Gradient in general co-ordinates

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Homework Statement



I know that for a general co-ordinate system, the gradient can be expressed as it is at the bottom of this page:

http://en.wikipedia.org/wiki/Orthogo...ree_dimensions [Broken]

However, the book I am working from (A First Course in Continuum Mechanics by Gonzalez and Stuart) defines it as:

"Let {ei} be an arbitrary frame. Then grad phi(x)= (partial d phi by d xi)(x) ei where (x1,x2,x3) are the coordinates of x in ei"

I don't understand how this relates to the actual definition, for example given the definition in my book, how do I show that the gradient in cylindricals contains a 1/r in front of the unit theta term?

I presume that these ei are completely general so they are not unit vectors?

I need to understand this notation properly so I can apply it to deformation gradients etc



Homework Equations





The Attempt at a Solution



From reading http://en.wikipedia.org/wiki/Curvilinear_coordinates where the same formula as the book is given, I presume it has something to do with covariant and contravariant basis but this isn't really mentioned in the book.
 
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HallsofIvy
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A circular arc, from point [itex](r, \theta)[/itex] to [itex](r, \theta+ d\theta)[/itex] will have length [itex]rd\theta[/itex]. A vector, in the [itex]\theta[/itex] direction, at the point with coordinates [itex](r, \theta_0)[/itex] will, similarly, have length given by [itex]r d\theta[/itex]. The "1/r" is required to remove the dependence on r.
 
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How would I know, merely from the definition in the book that I need that, for example if we were dealing with some totally arbitrary coordinates? (I completely understand why it is there in this case and know how to derive it)
 

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