Gradient, Divergence and Curl

Section 4.1 Gradient, Divergence and Curl

“Gradient, divergence and curl”, commonly called “grad, div and curl”, refer to a very widely used family of differential operators and related notations that we’ll get to shortly. We will later see that each has a “physical” significance. But even if they were only shorthand

Good shorthand is not only more brief, but also aids understanding “of the forest by hiding the trees”.

, they would be worth using.

For example, one of Maxwell’s equations (relating the electric field $\vE$ and the magnetic field $\vB$) written without the use of this notation is

\begin{align*} &\Big(\frac{\partial E_3}{\partial y} -\frac{\partial E_2}{\partial z}\Big)\hi -\Big(\frac{\partial E_3}{\partial x} -\frac{\partial E_1}{\partial z}\Big)\hj +\Big(\frac{\partial E_2}{\partial x} -\frac{\partial E_1}{\partial y}\Big)\hk\\ &\hskip2.5in=-\frac{1}{c}\Big(\frac{\partial B_1}{\partial t} \hi +\frac{\partial B_2}{\partial t} \hj +\frac{\partial B_3}{\partial t} \hk\Big) \end{align*}

The same equation written using this notation is

\begin{equation*} \vnabla\times\vE = -\frac{1}{c}\frac{\partial \vB}{\partial t} \end{equation*}

The shortest way to write (and easiest way to remember) gradient, divergence and curl uses the symbol “$\vnabla$” which is a differential operator like $\frac{\partial }{\partial x}\text{.}$ It is defined by

\begin{equation*} \vnabla = \hi\frac{\partial }{\partial x} +\hj\frac{\partial }{\partial y} +\hk\frac{\partial }{\partial z} \end{equation*}

and is called “del” or “nabla”. Here are the definitions.

Definition 4.1.1.

The gradient of a scalar-valued function $f(x,y,z)$ is the vector field
\begin{equation*} \text{grad}\,f=\vnabla f = \frac{\partial f}{\partial x}\hi +\frac{\partial f}{\partial y}\hj +\frac{\partial f}{\partial z} \hk \end{equation*}
Note that the input, $f\text{,}$ for the gradient is a scalar-valued function, while the output,$\vnabla f\text{,}$ is a vector-valued function.
The divergence of a vector field $\vF(x,y,z)$ is the scalar-valued function
\begin{equation*} \text{div}\,\vF=\vnabla\cdot\vF = \frac{\partial F_1}{\partial x} +\frac{\partial F_2}{\partial y} +\frac{\partial F_3}{\partial z} \end{equation*}
Note that the input, $\vF\text{,}$ for the divergence is a vector-valued function, while the output, $\vnabla\cdot\vF\text{,}$ is a scalar-valued function.
The curl of a vector field $\vF(x,y,z)$ is the vector field
\begin{equation*} \text{curl}\,\vF= \vnabla\times\vF = \Big(\frac{\partial F_3}{\partial y} -\frac{\partial F_2}{\partial z}\Big)\hi -\Big(\frac{\partial F_3}{\partial x} -\frac{\partial F_1}{\partial z}\Big)\hj +\Big(\frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y}\Big)\hk \end{equation*}
Note that the input, $\vF\text{,}$ for the curl is a vector-valued function, and the output, $\vnabla\times\vF\text{,}$ is a again a vector-valued function.
The Laplacian
²
Pierre-Simon Laplace (1749–1827) was a French mathematician and astronomer. He is also the Laplace of Laplace’s equation, the Laplace transform, and the Laplace-Bayes estimator. He was Napoleon’s examiner when Napoleon attended the Ecole Militaire in Paris.
of a scalar-valued function $f(x,y,z)$ is the scalar-valued function
\begin{equation*} \Delta f= \vnabla^2 f =\vnabla\cdot\vnabla f = \frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2} +\frac{\partial^2 f}{\partial z^2} \end{equation*}
The Laplacian of a vector field $\vF(x,y,z)$ is the vector field
\begin{equation*} \Delta\vF= \vnabla^2\vF =\vnabla\cdot\vnabla \vF = \frac{\partial^2 \vF}{\partial x^2} +\frac{\partial^2 \vF}{\partial y^2} +\frac{\partial^2 \vF}{\partial z^2} \end{equation*}
Note that the Laplacian maps either a scalar-valued function to a scalar-valued function, or a vector-valued function to a vector-valued function.

The gradient, divergence and Laplacian all have obvious generalizations to dimensions other than three. That is not the case for the curl. It does have a, far from obvious, generalization, which uses differential forms. Differential forms are well beyond our scope, but are introduced in the optional §4.7.

Example 4.1.2.

As an example of an application in which both the divergence and curl appear, we have Maxwell’s equations

To be picky, these are Maxwell’s equations in the absence of a material medium and in Gaussian units.

⁴

One important consequence of Maxwell’s equations is that electromagnetic radiation, like light, propagate at the speed of light.

⁵

James Clerk Maxwell (1831–1879) was a Scottish mathematical physicist. In a poll of prominent physicists, Maxwell was voted the third greatest physicist of all time. Only Newton and Einstein beat him.

, which form the foundation of classical electromagnetism.

\begin{align*} \vnabla\cdot\vE &= 4\pi\rho\\ \vnabla\cdot\vB &= 0\\ \vnabla\times\vE +\frac{1}{c}\frac{\partial\vB}{\partial t}&=0\\ \vnabla\times\vB -\frac{1}{c}\frac{\partial\vE}{\partial t}&=\frac{4\pi}{c}\vJ \end{align*}

Here $\vE$ is the electric field, $\vB$ is the magnetic field, $\rho$ is the charge density, $\vJ$ is the current density and $c$ is the speed of light.

Subsection 4.1.1 Vector Identities

Two computationally extremely important properties of the derivative $\diff{\ }{x}$ are linearity and the product rule.

\begin{align*} \diff{\ }{x}\big(af(x)+bg(x)\big) &=a\diff{f}{x}(x)+b\diff{g}{x}(x)\\ \diff{\ }{x}\big(f(x)\,g(x)\big) &=g(x)\,\diff{f}{x}(x)+f(x)\,\diff{g}{x}(x) \end{align*}

Gradient, divergence and curl also have properties like these, which indeed stem (often easily) from them. First, here are the statements of a bunch of them. (A memory aid and proofs will come later.) In fact, here are a very large number of them. Many are included just for completeness. Only a relatively small number are used a lot. They are in red.

Theorem 4.1.3. Gradient Identities.

$\displaystyle \color{red}{\vnabla(f+g)=\vnabla f+\vnabla g}$
$\color{red}{\vnabla(cf)=c\,\vnabla f}\text{,}$ for any constant $c$
$\displaystyle \color{red}{\vnabla(fg)=(\vnabla f)g+ f(\vnabla g)}$
$\vnabla(f/g)=\big(g\,\vnabla f-f\,\vnabla g\big)/g^2$ at points $\vx$ where $g(\vx)\ne 0\text{.}$
$\vnabla(\vF\cdot\vG)=\vF\times(\vnabla\times\vG)- (\vnabla\times\vF)\times\vG+(\vG\cdot\vnabla)\vF+(\vF\cdot\vnabla)\vG$

Here
⁶
This is really the only definition that makes sense. For example $\vG\cdot(\vnabla\vF)$ does not make sense because you can’t take the gradient of a vector-valued function.

\begin{equation*} (\vG\cdot\vnabla)\vF =\vG_1\frac{\partial\vF}{\partial x} +\vG_2\frac{\partial\vF}{\partial y} +\vG_3\frac{\partial\vF}{\partial z} \end{equation*}

Theorem 4.1.4. Divergence Identities.

$\displaystyle \color{red}{\vnabla\cdot(\vF+\vG)=\vnabla\cdot\vF+\vnabla \cdot\vG}$
$\color{red}{\vnabla\cdot(c\vF)=c\,\vnabla\cdot\vF}\text{,}$ for any constant $c$
$\displaystyle \color{red}{\vnabla\cdot(f\vF)=(\vnabla f)\cdot\vF+ f\,\vnabla\cdot\vF}$
$\displaystyle \vnabla\cdot(\vF\times\vG)=(\vnabla\times\vF)\cdot\vG - \vF\cdot(\vnabla\times\vG)$

Theorem 4.1.5. Curl Identities.

$\displaystyle \color{red}{\vnabla\times(\vF+\vG)=\vnabla\times\vF+\vnabla \times\vG} $
$\color{red}{\vnabla\times(c\vF)=c\,\vnabla\times\vF} \text{,}$ for any constant $c$
$\displaystyle \color{red}{\vnabla\times(f\vF)=(\vnabla f)\times\vF+f\,\vnabla\times\vF} $
$\vnabla\times(\vF\times\vG) =\vF(\vnabla\cdot\vG)-(\vnabla\cdot\vF)\vG +(\vG\cdot\vnabla)\vF-(\vF\cdot\vnabla)\vG$

Here

\begin{equation*} (\vG\cdot\vnabla)\vF =\vG_1\frac{\partial\vF}{\partial x} +\vG_2\frac{\partial\vF}{\partial y} +\vG_3\frac{\partial\vF}{\partial z} \end{equation*}

Theorem 4.1.6. Laplacian Identities.

$\displaystyle \color{red}{ \vnabla^2(f+g)=\vnabla^2 f+\vnabla^2 g }$
$\color{red}{\vnabla^2(cf)=c\,\vnabla^2 f}\text{,}$ for any constant $c$
$\displaystyle \vnabla^2(fg)=f\,\vnabla^2 g+2\vnabla f\cdot\vnabla g+g\,\vnabla^2 f$

Theorem 4.1.7. Degree Two Identities.

$\color{red}{\vnabla\cdot(\vnabla\times\vF)=0 \qquad}$ (divergence of curl)
$\color{red}{\vnabla\times(\vnabla f)=0 \qquad}$ (curl of gradient)
$\displaystyle \vnabla\cdot\big(f\{\vnabla g\times\vnabla h\}\big) =\vnabla f\cdot(\vnabla g\times\vnabla h)$
$\displaystyle \vnabla\cdot(f\vnabla g-g\vnabla f)=f\,\vnabla^2g-g\,\vnabla^2f$
$\vnabla\times(\vnabla\times\vF)=\vnabla(\vnabla\cdot\vF)-\vnabla^2\vF \qquad$ (curl of curl)

Memory Aid. Most of the vector identities (in fact all of them except Theorem 4.1.3.e, Theorem 4.1.5.d and Theorem 4.1.7) are really easy to guess. Just combine the conventional linearity and product rules with the facts that

if the left hand side is a vector (scalar), then the right hand side must also be a vector (scalar) and
the only valid products of two vectors are the dot and cross products and
the product of a scalar with either a scalar or a vector cannot be either a dot or cross product and
$\vA\times\vB = - \vB\times\vA\text{.}$ (The cross product is antisymmetric.)

