I Kronecker Delta summation (easy)

joshmccraney · Aug 27, 2016

Hi PF!

As outlined in my book ##\delta_{ij} \delta_{jk} = \delta_{ik}## but don't we sum over repeated indices (and the ##j## is repeated)? Can someone explain why we do not sum in this situation?

Thanks!

blue_leaf77 · Aug 27, 2016

Yes it is summed over ##j##. Written explicitly, the sum looks like
$$
\delta_{i1} \delta_{1k} + \delta_{i2} \delta_{2k} + \ldots + \delta_{iN} \delta_{Nk}
$$
You see that if ##i\neq k##, all terms above will vanish. If ##i=k## there will be only one term surviving, therefore that sum can be written compactly as ##\delta_{ik}##.

joshmccraney · Aug 27, 2016

Thanks for responding blue_leaf77! Let's assume ##i,j,k## vary from ##1,2,3##. Then we have $$\delta_{11} \delta_{11} + \delta_{22} \delta_{22} + \delta_{33} \delta_{33} = 3 \neq \delta_{ik}$$

Did I miss something?

PeroK · Aug 27, 2016

joshmccraney said: ↑

Thanks for responding blue_leaf77! Let's assume ##i,j,k## vary from ##1,2,3##. Then we have $$\delta_{11} \delta_{11} + \delta_{22} \delta_{22} + \delta_{33} \delta_{33} = 3 \neq \delta_{ik}$$

Did I miss something?

Why would ##i## and ##k## be summed?

And, if you sum them on the right hand side you would get equality.

joshmccraney · Aug 27, 2016

PeroK said: ↑

Why would ##i## and ##k## be summed?

That's what blue_leaf wrote.

blue_leaf77 said: ↑

$$
\delta_{i1} \delta_{1k} + \delta_{i2} \delta_{2k} + \ldots + \delta_{iN} \delta_{Nk}
$$

Since ##i## and ##k## can range from 1,2,3 we would have$$\delta_{11} \delta_{11} + \delta_{21} \delta_{11} + \delta_{31} \delta_{11}+ \delta_{12} \delta_{21}+ \delta_{22} \delta_{21}+ \delta_{32} \delta_{21}+\cdots + \delta_{33} \delta_{33}\\=\delta_{11} \delta_{11}+\delta_{22} \delta_{22}+\delta_{33} \delta_{33}$$ which is why I was summing.

PeroK · Aug 27, 2016

joshmccraney said: ↑

That's what blue_leaf wrote.

Since ##i## and ##k## can range from 1,2,3 we would have$$\delta_{11} \delta_{11} + \delta_{21} \delta_{11} + \delta_{31} \delta_{11}+ \delta_{12} \delta_{21}+ \delta_{22} \delta_{21}+ \delta_{32} \delta_{21}+\cdots + \delta_{33} \delta_{33}\\=\delta_{11} \delta_{11}+\delta_{22} \delta_{22}+\delta_{33} \delta_{33}$$ which is why I was summing.

blue leaf summed only ##j##, which is a repeated index.

joshmccraney · Aug 27, 2016

PeroK said: ↑

blue leaf summed only ##j##, which is a repeated index.

Ohhhhh I see now! Shoot, yea I was doing it all wrong! Thank you both!!!

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I Kronecker Delta summation (easy)

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