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Gradient in spherical coordinates problem

  1. Jul 27, 2008 #1

    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:

    [tex] \Phi = \Phi(r, \theta, \phi) [/tex]

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

    [tex] 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}[/tex]

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

    [tex] 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}[/tex]

    I would be thank you for helping :)


    Sorry for my bad english. I will practice and learn grammar for better english in the future ;)
  2. jcsd
  3. Jul 27, 2008 #2


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    In Cartesian coordinates
    [tex]\grad \phi= \frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}+ \frac{\partial \phi}{\partial z}\vec{k}[/tex]
    Now you have to use the chain rule to convert those derivatives to spherical coordinates:
    [tex]\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}[/tex]
    It's tedious but doable.
  4. Jul 27, 2008 #3
    This is what I don't saw. Thank you :)
  5. Aug 9, 2008 #4
    you can define the gradient operator such that :

    [tex] d \Phi = \left< grad \Phi , d\vec{r}\right > [/tex]

    knowing that in spherical coordinates :

    [tex] 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[/tex]

    then you should find what you want.
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