For example, consider Theorem 4.1.4.c, which says $\vnabla\cdot(f\vF)=(\vnabla f)\cdot\vF+ f\,\vnabla\cdot\vF\text{.}$

The left hand side, $\vnabla\cdot(f\vF)\text{,}$ is a scalar, so the right hand side must also be a scalar.
The left hand side, $\vnabla\cdot(f\vF)\text{,}$ is a derivative of the product of $f$ and $\vF\text{,}$ so, mimicking the product rule, the right hand side will be a sum of two terms, one with $\vF$ multiplying a derivative of $f\text{,}$ and one with $f$ multiplying a derivative of $\vF\text{.}$
The derivative acting on $f$ must be $\vnabla f\text{,}$ because $\vnabla\cdot f$ and $\vnabla\times f$ are not well-defined. To end up with a scalar, rather than a vector, we must take the dot product of $\vnabla f$ and $\vF\text{.}$ So that term is $(\vnabla f)\cdot\vF\text{.}$
The derivative acting on $\vF$ must be either $\vnabla\cdot\vF$ or $\vnabla\times\vF\text{.}$ We also need to multiply by the scalar $f$ and end up with a scalar. So the derivative must be a scalar, i.e. $\vnabla\cdot\vF$ and that term is $f\{\vnabla\cdot\vF\}\text{.}$
Our final guess is $\vnabla\cdot(f\vF)=(\vnabla f)\cdot\vF+ f\,\vnabla\cdot\vF\text{,}$ which, thankfully, is correct.

Proof of Theorems 4.1.3, 4.1.4, 4.1.5, 4.1.6 and 4.1.7.

All of the proofs (except for those of Theorem 4.1.7.c,d, which we will return to later) consist of

writing out the definition of the left hand side and
writing out the definition of the right hand side and
observing (possibly after a little manipulation) that they are the same.

For Theorem 4.1.3.a,b, Theorem 4.1.4.a,b, Theorem 4.1.5.a,b and Theorem 4.1.6.a,b, the computation is trivial — one line per identity, if one uses some efficient notation. Rename the coordinates $x,y,z$ to $x_1,x_2,x_3$ and the standard unit basis vectors $\hi\text{,}$ $\hj\text{,}$ $\hk$ to $\hi_1\text{,}$ $\hi_2\text{,}$ $\hi_3\text{.}$ Then $\vnabla = \sum_{n=1}^3\hi_n\frac{\partial\ }{\partial x_n}$ and the proof of, for example, Theorem 4.1.4.a is

\begin{align*} \vnabla\cdot(\vF+\vG) &=\sum_{n=1}^3\frac{\partial\ }{\partial x_n} \hi_n\cdot(\vF+\vG)\\ &=\sum_{n=1}^3\frac{\partial\ }{\partial x_n} \hi_n\cdot\vF +\sum_{n=1}^3\frac{\partial\ }{\partial x_n} \hi_n\cdot\vG =\vnabla\cdot\vF+\vnabla \cdot\vG \end{align*}

For Theorem 4.1.3.c,d, Theorem 4.1.4.c, Theorem 4.1.5.c and Theorem 4.1.6.c, the computation is easy — a few lines per identity. For example, the proof of Theorem 4.1.5.c is

\begin{align*} \vnabla\times(f\vF) &=\sum_{n=1}^3\frac{\partial }{\partial x_n}\hi_n\times(f \vF) =\sum_{n=1}^3\frac{\partial }{\partial x_n}\big(f\, \{\hi_n\times\vF\}\big)\\ &=\sum_{n=1}^3\frac{\partial f}{\partial x_n}\hi_n\times\vF +f\sum_{n=1}^3\frac{\partial }{\partial x_n}\hi_n\times\vF\\ &=(\vnabla f)\times\vF + f\,\vnabla\times\vF \end{align*}

In the second line we used Theorem 1.1.3.b. The similar verification of Theorems 4.1.3.c,d, 4.1.4.c, and 4.1.6.c are left as exercises. The latter two are parts (a) and (c) of Question 3 in Section 4.1 of the CLP-4 problembook.

For Theorem 4.1.4.d, the computation is also easy if one uses the fact that

\begin{equation*} \va\cdot(\vb\times\vc)=(\va\times\vb)\cdot\vc \end{equation*}

which is Lemma 4.1.8.a below. The verification of Theorem 4.1.4.d is part (b) of Question 3 in Section 4.1 of the CLP-4 problembook.

That leaves the proofs of Theorem 4.1.3.e, Theorem 4.1.5.d, Theorem 4.1.7.a,b,c,d,e, which we write out explicitly.

Theorem 4.1.3.e: First write out the left hand side as

\begin{gather*} \vnabla(\vF\cdot\vG) =\sum_{n=1}^3\hi_n\frac{\partial }{\partial x_n}(\vF\cdot\vG) =\sum_{n=1}^3\hi_n\Big(\frac{\partial\vF}{\partial x_n}\cdot\vG\Big) +\sum_{n=1}^3\hi_n\Big(\vF\cdot\frac{\partial\vG}{\partial x_n}\Big) \end{gather*}

Then rewrite $\va\times(\vb\times\vc)=(\vc\cdot\va)\vb-(\vb\cdot\va)\vc\text{,}$ which is Lemma 4.1.8.b below, as

\begin{equation*} (\vc\cdot\va)\vb=\va\times(\vb\times\vc)+(\vb\cdot\va)\vc \end{equation*}

Applying it once with $\vb=\hi_n\text{,}$ $\vc=\frac{\partial\vF}{\partial x_n}\text{,}$ $\va=\vG$ and once with $\vb=\hi_n\text{,}$ $\vc=\frac{\partial\vG}{\partial x_n}\text{,}$ $\va=\vF$ gives

\begin{align*} \vnabla(\vF\cdot\vG) &=\sum_{n=1}^3\bigg[ \vG\times\Big(\hi_n\times\frac{\partial\vF}{\partial x_n}\Big) +(\vG\cdot\hi_n)\frac{\partial\vF}{\partial x_n}\bigg]\\ &\hskip1in+\sum_{n=1}^3\bigg[ \vF\times\Big(\hi_n\times\frac{\partial\vG}{\partial x_n}\Big) +(\vF\cdot\hi_n)\frac{\partial\vG}{\partial x_n}\bigg]\\ &=\vG\times(\vnabla\times\vF) +(\vG\cdot\vnabla)\vF +\vF\times(\vnabla\times\vG) +(\vF\cdot\vnabla)\vG \end{align*}

Theorem 4.1.5.d: We use the same trick. Write out the left hand side as

\begin{align*} \vnabla\times(\vF\times\vG) &=\sum_{n=1}^3\hi_n\times\frac{\partial }{\partial x_n}(\vF\times\vG)\\ &=\sum_{n=1}^3\hi_n\times\Big(\frac{\partial\vF}{\partial x_n}\times\vG\Big) +\sum_{n=1}^3\hi_n\times\Big(\vF\times\frac{\partial\vG}{\partial x_n}\Big) \end{align*}

Applying $\va\times(\vb\times\vc)=(\vc\cdot\va)\vb-(\vb\cdot\va)\vc\text{,}$ which is Lemma 4.1.8.b below,

\begin{align*} \vnabla\times(\vF\times\vG)&= \sum_{n=1}^3\Big[G_n\frac{\partial \vF}{\partial x_n} -\frac{\partial F_n}{\partial x_n}\vG\Big] +\sum_{n=1}^3\Big[\frac{\partial G_n}{\partial x_n}\vF -F_n\frac{\partial \vG}{\partial x_n}\Big]\cr &=(\vG\cdot\vnabla)\vF -(\vnabla\cdot\vF)\vG+(\vnabla\cdot\vG)\vF -(\vF\cdot\vnabla)\vG \end{align*}

Theorem 4.1.7.a: Substituting in

\begin{align*} \vnabla\times\vF &=\Big(\frac{\partial F_3}{\partial y} -\frac{\partial F_2}{\partial z}\Big)\hi -\Big(\frac{\partial F_3}{\partial x} -\frac{\partial F_1}{\partial z}\Big)\hj +\Big(\frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y}\Big)\hk\\ \end{align*}

gives

\begin{align*} \vnabla\cdot\big(\vnabla\times F\big) &=\frac{\partial }{\partial x}\Big(\frac{\partial F_3}{\partial y} -\frac{\partial F_2}{\partial z}\Big) -\frac{\partial }{\partial y}\Big(\frac{\partial F_3}{\partial x} -\frac{\partial F_1}{\partial z}\Big) +\frac{\partial }{\partial z}\Big(\frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y}\Big)\cr &={\color{red}{\frac{\partial^2 F_3}{\partial x\partial y}}} -{\color{blue}{\frac{\partial^2 F_2}{\partial x\partial z}}} -{\color{red}{\frac{\partial^2 F_3}{\partial y\partial x}}} +\frac{\partial^2 F_1}{\partial y\partial z} +{\color{blue}{\frac{\partial^2 F_2}{\partial z\partial x}}} -\frac{\partial^2 F_1}{\partial z\partial y}\\ &=0 \end{align*}

because the two red terms have cancelled, the two blue terms have cancelled and the two black terms have cancelled.

