How to deduct the gradient in spherical coordinates?

1. Jul 30, 2013

igorronaldo

2. Jul 30, 2013

vanhees71

The one given in Wikipedia is correct, or what is your question?

3. Jul 30, 2013

yungman

Should be this one:

$$\nabla f(r, \theta, \phi) = \frac{\partial f}{\partial r}\mathbf{e}_r+ \frac{1}{r}\frac{\partial f}{\partial \theta}\mathbf{e}_\theta+ \frac{1}{r \sin\theta}\frac{\partial f}{\partial \phi}\mathbf{e}_\phi$$

4. Jul 30, 2013

vanhees71

That's correct. Maybe you want to know, how to derive it?

The point is to write
$$\mathrm{d} \phi=\mathrm{d} \vec{r} \cdot \vec{\nabla} \phi=\mathrm{d} r \partial_r \phi + \mathrm{d} \vartheta \partial_{\vartheta} \phi + \mathrm{d} \varphi \partial_{\varphi}\phi$$
in terms of the normalized coordinate basis $(\vec{e}_r,\vec{e}_{\vartheta},\vec{e}_{\varphi})$. The term from the variation of $r$ is
$$\mathrm{d} r \frac{\partial \vec{r}}{\partial r} \cdot \vec{\nabla} \phi=\mathrm{d} r \vec{e}_r \cdot \vec{\nabla} \phi.$$
Comparing the coefficients of $\mathrm{d} r$ gives
$$\vec{e}_r \cdot \vec{\nabla} \phi=\partial_r \phi.$$
For the $\vartheta$ component
$$\mathrm{d} \vartheta \frac{\partial \vec{r}}{\partial \vartheta} \cdot \vec{\nabla} \phi = r \vec{e}_{\vartheta} \cdot \vec{\nabla} \phi, \;\Rightarrow \; \vec{e}_{\vartheta} \cdot \vec{\nabla} \phi=\frac{1}{r} \partial_{\vartheta} \phi,$$
and for $\mathrm{d} \varphi$
$$\mathrm{d} \varphi \frac{\partial \vec{r}}{\partial \varphi} \cdot \vec{\nabla} \phi = r \sin \vartheta \vec{e}_{\vartheta} \cdot \vec{\nabla} \phi, \; \Rightarrow \; \vec{e}_{\varphi} \cdot \vec{\nabla} \phi=\frac{1}{r \sin \varphi} \partial_{\vartheta} \phi,$$

5. Jul 31, 2013

igorronaldo

Now clear, thanks.