\chapter{Vector calculus}
\label{vcalc}
\index{vector calculus}
\index{calculus!of the vector}

Chapter~\ref{vector} has introduced the algebra of the three-dimensional
geometrical vector.  Like the scalar, the vector is a continuous
quantity and as such has not only an algebra but also a calculus.  This
chapter develops the calculus of the vector.

% ----------------------------------------------------------------------

\section{Fields and their derivatives}
\label{vcalc:320}
\index{field}
\index{field!derivative of}
\index{vector!derivative of}
\index{derivative!of a field}

\index{function!of position}
A scalar quantity $\sigma(t)$ or vector quantity $\ve f(t)$ whose value
varies over time is ``a function of time~$t$.'' We can likewise call a
scalar quantity%
\footnote{
  This $\psi(\ve r)$ is unrelated to the Tait-Bryan and Euler roll
  angles~$\psi$ of \S~\ref{vector:210}, an unfortunate but tolerable
  overloading of the Greek letter~$\psi$ in the conventional notation of
  vector analysis.  In the unlikely event of confusion, you can use an
  alternate letter like~$\eta$ for the roll angle.  See
  appendix~\ref{greek}\@.
}
$\psi(\ve r)$ or vector quantity $\ve a(\ve r)$ whose
value varies over space ``a function of position~$\ve r$,'' but there is
a special, alternate name for such a quantity.  We call it a
\emph{field.}

\index{scalar field}
\index{vector field}
\index{pressure}
\index{air}
\index{wind}
\index{pressure}
\index{velocity}
\index{component!of a vector field}
A field is a quantity distributed over space or, if you prefer, a
function in which spatial position serves as independent variable.  Air
pressure $p(\ve r)$ is an example of a \emph{scalar field,} whose value
at a given location~$\ve r$ has amplitude but no direction.  Wind
velocity%
\footnote{
  As \S~\ref{vector:230}, this section also uses the letter~$q$ for
  velocity in place of the conventional~$v$ \cite[\S~18.4]{BSL}, which
  it needs for another purpose.
}
$\ve q(\ve r)$
is an example of a \emph{vector field,} whose value at a given
location~$\ve r$ has both amplitude and direction.  These are typical
examples.  Tactically, a vector field can be thought of as composed of
three scalar fields
\[
  \ve q(\ve r) =
      \vu x q_x(\ve r)
    + \vu y q_y(\ve r)
    + \vu z q_z(\ve r);
\]
but, since
\[
  \ve q(\ve r) =
      \vu x' q_{x'}(\ve r)
    + \vu y' q_{y'}(\ve r)
    + \vu z' q_{z'}(\ve r)
\]
for any orthogonal basis $[\ve x'\;\ve y'\;\ve z']$ as well, the
specific scalar fields $q_x(\ve r)$, $q_y(\ve r)$ and $q_z(\ve r)$ are
no more essential to the vector field $\ve q(\ve r)$ than the specific
scalars~$b_x$, $b_y$ and~$b_z$ are to a vector~$\ve b$.  As we said, the three
components come tactically; typically, such components are uninteresting
in themselves.  The field $\ve q(\ve r)$ as a whole is the interesting
thing.

Scalar and vector fields are of utmost use in the modeling of physical
phenomena.

\index{derivative!with respect to position}
As one can take the derivative $d\sigma/dt$ or $d\ve f/dt$ with respect
to time~$t$ of a function $\sigma(t)$ or $\ve f(t)$, one can likewise
take the derivative with respect to position~$\ve r$ of a field
$\psi(\ve r)$ or $\ve a(\ve r)$.  However, derivatives with respect to
position create a notational problem, for it is not obvious what
symbology like $d\psi/d\ve r$ or $d\ve a/d\ve r$ would actually mean.
The notation $d\sigma/dt$ means ``the rate of~$\sigma$ as time~$t$
advances,'' but if the notation $d\psi/d\ve r$ likewise meant ``the rate
of~$\psi$ as position~$\ve r$ advances'' then it would necessarily
prompt one to ask, ``advances in which direction?''  The notation offers
no hint.  In fact $d\psi/d\ve r$ and $d\ve a/d\ve r$ mean nothing very
distinct in most contexts and we shall avoid such notation.  If we will
speak of a field's derivative with respect to position~$\ve r$ then we
shall be more precise.

Section~\ref{vector:220} has given the vector three distinct kinds of
product.  This section gives the field no fewer than four distinct kinds
of derivative: the directional derivative; the gradient; the divergence;
and the curl.%
\footnote{
  Vector veterans may notice that the Laplacian is not listed.  This is
  not because the Laplacian were uninteresting but rather because the
  Laplacian is actually a second-order derivative---a derivative of a
  derivative.  We will address the Laplacian in \S~\ref{vcalc:350}.
}

\index{caveman}
\index{hut}
\index{roof}
\index{tree}
\index{antelope}
\index{sky}
\index{upstairs}
\index{downstairs}
\index{floor}
So many derivatives bring the student a conceptual difficulty one could
call ``the caveman problem.''  Imagine a caveman.  Suppose that you
tried to describe to the caveman a house or building of more than one
floor.  He might not understand.  You and I who already grasp the
concept of upstairs and downstairs do not find a building of two
floors, or three or even thirty, especially hard to envision, but our
caveman is used to thinking of the ground and the floor as more or less
the same thing.  To try to think of upstairs and downstairs might
confuse him with partly false images of sitting in a tree or of
clambering onto (and breaking) the roof of his hut.  ``There are many
trees and antelopes but only one sky and floor.  How can one speak of
many skies or many floors?''  The student's principal conceptual
difficulty with the several vector derivatives is of this kind.

\subsection{The~$\nabla$ operator}
\label{vcalc:320.10}
\index{del ($\nabla$)}
\index{$\nabla$ (del)}

\index{Tuesday}
\index{Wednesday}
\index{Thursday}
\index{week!days of}
\index{days of the week}
Consider a vector
\[
  \ve a = \vu x a_x + \vu y a_y + \vu z a_z.
\]
Then consider a ``vector''
\[
  \ve c = \vu x [\mbox{Tuesday}] + \vu y [\mbox{Wednesday}] + \vu z [\mbox{Thursday}].
\]
If you think that the latter does not look very much like a vector, then
the writer thinks as you do, but consider:
\[
  \ve c \cdot \ve a = [\mbox{Tuesday}] a_x + [\mbox{Wednesday}] a_y + [\mbox{Thursday}] a_z.
\]
The writer does not know how to interpret a nonsensical term like
\linebreak % bad break
``$[\mbox{Tuesday}] a_x$'' any more than the reader does, but the point
is that~$\ve c$ behaves as though it were a vector insofar as vector
operations like the dot product are concerned.  What matters in this
context is not that~$\ve c$ have amplitude and direction (it has
neither) but rather that it have the three orthonormal components it
needs to participate formally in relevant vector operations.  It has
these.  That the components' amplitudes seem nonsensical is beside the
point.  Maybe there exists a model in which ``$[\mbox{Tuesday}]$'' knows
how to operate on a scalar like~$a_x$.  (Operate on?  Yes.  Nothing in
the dot product's definition requires the component amplitudes of~$\ve
c$ \emph{to multiply} those of~$\ve a$.  Multiplication is what the
component amplitudes of true vectors do, but~$\ve c$ is not a true
vector, so ``$[\mbox{Tuesday}]$'' might do something to~$a_x$ other than
to multiply it.  Section~\ref{vcalc:320.15} elaborates the point.)  If
there did exist such a model, then the dot product $\ve c \cdot \ve a$
could be licit in that model.  As if this were not enough, the cross
product $\ve c \times \ve a$ too could be licit in that model, composed
according to the usual rule for cross products.  The model might allow
it.  The dot and cross products in and of themselves do not forbid it.

Now consider a ``vector''
\bq{vcalc:del}
  \nabla
    = \vu x \frac{\partial}{\partial x}
    + \vu y \frac{\partial}{\partial y}
    + \vu z \frac{\partial}{\partial z}.
\eq
This~$\nabla$ is not a true vector any more than~$\ve c$ is, maybe,
but if we treat it as one then we have that
\[
  \nabla \cdot \ve a
    = \frac{\partial a_x}{\partial x}
    + \frac{\partial a_y}{\partial y}
    + \frac{\partial a_z}{\partial z}.
\]
Such a dot product might or might not prove useful; but, unlike the
terms in the earlier dot product, at least we know what this one's terms
mean.

\index{partial derivative}
\index{derivative!partial}
\index{operator}
\index{vector!ersatz}
\index{invariance under the reorientation of axes}
\index{reorientation!invariance under}
\index{axes!invariance under the reorientation of}
Well, days of the week, partial derivatives, ersatz vectors---it all
seems rather abstract.  What's the point?  The answer is that there
wouldn't be any point if the only nonvector ``vectors'' in question were
of $\ve c$'s nonsensical kind.  The operator~$\nabla$ however shares more
in common with a true vector than merely having~$x$, $y$ and~$z$
components; for, like a true vector, the operator~$\nabla$ is amenable
to having its axes reoriented by~(\ref{vector:210:20}),
(\ref{vector:210:25}), (\ref{vector:210:40}) and~(\ref{vector:210:45}).
This is easier to see at first with respect the true vector~$\ve a$, as
follows.  Consider rotating the~$x$ and~$y$ axes through an angle~$\phi$
about the~$z$ axis.  There ensues
\bqb
  \ve a &=&
    \vu x a_x + \vu y a_y + \vu z a_z
  \\&=&
      (\vu x' \cos\phi - \vu y' \sin\phi)(a_{x'} \cos\phi - a_{y'} \sin\phi)
  \\&&\ \ \mbox{}
    + (\vu x' \sin\phi + \vu y' \cos\phi)(a_{x'} \sin\phi + a_{y'} \cos\phi)
    + \vu z' a_{z'}
  \\&=&
      \vu x'[ a_{x'} \cos^2\phi - a_{y'} \cos\phi\sin\phi
        + a_{x'} \sin^2\phi + a_{y'} \cos\phi\sin\phi ]
  \\&&\ \ \mbox{}
    + \vu y'[ -a_{x'} \cos\phi\sin\phi + a_{y'} \sin^2\phi
        + a_{x'} \cos\phi\sin\phi + a_{y'} \cos^2\phi ]
  \\&&\ \ \mbox{}
        + \vu z' a_{z'}
  \\&=&
    \vu x' a_{x'} + \vu y' a_{y'} + \vu z' a_{z'},
\eqb
where the final expression has different axes than the original but,
relative to those axes, exactly the same form.  Further rotation about
other axes would further reorient but naturally also would not alter the
form.  Now consider~$\nabla$.  The partial differential operators
$\partial/\partial x$, $\partial/\partial y$ and $\partial/\partial z$
change no differently under reorientation than the component
amplitudes~$a_x$, $a_y$ and~$a_z$ do.  Hence,
\bq{vcalc:delp}
  \nabla = \vui \frac{\partial}{\partial i}
    = \vu x' \frac{\partial}{\partial x'}
    + \vu y' \frac{\partial}{\partial y'}
    + \vu z' \frac{\partial}{\partial z'},
\eq
evidently the same operator regardless of the choice of basis $[\vu
x'\;\vu y'\;\vu z']$.  It is this invariance under reorientation that
makes the~$\nabla$ operator useful.

If~$\nabla$ takes the place of the ambiguous $d/d\ve r$, then what takes
the place of the ambiguous $d/d\ve r_o$, $d/d\tilde{\ve
r}$, $d/d\ve r^\dagger$, $d/d\ve r'$ and so on?  Answer:~$\nabla_{\!o}$,
$\tilde{\nabla}$, $\nabla^\dagger$, $\nabla'$ and so on.  Whatever mark
distinguishes the special~$\ve r$, the same mark distinguishes the
corresponding special~$\nabla$.  For example, where $\ve r_o = \vui
i_o$, there $\nabla_{\!o} = \vui\,\partial/\partial i_o$.  That is the
convention.%
\footnote{
  A few readers not fully conversant with the material of
  chapter~\ref{vector}, to whom this chapter had been making sense until the
  last two sentences, may suddenly find the notation incomprehensible.
  The notation is Einstein's.  It means
  \[
    \vui i_o = \!\!\!\sum_{i=x',y',z'}\!\!\! \vui i_o
      = \vu x' x_o' + \vu y' y_o' + \vu z' z_o',
  \]
  in the leftmost form of which the summation sign is implied not
  written.  Refer to \S~\ref{vector:240}.
}

\index{Hamilton, William Rowan (1805--1865)}
\index{Heaviside, Oliver (1850--1925)}
Introduced by William Rowan Hamilton and Oliver Heaviside, informally
pronounced ``del'' (in the author's country at least), the vector
differential operator~$\nabla$
finds extensive use in the modeling of physical phenomena.  After a
brief digression to discuss operator notation, the subsections that
follow will use the operator to develop and present the four basic kinds
of vector derivative.

\subsection{Operator notation}
\label{vcalc:320.15}
\index{operator notation}
\index{notation!of the operator}
\index{operator}

Section~\ref{vcalc:320.10} has introduced \emph{operator notation}
without explaining what it is or even what it concerns.  This subsection
digresses to explain.

\index{unary operator}
\index{operator!unary}
Operator notation concerns the representation of unary operators and the
operations they specify.  Section~\ref{integ:240} has already broadly
introduced the notion of the operator.  A \emph{unary operator} is a
mathematical agent that transforms a single discrete quantity, a single
distributed quantity, a single field, a single function or another
single mathematical object in some definite way.  For example,
$J \equiv \int_0^t dt$ is a unary operator, more fully written as
$J \equiv \int_0^t \cdot\, dt$ where the~``$\cdot$'' holds
the place of the thing operated upon,%
\footnote{\cite{Brown-conversation}}
whose effect is such that, for instance, $Jt = t^2/2$ and $J\cos\omega
t=(\sin \omega t)/\omega$.  Any letter might serve as well as the
example's~$J$; but what distinguishes operator notation is that, like
the matrix row operator~$A$ in matrix notation's product~$A\ve x$
(\S~\ref{matrix:120.10}), the operator~$J$ in operator notation's
operation~$Jt$ attacks from the left.  Thus, generally, $Jt \neq tJ$
if~$J$ is a unary operator, though the notation~$Jt$ usually formally
resembles multiplication in other respects as we shall see.

\index{matrix operator}
The matrix actually is a type of unary operator and matrix notation is a
specialization of operator notation, so we have met operator notation
before.  And, in fact, we have met operator notation much earlier than
that.  The product~$5t$ can if you like be regarded as the unary
operator~``5 times,'' operating on~$t$.  Perhaps you did not know
that~$5$ was an operator---and, indeed, the scalar~$5$ itself is no
operator but just a number---but where no other operation is defined
operator notation implies scalar multiplication by default.  Seen in
this way,~$5t$ and~$t5$ actually mean two different things; though
naturally in the specific case of scalar multiplication, which happens
to be commutative, it is true that $5t=t5$.

The $\ve a \cdot \mbox{}$ in the dot product $\ve a \cdot \ve b$ and
the $\ve a \times \mbox{}$ in the cross product $\ve a \times \ve b$ can
profitably be regarded as unary operators.

\index{unary linear operator}
\index{linear operator!unary}
\index{operator!unary linear}
\index{operator!linear}
\index{linearity!of an operator}
\index{commutivity!noncommutivity of unary linear operators}
\index{distributivity!of unary linear operators}
Whether operator notation can licitly represent any unary operation
whatsoever is a definitional question we will leave for the professional
mathematician to answer, but in normal usage operator notation
represents only \emph{linear unary operations,} unary operations that
honor \S~\ref{integ:240.05}'s rule of linearity.  The
operators~$J$ and~$A$ above are examples of linear unary operators; the
operator $K \equiv \cdot+3$ is not linear and almost certainly should
never be represented in operator notation as here, lest an expression
like~$Kt$ mislead an understandably unsuspecting audience.  Linear unary
operators often do not commute, so $J_1J_2 \neq J_2J_1$ generally; but
otherwise linear unary operators follow familiar rules of multiplication
like $(J_2+J_3)J_1=J_2J_1+J_3J_1$.  Linear unary operators obey a
definite algebra, the same algebra matrices obey.  It is precisely this
algebra that makes operator notation so useful.

