position vectorPositionVector2009-04-17 18:54:192009-04-17 18:54:19Definitionvector% this is the default PlanetPhysics preamble. as your knowledge
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\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
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%\usepackage{psfrag}
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%\usepackage{graphicx}
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%\usepackage{amsthm}
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%\usepackage{xypic}
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% define commands hereIn the space $\mathbb{R}^3$, the vector
$$\vec{r} \;:=\; (x,\,y,\,z) \;=\; x\vec{i}+y\vec{j}+z\vec{k}$$
directed from the origin to a point \,$(x,\,y,\,z)$\, is the {\em position vector} of this point.\, When the point is \PMlinkescapetext{variable}, $\vec{r}$ \PMlinkescapetext{represents} a vector field and its \PMlinkescapetext{length}
$$r \;:=\; \sqrt{x^2+y^2+z^2}$$
a scalar \PMlinkescapetext{field}.
The \PMlinkescapetext{simple formulae}
\begin{itemize}
\item $\nabla\!\cdot\vec{r} = 3$
\item $\nabla\!\times\!\vec{r} = \vec{0}$
\item $\displaystyle\nabla r = \frac{\vec{r}}{r} = \vec{r}^0$
\item $\displaystyle\nabla\frac{1}{r} = -\frac{\vec{r}}{r^3} = -\frac{\vec{r}^0}{r^2}$
\item $\displaystyle\nabla^2\frac{1}{r} = 0$
\end{itemize}
are valid, where $\vec{r}^0$ is the unit vector having the direction of $\vec{r}$.
If\, $\vec{c}$\, is a \PMlinkescapetext{constant} vector,\, $\vec{U}\!\!:\mathbb{R}^3\to\mathbb{R}^3$\, a vector function and\, $f\!\!:\mathbb{R}\to\mathbb{R}$\, is a twice differentiable function, then the formulae
\begin{itemize}
\item $\nabla(\vec{c}\cdot\!\vec{r}) = \vec{c}$
\item $\nabla\cdot(\vec{c}\times\vec{r}) = 0$
\item $(\vec{U}\!\cdot\!\nabla)\vec{r} = \vec{U}$
\item $(\vec{U}\!\times\!\nabla)\!\cdot\!\vec{r} = 0$
\item $(\vec{U}\!\times\!\nabla)\!\times\!\vec{r} = -2\vec{U}$
\item $\nabla f(r) = f'(r)\,\vec{r}^0$
\item $\displaystyle\nabla^2f(r) = f''(r)\!+\frac{2}{r}f'(r)$
\end{itemize}
hold.
\begin{thebibliography}{9}
\bibitem{VV}{\sc K. V\"ais\"al\"a:} {\em Vektorianalyysi}. \,Werner S\"oderstr\"om Osakeyhti\"o, Helsinki (1961).
\end{thebibliography}