Let $\mathbf{A}(x,y,z)$ be a vector field. The curl of this vector field is defined as

$$\nabla \times \mathbf{A} = \left(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}\right) \mathbf{i} + \left(\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}\right) \mathbf{j} + \left(\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\right) \mathbf{k}$$

Let's consider a simple example from Wikipedia.

$$\mathbf{A}(x,y,z) = y\mathbf{\hat{x}}-x\mathbf{\hat{y}}+0\mathbf{\hat{z}}$$

This corresponds to the following enter image description here

source: https://commons.wikimedia.org/wiki/File:Uniform_curl.svg

The curl is

$$\nabla \times \mathbf{A} =0\boldsymbol{\hat{x}}+0\mathbf{\hat{y}}+ \left({\frac{\partial}{\partial x}}(-x) -{\frac{\partial}{\partial y}} y\right)\mathbf{\hat{z}}=-2\mathbf{\hat{z}}$$

But there is where I am confused. I don't get what $\left({\frac{\partial}{\partial x}}(-x) -{\frac{\partial}{\partial y}} y\right)$ has to do with the rotation and size of the vectors in the picture shown above.

My Question

How do these derivatives tell me anything about the rotation? Any why must they be opposites? Why is it not $\frac{\partial A_x}{\partial x}$? I understand of course why, when you compute the cross product, it comes out that way, but I want to understand the intuition of why that is important for rotation.

  • $\begingroup$ Ah if I had it in front of me there was a really good sort of physical argument about it in a fluids book that made it feel intuitive if not super rigorous. $\endgroup$ – Triatticus Jan 21 '17 at 3:49

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