Hi All,(adsbygoogle = window.adsbygoogle || []).push({});

I think I have confused myself ... perhaps you can tell me where my reasoning is wrong. The idea is that in general coordinates the partial derivative of a vector,

[tex]\frac{\partial A^i}{\partial x^j},[/tex]

is not a tensor because an additional term arises (which is the motivation for defining the covariant derivative).

However, when I do the math that additional term always drops out as demonstrated below. I'm pretty sure my math is wrong, but I don't see where. Maybe you can help out.

Here's my thinking:

[tex] \frac{\partial A^i}{\partial x^j} = \frac{\partial }{\partial x^j} \left[\frac{\partial x^i}{\partial a^r}\bar{A^r} \right].[/tex]

Applying the product rule,

[tex] \frac{\partial A^i}{\partial x^j} = \frac{\partial^2 x^i}{\partial x^j \partial a^r} \bar{A}^r

+ \frac{\partial x^i}{\partial a^r} \frac{\partial \bar{A}^r}{\partial x^j}.[/tex]

Switching the order of the partial derivatives in the first term results in

[tex] \frac{\partial A^i}{\partial x^j} = \frac{\partial}{\partial a^r}\left[ \frac{\partial x^i}{\partial x^j}\right] \bar{A}^r

+ \frac{\partial x^i}{\partial a^r} \frac{\partial \bar{A}^r}{\partial x^j},[/tex]

which we may write as

[tex] \frac{\partial A^i}{\partial x^j} = \frac{\partial}{\partial a^r}\left[\delta^i_j \right] \bar{A}^r + \frac{\partial x^i}{\partial a^r} \frac{\partial \bar{A}^r}{\partial x^j}.[/tex]

Since the unit tensor,

[tex]\delta^i_j[/tex]

is constant, its derivative is zero, so the first term vanishes and we get, after applying the chain rule to the second term,

[tex] \frac{\partial A^i}{\partial x^j} = \frac{\partial x^i}{\partial a^r}\frac{\partial a^q}{\partial x^j} \frac{\partial \bar{A}^r}{\partial a^q}, [/tex]

which would imply that the lhs actually is a second-order tensor, which I think is wrong -- but where did I make the mistake?

Thanks so much for your help!

Best,

Johannes

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