I am attempting to understand how to derive $\vec{\nabla}.\vec{F}$ in both cylindrical and spherical polar co-ordinates.

I understand that in a Cartesian system we say: $$\vec{\nabla}\vec{F}=\frac{\partial{F}}{\partial{x}}\hat{x}+\frac{\partial{F}}{\partial{y}}\hat{y}+\frac{\partial{F}}{\partial{z}}\hat{z}.$$ In principle, it would be straightforward to 'translate' the vector derivitive into $r,\theta,\phi$ notation by the chain rule: $$\frac{\partial{F}}{\partial{x}}=\frac{\partial{F}}{\partial{r}}\left(\frac{\partial{r}}{\partial{x}}\right)+\frac{\partial{F}}{\partial{\theta}}\left(\frac{\partial{\theta}}{\partial{x}}\right)+\frac{\partial{F}}{\partial{\phi}}\left(\frac{\partial{\phi}}{\partial{x}}\right).$$ However, I understand this would take a very long time to derive via this method. The text-book that I am using states the above (Introduction to Electrodynamics- Griffiths), but it then suddenly jumps to the following with no explanation: $$\vec{\nabla}\vec{F}=\frac{\partial{F}}{\partial{r}}\hat{r}+\frac{1}{r}\frac{\partial{F}}{\partial{\theta}}\hat{\theta}+\frac{1}{rsin\theta}\frac{\partial{F}}{\partial{\phi}}\hat{\phi}.$$

My intuition tells me (which is obviously wrong) that the above should simply be: $$\vec{\nabla}\vec{F}=\frac{\partial{F}}{\partial{r}}\hat{r}+\frac{\partial{F}}{\partial{\theta}}\hat{\theta}+\frac{\partial{F}}{\partial{\phi}}\hat{\phi}.$$ As the gradient is the rate of change of F within each 'dimension'. I vaguely remember my lecturer saying that we can derive it simply due to unit consistency, but I have no clue why I am wrong, and from here, how to get to divergence and even curl.

If anyone can let me know, that'd be great, I am assuming it is a reasonably similar answer for cylindrical polars also. Thanks in advance.

  • 1
    $\begingroup$ On that note of dimensional consistency you can immediately tell you at least need a factor of $r$ for every angle, otherwise a term like $\frac{\partial}{\partial \theta}$ is dimensionless while $\frac{\partial}{\partial r}$ is in terms of $L^{-1}$ $\endgroup$ – Triatticus Jan 21 '18 at 18:02
  • $\begingroup$ ok, that makes sense, but how is it determined to be $\frac{1}{r}$ for the $\theta$ component and $\frac{1}{rsin\theta}$ for the $\phi$ component? $\endgroup$ – George Jan 22 '18 at 12:21

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