Theorem 4.1.7.b: Substituting in

\begin{align*} \vnabla f &=\frac{\partial f}{\partial x}\hi +\frac{\partial f}{\partial y}\hj +\frac{\partial f}{\partial z}\hk\cr\\ \end{align*}

gives

\begin{align*} \vnabla\times\big(\vnabla f\big) &=\Big(\frac{\partial }{\partial y}\frac{\partial f}{\partial z} -\frac{\partial }{\partial z}\frac{\partial f}{\partial y}\Big) \hi -\Big(\frac{\partial }{\partial x}\frac{\partial f}{\partial z} -\frac{\partial }{\partial z}\frac{\partial f}{\partial x}\Big) \hj\\ &\hskip1in+\Big(\frac{\partial }{\partial x}\frac{\partial f}{\partial y} -\frac{\partial }{\partial y}\frac{\partial f}{\partial x}\Big) \hk\\ &=0 \end{align*}

Theorem 4.1.7.c: By Theorem 4.1.4.c, followed by Theorem 4.1.4.d,

\begin{align*} \vnabla\cdot\big[f(\vnabla g\times\vnabla h)\big] &=\vnabla f\cdot(\vnabla g\times\vnabla h) +f\vnabla\cdot(\vnabla g\times\vnabla h)\\ &=\vnabla f\cdot(\vnabla g\times\vnabla h) +f\big[(\vnabla\times\vnabla g)\cdot\vnabla h-\vnabla g\cdot(\vnabla\times\vnabla h)\big] \end{align*}

By Theorem 4.1.7.b, $\vnabla\times\vnabla g=\vnabla\times\vnabla h=0\text{,}$ so

\begin{equation*} \vnabla\cdot\big[f(\vnabla g\times\vnabla f)\big] =\vnabla f\cdot(\vnabla g\times\vnabla h) \end{equation*}

Theorem 4.1.7.d: By Theorem 4.1.4.c,

\begin{align*} \vnabla\cdot(f\vnabla g-g\vnabla f) &= (\vnabla f)\cdot(\vnabla g) +f\,\vnabla\cdot(\vnabla g) -(\vnabla g)\cdot(\vnabla f) +g\,\vnabla\cdot(\vnabla f)\\ &=f\,\vnabla^2g-g\,\vnabla^2f \end{align*}

Theorem 4.1.7.e:

\begin{align*} \vnabla\times(\vnabla\times\vF) &=\sum_{\ell=1}^3\hi_\ell\frac{\partial } {\partial x_\ell} \times\bigg(\sum_{m=1}^3 \hi_m\frac{\partial } {\partial x_m} \times\sum_{n=1}^3\hi_n F_n\bigg)\\ &=\sum_{\ell,m,n=1}^3\hi_\ell\times \big(\hi_m\times\hi_n\big)\ \frac{\partial^2 F_n } {\partial x_\ell\partial x_m} \end{align*}

Using $\va\times(\vb\times\vc)=(\vc\cdot\va)\vb-(\vb\cdot\va)\vc\text{,}$ we have

\begin{equation*} \hi_\ell\times \big(\hi_m\times\hi_n\big) =(\hi_\ell\cdot\hi_n)\hi_m -(\hi_\ell\cdot\hi_m)\hi_n =\delta_{\ell,n}\hi_m -\delta_{\ell,m}\hi_n \end{equation*}

where

⁷

$\delta_{m,n}$ is called the Kronecker delta function. It is named after the German number theorist and logician Leopold Kronecker (1823–1891). He is reputed to have said “God made the integers. All else is the work of man.”

\begin{equation*} \delta_{m,n} =\begin{cases} 1& \text{if } m=n \\ 0& \text{if } m\ne n \end{cases} \end{equation*}

Hence

\begin{align*} \vnabla\times(\vnabla\times\vF) &=\sum_{\ell,m,n=1}^3\delta_{\ell,n}\hi_m\ \frac{\partial^2 F_n } {\partial x_\ell\partial x_m} -\sum_{\ell,m,n=1}^3\delta_{\ell,m}\hi_n\ \frac{\partial^2 F_n } {\partial x_\ell\partial x_m}\\ &=\sum_{m,n=1}^3\hi_m\ \frac{\partial }{\partial x_m} \frac{\partial F_n }{\partial x_n} -\sum_{m,n=1}^3\hi_n\ \frac{\partial^2 F_n}{\partial x_m^2}\cr &= \vnabla(\vnabla\cdot\vF)-\vnabla^2\vF \end{align*}

Lemma 4.1.8.

$\displaystyle \va\cdot(\vb\times\vc)=(\va\times\vb)\cdot\vc$
$\displaystyle \va\times(\vb\times\vc)=(\vc\cdot\va)\vb-(\vb\cdot\va)\vc$

Proof.

(a) Here are two proofs. For the first, just write out both sides

\begin{align*} \va\cdot(\vb\times\vc) &=(a_1,a_2,a_3)\cdot(b_2c_3-b_3c_2\,,\,b_3c_1-b_1c_3\,,\,b_1c_2-b_2c_1)\\ &=a_1b_2c_3-a_1b_3c_2+a_2b_3c_1-a_2b_1c_3+a_3b_1c_2-a_3b_2c_1\\ (\va\times\vb)\cdot\vc &=(a_2b_3-a_3b_2\,,\,a_3b_1-a_1b_3\,,\,a_1b_2-a_2b_1)\cdot(c_1,c_2,c_3)\\ &=a_2b_3c_1-a_3b_2c_1+a_3b_1c_2-a_1b_3c_2+a_1b_2c_3-a_2b_1c_3 \end{align*}

and observe that they are the same.

For the second proof, we again write out both sides, but this time we express them in terms of determinants.

\begin{align*} \va\cdot\vb\times\vc &=(a_1,a_2,a_3)\cdot\det\left[\begin{matrix}\hi&\hj&\hk \\ b_1&b_2&b_3 \\ c_1&c_2&c_3\end{matrix}\right]\\ &=a_1\det\left[\begin{matrix} b_2&b_3 \\ c_2&c_3\end{matrix}\right] -a_2\det\left[\begin{matrix} b_1&b_3 \\ c_1&c_3\end{matrix}\right] +a_3\det\left[\begin{matrix} b_1&b_2 \\ c_1&c_2\end{matrix}\right]\\ &=\det\left[\begin{matrix}a_1&a_2&a_3 \\ b_1&b_2&b_3 \\ c_1&c_2&c_3\end{matrix}\right]\\ \va\times\vb\cdot\vc &=\det\left[\begin{matrix}\hi&\hj&\hk \\ a_1&a_2&a_3 \\ b_1&b_2&b_3\end{matrix}\right]\cdot(c_1,c_2,c_3)\\ &=c_1\det\left[\begin{matrix} a_2&a_3 \\ b_2&b_3\end{matrix}\right] -c_2\det\left[\begin{matrix} a_1&a_3 \\ b_1&b_3\end{matrix}\right] +c_3\det\left[\begin{matrix} a_1&a_2 \\ b_1&b_2\end{matrix}\right]\\ &=\det\left[\begin{matrix}c_1&c_2&c_3\cr a_1&a_2&a_3 \\ b_1&b_2&b_3 \end{matrix}\right] \end{align*}

Exchanging two rows in a determinant changes the sign of the determinant. Moving the top row of a $3\times 3$ determinant to the bottom row requires two exchanges of rows. So the two $3\times 3$ determinants are equal.

(b) The proof is not exceptionally difficult — just write out both sides and grind. Substituting in

\begin{equation*} \vb\times\vc \ =\ (b_2c_3-b_3c_2)\hi-(b_1c_3-b_3c_1)\hj + (b_1c_2-b_2c_1)\hk \end{equation*}

gives, for the left hand side,

\begin{align*} \va\times(\vb\times\vc) =\phantom{-}&\!\!\!\det\left[\begin{matrix}\hi&\hj &\hk \\ a_1&a_2&a_3 \\ b_2c_3-b_3c_2&-b_1c_3+b_3c_1&b_1c_2-b_2c_1 \end{matrix}\right]\\ =\phantom{-}&\hi\big[a_2(b_1c_2-b_2c_1)-a_3(-b_1c_3+b_3c_1)\big]\\ -&\hj\big[a_1(b_1c_2-b_2c_1)-a_3(b_2c_3-b_3c_2)\big]\\ +&\hk\big[a_1(-b_1c_3+b_3c_1)-a_2(b_2c_3-b_3c_2)\big] \end{align*}

On the other hand, the right hand side

\begin{align*} (\va\cdot\vc)\vb-(\va\cdot\vb)\vc \ =\ &(a_1c_1+a_2c_2+a_3c_3)(b_1\hi+b_2\hj+b_3\hk)\\ &\hskip0.5in-(a_1b_1+a_2b_2+a_3b_3)(c_1\hi+c_2\hj+c_3\hk)\\ =\ & \hi\ \big[{\color{blue}{a_1b_1c_1}} +a_2b_1c_2+a_3b_1c_3- {\color{blue}{a_1b_1c_1}} -a_2b_2c_1-a_3b_3c_1\big]\\ {+}&\hj\ \big[a_1b_2c_1 +{\color{blue}{a_2b_2c_2}} +a_3b_2c_3-a_1b_1c_2 -{\color{blue}{a_2b_2c_2}} -a_3b_3c_2\big]\\ {+}&\hk\ \big[a_1b_3c_1+a_2b_3c_2 +{\color{blue}{a_3b_3c_3}} -a_1b_1c_3-a_2b_2c_3 -{\color{blue}{a_3b_3c_3}}\big]\\ {=}\ & \hi\ [a_2b_1c_2+a_3b_1c_3-a_2b_2c_1-a_3b_3c_1]\\ {+}&\hj\ [a_1b_2c_1+a_3b_2c_3-a_1b_1c_2-a_3b_3c_2]\\ {+}&\hk\ [a_1b_3c_1+a_2b_3c_2-a_1b_1c_3-a_2b_2c_3] \end{align*}

The last formula that we had for the left hand side is the same as the last formula we had for the right hand side.

Example 4.1.9. Screening tests.

We have seen the vector identity Theorem 4.1.7.b before. It says that if a vector field $\vF$ is of the form $\vF = \vnabla\varphi$ for some some function $\varphi$ (that is, if $\vF$ is conservative), then

\begin{equation*} \vnabla\times\vF = \vnabla\times(\vnabla\varphi) = 0 \end{equation*}

Conversely, we have also seen, in Theorem 2.4.8, that, if $\vF$ is defined and has continuous first order partial derivatives on all of $\bbbr^3\text{,}$ and if $\vnabla\times\vF=0\text{,}$ then $\vF$ is conservative. The vector identity Theorem 4.1.7.b is our screening test for conservativeness.

