# How is the Laplacian in spherical derived?

Suppose $$\Phi$$ is a function of $$r, \theta$$ and $$\phi$$. If I want to derive the Laplacian for this function, I would assume that..

$$\nabla ^2 \Phi = \nabla \cdot \nabla \Phi$$

And as, in spherical:

$$\nabla = \frac{\partial}{\partial r} \hat r + \frac{1}{r} \frac{\partial}{\partial \theta} \hat \theta + \frac{1}{r \sin \ \theta} \frac{\partial}{\partial \phi} \hat \phi$$

Which implies

$$\nabla \Phi= \frac{\partial \Phi}{\partial r} \hat r + \frac{1}{r} \frac{\partial \Phi}{\partial \theta} \hat \theta + \frac{1}{r \sin \ \theta} \frac{\partial \Phi}{\partial \phi} \hat \phi$$

Thus, the Laplacian would seem to me to be:

$$\large\begin{bmatrix} \frac{\partial}{\partial r} \\ \frac{1}{r} \frac{\partial}{\partial \theta} \\ \frac{1}{r \sin \ \theta} \frac{\partial}{\partial \phi} \\ \end{bmatrix} \cdot \begin{bmatrix} \frac{\partial \Phi}{\partial r} \\ \frac{1}{r} \frac{\partial \Phi}{\partial \theta} \\ \frac{1}{r \sin \ \theta} \frac{\partial \Phi}{\partial \phi} \\ \end{bmatrix}$$

Which will not give me the form for the Laplacian in spherical coordinates in my lecture notes or the internet. Where do I have it wrong?

• This is a point in which the notation $\nabla\cdot$ for the divergence is slightly misleading. Don't interpret it as a true scalar product, it is not, just a mnemonic. Nov 25 '18 at 16:48
• Well... that is indeed misleading. What am I supposed to do instead? Nov 25 '18 at 16:52
• See youtube.com/watch?v=AOT719oYmt0 The issue is that when you take the derivatives of $\hat r, \hat\theta, \hat \phi$, as opposed to the Cartesian case, those are not zero Nov 25 '18 at 16:53

The Laplacian is the divergence of the gradient. And the divergence in spherical coordinates is: $$\nabla\cdot \mathbf A = {1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r} + {1 \over r\sin\theta}{\partial \over \partial \theta} \left( A_\theta\sin\theta \right) + {1 \over r\sin\theta}{\partial A_\varphi \over \partial \varphi}$$
Now substitute the $$\nabla\Phi$$ that you already have for $$\mathbf A$$.