I wanted to calculate the Laplacian operator $\Delta$ based on $$\nabla=\left(\frac{\partial f}{\partial r},\frac{1}{r}\frac{\partial f}{\partial \theta},\frac{1}{r\sin(\theta)}\frac{\partial f}{\partial \phi}\right)$$
My calculation was: $$\nabla \cdot \nabla=\left(\frac{\partial f}{\partial r},\frac{1}{r}\frac{\partial f}{\partial \theta},\frac{1}{r\sin(\theta)}\frac{\partial f}{\partial \phi}\right)\cdot \begin{pmatrix}\frac{\partial f}{\partial r}\\ \frac{1}{r}\frac{\partial f}{\partial \theta}\\ \frac{1}{r\sin(\theta)}\frac{\partial f}{\partial \phi}\end{pmatrix} \\ = \frac{\partial^2 f}{\partial^2 r}+\frac{1}{r^2}\frac{\partial^2 f}{\partial^2 \theta}+\frac{1}{r^2\sin^2{\theta}}\frac{\partial^2 f}{\partial^2 \phi}(i) $$
but according to sources online the $\Delta$ is given by: $$ \Delta=\nabla^2=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial f}{\partial r}\right)+\frac{1}{r^2\sin(\theta)}\frac{\partial }{\partial \theta}(\sin(\theta)\frac{\partial f }{\partial \theta})+\frac{1}{r^2\sin^2{\theta}}\frac{\partial^2 f}{\partial^2 \phi}(ii)$$
I don't understand, why that is the case. Why ist the simple calculation in (i) wrong?
On the first glace I thought that I just need to expand equation (i) to reach equality to equation (ii), but that is obviously not possible.
$\Delta = \nabla\cdot\nabla$ does not mean that the Laplacian is the dot product of $\nabla$ with itself, it means $\Delta f = \nabla \cdot \nabla f$. I.e. the Laplacian is the divergence applied after the gradient, so you should take an arbitrary function and calculate its gradient and its divergence afterwards.