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

% This chapter's title is named again later in the book.  Thus this hack
% (which will not work if the title is named again earlier in the book;
% I should look up the right way to do this).  ---THB---
\newcommand\chaptertitlevector{Vector analysis}

\chapter{\chaptertitlevector}
\label{vector}
\index{vector}
\index{vector analysis}
\index{vector!three-dimensional geometrical}
\index{three-dimensional geometrical vector}
\index{geometrical vector}
\index{vector algebra}
\index{algebra!of the vector}

\index{physical world}
\index{world, physical}
Leaving the matrix, this chapter and the next turn to an
agent of applied mathematics that, though ubiquitous in some fields of
study like physics, remains curiously underappreciated in other fields
that should use it more.  This agent is the \emph{three-dimensional
geometrical vector,} first met in \S\S~\ref{trig:230}, \ref{trig:240}
and~\ref{trig:277}.  Seen from one perspective, the three-dimensional
geometrical vector is the $n=3$ special case of the general,
$n$-dimensional vector of chapters~\ref{matrix} through~\ref{eigen}.
Because its three elements represent the three dimensions of
the physical world, however, the three-dimensional geometrical vector merits
closer attention and special treatment.%
\footnote{\cite[chapter~2]{Cheng}}

\index{vector!matrix}
\index{matrix vector}
\index{vector!$n$-dimensional}
\index{$n$-dimensional vector}
It also merits a shorter name.  Where the geometrical context
is clear---as it is in this chapter and the next---we will call the
three-dimensional geometrical vector just a \emph{vector.}  A name like
``matrix vector'' or ``$n$-dimensional vector'' can disambiguate the
vector of chapters~\ref{matrix} through~\ref{eigen} where necessary but,
since the three-dimensional geometrical vector is in fact a vector
in the broader sense, to disambiguate is usually unnecessary.  The lone
word \emph{vector} serves.

\index{amplitude}
\index{direction}
\index{coordinate}
\index{rectangular coordinates}
\index{cylindrical coordinates}
\index{spherical coordinates}
In the present chapter's context and according to \S~\ref{trig:230}, a
vector consists of an amplitude of some kind plus a direction.  Per
\S~\ref{trig:277}, three scalars called \emph{coordinates} suffice
together to specify the amplitude and direction and thus the vector, the
three being $(x,y,x)$ in the rectangular coordinate system,
$(\rho;\phi,z)$ in the cylindrical coordinate system, or
$(r;\theta;\phi)$ in the spherical spherical coordinate system---as
Fig.~\ref{vector:fig-sphere} illustrates and Table~\ref{trig:277:20} on
page~\pageref{trig:277:20} interrelates---among other, more exotic
possibilities (\S~\ref{vector:280}).%
\begin{figure}
  \caption[A point on a sphere.]{%
    A point on a sphere, in spherical $(r;\theta;\phi)$ and cylindrical
    $(\rho;\phi,z)$ coordinates.  (The axis labels bear circumflexes in
    this figure only to disambiguate the~$\hat z$ axis from the
    cylindrical coordinate~$z$.  See also
    Fig.~\ref{vector:230:fig-spherical-basis}.)%
  }%
  \label{vector:fig-sphere}
  \index{sphere}
  \sphere
\end{figure}

\index{vector notation}
\index{notation!of the vector}
\index{rotation}
\index{axes!rotation of}
The vector brings an elegant notation.  This chapter and chapter~\ref{vcalc}
detail it.  Without the notation, one would write an expression like
\[
    \frac{
      (z-z')
      -\left[\partial z'/\partial x\right]_{x=x',y=y'}
      (x-x')
      -\left[\partial z'/\partial y\right]_{x=x',y=y'}
      (y-y')
    }{
      \sqrt{
        \left[ 1
          + (\partial z'/\partial x)^2
          + (\partial z'/\partial y)^2
        \right]_{x=x',y=y'}
        \left[(x-x')^2 + (y-y')^2 + (z-z')^2\right]
      }
    }
\]
for the aspect coefficient relative to a local surface normal (and if
the sentence's words do not make sense to you yet, don't worry; just
look the symbols over and appreciate the expression's bulk).  The same
coefficient in standard vector notation is
\[
  \vu n \cdot \Delta \vu r.
\]
Besides being more evocative (once one has learned to read it) and much
more compact, the standard vector notation brings the major advantage
of freeing a model's geometry from reliance on any particular coordinate
system.  Reorienting axes (\S~\ref{vector:210}) for example knots the
former expression like spaghetti but does not disturb the latter
expression at all.

\index{vector!two-dimensional geometrical}
\index{two-dimensional geometrical vector}
Two-dimensional geometrical vectors arise in practical modeling about as
often as three-dimensional geometrical vectors do.  Fortunately, the
two-dimensional case needs little special treatment, for it is
just the three-dimensional with $z=0$ or $\theta = 2\pi/4$ (see
however \S~\ref{vector:270}).

\index{number!real}
\index{real number}
\index{coordinate!real}
\index{coordinate!complex}
\index{real coordinate}
\index{complex coordinate}
Here at the outset, a word on complex numbers seems in
order.  Unlike most of the rest of the
book this chapter and the next will work chiefly in real numbers, or at
least in real coordinates.  Notwithstanding, complex coordinates are
possible.  Indeed, in the rectangular coordinate system complex
coordinates are perfectly appropriate and are straightforward enough to
handle.  The cylindrical and spherical systems however, which these
chapters also treat, were not conceived with complex
coordinates in mind; and, although it might with some theoretical
subtlety be possible to treat complex radii, azimuths and elevations
consistently as three-dimensional coordinates, these chapters will not
try to do so.%
\footnote{
  The author would be interested to learn if there existed an
  uncontrived scientific or engineering application that actually used
  complex, nonrectangular coordinates.
}
(This is not to say that you cannot have a complex vector like, say,
$\wu\rho[3+j2] - \wu\phi[1/4]$ in a nonrectangular basis.  You can have
such a vector, it is fine, and these chapters will not avoid it.  What
these chapters will avoid are complex nonrectangular \emph{coordinates}
like $[3+j2;-1/4,0]$.)

\index{vector!addition of}
\index{addition!of vectors}
Vector addition will already be familiar to the reader from
chapter~\ref{trig} or (quite likely) from earlier work outside this book.
This chapter therefore begins with the reorientation of axes in
\S~\ref{vector:210} and vector multiplication in \S~\ref{vector:220}.

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

\section{Reorientation}
\label{vector:210}
\index{reorientation}
\index{axes!reorientation of}

Matrix notation expresses the rotation of axes~(\ref{trig:240:10}) as
\[
  \mf{c}{\vu x' \\ \vu y' \\ \vu z'}
  = \mf{ccc}{
     \cos\phi & \sin\phi & 0 \\
    -\sin\phi & \cos\phi & 0 \\
     0 & 0 & 1
  }
  \mf{c}{\vu x \\ \vu y \\ \vu z}.
\]
In three dimensions however one can do more than just to rotate the~$x$
and~$y$ axes about the~$z$.  One can reorient the three axes generally
as follows.

\subsection{The Tait-Bryan rotations}
\label{vector:210.20}
\index{rotation!Tait-Bryan or Cardan}
\index{Tait-Bryan rotations}
\index{Cardan, Girolamo
  (also known as Cardano or Cardanus, 1501--1576)}
\index{Cardan rotations}
\index{Tait, Peter Guthrie (1831--1901)}
\index{Bryan, George Hartley (1864--1928)}

\index{yaw}
\index{pitch}
\index{roll}
With a \emph{yaw} and a \emph{pitch} to
point the~$x$ axis in the desired direction plus a \emph{roll} to
position the~$y$ and~$z$ axes as desired about the new~$x$ axis,%
\footnote{
  The English maritime verbs \emph{to yaw, to pitch} and \emph{to roll}
  describe the rotational motion of a vessel at sea.  For a vessel to
  yaw is for her to rotate about her vertical axis, so her bow (her
  forwardmost part) yaws from side to side.  For a vessel to pitch is
  for her to rotate about her ``beam axis,'' so her bow pitches up and
  down.  For a vessel to roll is for her to rotate about her ``fore-aft
  axis'' such that she rocks or lists (leans) without changing the
  direction she points~\cite[``Glossary of nautical terms,'' 23:00, 20
  May 2008]{wikip}\@.  In the Tait-Bryan rotations as explained in
  this book, to yaw is to rotate about the~$z$ axis, to pitch about
  the~$y$, and to roll about the~$x$~\cite{Jones/Fjeld}.  In the Euler
  rotations as explained in this book later in the present section,
  however, the axes are assigned to the vessel differently such that to
  yaw is to rotate about the~$x$ axis, to pitch about the~$y$, and to
  roll about the~$z$.  This implies that the Tait-Bryan vessel points
  $x$-ward whereas the Euler vessel points $z$-ward.
  The reason to shift perspective so is to maintain the semantics of the
  symbols~$\theta$ and~$\phi$ (though not~$\psi$) according to
  Fig.~\ref{vector:fig-sphere}.

  If this footnote seems confusing, then read~(\ref{vector:210:20})
  and~(\ref{vector:210:40}) which are correct.
}
one can reorient the three axes generally:
\bq{vector:210:20}
  \mf{c}{\vu x' \\ \vu y' \\ \vu z'} =
  \mf{ccc}{
     1 & 0 & 0 \\
     0 &  \cos\psi & \sin\psi \\
     0 & -\sin\psi & \cos\psi
  }
  \mf{ccc}{
     \cos\theta & 0 & -\sin\theta \\
     0 & 1 & 0 \\
     \sin\theta & 0 &  \cos\theta
  }
  \mf{ccc}{
     \cos\phi & \sin\phi & 0 \\
    -\sin\phi & \cos\phi & 0 \\
     0 & 0 & 1
  }
  \mf{c}{\vu x \\ \vu y \\ \vu z};
\eq
or, inverting per~(\ref{trig:240:12}),
\bq{vector:210:25}
  \mf{c}{\vu x \\ \vu y \\ \vu z} =
  \mf{ccc}{
     \cos\phi & -\sin\phi & 0 \\
     \sin\phi &  \cos\phi & 0 \\
     0 & 0 & 1
  }
  \mf{ccc}{
     \cos\theta & 0 & \sin\theta \\
     0 & 1 & 0 \\
    -\sin\theta & 0 & \cos\theta
  }
  \mf{ccc}{
     1 & 0 & 0 \\
     0 & \cos\psi & -\sin\psi \\
     0 & \sin\psi &  \cos\psi
  }
  \mf{c}{\vu x' \\ \vu y' \\ \vu z'}.
\eq
These are called the \emph{Tait-Bryan rotations,} or alternately the
\emph{Cardan rotations.}%
\footnote{%
  The literature seems to agree on no standard order among the three
  Tait-Bryan rotations; and, though the rotational angles are usually
  named~$\phi$, $\theta$ and~$\psi$, which angle gets which name
  admittedly depends on the author.  If unsure, prefer the names given
  here.%
}$\mbox{}^{,}$%
\footnote{%
  \cite{CERN-Alice-delta}%
}

Notice in~(\ref{vector:210:20}) and~(\ref{vector:210:25}) that the
transpose (though curiously not the adjoint) of each $3 \times 3$
Tait-Bryan factor is also its inverse.

In concept, the Tait-Bryan equations~(\ref{vector:210:20})
and~(\ref{vector:210:25}) say nearly all one needs to say about
reorienting axes in three dimensions; but, still, the equations can
confuse the uninitiated.  Consider a vector
\bq{vector:210:v}
  \ve v = \vu x x + \vu y y + \vu z z.
\eq
It is not the vector one reorients but rather the axes used to describe
the vector.  Envisioning the axes as in Fig.~\ref{vector:fig-sphere}
with the~$z$ axis upward, one first yaws the~$x$ axis through an
angle~$\phi$ toward the~$y$ then pitches it downward through an
angle~$\theta$ away from the~$z$.  Finally, one rolls the~$y$ and~$z$
axes through an angle~$\psi$ about the new~$x$, all the while
maintaining the three axes rigidly at right angles to one another.
These three Tait-Bryan rotations can orient axes any way.  Yet, even
once one has clearly visualized the Tait-Bryan sequence, the prospect of
applying~(\ref{vector:210:25}) (which inversely represents the sequence)
to~(\ref{vector:210:v}) can still seem daunting until one rewrites the
latter equation in the form
\bq{vector:210:28}
  \ve v = \left[
    \br{ccc}
      \vu x & \vu y & \vu z
    \er
  \right]
  \left[
    \br{c}
      x\\y\\z
    \er
  \right],
\eq
after which the application is straightforward.  There results
\[
  \ve v' = \vu x' x' + \vu y' y' + \vu z' z',
\]
where
\bq{vector:210:30}
  \mf{c}{x' \\ y' \\ z'} \equiv
  \mf{ccc}{
     1 & 0 & 0 \\
     0 &  \cos\psi & \sin\psi \\
     0 & -\sin\psi & \cos\psi
  }
  \mf{ccc}{
    \cos\theta & 0 & -\sin\theta \\
     0 & 1 & 0 \\
    \sin\theta & 0 &  \cos\theta
  }
  \mf{ccc}{
      \cos\phi & \sin\phi & 0 \\
     -\sin\phi & \cos\phi & 0 \\
     0 & 0 & 1
  }
  \mf{c}{x \\ y \\ z},
\eq
and where Table~\ref{trig:277:20} converts to cylindrical or spherical
coordinates if and as desired.  Since~(\ref{vector:210:30})
resembles~(\ref{vector:210:20}), it comes as no surprise that its
inverse,
\bq{vector:210:35}
  \mf{c}{x \\ y \\ z} =
  \mf{ccc}{
     \cos\phi & -\sin\phi & 0 \\
     \sin\phi &  \cos\phi & 0 \\
     0 & 0 & 1
  }
  \mf{ccc}{
     \cos\theta & 0 & \sin\theta \\
     0 & 1 & 0 \\
    -\sin\theta & 0 & \cos\theta
  }
  \mf{ccc}{
     1 & 0 & 0 \\
     0 & \cos\psi & -\sin\psi \\
     0 & \sin\psi &  \cos\psi
  }
  \mf{c}{x' \\ y' \\ z'},
\eq
resembles~(\ref{vector:210:25}).

