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, 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.


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.


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 $$ \shaded{ \begin{align*} \rm{curl}\left(\vec F\right) &= \nabla\times\vec F \\ &= \left| \begin{array}{ccc} \hat\imath & \hat\jmath & \hat k \\ \pdv{}{x} & \pdv{}{y} & \pdv{}{z} \\ P & Q & R \end{array} \right| \\ &= \left\langle R_y – Q_z, P_z-R_x, Q_x-P_y\right\rangle \end{align*} } \label{eq:curl2} $$


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 (\(\omega\)).



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).


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 \(0\).


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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