\index{associativity!of unary linear operators}
Operators associate from right to left (\S~\ref{alggeo:222.1}) so that,
in operator notation, $J\omega t = J(\omega t)$, not $(J\omega)t$.
Observe however that the perceived need for parentheses comes only of
the purely notational ambiguity as to
whether~$\omega t$ bears the semantics of ``the product of~$\omega$
and~$t$'' or those of ``the unary operator~`$\omega$ times,' operating
on~$t$.'' The perceived need and any associated confusion would vanish if $\Omega
\equiv (\omega)(\cdot)$ were unambiguously an operator, in which case
the product~$J\Omega$ would itself be an operator, whereupon
$(J\Omega) t = J\Omega t = J(\Omega t) = J(\omega t)$.  Indeed, one can
compare the distinction in \S~\ref{matrix:180.35} between~$\lambda$
and~$\lambda I$ against the distinction between~$\omega$ and~$\Omega$
here, for a linear unary operator enjoys the same
associativity~(\ref{matrix:000:42}) a matrix enjoys, and for the same
reason.  Still, rather than go to the trouble of defining extra symbols
like~$\Omega$, it is usually easier just to write the parentheses, which
take little space on the page and are universally understood; or,
better, to rely on the right-to-left convention that $J\omega t =
J(\omega t)$.  Modern conventional applied mathematical notation though
generally excellent remains imperfect; so, notationally, when it
matters, operators associate from right to left except where parentheses
group otherwise.

\index{unresolved operator}
\index{operator!unresolved}
\index{unary operator!unresolved}
One can speak of a unary operator like~$J$, $A$ or~$\Omega$ without
giving it anything in particular to operate upon.  One can leave an
operation unresolved.  For example,~$tJ$ is itself a unary operator---it
is the operator $t\int_0^t dt$---though one can assign no particular
value to it until it actually operates on something.  The
operator~$\nabla$ of~(\ref{vcalc:delp}) is an unresolved unary operator
of the same kind.

\subsection{The directional derivative and the gradient}
\label{vcalc:320.20}
\index{directional derivative}
\index{derivative!directional}
\index{field!directional derivative of}
\index{scalar field!directional derivative of}
\index{vector field!directional derivative of}
\index{gradient}
\index{field!gradient of}
\index{scalar field!gradient of}

\index{reference vector}
\index{vector!reference}
In the calculus of vector fields, the derivative notation $d/d\ve r$ is
ambiguous because, as the section's introduction has observed, the
notation gives~$\ve r$ no specific direction in which to advance.  In
operator notation, however, given~(\ref{vcalc:delp}) and accorded a
reference vector~$\ve b$ to supply a direction and a scale, one can
compose the \emph{directional derivative} operator
\bq{vcalc:320:21}
  (\ve b \cdot \nabla) = b_i\frac{\partial}{\partial i}
\eq
to express the derivative unambiguously.  This operator applies equally
to the scalar field,
\[
  (\ve b \cdot \nabla)\psi(\ve r) = b_i\frac{\partial\psi}{\partial i},
\]
as to the vector field,
\bq{vcalc:ddvf}
  (\ve b \cdot \nabla)\ve a(\ve r) = b_i\frac{\partial\ve a}{\partial i}
  = \vuj b_i \frac{\partial a_j}{\partial i}.
\eq
For the scalar field the parentheses are unnecessary and conventionally
are omitted, as
\bq{vcalc:ddsf}
  \ve b \cdot \nabla\psi(\ve r) = b_i\frac{\partial\psi}{\partial i}.
\eq
In the case~(\ref{vcalc:ddvf}) of the vector field, however, $\nabla\ve
a(\ve r)$ itself means nothing coherent%
\footnote{
  Well, it does mean something coherent in \emph{dyadic analysis}
  \cite[appendix~B]{vanBladel}, but this book won't treat that.
}
so the parentheses usually are retained.  Equations~(\ref{vcalc:ddvf})
and~(\ref{vcalc:ddsf}) define the directional derivative.

Note that the directional derivative is the derivative not of the
reference vector~$\ve b$ but only of the field $\psi(\ve r)$ or $\ve
a(\ve r)$.  The vector~$\ve b$ just directs and scales the derivative; it is not
the object of it.  Nothing requires~$\ve b$ to be constant, though.  It
can be a vector field $\ve b(\ve r)$ that varies from place to place;
the directional derivative does not care.

\index{steepest rate}
Within~(\ref{vcalc:ddsf}), the quantity
\bq{vcalc:grad}
  \nabla \psi(\ve r) = \vui \frac{\partial\psi}{\partial i}
\eq
is called the \emph{gradient} of the scalar field $\psi(\ve r)$.
Though both scalar and vector fields have directional derivatives, only
scalar fields have gradients.  The gradient represents the amplitude and
direction of a scalar field's locally steepest rate.

Formally a dot product, the directional derivative operator $\ve b \cdot
\nabla$ is invariant under reorientation of axes, whereupon the
directional derivative is invariant, too.  The result of a~$\nabla$
operation, the gradient $\nabla\psi(\ve r)$ is likewise invariant.

\subsection{Divergence}
\label{vcalc:320.30}
\index{divergence}
\index{field!divergence of}
\index{vector field!divergence of}

There exist other vector derivatives than the directional derivative and
gradient of \S~\ref{vcalc:320.20}.  One of these is divergence.  It is
not easy to motivate divergence directly, however, so we will approach
it indirectly, through the concept of flux as follows.

\index{flux}
\index{surface!closed}
\index{football}
\index{goal post and goal line}
\index{air}
\index{barometric pressure}
The \emph{flux} of a vector field $\ve a(\ve r)$ outward from a region
in space is
\bq{vcalc:flux}
  \Phi \equiv \oint_S \ve a(\ve r) \cdot d\ve s,
\eq
where
\bq{vcalc:ds}
  d\ve s \equiv \vu n \cdot ds
\eq
is a vector infinitesimal of amplitude~$ds$, directed normally outward
from the closed surface bounding the region---$ds$ being the area of an
infinitesimal element of the surface, the area of a tiny patch.  Flux is
flow through a surface: in this case, net flow outward from the region
in question.  (The paragraph says much in relatively few words.  If it
seems opaque then try to visualize eqn.~\ref{vcalc:flux}'s dot product
$\ve a[\ve r] \cdot d\ve s$, in which the vector~$d\ve s$ represents the area
and orientation of a patch of the region's enclosing surface.  When
something like air flows through any surface---not necessarily a
physical barrier but an imaginary surface like the goal line's vertical
plane in a football game%
\footnote{
  The author has American football in mind but other football games
  have goal lines and goal posts, too.  Pick your favorite brand of
  football.
}%
---what matters is not the surface's area as such but rather the area
the surface presents to the flow.  The surface presents its full area to
a perpendicular flow but otherwise the flow sees a foreshortened
surface, \emph{as though the surface were projected onto a plane
perpendicular to the flow.}  Refer to Fig.~\ref{vector:220:fig-dot}.
Now realize that eqn.~\ref{vcalc:flux} actually describes flux not
through an open surface but through a closed---it could be the imaginary
rectangular box
enclosing the region of football play to goal-post height; where wind
blowing through the region, entering and leaving, would constitute zero
net flux; but where a positive net flux would have barometric pressure
falling and air leaving the region maybe because a storm is coming---and
you've got the idea.)

\index{source}
\index{sink}
A region of positive flux is a \emph{source;} of negative flux, a
\emph{sink.}  One can contemplate the flux
$\Phi_\mr{open} = \int_S \ve a(\ve r) \cdot d\ve s$
through an open surface as well as through a closed,
but it is the outward flux~(\ref{vcalc:flux}) through a closed surface
that will concern us here.

\index{derivative!constant}
\index{volume!enclosed by a surface}
\index{surface!closed}
The outward flux~$\Phi$ of a vector field $\ve a(\ve r)$ through a
closed surface bounding some definite region in space is evidently
\[
  \Phi =
    \intsx \Delta a_x(y,z) \,dy\,dz
  + \intsx \Delta a_y(z,x) \,dz\,dx
  + \intsx \Delta a_z(x,y) \,dx\,dy,
\]
where
\[
  \begin{split}
    \Delta a_x(y,z) &=
      \int_{x_\mr{min}(y,z)}^{x_\mr{max}(y,z)}
      \frac{\partial a_x}{\partial x} \,dx, \\
    \Delta a_y(z,x) &=
      \int_{y_\mr{min}(z,x)}^{y_\mr{max}(z,x)}
      \frac{\partial a_y}{\partial y} \,dy, \\
    \Delta a_z(x,y) &=
      \int_{z_\mr{min}(x,y)}^{z_\mr{max}(x,y)}
      \frac{\partial a_z}{\partial z} \,dz
  \end{split}
\]
represent the increase across the region respectively of~$a_x$, $a_y$
or~$a_z$ along an $\vu x$-, $\vu y$- or $\vu z$-directed line.%
\footnote{
  Naturally, if the region's boundary happens to be concave, then some
  lines might enter and exit the region more than once, but this merely
  elaborates the limits of integration along those lines.  It changes
  the problem in no essential way.
}
If the field has constant derivatives $\partial\ve a/\partial i$, or
equivalently if the region in question is small enough that the
derivatives are practically constant through it, then these increases
are simply
\[
  \begin{split}
    \Delta a_x(y,z) &=
      \frac{\partial a_x}{\partial x} \,\Delta x(y,z), \\
    \Delta a_y(z,x) &=
      \frac{\partial a_y}{\partial y} \,\Delta y(z,x), \\
    \Delta a_z(x,y) &=
      \frac{\partial a_z}{\partial z} \,\Delta z(x,y),
  \end{split}
\]
upon which
\bqb
  \Phi &=&
    \frac{\partial a_x}{\partial x} \intsx \Delta x(y,z) \,dy\,dz
  + \frac{\partial a_y}{\partial y} \intsx \Delta y(z,x) \,dz\,dx
  \\&&\ \ \mbox{}
  + \frac{\partial a_z}{\partial z} \intsx \Delta z(x,y) \,dx\,dy.
\eqb
But each of the last equation's three integrals represents the region's
volume~$V$, so
\[
  \Phi = (V)\left(
      \frac{\partial a_x}{\partial x}
    + \frac{\partial a_y}{\partial y}
    + \frac{\partial a_z}{\partial z}
  \right);
\]
or, dividing through by the volume,
\bq{vcalc:320:35}
  \frac{\Phi}{V} =
      \frac{\partial a_x}{\partial x}
    + \frac{\partial a_y}{\partial y}
    + \frac{\partial a_z}{\partial z}
  = \frac{\partial a_i}{\partial i}
  = \nabla \cdot \ve a(\ve r).
\eq
We give this ratio of outward flux to volume,
\bq{vcalc:div}
  \nabla \cdot \ve a(\ve r)
  = \frac{\partial a_i}{\partial i},
\eq
the name \emph{divergence,} representing the intensity of a source or
sink.

Formally a dot product, divergence is invariant under reorientation of
axes.

\subsection{Curl}
\label{vcalc:320.40}
\index{curl}
\index{field!curl of}
\index{vector field!curl of}

\index{circulation}
Curl is to divergence as the cross product is to the dot product.  Curl
is a little trickier to visualize, though.  It needs first the concept
of circulation as follows.

\index{contour!closed planar}
The \emph{circulation} of a vector field $\ve a(\ve r)$ about a closed
contour in space is
\bq{vcalc:circulation}
  \Gamma \equiv \oint \ve a(\ve r) \cdot d\we\ell,
\eq
where, unlike the~$\oint_S$ of~(\ref{vcalc:flux}) which represented a
double integration over a surface, the~$\oint$ here represents only a
single integration.  One can in general contemplate circulation about
any closed contour, but it suits our purpose here to consider
specifically a closed contour that happens not to depart from a single,
flat plane in space.

\index{curl!directional}
\index{directional curl}
\index{area!enclosed by a contour}
Let $[\vu u\;\vu v\;\vu n]$ be an orthogonal basis with~$\vu n$
normal to the contour's plane such that travel positively along the
contour tends from~$\vu u$ toward~$\vu v$ rather than the reverse.
The circulation~$\Gamma$ of a vector field $\ve a(\ve r)$ about this
contour is evidently
\[
  \Gamma =
      \int \Delta a_v(v) \,dv
    - \int \Delta a_u(u) \,du,
\]
where
\[
  \begin{split}
    \Delta a_v(v) &= \int_{u_\mr{min}(v)}^{u_\mr{max}(v)} \frac{\partial a_v}{\partial u} \,du, \\
    \Delta a_u(u) &= \int_{v_\mr{min}(u)}^{v_\mr{max}(u)} \frac{\partial a_u}{\partial v} \,dv
  \end{split}
\]
represent the increase across the contour's interior respectively
of~$a_v$ or~$a_u$ along a $\vu u$- or $\vu v$-directed line.
If the field has constant derivatives $\partial\ve a/\partial i$, or
equivalently if the contour in question is short enough that the
derivatives are practically constant over it, then these increases
are simply
\[
  \begin{split}
    \Delta a_v(v) &= \frac{\partial a_v}{\partial u} \,\Delta u(v), \\
    \Delta a_u(u) &= \frac{\partial a_u}{\partial v} \,\Delta v(u),
  \end{split}
\]
upon which
\[
  \Gamma =
      \frac{\partial a_v}{\partial u} \int \,\Delta u(v) \,dv
    - \frac{\partial a_u}{\partial v} \int \,\Delta v(u) \,du.
\]
But each of the last equation's two integrals represents the area~$A$
within the contour, so
\[
  \Gamma = (A)\left(
      \frac{\partial a_v}{\partial u}
    - \frac{\partial a_u}{\partial v}
  \right);
\]
or, dividing through by the area,
\bqa
  \frac{\Gamma}{A} &=&
      \frac{\partial a_v}{\partial u}
    - \frac{\partial a_u}{\partial v}
  \xn\\&=&
    \vu n \cdot \left[ \ep_{ijk} \vui \frac{\partial a_k}{\partial j} \right] =
    \vu n \cdot \left|
      \br{ccc}
        \vu u & \vu v & \vu n \\
        \partial/\partial u &
        \partial/\partial v &
        \partial/\partial n \\
        a_u & a_v & a_n
      \er
    \right|
  \xn\\&=& \vu n \cdot [ \nabla \times \ve a(\ve r) ].
  \label{vcalc:320:45}
\eqa
We give this ratio of circulation to area,
\bq{vcalc:320:43}
  \vu n \cdot [ \nabla \times \ve a(\ve r) ]
  = \vu n \cdot \left[ \ep_{ijk} \vui \frac{\partial a_k}{\partial j} \right]
  = \frac{\partial a_v}{\partial u} - \frac{\partial a_u}{\partial v},
\eq
the name \emph{directional curl,} representing the intensity of
circulation, the degree of twist so to speak, about a specified axis.
The cross
product in~(\ref{vcalc:320:43}),
\bq{vcalc:curl}
  \nabla \times \ve a(\ve r)
  = \ep_{ijk} \vui \frac{\partial a_k}{\partial j},
\eq
we call \emph{curl.}

Curl~(\ref{vcalc:curl}) is an interesting quantity.  Although it emerges from directional
curl~(\ref{vcalc:320:43}) and although we have developed directional
curl with respect to a contour in some specified
plane, curl~(\ref{vcalc:curl}) itself turns out to be altogether
independent of any particular plane.  We might have chosen another plane
and though~$\vu n$ would then be different the same~(\ref{vcalc:curl})
would necessarily result.  Directional curl, a scalar, is a property of
the field and the plane.  Curl, a vector, unexpectedly is a property of
the field only.  Directional curl evidently cannot exceed curl in
magnitude, but will equal it when~$\vu n$ points in its direction, so it
may be said that curl is the locally greatest directional curl, oriented
normally to the locally greatest directional curl's plane.

We have needed~$\vu n$ and~(\ref{vcalc:320:43}) to motivate and develop
the concept
% bad break
(\ref{vcalc:curl})
of curl.  Once developed, however, the concept of curl stands on its
own, whereupon one can return to define directional curl more generally
than~(\ref{vcalc:320:43}) has defined it.  As in~(\ref{vcalc:ddvf})
here too any reference vector~$\ve b$ or vector field $\ve b(\ve r)$ can
serve to direct the curl, not only~$\vu n$.  Hence,
\bq{vcalc:dcurl}
  \ve b \cdot [ \nabla \times \ve a(\ve r) ]
  = \ve b \cdot \left[ \ep_{ijk} \vui \frac{\partial a_k}{\partial j} \right]
  = \ep_{ijk} b_i \frac{\partial a_k}{\partial j}.
\eq
This would be the actual definition of \emph{directional curl.}  Note
however that directional curl so defined is not a distinct kind of
derivative but rather is just curl, dot-multiplied by a reference
vector.

Formally a cross product, curl is invariant under reorientation of axes.
An ordinary dot product, directional curl is likewise invariant.