\subsection{The Euler rotations}
\label{vector:210.25}
\index{rotation!Euler}
\index{Euler rotations}
\index{Euler, Leonhard (1707--1783)}

A useful alternative to the Tait-Bryan rotations are the \emph{Euler
rotations,} which view the problem of reorientation from the perspective
of the~$z$ axis rather than of the~$x$.  The Euler rotations consist of
a roll and a pitch followed by another roll, without any explicit yaw:%
\footnote{
  As for the Tait-Bryan, for the Euler also the literature agrees on no
  standard sequence.  What one author calls a pitch, another might call
  a yaw, and some prefer to roll twice about the~$x$ axis rather than
  the~$z$.  What makes a reorientation an Euler rather than a Tait-Bryan
  is that the Euler rolls twice.
}
\bq{vector:210:40}
  \mf{c}{\vu x' \\ \vu y' \\ \vu z'} =
  \mf{ccc}{
     \cos\psi & \sin\psi & 0 \\
    -\sin\psi & \cos\psi & 0 \\
     0 & 0 & 1
  }
  \mf{ccc}{
     \cos\theta & 0 & -\sin\theta \\
     0 & 1 & 0 \\
     \sin\theta & 0 &  \cos\theta
  }
  \mf{ccc}{
     \cos\phi & \sin\phi & 0 \\
    -\sin\phi & \cos\phi & 0 \\
     0 & 0 & 1
  }
  \mf{c}{\vu x \\ \vu y \\ \vu z};
\eq
and inversely
\bq{vector:210:45}
  \mf{c}{\vu x \\ \vu y \\ \vu z} =
  \mf{ccc}{
     \cos\phi & -\sin\phi & 0 \\
     \sin\phi &  \cos\phi & 0 \\
     0 & 0 & 1
  }
  \mf{ccc}{
     \cos\theta & 0 & \sin\theta \\
     0 & 1 & 0 \\
    -\sin\theta & 0 & \cos\theta
  }
  \mf{ccc}{
     \cos\psi & -\sin\psi & 0 \\
     \sin\psi &  \cos\psi & 0 \\
     0 & 0 & 1
  }
  \mf{c}{\vu x' \\ \vu y' \\ \vu z'}.
\eq
Whereas the Tait-Bryan point the~$x$ axis first, the Euler tactic is to
point first the~$z$.

So, that's it.  One can reorient three axes arbitrarily by rotating them
in pairs about the~$z$, $y$ and~$x$ or the~$z$, $y$ and~$z$ axes in
sequence---or, generalizing, in pairs about any of the three axes so
long as the axis of the middle rotation differs from the axes
(Tait-Bryan) or axis (Euler) of the first and last.  A firmer grasp of
the reorientation of axes in three dimensions comes with practice, but
those are the essentials of it.

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

\section{Multiplication}
\label{vector:220}
\index{vector!multiplication of}
\index{multiplication!of a vector}
\index{product!of vectors}

\index{multiplication!of a vector by a scalar}
\index{scalar multiplication!of a vector}
\index{vector!scalar multiplication of}
\index{product!of a vector and a scalar}
One can multiply a vector in any of three ways.  The first, scalar
multiplication, is trivial: if a vector~$\ve v$ is as defined
by~(\ref{vector:210:v}), then
\bq{vector:220:02}
  \psi \ve v = \vu x \psi x
    + \vu y \psi y + \vu z \psi z.
\eq
Such scalar multiplication evidently scales a vector's length without
diverting its direction.  The other two forms of vector multiplication
involve multiplying a vector by another vector and are the subjects of
the two subsections that follow.

\subsection{The dot product}
\label{vector:220.20}
\index{dot product}
\index{vector!dot or inner product of two}

\index{cosine}
\index{angle}
\index{invariance under the reorientation of axes}
\index{reorientation!invariance under}
\index{axes!invariance under the reorientation of}
\index{orthogonal vector}
\index{vector!orthogonal}
\index{vector!angle between two}
We first met the dot product in \S~\ref{mtxinv:445}.  It works similarly
for the geometrical vectors of this chapter
as for the matrix vectors of chapter~\ref{mtxinv}:
\bq{vector:dot1}
  \ve v_1 \cdot \ve v_2 = x_1 x_2 + y_1 y_2 + z_1 z_2,
\eq
which, if the vectors~$\ve v_1$ and~$\ve v_2$ are real, is the product
of the two vectors to the extent to which they run in the same
direction.  It is the product to the extent to which the vectors run in
the same direction because one can reorient axes to point~$\vu x'$ in
the direction of~$\ve v_1$, whereupon $\ve v_1 \cdot \ve v_2 = x_1'x_2'$
since~$y_1'$ and~$z_1'$ have vanished.

Naturally, to be valid, the
dot product must not vary under a reorientation of axes; and indeed if
we write~(\ref{vector:dot1}) in matrix notation,
\bq{vector:dotm}
  \ve v_1 \cdot \ve v_2 = \left[
    \br{ccc}
      x_1 & y_1 & z_1
    \er
  \right]
  \left[
    \br{c}
      x_2 \\ y_2 \\ z_2
    \er
  \right],
\eq
and then expand each of the two factors on the right according
to~(\ref{vector:210:35}), we see that the dot product does not in fact
vary.
As in~(\ref{mtxinv:445:35}) of \S~\ref{mtxinv:445}, here
too the relationship
\bq{vector:220:23}
  \begin{split}
    \ve v_1^{*} \cdot \ve v_2 &= v_1^{*} v_2 \cos \theta, \\
    \vu v_1^{*} \cdot \vu v_2 &= \cos \theta,
  \end{split}
\eq
gives the angle~$\theta$ between two vectors according
Fig.~\ref{trig:226:f1}'s cosine if the vectors are real, by definition
hereby if complex.  Consequently, the two vectors are mutually
orthogonal---that is, the vectors run at right angles $\theta = 2\pi/4$
to one another---if and only if
\[
  \ve v_1^{*} \cdot \ve v_2 = 0.
\]

\index{commutivity!of the dot product}
\index{dot product!commutivity of}
That the dot product is commutative,
\bq{vector:220:26}
  \ve v_2 \cdot \ve v_1 = \ve v_1 \cdot \ve v_2,
\eq
is obvious from~(\ref{vector:dot1}).
Fig.~\ref{vector:220:fig-dot} illustrates the dot product.
\begin{figure}
  \caption{The dot product.}
  \label{vector:220:fig-dot}
  \bc
    \nc\fxa{-0.7} \nc\fxb{6.1}
    \nc\fya{-0.8} \nc\fyb{2.0}
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      {
        \small
        \psset{dimen=middle}
        \psline[linewidth=2.0pt]{<->}(4,0)(0,0)(2.7713,1.6)
        {
          \psset{linewidth=0.5pt}
          \psarc{->}(0,0){0.8}{0}{30}
          \rput(1.00,0.27){$\theta$}
          \rput(2.7713,0){
            \nc\xxa{0.3}
            \psline[linestyle=dashed](0,1.6)(0,0)
            \psline(-\xxa,0)(-\xxa,\xxa)(0,\xxa)
          }
          \rput(1.3256,-0.25){
            \nc\xxa{1.4457}
            \nc\xxb{0.1}
            \nc\xxc{0.65}
            \psline{->}( \xxc,0)( \xxa,0)
            \psline{->}(-\xxc,0)(-\xxa,0)
            \psline( \xxa,\xxb)( \xxa,-\xxb)
            \psline(-\xxa,\xxb)(-\xxa,-\xxb)
            \rput[c](0,0){$b\cos\theta$}
          }
        }
        \rput(4.25,0.00){$\ve a$}
        \rput(1.45,1.20){$\ve b$}
        \rput[l](3.50,0.90){$\ve a \cdot \ve b = ab \cos \theta$}
      }
    \end{pspicture}
  \ec
\end{figure}

\subsection{The cross product}
\label{vector:220.30}
\index{cross product}

\index{mnemonic}
The dot product of two vectors according to \S~\ref{vector:220.20} is a
scalar.  One can also multiply two vectors to obtain a vector, however,
and it is often useful to do so.  As the dot product is the product of
two vectors to the extent to which they run in the same direction, the
\emph{cross product} is the product of two vectors to the extent to which
they run in different directions.  Unlike the dot product the cross
product is a vector, defined in rectangular coordinates as
\bqa
  \ve v_1 \times \ve v_2 &=& \left|
      \br{ccc}
        \vu x & \vu y & \vu z \\
        x_1 & y_1 & z_1 \\
        x_2 & y_2 & z_2
      \er
    \right|
    \label{vector:cross1}
    \\&\equiv&
      % The following throws the above off-center, which visually does
      % not look right, but available alternatives seem worse.
      \vu x (y_1z_2 - z_1y_2)
      + \vu y (z_1x_2 - x_1z_2)
      + \vu z (x_1y_2 - y_1x_2),
    \xn
\eqa
where the~$\left|\cdot\right|$ notation is a mnemonic (actually a
pleasant old determinant notation \S~\ref{eigen:310} could have but did
not happen to use) whose semantics are as shown.

\index{invariance under the reorientation of axes}
\index{reorientation!invariance under}
\index{axes!invariance under the reorientation of}
As the dot product, the cross product too is invariant under
reorientation.  One could demonstrate this fact by multiplying
out~(\ref{vector:210:25}) and~(\ref{vector:210:35}) then substituting
the results into~(\ref{vector:cross1}): a lengthy, unpleasant exercise.
Fortunately, it is also an unnecessary exercise, forasmuch as an
arbitrary reorientation consists of three rotations
(\S~\ref{vector:210}) in sequence it suffices
merely that rotation about one axis not alter the cross product.  One
proves the proposition in the latter form by setting any two of~$\phi$,
$\theta$ and~$\psi$ to zero before multiplying out and substituting.
For instance, setting~$\theta$ and~$\psi$ to zero,
\bqb
  \ve v_1 \times \ve v_2
    &=&
    \left|
      \br{ccc}
        \vu x & \vu y & \vu z \\
        x_1 & y_1 & z_1 \\
        x_2 & y_2 & z_2
      \er
    \right|
    \\&=&
    \left|
      \br{ccc}
        \vu x'\cos\phi-\vu y'\sin\phi & \vu x'\sin\phi+\vu y'\cos\phi & \vu z \\
        x_1 & y_1 & z_1 \\
        x_2 & y_2 & z_2
      \er
    \right|
    \\&=&
      \vu
      x'\left[y_1z_2\cos\phi-z_1y_2\cos\phi+z_1x_2\sin\phi-x_1z_2\sin\phi\right]
      \\&&\quad\mbox{}
      + \vu y'\left[-y_1z_2\sin\phi+z_1y_2\sin\phi+z_1x_2\cos\phi-x_1z_2\cos\phi\right]
      \\&&\quad\mbox{}
      + \vu z\left[x_1y_2-y_1x_2\right]
    \\&=&
      \vu
      x'\left[(-x_1\sin\phi+y_1\cos\phi)z_2-z_1(-x_2\sin\phi+y_2\cos\phi)\right]
      \\&&\quad\mbox{}
      + \vu y'\left[z_1(y_2\sin\phi+x_2\cos\phi)-(y_1\sin\phi+x_1\cos\phi)z_2\right]
      \\&&\quad\mbox{}
      + \vu z\left[x_1y_2-y_1x_2\right]
    \\&=&
      \vu x'\left[y_1'z_2-z_1y_2'\right] + \vu y'\left[z_1x_2'-z_2x_1'\right]
      \\&&\quad\mbox{}
      + \vu z[(x_1'\cos\phi-y_1'\sin\phi)(x_2'\sin\phi+y_2'\cos\phi)
      \\&&\qquad\qquad\mbox{}
      -(x_1'\sin\phi+y_1'\cos\phi)(x_2'\cos\phi-y_2'\sin\phi)].
\eqb
Since according to Pythagoras in Table~\ref{trig:228:table} $\cos^2\phi
+ \sin^2\phi = 1$,
\bqb
  \ve v_1 \times \ve v_2
    &=&
      \vu x'\left[y_1'z_2-z_1y_2'\right] + \vu y'\left[z_1x_2'-z_2x_1'\right]
      + \vu z[x_1'y_2'-y_1'x_2']
    \\&=&
    \left|
      \br{ccc}
        \vu x' & \vu y' & \vu z \\
        x_1' & y_1' & z_1 \\
        x_2' & y_2' & z_2
      \er
    \right|
\eqb
as was to be demonstrated.

Several facets of the cross product draw attention to themselves.
\bi
  \item
    \index{cyclic progression of coordinates}
    \index{progression of coordinates, cyclic}
    \index{coordinates!cyclic progression of}
    \index{parity}
    \index{right-hand rule}
    The cyclic progression
    \bq{vector:220:36}
      \cdots \ra x \ra y \ra z \ra x \ra y \ra z \ra x \ra y \ra \cdots
    \eq
    of~(\ref{vector:cross1}) arises again and again in vector analysis.
    Where the progression is honored, as in~$\vu z x_1y_2$, the
    associated term bears a~$+$ sign, otherwise a~$-$ sign, due to
    \S~\ref{matrix:322}'s parity principle and the right-hand rule.
  \item
    \index{commutivity!noncommutivity of the cross product}
    \index{noncommutivity!of the cross product}
    \index{cross product!noncommutivity of}
    The cross product is not commutative.  In fact,
    \bq{vector:220:32}
      \ve v_2 \times \ve v_1 = -\ve v_1 \times \ve v_2,
    \eq
    which is a direct consequence of the previous point regarding
    parity, or which can be seen more prosaically
    in~(\ref{vector:cross1}) by swapping the places of~$\ve v_1$
    and~$\ve v_2$.
  \item
    \index{associativity!nonassociativity of the cross product}
    \index{nonassociativity!of the cross product}
    \index{cross product!nonassociativity of}
    The cross product is not associative.  That is,
    \[
      (\ve v_1 \times \ve v_2) \times \ve v_3 \neq \ve v_1 \times (\ve v_2 \times \ve v_3),
    \]
    as is proved by a suitable counterexample like
    $\ve v_1 = \ve v_2 = \vu x$, $\ve v_3 = \vu y$.
  \item
    \index{cross product!perpendicularity of}
    The cross product runs perpendicularly to each of its two factors if
    the vectors involved are real.  That is,
    \bq{vector:220:35}
      \ve v_1 \cdot ( \ve v_1 \times \ve v_2 )
      = 0 =
      \ve v_2 \cdot ( \ve v_1 \times \ve v_2 ),
    \eq
    as is seen by substituting~(\ref{vector:cross1})
    into~(\ref{vector:dot1}) with an appropriate change of variables and
    simplifying.
  \item
    \index{space!three-dimensional}
    \index{three-dimensional space}
    \index{space!two-dimensional}
    \index{two-dimensional space}
    Unlike the dot product, the cross product is closely tied to
    three-dimensional space.  Two-dimensional space (a plane) can have a
    cross product so long as one does not mind that the product points off
    into the third dimension, but to speak of a cross product in
    four-dimensional space would require arcane definitions and would
    otherwise make little sense.  Fortunately, the physical world is
    three-dimensional (or, at least, the space in which we model all but
    a few, exotic physical phenomena is three-dimensional), so the cross
    product's limitation as defined here to three dimensions will seldom
    disturb us.
  \item
    \index{sine}
    \index{angle}
    \index{electromagnetic power}
    \index{power!electromagnetic}
    Section~\ref{vector:220.20}
    %Equation~(\ref{vector:220:23})
    has related the cosine of the angle
    between vectors to the dot product.  One can similarly relate the
    angle's sine to the cross product if the vectors involved are real,
    as
    \bq{vector:220:33}
      \begin{split}
        |\ve v_1 \times \ve v_2| &= v_1v_2 \sin \theta, \\
        |\vu v_1 \times \vu v_2| &= \sin \theta,
      \end{split}
    \eq
    demonstrated by reorienting axes such that $\vu v_1 = \vu x'$,
    that~$\vu v_2$ has no component in the~$\vu z'$ direction, and that~$\vu
    v_2$ has only a nonnegative component in the~$\vu y'$ direction; by
    remembering that reorientation cannot alter a cross product; and
    finally by applying~(\ref{vector:cross1}) and comparing the result
    against Fig.~\ref{trig:226:f1}'s sine.  (If the vectors involved are
    complex then nothing prevents the operation $|\ve v_1^{*} \times \ve
    v_2|$ by analogy with eqn.~\ref{vector:220:23}---in fact the
    operation $\ve v_1^{*} \times \ve v_2$ without the magnitude sign is
    used routinely to calculate electromagnetic power flow%
    \footnote{\cite[eqn.~1-51]{Harrington}}---%
    but each of the cross product's three rectangular components has its
    own complex phase which the magnitude operation flattens, so the
    result's relationship to the sine of an angle is not immediately
    clear.)
\ei
Fig.~\ref{vector:220:fig-cross} illustrates the cross product.
\begin{figure}
  \caption{The cross product.}
  \label{vector:220:fig-cross}
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    \nc\xxphir{0.45} % Angle dimension radius.
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          % function res = coords(r,phi,a)
          %   p = phi/180*pi;
          %   res(1)=0;
          %   res(2)=r;
          %   res(3)=r*cos(p);
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        \rput(2.25,0.85){$\ve b$}
        \rput[l](0.00,2.40){
          \setlength\arraycolsep{0.15em}
          $
            \br{rcl}
              \ve c &=& \ve a \times \ve b \\
                &=& \vu c ab \sin \theta
            \er
          $
        }
      }
    \end{pspicture}
  \ec
\end{figure}

