Differentiation in spherical coordinates.

[yungman]

1) If u(r,\theta,\phi)=\frac{1}{r}, is \frac{\partial{u}}{\partial {\theta}}=\frac{\partial{u}}{\partial {\phi}}=0 because u is independent of \theta and\;\phi?

2) If u(r,\theta,\phi)=\frac{1}{r}, is:
\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

Because \frac{\partial{u}}{\partial {\theta}}=\frac{\partial{u}}{\partial {\phi}}=0.

Thanks

 

[jasonRF]

1) If u(r,\theta,\phi)=\frac{1}{r}, is \frac{\partial{u}}{\partial {\theta}}=\frac{\partial{u}}{\partial {\phi}}=0 because u is independent of \theta and\;\phi?

Yep.

2) If u(r,\theta,\phi)=\frac{1}{r}, is:
\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

Because \frac{\partial{u}}{\partial {\theta}}=\frac{\partial{u}}{\partial {\phi}}=0.

Thanks

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

 

[jasonRF]

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.

 

[yungman]

Thank you Jason, I know I can count on you. Yes, I forgot to specify r>0.

 

[jasonRF]

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

Jason