\subsection{Cross-directional derivatives}
\label{vcalc:320.45}
\index{derivative!cross-directional}
\index{cross-directional derivative}

\index{curl!cross-directional}
\index{cross-directional curl}
The several directional derivatives of the $\ve b \cdot \nabla$ class,
including the scalar
% bad break
(\ref{vcalc:ddsf})
and vector~(\ref{vcalc:ddvf}) directional derivatives themselves and
also including directional curl~(\ref{vcalc:dcurl}), compute rates
with reference to some direction~$\ve b$.  Another class of directional
derivatives however is possible, that of the \emph{cross-directional
derivatives.}%
\footnote{
  The author is unaware of a conventional name for these derivatives.
  The name \emph{cross-directional} seems as apt as any.
}
These compute rates perpendicularly to~$\ve b$.  Unlike the vector
directional derivative~(\ref{vcalc:ddvf}), the cross-directional
derivatives are not actually new derivatives but are cross products
of~$\ve b$ with derivatives already familiar to us.  The
cross-directional derivatives are
\bq{vcalc:320:46}
  \begin{split}
    \ve b \times \nabla \psi &= \ep_{ijk} \vui b_j \pp{\psi}{k}, \\
    \ve b \times \nabla \times \ve a &= \vuj b_i \left( \pp{a_i}{j} - \pp{a_j}{i} \right).
  \end{split}
\eq
We call these respectively the \emph{cross-directional derivative}
(itself) and \emph{cross-directional curl,} the latter derived as
\bqb
  \ve b \times \nabla \times \ve a
  &=&
    \ve b \times \left( \ep_{ijk} \vui \pp{a_k}{j} \right)
  \\&=&
    \ep_{mni} \vu m b_n \left( \ep_{ijk} \vui \pp{a_k}{j} \right)_{\!i}
  =
    \ep_{mni} \ep_{ijk} \vu m b_n \pp{a_k}{j}
  \\&=&
    ( \delta_{mj}\delta_{nk} - \delta_{mk}\delta_{nj} ) \vu m b_n \pp{a_k}{j}
  \\&=&
    \vuj b_k \pp{a_k}{j} - \vu k b_j \pp{a_k}{j}
  =
    \ds\vuj b_i \pp{a_i}{j} - \ds\vuj b_i \pp{a_j}{i}
\eqb
where the Levi-Civita identity that $
  \ep_{mni} \ep_{ijk} = \ep_{imn} \ep_{ijk}
  = \delta_{mj}\delta_{nk} - \delta_{mk}\delta_{nj}
$ comes from Table~\ref{vector:240:tbl1}.

% ----------------------------------------------------------------------

\section{Integral forms}
\label{vcalc:330}
\index{integral forms of vector calculus}
\index{vector calculus!integral forms of}

The vector field's distinctive maneuver is the shift between integral forms,
which we are now prepared to treat.  This shift comes in two kinds.  The
two subsections that follow explain.

\subsection{The divergence theorem}
\label{vcalc:330.30}
\index{divergence theorem}

\index{surface!inner and outer}
\index{volume!enclosed by a surface}
\index{surface!closed}
\index{inner surface}
\index{outer surface}
\index{volume element}
\index{element!interior and boundary}
\index{interior element}
\index{boundary element}
\index{flux}
\index{divergence}
Section~\ref{vcalc:320.30} has contemplated the flux of a vector field
$\ve a(\ve r)$ from a volume small enough that the divergence $\nabla \cdot
\ve a$ were practically constant through the volume.  One would like to treat
the flux from larger, more general volumes as well.  According
to the definition~(\ref{vcalc:flux}), the flux from any volume is
\[
  \Phi = \oint_S \ve a \cdot d\ve s.
\]
If one subdivides a large volume into infinitesimal volume
elements~$dv$, then the flux from a single volume element is
\[
  \Phi_{\mr{element}} = \oint_{S_\mr{element}} \ve a \cdot d\ve s.
\]
Even a single volume element however can have two distinct kinds of
surface area: inner surface area shared with another element; and outer
surface area shared with no other element because it belongs to the
surface of the larger, overall volume.  Interior elements naturally have
only the former kind but boundary elements have both kinds of surface
area, so one can elaborate the last equation to read
\[
  \Phi_{\mr{element}} = \int_{S_\mr{inner}} \ve a \cdot d\ve s
  + \int_{S_\mr{outer}} \ve a \cdot d\ve s
\]
for a single element, where
% The following is probably obsolete:
% A \rule could be added here to leave a little space between the
% in-line formula, which ends the paragraph, and the first line of the
% paragraph following, for LaTeX crowds it too much.  However, the
% writer is afraid of what the rule might quietly do if future
% redactions earlier in the chapter left fewer lines on the page and
% thus a larger rubber space between the paragraphs.  It seems more
% prudent for the time being to let these paragraphs continue to crowd,
% for the result though slightly unsightly is still readable.
$\oint_{S_\mr{element}}=\int_{S_\mr{inner}}+\int_{S_\mr{outer}}$.
Adding all the elements together, we have that
\[
  \sum_\mr{elements} \Phi_{\mr{element}} =
  \sum_\mr{elements} \int_{S_\mr{inner}} \ve a \cdot d\ve s
  + \sum_\mr{elements} \int_{S_\mr{outer}} \ve a \cdot d\ve s;
\]
but the inner sum is null because it includes each interior surface
twice, because each interior surface is shared by two elements such that
$d\ve s_2 = -d\ve s_1$ (in other words, such that the one volume
element's~$d\ve s$ on the surface the two elements share points
oppositely to the other volume element's~$d\ve s$ on the same surface),
so
\[
  \sum_\mr{elements} \Phi_{\mr{element}}
  = \sum_\mr{elements} \int_{S_\mr{outer}} \ve a \cdot d\ve s
  = \oint_S \ve a \cdot d\ve s.
\]
In this equation, the last integration is over the surface of the
larger, overall volume, which surface after all consists of nothing
other than the several boundary elements' outer surface patches.
Applying~(\ref{vcalc:320:35}) to the equation's left side to express
the flux~$\Phi_{\mr{element}}$ from a single volume element yields that
\[
  \sum_\mr{elements} \nabla \cdot \ve a \,dv
  = \oint_S \ve a \cdot d\ve s.
\]
That is,
\bq{vcalc:divthm}
  \int_V \nabla \cdot \ve a \,dv =
  \oint_S \ve a \cdot d\ve s.
\eq

\index{calculus!fundamental theorem of}
\index{fundamental theorem of calculus}
\index{integral!swapping one vector form for another in}
\index{integrability}
Equation~(\ref{vcalc:divthm}) is the \emph{divergence theorem.}%
\footnote{\cite[eqn.~1.2.8]{Stratton}}
The divergence theorem, the
vector's version of the fundamental theorem of
calculus~(\ref{integ:antider}), neatly relates the divergence within a
volume to the flux from it.  It is an important result.  The
integral on the equation's left and the one on its right each arise in
vector analysis more often than one might expect.
% diagn: Add here one or more references to where it arises?
When they do,~(\ref{vcalc:divthm}) swaps the one integral for the
other, often a profitable maneuver.%
\footnote{%
  Where a wave propagates through a material interface, the associated
  field can be discontinuous and, consequently, the field's divergence can
  be infinite, which would seem to call assumptions
  underlying~(\ref{vcalc:divthm}) into question.  However, the infinite
  divergence at a material interface is normally \emph{integrable} in the
  same way the Dirac delta of \S~\ref{integ:670}, though infinite, is
  integrable.  One can integrate finitely through either infinity.  If one
  can conceive of an interface not as a sharp layer of zero thickness but
  rather as a thin layer of thickness~$\ep$, through which the associated
  field varies steeply but continuously, then the divergence theorem
  necessarily remains valid in the limit $\ep \ra 0$.
}

\subsection{Stokes' theorem}
\label{vcalc:330.40}
\index{Stokes' theorem}
\index{Stokes, Sir George Gabriel (1819--1903)}

\index{edge!inner and outer}
\index{inner edge}
\index{outer edge}
\index{surface element}
\index{element!interior and boundary}
\index{interior element}
\index{boundary element}
\index{contour!closed}
\index{circulation}
\index{curl!directional}
\index{directional curl}
\index{integral!swapping one vector form for another in}
Corresponding to the divergence theorem of \S~\ref{vcalc:330.30}
is a second, related theorem for directional curl, developed as follows.
If an open surface, whether the surface be confined to a plane or be
warped in three dimensions (as for example in bowl shape), is
subdivided into infinitesimal surface elements~$d\ve s$---each element
small enough not only to experience essentially constant curl but also
to be regarded as planar---then according to~(\ref{vcalc:circulation})
the circulation about the entire surface is
\[
  \Gamma
  = \oint \ve a \cdot d\we\ell
\]
and the circulation about any one surface element is
\[
  \Gamma_{\mr{element}}
  = \oint_{\mr{element}} \ve a \cdot d\we\ell.
\]
From this equation, reasoning parallel to that of
\S~\ref{vcalc:330.30}---only using~(\ref{vcalc:320:45}) in place
of~(\ref{vcalc:320:35})---concludes that
\bq{vcalc:stokesthm}
  \int_S
  ( \nabla \times \ve a ) \cdot d\ve s
  = \oint \ve a \cdot d\we\ell.
\eq
Equation~(\ref{vcalc:stokesthm}) is \emph{Stokes' theorem,}%
\footnote{\cite[eqn.~1.4.20]{Stratton}}$\mbox{}^{,}$%
\footnote{
  If~(\ref{vcalc:divthm}) is ``the divergence theorem,'' then
  should~(\ref{vcalc:stokesthm}) not be ``the curl theorem''?
  Answer: maybe it should be, but no one calls it that.  Sir George
  Gabriel Stokes is evidently not to be denied his fame!
}
neatly relating the directional curl over a (possibly nonplanar) surface
to the circulation about it.  Like the divergence
theorem~(\ref{vcalc:divthm}), Stokes' theorem~(\ref{vcalc:stokesthm})
serves to swap one vector integral for another where such a maneuver is
needed.

% ----------------------------------------------------------------------

\section[Summary of definitions and identities]%
  {Summary of definitions and identities of vector calculus}
\label{vcalc:340}
\index{identity!differential, of the vector}
\index{vector!differential identities of}

Table~\ref{vcalc:tbl-diffid} lists useful definitions and
identities of vector calculus,%
\footnote{%
  \cite[appendix~II.3]{Balanis}%
  \cite[appendix~II]{Stratton}%
  \cite[appendix~A]{Harrington}%
}
the first several of which it gathers from \S\S~\ref{vcalc:320}
and~\ref{vcalc:330}, the last several of which (exhibiting heretofore
unfamiliar symbols like~$\nabla^2$) it gathers from \S~\ref{vcalc:350}
to follow.  Of the identities in the middle of the table, a few are
statements of the~$\nabla$ operator's distributivity over summation.
The rest are vector derivative product rules
(\S~\ref{drvtv:250.30}).
\begin{table}
  \caption[Definitions and identities of vector calculus.]%
    {%
      Definitions and identities of vector calculus (see also
      Table~\ref{vector:tbl-algid} on page~\pageref{vector:tbl-algid}).%
    }
  \label{vcalc:tbl-diffid}
  \index{calculus!vector, definitions and identities of}
  \index{vector calculus!definitions and identities of}
  \index{identity!of vector calculus}
  \index{definition!of vector calculus}
  \[
    \nc\xxa{1.23}
    \nc\xxb{1.58}
    \nc\xxc{1.93}
    \renewcommand\arraystretch{\xxc}
    \br{c}
      \renewcommand\arraystretch{\xxc}
      \br{rclcrcl}
        \nabla &\equiv& \ds\vui\ppx{i} &\ \ &
        \ve b \cdot \nabla &=& \ds b_i\ppx{i} \\
        \nabla \psi &=& \ds\vui\pp{\psi}{i} &&
        \ve b \cdot \nabla \psi &=& \ds b_i\pp{\psi}{i} \\
        \nabla \cdot \ve a &=& \ds\pp{a_i}{i} &&
        (\ve b \cdot \nabla)\ve a &=& \ds b_i\pp{\ve a}{i} = \ds\vuj b_i\pp{a_j}{i} \\
        \nabla \times \ve a &=& \ds\ep_{ijk} \vui \pp{a_k}{j} &&
        \ve b \cdot \nabla \times \ve a &=& \ds\ep_{ijk} b_i \pp{a_k}{j} \\
        \ve b \times \nabla \psi &=& \ds\ep_{ijk} \vui b_j \pp{\psi}{k} &&
        \ve b \times \nabla \times \ve a &=& \ds\vuj b_i \left( \pp{a_i}{j} - \pp{a_j}{i} \right) \\
        \Phi &\equiv& \ds\int_S \ve a \cdot d\ve s &&
        \ds \int_V \nabla \cdot \ve a \,dv &=& \ds \oint_S \ve a \cdot d\ve s \\
        \Gamma &\equiv& \ds\int_C \ve a \cdot d\we\ell &&
        \ds \int_S ( \nabla \times \ve a ) \cdot d\ve s &=& \ds \oint \ve a \cdot d\we\ell
      \er \\
      \renewcommand\arraystretch{\xxa}
      \br{rcl}
        \nabla \cdot ( \ve a + \ve b ) &=& \nabla \cdot \ve a + \nabla \cdot \ve b \\
        \nabla \times ( \ve a + \ve b ) &=& \nabla \times \ve a + \nabla \times \ve b \\
        \nabla( \psi + \omega ) &=& \nabla \psi + \nabla \omega \\
        \nabla( \psi\omega ) &=& \omega \nabla \psi + \psi \nabla \omega \\
        \nabla \cdot ( \psi \ve a ) &=& \ve a \cdot \nabla \psi + \psi \nabla \cdot \ve a \\
        \nabla \times ( \psi \ve a ) &=& \psi \nabla \times \ve a - \ve a \times \nabla \psi \\
        \nabla ( \ve a \cdot \ve b ) &=&
          ( \ve b \cdot \nabla + \ve b \times \nabla \times \mbox{} ) \ve a +
          ( \ve a \cdot \nabla + \ve a \times \nabla \times \mbox{} ) \ve b \\
        \nabla \cdot ( \ve a \times \ve b ) &=&
          \ve b \cdot \nabla \times \ve a - \ve a \cdot \nabla \times \ve b \\
        \nabla \times ( \ve a \times \ve b ) &=&
          (\ve b \cdot \nabla + \nabla \cdot \ve b) \ve a -
          (\ve a \cdot \nabla + \nabla \cdot \ve a) \ve b
      \er \\
      \renewcommand\arraystretch{\xxc}
      % The columns of this next array step a bit on one another, but it
      % so happens that the row with the widest equation in the left
      % column meets some empty space on that row in the right column.
      % The columns are thus overlapped to fit the whole table within
      % margins.
      \br{rclrcl}
        \nabla^2 &\equiv& \ds\frac{\partial^2}{\partial i^2} &
        \nabla \nabla \cdot \ve a &=& \ds \vuj \frac{\partial^2 a_i}{\partial j \, \partial i} \\
        \nabla^2 \psi &=& \nabla \cdot \nabla \psi = \ds\frac{\partial^2\psi}{\partial i^2} &
        \nabla^2 \ve a &=& \ds\frac{\partial^2 \ve a}{\partial i^2}
          = \vuj\frac{\partial^2 a_j}{\partial i^2}
          = \vuj\nabla^2(\vuj\cdot\ve a) \\
        \nabla \times \nabla \psi &=& 0 &
        \!\nabla \cdot \nabla \times \ve a &=& 0
      \er \\
      \renewcommand\arraystretch{\xxb}
      \br{rcl}
        \nabla \times \nabla \times \ve a &=&
          \ds
            \vuj \ppx{i} \left(
                \frac{\partial a_i}{\partial j}
              - \frac{\partial a_j}{\partial i}
            \right) \\
        \nabla \nabla \cdot \ve a &=& \nabla^2 \ve a + \nabla \times \nabla \times \ve a
      \er
    \er
  \]
\end{table}

