1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Deriving laplacian in spherical coordinates

  1. Nov 17, 2006 #1
    Hey...

    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
  5. Nov 19, 2006 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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:
    [tex]\frac{\partial\vec{i}_{r}}{\partial{r}}=\frac{\partial\vec{i}_{\phi}}{\partial{r}}=\frac{\partial\vec{i}_{\theta}}{\partial{r}}=\vec{0}[/tex]
    [tex]\frac{\partial\vec{i}_{\theta}}{\partial\phi}=\vec{0},\frac{\partial\vec{i}_{\phi}}{\partial\phi}=-\vec{i}_{r}[/tex]
    [tex]\frac{\partial\vec{i}_{\phi}}{\partial\theta}=\cos\phi\vec{i}_{\theta},\frac{\partial\vec{i}_{\theta}}{\partial\theta}=-\sin\phi\vec{i}_{r}-\cos\phi\vec{i}_{\phi}[/tex]

    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:
    [tex]\nabla^{2}=\nabla\cdot\nabla=\vec{i}_{r}\cdot\frac{\partial\nabla}{\partial{r}}+\frac{1}{r}\vec{i}_{\phi}\cdot\frac{\partial\nabla}{\partial\phi}+\frac{1}{r\sin\phi}\vec{i}_{\theta}\cdot\frac{\partial\nabla}{\partial\theta}[/tex]
     
    Last edited: Nov 19, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?