Why does the curl vector point along the axis of rotation?

Question

Thanks for reading.

Say we have a vector-field $F=P(x,y,z)\hat{i}+Q(x,y,z)\hat{j}+R(x,y,z)\hat{z}$, and we compute the curl of $F$ at some point $(x,y,z)$by calculating $\nabla \times F$ at that point.

Geometrically, I intuitively understand the components of the resulting vector.

1. The $\hat{i}$ component will be a vector representing the curl of the vector field on the $(y,z)$ plane.

2. The $\hat{j}$ component will be a vector representing the curl of the vector field on the $(x,z)$ plane.

3. The $\hat{k}$ component will be a vector representing the curl of the vector field on the $(x,y)$ plane.

Additionally, I geometrically understand each of the resulting "curls" independently, and their equations independently (for example, I understand why $(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\hat{k}$ gives us the curl on the $xy$ plane at that point).

However, sources online (Khan Academy) say that the overall resulting vector points "orthogonal to the overall direction of rotation at $(x,y,z)$".

And that...I don't intuitively understand.

That is, why is it that the fact that each of the resulting components of $\nabla \times F$ represents the curl on one of the planes at a point $(x,y,z)$ imply that the overall resulting vector points perpendicular to the overall rotational direction at that point (following the right-hand rule)?

Thanks!

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Edit:

Take a look at this post from Wikipedia:

Suppose the vector field describes the velocity field of a fluid flow (such as a large tank of liquid or gas) and a small ball is located within the fluid or gas (the centre of the ball being fixed at a certain point). If the ball has a rough surface, the fluid flowing past it will make it rotate. The rotation axis (oriented according to the right hand rule) points in the direction of the curl of the field at the centre of the ball, and the angular speed of the rotation is half the magnitude of the curl at this point.

And then....

The curl of the vector at any point is given by the rotation of an infinitesimal area in the $xy$-plane (for $z$-axis component of the curl), $zx$-plane (for $y$-axis component of the curl) and $yz$-plane (for $x$-axis component of the curl vector). This can be clearly seen in the examples below.

In a nutshell, I'm trying to connect the two interpretations. I understand the second one. And I understand the equations for each of the components separately.

What I don't yet understand is how it gives us an overall vector pointing along an axis of rotation...

Thank you!

Answer 1

Firstly, the "intuitive explanation" from Wikipedia is not quite right (certainly the referred book of Gibbs does not say what the Wikipedia text says), because the actual ball would follow Newton's mechanics and the curl will affect its acceleration, rather than velocity; if the ball started from rest, its direction of rotation will be the same as one given by curl, so that part is correct, but its speed will depend on other factors, so that is false. The Gibbs book is talking about a very different thing, the motion of a spherical region of gas/fluid itself, which will be translated along with $F$ but also distorted into ellipsoid AND rotated, with axis and speed of rotation controlled by $curl F$.

(A side note: In first approximation, the translational part of the motion is controlled by $F$; the distortion of shape and rotation is controlled by parts of $DF$- the matrix of derivatives of 3 components of $F$ is 3 coordinate directions. The distortion is controlled by the symmetric part of $DF$ (the axis being its eigendirections; note that $div F$ is the trace of this symmetric part, aka the sum of eigenvalues aka the sum of axis lengths) and controls the growth of volume of the ellipsoid), and the rotation is controlled by the anti-symmetric part of $DF$, which contains precisely the 3 components of $curl F$ as entries sitting away from the (zero) diagonal.)

Be that as it may, the main point here is that to first order rotation is produced by (solutions to) a differential equation $\vec{r}'=\vec{r}\times L$ where $L$ is the angular momentum of the rotation (vector pointing along rotation axis and with length proportional to rotation speed). You seem to say that you are comfortable that in 2D "coordinate plane" setting, say the $x-y$ plane, $L$ will be proportional to $(curl F)_3 \hat{k}$. In whatever sense you believe this to be true, one may say that the overall rotational motion due to $F$ is obtained as combination of motions $\vec{r}'= L_1 \times \vec{r}$, $\vec{r}'= L_2 \times \vec{r}$ and $\vec{r}'= L_3 \times \vec{r}$ with $L_3$ proportional to $(curl F)_3 \hat{k} $, $L_2$ proportional to $(curl F)_2 \hat{j} $ and $L_1$ to $(curl F)_2 \hat{i}$. Because these equations are all linear, the overall motion will solve the equation $\vec{r}'=\vec{r}\times (L_1+L_2+L_3)$, that is the angular momentum of resulting rotation is proportional to $curl F$.

In other words (and without invoking the language of differential equations), how does one describe a rotational motion? By giving an axis and an angular velocity. Then, for an axis with direction $\hat{l}$ passing through the origin, and angular velocity $\omega$ we define $\vec{l}=\omega \hat{l}$ (this is proportional to angular momentum) and the velocity of a point at location $\vec{r}$ is $\vec{l} \times \vec{r}$ (to sanity check/convince yourself this is true observe that this is orthogonal to $\vec{l}$ and $\vec{r}$, grows linearly with $\omega$ and with distance from $\vec{r}$ to the axis of rotation).