\index{derivative!product rule for}
\index{vector!derivative product rule for}
\index{field!derivative product rule for}
\index{product rule, derivative!of the vector}
The product rules resemble the triple products of
Table~\ref{vector:tbl-algid}, only with the~$\nabla$ operator in place
of the vector~$\ve c$.  However, since~$\nabla$ is a differential
operator for which, for instance, $\ve b \cdot \nabla \neq \nabla \cdot
\ve b$, its action differs from a vector's in some cases, and there are
more distinct ways in which it can act.  Among the several product rules
the easiest to prove is that
\[
  \nabla(\psi\omega)
  = \vui \pp{(\psi\omega)}{i}
  = \omega \vui \pp{\psi}{i} + \psi \vui \pp{\omega}{i}
  = \omega \nabla \psi + \psi \nabla \omega.
\]
The hardest to prove is that
\bqb
  \nabla(\ve a \cdot \ve b)
  &=& \nabla(a_ib_i) = \vuj \pp{(a_ib_i)}{j}
   =  \vuj b_i \pp{a_i}{j} + \vuj a_i \pp{b_i}{j} \\
  &=& \vuj b_i \pp{a_j}{i}
    + \vuj b_i \left(\pp{a_i}{j} - \pp{a_j}{i}\right)
    + \vuj a_i \pp{b_j}{i}
    + \vuj a_i \left(\pp{b_i}{j} - \pp{b_j}{i}\right) \\
  &=& (\ve b \cdot \nabla) \ve a
    + \ve b \times \nabla \times \ve a
    + (\ve a \cdot \nabla) \ve b
    + \ve a \times \nabla \times \ve b \\
  &=& ( \ve b \cdot \nabla + \ve b \times \nabla \times \mbox{} ) \ve a +
      ( \ve a \cdot \nabla + \ve a \times \nabla \times \mbox{} ) \ve b,
\eqb
because to prove it one must recognize in it the cross-directional
curl of
% bad break
(\ref{vcalc:320:46}).
Also nontrivial to prove is that
\bqb
   \nabla \times ( \ve a \times \ve b )
   &=& \nabla \times \left( \ep_{ijk} \vui a_j b_k \right) \\
   &=& \ep_{mni} \vu m \pp{( \ep_{ijk} \vui a_j b_k )_i}{n}
    = \ep_{mni} \ep_{ijk} \vu m \pp{(a_j b_k)}{n} \\
   &=& (\delta_{mj}\delta_{nk} - \delta_{mk}\delta_{nj}) \vu m \pp{(a_j b_k)}{n} \\
   &=& \vuj \pp{(a_j b_k)}{k} - \vu k \pp{(a_j b_k)}{j}
    =  \vuj \pp{(a_j b_i)}{i} - \vuj \pp{(a_i b_j)}{i} \\
   &=&
     \left( \vuj b_i \pp{a_j}{i} + \vuj a_j \pp{b_i}{i} \right) -
     \left( \vuj a_i \pp{b_j}{i} + \vuj b_j \pp{a_i}{i} \right) \\
   &=&
     (\ve b \cdot \nabla + \nabla \cdot \ve b) \ve a -
     (\ve a \cdot \nabla + \nabla \cdot \ve a) \ve b.
\eqb
The others are less hard:%
\footnote{
  And probably should have been left as exercises, except that this book
  is not actually an instructional textbook.  The reader who wants
  exercises might hide the page from sight and work the three
  identities out with his own pencil.
}
\bqb
  \nabla \cdot ( \psi \ve a ) &=&
      \pp{( \psi a_i )}{i} = a_i\pp{\psi}{i} + \psi\pp{a_i}{i}
    = \ve a \cdot \nabla \psi + \psi \nabla \cdot \ve a; \\
  \nabla \times ( \psi \ve a ) &=&
      \ep_{ijk} \vui \pp{( \psi a_k )}{j}
    = \ep_{ijk} \vui \psi \pp{a_k}{j}
    + \ep_{ijk} \vui a_k \pp{\psi}{j}
    \\&=& \psi \nabla \times \ve a - \ve a \times \nabla \psi; \\
  \nabla \cdot ( \ve a \times \ve b ) &=&
      \pp{(\ep_{ijk} a_j b_k)}{i}
    = \ep_{ijk} b_k \pp{a_j}{i}
    + \ep_{ijk} a_j \pp{b_k}{i}
    \\&=& \ve b \cdot \nabla \times \ve a - \ve a \cdot \nabla \times \ve b.
\eqb

\index{invariance under the reorientation of axes}
\index{reorientation!invariance under}
\index{axes!invariance under the reorientation of}
Inasmuch as none of the derivatives or products within the table's
several product rules vary under rotation of axes, the product rules are
themselves invariant.  That the definitions and identities at the top of
the table are invariant, we have already seen; and \S~\ref{vcalc:350},
next, will give invariance to the definitions and identities at the
bottom.  The whole table is therefore invariant under rotation of axes.

% ----------------------------------------------------------------------

\section[The Laplacian, et al.]%
  {The Laplacian and other second-order derivatives}
\label{vcalc:350}
\index{Laplacian}
\index{Laplace, Pierre-Simon (1749--1827)}
\index{field!second-order derivative of}
\index{vector!second-order derivative of}
\index{derivative!of a field, second-order}

\index{curl!of a curl}
\index{gradient!of a divergence}
\index{divergence!of a gradient}
Table~\ref{vcalc:tbl-diffid} ends with second-order vector derivatives.
Like vector products and first-order vector derivatives, second-order
vector derivatives too come in several kinds, the simplest of which is
the \emph{Laplacian}%
\footnote{%
  Though seldom seen in applied usage in the author's country, the
  alternate symbol~$\Delta$ replaces~$\nabla^2$ in some books,
  especially some British books.  The author prefers the~$\nabla^2$,
  which better captures the sense of the thing and which leaves~$\Delta$
  free for other uses.%
}
\bq{vcalc:laplacian}
  \begin{split}
    \nabla^2 &\equiv \frac{\partial^2}{\partial i^2}, \\
    \nabla^2 \psi &= \nabla \cdot \nabla \psi
      = \ds\frac{\partial^2\psi}{\partial i^2}, \\
    \nabla^2 \ve a &= \ds\frac{\partial^2 \ve a}{\partial i^2}
      = \vuj\frac{\partial^2 a_j}{\partial i^2}
      = \vuj\nabla^2(\vuj\cdot\ve a).
  \end{split}
\eq
Other second-order vector derivatives include
\bq{vcalc:350:20}
  \begin{split}
    \nabla \nabla \cdot \ve a &= \vuj \frac{\partial^2 a_i}{\partial j \, \partial i}, \\
    \nabla \times \nabla \times \ve a &=
      \vuj \ppx{i} \left(
          \frac{\partial a_i}{\partial j}
        - \frac{\partial a_j}{\partial i}
      \right),
  \end{split}
\eq
the latter of which is derived as
\bqb
  \nabla \times \nabla \times \ve a
  &=&
    \nabla \times \left( \ep_{ijk} \vui \pp{a_k}{j} \right)
  \\&=&
    \ep_{mni} \vu m \ppx{n} \left( \ep_{ijk} \vui \pp{a_k}{j} \right)_{\!i}
  =
    \ep_{mni} \ep_{ijk} \vu m \frac{\partial^2 a_k}{\partial n \,\partial j}
  \\&=&
    ( \delta_{mj}\delta_{nk} - \delta_{mk}\delta_{nj} )
    \vu m \frac{\partial^2 a_k}{\partial n \,\partial j}
  \\&=&
      \vuj  \frac{\partial^2 a_k}{\partial k \,\partial j}
    - \vu k \frac{\partial^2 a_k}{\partial j^2}
  =
      \vuj  \frac{\partial^2 a_i}{\partial i \,\partial j}
    - \vuj \frac{\partial^2 a_j}{\partial i^2}.
\eqb
Combining the various second-order vector derivatives yields the useful
identity that
\bq{vcalc:350:30}
  \nabla \nabla \cdot \ve a = \nabla^2 \ve a + \nabla \times \nabla \times \ve a.
\eq
Table~\ref{vcalc:tbl-diffid} summarizes.

\index{curl!of a gradient}
\index{divergence!of a curl}
\index{null vector identity}
\index{field!irrotational}
\index{irrotational field}
\index{field!solenoidal}
\index{solenoidal field}
\index{field!source-free}
\index{source-free field}
\index{field!divergenceless}
\index{divergenceless field}
\index{Helmholtz, Hermann Ludwig Ferdinand von (1821--1894)}
The table includes two curious null identities,
\bq{vcalc:350:null}
  \begin{split}
    \nabla \times \nabla \psi &= 0, \\
    \nabla \cdot \nabla \times \ve a &= 0.
  \end{split}
\eq
In words,~(\ref{vcalc:350:null}) states that \emph{gradients do not
curl and curl does not diverge.}  This is unexpected but is a
direct consequence of the definitions of the gradient, curl and
divergence:
\bqb
  \nabla \times \nabla \psi
    = \nabla \times \left( \vui \pp{\psi}{i} \right)
    = \ep_{mni} \vu m \frac{ \partial^2 \psi }{\partial n \, \partial i}
    &=& 0; \\
  \nabla \cdot \nabla \times \ve a
    = \nabla \cdot \left( \ep_{ijk} \vui \pp{a_k}{j} \right)
    = \ep_{ijk} \frac{\partial^2 a_k}{\partial i \, \partial j}
    &=& 0.
\eqb
A field like~$\nabla\psi$ that does not curl is called an
\emph{irrotational} field.  A field like $\nabla\times\ve a$ that does
not diverge is called a \emph{solenoidal, source-free} or
(prosaically) \emph{divergenceless} field.%
\footnote{
  In the writer's country, the United States, there has been a mistaken
  belief afoot that, if two fields $\ve b_1(\ve r)$ and $\ve b_2(\ve r)$
  had everywhere the same divergence and curl, then the two fields could
  differ only by an additive constant.  Even at least one widely
  distributed textbook expresses this belief, naming it
  \emph{Helmholtz's theorem;} but it is not just the one textbook, for
  the writer has heard it verbally from at least two engineers,
  unacquainted with one other, who had earned Ph.D.s in different eras
  in different regions of the country.  So the belief must be correct,
  mustn't it?

  Well, maybe it is, but the writer remains unconvinced.  Consider the
  admittedly contrived counterexample of $\ve b_1 = \vu xy + \vu yx$,
  $\ve b_2 = 0$.

  \index{Helmholtz, Hermann Ludwig Ferdinand von (1821--1894)}
  On an applied level, the writer knows of literally no other false
  theorem so widely believed to be true, which leads the writer to
  suspect that he himself had somehow erred in judging the theorem
  false.  What the writer really believes however is that Hermann von Helmholtz
  probably originally had put some appropriate restrictions on~$\ve b_1$
  and~$\ve b_2$ which, if obeyed, made his theorem true but which at
  some time after his death got lost in transcription.  That a
  transcription error would go undetected so many years would tend to
  suggest that Helmholtz's theorem, though interesting, were not
  actually very necessary in practical applications.
  (Some believe the theorem necessary to establish a ``gauge'' in a wave
  equation,
  % diagn: Add a reference somewhere around here to the appropriate
  %   wave-equation section.
  but if they examine the use of their gauges closely then they will
  likely discover that one does not logically actually need to invoke the theorem
  to use the gauges.)

  Corrections by readers are invited.
}

\index{invariance under the reorientation of axes}
\index{reorientation!invariance under}
\index{axes!invariance under the reorientation of}
Each of this section's second-order vector derivatives---including the
vector Laplacian~$\nabla^2 \ve a$, according
to~(\ref{vcalc:350:30})---is or can be composed of first-order vector
derivatives already familiar to us from \S~\ref{vcalc:320}.  Therefore,
inasmuch as each of those first-order vector derivatives is invariant
under reorientation of axes, each second-order vector derivative is
likewise invariant.

% ----------------------------------------------------------------------

\section{Contour derivative product rules}
\label{vcalc:380}
\index{derivative!product rule for}
\index{field!derivative product rule for}
\index{product rule, derivative!of the contour}
\index{vector!derivative product rule for}
\index{contour!derivative product rule along}

Equation~(\ref{drvtv:prod2}) gives the derivative product rule for
functions of a scalar variable.  Fields---that is, functions of a vector
variable---obey product rules, too, several of which
Table~\ref{vcalc:tbl-diffid} lists.  The table's product rules however
are general product rules that regard full spatial derivatives.  What
about derivatives along an arbitrary contour?  Do they obey product
rules, too?  That is, one supposes that%
\footnote{
  The~$-$ sign in~(\ref{vcalc:380:10})'s last line is an artifact of
  ordering the line's factors in the style of
  Table~\ref{vcalc:tbl-diffid}.  Before proving the line, the narrative
  will reverse the order to kill the sign.  See below.
}
\bq{vcalc:380:10}
  \begin{split}
    \ppx\ell ( \psi\omega ) &= \omega \pp\psi\ell + \psi \pp\omega\ell, \\
    \ppx\ell ( \psi\ve a ) &= \ve a \pp\psi\ell + \psi \pp{\ve a}\ell, \\
    \ppx\ell ( \ve a\cdot\ve b ) &= \ve b \cdot \pp{\ve a}\ell + \ve a \cdot \pp{\ve b}\ell, \\
    \ppx\ell ( \ve a\times\ve b ) &= -\ve b \times \pp{\ve a}\ell + \ve a \times \pp{\ve b}\ell.
  \end{split}
\eq
where~$\ell$ is the distance along some arbitrary contour in space.  As
a hypothesis,~(\ref{vcalc:380:10}) is attractive.  But is it true?

That the first line of~(\ref{vcalc:380:10}) is true is clear, if you
think about it in the right way, because, in the restricted
case~(\ref{vcalc:380:10}) represents, one can treat the scalar
fields $\psi(\ve r)$ and $\omega(\ve r)$ as ordinary scalar
functions $\psi(\ell)$ and $\omega(\ell)$ of the scalar distance~$\ell$
along the contour, whereupon~(\ref{drvtv:prod2})
applies---for~(\ref{vcalc:380:10}) never evaluates $\psi(\ve r)$
or $\omega(\ve r)$ but along the contour.  The same naturally goes for
the vector fields $\ve a(\ve r)$ and $\ve b(\ve r)$, which one can treat
as vector functions $\ve a(\ell)$ and $\ve b(\ell)$ of the scalar
distance~$\ell$; so the second and third lines of~(\ref{vcalc:380:10})
are true, too, since one can write the second line in the form
\[
  \vui \left[ \ppx\ell ( \psi a_i ) \right]
    = \vui \left[ a_i \pp\psi\ell + \psi \pp{a_i}\ell \right]
\]
and the third line in the form
\[
  \ppx\ell ( a_i b_i ) = b_i \pp{a_i}\ell + a_i \pp{b_i}\ell,
\]
each of which, according to the first line, is true separately for
$i=x$, for $i=y$ and for $i=z$.

The truth of~(\ref{vcalc:380:10})'s last line is slightly less obvious.
Nevertheless, one can reorder factors to write the line as
\[
  \ppx\ell ( \ve a\times\ve b )
    = \pp{\ve a}\ell \times \ve b + \ve a \times \pp{\ve b}\ell,
\]
the Levi-Civita form (\S~\ref{vector:240.30}) of which is
\[
  \ep_{ijk} \vui \left[ \ppx\ell ( a_j b_k ) \right]
    = \ep_{ijk} \vui \left[ \pp{a_j}\ell b_k + a_j \pp{b_k}\ell \right].
\]
The Levi-Civita form is true separately for $(i,j,k)=(x,y,z)$, for
$(i,j,k)=(x,z,y)$, and so forth, so~(\ref{vcalc:380:10})'s last line as
a whole is true, too, which completes the proof of~(\ref{vcalc:380:10}).

% ----------------------------------------------------------------------

\section{Metric coefficients}
\label{vcalc:420}
\index{metric coefficient}
\index{coefficient!metric}

\index{scalar field}
\index{model}
\index{coordinates}
A scalar field $\psi(\ve r)$ is the same field whether expressed
as a function $\psi(x,y,z)$ of rectangular coordinates,
$\psi(\rho;\phi,z)$ of cylindrical coordinates or $\psi(r;\theta;\phi)$
of spherical coordinates,%
\footnote{
  For example, the field $\psi = x^2 + y^2$ in rectangular
  coordinates is $\psi = \rho^2$ in cylindrical coordinates.
  Refer to Table~\ref{trig:277:20}. % and~(\ref{trig:277:30}).
}
or indeed of coordinates in any three-dimensional system.  However,
cylindrical and spherical geometries normally recommend cylindrical and
spherical coordinate systems, systems which make some of the identities
of Table~\ref{vcalc:tbl-diffid} hard to use.

\index{cylindrical coordinates}
\index{spherical coordinates}
\index{coordinates!cylindrical}
\index{coordinates!spherical}
\index{angle}
\index{length}
The reason a cylindrical or spherical system makes some of the table's
identities hard to use is that some of the table's identities involve
derivatives $d/di$, notation which per \S~\ref{vector:240.20} stands for
$d/dx'$, $d/dy'$ or $d/dz'$ where the coordinates~$x'$, $y'$ and~$z'$
represent lengths.  But among the cylindrical and spherical coordinates
are~$\theta$ and~$\phi$, angles rather than lengths.  Because one cannot
use an angle as though it were a length, the notation $d/di$ cannot
stand for $d/d\theta$ or $d/d\phi$ and, thus, one cannot use the table
in cylindrical or spherical coordinates as the table stands.

