# Differentiation with respect to covariant component of a vector

1. May 12, 2015

### Telemachus

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. May 13, 2015

### Telemachus

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
3. May 13, 2015

### Telemachus

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
4. May 13, 2015

### Telemachus

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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