Calculus &
Linear Algebra II
Chapter 45
45 Curl of a vector field
By the end of this section, you should be able to answer the following questions:
- How do you calculate the curl of a given vector field?
- What is the significance of curl?
- How do you test whether or not a given three dimensional vector field is conservative?
45.1 Calculating curl
If $(x, y, z)$ is a right handed Cartesian coordinate system and $\v(x, y, z) = v_1~\i+v_2~\j +v_3~\k$ is a differentiable vector field, then define
| $\displaystyle \curl (\v) $ | $\displaystyle = \nabla \times \v $ $\displaystyle = \left| \begin{array}{ccc} \i & \j & \k \\ \dfrac{\partial}{\partial x} &\dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ v_1 & v_2 & v_3 \end{array} \right| $ |
| $\displaystyle = \left(\dfrac{\partial v_3}{\partial y} - \dfrac{\partial v_2}{\partial z} \right)\i $ $\displaystyle - \left( \dfrac{\partial v_3}{\partial x}- \dfrac{\partial v_1}{\partial z} \right)\j $ $\displaystyle + \left(\dfrac{\partial v_2}{\partial x} - \dfrac{\partial v_1}{\partial y} \right)\k $ | |
| $\displaystyle = \left(\dfrac{\partial v_3}{\partial y} - \dfrac{\partial v_2}{\partial z} \right)\i $ $\displaystyle + \left( \dfrac{\partial v_1}{\partial z}- \dfrac{\partial v_3}{\partial x} \right)\j $ $\displaystyle + \left(\dfrac{\partial v_2}{\partial x} - \dfrac{\partial v_1}{\partial y} \right)\k $ |
Note that $\curl(\v)$ is a vector field.
45.1.1 Example: let $\v = yz^2~\i + zx^2~\j + xy^2~\k.$ Find $\curl(\v).$
| $\nabla \times \v$ | $\displaystyle = \left| \begin{array}{ccc} \i & \j & \k \\ \dfrac{\partial}{\partial x} &\dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ yz^2 & zx^2 & xy^2 \end{array} \right| $ |
| $\displaystyle = \left| \begin{array}{cc} \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z}\\ zx^2 & xy^2 \end{array} \right| \i$ $- \left| \begin{array}{cc} \dfrac{\partial}{\partial x} &\dfrac{\partial}{\partial z}\\ yz^2 & xy^2 \end{array} \right| \j $ $+ \left| \begin{array}{cc} \dfrac{\partial}{\partial x} &\dfrac{\partial}{\partial y}\\ yz^2 & zx^2 \end{array} \right| \k $ |
45.1.1 Example: let $\v = yz^2~\i + zx^2~\j + xy^2~\k.$ Find $\curl(\v).$
| $\nabla \times \v$ | $\displaystyle = \left| \begin{array}{cc} \dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z}\\ zx^2 & xy^2 \end{array} \right| \i - \left| \begin{array}{cc} \dfrac{\partial}{\partial x} &\dfrac{\partial}{\partial z}\\ yz^2 & xy^2 \end{array} \right| \j + \left| \begin{array}{cc} \dfrac{\partial}{\partial x} &\dfrac{\partial}{\partial y}\\ yz^2 & zx^2 \end{array} \right| \k $ |
| $\displaystyle = \left(\dfrac{\partial}{\partial y}\left(xy^2\right) - \dfrac{\partial }{\partial z} \left(zx^2\right) \right)\i $ $- \left(\dfrac{\partial}{\partial x}\left(xy^2\right) -\dfrac{\partial }{\partial z}\left(yz^2\right) \right)\j$ | |
| $\displaystyle \quad\qquad + \left(\dfrac{\partial }{\partial x}\left(zx^2\right) - \dfrac{\partial}{\partial y} \left(yz^2\right) \right)\k $ | |
| $\displaystyle = \left( 2xy - x^2 \right)\i+ \left(2yz-y^2 \right)\j + \left(2xz-z^2 \right)\k.$ |
45.2 Understanding curl
For the rotation of a rigid body about a fixed axis with angular velocity $\w,$ the velocity at a point $P,$ whose position vector is $\r,$ is given by $\v = \w \times \r.$
45.2 Understanding curl
For the rotation of a rigid body about a fixed axis with angular velocity $\w,$ the velocity at a point $P,$ whose position vector is $\r,$ is given by $\v = \w \times \r.$
45.2 Understanding curl
For the rotation of a rigid body about a fixed axis with angular velocity $\w,$ the velocity at a point $P,$ whose position vector is $\r,$ is given by $\v = \w \times \r.$
If we choose the axis of rotation to be the $z$-axis, then $\w = \omega \k$. Calculate $\curl(\v)$:
We have that $\r = x~\i + y~\j + z~\k$. Then
| $\v$ | $= \w \times \r$ $\displaystyle = \left| \begin{array}{ccc} \i & \j & \k \\ 0 & 0 & \omega \\ x & y & z \end{array} \right| $ |
| $= -\omega \left( y~\i - x~\j \right)$ |
45.2 Understanding curl
Then $\v = -\omega \left( y~\i - x~\j \right)$
On the other hand, we have that
| $\curl \v$ | $= \nabla \times \v$ $\displaystyle = \left| \begin{array}{ccc} \i & \j & \k \\ \displaystyle \frac{\partial }{\partial x} & \displaystyle \frac{\partial }{\partial y} & \displaystyle \frac{\partial }{\partial z}\\ -\omega y & \omega x & 0 \end{array} \right| $ |
| $=(0) ~\i - (0)~\j + (\omega+\omega)~\k$ | |
| $=2\omega ~\k$ |
This implies that the $\curl \v $ is proportional to the angular vector $\w$.
