Gradient 2006-07-25 20:30:50 2006-07-25 21:02:21 Definition % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here The gradient is the vector sum of the resultant rate of increase of a scalar funcion $V$ and is denoted $\nabla V$. It represents a directed rate of change of $V$. A directed derivative or vector derivative of $V$, so to speak. In cartesian coordinates \begin{equation} \nabla V = \frac{ \partial V}{\partial x} {\bf \hat{i}} + \frac{ \partial V}{\partial y} {\bf \hat{j}} + \frac{ \partial V}{\partial z} {\bf \hat{k}} \end{equation} It is common to regard $\nabla$ as the gradient operator which obtains a vector $\nabla V$ from a scalar function $V$ of position in space. $$ \nabla V = \left ( \frac{ \partial }{\partial x} {\bf \hat{i}} + \frac{ \partial }{\partial y} {\bf \hat{j}} + \frac{ \partial }{\partial z} {\bf \hat{ik}} \right ) V $$ Thus it is easy to work with just the gradient operator \begin{equation} \nabla V = \frac{ \partial }{\partial x} {\bf \hat{i}} + \frac{ \partial }{\partial y} {\bf \hat{j}} + \frac{ \partial }{\partial z} {\bf \hat{k}} \end{equation} This symbolic operator $\nabla$ was introduced by Sir W. R. Hamilton. It has been found by experience that the monosyllable \emph{del} is so short and easy to pronounce that even in complicated formulas in which $\nabla$ occurs a number of times no inconvenience to the speaker or hearer arises from the repetition. $\nabla V$ is read simply as "del $V$." \subsection{Coordinate System Independence} Although this operator $\nabla$ has been defined as $$\nabla V = \frac{ \partial }{\partial x} {\bf \hat{i}} + \frac{ \partial }{\partial y} {\bf \hat{j}} + \frac{ \partial }{\partial z} {\bf \hat{k}} $$ so that it appears to depend upon the choice of the axes, it is in reality independent of them. This would be surmised from the interpretation of $\nabla$ as the magnitude and direction of the most rapid increase of $V$. To demonstrate the independence take another set of axes, ${\bf \hat{i}'}$, ${\bf \hat{j}'}$, ${\bf \hat{k}'}$ and a new set of variables $x'$, $y'$, $z'$ referred to them. Then $\nabla$ referred to this system is $$\nabla' = \frac{ \partial }{\partial x'} {\bf \hat{i'}} + \frac{ \partial }{\partial y'} {\bf \hat{j'}} + \frac{ \partial }{\partial z'} {\bf \hat{k'}} $$ (Please Insert PROOF here...) Leaving behind the proof of coordinate system independence, here is the gradient opertor in the most common coordinate systems. Cartesian Coordinates \begin{equation} \nabla V = \frac{ \partial V}{\partial x} {\bf \hat{i}} + \frac{ \partial V}{\partial y} {\bf \hat{j}} + \frac{ \partial V}{\partial z} {\bf \hat{k}} \end{equation} Cylindrical Coordinates \begin{equation} \nabla V = \frac{ \partial V}{\partial r} {\bf \hat{r}} + \frac{1}{r}\frac{ \partial V}{\partial \theta} {\bf \hat{\theta}} + \frac{ \partial V}{\partial z} {\bf \hat{z}} \end{equation} Spherical Coordinates \begin{equation} \nabla V = \frac{ \partial V}{\partial r} {\bf \hat{r}} + \frac{1}{r}\frac{ \partial V}{\partial \phi} {\bf \hat{\phi}} + \frac{1}{r \sin \phi}\frac{ \partial V}{\partial \theta} {\bf \hat{\theta}} \end{equation}