- #1

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I cannot see how this is? I like to be solid in the math I use for a derivation, so this would really help if someone could clear this up for me. Thanks!

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- Thread starter Master J
- Start date

- #1

- 226

- 0

I cannot see how this is? I like to be solid in the math I use for a derivation, so this would really help if someone could clear this up for me. Thanks!

- #2

HallsofIvy

Science Advisor

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[tex]\nabla^2 u= \frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)+ \frac{1}{r^2 sin^2(\phi)}\frac{\partial^2 u}{\partial\theta^2}+ \frac{1}{r^2 sin(\phi)}\frac{\partial }{\partial\phi}\left(sin(\phi)\frac{\partial u}{\partial \phi}\right)[/tex].

(See http://mathworld.wolfram.com/SphericalCoordinates.html)

With u= 1/r, the derivatives with respect to [itex]\theta[/itex] and [itex]\phi[/itex] will be 0, of course.

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