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Spherical Coordinates

  1. Apr 20, 2010 #1
    How do I find [tex]\nabla[/tex] in Spherical Coordinates. Please help.
  2. jcsd
  3. Apr 20, 2010 #2


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  4. Apr 21, 2010 #3
    How do I go about doing it from scratch.
    How do I find i, j, & k from the definition of [tex]\nabla[/tex]
    Last edited: Apr 21, 2010
  5. Apr 21, 2010 #4
    [tex]\nabla[/tex] = del/del(x) i + del/del(y) j + del/del(z) k
    I found del/del(x), del/del)(y), del/del(z) but how do I find i, j, k. Help please.
  6. Apr 22, 2010 #5


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    Have you looked at the site tiny-tim gives? No one is going to go through the whole thing just for you! It's not terribly deep but very tedious!

    Here's a start only:

    Since [itex]\nabla u= \frac{\partial u}{\partial x}\vec{i}+ \frac{\partial u}{\partial y}\vec{j}+ \frac{\partial u}{\partial z}\vec{k}[/itex]
    so you need to use the chain rule

    [tex]\frac{\partial u}{\partial x}= \frac{\partial u}{\partial \rho}\frac{\partial \rho}{\partial x}+ \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}+ \frac{\partial u}{\partial \phi}\frac{\partial \phi}{\partial x}[/tex]

    Since [itex]\rho= (x^2+ y^2+ z^2)^{1/2}[/itex],
    [tex]\frac{\partial \rho}{\partial x}= (1/2)(x^2+ y^2+ z^2)^{-1/2}(2x}= \frac{\rho cos(\theta)sin(\phi)}{\rho}= cos(\theta)sin(\phi)[/tex]
    and similarly for the others.
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