Gradient in general co-ordinate system

  • Thread starter sebb1e
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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/Orthogonal_coordinates#Differential_operators_in_three_dimensions

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
 
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Answers and Replies

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Yes, non Cartesian coordinates are not necessarily unit vectors, especially if one denotes the distance to the origin. The trick is to consider them as the Graßmann product of differential forms: ##dx\wedge dy \wedge dz##. From there you can substitute ##x,y,z## and recalculate the product.
 

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