Vector Field | Sahithyan's Notes

A field that assigns a vector to each point in its domain $D \subseteq \mathbb{R}^n$ ,

\boldsymbol{F}(x,y,z) = f(x,y,z)\hat{i} + g(x,y,z)\hat{j} + h(x,y,z)\hat{k}

Here $f$ , $g$ , $h$ are real-valued component functions of $\boldsymbol{F}$ on $D$ .

$\boldsymbol{F}$ is continuous iff each component function is continuous. $\boldsymbol{F}$ is differentiable iff each component function is differentiable.

Inverse-Square Field

Let $\boldsymbol{r} \ne \boldsymbol{0}$ be the position vector of a point and $c$ a constant. A vector field of the below form is an inverse-square field.

\boldsymbol{F}(\boldsymbol{r}) = \frac{c}{\lVert \boldsymbol{r} \rVert^3}\boldsymbol{r}

For example, the electric field given by Coulomb’s law is an inverse-square field

Divergence

Let $\boldsymbol{F}(x,y,z) = f\hat{i} + g\hat{j} + h\hat{k}$ . The divergence of $\boldsymbol{F}$ is

\text{div}\,\boldsymbol{F} = \nabla \cdot \boldsymbol{F} = \frac{\partial f}{\partial x} + \frac{\partial g}{\partial y} + \frac{\partial h}{\partial z}

$\text{div}\,\boldsymbol{F}$ is a scalar field.

Curl

Let $\boldsymbol{F}(x,y,z) = f\hat{i} + g\hat{j} + h\hat{k}$ . The curl of $\boldsymbol{F}$ is

\text{curl}\,\boldsymbol{F} = \nabla \times \boldsymbol{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[4pt] \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\[4pt] f & g & h \end{vmatrix}

$\text{curl}\,\boldsymbol{F}$ is a vector field.

Worked Examples

Divergence and Curl

Find $\text{div}\,\boldsymbol{F}$ and $\text{curl}\,\boldsymbol{F}$ for $\boldsymbol{F} = xy\,\hat{i} + yz\,\hat{j} + zx\,\hat{k}$ .

Components are $f = xy$ , $g = yz$ , $h = zx$ .

Divergence:

\text{div}\,\boldsymbol{F} = \frac{\partial (xy)}{\partial x} + \frac{\partial (yz)}{\partial y} + \frac{\partial (zx)}{\partial z} = y + z + x

Curl:

\text{curl}\,\boldsymbol{F} = \left(\frac{\partial h}{\partial y} - \frac{\partial g}{\partial z}\right)\hat{i} + \left(\frac{\partial f}{\partial z} - \frac{\partial h}{\partial x}\right)\hat{j} + \left(\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}\right)\hat{k}

= (0 - y)\,\hat{i} + (0 - z)\,\hat{j} + (0 - x)\,\hat{k} = -y\,\hat{i} - z\,\hat{j} - x\,\hat{k}

Curl of a Rotation Field

Find $\text{curl}\,\boldsymbol{F}$ for $\boldsymbol{F} = -y\,\hat{i} + x\,\hat{j}$ .

Components are $f = -y$ , $g = x$ , $h = 0$ .

\text{curl}\,\boldsymbol{F} = (0 - 0)\,\hat{i} + (0 - 0)\,\hat{j} + \left(\frac{\partial x}{\partial x} - \frac{\partial (-y)}{\partial y}\right)\hat{k} = 2\hat{k}

Curl of a Gradient Field

Find $\text{curl}\,\boldsymbol{F}$ for $\boldsymbol{F} = \nabla (xyz)$ .

\boldsymbol{F} = yz\,\hat{i} + xz\,\hat{j} + xy\,\hat{k}

\text{curl}\,\boldsymbol{F} = (x - x)\,\hat{i} + (y - y)\,\hat{j} + (z - z)\,\hat{k} = \boldsymbol{0}

Divergence of an Inverse-Square Field

Show that $\text{div}\,\boldsymbol{F} = 0$ for $\boldsymbol{F} = \dfrac{c}{r^3}\boldsymbol{r}$ , where $r = \lVert \boldsymbol{r} \rVert \ne 0$ .

The components are $f = \dfrac{cx}{r^3}$ , $g = \dfrac{cy}{r^3}$ , $h = \dfrac{cz}{r^3}$ .

Since $\dfrac{\partial r}{\partial x} = \dfrac{x}{r}$ :

\frac{\partial f}{\partial x} = c\left(\frac{1}{r^3} - \frac{3x^2}{r^5}\right)

By symmetry:

\frac{\partial g}{\partial y} = c\left(\frac{1}{r^3} - \frac{3y^2}{r^5}\right), \quad \frac{\partial h}{\partial z} = c\left(\frac{1}{r^3} - \frac{3z^2}{r^5}\right)

Sum:

\text{div}\,\boldsymbol{F} = c\left(\frac{3}{r^3} - \frac{3(x^2 + y^2 + z^2)}{r^5}\right) = c\left(\frac{3}{r^3} - \frac{3}{r^3}\right) = 0