Because its right hand side is zero, the vector identity Theorem 4.1.7.a is suggestive. It says that if a vector field $\vF$ is of the form $\vF = \vnabla\times\vA$ for some some vector field $\vA\text{,}$ then

\begin{equation*} \vnabla\cdot\vF = \vnabla\cdot(\vnabla\times\vA) = 0 \end{equation*}

When $\vF = \vnabla\times\vA\text{,}$ $\vA$ is called a vector potential for $\vF\text{.}$ We shall see in Theorem 4.1.16, below, that, conversely, if $\vF(\vx)$ is defined and has continuous first order partial derivatives on all of $\bbbr^3\text{,}$ and if $\vnabla\cdot\vF=0\text{,}$ then $\vF$ has a vector potential

⁸

Does this remind you of Theorem 2.4.8? It should.

. The vector identity Theorem 4.1.7.a is indeed another screening test.

As an example, consider the Maxwell’s equations

\begin{align*} \vnabla\cdot\vB &= 0\\ \vnabla\times\vE +\frac{1}{c}\frac{\partial\vB}{\partial t}&=0 \end{align*}

that we saw in Example 4.1.2. The first equation implies that (assuming $\vB$ is sufficiently smooth) there is a vector field $\vA\text{,}$ called the magnetic potential, with $\vB=\vnabla\times\vA\text{.}$ Substituting this into the second equation gives

\begin{equation*} 0= \vnabla\times\vE +\frac{1}{c}\frac{\partial\ }{\partial t}\vnabla\times\vA =\vnabla\times\Big(\vE+\frac{1}{c}\frac{\partial\vA }{\partial t}\Big) \end{equation*}

So $\vE+\frac{1}{c}\frac{\partial\vA }{\partial t}$ passes the screening test of Theorem 4.1.7.b and there is a function $\varphi$ (called the electric potential) with

\begin{equation*} \vE+\frac{1}{c}\frac{\partial\vA }{\partial t} = -\nabla \varphi \end{equation*}

We have put in the minus sign just to provide compatibility with the usual physics terminology.

Example 4.1.10.

Let $\vr(x,y,z) = x\,\hi+y\,\hj+z\,\hk$ and let $\psi(x,y,z)$ be an arbitrary function. Verify that

\begin{equation*} \vnabla\cdot\big(\vr\times\vnabla\psi\big) = 0 \end{equation*}

Solution.

By the vector identity Theorem 4.1.4.d,

\begin{gather*} \vnabla\cdot\big(\vr\times\vnabla\psi\big) =(\vnabla\times\vr)\cdot\vnabla \psi -\vr\cdot\big(\vnabla\times(\vnabla\psi)\big) \end{gather*}

By the vector identity Theorem 4.1.7.b, the second term is zero. Now since

\begin{equation*} \vnabla\times\vr = \Big(\frac{\partial z}{\partial y} -\frac{\partial y}{\partial z}\Big)\hi -\Big(\frac{\partial z}{\partial x} -\frac{\partial x}{\partial z}\Big)\hj +\Big(\frac{\partial y}{\partial x} -\frac{\partial x}{\partial y}\Big)\hk =\vZero \end{equation*}

the first term is also zero. Indeed $\vnabla\cdot\big(\vr\times\vnabla\psi\big) = 0$ holds for any curl free $\vr(x,y,z)\text{.}$

Subsection 4.1.2 Vector Potentials

We’ll now further explore the vector potentials that were introduced in Example 4.1.9. First, here is the formal definition.

Definition 4.1.11.

The vector field $\vA$ is said to be a vector potential for the vector field $\vB$ if

\begin{equation*} \vB=\vnabla\times\vA \end{equation*}

As we saw in Example 4.1.9, if a vector field $\vB$ has a vector potential, then the vector identity Theorem 4.1.7.a implies that $\vnabla\cdot\vB=0\text{.}$ This fact deserves to be called a theorem.

Theorem 4.1.12. Screening test for vector potentials.

If there exists a vector potential for the vector field $\vB\text{,}$ then

\begin{equation*} \vnabla\cdot\vB=0 \end{equation*}

Of course, we’ll consider the converse soon. Also note that the vector potential, when it exists, is far from unique. Two vector fields $\vA$ and $\tilde\vA$ are both vector potentials for the same vector field if and only if

\begin{equation*} \vnabla\times\vA=\vnabla\times\tilde\vA \iff \vnabla\times(\vA-\tilde\vA)=\vZero \end{equation*}

That is, if and only if the difference $\vA-\tilde\vA$ passes the conservative field screening test of Theorems 2.3.9 and 2.4.8. In particular, if $\vA$ is one vector potential for a vector field $\vB$ (i.e. if $\vB=\vnabla\times\vA$), and if $\psi$ is any function, then

\begin{equation*} \vnabla\times(\vA+\vnabla\psi) =\vnabla\times\vA + \vnabla\times\vnabla\psi =\vB \end{equation*}

by the vector identity Theorem 4.1.7.b. That is, $\vA+\vnabla\psi$ is another vector potential for $\vB\text{.}$

To simplify computations, we can always choose $\psi$ so that, for example, the third component of $\vA+\vnabla\psi\text{,}$ namely $\big(\vA+\vnabla\psi\big)\cdot\hk= \vA_3+\frac{\partial\psi}{\partial z}\text{,}$ is zero — just choose $\psi = -\int \vA_3\, \dee{z}\text{.}$ We have just proven

Lemma 4.1.13.

If the vector field $\vB$ has a vector potential, then, in particular, there is a vector potential $\vA$ for $\vB$ with

⁹

There is nothing special about the subscript $3$ here. By precisely the same argument, we could come up with another vector potential whose second component is zero, and with a third vector potential whose first component is zero.

$\vA_3=0\text{.}$

Here is an example which exploits this choice to simplify the computations used to find a vector potential.

Example 4.1.14.

Let

\begin{equation*} \vB = yz\,\hi + zx\,\hj + xy\,\hk \end{equation*}

This vector field has been set up carefully to obey

\begin{equation*} \vnabla\cdot\vB = \frac{\partial }{\partial x}(yz) +\frac{\partial }{\partial y}(zx) +\frac{\partial }{\partial z}(xy) =0 \end{equation*}

and so passes the screening test of Theorem 4.1.12.

Let’s try and find a vector potential for $\vB\text{.}$ That is, let’s try and find a vector field $\vA= A_1\,\hi + A_2\,\hj + A_3\,\hk$ that obeys $\vnabla\times\vA = \vB\text{,}$ or equivalently,

\begin{align*} \frac{\partial A_3}{\partial y} -\frac{\partial A_2}{\partial z} &= B_1 =yz\\ -\frac{\partial A_3}{\partial x} +\frac{\partial A_1}{\partial z}&=B_2=zx\\ \frac{\partial A_2}{\partial x} -\frac{\partial A_1}{\partial y}&=B_3=xy \end{align*}

This system is nasty to solve because every equation contains more than one of the three unknowns, $A_1\text{,}$ $A_2\text{,}$ $A_3\text{.}$ Let us take advantage of our observation above that, if any vector potential exists, then, in particular, a vector potential $\vA$ exists that also obeys $A_3=0\text{.}$ So let’s also require that $A_3=0\text{.}$ Then the equations above simplify to

\begin{align*} -\frac{\partial A_2}{\partial z} &=yz\\ \frac{\partial A_1}{\partial z} &=zx\\ \frac{\partial A_2}{\partial x} -\frac{\partial A_1}{\partial y}&=xy \end{align*}

This system is much easier because, now that we have chosen $A_3=0\text{,}$ the first equation contains only a single unknown, namely $A_2$ and we can find all $A_2$’s that obey the first equation simply by integrating with respect to $z\text{:}$

\begin{equation*} A_2 = -\frac{yz^2}{2} + N(x,y) \end{equation*}

Note that, because $\frac{\partial }{\partial z}$ treats $x$ and $y$ as constants, the constant of integration $N$ is allowed to depend on $x$ and $y\text{.}$

Similarly, the second equation contains only a single unknown, $A_1\text{,}$ and is easily solved by integrating with respect to $z\text{.}$ The second equation is satisfied if and only if

\begin{equation*} A_1 = \frac{xz^2}{2} + M(x,y) \end{equation*}

for some function $M\text{.}$

Finally, the third equation is also satisfied if and only if $M(x,y)$ and $N(x,y)$ obey

\begin{equation*} \frac{\partial }{\partial x}\Big(-\frac{yz^2}{2} + N(x,y)\Big) -\frac{\partial }{\partial y}\Big(\frac{xz^2}{2} + M(x,y)\Big)=xy \end{equation*}

which simplifies to

\begin{equation*} \frac{\partial N}{\partial x}(x,y) -\frac{\partial M}{\partial y}(x,y) =xy \end{equation*}

This is one linear equation in two unknowns, $M$ and $N\text{.}$ Typically, we can easily solve one linear equation in one unknown. So we are free to eliminate one of the unknowns by setting, for example, $M=0\text{,}$ and then choose any $N$ that obeys

\begin{equation*} \frac{\partial N}{\partial x}(x,y) = xy \end{equation*}

Integrating with respect to $x$ gives, as one possible choice, $N(x,y) = \frac{x^2y}{2}\text{.}$ So we have found a vector potential. Namely

\begin{equation*} \vA = \frac{xz^2}{2} \hi +\Big(-\frac{yz^2}{2} + \frac{x^2y}{2}\Big)\hj \end{equation*}

One can, and indeed should, quickly check that $\vnabla\times\vA=\vB\text{.}$

Let’s do another.

Example 4.1.15.

