Curl (in space)

\( \newcommand{dv}[2]{\frac{d #1}{d #2}} \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \newcommand{ppdv}[3]{\frac{\partial^2 #1}{\partial #2\partial #3}} \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \newcommand{ppdv}[2]{\frac{\partial^2 #1}{\partial #2^2}} \)

My notes of the excellent lecture 30 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

The curl measures the value of the vector field to be conservative. For a velocity field, curl measures the rotation component of the motion.

Definition

Let \(\vec F\) be a vector field with components \(P(x,y,z)\), \(Q(x,y,z)\) and \(R(x,y,z)\), then $$ \begin{align*} \vec F &= \left\langle P,Q,R\right\rangle \\ &= P\hat\imath+Q\hat\jmath+R\hat k \end{align*} $$

Then $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align} \rm{curl}\left(\vec F\right) &= \left\langle \pdv{R}{y} – \pdv{Q}{z}\,,\, \pdv{P}{z}-\pdv{R}{x}\,,\, \pdv{Q}{x}-\pdv{P}{y} \right\rangle \nonumber \\[.5em] &= \left\langle R_y – Q_z\,,\, P_z-R_x\,,\, Q_x-P_y \right\rangle \nonumber \\ &= (R_y – Q_z)\hat\imath + (P_z-R_x)\hat\jmath + (Q_x-P_y)\hat k \label{eq:curl1} \end{align} $$

If \(\vec F\) is defined in a simply-connected region \(S\), and differentiable everywhere on \(S\), then $$ \shaded{ \vec F \text{ is conservative} \Leftrightarrow \rm{curl}\left(\vec F\right)=0 } \nonumber $$

The difference with curl in the plane, is that in space the curl is again a vector field, not a scalar.

Each of the terms has to be \(0\) for the field to be conservative.

\(\nabla\)-notation

We have seen

the symbolic \(\nabla\) “del” notation for the operator used for the gradient and divergence $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \nabla &= \left\langle\pdv{}{x}, \pdv{}{y}, \pdv{}{z}\right\rangle \\[0.5em] \nabla f &= \left\langle\pdv{f}{x}, \pdv{f}{y}, \pdv{f}{z}\right\rangle & \text{gradient} \\[0.5em] \nabla\cdot\vec F &= \left\langle\pdv{}{x}, \pdv{}{y}, \pdv{}{z}\right\rangle \cdot \left\langle P,Q,R\right\rangle \\ &= \pdv{P}{x} + \pdv{Q}{y} + \pdv{R}{z} & \text{divergence} \end{align*} $$

Let’s try to do the cross-product using the pseudo-determinant $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \nabla\times\vec F &= \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ \pdv{}{x} & \pdv{}{y} & \pdv{}{z} \\ P & Q & R \end{array} \right| \\ &= \hat\imath \left| \begin{array}{cc} \pdv{}{y} & \pdv{}{z} \\ Q & R \end{array} \right| – \hat\jmath \left| \begin{array}{cc} \pdv{}{x} & \pdv{}{z} \\ P & Q \end{array} \right| + \hat k \left| \begin{array}{cc} \pdv{}{x} & \pdv{}{y} \\ P & Q \end{array} \right| \\ &= \left( \pdv{R}{y} – \pdv{Q}{z} \right) \hat\imath – \left( \pdv{R}{x} – \pdv{P}{z} \right) \hat\jmath + \left( \pdv{Q}{x} – \pdv{P}{y} \right) \hat k \\ &= (R_y – Q_z)\hat\imath + (P_z-R_x)\hat\jmath + (Q_x-P_y)\hat k \end{align*} $$

This matches equation \(\eqref{eq:curl1}\). So, the best way to remember the formula for curl is using this \(\nabla\)-notation $$ \shaded{ \begin{align*} \rm{curl}\left(\vec F\right) &= \nabla\times\vec F \\ &= \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ \pdv{}{x} & \pdv{}{y} & \pdv{}{z} \\ P & Q & R \end{array} \right| \\ &= \left\langle R_y – Q_z, P_z-R_x, Q_x-P_y\right\rangle \end{align*} } \label{eq:curl2} $$

Geometrically

Curl measures the rotation component of a velocity field:

  • the direction corresponds to the axis of rotation, and
  • the magnitude corresponds to twice the angular velocity (\(\omega\)).

Examples

One

Let \(\vec v\) be a fluid that is rotating with angular velocity \(\omega\) around the \(z\)-axis

So $$ \vec v = \left\langle -\omega y, \omega x,0 \right\rangle \nonumber $$

The curl $$ \newcommand{pdv}[2]{\frac{\partial #1}{\partial #2}} \begin{align*} \nabla\times\vec v &= \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ \pdv{}{x} & \pdv{}{y} & \pdv{}{z} \\ -\omega y & \omega x & 0 \end{array} \right| \\ &= (0 – 0)\hat\imath – (0-0)\hat\jmath + (\omega-(-\omega)))\hat k \\ &= 2\omega\hat k = \left\langle 0,0,2\omega \right\rangle \end{align*} $$

The curl gives you the angular rotation (\(\times 2\)), and the axis of rotation (the vertical axis).

Two

Let \(\vec v\) be a fluid moving in a straight direction

So $$ \vec F = \left\langle a,b,c \right\rangle \nonumber $$

All the partial derivatives are \(0\), so the curl is \(0\).

Three

Let \(\vec v\) be a vector field that stretches things along the \(x\)-axis (expanding)

So $$ \vec F = \left\langle x,0,0 \right\rangle \nonumber $$

Again, the curl will be \(0\). But the \(\rm{div}(\vec F)=1\), because that measures stretching of a vector field.

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