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%TCIDATA{<META NAME="Title" CONTENT="Scalar and Vector Potentials">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Thu Feb 13 17:12:34 1997}
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\section{\label{Scalar potential}Scalar and Vector Potentials}

\textsf{Scalar Potential }on the \textsf{Vector Calculus} menu is the \emph{%
inverse} of the gradient in the sense that it finds a scalar function whose
gradient is the given vector field, or it tells you that such a function
does not exist. The vector potential has an analogous interpretation in
terms of the curl.

\subsection{\ Scalar Potentials}

The following are examples of scalar potential with the standard basis
variables.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Vector Calculus + Scalar Potential}
\end{quote}

\begin{quotation}
$(x,y,z)$, Scalar potential is $\frac{1}{2}z^{2}+\frac{1}{2}y^{2}+\frac{1}{2}%
x^{2}$

$(x,z,y)$, Scalar potential is $yz+\frac{1}{2}x^{2}$

$(y,z,x)$, Scalar potential does not exist\medskip
\end{quotation}

In the next example, choose \textsf{Evaluate}, and then from the \textsf{%
Vector Calculus} submenu choose \textsf{Scalar Potential}. Because the
vector field is a gradient, it has the original function as a scalar
potential.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate,} \textsf{Vector Calculus + Scalar
Potential}
\end{quote}

\begin{quotation}
$\nabla \left( xy^{2}+yz^{3}\right) =\left( y^{2},2xy+z^{3},3yz^{2}\right) $%
, Scalar potential is $xy^{2}+yz^{3}\medskip $

\emph{Note\quad }When the vector field is a gradient, it has the original
function as a scalar potential.\medskip
\end{quotation}

You would normally expect the scalar potential of the vector field$\ $ $%
\left( cv,cu+2vw,v^2\right) $ to be $ucv+v^2w$; that is, you expect $c$ to
be treated as a constant. When the number of variables differs from the
number of components in the field vector, a dialog box asks for the field
variables. In this case, you can enter \QTR{UserInput}{u,v,w} to get the
expected result. The dialog box also appears when you ask for the scalar
potential of a vector field that specifies fewer than three variables, such
as $\left( y,x,0\right) $. Enter \QTR{UserInput}{x,y,z} in the dialog box to
get the expected result $xy$ for the scalar potential of this vector field.

\subsection{\ Vector Potential\label{Vector, potential}}

Unless otherwise specified, the 
\index{Vector@Vector!potential@potential}operators \emph{curl}\textsl{\ } 
\index{Vector calculus@Vector calculus!vector potential@vector potential}and 
\emph{vector potential} apply to scalar or vector functions of a set of
exactly three 
\index{Standard basis variables@Standard basis variables}\emph{standard
basis variables}. \label{Standard basis variables}The default is $x,y,z$,
but you can use other sets of basis variables by choosing \textsf{Set Basis
Variables} from the \textsf{Vector Calculus} submenu and changing the
default variables.

Start with $\nabla \times (xy,yz,zx)=\,\left[ 
\begin{array}{c}
-y \\ 
-z \\ 
-x
\end{array}
\right] $ to get the following.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Vector Calculus + Vector Potential}
\end{quote}

\begin{quotation}
$\,\left[ 
\begin{array}{c}
-y \\ 
-z \\ 
-x
\end{array}
\right] $, Vector potential is $\left[ 
\begin{array}{c}
-%
\frac{1}{2}z^{2}+xy \\ 
yz \\ 
0
\end{array}
\right] \medskip $
\end{quotation}

Notice that we did not get the original vector field when we asked for a
vector potential of its curl. That is because the vector potential is
determined only up to a field whose curl is zero. You can verify that this
is the case. First, calculate the difference of the two vectors.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{c}
xy \\ 
yz \\ 
zx
\end{array}
\right] -\left[ 
\begin{array}{c}
-\frac{1}{2}z^{2}+xy \\ 
yz \\ 
0
\end{array}
\right] =\,\left[ 
\begin{array}{c}
\frac{1}{2}z^{2} \\ 
0 \\ 
xz
\end{array}
\right] \medskip $
\end{quotation}

Then compute the curl of the difference.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\nabla \times \left[ 
\begin{array}{c}
\frac{1}{2}z^{2} \\ 
0 \\ 
xz
\end{array}
\right] =\,\left[ 
\begin{array}{c}
0 \\ 
0 \\ 
0
\end{array}
\right] \medskip $
\end{quotation}

Try the same experiment after changing the basis variables to $u,v,w$ with 
\textsf{Vector Calculus + Set Basis Variables.\medskip }

\begin{quote}
$\blacktriangleright $ \textsf{Vector Calculus + Vector Potential}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{c}
-v \\ 
-w \\ 
-u
\end{array}
\right] $, Vector potential is $\left[ 
\begin{array}{c}
-\frac{1}{2}w^{2}+uv \\ 
vw \\ 
0
\end{array}
\right] \medskip $
\end{quotation}

A vector field can be written either as the triple $(v,w,u)$ or as a column
matrix. Try choosing \textsf{Vector Potential} for $(v,w,u)$.\medskip

\begin{quote}
$\blacktriangleright $ \textsf{Vector Calculus + Vector Potential}
\end{quote}

\begin{quotation}
$(v,w,u)$, Vector potential is $\left( \frac{1}{2}w^{2}-uv,-vw,0\right)
\medskip $
\end{quotation}

\subsection{\ Related topics}

\begin{itemize}
\item  \hyperref{Vectors}{}{}{DM9-1.tex#Vectors}

\item  \hyperref{Gradient, Divergence, and Curl}{}{}{DM9-2.tex#Gradient}

\item  \hyperref{Plots of Vector Fields and Gradients}{}{}{DM9-3.tex#Vector
field}

\item  \hyperref{Matrix-Valued Operators}{}{}{DM9-5.tex#Hessian}

\item  \hyperref{Plots of Complex Functions}{}{}{DM9-6.tex#Conformal plot}

\item  \hyperref{Exercises and Solutions}{}{}{DM9-7.tex#Exercises}
\end{itemize}

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