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Homework Help: Deriving laplacian in spherical coordinates

  1. Nov 17, 2006 #1

    Could someone help me out with deriving the LaPlacian in spherical coordinates? I tried using the chain rule but it just isn't working out well.. any sort of hint would be appriciated. :)

    [tex] \nabla^2 = \frac{1}{r^2} [ \frac{\partial}{\partial r} ( r^2 \frac{\partial}{\partial r} ) + \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} ( \sin \theta \frac{\partial}{\partial \theta\ ) + \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} ] [/tex]
    Last edited: Nov 17, 2006
  2. jcsd
  3. Nov 17, 2006 #2
    I want to understand the derivation too. I understand the gradient in spherical coordinate, but the divergence formula bothers me. I got some info from some website saying that one can find a basis of vectors where its divergence is zero then take the divergence of the function based on those vectors....
  4. Nov 19, 2006 #3
    Last edited by a moderator: Apr 22, 2017
  5. Nov 19, 2006 #4


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    That's one way of doing it.
    Here's another one:
    [tex]\vec{i}_{r}=\sin\phi\cos\theta\vec{i}+\sin\phi\sin\theta\vec{j}+\cos\phi\vec{k}, \vec{i}_\phi}=\frac{\partial\vec{i}_{r}}{\partial\phi}=\cos\phi\cos\theta\vec{i}+\cos\phi\sin\theta\vec{j}-\sin\phi\vec{k},\vec{i}_{\theta}=\frac{1}{\sin\phi}\frac{\partial\vec{i}_{r}}{\partial\theta}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/tex]
    along with:

    Now, given these relations, along with the expression for the gradient in spherical coordinates [tex]\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\phi}\frac{1}{r}\frac{\partial}{\partial\phi}+\vec{i}_{\theta}\frac{1}{r\sin\phi}\frac{\partial}{\partial\theta}[/tex], we may easily derive the expression for the Laplacian by differentiating, and performing the dot products:
    Last edited: Nov 19, 2006
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