# Evaluating Kronecker Deltas

• Oct 7th 2012, 10:45 AM
bilalsaeedkhan
Evaluating Kronecker Deltas
Need help with evaluating:

$\displaystyle \delta _{i,j}\delta _{i,j}$

$\displaystyle \delta _{i,j}\delta _{j,k}$

Thanks
• Oct 7th 2012, 11:01 AM
Plato
Re: Evaluating Kronecker Deltas
Quote:

Originally Posted by bilalsaeedkhan
Need help with evaluating:

$\displaystyle \delta _{i,j}\delta _{i,j}$

$\displaystyle \delta _{i,j}\delta _{j,k}$

• Oct 7th 2012, 11:19 AM
bilalsaeedkhan
Re: Evaluating Kronecker Deltas
My attempt at expanding $\displaystyle \delta _{i,j}\delta _{i,j}$:

When i=j
$\displaystyle \delta _{i,i}\delta _{i,i}$
$\displaystyle (\delta _{1,1}+\delta _{2,2}+\delta _{3,3})$ *$\displaystyle (\delta _{1,1}+\delta _{2,2}+\delta _{3,3})$
(1+1+1)x(1+1+1)
9
• Oct 7th 2012, 11:52 AM
Plato
Re: Evaluating Kronecker Deltas
Quote:

Originally Posted by bilalsaeedkhan
My attempt at expanding $\displaystyle \delta _{i,j}\delta _{i,j}$:
When i=j
$\displaystyle \delta _{i,i}\delta _{i,i}$
$\displaystyle (\delta _{1,1}+\delta _{2,2}+\delta _{3,3})$ *$\displaystyle (\delta _{1,1}+\delta _{2,2}+\delta _{3,3})$
(1+1+1)x(1+1+1)
9

Have you changed the question?
Where did $\displaystyle (\delta _{1,1}+\delta _{2,2}+\delta _{3,3})$ come from?
That was not part of the OP!

The OP asked for $\displaystyle \delta _{i,j}\delta _{i,j}=\left\{ {\begin{array}{lr} {1,} & {i = j} \\ {0,} & {i \ne j} \\\end{array}} \right.$

The second part of the OP asked $\displaystyle \delta _{ij} \delta _{jk} = \left\{ {\begin{array}{lr} {1,} & {i = j = k} \\ {0,} & {else} \\\end{array}}\right.$

So why did you change it?
• Oct 7th 2012, 12:07 PM
bilalsaeedkhan
Re: Evaluating Kronecker Deltas
Quote:

Originally Posted by Plato
Have you changed the question?
Where did $\displaystyle (\delta _{1,1}+\delta _{2,2}+\delta _{3,3})$ come from?
That was not part of the OP!

The OP asked for $\displaystyle \delta _{i,j}\delta _{i,j}=\left\{ {\begin{array}{lr} {1,} & {i = j} \\ {0,} & {i \ne j} \\\end{array}} \right.$

The second part of the OP asked $\displaystyle \delta _{ij} \delta _{jk} = \left\{ {\begin{array}{lr} {1,} & {i = j = k} \\ {0,} & {else} \\\end{array}}\right.$

So why did you change it?

I never asked for a definition of Kronecker Delta. I need help with evaluating them. I am asked to evaluate.
• Oct 7th 2012, 12:40 PM
Plato
Re: Evaluating Kronecker Deltas
Quote:

Originally Posted by bilalsaeedkhan
I never asked for a definition of Kronecker Delta. I need help with evaluating them. I am asked to evaluate.

Quote:

Originally Posted by bilalsaeedkhan
Need help with evaluating:
$\displaystyle \delta _{i,j}\delta _{i,j}$
$\displaystyle \delta _{i,j}\delta _{j,k}$

You did ask us to evaluate those two expressions.
You cannot say you did not.
I did exactly that in reply #4.

Why not post the actual question then?
• Oct 7th 2012, 01:01 PM
bilalsaeedkhan
Re: Evaluating Kronecker Deltas
Quote:

Originally Posted by Plato

You did ask us to evaluate those two expressions.
You cannot say you did not.
I did exactly that in reply #4.

Why not post the actual question then?

I did post the question word for word as it is given to me and I did ask for help on evaluating them. I guess dropping the summation notation is what is throwing you off. We were told that the index notation implies that the quantities are being summed. Let me rewrite the problem although this is not how the question is given to us.

$\displaystyle \sum_{i,j,k=1}^{i,j,k=3}\delta _{i,j}\delta _{i,j}$

Similarly the second problem is being summed too.
• Oct 7th 2012, 01:36 PM
Plato
Re: Evaluating Kronecker Deltas
Quote:

Originally Posted by bilalsaeedkhan
I did post the question word for word as it is given to me and I did ask for help on evaluating them. I guess dropping the summation notation is what is throwing you off. We were told that the index notation implies that the quantities are being summed. Let me rewrite the problem although this is not how the question is given to us.

$\displaystyle \sum_{i,j,k=1}^{i,j,k=3}\delta _{i,j}\delta _{i,j}$

Similarly the second problem is being summed too.

$\displaystyle \sum_{i,j,k=1}^{i,j,k=3}\delta _{i,j}\delta _{i,j}=3$
There are twenty-seven triples $\displaystyle (i,j,k)$ where $\displaystyle i=1,2,3~j=1,2,3~k=1,2,3$.
Only three of them have $\displaystyle i=j=k$ (see reply #4) so you three 1's and twenty-four 0's in that sum.