Let

\begin{equation*} \vB = (2x)\,\hi+(2z-2x)\,\hj +(2x-2z)\,\hk \end{equation*}

This vector field obeys

\begin{equation*} \vnabla\cdot\vB = \frac{\partial }{\partial x}(2x) +\frac{\partial }{\partial y}(2z-2x) +\frac{\partial }{\partial z}(2x-2z) =0 \end{equation*}

and so passes the screening test of Theorem 4.1.12. We’ll now find a vector potential $\vA= A_1\,\hi + A_2\,\hj + A_3\,\hk$ for $\vB\text{.}$ As in the last example, we’ll simplify the computations by further requiring

¹⁰

Of course, we could equally well pick $A_1=0$ or $A_2=0\text{.}$

that $A_3=0\text{.}$

The requirements that $\vnabla\times\vA = \vB$ and $A_3=0$ come down to

\begin{align*} -\frac{\partial A_2}{\partial z} &=2x\\ \frac{\partial A_1}{\partial z} &=2z-2x\\ \frac{\partial A_2}{\partial x} -\frac{\partial A_1}{\partial y}&=2x-2z \end{align*}

Because $\frac{\partial }{\partial z}$ treats $x$ and $y$ as constants, the first equation is satisfied if and only if there is a function $N(x,y)$

\begin{equation*} A_2 = -2xz + N(x,y) \end{equation*}

and second equation is satisfied if and only if there is a function $M(x,y)$

\begin{equation*} A_1 = z^2-2xz + M(x,y) \end{equation*}

Finally, the third equation is also satisfied if and only if $M(x,y)$ and $N(x,y)$ obey

\begin{align*} &\frac{\partial }{\partial x}\Big(-2xz + N(x,y)\Big) -\frac{\partial }{\partial y}\Big(z^2-2xz + M(x,y))\Big)=2x-2z\\ &\iff -2z+\frac{\partial N}{\partial x}(x,y) -\frac{\partial M}{\partial y}(x,y) =2x -2z\\ &\iff \phantom{-2z+\ \,} \frac{\partial N}{\partial x}(x,y) -\frac{\partial M}{\partial y}(x,y) =2x \end{align*}

All of the $z$’s in this equation have cancelled out

¹¹

If the $z$’s had not cancelled out, no $N(x,y)$ and $M(x,y)\text{,}$ which after all are independent of $z\text{,}$ could satisfy the equation. That would have been a sure sign of a user error.

, and we can choose, for example, $M(x,y)=0$ and $N(x,y) = x^2\text{.}$ So we have found a vector potential. Namely

\begin{equation*} \vA = (z^2-2xz) \hi +(x^2-2xz)\hj \end{equation*}

Again it is a good idea to check that $\vnabla\times\vA=\vB\text{.}$

We can use exactly the strategy of the last examples to prove

Theorem 4.1.16.

Let $\vB$ be a vector field that is defined and has all of its first order partial derivatives continuous on all of $\bbbr^3\text{.}$ Then there exists a vector potential for $\vB$ if and only if it passes the screening test $\vnabla\cdot\vB=0\text{.}$

Proof.

We already know that the existence of a vector potential implies that $\vnabla\cdot\vB=0\text{.}$ So we just have to assume that $\vnabla\cdot\vB=0$ and prove that this implies the existence of a vector field $\vA$ that obeys $\vnabla\times\vA = \vB\text{.}$ Hence we need to solve

\begin{align*} \frac{\partial A_3}{\partial y} -\frac{\partial A_2}{\partial z} &= B_1(x,y,z)\\ -\frac{\partial A_3}{\partial x} +\frac{\partial A_1}{\partial z}&=B_2(x,y,z)\\ \frac{\partial A_2}{\partial x} -\frac{\partial A_1}{\partial y}&=B_3(x,y,z) \end{align*}

We’ll explicitly find such an $\vA$ using exactly the strategy of Example 4.1.14. In particular, we’ll look for an $\vA$ that also has $A_3=0\text{.}$ Then the equations simplify to

\begin{align*} -\frac{\partial A_2}{\partial z} &=B_1(x,y,z)\\ \frac{\partial A_1}{\partial z} &=B_2(x,y,z)\\ \frac{\partial A_2}{\partial x} -\frac{\partial A_1}{\partial y}&=B_3(x,y,z) \end{align*}

The first equation is satisfied if and only if

\begin{equation*} A_2(x,y,z) = -\int_0^z B_1(x,y,\tilde z)\ \dee{\tilde z} +N(x,y) \end{equation*}

for some function $N(x,y)\text{.}$ And the second equation is satisfied if and only if

\begin{equation*} A_1(x,y,z) = \int_0^z B_2(x,y,\tilde z)\ \dee{\tilde z} +M(x,y) \end{equation*}

So all three equations are satisfied if and only only if we can find $M(x,y)$ and $N(x,y)$ that obey

\begin{align*} &\frac{\partial }{\partial x}\Big( \overbrace{-\int_0^z B_1(x,y,\tilde z)\ \dee{\tilde z} +N(x,y)}^{A_2(x,y,z)}\Big)\\ &\hskip1in -\frac{\partial }{\partial y}\Big( \overbrace{\int_0^z B_2(x,y,\tilde z)\ \dee{\tilde z} +M(x,y)}^{A_1(x,y,z)}\Big) =B_3(x,y,z) \end{align*}

which is the case if and only if

\begin{gather*} \frac{\partial N}{\partial x}(x,y) -\frac{\partial M}{\partial y}(x,y) = B_3(x,y,z) + \int_0^z\Big(\frac{\partial B_1}{\partial x}(x,y,\tilde z) + \frac{\partial B_2}{\partial y}(x,y,\tilde z)\Big) \ \dee{\tilde z} \end{gather*}

Oof! At first sight, it looks like we have a very big problem here. No matter what $N$ and $M$ we pick the left hand side will depend on $x$ and $y$ only — not on $z\text{.}$ But it appears like the right hand side depends on $z$ too. Fortunately the screening test (which we have not used to this point in the proof) rides to the rescue and ensures that the right hand actually does not depend on $z\text{.}$ By the screening test,

\begin{equation*} \vnabla\cdot\vB =\frac{\partial B_1}{\partial x} +\frac{\partial B_2}{\partial y} +\frac{\partial B_3}{\partial z}=0 \end{equation*}

and we have

\begin{equation*} \frac{\partial B_1}{\partial x} +\frac{\partial B_2}{\partial y} =-\frac{\partial B_3}{\partial z} \end{equation*}

so that the right hand side is

\begin{align*} B_3(x,y,z) + \int_0^z\Big(-\frac{\partial B_3}{\partial z}(x,y,\tilde z) \Big) \ \dee{\tilde z} &= B_3(x,y,z) +\Big[-B_3(x,y,\tilde z)\Big]^{\tilde z=z}_{\tilde z=0}\\ &=B_3(x,y,0) \end{align*}

by the fundamental theorem of calculus. So we just have to choose $M$ and $N$ to obey

\begin{gather*} \frac{\partial N}{\partial x}(x,y) -\frac{\partial M}{\partial y}(x,y) = B_3(x,y,0) \end{gather*}

For example, $M=0\text{,}$ $N(x,y) = \int _0^x B_3(\tilde x,y,0)\ \dee{\tilde x}$ work. So not only have we proven that a vector potential exists, but we have found a formula for it.

Warning 4.1.17.

Note that in Theorem 4.1.16 we are assuming that $\vB$ passes the screening test on all of $\bbbr^3\text{.}$ If that is not the case, for example because the vector field is not defined on all of $\bbbr^3\text{,}$ then $\vB$ can fail to have a vector potential. An example (the point source) is provided in Example 4.4.8.

Subsection 4.1.3 Interpretation of the Gradient

In this section we’ll develop an interpretation of the gradient $\vnabla f(\vr_0)\text{.}$ This should just be a review of material that you have seen before.

Suppose that you are moving through space and that your position at time $t$ is $\vr(t)=\big(x(t),y(t),z(t)\big)\text{.}$ As you move along, you measure, for example, the temperature. If the temperature at position $(x,y,z)$ is $f(x,y,z)\text{,}$ then the temperature that you measure at time $t$ is $f\big(x(t),y(t),z(t)\big)\text{.}$ So the rate of change of temperature that you feel is

\begin{align*} &\diff{ }{t}f\big(x(t),y(t),z(t)\big)\\ &\hskip0.3in =\frac{\partial f}{\partial x}\big(x(t),y(t),z(t)\big) \diff{x}{t}(t) +\frac{\partial f}{\partial y}\big(x(t),y(t),z(t)\big) \diff{y}{t}(t)\\ &\hskip1in +\frac{\partial f}{\partial z}\big(x(t),y(t),z(t)\big) \diff{z}{t}(t) \qquad \text{(by the chain rule)}\\ &\hskip0.3in=\vnabla f\big(\vr(t)\big)\cdot\vr'(t)\\ &\hskip0.3in=\big|\vnabla f\big(\vr(t)\big)\big| \,\big|\vr'(t)\big|\,\cos\theta \end{align*}

where $\theta$ is the angle between the gradient vector $\vnabla f\big(\vr(t)\big)$ and the velocity vector $\vr'(t)\text{.}$ This is the rate of change per unit time. We can get the rate of change per unit distance travelled by moving with speed one, so that $\big|\vr'(t)\big|=1$ and then

\begin{gather*} \diff{ }{t}f\big(\vr(t)\big) =\big|\vnabla f\big(\vr(t)\big)\big|\,\cos\theta \end{gather*}

If, at a given moment $t=t_0\text{,}$ you are at $\vr(t_0)=\vr_0\text{,}$ then

\begin{gather*} \diff{ }{t}f\big(\vr(t)\big)\Big|_{t=t_0} =\big|\vnabla f(\vr_0)\big|\,\cos\theta \end{gather*}

Recall that $\theta$ is the angle between our direction of motion and the gradient vector $\vnabla f(\vr_0)\text{.}$ So to maximize the rate of change of temperature that we feel, as we pass through $\vr_0\text{,}$ we should choose our direction of motion to be the direction of the gradient vector $\vnabla f(\vr_0)\text{.}$ In conclusion

Equation 4.1.18.

\begin{alignat*}{2} \vnabla f(\vr_0) & \text{ has direction } && \left\{ \begin{array}{l} \text{direction of maximum rate of change}\\ \text{of $f$ at $\vr_0$} \end{array}\right.\\ & \text{ has magnitude } && \left\{ \begin{array}{l} \text{magnitude of maximum rate of change} \\ \text{ (per unit distance) of $f$ at $\vr_0$} \end{array} \right. \end{alignat*}

Subsection 4.1.4 Interpretation of the Divergence

In this section we’ll develop an interpretation of the divergence $\vnabla\cdot\vv(\vr_0)$ of the vector field $\vv(\vr)$ at the point $\vr_0\text{.}$ We shall do so in two steps.

