How to deduct the gradient in spherical coordinates?

[igorronaldo]

http://en.wikipedia.org/wiki/Gradient#Cylindrical_and_spherical_coordinates

which formula do we apply to get the gradient in spherical coordinates?

 

[vanhees71]

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

 

[yungman]

Should be this one:

[tex] \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 [/tex]

 

[vanhees71]

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

The point is to write
[tex]\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
[/tex]
in terms of the normalized coordinate basis [itex](\vec{e}_r,\vec{e}_{\vartheta},\vec{e}_{\varphi})[/itex]. The term from the variation of [itex]r[/itex] is
[tex]\mathrm{d} r \frac{\partial \vec{r}}{\partial r} \cdot \vec{\nabla} \phi=\mathrm{d} r \vec{e}_r \cdot \vec{\nabla} \phi.[/tex]
Comparing the coefficients of [itex]\mathrm{d} r[/itex] gives
[tex]\vec{e}_r \cdot \vec{\nabla} \phi=\partial_r \phi.[/tex]
For the [itex]\vartheta[/itex] component
[tex]\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,[/tex]
and for [itex]\mathrm{d} \varphi[/itex]
[tex]\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,[/tex]

 

[igorronaldo]

Now clear, thanks.