1. Jun 2, 2014

joda80

Hi All,

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,
$$\frac{\partial A^i}{\partial x^j},$$
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:

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

Applying the product rule,

$$\frac{\partial A^i}{\partial x^j} = \frac{\partial^2 x^i}{\partial x^j \partial a^r} \bar{A}^r + \frac{\partial x^i}{\partial a^r} \frac{\partial \bar{A}^r}{\partial x^j}.$$

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

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

which we may write as

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

Since the unit tensor,

$$\delta^i_j$$

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

$$\frac{\partial A^i}{\partial x^j} = \frac{\partial x^i}{\partial a^r}\frac{\partial a^q}{\partial x^j} \frac{\partial \bar{A}^r}{\partial a^q},$$

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

2. Jun 2, 2014

Matterwave

Well， the problem is the term with the second derivative (the first term), that term shouldn't vanish. It is that term which prevents the expression from being a tensor.

Let's examine that term more explicitly. We know that the transformation of coordinates is $x^i=x^i(a^r)$ This is a function of the new coordinates. So, say we have a function y(x), what you have done in that first term is analogous to:

$$\frac{\partial}{\partial y}\left(\frac{\partial y}{\partial x}\right)= \frac{\partial}{\partial x}\left(\frac{\partial y}{\partial y}\right)=0$$

This is not kosher. You can't switch the derivative around in there like that. Equality of mixed partial derivatives works for multivariate functions like y(x,z) where I can exchange derivatives in x and z, but obviously we don't have functions like y(x,y) and try to exchange derivatives in x and y...

What you should have done is do the chain rule:
$$\frac{\partial}{\partial x^j}=\frac{\partial a^q}{\partial x^j}\frac{\partial}{\partial a^q}$$

This will yield the first term looking like:

$$\frac{\partial a^q}{\partial x^j}\frac{\partial^2 x^i}{\partial a^q \partial a^r}\bar{A}^r$$

Which is not 0.

3. Jun 5, 2014

joda80

That makes sense … thanks for helping out!!