Gradient in general co-ordinate system

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SUMMARY

The discussion centers on the expression of the gradient in a general coordinate system, specifically referencing the definition from "A First Course in Continuum Mechanics" by Gonzalez and Stuart. The gradient is defined as grad phi(x) = (partial d phi by d xi)(x) ei, where {ei} represents an arbitrary frame. The conversation highlights the challenge of relating this definition to cylindrical coordinates, particularly the presence of the 1/r term in the unit theta component. It is confirmed that the vectors {ei} are not necessarily unit vectors, especially in non-Cartesian systems.

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  • Familiarity with differential operators in three dimensions
  • Knowledge of cylindrical coordinates and their properties
  • Basic concepts of continuum mechanics and deformation gradients
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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
 
Last edited:
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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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