The Curl of a Vector Field

The Curl of a Vector Field

Recall from The Divergence of a Vector Field page that if $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ is a vector field on $\mathbb{R}^3$ and the partial derivatives $\frac{\partial P}{\partial x}$, $\frac{\partial Q}{\partial y}$ and $\frac{\partial R}{\partial z}$ all exist, then the divergence of $\mathbf{F}$ is a scalar field (function) given by the following formula:

(1)
\begin{align} \quad \mathrm{div} (\mathbf{F}) = \nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \end{align}

We will now look at another important function regarding vector fields known as the curl of $\mathbf{F}$.

 Definition: If $\mathbf{F} (x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ is a vector field on $\mathbb{R}^3$ and the appropriate partial derivatives of $P$, $Q$, and $R$ exist then the Curl of $\mathbf{F}$ is a vector field given by $\mathrm{curl} (\mathbf{F}) = \nabla \times \mathbf{F} = \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) \vec{i} + \left ( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right ) \vec{j} + \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \vec{k}$.

An important distinct to note is that $\mathrm{div} (\mathbf{F})$ produces a scalar field / function while $\mathrm{curl} (\mathbf{F})$ produces a vector field.

The notation $\mathrm{curl} (\mathbf{F}) = \nabla \times \mathbf{F}$ is important in remembering the formula for the curl of a vector field. Recall that $\nabla$ is the vector differential operator and $\nabla = \frac{\partial}{\partial x} \vec{i} + \frac{\partial}{\partial y} \vec{j} + \frac{\partial}{\partial z} \vec{k}$ and $\mathbf{F} = P \vec{i} + Q \vec{j} + R \vec{k}$ and so:

(2)
\begin{align} \quad \mathrm{curl} (\mathbf{F}) = \nabla \times \mathbf{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ P & Q & R \end{vmatrix} = \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) \vec{i} + \left ( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right ) \vec{j} + \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \vec{k} \end{align}

Geometrically, the curl of a vector field can be understood as follows. Suppose that the vector field $\mathbf{F}$ on $\mathbb{R}^3$ represents the velocity for fluid flow. An object at a point $P(x, y, z)$ will curl or rotate around the axis that points in the same direction as the curl of $\mathbf{F}$ at $P$, and the rate at which the object turns depends on the length of the curl vector.

If $\mathrm{curl}(\mathbf{F}) = 0$, then an important term defined below is used to describe $\mathbf{F}$ at $P$.

 Definition: Let $\mathbf{F}$ be a vector field on $\mathbb{R}^3$. If $\mathrm{curl} (\mathbf{F}) = 0$ at a point $P$ at $(x, y, z)$ then $\mathbf{F}$ is said to be Irrotational at $P$.
 Theorem 1: If $\mathbf{F} (x, y, z) = f(x) \vec{i} + g(y) \vec{j} + h(z) \vec{k}$ where $f$, $g$, and $h$ are differentiable functions then $\mathbf{F}$ is irrotational for all $(x, y, z) \in D(\mathbf{F})$.
• Proof: Let $\mathbf{F} (x, y, z) = f(x) \vec{i} + g(y) \vec{j} + h(z) \vec{k}$. Then the curl of $\mathbf{F}$ is:
(3)
\begin{align} \quad \quad \mathrm{curl} (\mathbf{F}) = \nabla \times \mathbf{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f(x) & g(y) & h(z) \end{vmatrix} = \left ( \frac{\partial}{\partial y} h(z) - \frac{\partial}{\partial z} g(y) \right ) \vec{i}+ \left ( \frac{\partial}{\partial z} f(x) - \frac{\partial}{\partial x} h(z) \right ) \vec{k} + \left ( \frac{\partial}{\partial x} g(y) - \frac{\partial}{\partial y} f(x) \right ) \vec{j} = 0\vec{i} + 0 \vec{j} + 0 \vec{k} \end{align}
• Therefore $\mathrm{curl} (\mathbf{F}) = \vec{0}$ so $\mathbf{F}$ is irrotational for all $(x, y, z) \in D(\mathbf{F})$. $\blacksquare$

Let's look at some examples of determining the curl of a vector field.

Example 1

Find the curl of the vector field $\mathbf{F}(x, y, z) = x^2y \vec{i} + xy^2z \vec{j} + xe^z \vec{k}$.

We will first calculate the necessary partial derivatives for the curl formula. We have that $\frac{\partial R}{\partial y} = 0$, $\frac{\partial Q}{\partial z} = xy^2$, $\frac{\partial P}{\partial z} = 0$, $\frac{\partial R}{\partial x} = e^z$, $\frac{\partial Q}{\partial x} = y^2z$, and $\frac{\partial P}{\partial y} = x^2$.

Applying our formula for the curl of $\mathbf{F}$ directly and we get that:

(4)
\begin{align} \quad \mathrm{curl} (\mathbf{F}) = (0 - xy^2) \vec{i} + (0 - e^z) \vec{j} + (y^2 - x^2) \vec{k} \\ \quad \mathrm{curl} (\mathbf{F}) = -xy^2 \vec{i} - e^z \vec{j} + (y^2z - x^2) \vec{k} \end{align}

Example 2

Fin the curl of the vector field $\mathbf{F}(x, y, z) = \sin (xz) \vec{i} + \cos(xyz) \vec{j} + xyz \vec{k}$.

We first calculate the necessary partial derivatives for the curl formula. We have that $\frac{\partial R}{\partial y} = xz$, $\frac{\partial Q}{\partial z} -xy \sin (xyz)$, $\frac{\partial P}{\partial z} = x \cos (xz)$, $\frac{\partial R}{\partial x} = yz$, $\frac{\partial Q}{\partial x} = -yz \sin (xyz)$, and $\frac{\partial P}{\partial y} = 0$.

Applying our formula for the curl of $\mathbf{F}$ directly and we get that:

(5)
\begin{align} \quad \mathrm{curl} (\mathbf{F}) = (xz + xy \sin (xyz)) \vec{i} + (x\cos(xz) - yz) \vec{j} + (-yz \sin (xyz)) \vec{k} \\ \end{align}