# Spherical Coordinates

1. Apr 20, 2010

### squenshl

How do I find $$\nabla$$ in Spherical Coordinates. Please help.

2. Apr 20, 2010

### tiny-tim

Last edited by a moderator: Apr 25, 2017
3. Apr 21, 2010

### squenshl

How do I go about doing it from scratch.
How do I find i, j, & k from the definition of $$\nabla$$

Last edited: Apr 21, 2010
4. Apr 21, 2010

### squenshl

$$\nabla$$ = del/del(x) i + del/del(y) j + del/del(z) k
I found del/del(x), del/del)(y), del/del(z) but how do I find i, j, k. Help please.

5. Apr 22, 2010

### HallsofIvy

Have you looked at the site tiny-tim gives? No one is going to go through the whole thing just for you! It's not terribly deep but very tedious!

Here's a start only:

Since $\nabla u= \frac{\partial u}{\partial x}\vec{i}+ \frac{\partial u}{\partial y}\vec{j}+ \frac{\partial u}{\partial z}\vec{k}$
so you need to use the chain rule

$$\frac{\partial u}{\partial x}= \frac{\partial u}{\partial \rho}\frac{\partial \rho}{\partial x}+ \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}+ \frac{\partial u}{\partial \phi}\frac{\partial \phi}{\partial x}$$

Since $\rho= (x^2+ y^2+ z^2)^{1/2}$,
$$\frac{\partial \rho}{\partial x}= (1/2)(x^2+ y^2+ z^2)^{-1/2}(2x}= \frac{\rho cos(\theta)sin(\phi)}{\rho}= cos(\theta)sin(\phi)$$
and similarly for the others.