# Differentiation in spherical coordinates.

1. Dec 11, 2013

### 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

2. Dec 11, 2013

### jasonRF

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

3. Dec 11, 2013

### 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.

4. Dec 11, 2013

### yungman

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

5. Dec 11, 2013

### 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