# Understanding Kronecker Delta

1. Sep 10, 2011

### Bellin12

1. The problem statement, all variables and given/known data

I am a bit confused with how to deal with Kronecker Delta.
I need to show that P,i = ( P*δij ),j

The i and j are subscripts.

2. Relevant equations

3. The attempt at a solution

P,i = ( P*δij ),j = P,j*δij + P*δij,j
I assumed I could get rid of P*δij,j leaving me with P,j*δij.
After this I am stuck. I'm not sure if I am supposed to consider Kronecker Delta as the Identity Matrix now or not. If I do, I end up with P,j which technically would equal P,i.

2. Sep 12, 2011

### vela

Staff Emeritus
What exactly is P here?

3. Sep 13, 2011

### quietrain

is it like this?

we have
{P * dij }j

since dij is 0 for all i,j except i=j where it is 1,

then {P * dij }j
= {P * 1 }i ==> since i = j
= Pi

i was taught to think of the kronecker delta as a sieving function, i.e, it 'sieves' out the index 'j' and replace it by 'i' , as per above

4. Sep 13, 2011

### vela

Staff Emeritus
There's no need to assume the second term is 0. δij is equal to 0 or 1, so if you differentiate it, you get 0.

The second term isn't equal to P,j. The index j is a dummy index you're summing over. Just expand the summation out to see what's happening:
$$P_{,j}\delta_{ij} = P_{,1}\delta_{i1} + P_{,2}\delta_{i2} + \cdots + P_{,n}\delta_{in}$$In only one term, the ith term, does the Kronecker delta not vanish, so you end up with P,i.