# Some vector multiplications

1. May 5, 2007

Are the following identities correct? (They are under integration btw... and we are using the summation convention...)

1.) $$\delta_{ij}\delta_{ji}=1$$

2.) $$\delta_{ij}a_j a_i = a^2$$

3.) $$q_i q_j \delta_{ji} = q^2$$

4.) $$q_i q_j a_j a_i = q^2 a^2$$

My input:

kronecker delta's are equal to one if i=j and zero other wise... ?

a_i a_i = a^2... ?

q_i q_j = q^2 delta_ij... ?

Cheers

Last edited: May 5, 2007
2. May 5, 2007

### Mentz114

Show us what you think and why. This is not an answer factory ( you should know that after 55 posts ).

3. May 5, 2007

lol yeah i realised that a few minutes after posting. I edited my first post so it shows my reasoning...

thanks

4. May 5, 2007

### Mentz114

Your notation is not clear to me. Do you sum over repeated indexes ?

For instance in question 2, how many $$a_i$$s are there ?

5. May 5, 2007

### nrqed

It depends. In how many dimensions are you working?
Right, assuming Einstein's summation convention.
No. Check again the indices.

6. May 5, 2007

Hmm not sure whether the summation convention is being used or not to be honest. I can give you two identities that I'm 100% sure of and perhaps it will be clear then...

$$\hat{r_i} \delta_{ij} \hat{r_j} = 1$$

$$\hat{r_i} q_i q_j \hat{r_j} = (q \cdot \hat{r})^2$$

As for how many a_i 's there are, I'm not sure I understand the question but perhaps it is answered if I say that i and j can take the values 1, 2 or 3?

7. May 5, 2007

hmm actually I do reckon I'm using the summation convention...

Ok so considering nrqed's post I guess for the first one the answer is 3...? (I'm working in 3 dimensions).

So the second and third are OK...

The last one I really have no idea how to tackle. Using the summation convention we are bound to end up with something with no indices (repeated indices cancel and are summed over) so perhaps I'm missing some unit vectors?

Thanks for the help so far guys.

Last edited: May 5, 2007
8. May 5, 2007

### Mentz114

Regarding 2 and 3, if the a_i are components of a vector and a^2 = a.a then they are correct. I think 4 is $$\vec{a}.\vec{q}$$

 nrqed is right (a.q)^2

Last edited: May 5, 2007
9. May 5, 2007

### nrqed

$a_i q_i$ is simply equal to $a \cdot q$, the inner product between the two vectors (the usual dot or scalar product if you are in the usual Euclidian three-dimensional space). So your question number 4 simply gives $(a \cdot q) (a \cdot q) = (a \cdot q)^2$.

Patrick

NOTE ADDED: notice that $\delta_{ij} \delta_{ij} = \delta_{ii} = N$ assuming that you are in Euclidian space and where N is the number of dimensions.

Last edited: May 5, 2007
10. May 5, 2007