I'm reading through this page and though I haven't checked all the computations I got the idea. I was wondering if the same result could be derived in a similar fascion
The Jacobian that transform rectangular into spherical coordinates is given by
$$ J = \frac{\partial(r,\theta,\phi)}{\partial(x,y,z)} = \begin{pmatrix} \sin(\theta)\cos(\phi) & \sin(\theta)\sin(\phi) & \cos(\theta) \\ \frac{1}{r}\cos(\theta)\cos(\phi) & \frac{1}{r}\cos(\theta)\sin(\phi) & -\frac{1}{r}\sin(\theta) \\ -\frac{1}{r}\frac{\sin(\phi)}{\sin(\theta)} & \frac{1}{r}\frac{\cos(\phi)}{\sin(\theta)} & 0 \end{pmatrix} $$
Therefore we have
$$ \nabla_c f = \left( J \cdot \nabla_s\right) f \Rightarrow \nabla_c = \left( J \cdot \nabla_s\right) $$
where $$ \nabla_c = \begin{pmatrix} \frac{\partial}{\partial y} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z}\end{pmatrix}, \nabla_s = \begin{pmatrix} \frac{\partial}{\partial r} \\ \frac{\partial}{\partial \theta} \\ \frac{\partial}{\partial \phi}\end{pmatrix}. $$
Therefore we have
$$ \left(\nabla_c^T \cdot \nabla_c \right) = \left( J \cdot \nabla_s\right)^T \cdot \left( J \cdot \nabla_s\right) = \left(\nabla_s^T \cdot J^T \cdot J \cdot \nabla_s \right) \Rightarrow \left(\nabla_c^T \cdot \nabla_c \right) = \left(\nabla_s^T \cdot J^T \cdot J \cdot \nabla_s \right) $$
Hence we have
$$ \nabla_c^2 f = \left(\nabla_s^T \cdot J^T \cdot J \cdot \nabla_s \right) f $$
Is there anything I could do to use the last equation to work out the laplacian in spherical coordinates? as it is usually stated I mean.