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How to deduct the gradient in spherical coordinates?

  1. Jul 30, 2013 #1
  2. jcsd
  3. Jul 30, 2013 #2

    vanhees71

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    The one given in Wikipedia is correct, or what is your question?
     
  4. Jul 30, 2013 #3
    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]
     
  5. Jul 30, 2013 #4

    vanhees71

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    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]
     
  6. Jul 31, 2013 #5
    Now clear, thanks.
     
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