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If F = P i + Q j + R k is a vector field on $\mathbb{R}^{3}$, then the curl of F is defined by

$$\operatorname{Curl}(F) = \nabla \times F$$

where $\nabla$ is the differential operator.

Is there a geometric intuition behind the curl of a vector function F? I would like to see a clear geometric diagram explaining the concept of the curl of F, as my book only gives the definition.

Any help is appreciated!

Lavar Ball
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  • If you look very closely on the field to see if there is any "swirl" ( like microscopic rotation ). Like the derivative version of something that rotates so it's only valid very locally in the field, or very "tiny" rotations. That is why curl is sometimes also called "rot" (or rotation) – mathreadler Jul 10 '17 at 20:42
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    I would start with wiki. Examples section in particular is very visual. – Ennar Jul 10 '17 at 20:47
  • i remembered founding elementary geometric intuition of curl in Steven Strogatz's "The Joy of X" – serg_1 Jul 10 '17 at 21:13

2 Answers2

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As @G-Tony-Jacobs, I will mostly restrict the discussion to dimension $2$. Let us assume $\vec{F}$ is of the form: $$ \vec{F}(x,y, z) = \vec{F}(x,y) = F_1(x,y) \vec{i} + F_2(x,y) \vec{j} + 0 \vec{k}. $$ Then, we have $$ \operatorname{curl} \vec{F} = \bigg( 0,0, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \bigg). $$ We can define a circulation density as $$ \text{circulation density } = \lim \frac{\text{flux around a small rectangle}}{\text{area}}, $$ where the limit is taken as the area goes to zero. Flux around a rectangle

We have

$\bullet$ flux in the direction of $S_1 \simeq \big( \vec{F}(x+\Delta x, y) \cdot \vec{j} \big) \Delta y = F_2(x+\Delta x, y) \Delta y$

$\bullet$ flux in the direction of $S_2 \simeq - \big( \vec{F}(x, y+\Delta y) \cdot \vec{i} \big) \Delta x = - F_1(x, y+\Delta y)\Delta x$

$\bullet$ flux in the direction of $S_3 \simeq - \big( \vec{F}(x, y) \cdot \vec{j} \big) \Delta y = - F_2(x, y)\Delta y$

$\bullet$ flux in the direction of $S_4 \simeq \big( \vec{F}(x, y) \cdot \vec{i} \big) \Delta x = F_1(x, y) \Delta x$

Thus, $$ \begin{split} \text{circulation density at } (x,y) & = \lim_{\Delta x, \Delta y \to 0} \frac{F_2(x+\Delta x, y) \Delta y - F_1(x, y+\Delta y)\Delta x - F_2(x, y)\Delta y + F_1(x, y) \Delta x}{\Delta x \Delta y} \\ & = \lim_{\Delta x \to 0} \frac{F_2(x+\Delta x, y) - F_2(x, y)}{\Delta x} + \lim_{\Delta y \to 0} \frac{F_1(x, y+\Delta y) - F_1(x, y)}{\Delta y} \\ & = \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}, \end{split} $$

More generally (and much in the same way), we can think that, for any vector field $\vec{F}$:

$\bullet$ The direction of $\operatorname{curl} \vec{F}$ is the rotation axis and the circulation obeys the right hand rule.

$\bullet$ A vector field is irrotational (borrowed terminology from fluid dynamics) when $\operatorname{curl} \vec{F} = \vec{0}$.

Diego Marcon
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I have a good picture for a 2-dimensional vector field, which can be adapted for 3 dimensions. For the curl at a point in $\mathbb{R}^2$, I imagine placing a little flywheel at that point, so that it can be made to spin by the vectors around it. If you have a field given by, say $F(x,y) = \langle 0,x\rangle$ then a flywheel sitting at $(1,0)$ would have a little more upward push on its right side than on its left side, so it would rotate counterclockwise. On the other hand, with $F(x,y)=\langle 0,1\rangle$, the forces would be equal on either side, so the wheel would remain stationary.

Now, in $\mathbb{R}^3$, imagine locating that flywheel somehow in space so it is free to move itself to whatever orientation results in the fastest spin. Alternatively, let the flywheel be a ping pong ball, staying in its location but free to spin, and let the vector field be wind. Which way would the ping pong ball spin, and how quickly?

G Tony Jacobs
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