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Gradient in spherical coordinates

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Given the gradient

    del = x-hat d/dx + y-hat d/dy + z-hat d/dz

    in rectangular coordinates, how would you write that in spherical coordinates. I can transform the derivatives into spherical coordinates. But then I need to express the rectangular basis vectors in terms of the spherical basis vectors. Is that possible?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 17, 2007 #2
    Anyone?
     
  4. Dec 17, 2007 #3

    quasar987

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    Sure it's possible!

    If you're working on Zwiebach, I'm sure you have a mechanics texbook lying around somewhere! Or a Griffith's E&M ! It's done in there if you ever get tired of looking. But it's not very difficult.
     
  5. Dec 19, 2007 #4
    del = x-hat d/dx + y-hat d/dy + z-hat d/dz

    I am sorry. This is really bothering me. Can someone please walk me through how you would convert this to spherical coordinates? All of my books just say something like "after tedious calculations," and then show the gradient and the Laplacian in spherical coordinates.

    You do not need to show me the tedious calculations. Just tell me what to do!
     
  6. Dec 19, 2007 #5

    arildno

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    Okay, so you wish to represent rectangular coordinates in terms of spherical ones.

    We have:
    [tex]\vec{i}_{r}=\sin\phi(\cos\theta\vec{i}+\sin\theta\vec{j})+\cos\phi\vec{k}[/tex]
    [tex]\vec{i}_{\phi}=\cos\phi(\cos\theta\vec{i}+\sin\theta\vec{j})-\sin\phi\vec{k}[/tex]
    [tex]\vec{i}_{\theta}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/tex]
    Now,
    Compute the following quantity:
    [tex]\sin\phi\cos\theta\vec{i}_{r}+\cos\phi\cos\theta\vec{i}_{\phi}-\sin\theta\vec{i}_{\theta}[/tex]
    see if that helps you
     
  7. Dec 19, 2007 #6
    That equals [tex] \vec{i}[/tex].
    So, the best way to find the expressions for [tex] \vec{j}[/tex] and [tex] \vec{k}[/tex] in terms of [tex] \vec{i_r},\vec{i_{\phi}},\vec{i_{\theta}}[/tex] is probably by inverting that matrix, right?
     
  8. Dec 19, 2007 #7

    arildno

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    Sure, but you should be able to do that without a full row reduction procedure.

    Remember that the square of cosine plus the sine square equals 1.
    What should you therefore multiply your vectors with to get, say, k?
     
  9. Dec 19, 2007 #8
    OK. Got the gradient. Now, to calculate the Laplacien, I just dot that with itself. But in the equation of the Laplacian, it seems like there are cross-terms between the coefficients of different spherical basis vectors. Does that make sense. Why do you not just square the coefficients of [tex] \vec{i_r},\vec{i_{\phi}},\vec{i_{\theta}}[/tex] in the gradient and add them up? Since [tex] \vec{i_r},\vec{i_{\phi}},\vec{i_{\theta}}[/tex] I thought you could do that.
     
  10. Dec 20, 2007 #9
    Does my question make sense to anyone?
     
  11. Dec 20, 2007 #10
    Please?
     
  12. Dec 20, 2007 #11

    Dick

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    Ok. Sure. You don't dot a gradient with itself to get the laplacian. You take the divergence of the gradient to get the laplacian.
     
  13. Dec 21, 2007 #12

    arildno

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    Sure you can "dot" the gradient with itself to get the Laplacian operator, as long as you do so PROPERLY.

    The gradient in spherical coordinates is:
    [tex]\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\phi}\frac{\partial}{r\partial\phi}+\vec{i}_{\theta}\frac{\partial}{r\sin\phi\partial\theta}[/tex]
    The Laplacian is now gained as follows:
    [tex]\nabla^{2}=\vec{i}_{r}\cdot\frac{\partial\nabla}{\partial{r}}+\vec{i}_{\phi}\cdot\frac{\partial\nabla}{r\partial\phi}+\vec{i}_{\theta}\cdot\frac{\partial\nabla}{r\sin\phi\partial\theta}[/tex]
    Remember that unit vectors are varying functions of the angles.
     
    Last edited: Dec 21, 2007
  14. Dec 23, 2007 #13
    I see. Thanks.
     
  15. Feb 2, 2010 #14
    could anyone help me in preparing my presentation?
    who introduce the spherical coordinates? its back ground?
     
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