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Homework Help: Laplace Operator

  1. Oct 7, 2005 #1
    I have read in an electrodynamics book that
    [tex]
    \int d^3r \frac{\Delta\Phi(r)}{r}=\int d^3r \Phi(r)}\Delta \frac{1}{r}
    [/tex]
    is possible through partial integration. But how?? Can some one help me on this by showing me the steps? Thanks
     
  2. jcsd
  3. Oct 7, 2005 #2

    Physics Monkey

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    Start with the following two identities:
    [tex] \int d^3 r \nabla \cdot \left( \frac{1}{r} \nabla \Phi \right) = \int d^3 r \left( \nabla \frac{1}{r} \cdot \nabla \Phi + \frac{1}{r} \Delta \Phi \right)[/tex]

    [tex] \int d^3 r \nabla \cdot \left( \Phi \nabla \frac{1}{r} \right) = \int d^3 r \left( \nabla \frac{1}{r} \cdot \nabla \Phi + \Phi \Delta \frac{1}{r}\right) [/tex]

    Now the left hand side of each equation can written as a surface integral which gives zero. This gives
    [tex]
    \int d^3 r \left( \nabla \frac{1}{r} \cdot \nabla \Phi + \frac{1}{r} \Delta \Phi \right) = 0
    [/tex]
    [tex]
    \int d^3 r \left( \nabla \frac{1}{r} \cdot \nabla \Phi + \Phi \Delta \frac{1}{r} \right) = 0
    [/tex]
    but since these two expressions have the same first term you can easily see that
    [tex]
    \int d^3 r \frac{1}{r} \Delta \Phi = \int d^3 r \Phi \Delta \frac{1}{r}
    [/tex]
     
  4. Oct 7, 2005 #3
    Thanks for your reply. Just one thing I am not able to see yet. How are these equal to zero?
    [tex]
    \int d^3 r \nabla \cdot \left( \frac{1}{r} \nabla \Phi \right)=\oint dA \frac{1}{r} \nabla \Phi=0?
    [/tex]
    [tex]
    \int d^3 r \nabla \cdot \left( \Phi \nabla \frac{1}{r} \right)=\oint dA \Phi \nabla \frac{1}{r}=0?
    [/tex]
     
    Last edited: Oct 7, 2005
  5. Oct 7, 2005 #4

    Physics Monkey

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    You integrate on a surface at infinity, and with very weak assumptions about the fall off of [tex] \Phi [/tex] with distance, you can show those area integrals go to zero.
     
  6. Oct 7, 2005 #5
    That's clever. Thanks!
     
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