# Kronecker delta

1. May 8, 2007

### MadMax

Using the Einstein summation convention...

Why is

$$\mathbf{a}^2 \mathbf{b}^2$$

not the same as

$$3 a_i a_j b_j b_i = 3(\mathbf{a} \cdot \mathbf{b})^2$$

given that

$$\mathbf{a}^2 = a_i \cdot a_i = a_i a_j \delta_{ij}$$

$$\mathbf{b}^2 = b_i \cdot b_i = b_i b_j \delta_{ij}$$

and

$$\delta_{ij} \delta_{ji} = 3$$

-> $$\mathbf{a}^2 \mathbf{b}^2 = a_i a_j \delta_{ij} b_i b_j \delta_{ij} = 3(\mathbf{a} \cdot \mathbf{b})^2$$

??

Any help would be much appreciated.

Last edited: May 8, 2007
2. May 8, 2007

### MadMax

Hmm I think i just figured it out maybe...

Is it 'cos you can't split the a^2 and b^2 up?

3. May 8, 2007

### Hurkyl

Staff Emeritus
You labelled two different dummy variables with the letter i. By doing so, you tricked yourself into thinking they were the same dummy variable.

P.S. your first clue that something is horribly wrong should have been when you had the same letter appear four times in that term as an index.

Last edited: May 8, 2007
4. May 8, 2007

### MadMax

Ahh yeah, good point :P Forgot about that. Thanks Hurkyl.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?