\index{sweeping out a length}
We therefore want factors to convert the angles in question to
lengths (or, more generally, when special coordinate systems like the
parabolic systems of \S~\ref{vector:280} come into play, to convert
coordinates other than lengths to lengths).  Such factors are called
\emph{metric coefficients} and Table~\ref{vcalc:420:tbl} lists them.%
\footnote{\cite[\S~2-4]{Cheng}}
\begin{table}
  \caption[Metric coefficients.]{The metric coefficients of
    the rectangular, cylindrical and spherical coordinate systems.}
  \label{vcalc:420:tbl}
  \index{metric coefficient}
  \index{coefficient!metric}
  \[
    \setlength\arraycolsep{0.30\arraycolsep}
    \br{rcl}
      &\makebox[0pt][c]{RECT.}& \\
      h_x &=& 1 \\
      h_y &=& 1 \\
      h_z &=& 1
    \er
    \ \ \ \
    \br{rcl}
      &\makebox[0pt][c]{CYL.}& \\
      h_\rho &=& 1 \\
      h_\phi &=& \rho \\
      h_z &=& 1
    \er
    \ \ \ \
    \br{rcl}
      &\makebox[0pt][c]{SPHER.}& \\
      h_r &=& 1 \\
      h_\theta &=& r \\
      h_\phi &=& \rho = r\sin\theta
    \er
  \]
\end{table}
The use of the table is this: that for any metric coefficient~$h_\alpha$
a change~$d\alpha$ in its coordinate~$\alpha$ sweeps out a
length~$h_\alpha\,d\alpha$.  For example, in cylindrical
coordinates $h_\phi = \rho$ according to table, so a change~$d\phi$ in
the azimuthal coordinate~$\phi$ sweeps out a length~$\rho\,d\phi$---a
fact we have already informally observed as far back as
\S~\ref{integ:241.10}, which the table now formalizes.

\index{$\rho$}
Incidentally, the metric coefficient~$h_\phi$ seems to have two
different values in the table, one value in cylindrical coordinates and
another in spherical.  The two are really the same value, though,
since $\rho = r\sin\theta$ per Table~\ref{trig:277:20}.

\subsection{Displacements, areas and volumes}
\label{vcalc:420.20}
\index{displacement}
\index{area}
\index{volume}

\index{infinitesimal!surface}
\index{surface infinitesimal}
\index{spherical surface}
\index{surface!spherical}
In any orthogonal, right-handed, three-dimensional coordinate system
\linebreak % bad break
$(\alpha;\beta;\gamma)$---whether the symbols
$(\alpha;\beta;\gamma)$ stand for $(x,y,z)$, $(y,z,x)$, $(z,x,y)$,
$(x',y',z')$, $(\rho;\phi,z)$, $(r;\theta;\phi)$, $(\phi^x,r;\theta^x)$,
etc.,%
\footnote{\label{vcalc:420:fn1}%
  The book's admittedly clumsy usage of semicolons~``;'' and
  commas~``,'' to delimit coordinate triplets, whereby a semicolon
  precedes an angle (or, in this section's case, precedes a generic
  coordinate like~$\alpha$ that could stand for an angle), serves well
  enough to distinguish the three principal coordinate systems
  $(x,y,z)$, $(\rho;\phi,z)$ and $(r;\theta;\phi)$ visually from one
  another but ceases to help much when further coordinate systems such
  as $(\phi^x,r;\theta^x)$ come into play.  Logically, maybe, it would make
  more sense to write in the manner of $(,x,y,z)$, but to do so seems
  overwrought and fortunately no one the author knows of does it in that
  way.  The delimiters just are not that important.

  The book adheres to the semicolon convention not for any deep reason
  but only for lack of a better convention.  See also chapter~\ref{vector}'s
  footnote~\ref{vector:280:fn1}.
}
or even something exotic like the parabolic $(\sigma,\tau,z)$ of
\S~\ref{vector:280}---the product
\bq{vcalc:420:21}
  d\ve s = \wu \alpha h_{\beta} h_{\gamma} \,d\beta \,d\gamma
\eq
represents an area infinitesimal normal to~$\wu\alpha$.  For
example, the area infinitesimal on a spherical surface of radius~$r$ is
$d\ve s
= \vu r h_{\theta} h_{\phi} \,d\theta \,d\phi
= \vu r r^2\sin\theta \,d\theta \,d\phi$.

\index{infinitesimal!volume}
\index{volume infinitesimal}
\index{volume!in a spherical geometry}
Again in any orthogonal, right-handed, three-dimensional coordinate
system $(\alpha;\beta;\gamma)$, the product
\bq{vcalc:420:22}
  dv = h_{\alpha} h_{\beta} h_{\gamma} \,d\alpha \,d\beta \,d\gamma
\eq
represents a volume infinitesimal.  For example,
the volume infinitesimal in a spherical geometry is
$dv
= h_r h_{\theta} h_{\phi} \,dr \,d\theta \,d\phi
= r^2\sin\theta \,dr \,d\theta \,d\phi$.

Notice that \S~\ref{integ:241} has already calculated several integrals
involving area and volume infinitesimals of these kinds.

\index{infinitesimal!displacement}
\index{displacement infinitesimal}
\index{length infinitesimal}
\index{contour infinitesimal}
A volume infinitesimal~(\ref{vcalc:420:22})
cannot meaningfully in three dimensions be
expressed as a vector as an area infinitesimal~(\ref{vcalc:420:21}) can,
since in three dimensions a volume has no orientation.  Naturally
however, a length or \emph{displacement} infinitesimal can indeed be
expressed as a vector, as
\bq{vcalc:420:23}
  d\we\ell = \wu\alpha h_\alpha \,d\alpha.
\eq

Section~\ref{vcalc:520} will have more to say about vector
infinitesimals in nonrectangular coordinates.

\subsection{The vector field and its scalar components}
\label{vcalc:420.30}
\index{vector field!decomposition of}

Like a scalar field $\psi(\ve r)$, a vector field $\ve a(\ve r)$ too is
the same field whether expressed as a function $\ve a(x,y,z)$ of
rectangular coordinates, $\ve a(\rho;\phi,z)$ of cylindrical coordinates
or $\ve a(r;\theta;\phi)$ of spherical coordinates, or indeed of
coordinates in any three-dimensional system.  A vector field however is
the sum of three scalar fields, each scaling an appropriate unit vector.
In rectangular coordinates,
\[
  \ve a(\ve r) = \vu x a_x(\ve r) + \vu y a_y(\ve r) +  \vu z a_z(\ve r);
\]
in cylindrical coordinates,
\[
  \ve a(\ve r) = \wu\rho a_\rho(\ve r) + \wu\phi a_\phi(\ve r) +  \vu z a_z(\ve r);
\]
and in spherical coordinates,
\[
  \ve a(\ve r) = \vu r a_r(\ve r) + \wu\theta a_\theta(\ve r) + \wu\phi a_\phi(\ve r).
\]

\index{vector field!components of}
\index{vector!unit basis, cylindrical}
\index{vector!unit basis, spherical}
\index{vector!unit basis, variable}
\index{unit basis vector!cylindrical}
\index{unit basis vector!spherical}
\index{unit basis vector!variable}
\index{vector!unit}
\index{unit vector}
The scalar fields $a_\rho(\ve r)$, $a_r(\ve r)$, $a_\theta(\ve r)$ and
$a_\phi(\ve r)$ in and of themselves do not differ in nature from
$a_x(\ve r)$, $a_y(\ve r)$, $a_z(\ve r)$, $\psi(\ve r)$ or any other
scalar field.  One does tend to use them differently, though, because
constant unit vectors~$\vu x$, $\vu y$ and~$\vu z$ exist to combine the
scalar fields $a_x(\ve r)$, $a_y(\ve r)$, $a_z(\ve r)$ to compose the
vector field $\ve a(\ve r)$ whereas no such constant unit vectors exist
to combine the scalar fields $a_\rho(\ve r)$, $a_r(\ve r)$,
$a_\theta(\ve r)$ and $a_\phi(\ve r)$.  Of course there are the variable
unit vectors $\wu\rho(\ve r)$, $\vu r(\ve r)$, $\wu\theta(\ve r)$ and
$\wu\phi(\ve r)$, but the practical and philosophical differences between these and the
constant unit vectors is greater than it might seem.  For instance, it
is true that $\wu\rho \cdot \wu\phi = 0$, so long as what is meant by
this is that $\wu\rho(\ve r) \cdot \wu\phi(\ve r) = 0$.  However,
$\wu\rho(\ve r_1) \cdot \wu\phi(\ve r_2) \neq 0$, an algebraic error
fairly easy to commit.  On the other hand, that $\vu x \cdot \vu y = 0$
is always true.

(One might ask why such a subsection as this would appear in a section on
metric coefficients.  The subsection is here because no obviously better
spot for it presents itself, but moreover because we shall need the
understanding the subsection conveys to apply metric coefficients
consistently and correctly in \S~\ref{vcalc:440} to come.)

% ----------------------------------------------------------------------

\section{Nonrectangular notation}
\label{vcalc:425}
\index{vector!nonrectangular notation for}
\index{notation!for the vector, nonrectangular}
\index{nonrectangular notation}

\index{Einstein's summation convention}
\index{summation convention, Einstein's}
\index{Einstein notation}
Section~\ref{vector:240} has introduced Einstein's summation convention,
the Kronecker delta~$\delta_{ij}$ and the Levi-Civita
epsilon~$\ep_{ijk}$ together as notation for use in the definition of
vector operations and in the derivation of vector identities.  The
notation relies on symbols like~$i$, $j$ and~$k$ to stand for
unspecified coordinates, and Tables~\ref{vector:tbl-algid}
and~\ref{vcalc:tbl-diffid} use it extensively.  Unfortunately, the
notation fails in the nonrectangular coordinate systems when derivatives
come into play, as they do in Table~\ref{vcalc:tbl-diffid}, because
$\pl/\pl i$ is taken to represent a derivative specifically with respect
to a length whereas nonrectangular coordinates like~$\theta$ and~$\phi$
are not lengths.  Fortunately, this failure is not hard to redress.

Whereas the standard Einstein symbols~$i$, $j$ and~$k$ can stand only
for lengths, the modified Einstein symbols~$\tdi$, $\tdj$ and~$\tdk$,
which this section now introduces, can stand for any coordinates, even
for coordinates like~$\theta$ and~$\phi$ that are not lengths.  The
tilde~``$\tilde{\ \ }$'' atop the symbol~$\tdi$ warns readers that the
coordinate it represents is not necessarily a length and that, if one
wants a length, one must multiply~$\tdi$ by an appropriate metric
coefficient~$h_\tdi$ (\S~\ref{vcalc:420}).  The products~$h_\tdi \tdi$,
$h_\tdj \tdj$
% The \tdk subscript slightly crowds the symbol that follows it.
% The \hspace remedies.
and~$h_\tdk\hspace{0.5pt}\tdk$ always represent lengths.

The symbols~$\vui$, $\vuj$ and~$\vu k$ need no modification even when
modified symbols like~$\tdi$, $\tdj$ and~$\tdk$ are in use.  This is
because~$\vui$, $\vuj$ and~$\vu k$ are taken to represent unit
vectors---and $[\vui\;\vuj\;\vu k]$, to represent a proper orthogonal
basis---irrespective of the coordinate system; so long, naturally, as
the coordinate system is an orthogonal, right-handed coordinate system
% diagn: in the final release, review the claim the next line makes.
as are all the coordinate systems in this book.

The modified notation will find use in \S~\ref{vcalc:440.geo}.

% ----------------------------------------------------------------------

\section{Derivatives of the basis vectors}
\label{vcalc:430}
\index{unit basis vector!derivative of}
\index{vector!unit basis, derivative of}
\index{derivative!of a unit basis vector}

The derivatives of the various unit basis vectors with respect to the
several coordinates of their respective coordinate systems are not hard
to compute.  In fact, looking at
Fig.~\ref{vector:fig-sphere} on page~\pageref{vector:fig-sphere},
Fig.~\ref{vector:230:fig-cylindrical-basis} on page~\pageref{vector:230:fig-cylindrical-basis}, and
Fig.~\ref{vector:230:fig-spherical-basis} on page~\pageref{vector:230:fig-spherical-basis},
one can just write them down.
Table~\ref{vcalc:430:tbl} records them.
\begin{table}
  \caption{Derivatives of the basis vectors.}
  \label{vcalc:430:tbl}
  \renewcommand\arraystretch{2.0}
  \setlength\arraycolsep{0.30\arraycolsep}
  \[
    \br{rclcrclcrcl}
      \multicolumn{11}{c}{\mbox{RECTANGULAR}} \\
      \ds\pp{\vu x}{x} &=& 0 &\ \ \ &
      \ds\pp{\vu x}{y} &=& 0 &\ \ \ &
      \ds\pp{\vu x}{z} &=& 0 \\
      \ds\pp{\vu y}{x} &=& 0 &&
      \ds\pp{\vu y}{y} &=& 0 &&
      \ds\pp{\vu y}{z} &=& 0 \\
      \ds\pp{\vu z}{x} &=& 0 &&
      \ds\pp{\vu z}{y} &=& 0 &&
      \ds\pp{\vu z}{z} &=& 0
    \er
  \]
  \[
    \br{rclcrclcrcl}
      \multicolumn{11}{c}{\mbox{CYLINDRICAL}} \\
      \ds\pp{\wu\rho}{\rho} &=& 0 &\ \ \ &
      \ds\pp{\wu\rho}{\phi} &=& +\wu\phi &\ \ \ &
      \ds\pp{\wu\rho}{z} &=& 0 \\
      \ds\pp{\wu\phi}{\rho} &=& 0 &&
      \ds\pp{\wu\phi}{\phi} &=& -\wu\rho &&
      \ds\pp{\wu\phi}{z} &=& 0 \\
      \ds\pp{\vu z}{\rho} &=& 0 &&
      \ds\pp{\vu z}{\phi} &=& 0 &&
      \ds\pp{\vu z}{z} &=& 0
    \er
  \]
  \[
    \br{rclcrclcrcl}
      \multicolumn{11}{c}{\mbox{SPHERICAL}} \\
      \ds\pp{\vu r}{r} &=& 0 &\ \ \ &
      \ds\pp{\vu r}{\theta} &=& +\wu\theta &\ \ \ &
      \ds\pp{\vu r}{\phi} &=& +\wu\phi \sin\theta \\
      \ds\pp{\wu\theta}{r} &=& 0 &&
      \ds\pp{\wu\theta}{\theta} &=& -\vu r &&
      \ds\pp{\wu\theta}{\phi} &=& +\wu\phi \cos\theta \\
      \ds\pp{\wu\phi}{r} &=& 0 &&
      \ds\pp{\wu\phi}{\theta} &=& 0 &&
      \ds\pp{\wu\phi}{\phi} &=& -\wu\rho = \makebox[0pt][l]{$-\vu r \sin\theta - \wu\theta \cos\theta$}
    \er
  \]
\end{table}

\index{sketch, proof by}
\index{proof!by sketch}
Naturally, one can compute the table's derivatives symbolically,
instead, as for example
\[
  \frac{\pl\wu\rho}{\pl\phi}
  = \frac{\pl}{\pl\phi}(\vu x\cos\phi + \vu y\sin\phi)
  = -\vu x\sin\phi + \vu y\cos\phi = +\wu\phi.
\]
Such an approach prospers in special coordinate systems like the
parabolic systems of Tables~\ref{vector:280:t30}
and~\ref{vector:280:t56}, but in cylindrical and spherical coordinates
it is probably easier just to look at the figures.

% ----------------------------------------------------------------------

\section{Derivatives in the nonrectangular systems}
\label{vcalc:440}
\index{field!derivative of, nonrectangular}
\index{vector!derivative of, nonrectangular}
\index{derivative!of a field, nonrectangular}

This section develops vector derivatives in cylindrical and spherical
coordinates.

