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Differentiation with respect to covariant component of a vector

  1. May 12, 2015 #1
    I want to prove that differentiation with respec to covariant component gives a contravariant vector operator. I'm following Jackson's Classical Electrodynamics. In the first place he shows that differentiation with respecto to a contravariant component of the coordinate vector transforms as the component of a covariant vector operator.

    For this, he implicitly uses a change of variables: ##\displaystyle x^{\alpha}=x^{\alpha} (x^{ \beta } )## So using the rule for implicit differentiation:

    ##\displaystyle \frac{ \partial}{ \partial x^{ \alpha} }= \frac{ \partial x^{\beta} }{ \partial x^{\alpha} } \frac{\partial}{ \partial x^{ \beta} }##

    Now I want to show that: ##\displaystyle \frac{\partial}{\partial x_{\alpha}}=\frac{\partial x^{\alpha}}{\partial x^{\beta}} \frac{\partial}{\partial x_{ \beta}}## I think that this is the expresion I should find, but I'm not sure, Jackson didn't give it explicitly.

    He suggests to use that ##x_{\alpha}=g_{\alpha \beta}x^{\beta}##, g is the metric tensor.


    I need some guidance for this.

    Thanks in advance.
     
    Last edited: May 12, 2015
  2. jcsd
  3. May 13, 2015 #2
    Ok, I worked it this way. I supposed I had a scalar function of ##x_{\alpha}##, ##f(x_{\alpha})##. And because ##x_{\alpha}=g_{\alpha \beta}x^{\beta}## I have that ##x_{\alpha}=x_{\alpha}(x^{\beta})##.

    Then I differentiated f:

    ##\displaystyle \frac{df}{dx_{\alpha}}= \frac{\partial f}{\partial x_{\alpha}} \frac{\partial x_{\alpha}}{\partial x^{\beta}}=g_{\alpha \beta}\frac{\partial f}{\partial x_{\alpha}}## I've used that the metric tensor is independt of the coordinates of the vectors, I'm not sure if thats right in general, I'm working with special relativity and I think that this hold in this particular case.

    Then I have

    ##\displaystyle \frac{\partial }{\partial x_{\alpha}}=g_{\alpha \beta}\frac{\partial }{\partial x_{\alpha}}##

    This isn't the law of transformation I'm looking for. Anyway, it seems pretty intuitive that it should hold that if the differentiation with respect to a contravariant component of the coordinate vector transforms as the component of a covariant vector operator, then differentiation with respect to a covariant component should transform as a contravariant vector, because I think there must be some sort of symmetry between the space and its dual space.
     
    Last edited: May 13, 2015
  4. May 13, 2015 #3
    An other way would be just to consider a change of variables ##x_{\alpha}=x_{\alpha}(x_{\beta})## and a function ##f(x_{\alpha})##, and differentiate using the chain rule.

    Then I get the desired result, but is not clear to me why in the first case, when deriving with respect to a contravariant component, I should use implicit differentiation, and why in this other case I get the result just by direct differentiation and using the chain rule.

    I was wrong, I don't get the desired result, I get: ##\displaystyle \frac{df}{dx_{\alpha}}= \frac{\partial f}{\partial x_{\alpha}} \frac{\partial x_{\alpha}}{\partial x_{\beta}}##

    And I should have: ##\displaystyle \frac{df}{dx_{\alpha}}= \frac{\partial f}{\partial x_{\beta}} \frac{\partial x^{\alpha}}{\partial x^{\beta}}##
     
    Last edited: May 13, 2015
  5. May 13, 2015 #4
    I think I was making a bad use of the chain rule. So, here I go again: ##x_{\alpha}=x_{\alpha}(x_{\beta})##. ##f=f(x_{\alpha})##

    ##\displaystyle \frac{df}{dx_{\beta}}=\frac{\partial f}{\partial x_{\alpha}} \frac{\partial x_{\alpha}}{\partial x_{\beta}}##

    Now I've used that ##g_{\alpha \beta}=g_{\beta \alpha}## and that ##dx^{\alpha}=g^{\alpha \beta}dx_{\beta}##

    Then:

    ##\displaystyle \frac{\partial f}{\partial x_{\alpha}} \frac{\partial x_{\alpha}}{\partial x_{\beta}}=\frac{g^{\beta \alpha}}{g^{\alpha \beta}}\frac{\partial x_{\alpha}}{\partial x_{\beta}} \frac{\partial f}{\partial x_{\alpha}}=\frac{\partial x^{\beta}}{\partial x^{\alpha}}\frac{\partial f}{\partial x_{\alpha}}##

    So that

    ##\displaystyle \frac{\partial }{\partial x_{\beta}}= \frac{\partial x^{\beta}}{\partial x^{\alpha}} \frac{\partial }{\partial x_{\alpha}}##

    As I wanted to show.

    EDIT: I see I have a problem here too, I have repited indices more than twice, so this approach is wrong too.
     
    Last edited: May 13, 2015
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