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Where does Laplace's equation in spherical polars come from

  1. Apr 9, 2009 #1
    Where does Laplace's equation in spherical polars come from

    [itex]\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}=0[/itex]

    ???

    i can derive from scratch the expression for the laplacian in spherical polars but this bears no resemblance to the above, even if i decide to ignore the [itex]\phi[/itex] dependence?
     
  2. jcsd
  3. Apr 9, 2009 #2

    robphy

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    Re: Laplacian

    Those are in 2D-polar coordinates.
     
  4. Apr 9, 2009 #3
    Re: Laplacian

    hmm, im guessing that's meant to be easier but we haven't covered that in class (despite doing the 3D case) - could you possibly run over the derivation or post a link to a website (whenever i google it i only get the 3d case) please?
     
  5. Apr 9, 2009 #4

    robphy

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  6. Apr 9, 2009 #5

    Ben Niehoff

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    Re: Laplacian

    When deriving it from scratch, remember that in spherical coordinates, the unit vectors are not constants anymore!

    The gradient operator is

    [tex]\vec \nabla = \hat r \frac{\partial}{\partial r} + \hat \theta \frac1{r} \frac{\partial}{\partial \theta} + \hat \phi \frac1{r \sin \theta} \frac{\partial}{\partial \phi}[/tex]

    what you want to find is

    [tex]\vec \nabla \cdot \vec \nabla[/tex]

    but you must take into account the fact that the differential operators also act on the unit vectors.
     
  7. May 10, 2009 #6
    Re: Laplacian

    From my weblog
    http://buyanik.wordpress.com/2009/05/02/laplacian-in-spherical-coordinates/" [Broken]
     
    Last edited by a moderator: May 4, 2017
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