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

\section{Orthogonal bases}
\label{vector:230}
\index{orthogonal basis}
\index{basis!orthogonal}

A vector exists independently of the components by which one expresses
it, for, whether $\ve q = \vu x x + \vu y y + \vu z z$ or $\ve q = \vu
x' x' + \vu y' y' + \vu z' z'$, it remains the same vector~$\ve q$.
However, where a model involves a circle, a cylinder or a sphere, where
a model involves a contour or a curved surface of some kind, to
choose~$\vu x'$, $\vu y'$ and~$\vu z'$ wisely can immensely simplify the
model's analysis.  Normally one requires that~$\vu x'$, $\vu y'$
and~$\vu z'$ each retain unit length, run perpendiclarly to one another,
and obey the right-hand rule (\S~\ref{trig:230}), but otherwise any~$\vu
x'$, $\vu y'$ and~$\vu z'$ can serve.
Moreover, various parts of a model can specify various~$\vu x'$, $\vu
y'$ and~$\vu z'$, or various substitutes therefor, under various
conditons.
%for nothing requires the three to be uniform.

\index{winding road}
\index{road!winding}
\index{automobile}
\index{speed}
\index{velocity}
\index{wind}
\index{headwind}
\index{crosswind}
Recalling the constants and variables of \S~\ref{alggeo:231}, such a
concept is flexible enough to confuse the uninitiated severely and soon.
As in \S~\ref{alggeo:231}, here too an example affords perspective.
Imagine driving your automobile down a winding road, where~$q$
represented your speed%
\footnote{\label{vector:230:fn1}%
  Conventionally one would prefer the letter~$v$ to represent speed,
  with velocity as~$\ve v$ which in the present example would happen to
  be $\ve v = \wu\ell v$.  However, this section will require the
  letter~$v$ for an unrelated purpose.
}
and~$\wu\ell$ represented the direction the road ran, not generally, but
just at the spot along the road at which your automobile momentarily
happened to be.  That your velocity were~$\wu\ell q$ meant that you kept
skilfully to your lane; on the other hand, that your velocity were
$(\wu\ell \cos\psi + \vu v \sin\psi)q$---where~$\vu v$, at right
angles to~$\wu\ell$, represented the direction right-to-left across the
road---would have you drifting out of your lane at an angle~$\psi$.  A
headwind had velocity~$-\wu\ell q_{\mr{wind}}$; a crosswind,~$\pm\vu v
q_{\mr{wind}}$.  A car a mile ahead of you had velocity $\wu\ell_2q_2=(\wu\ell
\cos\beta + \vu v \sin\beta)q_2$, where~$\beta$ represented the difference
(assuming that the other driver kept skilfully to his own lane) between
the road's direction a mile ahead and its direction at your spot.
For all these purposes the unit vector~$\wu\ell$ would remain constant.
However, fifteen seconds later, after you had rounded a bend in the
road, the symbols~$\wu\ell$ and~$\vu v$ would by definition represent
different vectors than before, with respect to which one would express
your new velocity as~$\wu\ell q$ but would no longer express the
headwind's velocity as~$-\wu\ell q_{\mr{wind}}$ because, since the road had
turned while the wind had not, the wind would no longer be a headwind.
And this is where confusion can arise: your own velocity had changed
while the expression representing it had not; whereas the wind's
velocity had not changed while the expression representing \emph{it}
had.  This is not because~$\wu\ell$ differs from place to place at a
given moment, for like any other vector the vector~$\wu\ell$ (as defined
in this particular example) is the same vector everywhere.  Rather, it
is because~$\wu\ell$ is defined relative to the road at your
automobile's location, which location changes as you drive.

\index{basis!constant}
\index{basis!variable}
\index{orthogonal basis!variable}
\index{model}
If a third unit vector~$\vu w$ were defined, perpendicular both
to~$\wu\ell$ and to~$\vu v$ such that $[\wu\ell\;\vu v\;\vu w]$ obeyed
the right-hand rule, then the three together would constitute an
\emph{orthogonal basis.}  Any three real,%
\footnote{
  A complex orthogonal basis is also theoretically possible but is
  normally unnecessary in geometrical applications and involves
  subtleties in the cross product.  This chapter, which specifically
  concerns three-dimensional geometrical vectors rather than the
  general, $n$-dimensional vectors of chapter~\ref{matrix}, is content to
  consider real bases only.  Note that one can express a complex vector
  in a real basis.
}
right-handedly mutually perpendicular unit vectors
$[\vu x'\;\vu y'\;\vu z']$ in three dimensions, whether constant or
variable, for which
\bq{vector:230:10}
  \index{right-hand rule}
  \setlength\arraycolsep{0.30\arraycolsep}
  \br{rclcrclcrcl}
    \vu y' \cdot \vu z' &=& 0, &\ \ \ \ & \vu y' \times \vu z' &=& \vu x', &\ \ \ \ & \Im(\vu x') &=& 0,\\
    \vu z' \cdot \vu x' &=& 0, && \vu z' \times \vu x' &=& \vu y', && \Im(\vu y') &=& 0,\\
    \vu x' \cdot \vu y' &=& 0, && \vu x' \times \vu y' &=& \vu z', && \Im(\vu z') &=& 0,
  \er
\eq
constitutes such an orthogonal basis, from which other vectors can be
built.  The geometries of some models suggest no particular basis, when
one usually just uses a constant $[\vu x\;\vu y\;\vu z]$.  The
geometries of other models however do suggest a particular basis, often
a variable one.
\bi
  \item
    \index{contour}
    Where the model features a contour like the example's winding
    road,
    an $[\wu\ell\;\vu v\;\vu w]$ basis (or
    a $[\vu u\;\vu v\;\wu\ell]$ basis or even
    a $[\vu u\;\wu\ell\;\vu w]$ basis)
    can be used, where~$\wu\ell$ locally follows the contour.  The
    variable unit vectors~$\vu v$ and~$\vu w$ (or~$\vu u$ and~$\vu v$,
    etc.)\ can be defined in any convenient way so long as they remain
    perpendicular to one another and to~$\wu\ell$---such that $(\vu z
    \times \wu\ell) \cdot \vu w = 0$ for instance (that is, such that~$\vu w$
    lay in the plane of~$\vu z$ and~$\wu\ell$)---but if the geometry
    suggests a particular~$\vu v$ or~$\vu w$ (or~$\vu u$), like the
    direction right-to-left across the example's road, then that~$\vu v$
    or~$\vu w$ should probably be used.  The letter~$\ell$ here stands for
    ``longitudinal.''%
    \footnote{The assertion wants a citation, which the author lacks.}
  \item
    \index{normal unit vector}
    \index{perpendicular unit vector}
    \index{unit vector!normal or perpendicular}
    \index{surface!orientation of}
    \index{surface normal}
    \index{unit normal}
    \index{sea!wavy surface of}
    \index{wavy sea}
    Where the model features a curved surface like the surface of a
    wavy sea,%
    \footnote{\cite{PM:1964}}
    a $[\vu u\;\vu v\;\vu n]$ basis (or
    a $[\vu u\;\vu n\;\vu w]$ basis, etc.)\ %
    can be used, where~$\vu n$ points locally perpendicularly to the
    surface.  The letter~$n$ here stands for ``normal,'' a synonym for
    ``perpendicular.''  Observe, incidentally but significantly, that
    such a unit normal~$\vu n$ tells one everything one needs to know
    about its surface's local orientation.
  \item
    Combining the last two, where the model features a contour along
    a curved surface,
    an $[\wu\ell\;\vu v\;\vu n]$ basis
    can be used.  One need not worry about choosing a direction for~$\vu
    v$ in this case since necessarily $\vu v = \vu n \times \wu\ell$.
  \item
    \index{circle}
    \index{cylinder}
    \index{cylindrical basis}
    \index{basis!cylindrical}
    \index{axis!of a cylinder or circle}
    \index{azimuth}
    \index{up}
    \index{south}
    \index{east}
    Where the model features a circle or cylinder,
    a $[\wu\rho\;\wu\phi\;\vu z]$ basis
    can be used, where~$\vu z$ is constant and runs along the cylinder's
    axis (or perpendicularly through the circle's center),~$\wu\rho$ is
    variable and points locally away from the axis, and~$\wu\phi$ is
    variable and runs locally along the circle's perimeter in the
    direction of increasing azimuth~$\phi$.  Refer to \S~\ref{trig:277}
    and Fig.~\ref{vector:230:fig-cylindrical-basis}.
    \begin{figure}
      \caption[The cylindrical basis.]{%
        The cylindrical basis.  (The conventional symbols~\vectortoward\
        and~\vectoraway\ respectively represent vectors pointing out of
        the page toward the reader and into the page away from the
        reader.  Thus, this figure shows the constant basis vector~$\vu
        z$ pointing out of the page toward the reader.  The dot in the
        middle of the~\vectortoward\ is supposed to look like the tip of
        an arrowhead.)%
      }
      \label{vector:230:fig-cylindrical-basis}
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    \end{figure}
  \item
    \index{sphere}
    \index{spherical basis}
    \index{basis!spherical}
    \index{azimuth}
    \index{elevation}
    Where the model features a sphere,
    an $[\vu r\;\wu\theta\;\wu\phi]$ basis
    can be used, where~$\vu r$ is variable and points locally away from
    the sphere's center,~$\wu\theta$ is variable and runs locally
    tangentially to the sphere's surface in the direction of increasing
    elevation~$\theta$ (that is, though not usually in the~$-\vu z$
    direction itself, as nearly as possible to the~$-\vu z$ direction
    without departing from the sphere's surface), and~$\wu\phi$ is
    variable and runs locally tangentially to the sphere's surface in
    the direction of increasing azimuth~$\phi$ (that is, along the
    sphere's surface perpendicularly to~$\vu z$).  Standing on the
    earth's surface, with the earth as the sphere,~$\vu r$ would be
    up,~$\wu\theta$ south, and~$\wu\phi$ east.  Refer to
    \S~\ref{trig:277} and Fig.~\ref{vector:230:fig-spherical-basis}.
    \begin{figure}
      \caption[The spherical basis.]{%
        The spherical basis (see also Fig.~\ref{vector:fig-sphere}).%
      }
      \label{vector:230:fig-spherical-basis}
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    \end{figure}
  \item
    \index{secondary circle}
    \index{secondary cylindrical basis}
    \index{basis!secondary cylindrical}
    \index{circle!secondary}
    Occasionally a model arises with two circles that share a center
    but whose axes stand
    perpendicular to one another.  In such a model one conventionally
    establishes~$\vu z$ as the direction of the principal circle's axis
    but then is left with~$\vu x$ or~$\vu y$ as the direction of the
    secondary circle's axis, upon which an
    $[\vu x\;\wu\rho^x\;\wu\phi^x]$,
    $[\wu\phi^x\;\vu r\;\wu\theta^x]$,
    $[\wu\phi^y\;\vu y\;\wu\rho^y]$ or
    $[\wu\theta^y\;\wu\phi^y\;\vu r]$ basis
    can be used locally as appropriate.  Refer to \S~\ref{trig:277}.
\ei
Many other orthogonal bases are possible (as in \S~\ref{vector:280}, for
instance) but the foregoing are the most common.  Whether listed here
or not, each orthogonal basis orders its three unit vectors by the
right-hand rule~(\ref{vector:230:10}).

\index{quiz}
\index{azimuth}
Quiz: what does the vector expression $\wu\rho 3 - \wu\phi(1/4) + \vu z
2$ mean?  Wrong answer: it meant the cylindrical coordinates
$(3;-1/4,2)$; or, it meant the position vector $\vu x 3 \cos(-1/4) + \vu
y 3 \sin(-1/4) + \vu z 2$ associated with those coordinates.  Right
answer: the expression means nothing certain in itself but acquires a
definite meaning only when an azimuthal coordinate~$\phi$ is also
supplied, after which the expression indicates the ordinary rectangular
vector $\vu x' 3 - \vu y'(1/4) + \vu z' 2$, where $\vu x'=\wu\rho=\vu
x\cos\phi + \vu y\sin\phi$, $\vu y'=\wu\phi=-\vu x\sin\phi + \vu
y\cos\phi$, and $\vu z'=\vu z$.  But, if this is so---if the
cylindrical basis $[\wu\rho\;\wu\phi\;\vu z]$ is used solely to express
\emph{rectangular} vectors---then why should we name this basis
``cylindrical''?  Answer: only because cylindrical coordinates (supplied
somewhere) determine the actual directions of its basis vectors.  Once
directions are determined, such a basis is used rectangularly
like any other orthogonal basis.

