Like 2-dimensional curl, one way to visualize 3-dimensional curl is to say that it measures how a paddle-wheel placed in a fluid spins. However, this idea is more complicated in \(\R^3\text{.}\)
To see why, look at the vector field below; the arrows are colored by length.
Figure7.1
Example7.2
Click the “Show paddlewheel” checkbox, and you'll see a paddle-wheel with its stick pointing in the direction \(\langle 0, 1, 0 \rangle\text{.}\) (We should really be thinking about an incredibly tiny paddle-wheel, but we've made it fairly large so that you can see it.) Imagine that you held the stick steady and let the blades be pushed by the fluid represented by the vector field. Would the paddle-wheel spin? Rotate your view of the vector field and paddle-wheel so that you get a clear view that helps you decide.
As you should see, the fluid moves in a direction tangent to the blades, so it doesn't push on the blades at all. Therefore, the paddle-wheel wouldn't spin.
Example7.3
Change the direction of the stick to \(\langle 1, 0, 0 \rangle\) (in the applet, enter this as (1, 0, 0)). Now, would the paddlewheel spin?
Now, you should see that the paddle-wheel would spin, because the fluid at higher \(z\)-values pushes against the blades harder. If you orient the picture so the stick is pointing at you, the paddle-wheel would spin clockwise. (And if you orient the picture so the stick is pointing away from you, the paddle-wheel would spin counter-clockwise. So, when we use the words “clockwise” and “counter-clockwise” in \(\R^3\text{,}\) we must also specify which direction we're looking in!)
So, we can see that how the paddle-wheel would spin depends not only on where it is, but also on on how it's oriented. Therefore, we can't hope to capture the spinning of the paddle-wheel at a point with a single number; that's why the 3-dimensional curl is a vector rather than a scalar like the 2-dimensional curl.
Here's the definition of the (3-dimensional) curl.
Definition7.4
If \(\vec{F}\) is a vector field on \(\R^3\text{,}\) the curl of \(\vec{F}\) (written \(\curl \vec{F}\)) is the vector field with the property that, if you put a paddle-wheel in a fluid of velocity \(\vec{F}\) with its stick pointing in a direction \(\vec{u}\) (where \(\vec{u}\) is a unit vector), the scalar \((\curl \vec{F}) \cdot \vec{u}\) will describe how quickly the paddle-wheel spins, with positive values indicating that the wheel is spinning counter-clockwise when viewed from the end of the stick.
This description looks incredibly complicated when you first encounter it, but it actually makes it very easy to write down a formula for the curl. Let's say we have a vector field \(\vec{F} = \langle P, Q, R \rangle\) and we put the paddle-wheel at a point with its stick pointing in the direction \(\vec{u} = \langle 0, 0, 1 \rangle\text{:}\)
How will it spin? Because of the direction the stick is pointing, only the components \(P\) and \(Q\) of the vector field affect how it spins, so this is really a 2-dimensional question. In fact, if we look from the end of the stick down at the part of vector field that affects the paddle-wheel, it's like we're looking at the vector field \(\langle P, Q \rangle\) on \(\R^2\text{,}\) so the paddle-wheel's spin is described exactly by the 2-dimensional curl \(Q_x - P_y\text{.}\)
Just the vectors in the same plane as the paddle-wheel
View from the end of the stick
That tells us that \(\curl \vec{F} \cdot \langle 0, 0, 1 \rangle = Q_x - P_y\text{,}\) which means the \(z\)-component of \(\curl \vec{F}\) must be \(Q_x - P_y\text{.}\) So, we know \(\curl \vec{F} = \langle \text{something}, \text{something}, Q_x - P_y
\rangle\text{.}\)
We can reason similarly about what happens if the stick instead points in the direction \(\langle 1, 0, 0 \rangle\) or \(\langle 0, 1, 0 \rangle\text{.}\) Putting these together, we arrive at the formula for the curl.
If \(\vec{F} = \langle P, Q, R \rangle\) is a vector field on \(\R^3\text{,}\) \(\curl \vec{F} = \langle R_y - Q_z, P_z - R_x, Q_x - P_y \rangle\text{.}\)
