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Differentiation in spherical coordinates.

  1. Dec 11, 2013 #1
    1) If [itex]u(r,\theta,\phi)=\frac{1}{r}[/itex], is [itex]\frac{\partial{u}}{\partial {\theta}}=\frac{\partial{u}}{\partial {\phi}}=0[/itex] because [itex]u[/itex] is independent of [itex]\theta[/itex] and[itex] \;\phi[/itex]?

    2) If [itex]u(r,\theta,\phi)=\frac{1}{r}[/itex], is:
    [tex]\nabla^2u(r,\theta,\phi)=\frac{\partial^2{u}}{\partial {r}^2}+\frac{2}{r}\frac{\partial{u}}{\partial {r}}+\frac{1}{r^2}\frac{\partial^2{u}}{\partial {\theta}^2}+\frac{\cot\theta}{r^2}\frac{\partial{u}}{\partial {\theta}}+\frac{1}{r^2\sin\theta}\frac{\partial^2{u}}{\partial {\phi}^2}=\frac{\partial^2{u}}{\partial {r}^2}+\frac{2}{r}\frac{\partial{u}}{\partial {r}}=0[/tex]

    Because [itex]\frac{\partial{u}}{\partial {\theta}}=\frac{\partial{u}}{\partial {\phi}}=0[/itex].

    Thanks
     
  2. jcsd
  3. Dec 11, 2013 #2

    jasonRF

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    Yep.
    Hint: what is the electrostatic potential of a point charge? What is the charge density of a point charge?

    EDIT: hint: Poisson's equation

    jason
     
  4. Dec 11, 2013 #3

    jasonRF

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    I guess I should add: your result for question 2 is completely correct as long as r is not zero. But for the kinds of applications I know you care about you will want to care about what happens at zero.
     
  5. Dec 11, 2013 #4
    Thank you Jason, I know I can count on you. Yes, I forgot to specify [itex]r>0[/itex].
     
  6. Dec 11, 2013 #5

    jasonRF

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    I didn't really think about whether you stated r>0; I am just thinking about how [itex]\nabla^2 \frac{1}{r}[/itex] is proportional to [itex]\delta(r)[/itex], which is of course zero away from r=0, as you say. But for E&M the delta function is key.

    Jason
     
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