1. Jul 27, 2008

### Theoretiker

Hello,

I need help. The topic is a gradient in spherical coordinates. In cartesian it is clear but in spherical coordinates I have two terms which I don't understand from where they come.

Okay, I have a scalar field in spherical coordinates:

$$\Phi = \Phi(r, \theta, \phi)$$

I thought that this is the gradient but it is wrong and I don't know why :(

$$grad \Phi = \frac{\partial \phi}{\partial r} \vec{e}_{r} + \frac{\partial \phi}{\partial \theta} \vec{e}_{\theta} + \frac{\partial \phi}{\partial \phi} \vec{e}_{\phi}$$

My mathbook tells me that this is the gradient in spherical coordinates but I don't understand the terms $$\frac{1}{r}$$ and $$\frac{1}{r \sin(\theta)}$$

$$grad \Phi = \frac{\partial \phi}{\partial r} \vec{e}_{r} + \frac{1}{r} ~ \frac{\partial \phi}{\partial \theta} \vec{e}_{\theta} + \frac{1}{r \sin(\theta)} ~ \frac{\partial \phi}{\partial \phi} \vec{e}_{\phi}$$

I would be thank you for helping :)

greetings

P.S.
Sorry for my bad english. I will practice and learn grammar for better english in the future ;)

2. Jul 27, 2008

### HallsofIvy

Staff Emeritus
In Cartesian coordinates
$$\grad \phi= \frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}+ \frac{\partial \phi}{\partial z}\vec{k}$$
Now you have to use the chain rule to convert those derivatives to spherical coordinates:
$$\frac{\partial\Phi}{\partial x}= \frac{\partial \Phi}{\partial \rho}\frac{\partial \rho}{\partial x}+ \frac{\partial \Phi}{\partial \theta}\frac{\partial \theta}{\partial x}+ \frac{\partial \Phi}{\partial \phi}\frac{\partial \phi}{\partial x}$$
It's tedious but doable.

3. Jul 27, 2008

### Theoretiker

This is what I don't saw. Thank you :)

4. Aug 9, 2008

### deiki

you can define the gradient operator such that :

$$d \Phi = \left< grad \Phi , d\vec{r}\right >$$

knowing that in spherical coordinates :

$$d\vec{r}\right = \vec{e}_{r} dr + \vec{e}_{\theta} \cdot r d\theta + \vec{e}_{\phi} \cdot r \cdot sin(\theta) d\phi$$

then you should find what you want.