\subsection{Derivatives in cylindrical coordinates}
\label{vcalc:440.cyl}
\index{field!derivative of, in cylindrical coordinates}
\index{vector!derivative of, in cylindrical coordinates}
\index{derivative!of a field in cylindrical coordinates}

\index{directional derivative!in cylindrical coordinates}
\index{gradient!in cylindrical coordinates}
According to Table~\ref{vcalc:tbl-diffid},
\[
  \nabla \psi = \vui\pp{\psi}{i},
\]
but as \S~\ref{vcalc:420} has observed Einstein's symbol~$i$ must stand
for a length not an angle, whereas one of the three cylindrical
coordinates---the azimuth~$\phi$---is an angle.  The cylindrical metric
coefficients of Table~\ref{vcalc:420:tbl} make the necessary conversion,
the result of which is
\bq{vcalc:440.cyl:grad}
  \nabla \psi =
    \wu\rho \pp{\psi}{\rho}
    + \wu\phi \frac{\pl\psi}{\rho\,\pl\phi}
    + \vu z \pp{\psi}{z}.
\eq
%
Again according to Table~\ref{vcalc:tbl-diffid},
\[
  (\ve b \cdot \nabla)\ve a = b_i\pp{\ve a}{i}.
\]
Applying the cylindrical metric coefficients, we have that
\bq{vcalc:440.cyl:dird}
  (\ve b \cdot \nabla)\ve a =
    b_\rho \pp{\ve a}{\rho}
    + b_\phi \frac{\pl\ve a}{\rho\,\pl\phi}
    + b_z \pp{\ve a}{z}.
\eq
Expanding the vector field~$\ve a$ in the cylindrical basis,
\[
  (\ve b \cdot \nabla)\ve a = \left\{
    b_\rho \ppx{\rho}
    + b_\phi \frac{\pl}{\rho\,\pl\phi}
    + b_z \ppx{z}
  \right\} \left(
    \wu\rho a_\rho + \wu\phi a_\phi + \vu z a_z
  \right).
\]
Here are three derivatives of three terms, each term of two factors.
Evaluating the derivatives according to the contour derivative product
rule~(\ref{vcalc:380:10}) yields $(3)(3)(2) = \mbox{0x12}$ (eighteen)
terms in the result.  Half the~0x12 terms involve derivatives of the
basis vectors, which Table~\ref{vcalc:430:tbl} computes.  Some of
the~0x12 terms turn out to be null.  The result is that
\bqa
  (\ve b \cdot \nabla)\ve a &=&
    b_\rho \left[
      \wu\rho \frac{\pl a_\rho}{\pl\rho}
      + \wu\phi \frac{\pl a_\phi}{\pl\rho}
      + \vu z \frac{\pl a_z}{\pl\rho}
    \right]
  \xn\\&&\ \ \mbox{}
    + \frac{b_\phi}{\rho} \left[
      \wu\rho\left( \frac{\pl a_\rho}{\pl\phi} - a_\phi \right)
      + \wu\phi\left( \frac{\pl a_\phi}{\pl\phi} + a_\rho \right)
      + \vu z \frac{\pl a_z}{\pl\phi}
    \right]
  \xn\\&&\ \ \mbox{}
    + b_z \left[
      \wu\rho \frac{\pl a_\rho}{\pl z}
      + \wu\phi \frac{\pl a_\phi}{\pl z}
      + \vu z \frac{\pl a_z}{\pl z}
    \right].
  \label{vcalc:440.cyl:dird3}
\eqa

\index{orthogonal basis!constant}
\index{basis!constant}
To evaluate divergence and curl wants more care.  It also
wants a constant basis to work in, whereas $[\vu x\;\vu y\;\vu
z]$ is awkward in a cylindrical geometry and $[\wu\rho\;\wu\phi\;\vu z]$
is not constant.  Fortunately, nothing prevents us from defining a
constant basis $[\wu\rho_o\;\wu\phi_o\;\vu z]$ such that
$[\wu\rho\;\wu\phi\;\vu z] = [\wu\rho_o\;\wu\phi_o\;\vu z]$ \emph{at the
point $\ve r = \ve r_o$ at which the derivative is evaluated.}  If this
is done, then the basis $[\wu\rho_o\;\wu\phi_o\;\vu z]$ is constant like
$[\vu x\;\vu y\;\vu z]$ but not awkward like it.

\index{divergence!in cylindrical coordinates}
\index{curl!in cylindrical coordinates}
\index{cleverness}
According to Table~\ref{vcalc:tbl-diffid},
\[
  \nabla \cdot \ve a = \ds\pp{a_i}{i}
\]
In cylindrical coordinates and the $[\wu\rho_o\;\wu\phi_o\;\vu z]$
basis, this is%
\footnote{
  Mistakenly to write here that
  \[
    \nabla \cdot \ve a =
      \pp{a_\rho}{\rho}
      + \frac{\pl a_\phi}{\rho\,\pl\phi}
      + \pp{a_z}{z},
  \]
  which is not true, would be a ghastly error, leading to any number of
  hard-to-detect false conclusions.  Refer to \S~\ref{vcalc:420.30}.
}
\[
  \nabla \cdot \ve a =
    \pp{(\wu\rho_o\cdot\ve a)}{\rho}
    + \frac{\pl(\wu\phi_o\cdot\ve a)}{\rho\,\pl\phi}
    + \pp{(\vu z \cdot \ve a)}{z}.
\]
Applying the contour derivative product rule~(\ref{vcalc:380:10}),
\[
  \nabla \cdot \ve a =
    \wu\rho_o \cdot \pp{\ve a}{\rho}
    + \pp{\wu\rho_o}{\rho} \cdot \ve a
    + \wu\phi_o \cdot \frac{\pl \ve a }{\rho\,\pl\phi}
    + \frac{\pl\wu\phi_o}{\rho\,\pl\phi} \cdot\ve a
    + \vu z \cdot \pp{\ve a}{z}
    + \pp{\vu z}{z} \cdot \ve a.
\]
But $[\wu\rho_o\;\wu\phi_o\;\ve z]$ are constant unit vectors, so
\[
  \nabla \cdot \ve a =
    \wu\rho_o \cdot \pp{\ve a}{\rho}
    + \wu\phi_o \cdot \frac{\pl \ve a }{\rho\,\pl\phi}
    + \vu z \cdot \pp{\ve a}{z}.
\]
That is,
\[
  \nabla \cdot \ve a =
    \wu\rho \cdot \pp{\ve a}{\rho}
    + \wu\phi \cdot \frac{\pl \ve a }{\rho\,\pl\phi}
    + \vu z \cdot \pp{\ve a}{z}.
\]
Expanding the field in the cylindrical basis,
\[
  \nabla \cdot \ve a = \left\{
    \wu\rho \cdot \ppx{\rho}
    + \wu\phi \cdot \frac{\pl}{\rho\,\pl\phi}
    + \vu z \cdot \ppx{z}
  \right\} \left( \wu\rho a_\rho + \wu\phi a_\phi + \vu z a_z \right).
\]
As above, here again the expansion yields~0x12 (eighteen) terms.
Fortunately, this time most of the terms turn out to be null.  The
result is that
\[
  \nabla \cdot \ve a =
    \pp{a_\rho}{\rho}
    + \frac{a_\rho}{\rho}
    + \frac{\pl a_\phi}{\rho\,\pl\phi}
    + \pp{a_z}{z},
\]
or, expressed more cleverly in light of~(\ref{drvtv:250:35}), that
\bq{vcalc:440.cyl:divg}
  \nabla \cdot \ve a =
    \frac{\pl(\rho a_\rho)}{\rho\,\pl\rho}
    + \frac{\pl a_\phi}{\rho\,\pl\phi}
    + \pp{a_z}{z}.
\eq
%
Again according to Table~\ref{vcalc:tbl-diffid},
\bqb
  \nabla \times \ve a &=& \ep_{ijk} \vui \pp{a_k}{j}
  \\&=&
    \wu\rho_o \Bigg[
      \frac{\pl(\vu z\cdot\ve a)}{\rho\,\pl\phi}
      - \frac{\pl(\wu\phi_o\cdot\ve a)}{\pl z}
    \Bigg]
    + \wu\phi_o \Bigg[
      \frac{\pl(\wu\rho_o\cdot\ve a)}{\pl z}
      - \frac{\pl(\vu z\cdot\ve a)}{\pl\rho}
    \Bigg]
  \\&&\ \ \mbox{}
    + \vu z \Bigg[
      \frac{\pl(\wu\phi_o\cdot\ve a)}{\pl\rho}
      - \frac{\pl(\wu \rho_o\cdot\ve a)}{\rho\,\pl\phi}
    \Bigg].
\eqb
That is,
\bqb
  \nabla \times \ve a &=&
    \wu\rho \left[
      \vu z \cdot \frac{\pl\ve a}{\rho\,\pl\phi}
      - \wu\phi \cdot \frac{\pl\ve a}{\pl z}
    \right]
    + \wu\phi \left[
      \wu\rho \cdot \frac{\pl\ve a}{\pl z}
      - \vu z \cdot \frac{\pl\ve a}{\pl\rho}
    \right]
  \\&&\ \ \mbox{}
    + \vu z \left[
      \wu\phi \cdot \frac{\pl\ve a}{\pl\rho}
      - \wu \rho \cdot \frac{\pl\ve a}{\rho\,\pl\phi}
    \right].
\eqb
Expanding the field in the cylindrical basis,
\bqb
  \nabla \times \ve a &=& \bigg\{
    \wu\rho \left[
      \vu z \cdot \frac{\pl}{\rho\,\pl\phi}
      - \wu\phi \cdot \frac{\pl}{\pl z}
    \right]
    + \wu\phi \left[
      \wu\rho \cdot \frac{\pl}{\pl z}
      - \vu z \cdot \frac{\pl}{\pl\rho}
    \right]
  \\&&\ \ \ \ \mbox{}
    + \vu z \left[
      \wu\phi \cdot \frac{\pl}{\pl\rho}
      - \wu \rho \cdot \frac{\pl}{\rho\,\pl\phi}
    \right]
  \bigg\} \left( \wu\rho a_\rho + \wu\phi a_\phi + \vu z a_z \right).
\eqb
Here the expansion yields~0x24 (thirty-six) terms, but fortunately as
last time this
time most of the terms again turn out to be null.  The result is that
\[
  \nabla \times \ve a = \wu\rho \left[
    \frac{\pl a_z}{\rho\,\pl\phi} - \frac{\pl a_\phi}{\pl z}
  \right]
  + \wu\phi \left[
    \frac{\pl a_\rho}{\pl z} - \frac{\pl a_z}{\pl\rho}
  \right]
  + \vu z \left[
    \frac{\pl a_\phi}{\pl\rho} + \frac{a_\phi}{\rho} - \frac{\pl a_\rho}{\rho\,\pl\phi}
  \right],
\]
or, expressed more cleverly, that
\bq{vcalc:440.cyl:curl}
  \nabla \times \ve a = \wu\rho \left[
    \frac{\pl a_z}{\rho\,\pl\phi} - \frac{\pl a_\phi}{\pl z}
  \right]
  + \wu\phi \left[
    \frac{\pl a_\rho}{\pl z} - \frac{\pl a_z}{\pl\rho}
  \right]
  + \frac{\vu z}{\rho} \left[
    \frac{\pl (\rho a_\phi)}{\pl\rho} - \frac{\pl a_\rho}{\pl\phi}
  \right].
\eq

Table~\ref{vcalc:440.cyl:tbl} summarizes.\footnote{\cite[appendix~II.2.2]{Balanis}}%
\begin{table}
  \caption{Vector derivatives in cylindrical coordinates.}
  \label{vcalc:440.cyl:tbl}
  \bqb
    \nabla \psi &=&
      \wu\rho \pp{\psi}{\rho}
      + \wu\phi \frac{\pl\psi}{\rho\,\pl\phi}
      + \vu z \pp{\psi}{z} \\
    (\ve b \cdot \nabla)\ve a &=&
      b_\rho \pp{\ve a}{\rho}
      + b_\phi \frac{\pl\ve a}{\rho\,\pl\phi}
      + b_z \pp{\ve a}{z}
    \\&=&
      b_\rho \left[
        \wu\rho \frac{\pl a_\rho}{\pl\rho}
        + \wu\phi \frac{\pl a_\phi}{\pl\rho}
        + \vu z \frac{\pl a_z}{\pl\rho}
      \right]
    \xn\\&&\ \ \mbox{}
      + \frac{b_\phi}{\rho} \left[
        \wu\rho\left( \frac{\pl a_\rho}{\pl\phi} - a_\phi \right)
        + \wu\phi\left( \frac{\pl a_\phi}{\pl\phi} + a_\rho \right)
        + \vu z \frac{\pl a_z}{\pl\phi}
      \right]
    \xn\\&&\ \ \mbox{}
      + b_z \left[
        \wu\rho \frac{\pl a_\rho}{\pl z}
        + \wu\phi \frac{\pl a_\phi}{\pl z}
        + \vu z \frac{\pl a_z}{\pl z}
      \right] \\
    \nabla \cdot \ve a &=&
      \frac{\pl(\rho a_\rho)}{\rho\,\pl\rho}
      + \frac{\pl a_\phi}{\rho\,\pl\phi}
      + \pp{a_z}{z} \\
    \nabla \times \ve a &=& \wu\rho \left[
      \frac{\pl a_z}{\rho\,\pl\phi} - \frac{\pl a_\phi}{\pl z}
    \right]
    + \wu\phi \left[
      \frac{\pl a_\rho}{\pl z} - \frac{\pl a_z}{\pl\rho}
    \right]
    + \frac{\vu z}{\rho} \left[
      \frac{\pl (\rho a_\phi)}{\pl\rho} - \frac{\pl a_\rho}{\pl\phi}
    \right]
  \eqb
\end{table}

\index{derivative!of a field in cylindrical coordinates, second-order}
One can compute a second-order vector derivative in cylindrical
coordinates as a sequence of two first-order cylindrical vector derivatives.  For
example, because Table~\ref{vcalc:tbl-diffid} gives the scalar Laplacian
as $\nabla^2\psi = \nabla \cdot \nabla \psi$, one can
calculate~$\nabla^2\psi$ in cylindrical coordinates by taking the
divergence of $\psi$'s gradient.%
\footnote{
  A concrete example: if $\psi(\ve r) = e^{i\phi}/\rho$, then
  $\nabla\psi = (-\wu\rho +i\wu\phi)e^{i\phi}/\rho^2$
  per Table~\ref{vcalc:440.cyl:tbl}, whereupon
  \[
    \nabla^2\psi = \nabla\cdot\left[
      \left( -\wu\rho + i\wu\phi \right)\frac{e^{i\phi}}{\rho^2}
    \right]
    = \left( -\wu\rho +i\wu\phi \right) \cdot \nabla \left( \frac{e^{i\phi}}{\rho^2} \right)
    + \frac{e^{i\phi}}{\rho^2} \nabla \cdot \left(-\wu\rho +i\wu\phi \right).
  \]
  To finish the example is left as an exercise.
}
To calculate the vector Laplacian~$\nabla^2\ve a$ in cylindrical
coordinates is tedious but nonetheless can with care be done accurately
by means of Table~\ref{vcalc:tbl-diffid}'s identity that $\nabla^2\ve a
= \nabla\nabla\cdot\ve a - \nabla\times\nabla\times\ve a$.  (This means
that to calculate the vector Laplacian~$\nabla^2\ve a$ in cylindrical
coordinates takes not just two but actually four first-order cylindrical
vector derivatives, for the author regrettably knows of no valid
shortcut---the clumsy alternative, less proper, less insightful, even
more tedious and not recommended, being to take the Laplacian in
rectangular coordinates and then to convert back to the cylindrical
domain; for to work cylindrical problems directly in
cylindrical coordinates is almost always advisable.)

