Laplacian as the divergence of the gradient - in spherical coordinates

Question

Say I have an arbitrary function $f(r,\theta,\phi)$. If I take the gradient in spherical coordinates it reads \begin{equation} \nabla f= \hat{\bf r}\frac{\partial f}{\partial r} + {\bf \hat{\theta}}\frac1r\frac{\partial f}{\partial\theta} + {\bf \hat{\phi}} \frac1{r\sin\theta}\frac{\partial f}{\partial\phi}. \end{equation} By definition, $\Delta f \equiv \nabla \cdot\nabla f$, so naively one would conclude: \begin{align} \Delta f &= \left[\hat{\bf r}\frac{\partial}{\partial r} + {\bf \hat{\theta}}\frac1r\frac{\partial}{\partial\theta} + {\bf \hat{\phi}} \frac1{r\sin\theta}\frac{\partial}{\partial\phi}\right] \bullet\left[\hat{\bf r}\frac{\partial f}{\partial r} + {\bf \hat{\theta}}\frac1r\frac{\partial f}{\partial\theta} + {\bf \hat{\phi}} \frac1{r\sin\theta}\frac{\partial f}{\partial\phi}\right]\\ &=\frac{\partial^2 f}{\partial r^2} + \frac1{r^2}\frac{\partial^2 f}{\partial\theta^2} + {\bf \hat{\phi}} \frac1{r^2\sin^2\theta}\frac{\partial^2 f}{\partial\phi^2}, \end{align} however this is incorrect: \begin{equation} \Delta f = \frac1{r^2}\frac{\partial}{\partial r}\left[r^2\frac{\partial f}{\partial f}\right] + \frac1{r^2\sin\theta}\frac{\partial}{\partial \theta}\left[\sin\theta\frac{\partial f}{\partial\theta}\right] + \frac1{r^2\sin^2\theta} \frac{\partial^2f}{\partial\phi}. \end{equation} Why does this not hold, i.e. what assumption has been omitted?

Answer 1

U have the incorrect expression for the divergence operator!!!! It reads as $$\nabla\cdot\bf{G}=\frac{1}{h_{1}h_{2}h_{3}}\Big(\frac{\partial(h_{2}h_{3}G_{1})}{\partial{x_{1}}}+\frac{\partial(h_{1}h_{3}G_{2})}{\partial{x_{2}}}+\frac{\partial(h_{2}h_{1}G_{3})}{\partial{x_{3}}}\Big)$$ Where $h_{i}=\sqrt{g_{ii}}$ are scale factors of the coordinate system. For spherical coordinates $h_{r}=1$, $h_{\theta}=r\sin\varphi$ and $h_{\varphi}=r$. However, your expression of the gradient is correct. Take the correct operator and you'll get the result!