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I was trying to understand why curl measures a vector field's tendency of rotation. Two examples from physics seem to answer my question:

  1. Curl of the velocity field is twice the angular velocity
  2. Curl of the force field is the torque.

But I can only prove the first one when the velocity field describes a uniform circular motion. How can I show that the two examples are true in general to show that curl is really the measure of rotation.

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    $\begingroup$ Try computing the line integral of $F$ on a small circular path centered at the point $x$ in the plane perpendicular to $(\nabla \times F)(x)$. $\endgroup$
    – Ian
    Commented Jan 17, 2016 at 20:04

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Please note that this is a conceptual, and NOT a technical, approach to answering your question, as your question itself seemed to me to be conceptual in nature.

First, recall that the curl is a cross product, and is hence a vector itself. Since the $\nabla$ operator is not itself a vector quantity, but rather is an operator on vector functions tracking the vector changes in your field, the cross product, which must be perpendicular to both components, is thus perpendicular to the plane containing the rotation of the rigid, "small" area in your vector field which you are examining. This is why the curl is normal or perpendicular to the action, so to speak, and why we have to bring in the z-axis when discussing fields in the xy-plane.

Remember, we are dealing with very "small" quantities here, since we are involving differentials. So, imagine a rigid object at a particular point of interest in your field - for instance, a small piece of driftwood flowing in your field (rather than a formation of airplanes, which could itself stretch or tighten). Imagine standing on the center of mass of this piece of driftwood you have isolated in your mind, and imagine also that you can not see the nature of the vector field responsible for the velocity (or force) driving your piece of driftwood.

With this setup, we can attack your questions conceptually. For number 1, if, for instance, your driftwood was in a river with straight-line flow of equal magnitude across your board, you will not be able to distinguish your motion from remaining still (since you cannot see the nature of what is moving you, by the above assumption). However, if the velocity field was stronger just to your right than to your left, you would experience the driftwood spinning around you in a counterclockwise way; likewise, if the velocity had the same magnitude across your board (no shear, in other words), but your board was in a counterclockwise circulation, your right hand side would appear to be moving faster than your left hand side, meaning you would also view a relative counterclockwise motion.

The velocity vector, recall, can be decomposed relative to any point to a vector component parallel to motion directly toward or away from that point (linear) and a component motion perpendicular to linear motion away or toward that point (rotational). Likewise, the only component of your force field which will tend to initiate rotation on a rigid object fixed on one end at a point (eg. a restaurant swinging door) will be the component perpendicular to the object. The displacement of action relative to the fixed point is important - pure rotation acting on the center of mass and nowhere else will "spin in place," just as there will be no torque if there is only a perpendicular component of force at the fixed point of the axis (pushing the restaurant door at the hinges won't open it even if you're perpendicular to the rest of the door).

Regarding integration, imagine a small collection of connected rotating gears with a piece of tape around its edge (or surface, if you mean 3-D). The rotation of the tape around the edge, i.e. the total sum of the of the velocity oriented in the direction of the border's components, corresponds to the sum of the rotational (oriented) speeds inside of the area, i.e. the net tendency of the velocity field to rotate around centers of masses of small area (or volume) components within the border.

While rough and mathematically inexact, we've really described Stokes Theorem here:

$$\iint\limits_S {(\nabla \times V)\cdot dS}=\int\limits_{\delta S} {(V \cdot dr)}$$

Again in basic terms, the left-hand side is a measurement of the summation across an area (or volume) of the magnitude of the flow perpendicular to the components that are linear relative to the center of mass of "small" sections in a partition of this area (and hence captures ONLY the rotational component of the field over this area). The right hand side is the net sum of the component of velocity in the direction of the boundary of the area (or volume), which results in how the entire area (or volume) rotates within the field.

Hence the problem you ask will reduce to your solution for number 1 that you have already obtained. Since curl is set up to "ignore" relative linear components of your flow, you'll have that its integral only pays attention to the parts of your flow that "rotate" anyway.

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