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!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Laplace Operator
  1. Laplace Operator (Replies: 2)

  2. Laplace operator help (Replies: 3)

  3. Laplace Operator (Replies: 3)

Loading...