We give the formulas for the gradient expressed in various curvilinear coordinate systems. We also show the metric tensors $g_{ij}$ so that the reader may verify the results by working from the basic formulas for the gradient.

## 1 Cylindrical coordinate system

In the cylindrical system of coordinates $(r,\theta,z)$ we have

 $g_{ij}=\begin{pmatrix}1&0&0\\ 0&r^{2}&0\\ 0&0&1\end{pmatrix}\,.$

So that

 $\displaystyle\nabla f$ $\displaystyle=\frac{\partial f}{\partial r}\,\mathbf{e}_{r}+\frac{1}{r}\frac{% \partial f}{\partial\theta}\,\mathbf{e}_{\theta}+\frac{\partial f}{\partial z}% \,\mathbf{k}\,,$

where

 $\displaystyle\mathbf{e}_{r}$ $\displaystyle=\frac{\partial}{\partial r}=\frac{x}{r}\,\mathbf{i}+\frac{y}{r}% \,\mathbf{j}$ $\displaystyle\mathbf{e}_{\theta}$ $\displaystyle=\frac{1}{r}\frac{\partial}{\partial\theta}=-\frac{y}{r}\,\mathbf% {i}+\frac{x}{r}\,\mathbf{j}$

are the unit vectors in the direction of increase of $r$ and $\theta$. Of course, $\mathbf{i},\mathbf{j},\mathbf{k}$ denote the unit vectors along the positive $x,y,z$ axes respectively.

The notations $\partial/\partial r,\partial/\partial\theta$, etc., denote the tangent vectors corresponding to infinitesimal changes in $r,\theta$, etc. respectively. Concretely, in terms of Cartesian coordinates, $\partial/\partial r$ is the vector $\mathbf{i}\,\partial x/\partial r+\mathbf{j}\,\partial y/\partial r+\mathbf{k}% \,\partial z/\partial r$. And similarly for the other variables. (There is a deep reason for using the seemingly strange notation: see Leibniz notation for vector fields for details.)

## 2 Polar coordinate system

This is just the special case of the cylindrical coordinate system where we chop off the $z$ coordinate. Thus

 $\displaystyle\nabla f$ $\displaystyle=\frac{\partial f}{\partial r}\,\mathbf{e}_{r}+\frac{1}{r}\frac{% \partial f}{\partial\theta}\,\mathbf{e}_{\theta}\,.$

## 3 Spherical coordinate system

To stave off confusion, note that this is the “mathematicians’ ” convention for the spherical coordinate system $(\rho,\phi,\theta)$. That is, $\phi$ is the co-latitude angle, and $\theta$ is the longitudinal angle.

 $g_{ij}=\begin{pmatrix}1&0&0\\ 0&\rho^{2}&0\\ 0&0&\rho^{2}\sin^{2}\phi\end{pmatrix}.$
 $\displaystyle\nabla f$ $\displaystyle=\frac{\partial f}{\partial\rho}\,\mathbf{e}_{\rho}+\frac{1}{\rho% }\frac{\partial f}{\partial\phi}\,\mathbf{e}_{\phi}+\frac{1}{\rho\sin\phi}% \frac{\partial f}{\partial\theta}\,\mathbf{e}_{\theta}\,,$

where

 $\displaystyle\mathbf{e}_{\rho}$ $\displaystyle=\frac{\partial}{\partial\rho}=\frac{x}{\rho}\,\mathbf{i}+\frac{y% }{\rho}\,\mathbf{j}+\frac{z}{\rho}\,\mathbf{k}$ $\displaystyle\mathbf{e}_{\phi}$ $\displaystyle=\frac{1}{\rho}\frac{\partial}{\partial\phi}=\frac{zx}{r\rho}\,% \mathbf{i}+\frac{zy}{r\rho}\,\mathbf{j}-\frac{r}{\rho}\,\mathbf{k}$ $\displaystyle\mathbf{e}_{\theta}$ $\displaystyle=\frac{1}{\rho\sin\theta}\frac{\partial}{\partial\theta}=-\frac{y% }{r}\,\mathbf{i}+\frac{x}{r}\,\mathbf{j}$

are the unit vectors in the direction of increase of $\rho,\phi,\theta$, respectively.

Title gradient in curvilinear coordinates GradientInCurvilinearCoordinates 2013-03-22 15:27:32 2013-03-22 15:27:32 stevecheng (10074) stevecheng (10074) 5 stevecheng (10074) Result msc 26B12 msc 26B10 gradient Gradient