# Kronecker delta as tensor proof

1. Jun 18, 2009

### faklif

1. The problem statement, all variables and given/known data
The problem straight out of the book reads:
Prove that the Kronecker delta has the tensor character indicated.
Prove also that it is a constant or numerical tensor, that is, it has
the same components in all coordinate systems.

Without a context the first sentence might be a bit weird but to me it is still weird with a context becuase I can't figure out what it means. In the passage that the question is related to I find no clues so any help is appreciated.

2. Relevant equations

3. The attempt at a solution
Since I don't even really understand the first part I can'r really do much about that. As for the second part I really don't know much about tensors yet and I can't fit the Kronecker delta with transformations properly so I'm at a loss. Hope someone can help!

2. Jun 18, 2009

### EnumaElish

What is the definition ("character") of a tensor stated in your book?

3. Jun 18, 2009

### faklif

Thanks for your reply! What looks most like a definition to me is:

A contravariant tensor of rank 1 is a set of quantities, written X^a in the x^a coordinate system, associated with a point P, which transforms under a change of coordinates according to
$$X'^a=\frac{{\partial}x'^a}{{\partial}x^b}X^b$$
where the transformation matrix is evaluated at P.

Then a covariant tensor is defined in a similar way but with
$$X'_a=\frac{{\partial}x^b}{{\partial}x'^a}X_b$$

Mixed tensors are also presented and I'm thinking that maybe I'm supposed to show that the delta is a (1,1) tensor which it looks like it would be according to it's indices? Don't really know how to do that though.

4. Jun 18, 2009

### Avodyne

Yes, that's what you're supposed to do. So make the appropriate transformation on each index. Then you should see a nice way to make use of the chain rule for derivatives.

5. Jun 18, 2009

### faklif

Thank you!

$$\delta'^a_b=\frac{{\partial}x'^a}{{\partial}x^c}\frac{{\partial}x^d}{{\partial}x'^b}\delta^c_d=\frac{{\partial}x'^a}{{\partial}x^c}\frac{{\partial}x^c}{{\partial}x'^b}=\frac{{\partial}x'^a}{{\partial}x'^b}=\delta^a_b$$

So that would do it?

Two main things get me here. I have a very poor understanding of what a tensor is. Especially that I don't know much of the difference between a contravariant and covariant one, except how they transform. Then there's my lack of understanding when it comes to proofs. My background being in engineering I'm mostly just used to understanding theorems so that I can use them. So with this as an example what is it that proves that it is a (1,1) tensor? Is it the fact that in the transformation I get back the same thing I started out with? Something like, since it behaves like a tensor it is a tensor? And is what I wrote down above a definition of a tensor?

6. Jun 18, 2009

### Dick

Yes, that's exactly correct. That proves it's a (1,1) tensor. It has the correct coordinate transformation. If it walks like a duck and talks like a duck, it is a duck. If it behaves like a tensor, it is a tensor.

7. May 6, 2010

### particlemania

the last step -

$$\frac{{\partial}x'^a}{{\partial}x^c}\frac{{\partial}x^c}{{\partial}x'^b}=\frac{{\partial}x'^a}{{\partial}x'^b}=\delta^a_b$$

seems quite incorrect to me. Since 'c' is a dummy variable here, over which the expression is summed, hence shouldnt it be :

$$\frac{{\partial}x'^a}{{\partial}x^c}\frac{{\partial}x^c}{{\partial}x'^b}=\frac{{\partial}x'^a}{{\partial}x'^b} \times D$$

where D is the Dimension number of tensor space.