The components of the curl

Recall that one can visualize the curl of a three-dimensional vector field F = (F₁, F₂, F₃) by inserting a small sphere into a fluid with flow given by F, fixing the center of the sphere, and allowing the sphere to rotate in any direction. The direction of the vector curl F is given using the right hand rule, and the magnitude || curl F|| is given by the speed at which the sphere rotates. In the below animation, curl F is given by the green arrow.

Since curl F is a vector, we can talk about the components of the vector. If we let v = curl F, then we could write curl F in terms of components as

We can also use our rotating sphere to visualize the components of the curl. For example, to visualize v₁, the component of the curl in the x direction, we could place our sphere in the fluid as before, only with one important difference. Rather than allowing the sphere to rotate in any direction, we place the sphere on a rod that is parallel to the x-axis. The rod allows the sphere to rotate only around the axis of the rod. Using the right hand rule, we see that the "direction vector" corresponding to the rotation points along the rod, i.e., it is parallel to the x-axis. In fact, this rotation corresponds v₁i, where v₁ is the component of the curl in the x direction. (If, from the right hand rule, the direction of this rotation points in the positive-x direction, the speed of the sphere's rotation corresponds to v₁. If this direction points in the negative-x direction, then v₁ is negative and the speed of the sphere's rotation corresponds to | v₁|.)

From the above animation, we can see that the component of the curl in the x direction is positive (using the right hand rule). If the sphere were rotating in the other direction, the component of the curl in the x direction would be negative. Since curl F is not parallel to the x-axis, the sphere is rotating more slowly than it did when it was allowed to rotate freely. You can visualize that there is also "macroscopic circulation" in this same direction by rotating the figure so that the x-axis is coming straight out of the screen. Viewed from that axis, the vector field still appears to circulate.

With some knowledge about line integrals, you would be able to see why this microscopic circulation in the x direction is

$${\frac{{\partial F_3}}{{\partial y}}} {\frac{{\partial F_2}}{{\partial z}}}$$

Glancing at the green arrow in the original animation, we can see that it is perpendicular to the y-axis. So, for our particular example, we can see the component curl F in the y direction is zero. If we placed the sphere on a rod that is parallel to y axis, the fluid flow would not make the sphere rotate.

If you rotate the above figure so that the y-axis is coming straight out of the screen, you can see that the vector field no longer looks like it is rotating. (Though, remember, this isn't a true test of curl, which is "microscopic circulation.") We can still calculate the component of curl F parallel to the y-axis. Using the same calculation as for the x-component, we would conclude that the circulation parallel to the xz-plane is

$${\frac{{\partial F_1}}{{\partial z}}} {\frac{{\partial F_3}}{{\partial x}}}$$

In our example, v₂ is zero.

Lastly, we can visualize the component of curl F parallel to the z-axis by placing the sphere on a rod parallel to the z-axis. Looking at the green arrow in the first animation, we see this should be nonzero for our example, and the below animation verifies this. As above, we could calculate this component of the curl as

$${\frac{{\partial F_2}}{{\partial x}}} {\frac{{\partial F_1}}{{\partial y}}}$$

In summary, we can see that the curl F is a vector whose components indicate the "microscopic circulation" in the x, y, and z directions, respectively:

$$\left(\vphantom{\frac{\partial F_3}{\partial y}-\frac{\partial F_... ... x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} }\right. {\frac{{\partial F_3}}{{\partial y}}} {\frac{{\partial F_2}}{{\partial z}}} {\frac{{\partial F_1}}{{\partial z}}} {\frac{{\partial F_3}}{{\partial x}}} {\frac{{\partial F_2}}{{\partial x}}} {\frac{{\partial F_1}}{{\partial y}}} \left.\vphantom{\frac{\partial F_3}{\partial y}-\frac{\partial F_... ... x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} }\right)$$

It also turns out that there isn't anything special about the directions i, j, and k. For any unit vector u (i.e., ||u|| = 1), the projection curl F ^. u gives the "microscopic circulation" in the direction of u. This microscopic circulation would correspond to the rotation of a sphere around a rod parallel to u. (We'll need this idea to understand Stokes' Theorem.)

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