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?
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- $\begingroup$ No $\hat \phi$ on the second expression. Also $\frac{\partial f}{\partial f}$ on the real laplacian should probably be something else. $\endgroup$mathreadler– mathreadler2017-02-20 11:42:25 +00:00Commented Feb 20, 2017 at 11:42
- $\begingroup$ The second nabla works on the first with the product and chain rules of differentiation so you can't just group them up like that. $\endgroup$mathreadler– mathreadler2017-02-20 11:47:46 +00:00Commented Feb 20, 2017 at 11:47
- $\begingroup$ @mathreadler That I can see. But what formal steps should I take to derive the right expression? $\endgroup$user55789– user557892017-02-20 11:51:49 +00:00Commented Feb 20, 2017 at 11:51
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2 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!
- $\begingroup$ Thank you. Could you clarify on why the notation $\nabla\cdot{\bf G}$ is allowed? This is clearly not the "inner product of $\nabla$ and $\bf G$"??? $\endgroup$user55789– user557892017-02-20 11:46:25 +00:00Commented Feb 20, 2017 at 11:46
- $\begingroup$ It is miss conceptual really. This is not an inner product, as $\nabla$ is not a vector in a sense. The proper way to write is $div[\bf{G}]$. I think this notation is popular because of some educative purposes, say to emphasize that the result is scalar.... $\endgroup$Kiryl Pesotski– Kiryl Pesotski2017-02-20 11:51:00 +00:00Commented Feb 20, 2017 at 11:51