First we’ll express $\vnabla\cdot\vv(\vr_0)$ in terms of flux integrals.
Then we’ll use the interpretation of flux integrals given in Lemma 3.4.1 to get an interpretation of $\vnabla\cdot\vv(\vr_0)\text{.}$

Think of $\vv(x,y,z)$ as the velocity of a fluid at $(x,y,z)$ and fix any point $\vr_0=(x_0,y_0,z_0)\text{.}$ Let, for any $\veps \gt 0\text{,}$ $S_\veps$ be the sphere

centered at $\vr_0$
of radius $\veps\text{.}$
Denote by $\hn(x,y,z)$ the outward normal to $S_\veps$ at $(x,y,z)\text{.}$

We shall prove, in Lemma 4.1.20, below, that we can write $\vnabla\cdot\vv(\vr_0)$ as the limit

\begin{gather*} \vnabla\cdot\vv(x_0,y_0,z_0) =\lim_{\veps\rightarrow 0} \frac{1}{\frac{4}{3}\pi\veps^3} \dblInt_{S_\veps}\vv(x,y,z)\cdot\hn(x,y,z)\,\dee{S} \end{gather*}

Once we have that lemma we can use that

$\frac{4}{3}\pi\veps^3$ is the volume of the interior of the sphere $S_\veps$ and
by Lemma 3.4.1, $\dblInt_{S_\veps}\vv(x,y,z)\cdot\hn(x,y,z)\,\dee{S}$ is the rate
¹²
Lemma 3.4.1 is being applied with the density $\rho$ set equal to one, so, more precisely, the rate is the number of units of volume of fluid exiting $S_\veps$ per unit time
at which fluid is exiting $S_\veps$

to conclude that

Equation 4.1.19.

\begin{align*} \vnabla\cdot\vv(\vr_0) & =\left\{ \begin{array}{l} \text{rate at which fluid is exiting an }\\ \text{infinitesimal sphere centred }\\ \text{at $\vr_0$, per unit time, per unit volume} \end{array} \right.\\ &=\text{strength of the source at $\vr_0$} \end{align*}

Here is the critical computation.

Lemma 4.1.20.

\begin{gather*} \vnabla\cdot\vv(x_0,y_0,z_0) =\lim_{\veps\rightarrow 0} \frac{1}{\frac{4}{3}\pi\veps^3} \dblInt_{S_\veps}\vv(x,y,z)\cdot\hn(x,y,z)\,\dee{S} \end{gather*}

Proof. (Optional).

Here is one proof

¹³

There is another, easier to understand, proof of this result given in §4.4.1. We cannot give that proof here because it uses the divergence theorem, which we will get to later in the chapter.

of Lemma 4.1.20.

By translating our coordinate system, it suffices to consider $\vr_0=(x_0,y_0,z_0) = (0,0,0)\text{.}$ Then

\begin{gather*} S_\veps = \Set{(x,y,z)}{|(x,y,z)|=\veps}\qquad \hn(x,y,z) = \frac{1}{\veps}(x,y,z) \end{gather*}

We expand $\vv(x,y,z)$ in a Taylor expansion in powers of $x\text{,}$ $y\text{,}$ and $z\text{,}$ to first order, with second order error term.

\begin{gather*} \vv(x,y,z) = \vA + \vB\,x +\vC\,y +\vD\, z +\vR(x,y,z) \end{gather*}

where

\begin{gather*} \vA=\vv(0,0,0) \qquad \vB=\frac{\partial \vv}{\partial x}(0,0,0) \qquad \vC=\frac{\partial \vv}{\partial y}(0,0,0) \qquad \vD=\frac{\partial \vv}{\partial z}(0,0,0) \end{gather*}

and the error term $\vR(x,y,z)$ is bounded by a constant times

¹⁴

Terms like $xy\text{,}$ $xz$ and $yz$ are not needed because, for example, $|xy|\le \frac{1}{2}(x^2+y^2)\text{.}$ This inequality is equivalent to $\big(|x|-|y|\big)^2\ge 0\text{.}$

$x^2+y^2+z^2\text{.}$ In particular there is a constant $K$ so that, on $S_\veps\text{,}$

\begin{equation*} |\vR(x,y,z)|\le K\veps^2 \end{equation*}

\begin{align*} &\dblInt_{S_\veps}\vv(x,y,z)\cdot\hn(x,y,z)\,\dee{S}\\ &\hskip1in= \frac{1}{\veps} \dblInt_{S_\veps}\big( \vA + \vB\,x +\vC\,y +\vD\, z +\vR(x,y,z) \big)\cdot (x,y,z)\,\dee{S} \end{align*}

Multiply out the dot product so that the integrand becomes

\begin{alignat*}{3} &\phantom{+}\ \vA\cdot\hi\,x &&+\ \vA\cdot\hj\,y &&+\vA\cdot\hk\,z\\ &+\vB\cdot\hi\,x^2 &&+\ \vB\cdot\hj\,xy &&+\vB\cdot\hk\,xz\\ &+\vC\cdot\hi\,xy &&+\ \vC\cdot\hj\,y^2 &&+\vC\cdot\hk\,yz\\ &+\vD\cdot\hi\,xz\ &&+\ \vD\cdot\hj\,yz &&+\vD\cdot\hk\,z^2\\ &+\vR(x,y,z)\cdot (x,y,z) \end{alignat*}

That’s a lot of terms. But most of them integrate to zero, simply because the integral of an odd function over an even domain is zero. Because $S_\veps$ is invariant under $x\rightarrow -x$ and under $y\rightarrow -y$ and under $z\rightarrow -z$ we have

\begin{gather*} \dblInt_{S_\veps}\!\! x\,\dee{S} =\dblInt_{S_\veps}\!\! y\,\dee{S} =\dblInt_{S_\veps}\!\! z\,\dee{S} =\dblInt_{S_\veps}\!\! xy\,\dee{S} =\dblInt_{S_\veps}\!\! xz\,\dee{S} =\dblInt_{S_\veps}\!\! yz\,\dee{S} =0 \end{gather*}

which is a relief. We are now left with

\begin{align*} \dblInt_{S_\veps}\vv(x,y,z)\cdot\hn(x,y,z)\,\dee{S} &= \frac{1}{\veps} \dblInt_{S_\veps}\big( \vB\cdot\hi\,x^2 +\vC\cdot\hj\,y^2 +\vD\cdot\hk\, z^2 \big)\,\dee{S}\\ &\hskip1in +\frac{1}{\veps} \dblInt_{S_\veps}\vR(x,y,z)\cdot (x,y,z)\,\dee{S} \end{align*}

As well $S_\veps$ is invariant

¹⁵

Spheres have lots of symmetry!

under the interchange of $x$ and $y$ and also under the interchange of $x$ and $z\text{.}$ Consequently

\begin{align*} \dblInt_{S_\veps} x^2\,\dee{S} &=\dblInt_{S_\veps} y^2\,\dee{S} =\dblInt_{S_\veps} z^2\,\dee{S} =\frac{1}{3}\dblInt_{S_\veps}\big[x^2+y^2+ z^2\big]\,\dee{S}\\ &=\frac{1}{3}\dblInt_{S_\veps}\veps^2\,\dee{S} \qquad\text{since $x^2+y^2+z^2=\veps^2$ on } S_\veps\\ &=\frac{4}{3}\pi\veps^4 \end{align*}

since the surface area of the sphere $S_\veps$ is $4\pi\veps^2\text{.}$ So far, we have

\begin{align*} &\dblInt_{S_\veps}\vv(x,y,z)\cdot\hn(x,y,z)\,\dee{S} = \frac{4}{3}\pi\veps^3 \big(\vB\cdot\hi +\vC\cdot\hj +\vD\cdot\hk\big)\\ &\hskip2in+\frac{1}{\veps} \dblInt_{S_\veps}\vR(x,y,z)\cdot (x,y,z)\,\dee{S}\\ &\hskip0.75in= \frac{4}{3}\pi\veps^3 \vnabla\cdot\vv(\vZero) +\frac{1}{\veps} \dblInt_{S_\veps}\vR(x,y,z)\cdot (x,y,z)\,\dee{S}\\ &\hskip2in\text{(review the definitions of } \vB, \vC, \vD) \end{align*}

which implies

\begin{align*} &\lim_{\veps\rightarrow 0} \frac{1}{\frac{4}{3}\pi\veps^3} \dblInt_{S_\veps}\vv(x,y,z)\cdot\hn(x,y,z)\,\dee{S}\\ &\hskip1in= \vnabla\cdot\vv(\vZero) +\lim_{\veps\rightarrow 0} \frac{3}{4\pi\veps^4} \dblInt_{S_\veps}\vR(x,y,z)\cdot (x,y,z)\,\dee{S} \end{align*}

Finally, it suffices to recall that $|\vR(x,y,z)|\le K\veps^2$ and, on $S_\veps\text{,}$ $|(x,y,z)|=\veps\text{,}$ so that

\begin{align*} \frac{3}{4\pi\veps^4}\bigg| \dblInt_{S_\veps}\vR(x,y,z)\cdot (x,y,z)\,\dee{S} \bigg| &\le \frac{3}{4\pi\veps^4} \dblInt_{S_\veps}|\vR(x,y,z)|\,|(x,y,z)|\,\dee{S}\\ &\le \frac{3}{4\pi\veps^4} \dblInt_{S_\veps}K\veps^3\,\dee{S} = \frac{3}{4\pi\veps^4} \ K\veps^3\,\big(4\pi \veps^2)\\ &= 3K\veps \end{align*}

converges to zero as $\veps\rightarrow 0\text{.}$ So we are left with the desired result.

Example 4.1.21.

Here is a sketch of the vector field $\vv(x,y,z) = x\,\hi+y\,\hj+z\,\hk$ and a sphere centered on the origin, like $S_\veps\text{.}$

This velocity field has fluid being created and pushed out through the sphere. We have

\begin{gather*} \vnabla\cdot\vv(\vZero) = 3 \end{gather*}

consistent with our interpretation 4.1.19.

Example 4.1.22.

Here is a sketch of the vector field $\vv(x,y,z) = -y\,\hi+x\,\hj$ and a sphere centered on the origin, like $S_\veps\text{.}$

This velocity field just has fluid going around in circles. No fluid actually crosses the sphere. The divergence

\begin{gather*} \vnabla\cdot\vv(\vZero) = 0 \end{gather*}

consistent with our interpretation 4.1.19.

Example 4.1.23.

