I recently learned about curl and I was wondering what the reasoning is in taking nabla cross the vector-valued function.
What do you mean, what’s the reasoning? Are you asking why that’s what gets called “curl”, or why it’s a useful operation to look at?
One practical application is that it comes up in multiple places in Maxwell’s equations of electromagnetism. It’s also relevant to fluid flows, though I’m not as familiar with those.
Sorry about lack of specification. I mean it in a way like “why is the area of a circle pi*r^2?”
Why are we taking the cross product between a vector valued function and nabla to find curl? What is the logical reason for doing so?
The definition of curl (from one point of view) is that it’s the axis of rotation for each point in a vector field (say, a fluid flow), with the magnitude being the angular velocity. In 3D, a rotation axis is a 3-element vector. A cross product is the main way of producing a new vector given some input vectors. del (nabla) cross F produces a vector, so it’s at least a candidate for producing an axis of rotation.
But why that specifically? It’s probably easier to consider a single axis at a time. For the Z axis, we get \frac{\partial{F_y}}{\partial{x}}-\frac{\partial{F_x}}{\partial{y}}. If we reduce ourselves to 2D, that’s the (scalar) angular velocity of a moving vector. To gain some intuition why the value should look like that, consider a velocity vector and move a tiny bit in the +x direction. If the vertical velocity increases, then we have a counter-clockwise (right-hand) rotation. That accounts for the \frac{\partial{F_y}}{\partial{x}} term. Similarly, move a bit in the +y direction. If the horizontal velocity is negative, then again we have a counter-clockwise rotation. That accounts for the -\frac{\partial{F_x}}{\partial{y}} term. Not exactly a formal proof but maybe it’ll help with the intuition.
So, “How do we prove that the cross product of nabla with the vector field gives us the thing we want”? Well, then, give us a mathematical description of what “the thing we want” is.
The synopsis is that if you re-write your curl (possibly using the Riemannian structure and orientation of \mathbb{R}^3 if necessary), the resulting operation is nothing more than taking the differential of a 1-form, which looks much more straightforward.
In fact, we can see for ourselves, thanks to the magic of MathJax:
if \omega = A\, dx + B\, dy + C\, dz, then
d\omega = \left(\dfrac{\partial B}{\partial x}-\dfrac{\partial A}{\partial y}\right)\,dx\wedge dy + \left(\dfrac{\partial C}{\partial x}-\dfrac{\partial A}{\partial z}\right)\,dx\wedge dz + \left(\dfrac{\partial C}{\partial y}-\dfrac{\partial B}{\partial z}\right)\,dy\wedge dz.
So, stick to modern notation and Stokes’s theorem and such will be vastly less confusing.