\index{jet engine}
\index{axle}
\index{velocity!local}
\index{vector!local}
\index{vector!position}
\index{position vector}
This can seem confusing until one has grasped what the so-called
nonrectangular bases are for.  Consider the problem of air flow in a jet
engine.  It may suit such a problem that instantaneous local air
velocity within the engine cylinder be expressed in cylindrical
coordinates, with the~$z$ axis oriented along the engine's axle; but
this does not mean that the air flow within the engine cylinder were
everywhere $\vu z$-directed.  On the contrary, a local air velocity of
$\ve q = [-\wu\rho 5.0 + \wu\phi 30.0 - \vu z 250.0]$ m/s would have air
moving through the point in question at 250.0~m/s aftward along the
axle, 5.0~m/s inward toward the axle and 30.0~m/s circulating about the
engine cylinder.

In this model, it is true that the basis vectors~$\wu\rho$ and~$\wu\phi$
indicate different directions at different positions within the
cylinder, but at a particular position the basis vectors are still used
rectangularly to express~$\ve q$, the instantaneous local air velocity
at that position.  It's just that the ``rectangle'' is rotated locally
to line up with the axle.

Naturally, you cannot make full sense of an air-velocity vector~$\ve q$
unless you have also the coordinates $(\rho;\phi,z)$ of the position
within the engine cylinder at which the air has the velocity the vector
specifies---yet this is when confusion can arise, for besides the
air-velocity vector there is also, separately, a position vector $\ve r
= \vu x \rho\cos\phi + \vu y \rho\sin\phi + \vu z z$.  One may denote
the air-velocity vector as%
\footnote{
  Conventionally, one is much more likely to denote a velocity vector as
  $\ve u(\ve r)$ or $\ve v(\ve r)$, except that the present chapter is
  (as footnote~\ref{vector:230:fn1} has observed) already using the
  letters~$u$ and~$v$ for an unrelated purpose.  To denote position
  as~$\ve r$ however is entirely standard.
}
$\ve q(\ve r)$, a function of position; yet, though the position vector
is as much a vector as the velocity vector is, one nonetheless handles
it differently.  One will not normally express the position vector~$\ve
r$ in the cylindrical basis.

\index{tautology}
\index{coordinates}
\index{basis}
It would make little sense to try to express the position vector~$\ve r$
in the cylindrical basis because the position vector is the very thing
that \emph{determines} the cylindrical basis.  In the cylindrical basis,
after all, the position vector is necessarily $\ve r = \wu\rho\rho + \vu
z z$ (and consider: in the spherical basis it is the even more cryptic
$\ve r = \vu r r$), and how useful is that, really?  Well, maybe it is
useful in some situations, but for the most part to express the position
vector in the cylindrical basis would be as to say, ``My house is zero
miles away from home.''  Or, ``The time is presently now.''
Such statements may be tautologically true, perhaps, but they
are confusing because they only seem to give information.  The position
vector~$\ve r$ determines the basis, after which one expresses things
other than position, like instantaneous local air velocity~$\ve q$, in
that basis.  In fact, the only basis normally suitable to express a
position vector is a fixed rectangular basis like $[\vu x\;\vu y\;\vu
z]$.  Otherwise, one uses cylindrical \emph{coordinates} $(\rho;\phi,z)$, but
not a cylindrical \emph{basis} $[\wu\rho\;\wu\phi\;\vu z]$, to express a
position~$\ve r$ in a cylindrical geometry.

Maybe the nonrectangular bases were more precisely called ``rectangular
bases of the nonrectangular coordinate systems,'' but those are too many
words and, anyway, that is not how the usage has evolved.
Chapter~\ref{vcalc} will elaborate the story by considering spatial
derivatives of quantities like air velocity, when one must take the
variation in~$\wu\rho$ and~$\wu\phi$ from point to point into account,
but the foregoing is the basic story nevertheless.

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

\section{Notation}
\label{vector:240}
\index{vector!concise notation for}
\index{notation!for the vector, concise}

\index{concision}
\index{prolixity}
The vector notation of \S\S~\ref{vector:210} and~\ref{vector:220} is
correct, familiar and often expedient but sometimes inconveniently
prolix.  This admittedly difficult section augments the notation to
render it much more concise.

\subsection{Components by subscript}
\label{vector:240.10}
\index{components of a vector by subscript}
\index{subscript!indicating the components of a vector by}

\index{dot product!abbrevated notation for}
The notation
\[
  \settowidth\tla{$\wu\phi$}
  \nc\xxa[1]{\makebox[\tla][c]{$#1$} \cdot \ve a}
  \setlength\arraycolsep{0.30\arraycolsep}
  \br{lclclcl}
    a_x &\equiv& \xxa{\vu x}, &\ \ \ \ &
    a_\rho &\equiv& \xxa{\wu\rho}, \\
    a_y &\equiv& \xxa{\vu y}, &&
    a_r &\equiv& \xxa{\vu r}, \\
    a_z &\equiv& \xxa{\vu z}, &&
    a_\theta &\equiv& \xxa{\wu\theta}, \\
    a_n &\equiv& \xxa{\vu n}, &&
    a_\phi &\equiv& \xxa{\wu\phi},
  \er
\]
and so forth abbreviates the indicated dot product.  That is to say, the
notation represents the component of a vector~$\ve a$ in the indicated
direction.  Generically,
\bq{vector:240:10}
  a_\alpha \equiv \wu\alpha \cdot \ve a.
\eq
Applied mathematicians use subscripts for several unrelated or vaguely
related purposes, so the full dot-product notation $\wu\alpha \cdot \ve
a$ is often clearer in print than the abbreviation~$a_\alpha$ is, but
the abbreviation especially helps when several such dot products occur
together in the same expression.

\index{prime mark ($'$)}
\index{$'$}
\index{primed coordinate}
\index{coordinate!primed and unprimed}
\index{clutter, notational}
Since%
\footnote{
  ``Wait!'' comes the objection.  ``I thought that you said that~$a_x$
  meant $\vu x \cdot \ve a$.  Now you claim that it means the~$x$
  component of~$\ve a$?''

  But there is no difference between $\vu x \cdot \ve a$ and the~$x$
  component of~$\ve a$.  The two are one and the same.
}
\[
  \begin{split}
    \vu a &= \vu x a_x + \vu y a_y + \vu z a_z, \\
    \vu b &= \vu x b_x + \vu y b_y + \vu z b_z,
  \end{split}
\]
the abbreviation lends a more amenable notation to the dot and cross
products of~(\ref{vector:dot1}) and~(\ref{vector:cross1}):
\bqa
  \ve a \cdot \ve b &=& a_x b_x + a_y b_y + a_z b_z;
    \label{vector:dot} \\
  \ve a \times \ve b &=& \left|
    \br{ccc}
        \vu x & \vu y & \vu z \\
        a_x & a_y & a_z \\
        b_x & b_y & b_z
      \er
    \right|.
    \label{vector:cross}
\eqa
In fact---because, as we have seen, reorientation of axes cannot alter
the dot and cross products---any orthogonal basis $[\vu x'\;\vu y'\;\vu
z']$ (\S~\ref{vector:230}) can serve here,
so one can write more generally that
\bqa
  \ve a \cdot \ve b &=& a_{x'} b_{x'} + a_{y'} b_{y'} + a_{z'} b_{z'};
    \label{vector:dotp} \\
  \ve a \times \ve b &=& \left|
    \br{ccc}
        \vu x' & \vu y' & \vu z' \\
        a_{x'} & a_{y'} & a_{z'} \\
        b_{x'} & b_{y'} & b_{z'}
      \er
   \right|.
    \label{vector:crossp}
\eqa
Because all those prime marks burden the notation and for professional
mathematical reasons, the general
forms~(\ref{vector:dotp}) and~(\ref{vector:crossp}) are sometimes
rendered
\bqb
  \ve a \cdot \ve b &=& a_1 b_1 + a_2 b_2 + a_3 b_3, \\
  \ve a \times \ve b &=& \left|
    \br{ccc}
        \vu e_1 & \vu e_2 & \vu e_3 \\
        a_1 & a_2 & a_3 \\
        b_1 & b_2 & b_3
      \er
    \right|,
\eqb
but you have to be careful about that in applied usage
because people are not always sure
whether a symbol like~$a_3$ means ``the third component of the
vector~$\ve a$'' (as it does here) or ``the third vector's component in
the~$\vu a$ direction'' (as it would in eqn.~\ref{vector:dot1}).
Typically, applied mathematicians will write in the manner
of~(\ref{vector:dot}) and~(\ref{vector:cross}) with the implied
understanding that they really mean~(\ref{vector:dotp})
and~(\ref{vector:crossp}) but prefer not to burden the notation with
extra little strokes---that is, with the implied understanding that~$x$,
$y$ and~$z$ could just as well be~$\rho$, $\phi$ and~$z$ or the
coordinates of any other orthogonal, right-handed, three-dimensional
basis.

Some pretty powerful confusion can afflict the student regarding the
roles of the cylindrical symbols~$\rho$, $\phi$ and~$z$; or, worse,
of the spherical symbols~$r$, $\theta$ and~$\phi$.  Such confusion
reflects a pardonable but remediable lack of understanding of the
relationship between coordinates like~$\rho$, $\phi$ and~$z$ and their
corresponding unit vectors~$\wu\rho$, $\wu\phi$ and~$\vu z$.
Section~\ref{vector:230} has already written of the matter; but, further to
dispel the confusion, one can now ask the student what the cylindrical
coordinates of the vectors~$\wu\rho$, $\wu\phi$ and~$\vu z$ are.  The
correct answer: $(1;\phi,0)$, $(1;\phi+2\pi/4,0)$ and $(0;0,1)$,
respectively.  Then, to reinforce, one can ask the student which cylindrical
coordinates the variable vectors~$\wu\rho$ and~$\wu\phi$ are functions
of.  The correct answer: both are functions of the coordinate~$\phi$
only ($\vu z$, a constant vector, is not a function of anything).
What the student needs to understand is that, among the cylindrical
coordinates,~$\phi$ is a different kind of thing than~$z$ and~$\rho$
are:
\bi
  \item
    $z$ and~$\rho$ are lengths whereas~$\phi$ is an angle;
  \item
    but~$\wu\rho$, $\wu\phi$ and~$\vu z$ are all the same kind of
    thing, unit vectors;
  \item
    and, separately,~$a_\rho$, $a_\phi$ and~$a_z$ are all the same
    kind of thing, lengths.
\ei
Now to ask a harder question: in the cylindrical basis,
what is the vector representation of $(\rho_1;\phi_1,z_1)$?  The
correct answer: $\wu\rho\rho_1\cos(\phi_1-\phi) +
\wu\phi\rho_1\sin(\phi_1-\phi) + \vu z z_1$.  The student that gives
this answer probably grasps the cylindrical symbols.

If the reader feels that the notation begins to confuse more than it
describes, the writer empathizes but regrets to inform that
the rest of the section, far from granting the reader a comfortable
respite to absorb the elaborated notation as it stands, will not delay
to elaborate the notation yet further!  The confusion however is
subjective.  The trouble with vector work is that one has to learn to
abbreviate or the expressions involved grow repetitive and unreadably
long.  For vectors, the abbreviated notation really is the proper
notation.  Eventually one accepts the need and takes the trouble to
master the conventional vector abbreviation this section presents; and,
indeed, the abbreviation is rather elegant once one has grown used to it.
So, study closely and take heart!\footnote{%
  What to study?  Besides this book, one can study any good,
  introductory undergraduate textbook in fluid dynamics,
  electromagnetics, quantum mechanics or the like.  For
  example,~\cite{Cheng} is not bad.%
}
The notation is not actually as impenetrable as it at first will seem.

\subsection{Einstein's summation convention}
\label{vector:240.20}
\index{Einstein, Albert (1879--1955)}
\index{Einstein's summation convention}
\index{summation convention, Einstein's}
\index{Einstein notation}

\emph{Einstein's summation convention} is this: \emph{that repeated
indices are implicitly summed over.}%
\footnote{\cite{Hopman}}
For instance, where the convention is in force, the equation%
\footnote{
  Some professional mathematicians now write a superscript~$a^i$ in
  certain cases in place of a subscript~$a_i$, where the superscript
  bears some additional semantics~%
  \cite[``Einstein notation,'' 05:36, 10~Feb.\ 2008]{wikip}.
  Scientists and engineers however tend to prefer Einstein's original,
  subscript-only notation.
}
\bq{vector:240:20}
  \ve a \cdot \ve b = a_i b_i
\eq
means that
\[
  \ve a \cdot \ve b = \sum_{i} a_i b_i
\]
or more fully that
\[
  \ve a \cdot \ve b = \!\!\!\sum_{i=x',y',z'}\!\!\! a_i b_i
  = a_{x'} b_{x'} + a_{y'} b_{y'} + a_{z'} b_{z'},
\]
which is~(\ref{vector:dotp}), except that Einstein's
form~(\ref{vector:240:20}) expresses it more succinctly.  Likewise,
\bq{vector:240:22}
  \ve a \times \ve b = \vui ( a_{i+1} b_{i-1} - b_{i+1} a_{i-1} )
\eq
is~(\ref{vector:crossp})---although an experienced applied mathematician
would probably apply the Levi-Civita epsilon of \S~\ref{vector:240.30},
below, to further abbreviate this last equation to the form
of~(\ref{vector:240:32}) before presenting it.

Einstein's summation convention is also called the \emph{Einstein
notation,} a term sometimes taken loosely to include also the Kronecker
delta and Levi-Civita epsilon of \S~\ref{vector:240.30}.

\index{accountant}
What is important to understand about Einstein's summation convention is
that, in and of itself, it brings no new mathematics.  It is rather a
notational convenience.%
\footnote{\cite[``Einstein summation'']{EWW-web}}
It asks a reader to regard a repeated index like the~$i$ in~``$a_ib_i$''
as a dummy index (\S~\ref{alggeo:227}) and thus to read~``$a_ib_i$'' as
``$\sum_i a_ib_i$.''  It does not magically create a summation where
none existed; it just hides the summation sign to keep it from
cluttering the page.  It is the kind of notational trick an accountant
might appreciate.  Under the convention, the summational
operator~$\sum_i$ is implied not written, but the operator is still
there.  Admittedly confusing on first encounter, the convention's
utility and charm are felt after only a little practice.