45.2 Understanding curl
This implies that the $\curl \v $ is proportional to the angular vector $\w$.
45.2 Understanding curl
$\curl \v \neq \mathbf 0$ implies that a paddle wheel in a fluid with velocity field $\v$ will rotate around its axis.
45.2 Understanding curl
In general, $\curl \v$ characterises the rotation of a vector field.
45.3 Conservative fields revisited
It turns out that the curl of a vector field is exactly what we need to generalise the result mentioned in Chapter 36.3 to three dimensions.
Show that if $\F$ is a conservative vector field, then $\curl \F = \mathbf 0$.
Proof: If $\F$ is conservative, then there exists $f$ such that \[ \F = \nabla f. \] That is \[ \F = \frac{\partial f}{\partial x}~\i+ \frac{\partial f}{\partial y}~\j+ \frac{\partial f}{\partial z}~\k. \]
45.3 Conservative fields revisited
Proof: Since $ \F = \frac{\partial f}{\partial x}~\i+ \frac{\partial f}{\partial y}~\j+ \frac{\partial f}{\partial z}~\k $, then
| $\curl \F$ | $\displaystyle = \left| \begin{array}{ccc} \i & \j & \k \\ \dfrac{\partial}{\partial x} &\dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ \dfrac{\partial f}{\partial x} & \dfrac{\partial f}{\partial y} & \dfrac{\partial f}{\partial z} \end{array} \right| $ |
| $\displaystyle = \left(\frac{\partial ^2 f}{\partial y \partial z} - \frac{\partial ^2 f}{\partial z \partial y} \right)\i \, $ $\displaystyle + \left(\frac{\partial ^2 f}{\partial z \partial x} - \frac{\partial ^2 f}{\partial x \partial z} \right)\j \,$ | |
| $\displaystyle+ \left(\frac{\partial ^2 f}{\partial x \partial y} - \frac{\partial ^2 f}{\partial y \partial x} \right)\k$ $ = \mathbf 0.$ |
45.3 Conservative fields revisited
Proof: Since $ \F = \frac{\partial f}{\partial x}~\i+ \frac{\partial f}{\partial y}~\j+ \frac{\partial f}{\partial z}~\k $, then
| $\curl \F$ | $\displaystyle = \left| \begin{array}{ccc} \i & \j & \k \\ \dfrac{\partial}{\partial x} &\dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ \dfrac{\partial f}{\partial x} & \dfrac{\partial f}{\partial y} & \dfrac{\partial f}{\partial z} \end{array} \right| =\mathbf 0.$ |
In other words, $\nabla \times \F = \mathbf 0.$
45.3 Conservative fields revisited
Indeed, the diagram on Chapter 36 that outlines our logic can be extended directly to the three dimensional case. The only difference is the condition which will serve as our test for conservative fields, namely $\curl \F = \mathbf 0$.
45.3 Conservative fields revisited
The proofs of the links in the diagram for the three dimensional case below are very similar to those used in the two dimensional case. The only detail that is significantly different is showing that \[ \text{if }\,\curl \F = \mathbf 0 \,\text{ then }\, \oint_C\F \pd d\r = 0. \]
45.3 Conservative fields revisited
The proofs of the links in the diagram for the three dimensional case below are very similar to those used in the two dimensional case. The only detail that is significantly different is showing that \[ \text{if }\,\curl \F = \mathbf 0 \,\text{ then }\, \oint_C\F \pd d\r = 0. \]
Note also that $\F$ must be a vector field defined everywhere in $\R^3$ with continuous partial derivatives.
45.3 Conservative fields revisited
Note also that $\F$ must be a vector field defined everywhere in $\R^3$ with continuous partial derivatives.
The proof of that part of the diagram requires a generalisation of Green's theorem known as Stokes' theorem, which we will investigate in the next section.
45.3 Conservative fields revisited
The main consequence of this diagram is that we have the following test for a conservative vector field in three dimensions:
A vector field $\F$ is conservative if and only if $\curl \F = \mathbf 0$.
45.3.1 Determine whether or not the vector field $\,\F = (1+yz)~\i+(1+xz)~\j+xy~\k\,$ is conservative.
Recall that $\F$ is conservative if and only if $\curl \F=\mathbf 0$. Then we compute
| $\curl \F$ | $\displaystyle = \;\left| \begin{array}{ccc} \i & \j & \k \\ \dfrac{\partial}{\partial x} &\dfrac{\partial}{\partial y} &\dfrac{\partial}{\partial z} \\ 1+yz & 1+ xz & xy \end{array} \right| $ |
| $\displaystyle = \; \left( x - x \right)\i \, $ $\displaystyle + \left(y -y \right)\j \,$ $\displaystyle +\left(z - z \right)\k$ | |
| $=\mathbf 0.$ |
Therefore $\F$ is conservative.