\subsection{Derivatives in spherical coordinates}
\label{vcalc:440.sph}
\index{field!derivative of, in spherical coordinates}
\index{vector!derivative of, in spherical coordinates}
\index{derivative!of a field in spherical coordinates}

\index{directional derivative!in spherical coordinates}
\index{gradient!in spherical coordinates}
One can compute vector derivatives in spherical coordinates as in
cylindrical coordinates (\S~\ref{vcalc:440.cyl}), only the spherical
details though not essentially more complicated are messier.
According to Table~\ref{vcalc:tbl-diffid},
\[
  \nabla \psi = \vui\pp{\psi}{i}.
\]
Applying the spherical metric coefficients of Table~\ref{vcalc:420:tbl},
we have that
\bq{vcalc:440.sph:grad}
  \nabla \psi =
    \vu r \frac{\pl\psi}{\pl r}
    + \wu\theta \frac{\pl\psi}{r\,\pl\theta}
    + \wu\phi \frac{\pl\psi}{(r\sin\theta)\,\pl\phi}.
\eq
%
Again according to Table~\ref{vcalc:tbl-diffid},
\[
  (\ve b \cdot \nabla)\ve a = b_i\pp{\ve a}{i}.
\]
Applying the cylindrical metric coefficients, we have that
\bq{vcalc:440.sph:dird}
  (\ve b \cdot \nabla)\ve a =
    b_r \frac{\pl\ve a}{\pl r}
    + b_\theta \frac{\pl\ve a}{r\,\pl\theta}
    + b_\phi \frac{\pl\ve a}{(r\sin\theta)\,\pl\phi}.
\eq
Expanding the vector field~$\ve a$ in the spherical basis,
\[
  (\ve b \cdot \nabla)\ve a = \left\{
    b_r \frac{\pl}{\pl r}
    + b_\theta \frac{\pl}{r\,\pl\theta}
    + b_\phi \frac{\pl}{(r\sin\theta)\,\pl\phi}
  \right\} \left(
    \vu r a_r + \wu\theta a_\theta + \wu\phi a_\phi
  \right).
\]
Evaluating the derivatives,
\bqa
  (\ve b \cdot \nabla)\ve a &=&
    b_r \left[
      \vu r \frac{\pl a_r}{\pl r}
      + \wu\theta \frac{\pl a_\theta}{\pl r}
      + \wu\phi \frac{\pl a_\phi}{\pl r}
    \right]
  \xn\\&&\ \ \mbox{}
    + \frac{b_\theta}{r} \left[
      \vu r \left( \frac{\pl a_r}{\pl\theta} - a_\theta \right)
      + \wu\theta \left( \frac{\pl a_\theta}{\pl\theta} + a_r \right)
      + \wu\phi \frac{\pl a_\phi}{\pl\theta}
    \right]
  \xn\\&&\ \ \mbox{}
    + \frac{b_\phi}{r\sin\theta} \bigg[
      \vu r \left( \frac{\pl a_r}{\pl\phi} - a_\phi \sin\theta \right)
      + \wu\theta \left( \frac{\pl a_\theta}{\pl\phi} - a_\phi \cos\theta \right)
  \xn\\&&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{}
      + \wu\phi \left( \frac{\pl a_\phi}{\pl\phi} + a_r \sin\theta + a_\theta \cos\theta \right)
    \bigg].
  \label{vcalc:440.sph:dird3}
\eqa

\index{divergence!in spherical coordinates}
\index{curl!in spherical coordinates}
According to Table~\ref{vcalc:tbl-diffid}, reasoning as in
\S~\ref{vcalc:440.cyl},
\[
  \nabla \cdot \ve a = \ds\pp{a_i}{i} =
    \vu r \cdot \frac{\pl\ve a}{\pl r}
    + \wu\theta \cdot \frac{\pl\ve a}{r\,\pl\theta}
    + \wu\phi \cdot \frac{\pl\ve a}{(r\sin\theta)\,\pl\phi}.
\]
Expanding the field in the spherical basis,
\[
  \nabla \cdot \ve a = \left\{
    \vu r \cdot \frac{\pl}{\pl r}
    + \wu\theta \cdot \frac{\pl}{r\,\pl\theta}
    + \wu\phi \cdot \frac{\pl}{(r\sin\theta)\,\pl\phi}
  \right\} \left(
    \vu r a_r + \wu\theta a_\theta + \wu\phi a_\phi
  \right).
\]
Evaluating the derivatives, the result is that
\[
  \nabla \cdot \ve a =
    \frac{\pl a_r}{\pl r}
    + \frac{2a_r}{r}
    + \frac{\pl a_\theta}{r\,\pl\theta}
    + \frac{a_\theta}{r\tan\theta}
    + \frac{\pl a_\phi}{(r\sin\theta)\,\pl\phi},
\]
or, expressed more cleverly, that
\bq{vcalc:440.sph:divg}
  \nabla \cdot \ve a = \frac{1}{r}\left[
    \frac{\pl(r^2a_r)}{r\,\pl r}
    + \frac{\pl(a_\theta\sin\theta)}{(\sin\theta)\,\pl\theta}
    + \frac{\pl a_\phi}{(\sin\theta)\,\pl\phi}
  \right].
\eq
%
Again according to Table~\ref{vcalc:tbl-diffid}, reasoning as in
\S~\ref{vcalc:440.cyl},
\bqb
  \nabla \times \ve a &=& \ep_{ijk} \vui \pp{a_k}{j}
  \\&=&
    \vu r \left[
      \wu\phi \cdot \frac{\pl\ve a}{r\,\pl\theta}
      - \wu\theta \cdot \frac{\pl\ve a}{(r\sin\theta)\,\pl\phi}
    \right]
    + \wu\theta \left[
      \vu r \cdot \frac{\pl\ve a}{(r\sin\theta)\,\pl\phi}
      - \wu\phi \cdot \frac{\pl\ve a}{\pl r}
    \right]
  \\&&\ \ \mbox{}
    + \wu\phi \left[
      \wu\theta \cdot \frac{\pl\ve a}{\pl r}
      - \vu r \cdot \frac{\pl\ve a}{r\,\pl\theta}
    \right].
\eqb
Expanding the field in the spherical basis,
\bqb
  \nabla \times \ve a &=& \bigg\{
    \vu r \left[
      \wu\phi \cdot \frac{\pl}{r\,\pl\theta}
      - \wu\theta \cdot \frac{\pl}{(r\sin\theta)\,\pl\phi}
    \right]
    + \wu\theta \left[
      \vu r \cdot \frac{\pl}{(r\sin\theta)\,\pl\phi}
      - \wu\phi \cdot \frac{\pl}{\pl r}
    \right]
  \\&&\ \ \mbox{}
    + \wu\phi \left[
      \wu\theta \cdot \frac{\pl}{\pl r}
      - \vu r \cdot \frac{\pl}{r\,\pl\theta}
    \right]
  \bigg\} \left( \vu r a_r + \wu\theta a_\theta + \wu\phi a_\phi \right).
\eqb
Evaluating the derivatives, the result is that
\bqb
  \nabla \times \ve a &=& \vu r \left[
    \frac{\pl a_\phi}{r\,\pl\theta}
    + \frac{a_\phi}{r\tan\theta}
    - \frac{\pl a_\theta}{(r\sin\theta)\,\pl\phi}
  \right]
  + \wu\theta \left[
    \frac{\pl a_r}{(r\sin\theta)\,\pl\phi}
    - \frac{\pl a_\phi}{\pl r}
    - \frac{a_\phi}{r}
  \right]
  \xn\\&&\ \ \mbox{}
  + \wu\phi \left[
    \frac{\pl a_\theta}{\pl r}
    + \frac{a_\theta}{r}
    - \frac{\pl a_r}{r\,\pl\theta}
  \right],
  \xn
\eqb
or, expressed more cleverly, that
\bqa
  \nabla \times \ve a &=& \frac{\vu r}{r\sin\theta} \left[
    \frac{\pl(a_\phi\sin\theta)}{\pl\theta}
    - \frac{\pl a_\theta}{\pl\phi}
  \right]
  + \frac{\wu\theta}{r} \left[
    \frac{\pl a_r}{(\sin\theta)\,\pl\phi}
    - \frac{\pl(r a_\phi)}{\pl r}
  \right]
  \xn\\&&\ \ \mbox{}
  + \frac{\wu\phi}{r} \left[
    \frac{\pl(r a_\theta)}{\pl r}
    - \frac{\pl a_r}{\pl\theta}
  \right].
  \label{vcalc:440.sph:curl}
\eqa

Table~\ref{vcalc:440.sph:tbl} summarizes.\footnote{\cite[appendix~II.2.3]{Balanis}}%
\begin{table}
  \caption{Vector derivatives in spherical coordinates.}
  \label{vcalc:440.sph:tbl}
  \bqb
    \nabla \psi &=&
      \vu r \frac{\pl\psi}{\pl r}
      + \wu\theta \frac{\pl\psi}{r\,\pl\theta}
      + \wu\phi \frac{\pl\psi}{(r\sin\theta)\,\pl\phi} \\
    (\ve b \cdot \nabla)\ve a &=&
      b_r \frac{\pl\ve a}{\pl r}
      + b_\theta \frac{\pl\ve a}{r\,\pl\theta}
      + b_\phi \frac{\pl\ve a}{(r\sin\theta)\,\pl\phi}
    \\&=&
      b_r \left[
        \vu r \frac{\pl a_r}{\pl r}
        + \wu\theta \frac{\pl a_\theta}{\pl r}
        + \wu\phi \frac{\pl a_\phi}{\pl r}
      \right]
    \\&&\ \ \mbox{}
      + \frac{b_\theta}{r} \left[
        \vu r \left( \frac{\pl a_r}{\pl\theta} - a_\theta \right)
        + \wu\theta \left( \frac{\pl a_\theta}{\pl\theta} + a_r \right)
        + \wu\phi \frac{\pl a_\phi}{\pl\theta}
      \right]
    \\&&\ \ \mbox{}
      + \frac{b_\phi}{r\sin\theta} \bigg[
        \vu r \left( \frac{\pl a_r}{\pl\phi} - a_\phi \sin\theta \right)
        + \wu\theta \left( \frac{\pl a_\theta}{\pl\phi} - a_\phi \cos\theta \right)
    \\&&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{}
        + \wu\phi \left( \frac{\pl a_\phi}{\pl\phi} + a_r \sin\theta + a_\theta \cos\theta \right)
      \bigg] \\
    \nabla \cdot \ve a &=& \frac{1}{r}\left[
      \frac{\pl(r^2a_r)}{r\,\pl r}
      + \frac{\pl(a_\theta\sin\theta)}{(\sin\theta)\,\pl\theta}
      + \frac{\pl a_\phi}{(\sin\theta)\,\pl\phi}
    \right] \\
    \nabla \times \ve a &=& \frac{\vu r}{r\sin\theta} \left[
      \frac{\pl(a_\phi\sin\theta)}{\pl\theta}
      - \frac{\pl a_\theta}{\pl\phi}
    \right]
    + \frac{\wu\theta}{r} \left[
      \frac{\pl a_r}{(\sin\theta)\,\pl\phi}
      - \frac{\pl(r a_\phi)}{\pl r}
    \right]
    \\&&\ \ \mbox{}
    + \frac{\wu\phi}{r} \left[
      \frac{\pl(r a_\theta)}{\pl r}
      - \frac{\pl a_r}{\pl\theta}
    \right]
  \eqb
\end{table}

\index{derivative!of a field in spherical coordinates, second-order}
One can compute a second-order vector derivative in spherical
coordinates as in cylindrical coordinates, as a sequence of two
first-order vector derivatives.  Refer to \S~\ref{vcalc:440.cyl}.

\subsection{Finding the derivatives geometrically}
\label{vcalc:440.geo}
\index{geometrical argument}
\index{geometrical visualization}

\index{divergence!in nonrectangular coordinates}
The method of \S\S~\ref{vcalc:440.cyl} and~\ref{vcalc:440.sph} is
general, reliable and correct, but there exists an alternate, arguably
neater method to derive nonrectangular formulas for most vector
derivatives.  Adapting the notation to this subsection's purpose we can
write~(\ref{vcalc:320:35}) as
\bq{vcalc:440.geo:20}
  \nabla \cdot \ve a(\ve r) \equiv
    \lim_{\Delta V\ra 0}\frac{\Phi}{\Delta V},
\eq
thus defining a vector's divergence fundamentally as in
\S~\ref{vcalc:320.30}, geometrically, as the ratio of flux~$\Phi$
from a vanishing test volume~$\Delta V$ to the volume itself; where
per~(\ref{vcalc:flux}) $\Phi = \oint_S \ve a(\ve r') \cdot d\ve
s$, where~$\ve r'$ is a position on the test volume's surface, and where
$d\ve s = d\ve s(\ve r')$ is the corresponding surface patch.  So
long as the test volume~$\Delta V$ includes the point~$\ve r$ and is
otherwise infinitesimal in extent, we remain free
to shape the volume as we like,%
\footnote{
  A professional mathematician would probably enjoin the volume's shape
  to obey certain technical restrictions, such as that it remain wholly
  enclosed within a sphere of vanishing radius, but we will not try for
  such a level of rigor here.
}
so let us give it six sides and shape it as an almost rectangular box
that conforms precisely to the coordinate system $(\alpha;\beta;\gamma)$
in use:
\[
  \settowidth\tla{$\alpha$}
  \settowidth\tlc{$\alpha'$}
  \settowidth\tlb{$\Delta \alpha$}
  \begin{split}
    \alpha - \frac{\makebox[\tlb][l]{$\Delta \alpha$}}{2} \le \makebox[\tlc][r]{$\alpha'$}
      &\le \makebox[\tla][l]{$\alpha$} + \frac{\makebox[\tlb][l]{$\Delta \alpha$}}{2}; \\
    \beta  - \frac{\makebox[\tlb][l]{$\Delta \beta $}}{2} \le \makebox[\tlc][r]{$\beta '$}
      &\le \makebox[\tla][l]{$\beta $} + \frac{\makebox[\tlb][l]{$\Delta \beta $}}{2}; \\
    \gamma - \frac{\makebox[\tlb][l]{$\Delta \gamma$}}{2} \le \makebox[\tlc][r]{$\gamma'$}
      &\le \makebox[\tla][l]{$\gamma$} + \frac{\makebox[\tlb][l]{$\Delta \gamma$}}{2}.
  \end{split}
\]
The fluxes outward through the box's $+\alpha$- and $-\alpha$-ward sides
will then be%
\footnote{
  More rigorously, one might digress from this point to expand the field
  in a three-dimensional Taylor series (\S~\ref{taylor:370}) to account
  for the field's variation over a single side of the test volume.  So
  lengthy a digression however would only formalize what we already
  knew; namely, that one can approximate to first order the integral of
  a well-behaved quantity over an infinitesimal domain by the quantity's
  value at the domain's midpoint.  If you will believe that
  $\lim_{\Delta\tau \ra 0}\int_{\tau-\Delta\tau/2}^{\tau+\Delta\tau/2}
  f(\tau')\,d\tau' = f(\tau) \,\Delta\tau$ for any~$\tau$ in the
  neighborhood of which $f(\tau)$ is well behaved, then you will
  probably also believe its three-dimensional analog in the narrative.
  (If the vagueness in this context of the adjective ``well-behaved''
  deeply troubles any reader then that reader may possess the worthy
  temperament of a professional mathematician; he might review
  chapter~\ref{taylor} and then seek further illumination in the
  professional mathematical literature.  Other readers, of more
  practical temperament, are advised to visualize test volumes in
  rectangular, cylindrical and spherical coordinates and to ponder the
  matter a while.  Consider: if the field grows in strength across a
  single side of the test volume and if the test volume is small enough
  that second-order effects can be ignored, then what single value
  ought one to choose to represent the field over the whole side but its
  value at the side's midpoint?  Such visualization should soon clear up
  any confusion and is what the writer recommends.  Incidentally, the
  contrast between the two modes of thought this footnote reveals is
  exactly the sort of thing Courant and Hilbert were talking about in
  \S~\ref{intro:284.2}.)
}
\[
  \begin{split}
    \Phi_{+\alpha} &= \left.
      (+a_\alpha)
      (h_\beta h_\gamma \,\Delta\beta\,\Delta\gamma)
    \right|_{\ve r' = \ve r(\alpha+\Delta\alpha/2;\beta;\gamma)}
    \\&\ \ \ \ \ \ \ \ = \left.
      +a_\alpha h_\beta h_\gamma
    \right|_{\ve r' = \ve r(\alpha+\Delta\alpha/2;\beta;\gamma)}
    \,\Delta\beta\,\Delta\gamma,
    \\
    \Phi_{-\alpha} &= \left.
      (-a_\alpha)
      (h_\beta h_\gamma \,\Delta\beta\,\Delta\gamma)
    \right|_{\ve r' = \ve r(\alpha-\Delta\alpha/2;\beta;\gamma)}
    \\&\ \ \ \ \ \ \ \ = \left.
      -a_\alpha h_\beta h_\gamma
    \right|_{\ve r' = \ve r(\alpha-\Delta\alpha/2;\beta;\gamma)}
    \,\Delta\beta\,\Delta\gamma,
  \end{split}
\]
products of the outward-directed field components and the
areas~(\ref{vcalc:420:21}) of the sides through which the fields pass.
Thence by successive steps, the net flux outward through the pair of
opposing sides will be
\bqb
  \Phi_\alpha &=& \Phi_{+\alpha}+\Phi_{-\alpha}
  \\&=& \left[
      \left.
        a_\alpha h_\beta h_\gamma
      \right|_{\ve r' = \ve r(\alpha+\Delta\alpha/2;\beta;\gamma)}
    - \left.
        a_\alpha h_\beta h_\gamma
      \right|_{\ve r' = \ve r(\alpha-\Delta\alpha/2;\beta;\gamma)}
  \right] \,\Delta\beta\,\Delta\gamma
  \\&=& \left[
    \frac{\pl(a_\alpha h_\beta h_\gamma)}{\pl\alpha} \,\Delta\alpha
  \right] \,\Delta\beta\,\Delta\gamma
  = \Delta\alpha\,\Delta\beta\,\Delta\gamma\,
    \frac{\pl(a_\alpha h_\beta h_\gamma)}{\pl\alpha}
  \\&=& \frac{\Delta V \,\pl(a_\alpha h_\beta h_\gamma)}
    {h_\alpha h_\beta h_\gamma \,\pl\alpha}.
\eqb
Naturally, the same goes for the other two pairs of sides:
\[
  \begin{split}
    \Phi_\beta &= \frac{\Delta V \,\pl(a_\beta h_\gamma h_\alpha)}
      {h_\alpha h_\beta h_\gamma\,\pl\beta}; \\
    \Phi_\gamma &= \frac{\Delta V \,\pl(a_\gamma h_\alpha h_\beta)}
      {h_\alpha h_\beta h_\gamma \,\pl\gamma}.
  \end{split}
\]
The three equations are better written%
{%
  \settowidth\tla{$\pl\alpha$}%
  \settowidth\tlb{$a_\alpha$}%
  \[
    \begin{split}
      \Phi_\alpha &= \frac{\Delta V \,\pl}{h^3 \,\makebox[\tla][l]{$\pl\alpha$}}
        \left(\frac{h^3 \makebox[\tlb][l]{$a_\alpha$}}{h_\alpha}\right), \\
      \Phi_\beta  &= \frac{\Delta V \,\pl}{h^3 \,\makebox[\tla][l]{$\pl\beta $}}
        \left(\frac{h^3 \makebox[\tlb][l]{$a_\beta $}}{h_\beta }\right), \\
      \Phi_\gamma &= \frac{\Delta V \,\pl}{h^3 \,\makebox[\tla][l]{$\pl\gamma$}}
        \left(\frac{h^3 \makebox[\tlb][l]{$a_\gamma$}}{h_\gamma}\right),
    \end{split}
  \]%
}%
where
\bq{vcalc:440.geo:23}
  h^3 \equiv h_\alpha h_\beta h_\gamma.
\eq
The total flux from the test volume then is
\bqb
  \Phi &=& \Phi_\alpha + \Phi_\beta + \Phi_\gamma
  \\&=& \frac{\Delta V}{h^3} \left[
      \frac{\pl}{\pl\alpha}
        \left(\frac{h^3 a_\alpha}{h_\alpha}\right)
    + \frac{\pl}{\pl\beta }
        \left(\frac{h^3 a_\beta }{h_\beta }\right)
    + \frac{\pl}{\pl\gamma}
        \left(\frac{h^3 a_\gamma}{h_\gamma}\right)
  \right];
\eqb
or, invoking Einstein's summation convention in \S~\ref{vcalc:425}'s
modified style,
\[
  \Phi = \frac{\Delta V \,\pl}{h^3 \,\pl\tdi}
    \left(\frac{h^3 a_\tdi}{h_\tdi}\right).
\]
Finally, substituting the last equation into~(\ref{vcalc:440.geo:20}),
\bq{vcalc:440.geo:27}
  \nabla \cdot \ve a =
    \frac{\pl}{h^3 \,\pl\tdi}
    \left(\frac{h^3 a_\tdi}{h_\tdi}\right).
\eq