Here is a sketch of the vector field $\vv(x,y,z) = \hi$ and a sphere centered on the origin, like $S_\veps\text{.}$

This velocity field just has fluid moving uniformly to the right. Fluid enters the sphere from the left and leaves through the right at precisely the same rate, so that the net rate at fluid crosses the sphere is zero. The divergence

\begin{gather*} \vnabla\cdot\vv(\vZero) = 0 \end{gather*}

again consistent with our interpretation 4.1.19.

Subsection 4.1.5 Interpretation of the Curl

We’ll now develop the interpretation of the curl, or more precisely, of $\vnabla\times\vv(\vr_0)\cdot\hn$ for any unit vector $\hn\text{.}$ As we did in developing the interpretation of divergence, we’ll

first express $\vnabla\times\vv(\vr_0)\cdot\hn$ as a limit of integrals, and
then we’ll interpret the integrals.

To specify the integrals involved, let $C_\veps$ be the circle which

is centered at $\vr_0$
has radius $\veps$
lies in the plane through $\vr_0$ perpendicular to $\hn$
is oriented in the standard way with respect to $\hn\text{.}$ Imagine standing on the circle with your feet on the plane through $\vr_0$ perpendicular to $\hn\text{,}$ with the vector from your feet to your head in the same direction as $\hn$ and with your left arm pointing towards $\vr_0\text{.}$ Then you are facing in the positive direction for $C_\veps\text{.}$

We shall show in Lemma 4.1.25, below, that

\begin{equation*} \vnabla\times \vv(\vr_0)\cdot \hn =\lim_{\veps\rightarrow 0}\frac{1}{\pi\veps^2} \oint_{C_\veps} \vv(\vr)\cdot \dee{\vr} \end{equation*}

Now let’s work on interpreting the right hand side, and in particular on interpreting the integral $\oint_{C_\veps} \vv(\vr)\cdot \dee{\vr}\text{,}$ which is called the circulation of $\vv$ around $C_\veps\text{.}$ Place a tiny paddlewheel in the fluid with its axle running along $\hn$ and its paddles along $C_\veps\text{,}$ as in the figure below, except that

the paddlewheel is really expensive and has a lot more than just four paddles. Pretend

¹⁶

Method acting might help you here.

that you are one of the paddles.

If the paddlewheel is rotating at $\Om$ radians per unit time, then in one unit of time you sweep out an arc of a circle of radius $\veps$ that subtends an angle $\Om\text{.}$ That arc has length $\Om\veps\text{.}$ So you are moving at speed $\Om \veps\text{.}$
If you are at $\vr\text{,}$ the component of the fluid velocity in your direction of motion, i.e. tangential to $C_\veps\text{,}$ is $\vv(\vr)\cdot\diff{\vr}{s}\text{,}$ because $\hvt=\diff{\vr}{s}\text{,}$ with $s$ denoting arc length along the circle, is a unit vector tangential to $C_\veps\text{.}$
All paddles have to move at the same speed. So the speed of the paddles, $\Om\veps\text{,}$ should be the average value of $\vv(\vr)\cdot\diff{\vr}{s}$ around the circle.

Thus the rate of rotation, $\Om\text{,}$ of the paddlewheel should be determined by

\begin{align*} \Om \veps &= \frac{\oint_{C_\veps} \vv(\vr)\cdot\diff{\vr}{s}\dee{s}} {\oint_{C_\veps}\dee{s}} = \frac{\oint_{C_\veps} \vv(\vr)\cdot\dee{\vr}} {2\pi\veps} \end{align*}

Consequently, $\vnabla\times v(\vr_0)\cdot \hn$ is the limit as $\veps$ (the radius of the paddlewheel) tends to zero of

\begin{gather*} \frac{1}{\pi\veps^2} \oint_{C_\veps} \vv(\vr)\cdot \dee{\vr} =2\Om \end{gather*}

That’s our interpretation.

Equation 4.1.24.

If a fluid has velocity field $\vv$ and you place an infinitesimal paddlewheel at $\vr_0$ with its axle in direction $\vn\text{,}$ then it rotates at $\frac{1}{2}\vnabla\times \vv(\vr_0)\cdot \hn$ radians per unit time. In particular, to maximize the rate of rotation, orient the paddlewheel so that $\hn\parallel\vnabla\times \vv(\vr_0)\text{.}$

There will be some examples at the end of this section. First, we show

Lemma 4.1.25.

\begin{equation*} \vnabla\times \vv(\vr_0)\cdot \hn =\lim_{\veps\rightarrow 0}\frac{1}{\pi\veps^2} \oint_{C_\veps} \vv(\vr)\cdot \dee{\vr} \end{equation*}

Proof. (Optional).

Here is one proof

¹⁷

There is another, easier to understand, proof of this result given in §4.4.1. We cannot give that proof here because it uses Stokes’ theorem, which we will get to later in the chapter.

of Lemma 4.1.25.

Just as we did in the proof of Lemma 4.1.20, we can always translate our coordinate system so that $\vr_0=(x_0,y_0,z_0) = (0,0,0)\text{.}$ We can also rotate our coordinate system so that $\hn=\hk\text{.}$ Because $\vr_0=(0,0,0)$ and $\hn=\hk\text{,}$ so that $C_\veps$ lies in the $xy$-plane, we can parametrize $C_\veps$ by

\begin{equation*} \vr(t) =\veps\cos t\,\hi +\veps\sin t\,\hj \end{equation*}

Again as we did in the proof of Lemma 4.1.20, expand $\vv(x,y,z)$ in a Taylor expansion in powers of $x\text{,}$ $y\text{,}$ and $z\text{,}$ to first order, with second order error term.

\begin{gather*} \vv(x,y,z) = \vA + \vB\,x +\vC\,y +\vD\, z +\vR(x,y,z) \end{gather*}

where

\begin{gather*} \vA=\vv(0,0,0) \qquad \vB=\frac{\partial \vv}{\partial x}(0,0,0) \qquad \vC=\frac{\partial \vv}{\partial y}(0,0,0) \qquad \vD=\frac{\partial \vv}{\partial z}(0,0,0) \end{gather*}

and the error term $\vR(x,y,z)$ is bounded by a constant times $x^2+y^2+z^2\text{.}$ In particular there is a constant $K$ so that, on $C_\veps\text{,}$

\begin{equation*} |\vR(x,y,z)|\le K\veps^2 \end{equation*}

\begin{align*} &\oint_{C_\veps} \vv(\vr)\cdot \dee{\vr}\\ &\hskip0.25in=\int_0^{2\pi} \big(\vA + \vB\,\veps\,\cos t +\vC\,\veps\,\sin t +\vR(\vr(t))\big) \cdot \big(-\veps\sin t\,\hi +\veps\cos t\,\hj\big)\ \dee{t} \end{align*}

Again, multiply out the dot product so that the integrand becomes

\begin{alignat*}{2} & -\veps\vA\cdot\hi\,\sin t &&+\ \veps\vA\cdot\hj\,\cos t\\ &-\veps^2\vB\cdot\hi\,\sin t\cos t &&+\veps^2\vB\cdot\hj\,\cos^2t\\ &-\veps^2\vC\cdot\hi\,\sin^2 t &&+\ \veps^2\vC\cdot\hj\,\sin t\cos t\\ &+\vR(\vr(t))\cdot \big(-\veps\sin t\,\hi +\veps\cos t\,\hj\big) \end{alignat*}

Again most of these terms integrate to zero, because

\begin{alignat*}{2} \int_0^{2\pi}\sin t\ \dee{t} &=\hskip10pt\int_0^{2\pi}\cos t\ \dee{t} &&=0\\ \int_0^{2\pi}\sin t\cos t\ \dee{t} &= \frac{1}{2}\int_0^{2\pi}\sin(2t)\ \dee{t} &&=0 \end{alignat*}

and the $\sin^2t$ and $\cos^2 t$ terms are easily integrated using (see Example 2.4.4)

\begin{equation*} \int_0^{2\pi}\sin^2 t\ \dee{t}=\int_0^{2\pi}\cos^2 t\ \dee{t} =\frac{1}{2}\int_0^{2\pi}\big[\sin^2t+\cos^2 t\big]\ \dee{t}=\pi \end{equation*}

So we are left with

\begin{align*} \oint_{C_\veps} \vv(\vr)\cdot \dee{\vr} &= \pi\veps^2\vB\cdot\hj - \pi\veps^2\vC\cdot\hi +\int_0^{2\pi} \vR(\vr(t)) \cdot \big(-\veps\sin t\,\hi +\veps\cos t\,\hj\big)\ \dee{t} \end{align*}

which implies that

\begin{align*} \lim_{\veps\rightarrow 0}\frac{1}{\pi\veps^2} \oint_{C_\veps} \vv(\vr)\cdot \dee{\vr} &= \frac{\partial \vv_2}{\partial x}(0,0,0) - \frac{\partial \vv_1}{\partial y}(0,0,0)\\ &\hskip0.5in +\lim_{\veps\rightarrow 0}\! \frac{1}{\pi\veps^2} \int_0^{2\pi}\!\!\!\! \vR(\vr(t)) \cdot \big(-\veps\sin t\,\hi +\veps\cos t\,\hj\big)\,\dee{t}\\ &= \big(\vnabla\times\vv(0,0,0)\big)\cdot\hk\\ &\hskip0.5in +\lim_{\veps\rightarrow 0} \frac{1}{\pi\veps^2} \int_0^{2\pi} \vR(\vr(t)) \cdot \big(-\veps\sin t\,\hi +\veps\cos t\,\hj\big)\ \dee{t} \end{align*}

Finally, it suffices to recall that $|\vR(x,y,z)|\le K\veps^2\text{,}$ so that

\begin{align*} \frac{1}{\pi\veps^2}\bigg| \int_0^{2\pi} \vR(\vr(t)) \cdot \big(-\veps\sin t\,\hi +\veps\cos t\,\hj\big)\ \dee{t} \bigg| &\le \frac{1}{\pi\veps} \int_0^{2\pi}|\vR(\vr(t))|\,\dee{t}\\ &\le \frac{1}{\pi\veps} \int_0^{2\pi}K\veps^2\,\dee{t}\\ &= \frac{1}{\pi\veps} \ K\veps^2\,(2\pi)\\ &= 2K\veps \end{align*}

converges to zero as $\veps\rightarrow 0\text{.}$

Here are some examples. We will use the same vector fields as in Examples 4.1.21, 4.1.22 and 4.1.23. In all examples, we shall orient the paddlewheel so that $\hn=\hk$ and sketch the top view, so that the paddlewheel looks like

Example 4.1.26.

