Responsible party: Ekrpat, chap0326

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* Making sense of the Del Operator

* Beginning spin

When shown that

<math>\mathbf{\vec{ \nabla}} = \mathbf{\hat r} \partial r + \mathbf{\hat \theta} {\partial \theta \over r} + \mathbf{\hat \phi}{\partial \phi \over r\sin(\theta)} </math>

we were asked “does this make sense?” (DTMS)

We then recalled that <math>\mathbf{\vec{ \nabla}}</math> comes from

<math>\mathbf{\vec{ \nabla}} = \mathbf{\hat x}\partial x + \mathbf{\hat y}\partial y + \mathbf{\hat z}\partial z</math>

and that you could use a brute force type method to get the solution we are looking for from that. Also explained today was another method that involved introducing another Cartesian system: x',y',z', where

<math>\mathbf{\vec{ \nabla}} = \mathbf{\hat x'}\partial x' + \mathbf{\hat y'}\partial y' + \mathbf{\hat z'}\partial z'</math>,

and rotating the coordinates such that it lined up with the r-vector, <math>\theta</math> and <math>\phi</math>. Then it is possible to make a direct transformation from the [x',y',z'] coordinates to the [<math>\mathbf{\hat r}, \theta, \phi</math>] coordinates.

We find that <math>\partial x' = \partial r</math>, <math>\partial y' = {\partial \theta \over r}</math>, and <math>\partial z' = {\partial \phi \over r\sin(\theta)}</math>

Thus in a more elegant, though perhaps not as readily apparent way, we arrive at the desired expression:

<math>\mathbf{\vec{ \nabla}} = \mathbf{\hat r} \partial r + \mathbf{\hat \theta} {\partial \theta \over r} + \mathbf{\hat \phi}{\partial \phi \over r\sin(\theta)} </math>

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