Incidentally, nothing requires you to invoke Einstein's summation convention
everywhere and for all purposes.  You can waive the convention, writing
the summation symbol out explicitly whenever you like.%
\footnote{\cite{physics321}}
In contexts outside vector analysis,
to invoke the convention at all may make little sense.
Nevertheless, you should indeed learn the convention---if only because
you must learn it to understand the rest of this chapter---but once
having learned it you should naturally use it only where it actually
serves to clarify.  Fortunately, in vector work, it often does just
that.

\index{quiz}
Quiz:%
\footnote{\cite{Hopman}}
if~$\delta_{ij}$ is the Kronecker delta of \S~\ref{matrix:150}, then
what does the symbol~$\delta_{ii}$ represent where Einstein's summation
convention is in force?

\subsection{The Kronecker delta and the Levi-Civita epsilon}
\label{vector:240.30}
\index{Kronecker delta}
\index{delta, Kronecker}
\index{Kronecker, Leopold (1823--1891)}
\index{Levi-Civita, Tullio (1873--1941)}
\index{Levi-Civita epsilon}
\index{epsilon, Levi-Civita}
\index{$\delta$}
\index{$\epsilon$}

\index{repetition, unseemly}
\index{parity!and the Levi-Civita epsilon}
Einstein's summation convention expresses the dot
product~(\ref{vector:240:20}) neatly but, as we have seen
in~(\ref{vector:240:22}), does not by
itself wholly avoid unseemly repetition in the cross product.  The
\emph{Levi-Civita epsilon}%
\footnote{
  Also called the \emph{Levi-Civita symbol, tensor,} or \emph{permutor.}
  For native English speakers who do not speak Italian, the ``ci'' in
  Levi-Civita's name is pronounced as the ``chi'' in ``children.''
}%
~$\ep_{ijk}$ mends this, rendering the
cross-product as
\bq{vector:240:32}
  \ve a \times \ve b = \ep_{ijk} \vui a_j b_k,
\eq
where%
\footnote{\cite[``Levi-Civita permutation symbol'']{planetm}}
\bq{vector:240:30}
  \settowidth\tla{$+1$}
  \ep_{ijk} \equiv \begin{cases}
    \makebox[\tla][r]{$+1$} & \ \mbox{if $(i,j,k) = (x',y',z')$, $(y',z',x')$ or $(z',x',y')$;} \\
    \makebox[\tla][r]{$-1$} & \ \mbox{if $(i,j,k) = (x',z',y')$, $(y',x',z')$ or $(z',y',x')$;} \\
    \makebox[\tla][r]{$ 0$} & \ \mbox{otherwise [for instance if $(i,j,k) = (x',x',y')$].}
  \end{cases}
\eq
In the language of \S~\ref{matrix:322}, the Levi-Civita epsilon
quantifies parity.  (Chapters~\ref{matrix} and~\ref{eigen} did not use
it, but the Levi-Civita notation applies in any number of dimensions,
not only three as in the present chapter.  In this more general sense
the Levi-Civita is the determinant of the permutor whose ones hold the
indicated positions---which is a formal way of saying that it's a~$+$
sign for even parity and a~$-$ sign for odd.  For instance, in the
four-dimensional, $4 \times 4$ case $\ep_{1234}=1$ whereas
$\ep_{1243}=-1$: refer to \S\S~\ref{matrix:322}, \ref{matrix:325.10}
and~\ref{eigen:310}.  Table~\ref{vector:240:tbl1}, however, as the rest
of this section and chapter, concerns the three-dimensional case only.)

Technically, the Levi-Civita epsilon and Einstein's summation convention
are two separate, independent things, but a canny reader takes the
Levi-Civita's appearance as a hint that Einstein's convention is
probably in force, as in~(\ref{vector:240:32}).  The two tend to go
together.%
\footnote{
  The writer has heard the apocryphal belief expressed that the
  letter~$\epsilon$, a Greek~$e$, stood in this context for
  ``Einstein.''  As far as the writer knows,~$\epsilon$ is merely the
  letter after~$\delta$, which represents the name of Paul
  Dirac---though the writer does not claim his indirected story to be
  any less apocryphal than the other one (the capital letter~$\Delta$
  has a point on top that suggests the pointy nature of the Dirac delta
  of Fig.~\ref{integ:670:fig-d}, which makes for yet another plausible
  story).  In any event, one sometimes hears Einstein's summation
  convention, the Kronecker delta and the Levi-Civita epsilon together
  referred to as ``the Einstein notation,'' which though maybe not quite
  terminologically correct is hardly incorrect enough to argue over and
  is clear enough in practice.
}

The Levi-Civita epsilon~$\ep_{ijk}$ relates to the Kronecker
delta~$\delta_{ij}$ of \S~\ref{matrix:150} approximately as the
cross product relates to the dot product.  Both delta and epsilon find
use in vector work.  For example, one can write~(\ref{vector:240:20})
alternately in the form
\[
  \ve a \cdot \ve b = \delta_{ij} a_i b_j.
\]

Table~\ref{vector:240:tbl1} lists several relevant properties,%
\footnote{\cite{physics321}}
each as with Einstein's summation convention in force.%
\footnote{The table incidentally answers \S~\ref{vector:240.20}'s quiz.}
\begin{table}
  \caption[Properties of the Kronecker delta and Levi-Civita
    epsilon.]{%
    Properties of the Kronecker delta and the Levi-Civita epsilon, with
    Einstein's summation convention in force.
  }
  \label{vector:240:tbl1}
  \index{Kronecker delta!properties of}
  \index{delta, Kronecker!properties of}
  \index{Levi-Civita epsilon!properties of}
  \index{epsilon, Levi-Civita!properties of}
  \bqb
    \delta_{jk} &=& \delta_{kj} \\
    \delta_{ij}\delta_{jk} &=& \delta_{ik} \\
    \delta_{ii} &=& 3 \\
    \delta_{jk}\ep_{ijk} &=& 0 \\
    \delta_{nk}\ep_{ijk} &=& \ep_{ijn} \\
    \ep_{ijk} = \ep_{jki} = \ep_{kij}
      &=& -\ep_{ikj} = -\ep_{jik} = -\ep_{kji} \\
    \ep_{ijk}\ep_{ijk} &=& 6 \\
    \ep_{ijn}\ep_{ijk} &=& 2\delta_{nk} \\
    \ep_{imn}\ep_{ijk} &=& \delta_{mj}\delta_{nk} - \delta_{mk}\delta_{nj}
  \eqb
\end{table}
Of the table's several properties, the property that $\ep_{imn}\ep_{ijk}
= \delta_{mj}\delta_{nk} - \delta_{mk}\delta_{nj}$ is proved by
observing that, in the case that $i=x'$, either $(j,k)=(y',z')$ or
$(j,k)=(z',y')$, and also either $(m,n)=(y',z')$ or $(m,n)=(z',y')$; and
similarly in the cases that $i=y'$ and $i=z'$ (more precisely, in each
case the several indices can take any values, but combinations other
than the ones listed drive~$\ep_{imn}$ or~$\ep_{ijk}$, or both, to zero,
thus contributing nothing to the sum).  This implies that either
$(j,k)=(m,n)$ or $(j,k)=(n,m)$---which, when one takes parity into
account, is exactly what the property in question asserts.  The property
that $\ep_{ijn}\ep_{ijk} = 2\delta_{nk}$ is proved by observing that, in
any given term of the Einstein sum,~$i$ is either~$x'$ or~$y'$ or~$z'$
and that~$j$ is one of the remaining two, which leaves the third to be
shared by both~$k$ and~$n$.  The factor~$2$ appears because, for
$k=n=x'$, an $(i,j)=(y',z')$ term and an $(i,j)=(z',y')$ term both
contribute positively to the sum; and similarly for $k=n=y'$ and again
for $k=n=z'$.

Unfortunately, the last paragraph likely makes sense to few who do not
already know what it means.  A concrete example helps.  Consider the
compound product $\ve c \times (\ve a \times \ve b)$.  In this section's
notation and with the use of~(\ref{vector:240:32}), the compound product
is
\bqb
  \ve c \times (\ve a \times \ve b)
  &=&
    \ve c \times (\ep_{ijk}\vui a_j b_k)
  \xn\\&=&
    \ep_{mni} \vu m c_n (\ep_{ijk}\vui a_j b_k)_i
  \xn\\&=&
    \ep_{mni} \ep_{ijk} \vu m c_n a_j b_k
  \xn\\&=&
    \ep_{imn} \ep_{ijk} \vu m c_n a_j b_k
  \xn\\&=&
    (\delta_{mj}\delta_{nk} - \delta_{mk}\delta_{nj}) \vu m c_n a_j b_k
  \xn\\&=&
    \delta_{mj}\delta_{nk} \vu m c_n a_j b_k - \delta_{mk}\delta_{nj} \vu m c_n a_j b_k
  \xn\\&=&
    \vuj c_k a_j b_k - \vu k c_j a_j b_k
  \xn\\&=&
    (\vuj a_j)(c_k b_k) - (\vu k b_k)(c_j a_j).
  \xn
\eqb
That is, in light of~(\ref{vector:240:20}),
\bq{vector:240:35}
  \ve c \times (\ve a \times \ve b) =
    \ve a(\ve c \cdot \ve b) - \ve b(\ve c \cdot \ve a),
\eq
a useful vector identity.  Written without the benefit of Einstein's
summation convention, the example's central step would have been
\bqb
  \ve c \times (\ve a \times \ve b)
  &=&
    \sum_{i,j,k,m,n} \ep_{imn} \ep_{ijk} \vu m c_n a_j b_k
  \xn\\&=&
    \sum_{j,k,m,n} (\delta_{mj}\delta_{nk} - \delta_{mk}\delta_{nj}) \vu m c_n a_j b_k,
\eqb
which makes sense if you think about it hard enough,%
\footnote{
  If thinking about it hard enough does not work, then here it is in
  interminable detail:
  \bqb
    \lefteqn{\sum_{i,j,k,m,n} \ep_{imn} \ep_{ijk} f(j,k,m,n)}
    \\&=&
        \ep_{x'y'z'} \ep_{x'y'z'} f(y',z',y',z')
      + \ep_{x'y'z'} \ep_{x'z'y'} f(y',z',z',y')
    \\&&\ \ \mbox{}
      + \ep_{x'z'y'} \ep_{x'y'z'} f(z',y',y',z')
      + \ep_{x'z'y'} \ep_{x'z'y'} f(z',y',z',y')
    \\&&\ \ \mbox{}
      + \ep_{y'z'x'} \ep_{y'z'x'} f(z',x',z',x')
      + \ep_{y'z'x'} \ep_{y'x'z'} f(z',x',x',z')
    \\&&\ \ \mbox{}
      + \ep_{y'x'z'} \ep_{y'z'x'} f(x',z',z',x')
      + \ep_{y'x'z'} \ep_{y'x'z'} f(x',z',x',z')
    \\&&\ \ \mbox{}
      + \ep_{z'x'y'} \ep_{z'x'y'} f(x',y',x',y')
      + \ep_{z'x'y'} \ep_{z'y'x'} f(x',y',y',x')
    \\&&\ \ \mbox{}
      + \ep_{z'y'x'} \ep_{z'x'y'} f(y',x',x',y')
      + \ep_{z'y'x'} \ep_{z'y'x'} f(y',x',y',x')
    \\&=&
        f(y',z',y',z')
      - f(y',z',z',y')
      - f(z',y',y',z')
      + f(z',y',z',y')
    \\&&\ \ \mbox{}
      + f(z',x',z',x')
      - f(z',x',x',z')
      - f(x',z',z',x')
      + f(x',z',x',z')
    \\&&\ \ \mbox{}
      + f(x',y',x',y')
      - f(x',y',y',x')
      - f(y',x',x',y')
      + f(y',x',y',x')
    \\&=&
    \big[
        f(y',z',y',z')
      + f(z',x',z',x')
      + f(x',y',x',y')
    \\&&\ \ \ \ \ \ \ \ \mbox{}
      + f(z',y',z',y')
      + f(x',z',x',z')
      + f(y',x',y',x')
    \big]
    \\&&\ \ \mbox{}
    - \big[
        f(y',z',z',y')
      + f(z',x',x',z')
      + f(x',y',y',x')
    \\&&\ \ \ \ \ \ \ \ \mbox{}
      + f(z',y',y',z')
      + f(x',z',z',x')
      + f(y',x',x',y')
    \big]
    \\&=&
    \big[
        f(y',z',y',z')
      + f(z',x',z',x')
      + f(x',y',x',y')
    \\&&\ \ \ \ \ \ \ \ \mbox{}
      + f(z',y',z',y')
      + f(x',z',x',z')
      + f(y',x',y',x')
    \\&&\ \ \ \ \ \ \ \ \mbox{}
      + f(x',x',x',x')
      + f(y',y',y',y')
      + f(z',z',z',z')
    \big]
    \\&&\ \ \mbox{}
    - \big[
        f(y',z',z',y')
      + f(z',x',x',z')
      + f(x',y',y',x')
    \\&&\ \ \ \ \ \ \ \ \mbox{}
      + f(z',y',y',z')
      + f(x',z',z',x')
      + f(y',x',x',y')
    \\&&\ \ \ \ \ \ \ \ \mbox{}
      + f(x',x',x',x')
      + f(y',y',y',y')
      + f(z',z',z',z')
    \big]
    \\&=&
      \sum_{j,k,m,n} (\delta_{mj}\delta_{nk} - \delta_{mk}\delta_{nj}) f(j,k,m,n).
  \eqb
  That is for the property that $\ep_{imn}\ep_{ijk} =
  \delta_{mj}\delta_{nk} - \delta_{mk}\delta_{nj}$.
  For the property that $\ep_{ijn}\ep_{ijk} = 2\delta_{nk}$, the
  corresponding calculation is
  \bqb
    \lefteqn{\sum_{i,j,k,n} \ep_{ijn} \ep_{ijk} f(k,n)}
    \\&=&
        \ep_{y'z'x'} \ep_{y'z'x'} f(x',x')
      + \ep_{z'y'x'} \ep_{z'y'x'} f(x',x')
    \\&&\ \ \mbox{}
      + \ep_{z'x'y'} \ep_{z'x'y'} f(y',y')
      + \ep_{x'z'y'} \ep_{x'z'y'} f(y',y')
    \\&&\ \ \mbox{}
      + \ep_{x'y'z'} \ep_{x'y'z'} f(z',z')
      + \ep_{y'x'z'} \ep_{y'x'z'} f(z',z')
    \\&=&
        f(x',x')
      + f(x',x')
      + f(y',y')
      + f(y',y')
      + f(z',z')
      + f(z',z')
    \\&=& 2\big[
        f(x',x')
      + f(y',y')
      + f(z',z')
    \big]
    \\&=& 2\sum_{k,n} \delta_{nk} f(k,n).
  \eqb
  For the property that $\ep_{ijk}\ep_{ijk} = 6$,
  \[
    \sum_{i,j,k} \ep_{ijk} \ep_{ijk}
    =
        \ep_{x'y'z'}^2
      + \ep_{y'z'x'}^2
      + \ep_{z'x'y'}^2
      + \ep_{x'z'y'}^2
      + \ep_{y'x'z'}^2
      + \ep_{z'y'x'}^2
    = 6.
  \]
  It is precisely to encapsulate such interminable detail that we use
  the Kronecker delta, the Levi-Civita epsilon and the properties of
  Table~\ref{vector:240:tbl1}.
}
and justifies the table's claim that $\ep_{imn}\ep_{ijk} =
\delta_{mj}\delta_{nk} - \delta_{mk}\delta_{nj}$.  (Notice that the
compound Kronecker operator~$\delta_{mj}\delta_{nk}$ includes nonzero
terms for the case that $j=k=m=n=x'$, for the case that $j=k=m=n=y'$
and for the case that $j=k=m=n=z'$, whereas the compound Levi-Civita
operator~$\ep_{imn}\ep_{ijk}$ does not.  However, the compound Kronecker
operator~$-\delta_{mk}\delta_{nj}$ includes canceling terms for these
same three cases.  This is why the table's claim is valid as written.)

