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Vector Calculus - Laplacian on Scalar Field

  1. Apr 30, 2014 #1
    A scalar field [itex]\psi[/itex] is dependent only on the distance [itex]r = \sqrt{x^{2} + y^{2} + z^{2}}[/itex] from the origin.

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    [itex] \partial_{x}^{2}\psi = \left(\frac{1}{r} - \frac{x^{2}}{r^{3}}\right)\frac{d\psi}{dr} + \frac{x^{2}}{r^{2}}\frac{d^{2}\psi}{dr^{2}}[/itex]

    I've used the chain and product rules so far, but I'm unsure how to approach the problem to move onwards from this point. My working is as follows:

    [itex]\frac{\partial\psi}{\partial x} = \frac{\partial r}{\partial x}\frac{d\psi}{dr}[/itex] where the identity [itex] \frac{\partial r}{\partial x} = \frac{x}{r}[/itex] is given.
    [itex]\frac{\partial\psi}{\partial x} = \frac{x}{r}\frac{d\psi}{dr} (1) [/itex]
    [itex]\frac{\partial^{2}\psi}{\partial^{2}x} = \frac{\partial}{\partial x}\left(\frac{x}{r}\frac{d\psi}{dr}\right)[/itex]
    [itex]= \left(\frac{\partial}{\partial x}\frac{x}{r}\right)\left(\frac{d\psi}{dr}\right) + \left(\frac{\partial}{\partial x}\frac{d\psi}{dr}\right)\left(\frac{x}{r}\right)[/itex]

    At this point, I can solve the left term to give me:
    [itex]-\frac{x^{2}}{r^{3}}\frac{d\psi}{dr}[/itex]

    But I don't know how to properly manipulate the second term. I've tried re-arranging and substituting in (1), but this didn't work out. Any hints on how to proceed would be greatly appreciated.

    Thanks.
     
  2. jcsd
  3. Apr 30, 2014 #2

    LCKurtz

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    Good so far.

    You might find it easier to write it as ##\frac \partial {\partial x} (x\cdot \frac{\psi'(r} r)##. Now you have a product rule and when you calculate ##\frac \partial {\partial x}## of the second term (quotient rule), just use the chain rule like you did in the beginning.
     
  4. May 1, 2014 #3
    Aah, thanks alot for your help, i've got it now. (:
     
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