If I have a vector $\vec{a}_{cart} = a\hat{z}$ in cartesian coordinates in spherical coordinates that vector would be $\vec{a}_{sph} = a\hat{r}$ with $\theta = 0$ and $\phi =0$ (I must specify the angles for it to be unique).

If I then take the material derivative, $\vec{a}\cdot\nabla\vec{b}$ where $\vec{b} = (b_{r},b_{\theta},b_{\phi})$, does my result (taken from a Wikipedia identity) have to be evaluated at $\theta = 0$ and $\phi =0$ too?


If $\theta =0$, then $\hat z=\hat r$. But in general $\hat z=\hat r \cos(\theta)-\hat \theta \sin(\theta)$. Hence, if $\vec a=a\hat z $, then we have $$\vec a=a\left(\hat r \cos(\theta)-\hat \theta \sin(\theta)\right) \tag 1$$

Applying $(1)$, we see that

$$\vec (a\cdot \nabla) \vec b=a\cos(\theta)\frac{\partial \vec b}{\partial r}-\frac ar \sin(\theta)\frac{\partial \vec b}{\partial \theta}$$

Writing $\vec b=\hat r b_r+\hat \theta b_\theta+\hat \phi b_\phi$ and using $\frac{\partial \hat r}{\partial \theta}=\hat \theta$ and $\frac{\partial \hat \theta}{\partial \theta}=-\hat r$, we find that

$$\begin{align} \vec (a\cdot \nabla) \vec b&=a\cos(\theta)\frac{\partial \vec b}{\partial r}-\frac ar \sin(\theta)\frac{\partial \vec b}{\partial \theta}\\\\ &=a\cos(\theta)\left(\hat r \frac{\partial b_r}{\partial r}+\hat \theta \frac{\partial b_\theta}{\partial r}+\hat \phi \frac{\partial b_\phi}{\partial r}\right)\\\\ &-\frac ar\sin(\theta)\left(\hat r \left(-b_\theta+\frac{\partial b_r}{\partial \theta}\right)+\hat \theta \left(b_r+\frac{\partial b_\theta}{\partial \theta}\right)+\hat \phi \frac{\partial b_\phi}{\partial \theta}\right)\\\\ &=a\hat r \left(\cos(\theta)\frac{\partial b_r}{\partial r}-\frac{\sin(\theta)}{r}\left(-b_\theta+\frac{\partial b_r}{\partial \theta}\right)\right)\\\\ &+a\hat \theta \left(\cos(\theta)\frac{\partial b_\theta}{\partial r}-\frac{\sin(\theta)}{r}\left(b_r+\frac{\partial b_\theta}{\partial \theta}\right)\right)\\\\ &+a\hat \phi \left(\cos(\theta)\frac{\partial b_\phi}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial b_\phi}{\partial \theta}\right)\tag 2 \end{align}$$

If we evaluate $(2)$ at $\theta=0$, we find that

$$(\vec a\cdot \nabla) \vec b=\frac{\partial \vec b}{\partial r}=\frac{\partial \vec b}{\partial z}$$

as expected!

  • $\begingroup$ Nice! One comment and a few questions. I think you dropped the $a$ in the second equation you wrote and that carried through the whole derivation. Should I think about the general case as the spherical and cartesian coordinates are not necessarily aligned? I can always choose them to be aligned right? Also, should the $\frac{\partial\vec{b}}{\partial r}$ in the last equation you wrote be evaluated at $\theta=0$? $\endgroup$ – Alex Apr 25 '17 at 5:15
  • $\begingroup$ Yes, I dropped the $a$. I'll edit. I'm not sure what you mean by "coordinates are not necessarily aligned?" The systems provide bases that result in different representations of the same vectors. Yes the derivatives of $\vec b$ have been tacitly evaluated at $\theta =0$ for the last equation following $(2)$. $\endgroup$ – Mark Viola Apr 25 '17 at 5:23
  • $\begingroup$ Ah! I had to actually think about how the expression for $\hat{z}$ in terms of $\hat{r}$ and $\hat{\theta}$ worked out geometrically. That makes perfect sense now! $\endgroup$ – Alex Apr 25 '17 at 5:29
  • $\begingroup$ Please to hear that you have it now Alex! -Mark $\endgroup$ – Mark Viola Apr 25 '17 at 5:31

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