To belabor the topic further here would serve little purpose.  The
reader who does not feel entirely sure that he understands what is going
on might work out the table's several properties with his own pencil, in
something like the style of the example, until he is satisfied that he
adequately understands the several properties and their correct use.

\index{projection!onto a plane}
\index{plane!projection onto}
\index{vector!projection of onto a plane}
Section~\ref{vcalc:425} will refine the notation for use when
derivatives with respect to angles come into play but, before leaving
the present section, we might pause for a moment to
appreciate~(\ref{vector:240:35}) in the special case that $\ve b = \ve c
= \vu n$:
\bq{vector:240:36}
  -\vu n \times (\vu n \times \ve a) = \ve a - \vu n(\vu n \cdot \ve a).
\eq
The difference $\ve a - \vu n(\vu n \cdot \ve a)$ evidently projects a
vector~$\ve a$ onto the plane whose unit normal is~$\vu n$.
Equation~(\ref{vector:240:36}) reveals that the double cross product
$-\vu n \times (\vu n \times \ve a)$ projects the same vector onto the
same plane.  Figure~\ref{vector:240:fig-proj} illustrates.
\begin{figure}
  \caption{A vector projected onto a plane.}
  \label{vector:240:fig-proj}
  \bc
    \nc\fxa{-2.7} \nc\fxb{6.8}
    \nc\fya{-1.7} \nc\fyb{4.0}
    %\nc\xxs{0.20}   % Perspective ratio of the x-y circle.
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      \localscalebox{1.0}{1.0}{
        \small
        \psset{dimen=middle}
        % The slope of the plane's sides on the page is 4:1.
        \pspolygon[linewidth=0.5pt,linecolor=lightgray,fillstyle=solid,fillcolor=lightgray]
          (-1.7,1.1)(3.9,1.1)(3.3,-1.3)(-2.3,-1.3)
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          \psset{linewidth=0.5pt}
          \rput(2.00000,0.69282){%
            \psline(0.00000,0.45000)(-0.22500,0.37206)(-0.22500,-0.07794)
          }
          %\psline(0.00000,0.45000)(-0.22500,0.37206)(-0.22500,-0.07794)
          %\psline(0.00000,0.45000)(0.22500,0.52794)(0.22500,0.07794)
          \psline(0.00000,0.45000)(-0.45000,0.45000)(-0.45000,-0.02)
        }
        \rput(0.00,1.75){$\vu n$}
        \rput(1.20,2.80){$\ve a$}
        \rput[l](2.15,2.10){$\vu n(\vu n \cdot \ve a)$}
        \psline[linewidth=0.5pt](1.05,0.25)(1.15,0.15)(4.20,0.15)
        \rput[l](4.10,-0.05){
          $
            \br{l}
              -\vu n \times (\vu n \times \ve a)
              \\ = \ve a - \vu n(\vu n \cdot \ve a)
            \er
          $
        }
      }
    \end{pspicture}
  \ec
\end{figure}

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

\section{Algebraic identities}
\label{vector:250}
\index{identity!vector, algebraic}
\index{vector!algebraic identities of}

Vector algebra is not in principle very much harder than scalar algebra
is, but with three distinct types of product it has more rules
controlling the way its products and sums are combined.
Table~\ref{vector:tbl-algid} lists several of these.%
\footnote{\cite[appendix~II]{Stratton}\cite[appendix~A]{Harrington}}%
$\mbox{}^,$%
\footnote{
  Nothing in any of the table's identities requires the vectors involved
  to be real.  The table is equally as valid when vectors are complex.
}
\begin{table}
  \caption{Algebraic vector identities.}
  \label{vector:tbl-algid}
  \[
    \br{c}
      \br{rclcrclcrcl}
        \psi\ve a &=& \vui\psi a_i &\ \ &
        \ve a \cdot \ve b &\equiv& a_i b_i &\ \ &
        \ve a \times \ve b &\equiv& \ep_{ijk}\vui a_j b_k
      \er \\
      \br{rclcrcl}
        \ve a^{*} \cdot \ve a &=& \left| a \right|^2 &&
        (\psi)(\ve a + \ve b) &=& \psi \ve a + \psi \ve b \\
        \ve b \cdot \ve a &=& \ve a \cdot \ve b &&
        \ve b \times \ve a &=& -\ve a \times \ve b \\
        \ve c \cdot ( \ve a + \ve b ) &=& \ve c \cdot \ve a + \ve c \cdot \ve b &\ \ &
        \ve c \times ( \ve a + \ve b ) &=& \ve c \times \ve a + \ve c \times \ve b \\
        \ve a \cdot ( \psi \ve b ) &=& ( \psi ) ( \ve a \cdot \ve b ) &&
        \ve a \times ( \psi \ve b ) &=& ( \psi ) ( \ve a \times \ve b )
      \er \\
      \br{rclcl}
        \ve c \cdot ( \ve a \times \ve b ) &=&
        \ve a \cdot ( \ve b \times \ve c ) &=&
        \ve b \cdot ( \ve c \times \ve a ) \\
        \ve c \times ( \ve a \times \ve b ) &=&
        \multicolumn{3}{l}{\ve a ( \ve c \cdot \ve b) - \ve b ( \ve c \cdot \ve a)} \\
       -\vu n \times (\vu n \times \ve a) &=&
        \multicolumn{3}{l}{\ve a - \vu n(\vu n \cdot \ve a)}
      \er
    \er
  \]
\end{table}
Most of the table's identities are plain by the
formulas~(\ref{vector:220:02}), (\ref{vector:dot})
and~(\ref{vector:cross}) respectively for the scalar, dot and cross
products, and two were proved as~(\ref{vector:240:35})
and~(\ref{vector:240:36}).  The remaining identity is proved in the
notation of \S~\ref{vector:240} as
\[
  \br{rclclcl}
    \ep_{ijk} c_i a_j b_k &=&
      \ep_{ijk} c_i a_j b_k &=&
      \ep_{kij} c_k a_i b_j &=&
      \ep_{jki} c_j a_k b_i
    \\&=&
      \ep_{ijk} c_i a_j b_k &=&
      \ep_{ijk} a_i b_j c_k &=&
      \ep_{ijk} b_i c_j a_k
    \\&=&
      \ve c \cdot ( \ep_{ijk} \vui a_j b_k ) &=&
      \ve a \cdot ( \ep_{ijk} \vui b_j c_k ) &=&
      \ve b \cdot ( \ep_{ijk} \vui c_j a_k ).
  \er
\]
That is,
\bq{vector:250:30}
  \ve c \cdot ( \ve a \times \ve b ) =
  \ve a \cdot ( \ve b \times \ve c ) =
  \ve b \cdot ( \ve c \times \ve a ).
\eq
Besides the several vector identities, the table also includes the three
vector products in Einstein notation.%
\footnote{
  If the reader's native language is English, then he is likely to have heard
  of the unfortunate ``back cab rule,'' which actually is not a rule but
  an unhelpful mnemonic for one of Table~\ref{vector:tbl-algid}'s
  identities.  The mnemonic is mildly orthographically clever but, when
  learned, significantly impedes real understanding of the vector.  The
  writer recommends that the reader forget the rule if he has heard of
  it for, in mathematics, spelling-based mnemonics are seldom if ever a
  good idea.
}

Each definition and identity of Table~\ref{vector:tbl-algid} is
invariant under reorientation of axes.

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

\section{Isotropy}
\label{vector:270}
\index{isotropy}

\index{coordinates!isotropic}
A real,%
\footnote{
  The reader is reminded that one can licitly express a complex vector
  in a real basis.
}
three-dimensional coordinate system%
\footnote{
  This chapter's footnote~\ref{vector:280:fn1} and chapter~\ref{vcalc}'s
  footnote~\ref{vcalc:420:fn1} explain the usage of semicolons as
  coordinate delimiters.
}
$(\alpha;\beta;\gamma)$ is
\emph{isotropic} at a point $\ve r=\ve r_1$ if and only if
\bq{vector:270:10}
  \begin{split}
    \wu\beta(\ve r_1) \cdot \wu\gamma(\ve r_1) &= 0, \\
    \wu\gamma(\ve r_1) \cdot \wu\alpha(\ve r_1) &= 0, \\
    \wu\alpha(\ve r_1) \cdot \wu\beta(\ve r_1) &= 0,
  \end{split}
\eq
and
\bq{vector:270:15}
  \left|\frac{\pl\ve r}{\pl \alpha}\right|_{\ve r = \ve r_1}
  = \left|\frac{\pl\ve r}{\pl \beta}\right|_{\ve r = \ve r_1}
  = \left|\frac{\pl\ve r}{\pl \gamma}\right|_{\ve r = \ve r_1}.
\eq
That is, a three-dimensional system is isotropic if its three
coordinates advance locally at right angles to one another but at the
same rate.

% diagn: as more of the book is written, check that this paragraph's
% claim about "all ... coordinate systems this book treats"
% remains true
Of the three basic three-dimensional coordinate systems---indeed, of all
the three-dimensional coordinate systems this book treats---only the
rectangular is isotropic according to~(\ref{vector:270:10})
and~(\ref{vector:270:15}).%
\footnote{%
  Whether it is even possible to construct an isotropic, nonrectangular
  coordinate system in three dimensions is a question we will leave to
  the professional mathematician.  The author has not encountered such
  a system.%
}
Isotropy admittedly would not be a very interesting property if that
were all there were to it.  However, there is also \emph{two-dimensional
isotropy,} more interesting because it arises oftener.

\index{coordinates!logarithmic cylindrical}
\index{logarithmic cylindrical coordinates}
A real, two-dimensional coordinate system $(\alpha;\beta)$ is isotropic
at a point $\we\rho^\gamma=\we\rho^\gamma_1$ if and only if
\bq{vector:270:20}
  \wu \alpha(\we\rho^\gamma_1) \cdot \wu \beta(\we\rho^\gamma_1) = 0
\eq
and
\bq{vector:270:25}
  \left|\frac{\pl\we\rho^\gamma}{\pl \alpha}\right|_{\we\rho^\gamma = \we\rho^\gamma_1}
  = \left|\frac{\pl\we\rho^\gamma}{\pl \beta}\right|_{\we\rho^\gamma = \we\rho^\gamma_1},
\eq
where $\we\rho^\gamma = \wu\alpha\alpha + \wu\beta\beta$ represents
position in the~$\alpha$-$\beta$ plane.
(If the~$\alpha$-$\beta$ plane happens to be the~$x$-$y$ plane, as is
often the case, then $\we\rho^\gamma = \we\rho^z = \we\rho$ and per
eqn.~\ref{trig:277:40} one can omit the superscript.) The
two-dimensional rectangular system $(x,y)$ naturally is isotropic.
Because $\left|\pl\we\rho/\pl\phi\right| =
(\rho)\left|\pl\we\rho/\pl\rho\right|$ the standard two-dimensional
cylindrical system $(\rho;\phi)$ as such is nonisotropic, but
the change of coordinate
\bq{vector:270:30}
  \lambda \equiv \ln\frac{\rho}{\rho_o},
\eq
where~$\rho_o$ is some arbitrarily chosen reference radius, converts the
system straightforwardly into the \emph{logarithmic cylindrical} system
$(\lambda;\phi)$ which is isotropic everywhere in the plane except at
the origin $\rho = 0$.  Further two-dimensionally isotropic coordinate
systems include the parabolic system of \S~\ref{vector:280.10}, to
follow.

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

\section{Parabolic coordinates}
\label{vector:280}
\index{parabolic coordinates}
\index{coordinates!parabolic}

\index{coordinates!special}
Scientists and engineers find most spatial-geometrical problems they
encounter in practice to fall into either of two categories.  The first
category comprises problems of simple geometry conforming to any one
of the three basic coordinate systems---rectangular, cylindrical or
spherical.  The second category comprises problems of complicated
geometry, analyzed in the rectangular system not because the problems'
geometries fit that system but rather because they fit no system and
thus give one little reason to depart from the rectangular.  One however
occasionally encounters problems of a third category, whose geometries
are simple but, though simple, nevertheless fit none of the three basic
coordinate systems.  Then it may fall to the scientist or engineer to
devise a special coordinate system congenial to the problem.