Here is a sketch of the vector field $\vv(x,y,z) = x\,\hi+y\,\hj+z\,\hk$ and a circle centered on the origin, like $C_\veps\text{.}$

This velocity field has fluid moving parallel to the paddles, so the paddlewheel should not rotate at all. The computation

\begin{align*} \vnabla\times\vv(\vZero) = \det\left[\begin{matrix} \hi & \hj &\hk \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ x & y & z \end{matrix} \right] =\vZero \implies \vnabla\times\vv(\vZero)\cdot\hk = 0 \end{align*}

is consistent with our interpretation 4.1.24.

Example 4.1.27.

Here is a sketch of the vector field $\vv(x,y,z) = -y\,\hi+x\,\hj$ and a circle centered on the origin, like $C_\veps\text{.}$

This velocity field has fluid going around in circles, counterclockwise. So the paddlewheel should rotate counterclockwise too. That is, it should have positive angular velocity. Our interpretation 4.1.24 predicts an angular velocity of half

\begin{align*} \vnabla\times\vv(\vZero)\cdot\hk = \det\left[\begin{matrix} \hi & \hj &\hk \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ -y & x & 0 \end{matrix} \right] \cdot\hk =2\hk\cdot\hk =2 \end{align*}

which is indeed positive

¹⁸

Even for small values of $2\text{.}$

Example 4.1.28.

Here is a sketch of the vector field $\vv(x,y,z) = \hi$ and a circle centered on the origin, like $C_\veps\text{.}$

The fluid pushing on the top paddle tries to make the paddlewheel rotate clockwise. The fluid pushing on the bottom paddle tries to make the paddlewheel rotate counterclockwise, at the same rate. So the paddlewheel should not rotate at all. Our interpretation 4.1.24 predicts an angular velocity of

\begin{align*} \frac{1}{2}\cdot\vnabla\times\vv(\vZero)\cdot\hk =\frac{1}{2} \det\left[\begin{matrix} \hi & \hj &\hk \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ 1 & 0 & 0 \end{matrix} \right] \cdot\hk =\vZero\cdot\hk =0 \end{align*}

as expected.

Exercises 4.1.6 Exercises

Exercise Group.

Exercises — Stage 1

1. (✳).

Let $\vF = P\,\hi + Q\,\hj$ be the two dimensional vector field shown below.

Assuming that the vector field in the picture is a force field, the work done by the vector field on a particle moving from point $A$ to $B$ along the given path is:
1. Positive
2. Negative
3. Zero
4. Not enough information to determine.
Which statement is the most true about the line integral $\int_{C_2} \vF\cdot\dee{\vr}\text{:}$
1. $\displaystyle \int_{C_2} \vF\cdot\dee{\vr} \gt 0$
2. $\displaystyle \int_{C_2} \vF\cdot\dee{\vr}=0$
3. $\displaystyle \int_{C_2} \vF\cdot\dee{\vr} \lt 0$
4. Not enough information to determine.
$\vnabla\cdot\vF$ at the point $N$ (in the picture) is:
1. Positive
2. Negative
3. Zero
4. Not enough information to determine.
$Q_x - P_y$ at the point $Q$ is:
1. Positive
2. Negative
3. Zero
4. Not enough information to determine.
Assuming that $\vF = P\,\hi + Q\,\hj\text{,}$ which of the following statements is correct about $\frac{\partial P}{\partial x}$ at the point $D\text{?}$
1. $\frac{\partial P}{\partial x}=0$ at $D\text{.}$
2. $\frac{\partial P}{\partial x} \gt 0$ at $D\text{.}$
3. $\frac{\partial P}{\partial x} \lt 0$ at $D\text{.}$
4. The sign of $\frac{\partial P}{\partial x}$ at $D$ can not be determined by the given information.

2.

Does $\vnabla\times \vF$ have to be perpendicular to $\vF\text{?}$

3.

Verify the vector identities

$\displaystyle \vnabla\cdot(f\vF)=f\vnabla\cdot\vF+\vF\cdot\vnabla f$
$\displaystyle \vnabla\cdot(\vF\times\vG) =\vG\cdot(\vnabla\times\vF)- \vF\cdot(\vnabla\times\vG)$
$\displaystyle \vnabla^2(fg)=f\,\vnabla^2 g+2\vnabla f\cdot\vnabla g+g\,\vnabla^2 f$

Exercise Group.

Exercises — Stage 2

4.

Evaluate $\vnabla\cdot\vF$ and $\vnabla\times\vF$ for each of the following vector fields.

$\displaystyle \vF=x\,\hi+y\,\hj+z\,\hk$
$\displaystyle \vF=xy^2\hi-yz^2\hj+zx^2\hk$
$\vF=\frac{x\hi+y\hj}{\sqrt{x^2+y^2}}$ (the polar basis vector $\hat{\bf r}$ in 2d)
$\vF=\frac{-y\hi+x\hj}{\sqrt{x^2+y^2}}$ (the polar basis vector $\hat{\pmb{\theta}}$ in 2d)

5. (✳).

Compute and simplify $\vnabla\cdot\big(\frac{\vr}{r}\big)$ for $\vr=(x,y,z)$ and $r=|(x,y,z)|\text{.}$ Express your answer in terms of $r\text{.}$
Compute $\vnabla\times\big(yz\,\hi + 2xz\,\hj + e^{xy}\,\hk\big)\text{.}$

6. (✳).

In the following, we use the notation $\vr = x\,\hi + y\,\hj + z\,\hk\text{,}$ $r = |\vr|\text{,}$ and $k$ is some number $k = 0, 1, -1, 2, -2, \dots\text{.}$

Find the value $k$ for which
\begin{equation*} \vnabla (r^k) = -3\frac{\vr}{r^5} \end{equation*}
Find the value $k$ for which
\begin{equation*} \vnabla \cdot (r^k\vr) = 5r^2 \end{equation*}
Find the value $k$ for which
\begin{equation*} \vnabla^2 (r^k) = \frac{2}{r^4} \end{equation*}

7. (✳).

Let $\vr$ be the vector field $\vr = x\,\hi + y\,\hj + z\,\hk$ and let $r$ be the function $r = |\vr|\text{.}$ Let $\va$ be the constant vector $\va = a_1\,\hi + a_2\,\hj + a_3\,\hk\text{.}$ Compute and simplify the following quantities. Answers must be expressed in terms of $\va\text{,}$ $\vr\text{,}$ and $r\text{.}$ There should be no $x$’s, $y$’s, or $z$’s in your answers.

$\displaystyle \vnabla\cdot\vr$
$\displaystyle \vnabla(r^2)$
$\displaystyle \vnabla\times(\vr\times\va)$
$\displaystyle \vnabla\cdot\big(\vnabla(r)\big)$

8. (✳).

Let

\begin{equation*} \vr = x\,\hi + y\,\hj + z\,\hk,\qquad r = |\vr| \end{equation*}

Compute $a$ where $\vnabla\big(\frac{1}{r}\big) =- r^a\,\vr\text{.}$
Compute $a$ where $\vnabla\cdot\big(r\,\vr\big) = ar\text{.}$
Compute $a$ where $\vnabla\cdot\big(\vnabla(r^3)\big) = ar\text{.}$

9.

Find, if possible, a vector field $\vA$ that has $\hk$ component $A_3=0$ and that is a vector potential for

$\displaystyle \vF=(1+yz)\hi+(2y+zx)\hj+(3z^2+xy)\hk$
$\displaystyle \vG= yz\hi+zx\hj+xy\hk$

Exercise Group.

Exercises — Stage 3

10. (✳).

Let

\begin{gather*} \vF = \frac{-z}{x^2+z^2}\,\hi +y\,\hj +\frac{x}{x^2+z^2}\,\hk \end{gather*}

Determine the domain of $\vF\text{.}$
Determine the curl of $\vF\text{.}$ Simplify if possible.
Determine the divergence of $\vF\text{.}$ Simplify if possible.
Is $\vF$ conservative? Give a reason for your answer.

11. (✳).

A physicist studies a vector field $\vF$ in her lab. She knows from theoretical considerations that $\vF$ must be of the form $\vF=\nabla\times\vG\text{,}$ for some smooth vector field $\vG\text{.}$ Experiments also show that $\vF$ must be of the form

\begin{equation*} \vF(x,y,z)=(xz+xy)\hi+\alpha(yz-xy)\hj+\beta(yz+xz)\hk \end{equation*}

where $\alpha$ and $\beta$ are constant.

Determine $\alpha$ and $\beta\text{.}$
Further experiments show that $\vG=xyz\hi-xyz\hj+g(x,y,z)\hk\text{.}$ Find the unknown function $g(x,y,z)\text{.}$

12.

A rigid body rotates at an angular velocity of $\Om$ rad/sec about an axis that passes through the origin and has direction $\Ha\text{.}$ When you are standing at the head of $\Ha$ looking towards the origin, the rotation is counterclockwise. Set $\vOm=\Om\Ha\text{.}$

Show that the velocity of the point $\vr=(x,y,z)$ on the body is $\vOm\times\vr\text{.}$
Evaluate $\vnabla\times(\vOm\times\vr)$ and $\vnabla\cdot(\vOm\times\vr)\text{,}$ treating $\vOm$ as a constant.
Find the speed of the students in a classroom located at latitude $49^\circ$ N due to the rotation of the Earth. Ignore the motion of the Earth about the Sun, the Sun in the Galaxy and so on. The radius of the Earth is 6378 km.

13.

Suppose that the vector field $\vF$ obeys $\vnabla\cdot \vF=0$ in all of $\bbbr^3\text{.}$ Let

\begin{equation*} \vr(t)=tx\,\hi+ty\,\hj+tz\,\hk,\qquad 0\le t\le 1 \end{equation*}

be a parametrization of the line segment from the origin to $(x,y,z)\text{.}$ Define

\begin{equation*} \vG(x,y,z)=\int_0^1 t\,\vF\big(\vr(t)\big)\times\frac{d\vr}{dt}(t)\,dt \end{equation*}

Show that $\vnabla\times \vG=\vF$ throughout $\bbbr^3\text{.}$

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