Differentiation in spherical coordinates.

[yungman]

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

 

[jasonRF]

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]?

Yep.

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

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 [itex]r>0[/itex].

 

[jasonRF]

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