If I were to try to perform two rotations simultaneously, one around $\vec{l}_1=\omega_1 \hat{l}_1$ and another around $\vec{l}_2=\omega_2\hat{l}_2$ then velocity of any point $\vec{r}$ would be $$\vec{l}_1 \times \vec{r}+ \vec{l}_2 \times \vec{r}=(\vec{l}_1+\vec{l}_2) \times \vec{r}$$

That is, my motion is equivalent to a single rotation around an axis given by $\vec{l}=(\vec{l}_1+\vec{l}_2)$.

Answer 2

I'm thinking this is all due to the unstated assumption that the vector field is totally differentiable. It's analogous to how totally differentiable scalar fields have a direction of steepest ascent – the gradient – and how directional derivatives are projections of that gradient, even though they are computed along different lines. The essentially linear local nature of the field means all the vectors and 2D curls have to fit together nicely, in a way such that there exists a single direction perpendicular to which rotation is the greatest (what you're calling the overall direction of rotation), and all the others can be obtained by projection. Both in our case and in the scalar analogy the picture becomes inconsistent if we drop total differentiability.

Answer 3

Given a vector field ${\bf F}$ in ${\mathbb R}^3$ the associated field ${\rm curl}({\bf F})=:{\bf c}$ has a priori nothing to do with kinematic rotation. In the first place ${\bf c}$ measures the local nonconservativity of ${\bf F}$.

As measuring instrument for this nonconservativity at the point ${\bf p}$ we use a tiny circular wire $\gamma_{\bf n}$ of given radius $r\ll1$, centered at ${\bf p}$. Here ${\bf n}$ denotes the (correctly oriented) normal of the plane disc $B$ enclosed by $\gamma_{\bf n}$. The direction vector ${\bf n}$ can be chosen freely by the experimenter. The line integral $$\int_{\gamma_{\bf n}}{\bf F}\cdot d{\bf x}$$ usually depends on the chosen ${\bf n}$, and is $\equiv0$ when ${\bf F}$ is conservative. Now Stokes's theorem says that $$\int_{\gamma_{\bf n}}{\bf F}\cdot d{\bf x}=\int_B{\bf c}({\bf x})\cdot{\bf n}\>{\rm d}\omega\ ,$$ whereby ${\rm d}\omega$ is the (unsigned) area element on $B$. As $r\ll1$ we therefore can say that $$\int_{\gamma_{\bf n}}{\bf F}\cdot d{\bf x}\approx \pi r^2\ {\bf c}({\bf p})\cdot{\bf n} =\pi r^2\ {\rm curl}\,{\bf F}({\bf p})\cdot{\bf n}\ .$$ This shows that the value $\int_{\gamma_{\bf n}}{\bf F}\cdot d{\bf x}$ felt by our measuring instrument is largest when the normal ${\bf n}$ points in the direction of ${\rm curl}\,{\bf F}({\bf p})$.

Answer 4

Your first question is:

"Why does that mean that the curl vector's magnitude is the curl in a plane orthogonal to it, as you've said in your previous comment?"

If $\textbf{F}$ is a vector field, we know that the curl of $\textbf{F}$ at a point $p$ is defined as

\begin{equation} (\nabla \times \textbf{F}) \cdot \textbf{n} = \textrm{lim}_{A \rightarrow 0}\frac{\oint_{\partial S} \textbf{F} \cdot d\textbf{S}}{A} \; \; \; \textrm{(i)} \end{equation}

Here, $S$ is some closed area (containing $p$) with unit normal $\textbf{n}$ and $\partial S$ is its boundary curve.

We can rewrite equation (i) as

\begin{equation} |\nabla \times \textbf{F}| \cos(\theta) = \textrm{lim}_{A \rightarrow 0}\frac{\oint_{\partial S} \textbf{F} \cdot d\textbf{S}}{A} \; \; \; \textrm{(ii)} \end{equation}

Here, $\theta$ is the angle between $\nabla \times \textbf{F}$ and $\textbf{n}$.

Clearly, if $S$ is normal to $\nabla \times \textbf{F}$, we have that $\theta = 0$ (or $\pi$), so

\begin{equation} |\nabla \times \textbf{F}| = \pm \; \textrm{lim}_{A \rightarrow 0}\frac{\oint_{\partial S} \textbf{F} \cdot d\textbf{S}}{A} \; \; \; \textrm{(iii)} \end{equation}

In other words,

\begin{equation} |\nabla \times \textbf{F}| =| \; \textrm{lim}_{A \rightarrow 0}\frac{\oint_{\partial S} \textbf{F} \cdot d\textbf{S}}{A} | \; \; \; \textrm{(iv)} \end{equation}

That is, the curl vector's magnitude is the "curl" in a plane orthogonal to it. More precisely, the curl vector's magnitude is the magnitude of the rotation of $\textbf{F}$ around an infinitesimal loop in the plane orthogonal to the curl vector divided by the area of that loop.

Now, your second question is:

"Why is the curl as measured in that plane more than the curl around any other plane that contains $P$?"

Look at equation (ii) again. Since $|\nabla \times \textbf{F}|$ is fixed, it is clear that the RHS is maximized when $\theta = 0$ (or $\pi$). In other words, the "curl" as measured in the orthogonal plane is more than the "curl" around any other plane that contains $P$.