Kronecer delta tensor transformation

[Matterwave]

Homework Statement

Show that $$\delta_a^b$$ is a in fact a mixed tensor of valence (1,1).

Homework Equations

Definition of a (1,1) tensor:

$$\delta'_a^b=\frac{\partial x'^b}{\partial x^c}\frac{\partial x^d}{\partial x'^a}\delta_c^d$$

The Attempt at a Solution

So, I just explicitly put back the summations and I get a sum over c from 1 to 4 and a sum over d from 1 to 4 of the above expression. I did the sum over c first and got:

$$\sum_{d=1}^4 (\delta_1^d\frac{\partial x'^b}{\partial x^1}\frac{\partial x^d}{\partial x'^a}+...)$$

Where the ... had the $$\delta_2^d$$ terms etc. I did the summation over d, and I seem to be getting:

$$\frac{\partial x'^b}{\partial x^1}\frac{\partial x^1}{\partial x'^a}+\frac{\partial x'^b}{\partial x^2}\frac{\partial x^2}{\partial x'^a}+\frac{\partial x'^b}{\partial x^3}\frac{\partial x^3}{\partial x'^a}+\frac{\partial x'^b}{\partial x^4}\frac{\partial x^4}{\partial x'^a}$$

Which equals:
$$4\frac{\partial x'^b}{\partial x'^a}$$

I get the 4 because there's 4 instances where the kronecker deltas don't equal 0 (when d=1 the first term remains, when d=2 the second term remains, etc)...but I really shouldn't have that 4 there. What happened? =(

 

[gabbagabbahey]

Where the ... had the $$\delta_2^d$$ terms etc. I did the summation over d, and I seem to be getting:

$$\frac{\partial x'^b}{\partial x^1}\frac{\partial x^1}{\partial x'^a}+\frac{\partial x'^b}{\partial x^2}\frac{\partial x^2}{\partial x'^a}+\frac{\partial x'^b}{\partial x^3}\frac{\partial x^3}{\partial x'^a}+\frac{\partial x'^b}{\partial x^4}\frac{\partial x^4}{\partial x'^a}$$

This should look a lot like the chain rule to you

$x'^b$ can be thought of as being a function of $x_1$, $x_2$, $x_3$, and $x_4$; and each of those coordinates can be thought of as being a function of $x'^a$ (as well as the other 3 primed coordinates, but it's a partial derivative so that doesn't matter)

$$\frac{\partial x'^b}{\partial x^1}\frac{\partial x^1}{\partial x'^a}\neq\frac{\partial x'^b}{\partial x'^a}$$

 

[Matterwave]

Ok...but then I don't see how to do that summation then...>_>

So, is it true that:$$\frac{\partial x'^b}{\partial x^1}\frac{\partial x^1}{\partial x'^a}+\frac{\partial x'^b}{\partial x^2}\frac{\partial x^2}{\partial x'^a}+\frac{\partial x'^b}{\partial x^3}\frac{\partial x^3}{\partial x'^a}+\frac{\partial x'^b}{\partial x^4}\frac{\partial x^4}{\partial x'^a}=\frac{\partial x'^b}{\partial x'^a}$$

?

I don't see how that can be true...yet if I've done the steps right, that must be true...oh wait, is that just the reverse chain rule? If so, shouldn't it then be changed to a total derivative or something?

 

[gabbagabbahey]

$$\frac{\partial x'^b}{\partial x^1}\frac{\partial x^1}{\partial x'^a}+\frac{\partial x'^b}{\partial x^2}\frac{\partial x^2}{\partial x'^a}+\frac{\partial x'^b}{\partial x^3}\frac{\partial x^3}{\partial x'^a}+\frac{\partial x'^b}{\partial x^4}\frac{\partial x^4}{\partial x'^a}=\frac{\partial x'^b}{\partial x'^a}$$

?

Yes, this is the chain rule.

$$\frac{\partial}{\partial x'^a} x'^b(x^1,x^2,x^3,x^4)=\frac{\partial x'^b}{\partial x^1}\frac{\partial x^1}{\partial x'^a}+\frac{\partial x'^b}{\partial x^2}\frac{\partial x^2}{\partial x'^a}+\frac{\partial x'^b}{\partial x^3}\frac{\partial x^3}{\partial x'^a}+\frac{\partial x'^b}{\partial x^4}\frac{\partial x^4}{\partial x'^a}$$

 

[Matterwave]

Right, I looked it up on Wiki, and found the same. Thanks! =D

(looks like my astro-induced habit of canceling out differentials has backfired this time).