# Transformation behavior of the gradient

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!!