\index{root-length}
This section will treat the \emph{parabolic} coordinate systems which,
besides being arguably the most useful of the various special systems,
serve as good examples of the kind of special system a scientist or
engineer might be called upon to devise.  The two three-dimensional
parabolic systems are
the \emph{parabolic cylindrical} system $(\sigma,\tau,z)$ of \S~\ref{vector:280.51}
and the \emph{circular paraboloidal} system%
\footnote{\label{vector:280:fn1}%
  The reader will probably think nothing of it now, but later may wonder
  why the circular paraboloidal coordinates are $(\eta;\phi,\xi)$ rather
  than $(\xi;\phi,\eta)$ or $(\eta,\xi;\phi)$.  The peculiar ordering is
  to honor the right-hand rule (\S~\ref{trig:230}
  and eqn.~\ref{vector:230:10}), since
  $\wu\eta \times \wu\xi = -\wu\phi$ rather than~$+\wu\phi$.  See
  \S~\ref{vector:280.56}.  (Regarding the semicolon~``;'' delimiter, it
  doesn't mean much.  This book arbitrarily uses a semicolon when the
  following coordinate happens to be an angle, which helps to
  distinguish rectangular coordinates from cylindrical from spherical.
  Admittedly, such a notational convention ceases to help much when
  parabolic coordinates arrive, but we will continue to use it for
  inertia's sake.  See also chapter~\ref{vcalc}'s
  footnote~\ref{vcalc:420:fn1}.)
}
$(\eta;\phi,\xi)$ of \S~\ref{vector:280.56},
where the angle~$\phi$ and the length~$z$ are familiar to us
but~$\sigma$, $\tau$, $\eta$ and~$\xi$---neither angles nor lengths but
root-lengths (that is, coordinates having dimensions of
$[\mbox{length}]^{1/2}$)---are new.%
\footnote{
  The letters~$\sigma$, $\tau$, $\eta$ and~$\xi$ are available letters
  this section happens to use, not necessarily standard parabolic
  symbols.  See appendix~\ref{greek}\@.
}
Both three-dimensional parabolic systems derive from the
two-dimensional parabolic system $(\sigma,\tau)$ of
\S~\ref{vector:280.10}.%
\footnote{%
  \cite[\S~10.1]{Lebedev}%
  \cite[``Parabolic coordinates,'' 09:59, 19~July 2008]{wikip}%
}

However, before handling any parabolic system we ought formally to
introduce the parabola itself, next.

\subsection{The parabola}
\label{vector:280.05}
\index{parabola}

Parabolic coordinates are based on a useful geometrical curve called
the \emph{parabola,} which many or most readers will have met long
before opening this book's covers.  The parabola, simple but less
obvious than the circle, may however not be equally well known to all
readers, and even readers already acquainted with it (as from
\S\S~\ref{integ:241.07} and~\ref{integ:241.08}) might appreciate a
re\"examination.  This subsection reviews the parabola.

\index{focus}
\index{directrix}
\index{equidistance}
\emph{Given a point, called the \emph{focus,} and a line, called the
\emph{directrix,}%
\footnote{
  Whether the parabola's definition ought to forbid the directrix to
  pass through the focus is a stylistic question this book will leave
  unanswered.
}
plus the plane in which the focus and the directrix both lie, the
associated \emph{parabola} is that curve which lies in the plane
everywhere equidistant from both focus and directrix.}%
\footnote{\cite[\S~12-1]{Shenk}}
See Fig.~\ref{vector:280:fig-parabola}.
\begin{figure}
  \caption{The parabola.}
  \label{vector:280:fig-parabola}
  \bc
    \nc\fxa{-5.0} \nc\fxb{5.0}
    \nc\fya{-1.7} \nc\fyb{2.5}
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    \nc\xxd{0.12}
    \nc\xxe{1.5492}
    \nc\xxf{0.4}
    \nc\xxl{%
      \psline[linewidth=0.5pt]{<-}(0.06,0)(0.63,0)%
      \psline[linewidth=0.5pt]{<-}(1.60,0)(1.03,0)%
      \rput{*0}(0.83,0){$a$}%
    }
    \nc\xxm{2.4}
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
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      {
        \small
        \psset{dimen=middle}
        {
          \psset{linewidth=2.0pt}
          \psline[linewidth=1.0pt](-\xxc,-\xxb)(\xxc,-\xxb)
          \psline(0,-\xxd)(0,\xxd)
          \psline(-\xxd,0)(\xxd,0)
          \psdot(\xxe,\xxf)
        }
        {
          \psset{linewidth=0.5pt}
          \psline(-0.25,0)(-0.55,0)
          \psline{<-}(-0.4,0)(-0.4,-0.35)
          \psline{<-}(-0.4,-1.2)(-0.4,-0.85)
          \rput(-0.4,-0.6){$\sigma^2$}
        }
        \rput{-90}(\xxe,\xxf)\xxl
        \rput{-165.522}(\xxe,\xxf)\xxl
        {
          \psset{linewidth=2.0pt,plotpoints=80}
          \psplot{-\xxm}{-0.65}{
            0.6 dup 4.0 mul x dup mul exch div sub neg
          }
          \psplot{-0.15}{\xxe}{
            0.6 dup 4.0 mul x dup mul exch div sub neg
          }
          \psplot{\xxe}{\xxm}{
            0.6 dup 4.0 mul x dup mul exch div sub neg
          }
        }
      }
    \end{pspicture}
  \ec
\end{figure}

\index{projectile}
\index{parabolic arc}
Referring to the figure, if rectangular coordinates are established such
that~$\vu x$ and~$\vu y$ lie in the plane, that the parabola's focus
lies at $(x,y)=(0,k)$, and that the equation $y=k-\sigma^2$
describes the parabola's directrix, then the equation
\[
  x^2 + (y-k)^2 = (y-k+\sigma^2)^2
\]
evidently expresses the equidistance rule of the parabola's definition.
Solving for $y-k$ and then, from that solution, for~$y$, we have that
\bq{vector:280:05}
  y = \frac{x^2}{2\sigma^2} + \left( k - \frac{\sigma^2}{2} \right).
\eq
With the definitions that
\bq{vector:280:07}
  \begin{split}
    \mu &\equiv \frac{1}{2\sigma^2}, \\
    \kappa &\equiv k - \frac{\sigma^2}{2},
  \end{split}
\eq
given which
\bq{vector:280:08}
  \begin{split}
    \sigma^2 &= \frac{1}{2\mu}, \\
    k &= \kappa + \frac{1}{4\mu},
  \end{split}
\eq
eqn.~(\ref{vector:280:05}) becomes
\bq{vector:280:06}
  y = \mu x^2 + \kappa.
\eq
Equations fitting the general form~(\ref{vector:280:06}) often arise
in applications, for example in the equation that describes a
projectile's flight in the absence of air resistance.
%\footnote{
%  Though the chapter is not about projectiles, briefly:  Gravity~$g$
%  accelerates a projectile as
%  \[
%    \frac{d^2y}{dt^2} = -g,
%  \]
%  where~$y$ represents the projectile's altitude and~$t$ represents
%  time.  The equation's solution is that
%  \[
%    y = -\frac{gt^2}{2} + v_{yo}t + y_o,
%  \]
%  where the parameter~$v_{yo}$ represents the projectile's initial
%  vertical velocity and the parameter~$y_o$ represents its initial
%  height.  Neglecting air resistance, the projectile does not accelerate
%  in the~$x$ direction, so
%  \[
%    x = v_x t + x_o.
%  \]
%  Combining the last two equations, we have that
%  \[
%    y = -\frac{g}{2}\left(\frac{x-x_o}{v_x}\right)^2 + v_{yo}\left(\frac{x-x_o}{v_x}\right) + y_o,
%  \]
%  which, once its terms are distributed and collected and~$x$ is
%  appropriately offset, indeed fits the general parabolic
%  form~(\ref{vector:280:06}).
%}
Any equation that fits the form can be plotted as a parabola, which
is why projectiles fly in parabolic arcs.

\index{bisection}
\index{satellite dish antenna}
\index{dish antenna}
\index{parabolic antenna}
\index{antenna!satellite dish}
\index{antenna!parabolic}
\index{cross-section!parabolic}
\index{parabolic cross-section}
\index{mirror!parabolic}
\index{mirror!parabolic}
\index{parabolic mirror}
\index{ray}
\index{light ray}
Observe that the parabola's definition does not actually require the
directrix to be $\vu y$-oriented: the directrix can be $\vu x$-oriented
or, indeed, oriented any way.
%(though naturally in that case eqns.~\ref{vector:280:05}
%and~\ref{vector:280:06} would have to be modified).
Observe also the geometrical fact that \emph{the parabola's track
necessarily bisects the angle between the two line segments
% The following should be "labeled~$a$", but visually that does not work.
labeled~``$a$\!''
in Fig.~\ref{vector:280:fig-parabola}.}  One of the
consequences of this geometrical fact\footnote{%
  It seems better merely to let the reader visualize the fact than to
  try to justify in so many words.
  If words help nevertheless, some words: Consider that the two line
  segments labeled~$a$ in the figure run in the directions of increasing
  distance respectively from the focus and from the directrix.  If you
  want to draw away from the directrix at the same rate as you draw away
  from the focus, thus maintaining equal distances, then your track
  cannot but exactly bisect the angle between the two segments.

  Once you grasp the idea, the bisection is obvious, though to grasp the
  idea can take some thought.

  \emph{To bisect} a thing, incidentally---if the context has not
  already made the meaning plain---is to divide the thing at its middle
  into two equal parts.%
}%
---is that a parabolic mirror reflects precisely%
\footnote{
  Well, actually, physically, the ray model of light implied here is
  valid only insofar as $\lambda \ll \sigma^2$, where~$\lambda$
  represents the light's characteristic wavelength.  Also, regardless
  of~$\lambda$, the ray model breaks down in the immediate neighborhood
  of the mirror's focus.  However, we were not thinking of
  wave mechanics at the moment.  Insofar as
  rays are concerned, the focusing is precise.
}
toward its focus all light rays that arrive perpendicularly with respect
to the directrix (which for instance is why satellite dish antennas have
parabolic cross-sections).

\subsection{Parabolic coordinates in two dimensions}
\label{vector:280.10}
\index{parabolic coordinates!in two dimensions}
\index{coordinates!parabolic, in two dimensions}

\index{two-dimensional space}
\index{parabolic track}
Parabolic coordinates are most easily first explained in the
two-dimensional case that $z=0$.  In two dimensions, the
parabolic coordinates $(\sigma,\tau)$ represent the point in the~$x$-$y$
plane that lies equidistant
\bi
  \item from the line $y=-\sigma^2$,
  \item from the line $y=+\tau^2$, and
  \item from the point $\rho=0$,
\ei
where the parameter~$k$ of \S~\ref{vector:280.05} has been set to $k=0$.
Figure~\ref{vector:280:fig2} depicts the construction described.
\begin{figure}
  \caption[Locating a point by parabolic construction.]%
    {Locating a point in two dimensions by parabolic construction.}
  \label{vector:280:fig2}
  \bc
    \nc\fxa{-5.0} \nc\fxb{5.0}
    \nc\fya{-1.7} \nc\fyb{2.5}
    \nc\xxa{2.0}
    \nc\xxb{1.2}
    \nc\xxc{3.5}
    \nc\xxd{0.12}
    \nc\xxe{1.5492}
    \nc\xxf{0.4}
    \nc\xxl{%
      \psline[linewidth=0.5pt]{<-}(0.06,0)(0.63,0)%
      \psline[linewidth=0.5pt]{<-}(1.60,0)(1.03,0)%
      \rput{*0}(0.83,0){$a$}%
    }
    \nc\xxm{2.4}
    \nc\xxn{37.761}
    \nc\xxg{0.3}
    \nc\xxo{0.7}
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      %\localscalebox{1.0}{1.0}
      {
        \small
        \psset{dimen=middle}
        {
          \psset{linewidth=2.0pt}
          \psline(-\xxc,\xxa)(\xxc,\xxa)
          \psline(-\xxc,-\xxb)(\xxc,-\xxb)
          \psline(0,-\xxd)(0,\xxd)
          \psline(-\xxd,0)(\xxd,0)
          \psdot(\xxe,\xxf)
        }
        {
          \psset{linewidth=0.5pt}
          \psline(-0.25,0)(-0.55,0)
          \psline{<-}(-0.4,0)(-0.4,0.75)
          \psline{<-}(-0.4,2.0)(-0.4,1.25)
          \rput(-0.4,1.0){$\tau^2$}
          \psline{<-}(-0.4,0)(-0.4,-0.35)
          \psline{<-}(-0.4,-1.2)(-0.4,-0.85)
          \rput(-0.4,-0.6){$\sigma^2$}
          \psline(-0.10,0.10)(-0.20,0.20)(-0.20,0.55)(0.05,0.55)
          \rput[l](0.07,0.55){$\rho\!=\!0$}
          \rput{\xxn}(-\xxe,\xxf){%
            \psline(-\xxo,0)(\xxo,0)%
            \psline(0,-\xxo)(0,\xxo)%
            \psline(0,\xxg)(-\xxg,\xxg)(-\xxg,0)%
          }
        }
        \rput{90}(\xxe,\xxf)\xxl
        \rput{-90}(\xxe,\xxf)\xxl
        \rput{-165.522}(\xxe,\xxf)\xxl
        {
          \psset{linewidth=0.5pt,linestyle=dotted,plotpoints=80}
          \psplot{-\xxm}{-0.65}{
            0.6 dup 4.0 mul x dup mul exch div sub neg
          }
          \psplot{-0.15}{\xxe}{
            0.6 dup 4.0 mul x dup mul exch div sub neg
          }
          \psplot{\xxe}{\xxm}{
            0.6 dup 4.0 mul x dup mul exch div sub neg
          }
          \psplot{-\xxm}{-0.65}{
            -1.0 dup 4.0 mul x dup mul exch div sub neg
          }
          \psplot{-0.15}{\xxe}{
            -1.0 dup 4.0 mul x dup mul exch div sub neg
          }
          \psplot{\xxe}{\xxm}{
            -1.0 dup 4.0 mul x dup mul exch div sub neg
          }
        }
      }
    \end{pspicture}
  \ec
\end{figure}
In the figure are two dotted curves, one of which represents the
point's parabolic track if~$\sigma$ were varied while~$\tau$ were held
constant and the other of which represents the point's parabolic track
if~$\tau$ were varied while~$\sigma$ were held constant.
Observe according to \S~\ref{vector:280.05}'s bisection finding that
each parabola necessarily bisects the angle between two of the
three line segments labeled~$a$ in the figure.  Observe further that the
two angles' sum is the straight angle $2\pi/2$, from which one can
conclude, significantly, that \emph{the two parabolas cross at
right angles to one another.}

