over small dx - the change in force in X over small dy.

I understand the Z component of the force itself will not contribute to rotation about the Z axis, but if both the Y and X force vectors are both functions of X, Y, and Z, then shouldn't the net change in force in either direction have to be taken with respect to X Y and Z?

So when we look at the changes in Y force, we can't just look at the change in Y force as X changes, because although that contributes some, we also have changes in Y force as Y changes, and changes in Y force as Z changes, right?

7. Aug 21, 2016

FactChecker

No. If you look at the notation used in your link, it is dAy/dx - dAx/dy where Ay and Ax are not vectors. They are the coordinates of the A vector field in the y and x directions.

8. Aug 21, 2016

yosimba2000

But dAy/dx means the small change in Y vector field over small change x. So the change in vector field still has a direction?

9. Aug 21, 2016

FactChecker

It means a small change in the y-axis component of the vector field over a small change in x.

10. Aug 22, 2016

yosimba2000

Yes, sorry that's what I meant. So the change in the Y component of the field over small x still has a direction to it, right?

11. Aug 22, 2016

FactChecker

Sorry, I got sloppy. Ay and Ax are real valued, not vectors. So dAy/dx - dAx/dy can be 0. That was one of your original questions when you were wondering how dFy/dx - dFx/dy could be 0.