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:

\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

 

[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<br />
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,

 

[igorronaldo]

Now clear, thanks.