\index{curl!in nonrectangular coordinates}
An analogous formula for curl is not much harder to derive but is harder
to approach directly, so we will approach it by deriving first the
formula for $\wu\gamma$-directed directional curl.
Equation~(\ref{vcalc:320:45}) has it that%
\footnote{
  The appearance of both~$\ve a$ and~$A$ in~(\ref{vcalc:440.geo:30})
  is unfortunate but co\"incidental, as is the appearance of both~$\wu
  \gamma$ and~$\Gamma$.  The capital and minuscule symbols here
  represent unrelated quantities.
}
\bq{vcalc:440.geo:30}
  \wu\gamma \cdot \nabla \times \ve a(\ve r) \equiv
    \lim_{\Delta A\ra 0}\frac{\Gamma}{\Delta A},
\eq
where per~(\ref{vcalc:circulation}) $\Gamma = \oint_\gamma \ve
a(\ve r') \cdot d\we\ell$ and the notation~$\oint_\gamma$
reminds us that the contour of integration lies in the~$\alpha$-$\beta$
plane, perpendicular to~$\wu\gamma$.  In this case the contour of
integration bounds not a test volume but a test surface, which we give
four edges and an almost rectangular shape that conforms precisely to
the coordinate system $(\alpha;\beta;\gamma)$ in use:
\[
  \settowidth\tla{$\alpha$}
  \settowidth\tlc{$\alpha'$}
  \settowidth\tlb{$\Delta \alpha$}
  \begin{split}
    \alpha - \frac{\makebox[\tlb][l]{$\Delta \alpha$}}{2} \le \makebox[\tlc][r]{$\alpha'$}
      &\le \makebox[\tla][l]{$\alpha$} + \frac{\makebox[\tlb][l]{$\Delta \alpha$}}{2}; \\
    \beta  - \frac{\makebox[\tlb][l]{$\Delta \beta $}}{2} \le \makebox[\tlc][r]{$\beta '$}
      &\le \makebox[\tla][l]{$\beta $} + \frac{\makebox[\tlb][l]{$\Delta \beta $}}{2}; \\
    \makebox[\tlc][r]{$\gamma'$} &= \gamma.
  \end{split}
\]
The circulations along the $+\alpha$- and $-\alpha$-ward edges will be
\[
  \begin{split}
    \Gamma_{+\alpha} &=
    \left.
      +h_\beta a_\beta
    \right|_{\ve r' = \ve r(\alpha+\Delta\alpha/2;\beta;\gamma)}
    \,\Delta\beta,
    \\
    \Gamma_{-\alpha} &=
    \left.
      -h_\beta a_\beta
    \right|_{\ve r' = \ve r(\alpha-\Delta\alpha/2;\beta;\gamma)}
    \,\Delta\beta,
  \end{split}
\]
and likewise the circulations along the $-\beta$- and $+\beta$-ward
edges will be
\[
  \begin{split}
    \Gamma_{-\beta} &=
    \left.
      +h_\alpha a_\alpha
    \right|_{\ve r' = \ve r(\alpha;\beta-\Delta\beta/2;\gamma)}
    \,\Delta\alpha,
    \\
    \Gamma_{+\beta} &=
    \left.
      -h_\alpha a_\alpha
    \right|_{\ve r' = \ve r(\alpha;\beta+\Delta\beta/2;\gamma)}
    \,\Delta\alpha,
  \end{split}
\]
whence the total circulation about the contour is
\bqb
  \Gamma &=&
        \frac{\pl(h_\beta  a_\beta )}{\pl\alpha} \,\Delta\alpha \,\Delta\beta
      - \frac{\pl(h_\alpha a_\alpha)}{\pl\beta } \,\Delta\beta \,\Delta\alpha
  \\&=&
    \frac{h_\gamma \,\Delta A}{h^3} \left[
        \frac{\pl(h_\beta  a_\beta )}{\pl\alpha}
      - \frac{\pl(h_\alpha a_\alpha)}{\pl\beta }
    \right].
\eqb
Substituting the last equation into~(\ref{vcalc:440.geo:30}), we have that
\[
  \wu\gamma \cdot \nabla \times \ve a
  = \frac{h_\gamma}{h^3} \left[
        \frac{\pl(h_\beta  a_\beta )}{\pl\alpha}
      - \frac{\pl(h_\alpha a_\alpha)}{\pl\beta }
    \right].
\]
Likewise,%
{%
  \settowidth\tla{$h_\alpha$}%
  \settowidth\tlb{$(h_\alpha a_\alpha)$}%
  \[
    \begin{split}
      \wu\alpha \cdot \nabla \times \ve a
      &= \frac{\makebox[\tla][l]{$h_\alpha$}}{h^3} \left[
            \frac{\pl\makebox[\tlb][l]{$(h_\gamma a_\gamma)$}}{\pl\beta }
          - \frac{\pl\makebox[\tlb][l]{$(h_\beta  a_\beta )$}}{\pl\gamma}
        \right],
      \\
      \wu\beta  \cdot \nabla \times \ve a
      &= \frac{\makebox[\tla][l]{$h_\beta $}}{h^3} \left[
            \frac{\pl\makebox[\tlb][l]{$(h_\alpha a_\alpha)$}}{\pl\gamma}
          - \frac{\pl\makebox[\tlb][l]{$(h_\gamma a_\gamma)$}}{\pl\alpha}
        \right].
    \end{split}
  \]%
}%
But one can split any vector~$\ve v$ into locally rectangular components
as
$
  \ve v =
    \wu \alpha ( \wu \alpha \cdot \ve v )
  + \wu \beta  ( \wu \beta  \cdot \ve v )
  + \wu \gamma ( \wu \gamma \cdot \ve v ),
$
so
\bqb
  \nabla \times \ve a &=&
    \wu \alpha ( \wu \alpha \cdot \nabla \times \ve a )
  + \wu \beta  ( \wu \beta  \cdot \nabla \times \ve a )
  + \wu \gamma ( \wu \gamma \cdot \nabla \times \ve a )
  \\&=&
  \frac{\wu\alpha h_\alpha}{h^3} \left[
        \frac{\pl(h_\gamma a_\gamma)}{\pl\beta }
      - \frac{\pl(h_\beta  a_\beta )}{\pl\gamma}
  \right]
  + \frac{\wu\beta  h_\beta }{h^3} \left[
        \frac{\pl(h_\alpha a_\alpha)}{\pl\gamma}
      - \frac{\pl(h_\gamma a_\gamma)}{\pl\alpha}
  \right]
  \\&&\ \ \ \ \mbox{}
  + \frac{\wu\gamma h_\gamma}{h^3} \left[
      \frac{\pl(h_\beta  a_\beta )}{\pl\alpha}
    - \frac{\pl(h_\alpha a_\alpha)}{\pl\beta }
  \right]
  \\&=& \frac{1}{h^3} \left|
    \br{ccc}
      \wu\alpha h_\alpha & \wu\beta h_\beta & \wu\gamma h_\gamma \\
      \pl/\pl\alpha & \pl/\pl\beta & \pl/\pl\gamma \\
      h_\alpha a_\alpha & h_\beta a_\beta & h_\gamma a_\gamma
    \er
  \right|;
\eqb
or, in Einstein notation,%
\footnote{
  What a marvel mathematical notation is!  If you can
  read~(\ref{vcalc:440.geo:37}) and understand the message it conveys,
  then let us pause a moment to appreciate a few of the many concepts
  the notation implicitly encapsulates.  There are the vector, the unit
  vector, the field, the derivative, the integral, circulation, parity,
  rotational invariance, nonrectangular coordinates, three-dimensional
  geometry, the dummy variable and so on---each of which concepts itself
  yet encapsulates several further ideas---not to mention multiplication
  and division which themselves are not trivial.  It is doubtful that
  one could explain it even tersely to the uninitiated in fewer than
  fifty pages, and yet to the initiated one can express it all in half a
  line.
}
\bq{vcalc:440.geo:37}
  \nabla \times \ve a = \frac{
    \ep_{\tdi\hspace{0.5pt}\tdj\hspace{0.5pt}\tdk}\vui h_\tdi \,\pl(h_\tdk a_\tdk)
  }{
    h^3 \,\pl\tdj
  }.
\eq

\index{gradient!in nonrectangular coordinates}
Compared to the formulas~(\ref{vcalc:440.geo:27})
and~(\ref{vcalc:440.geo:37}) for divergence and curl, the corresponding
gradient formula seems almost trivial.  It is
\bq{vcalc:440.geo:47}
  \nabla\psi = \frac{\vui \,\pl\psi}{h_\tdi \,\pl\tdi}.
\eq

\index{parabolic coordinates}
\index{coordinates!parabolic}
One can generate most of the vector-derivative formulas of
Tables~\ref{vcalc:440.cyl:tbl} and~\ref{vcalc:440.sph:tbl} by means
of this subsection's~(\ref{vcalc:440.geo:27}), (\ref{vcalc:440.geo:37})
and~(\ref{vcalc:440.geo:47}).  One can generate additional
vector-derivative formulas for special coordinate systems like the
parabolic systems of \S~\ref{vector:280} by means of the same equations.

% ----------------------------------------------------------------------

\section{Vector infinitesimals}
\label{vcalc:520}
\index{infinitesimal!vector}
\index{vector infinitesimal}

\index{contour integration}
\index{contour integration!of a vector quantity}
\index{integral!contour}
\index{integral!vector contour}
\index{surface integration}
\index{integral!surface}
To integrate a field over a contour or surface is a typical maneuver of
vector calculus.  One might integrate in any of the forms
\[
  \renewcommand\arraystretch{2.0}
  \br{lllllll}
    \ds\int_C \psi \,d\ell          &\ \ &
    \ds\int_C \ve a \,d\ell         &\ \ &
    \ds\int_S \psi \,ds             &\ \ &
    \ds\int_S \ve a \,ds            \\
    \ds\int_C \psi \,d\we\ell       &&
    \ds\int_C \ve a \cdot d\we\ell  &&
    \ds\int_S \psi  \,d\ve s        &&
    \ds\int_S \ve a \cdot d\ve s    \\ &&
    \ds\int_C \ve a \times d\we\ell && &&
    \ds\int_S \ve a \times d\ve s
  \er
\]
among others.  Where the integration is over a contour, a pair of
functions $\alpha(\gamma)$ and $\beta(\gamma)$ typically can serve to
specify the contour.  Where over a surface, a single function
$\gamma(\alpha;\beta)$ can serve.
Given such functions and a field
integral to compute, one wants an expression for the integrand's
infinitesimal~$d\we\ell$ or~$d\ve s$ in terms respectively of the
contour functions $\alpha(\gamma)$ and $\beta(\gamma)$ or of the surface
function $\gamma(\alpha;\beta)$.

\index{infinitesimal!displacement}
\index{displacement infinitesimal}
\index{length infinitesimal}
\index{contour infinitesimal}
\index{infinitesimal!surface}
\index{surface infinitesimal}
\index{surface element}
\index{patch}
The contour infinitesimal is evidently
\bq{vcalc:520:20}
  d\we\ell = \left(
    \wu\gamma h_\gamma
    + \wu\alpha\frac{h_\alpha \,d\alpha}{d\gamma}
    + \wu\beta\frac{h_\beta \,d\beta}{d\gamma}
  \right) \,d\gamma,
\eq
consisting of a step in the~$\wu\gamma$ direction plus the corresponding
steps in the orthogonal~$\wu\alpha$ and~$\wu\beta$ directions.  This is
easy once you see how to do it.  Harder is the surface infinitesimal,
but one can nevertheless correctly construct it as the cross product
\bqa
  d\ve s &=&
  \left[
    \left(
      \wu\alpha h_\alpha
      + \wu\gamma\frac{h_\gamma \,\pl\gamma}{\pl\alpha}
    \right) \,d\alpha
  \right] \times \left[
    \left(
      \wu\beta h_\beta
      + \wu\gamma\frac{h_\gamma \,\pl\gamma}{\pl\beta}
    \right) \,d\beta
  \right]
  \xn\\&=&
  \left(
    \wu\gamma \frac{1}{h_\gamma}
    - \wu\alpha\frac{\pl\gamma}{h_\alpha \,\pl\alpha}
    - \wu\beta\frac{\pl\gamma}{h_\beta \,\pl\beta}
  \right) h^3 \,d\alpha \,d\beta
  \label{vcalc:520:30}
\eqa
of two vectors that lie on the surface, one vector normal to~$\wu\beta$
and the other to~$\wu\alpha$, edges not of a rectangular patch of the
surface but of a patch whose projection onto the~$\alpha$-$\beta$ plane
is an $(h_\alpha \,d\alpha)$-by-$(h_\beta \,d\beta)$ rectangle.

% ----------------------------------------------------------------------

So, that's it.  Those are the essentials of the three-dimensional
geometrical vector---of its analysis and of its calculus.  The
geometrical vector of chapters~\ref{vector} and~\ref{vcalc} and the matrix
of chapters~\ref{matrix} through~\ref{eigen} have in common that they
represent well-developed ways of marshaling several quantities together
to a common purpose: three quantities in the specialized case of the
geometrical vector;~$n$ quantities in the generalized case of the
matrix.  Matrices and vectors have admittedly not been easy for us to
treat but after a slow start, it must be said, they have proven
unexpectedly interesting.  In applications,
they are exceedingly significant.  Matrices and vectors vastly expand
the domain of physical phenomena a scientist or engineer can model.
Mathematically, one cannot well manage without them.

The time nevertheless has come to change the subject.  Turning the page,
we will begin from the start of the next chapter to introduce a series
of advanced topics that pick up where chapter~\ref{inttx} has left off,
entering first upon the broad topic of the Fourier transform.

