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I Del in index notation

  1. Sep 1, 2016 #1

    joshmccraney

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    Hi PF!

    Which way is appropriate for defining del in index notation: ##\nabla \equiv \partial_i()\vec{e_i}## or ##\nabla \equiv \vec{e_i}\partial_i()##. The two cannot be generally equivalent. Quick example.

    Let ##\vec{v}## and ##\vec{w}## be vectors. Then $$\nabla \vec{v} \cdot \vec{w} = \partial_i(v_j \vec{e_j})\vec{e_i} \cdot u_k \vec{e_k}\\ = \partial_i(v_j \vec{e_j}) u_i$$ yet using the other definition for del implies $$\nabla \vec{v} \cdot \vec{w} = \vec{e_i} \partial_i(v_j \vec{e_j}) \cdot u_k \vec{e_k}\\=\vec{e_i} v_ju_k (\partial_i(\vec{e_j}) \cdot \vec{e_k}) + \vec{e_i} u_j \partial_i(v_j)$$
     
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  3. Sep 1, 2016 #2

    pasmith

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    The second. In orthogonal coordinates [itex]\nabla = \sum_i \mathbf{e}_i h_i \partial _i [/itex] and in non-cartesian coordinates [itex]h_i[/itex] is generally a non-constant function of position.
     
  4. Sep 1, 2016 #3

    joshmccraney

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    So is ##h_i = |\partial_i \vec{r}|## where ##\vec{r}## is the position vector, expressed in cartesian coordinates as ##\vec{r} = x \hat{i} + y \hat{j} +z \hat{k}##?
     
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