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Radial component of del^2 in spherical coordinates? (again)

  1. Dec 22, 2008 #1
    I'm doing a question on a 3D isotropic harmonic oscillator. At one point I need to find write the radial component of del^2.

    Lecturer has written

    [tex]

    \frac{1}{r^{2}} \frac{d}{dr} \left( r^{2} \frac{d}{dr} \right)

    [/tex]

    where the del^2 used to be in the set of equations.

    Am I correct in saying the radial part of in spherical polar coordinates is just dr. Then del^2 would be dr^2? Well I'm not?

    I had a look at http://hyperphysics.phy-astr.gsu.edu/hbase/sphc.html but it didn't explain anything more for me.

    PS. I asked this question earlier but internet broke, (thanks virgin media), and I don't understand the answer given... https://www.physicsforums.com/showthread.php?p=2006618#post2006618
     
  2. jcsd
  3. Dec 22, 2008 #2

    tiny-tim

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    Hi philip041! :smile:

    (have a theta: θ and a phi: φ and a curly d: ∂ :wink:)

    isotropic means dθ = dφ = 0

    so in http://hyperphysics.phy-astr.gsu.edu/hbase/sphc.html at http://hyperphysics.phy-astr.gsu.edu/hbase/lapl.html#c2, the only non-zero terms in the last line are ∂2f/∂r2 + 2/r ∂f/∂r,

    which is the same as 1/r2 d/dr(r2d/dr) :smile:
     
  4. Dec 22, 2008 #3
    Thanks,

    With ∂^2f/∂r^2 + 2/r ∂f/∂r how come it doesn't go to ∂^2/∂r^2 -1/r^2?

    Sorry I don't understand how you get to that final line you wrote.
     
  5. Dec 22, 2008 #4

    tiny-tim

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    product rule

    uhhh? how did you get -1/r^2? :confused:

    Hint: use the product rule to find d/dr(r2df/dr) :smile:
     
  6. Dec 22, 2008 #5
    sweet. i also found an explanation of this in zetilli appendix b of quantum mechanics if anyone should ever search for this thread...
     
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