Figure~\ref{vector:280:fig1} lays out the parabolic coordinate grid.
\begin{figure}
  \caption{The parabolic coordinate grid in two dimensions.}
  \label{vector:280:fig1}
  \index{grid, parabolic coordinate}
  \index{coordinate grid!parabolic}
  \index{parabolic coordinate grid}
  \bc
    \nc\fxa{-5.0} \nc\fxb{5.0}
    \nc\fya{-3.6} \nc\fyb{3.6}
    \nc\tta{2.85}
    \nc\ttb{1.30}
    \nc\ttc{2.58}
    \nc\ttd{3.85}
    \nc\tte{5.00}
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      %\localscalebox{1.0}{1.0}
      {
        \small
        \psset{dimen=middle}
        {
          {
            \psset{linewidth=0.5pt,plotpoints=200}
            {
              \psset{linecolor=gray}
              \psplot{-2.925}{-1.950}{
                3.25 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                  3 2 roll dup mul 2.0 div sub mul
              }
              \psplot{-3.150}{-2.100}{
                3.50 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                  3 2 roll dup mul 2.0 div sub mul
              }
              \psplot{-3.375}{-2.250}{
                3.75 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                  3 2 roll dup mul 2.0 div sub mul
              }
              \psplot{-2.700}{-2.025}{
                2.25 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                  3 2 roll dup mul 2.0 div sub mul neg
              }
              \psplot{-3.000}{-2.250}{
                2.50 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                  3 2 roll dup mul 2.0 div sub mul neg
              }
              \psplot{-3.300}{-2.475}{
                2.75 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                  3 2 roll dup mul 2.0 div sub mul neg
              }
            }
            \psline(0,-\tta)(0,\tta)
            \psplot{-\ttb}{\ttb}{
              1.0 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                3 2 roll dup mul 2.0 div sub mul
            }
            \psplot{-\ttc}{\ttc}{
              2.0 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                3 2 roll dup mul 2.0 div sub mul
            }
            \psplot{-\ttd}{\ttd}{
              3.0 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                3 2 roll dup mul 2.0 div sub mul
            }
            \psplot{-\tte}{\tte}{
              4.0 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                3 2 roll dup mul 2.0 div sub mul
            }
            \psplot{-\ttb}{\ttb}{
              1.0 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                3 2 roll dup mul 2.0 div sub mul neg
            }
            \psplot{-\ttc}{\ttc}{
              2.0 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                3 2 roll dup mul 2.0 div sub mul neg
            }
            \psplot{-\ttd}{\ttd}{
              3.0 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                3 2 roll dup mul 2.0 div sub mul neg
            }
            \psplot{-\tte}{\tte}{
              4.0 0.30 x 1 index div dup mul 2 index dup mul 2.0 mul 1.0 exch div mul
                3 2 roll dup mul 2.0 div sub mul neg
            }
            \rput[r](0.08, 3.15){$\sigma = 0$}
            \rput[r](0.08,-3.15){$\tau = 0$}
            \rput(1.36, 2.99){$\pm 1$}
            \rput(1.36,-2.99){$\pm 1$}
            \rput(2.69, 2.51){$\pm 2$}
            \rput(2.69,-2.51){$\pm 2$}
            \rput(4.02, 1.70){$\pm 3$}
            \rput(4.02,-1.70){$\pm 3$}
          }
        }
      }
    \end{pspicture}
  \ec
\end{figure}
Notice in the figure that one of the grid's several cells is subdivided
at its quarter-marks for illustration's sake, to show how one can
subgrid at need to locate points like, for example, $(\sigma,\tau) =
(\frac 7 2,-\frac 9 4)$ visually.  (That the subgrid's cells approach
square shape implies that the parabolic system is isotropic, a
significant fact \S~\ref{vector:280.30} will formally demonstrate.)

Using the Pythagorean theorem, one can symbolically express the
equidistant construction rule above as
\bq{vector:280:11}
  \begin{split}
    a &= \sigma^2 + y = \tau^2 - y, \\
    a^2 &= \rho^2 = x^2 + y^2.
  \end{split}
\eq
From the first line of~(\ref{vector:280:11}),
\bq{vector:280:12}
  y = \frac{\tau^2 - \sigma^2}{2}.
\eq
On the other hand, combining the two lines of~(\ref{vector:280:11}),
\[
  (\sigma^2 + y)^2 = x^2 + y^2 = (\tau^2 - y)^2,
\]
or, subtracting~$y^2$,
\[
  \sigma^4 + 2\sigma^2y = x^2 = \tau^4 - 2\tau^2y.
\]
Substituting~(\ref{vector:280:12})'s expression for~$y$,
\[
  x^2 = (\sigma\tau)^2.
\]
That either $x = +\sigma\tau$ or $x = -\sigma\tau$ would satisfy this
equation.  Arbitrarily choosing the~$+$ sign gives us that
\bq{vector:280:13}
  x = \sigma\tau.
\eq
Also, since $\rho^2 = x^2 + y^2$,~(\ref{vector:280:12})
and~(\ref{vector:280:13}) together imply that
\bq{vector:280:19}
  \rho = \frac{\tau^2 + \sigma^2}{2}.
\eq
Combining~(\ref{vector:280:12}) and~(\ref{vector:280:19}) to
isolate~$\sigma^2$ and~$\tau^2$ yields that
\bq{vector:280:19a}
  \begin{split}
    \sigma^2 &= \rho - y, \\
    \tau^2 &= \rho + y.
  \end{split}
\eq

\subsection{Properties}
\label{vector:280.30}
\index{parabolic coordinates!properties of}
\index{coordinates!parabolic, properties of}

\index{isotropy!of parabolic coordinates}
\index{parabolic coordinates!isotropy of}
\index{coordinates!parabolic, isotropy of}
The derivatives of~(\ref{vector:280:13}), (\ref{vector:280:12})
and~(\ref{vector:280:19}) are
\bq{vector:280:21}
  \begin{split}
    dx &= \sigma\,d\tau + \tau\,d\sigma, \\
    dy &= \tau\,d\tau - \sigma\,d\sigma, \\
    d\rho &= \tau\,d\tau + \sigma\,d\sigma.
  \end{split}
\eq
Solving the first two lines of~(\ref{vector:280:21}) simultaneously
for~$d\sigma$ and~$d\tau$ and then collapsing the resultant
subexpression $\tau^2 + \sigma^2$ per~(\ref{vector:280:19}) yields that
\bq{vector:280:22}
  \begin{split}
    d\sigma &= \frac{\tau\,dx - \sigma\,dy}{2\rho}, \\
    d\tau &= \frac{\sigma\,dx + \tau\,dy}{2\rho},
  \end{split}
\eq
from which it is apparent that
\[
  \begin{split}
    \wu\sigma &= \frac{\vu x \tau - \vu y \sigma}{\sqrt{\tau^2 + \sigma^2}}, \\
    \wu\tau &= \frac{\vu x \sigma + \vu y \tau}{\sqrt{\tau^2 + \sigma^2}};
  \end{split}
\]
or, collapsing again per~(\ref{vector:280:19}), that
\bq{vector:280:17}
  \begin{split}
    \wu\sigma &= \frac{\vu x \tau - \vu y \sigma}{\sqrt{2\rho}}, \\
    \wu\tau &= \frac{\vu x \sigma + \vu y \tau}{\sqrt{2\rho}},
  \end{split}
\eq
of which the dot product
\bq{vector:280:17t}
  \wu\sigma \cdot \wu\tau = 0 \ \mbox{if} \ \rho \neq 0
\eq
is null, confirming our earlier finding that the
various grid parabolas cross always at right angles to one another.
Solving~(\ref{vector:280:17}) simultaneously for~$\vu x$ and~$\vu y$
then produces
\bq{vector:280:17a}
  \begin{split}
    \vu x = \frac{\wu\tau \sigma + \wu\sigma \tau}{\sqrt{2\rho}}, \\
    \vu y = \frac{\wu\tau \tau - \wu\sigma \sigma}{\sqrt{2\rho}}.
  \end{split}
\eq
One can express an infinitesimal change in position in the plane as
\bqb
  d\we\rho &=& \vu x \,dx + \vu y \,dy
  \\&=& \vu x (\sigma\,d\tau + \tau\,d\sigma)
    + \vu y (\tau\,d\tau - \sigma\,d\sigma)
  \\&=& (\vu x \tau - \vu y \sigma) \,d\sigma
    + (\vu x \sigma + \vu y \tau) \,d\tau,
\eqb
in which~(\ref{vector:280:21}) has expanded the differentials and from
which
\[
  \begin{split}
    \frac{\pl\we\rho}{\pl\sigma} &= \vu x \tau - \vu y \sigma, \\
    \frac{\pl\we\rho}{\pl\tau} &= \vu x \sigma + \vu y \tau,
  \end{split}
\]
and thus
\bq{vector:280:17u}
  \left|\frac{\pl\we\rho}{\pl\sigma}\right| =
  \left|\frac{\pl\we\rho}{\pl\tau}\right|.
\eq
Equations~(\ref{vector:280:17t}) and~(\ref{vector:280:17u})
respectively meet the requirements~(\ref{vector:270:20})
and~(\ref{vector:270:25}), implying that \emph{the two-dimensional
parabolic coordinate system is isotropic} except at $\rho = 0$.

Table~\ref{vector:280:t30} summarizes, gathering parabolic coordinate
properties from this subsection and \S~\ref{vector:280.10}.
\begin{table}
  \caption{Parabolic coordinate properties.}
  \label{vector:280:t30}
  \nc\tta{0.38}
  \nc\ttb{0.02}
  \bc
    \parbox[t]{\tta\textwidth}{
      \bqb
        x &=& \sigma\tau \\
        y &=& \frac{\tau^2 - \sigma^2}{2} \\
        \rho &=& \frac{\tau^2 + \sigma^2}{2} \\
        \rho^2 &=& x^2 + y^2 \\
        \sigma^2 &=& \rho - y \\
        \tau^2 &=& \rho + y
      \eqb
    }
    \hspace{\ttb\textwidth}
    \parbox[t]{\tta\textwidth}{
      \bqb
        \vu x &=& \frac{\wu\tau \sigma + \wu\sigma \tau}{\sqrt{2\rho}} \\
        \vu y &=& \frac{\wu\tau \tau - \wu\sigma \sigma}{\sqrt{2\rho}} \\
        \wu\sigma &=& \frac{\vu x \tau - \vu y \sigma}{\sqrt{2\rho}} \\
        \wu\tau &=& \frac{\vu x \sigma + \vu y \tau}{\sqrt{2\rho}} \\
        \wu\sigma \times \wu\tau &=& \vu z \\
        \wu\sigma \cdot \wu\tau &=& 0 \\
        \left|\frac{\pl\we\rho}{\pl\sigma}\right| &=& \left|\frac{\pl\we\rho}{\pl\tau}\right|
      \eqb
    }
  \ec
\end{table}

\subsection{The parabolic cylindrical coordinate system}
\label{vector:280.51}
\index{parabolic cylindrical coordinates}
\index{coordinates!parabolic cylindrical}
\index{cylindrical coordinates!parabolic}

\index{three-dimensional space}
\index{parabolic cylinder}
\index{cylinder!parabolic}
Two-dimensional parabolic coordinates are easily extended to three
dimensions by adding a~$z$ coordinate, thus constituting the
\emph{parabolic cylindrical} coordinate system $(\sigma,\tau,z)$.  The
surfaces of constant~$\sigma$ and of constant~$\tau$ in this system are
\emph{parabolic cylinders} (and the surfaces of constant~$z$ naturally
are planes).  All the properties of Table~\ref{vector:280:t30} apply.
Observe however that the system is isotropic only in two dimensions not
three.

The orthogonal parabolic cylindrical basis is $[\wu\sigma\;\wu\tau\;\vu
z]$.

\subsection{The circular paraboloidal coordinate system}
\label{vector:280.56}
\index{circular paraboloidal coordinates}
\index{paraboloidal coordinates}
\index{coordinates!circular paraboloidal}

Sometimes one would like to extend the parabolic system to three
dimensions by adding an azimuth~$\phi$ rather than a height~$z$.  This
is possible, but then one tends to prefer the parabolas,
foci and directrices of Figs.~\ref{vector:280:fig2}
and~\ref{vector:280:fig1} to run in the~$\rho$-$z$ plane rather than
in the~$x$-$y$.  Therefore, one defines the coordinates~$\eta$ and~$\xi$
to represent in the~$\rho$-$z$ plane what the letters~$\sigma$
and~$\tau$ have represented in the~$x$-$y$.  The properties of
Table~\ref{vector:280:t56} result, which are just the properties of
Table~\ref{vector:280:t30} with coordinates changed.
\begin{table}
  \caption{Circular paraboloidal coordinate properties.}
  \label{vector:280:t56}
  \nc\tta{0.42}
  \nc\ttb{0.02}
  \bc
    \parbox[t]{\tta\textwidth}{
      \bqb
        \rho &=& \eta\xi \\
        z &=& \frac{\xi^2 - \eta^2}{2} \\
        r &=& \frac{\xi^2 + \eta^2}{2} \\
        r^2 &=& \rho^2 + z^2 = x^2 + y^2 + z^2 \\
        \eta^2 &=& r - z \\
        \xi^2  &=& r + z
      \eqb
    }
    \hspace{\ttb\textwidth}
    \parbox[t]{\tta\textwidth}{
      \bqb
        \wu\rho &=& \frac{\wu\xi \eta + \wu\eta \xi}{\sqrt{2r}} \\
        \vu z &=& \frac{\wu\xi \xi - \wu\eta \eta}{\sqrt{2r}} \\
        \wu\eta &=& \frac{\wu\rho \xi - \vu z \eta}{\sqrt{2r}} \\
        \wu\xi &=& \frac{\wu\rho \eta + \vu z \xi}{\sqrt{2r}} \\
        \wu\eta \times \wu\xi &=& -\wu\phi \\
        \wu\eta \cdot \wu\xi &=& 0 \\
        \left|\frac{\pl\ve r}{\pl\eta}\right| &=& \left|\frac{\pl\ve r}{\pl\xi}\right|
      \eqb
    }
  \ec
\end{table}
The system is the \emph{circular paraboloidal system} $(\eta;\phi,\xi)$.

\index{paraboloid}
The surfaces of constant~$\eta$ and of constant~$\xi$ in the circular
paraboloidal system are \emph{paraboloids,} parabolas rotated about
the~$z$ axis (and the surfaces of constant~$\phi$ are planes, or
half planes if you like, just as in the cylindrical system).  Like
the parabolic cylindrical system, the circular paraboloidal system too
is isotropic in two dimensions.

\index{right-hand rule}
Notice that, given the usual definition of the~$\wu\phi$ unit basis
vector, $\wu\eta \times \wu\xi = -\wu\phi$ rather than~$+\wu\phi$ as one
might first guess.  The correct, right-handed sequence of the orthogonal
circular paraboloidal basis therefore would be
$[\wu\eta\;\wu\phi\;\wu\xi]$.%
\footnote{
  See footnote~\ref{vector:280:fn1}.
}

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

This concludes the present chapter on the algebra of vector analysis.
Chapter~\ref{vcalc}, next, will venture hence into the larger
and even more interesting realm of vector calculus.

