Ytukyg

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:

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

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?

The Laplacian is the divergence of the gradient. And the divergence in spherical coordinates is:
$$nablacdot mathbf A = {1 over r^2}{partial left( r^2 A_r right) over partial r}
+ {1 over rsintheta}{partial over partial theta} left( A_thetasintheta right)
+ {1 over rsintheta}{partial A_varphi over partial varphi}$$

Now substitute the $nablaPhi$ that you already have for $mathbf A$.

That leaves the question how we got that formula for the divergence. You can find the derivation here: Nabla in spherical coordinates. It explains it in terms of how divergence is defined.