# Deriving laplacian in spherical coordinates

1. Nov 17, 2006

### mooshasta

Hey...

Could someone help me out with deriving the LaPlacian in spherical coordinates? I tried using the chain rule but it just isn't working out well.. any sort of hint would be appriciated. :)

$$\nabla^2 = \frac{1}{r^2} [ \frac{\partial}{\partial r} ( r^2 \frac{\partial}{\partial r} ) + \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} ( \sin \theta \frac{\partial}{\partial \theta\ ) + \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} ]$$

Last edited: Nov 17, 2006
2. Nov 17, 2006

### tim_lou

I want to understand the derivation too. I understand the gradient in spherical coordinate, but the divergence formula bothers me. I got some info from some website saying that one can find a basis of vectors where its divergence is zero then take the divergence of the function based on those vectors....

3. Nov 19, 2006

### mooshasta

4. Nov 19, 2006

### arildno

That's one way of doing it.
Here's another one:
$$\vec{i}_{r}=\sin\phi\cos\theta\vec{i}+\sin\phi\sin\theta\vec{j}+\cos\phi\vec{k}, \vec{i}_\phi}=\frac{\partial\vec{i}_{r}}{\partial\phi}=\cos\phi\cos\theta\vec{i}+\cos\phi\sin\theta\vec{j}-\sin\phi\vec{k},\vec{i}_{\theta}=\frac{1}{\sin\phi}\frac{\partial\vec{i}_{r}}{\partial\theta}=-\sin\theta\vec{i}+\cos\theta\vec{j}$$
along with:
$$\frac{\partial\vec{i}_{r}}{\partial{r}}=\frac{\partial\vec{i}_{\phi}}{\partial{r}}=\frac{\partial\vec{i}_{\theta}}{\partial{r}}=\vec{0}$$
$$\frac{\partial\vec{i}_{\theta}}{\partial\phi}=\vec{0},\frac{\partial\vec{i}_{\phi}}{\partial\phi}=-\vec{i}_{r}$$
$$\frac{\partial\vec{i}_{\phi}}{\partial\theta}=\cos\phi\vec{i}_{\theta},\frac{\partial\vec{i}_{\theta}}{\partial\theta}=-\sin\phi\vec{i}_{r}-\cos\phi\vec{i}_{\phi}$$

Now, given these relations, along with the expression for the gradient in spherical coordinates $$\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\phi}\frac{1}{r}\frac{\partial}{\partial\phi}+\vec{i}_{\theta}\frac{1}{r\sin\phi}\frac{\partial}{\partial\theta}$$, we may easily derive the expression for the Laplacian by differentiating, and performing the dot products:
$$\nabla^{2}=\nabla\cdot\nabla=\vec{i}_{r}\cdot\frac{\partial\nabla}{\partial{r}}+\frac{1}{r}\vec{i}_{\phi}\cdot\frac{\partial\nabla}{\partial\phi}+\frac{1}{r\sin\phi}\vec{i}_{\theta}\cdot\frac{\partial\nabla}{\partial\theta}$$

Last edited: Nov 19, 2006