- #1

Telemachus

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**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 } )##

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}}##

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

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.

Thanks in advance.

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