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Homework Help: Covariant and contravariant analysis

  1. Oct 11, 2007 #1
    1. The problem statement, all variables and given/known data
    Can someone explain/help me prove the formulas
    [tex] \vec{e_a}' = \frac{ \partial{x^b}}{\partial x'^a} \vec{e_b} [/tex]

    [tex] \vec{e^a}' = \frac{ \partial{x'^a}}{\partial x^b} \vec{e^b} [/tex]

    I do not understand why the partial derivative flip?
    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 11, 2007 #2


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    I'll give you 2 hints: 1. What is a coordinate basis ? 2. What is the chain rule for partial derivatives ?
  4. Oct 11, 2007 #3


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    Basically, this is essentially a definition of what we mean by covariant and contravariant indices (sometimes people start from the components of a covariant or contravariant vector instead of the basis vectors in which case the transformations are the opposite of the ones given here)
  5. Oct 18, 2007 #4
    My book defines the coordinate basis vector at a point P as follows:

    [tex] \vec{e}_a = lim_{\delta x^a \to 0} \frac{ \delta \vec{s}}{\delta x^a} [/tex]

    where delta s is the infinitesimal vector displacement between P and a nearby point Q whose coordinate separation from P is delta x^a.

    I think you must be referring to perhaps a more useful definition!
  6. Oct 19, 2007 #5
    A vector V is a geometrical object, that can be written in a coordinate basis as
    where [tex]e_{a}[/tex] are basis vectors (I'm giving them a lower index). The [tex]v^{a}[/tex] are the coefficient functions of the vector. As a geometrical object, V cannot care about what coordinate basis you have chosen, so under changes of bases it must be left unchanged. So if the coefficient functions transform as
    [tex]v^{a}(x')=\frac{\partial x^{a}}{\partial x'^{b}}v^{b}(x)[/tex] the basis vectors must transform as
    [tex]e_{a}(x')=\frac{\partial x'^{c}}{\partial x^{a}}e_{c}(x)[/tex]
    [tex]\frac{\partial x'^{c}}{\partial x^{a}}\frac{\partial x^{a}}{\partial x'^{b}}=\delta^{c}_{b}[/tex]
    implies that

    Similar arguments go for co-vectors [tex]W=w_{a}e^{a}[/tex]
    where [tex]e^{a}[/tex] are a basis of co-vectors (also known as 1-forms) contracted with their coefficient functions. The location of the index on all these objects determines the form of the transformation. Note: There is confusion in the literature regarding the names covariant and contravariant as applies to upper vs. lower index objects (e.g. there are generally opposite naming conventions between maths and physics).
    Last edited: Oct 19, 2007
  7. Oct 20, 2007 #6
    OK. I see why the basis case holds. But for the dual basis case, we then need that:

    [tex] \frac{\partial x_a}{\partial x'_b} = \frac{\partial x'^b}{\partial x^a} [/tex]

    which is what I do not understand?
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