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Given: $$\vec{A}(r,\theta)$$ $$\vec{B}(r,\theta)$$ Is it always true that:

$$ \left(\vec{A}\frac{\partial}{\partial{\theta}}\right)\bullet\vec{B}\overset ? =\bigg(\vec{A}\bullet\frac{\partial\vec{B}}{\partial{\theta}}\bigg) $$

The reason I ask is because, when evaluating: $$\vec{\nabla} \bullet \vec{u}$$

it was assumed that: $$ \left(\hat{\theta} \frac{1}{r} \frac{\partial}{\partial \theta}\right) \bullet (u_r \hat{r}) =\hat{\theta} \bullet \left(\frac{1}{r} \frac{\partial (u_r \hat{r})}{\partial \theta} \right) $$ See image: Computation of the Divergence in Polar Form

Image Source:

John
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  • The left hand side of your equation $\stackrel{?}{=}$ does not "compile". Presumably $\vec{A}$ has more than one component. What is $(\vec{A}\frac{\partial}{\partial \theta}),?$ Please write this out to clarify. Never seen such a differential operator. And please do not ask us to watch youtube videos to understand the question. – Kurt G. Nov 12 '23 at 10:29
  • I scrapped the YouTube link, replaced it with an image. It comes from the definition of the gradient in polar form. – John Nov 12 '23 at 10:52
  • You mean divergence in polar form. 2D polar coordinates are not very different from 3D cylindrical coordinates. Besides: that image is not using an arbitrary vector field $\vec{A}$. It is using the basis vector field $\hat{\theta}\frac{1}{r}\partial_\theta,.$ Since those $\hat{\theta}$ thingies always lead to much confusion I prefer the shorter notation $\frac{1}{r}\partial_\theta,.$ For more details see this answer. – Kurt G. Nov 12 '23 at 11:23
  • When we apply the vector field $\hat{\theta}\frac{1}{r}\partial_\theta$ to a function $u$ we only differentiate $u$ w.r.t. $\theta$ and then multiply it by $1/r,.$ That's why I find my proposed notation in the previous comment much better. No $\hat{\theta}$ thingy and no $\bullet$ is necessary. – Kurt G. Nov 12 '23 at 11:39

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