Summation Convention – Substitution Rule

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1. Feb 16, 2016

FluidStu

Hello. I'm new to this forum. I'm starting a PhD – it's going to be a big long journey through the jungle that is CFD. I would like to arm myself with some tools before entering. The machete is Cartesian Tensors.

I know the rules regarding free suffix's and dummy suffixes, but I'm having trouble proving the substitution rule:
δi j δj k = δi k

Assuming a range of 3 for each component, I will choose that the free suffix's have the following values: i=3, k=1. Therefore, using the implicit summation rule for the dummy index:
δi j δj k = δ3 1 δ1 1 + δ3 2 δ2 1 + δ3 3 δ3 1 = ?

Is my expansion correct?

Many thanks.

Last edited by a moderator: Feb 17, 2016
2. Feb 16, 2016

blue_leaf77

You can visualize this equation by remembering that kronecker delta symbol actually represents the matrix elements of an identity matrix.

3. Feb 17, 2016

HallsofIvy

No, It is not because, since "i" and "k" are NOT repeated your result should have "i" and "k" in it, only "j" is replaced by numbers:
δijδjk= δi1δ1k+ δi2δ2k+ δi3δ3k

4. Feb 24, 2016

FluidStu

Ok, that makes sense (since the index which appears once simultaneously represents all the possible values of that index).

However, from your expansion, I do not understand how one could end up with δik. This was really the point in my question – I would like to see step-wise how δijδjk = δik.

5. Feb 24, 2016

blue_leaf77

What do you get when multiplying an identity matrix with another identity matrix?

6. Feb 24, 2016

FluidStu

I * I = I

7. Feb 24, 2016

blue_leaf77

Now, imagine you have a 3x3 matrices $A$, $B$, and $C$ such that $AB=C$. How do you write the element $c_{ik}$ of $C$ in terms of the elements of $A$ and $B$?

8. Feb 24, 2016

FluidStu

cik = aiibik + aijbjk + aikbkk

9. Feb 24, 2016

blue_leaf77

Almost, just that the index to be summed must be written as numbers, not as indices as you did there.

10. Feb 24, 2016

FluidStu

Do you mean:

cik = a11b13 + a12b23 + a13b33 ?

11. Feb 24, 2016

blue_leaf77

No. The left most and right most indices in each term in right hand side should be equal to the indices in the left hand side, which is $ik$. Only the two two middle indices changes due to the summation.
Having fixed this, you should be able to see how that equation becomes of when $A=B=C=I$.

12. Feb 24, 2016

HallsofIvy

I presume this is the "Kronecker delta"- $\delta_{ii}= 1$, $\delta_{ij}= 0$ if $i\ne j$.
Then $\delta_{ij}\delta_{jk}= \delta_{i1}\delta_{1k}+ \delta_{i2}\delta_{2j}+ \delta_{i3}\delta_{3j}$
If i= 1, k= 1, then only the first term is non-zero and it is 1: $\delta_{11}= 1+ 0+ 0= 1[/quote]. If i= 1, k= 2, then at least one of the factors in every term is 0: [itex]\delta_{12}= 0+ 0+ 0= 0[/quote]. If i= 1, k= 3, then at least one of the factors in every term is 0: [itex]\delta_{13}= 0+ 0+ 0= 0. If i= 2, k= 2, then only the second term is non-zero and it is 1: [itex]\delta_{22}= 0+ 1+ 0= 1. etc. 13. Feb 24, 2016 FluidStu Ok, so then cik = ai1b1k + ai2b2k + ai3b3k? In this way I can see that A=B=C=I. However, I still don't see how this leads to the proof that δijδjk= δik. If you want to expand on this for others visiting the thread, feel free. However, I can see it clearly from this: In all the cases where i = k, we end up with 1, while in all the cases where i ≠ k, we end up with 0. In other words, we end up with δik. Very good, thank you both for your help. 14. Feb 24, 2016 blue_leaf77 That the Kronecker delta is defined as $\delta_{ik}=1$ for $i=k$ and zero for $i\neq k$, doesn't it remind you of the elements of the identity matrix? 15. Feb 24, 2016 FluidStu Yes, it does. I can see that the Kronecker Delta in matrix form is simply an identity matrix. But how does this help me to prove: ? 16. Feb 24, 2016 blue_leaf77 \delta_{ij}\delta_{jk} = \delta_{i1}\delta_{1k} + \delta_{i2}\delta_{2k} + \delta_{i3}\delta_{3k} = \delta_{ik} Now you already have for the matrix equation $AB=C$. If A=B=C=I, you will only need to replace $a$, $b$, and $c$ with $\delta$ in that quoted equation. 17. Feb 24, 2016 FluidStu I see that replacing each with δ gives the required proof. But is it "ok" to replace a, b and c with δ? Aren't a, b and c simply elements of matrices? How can we just replace them with δ? 18. Feb 24, 2016 zinq At the risk of repetition: To put the essential point in one place: Let A = (aij), B = (bij), and C = (cij) be n-dimensional matrices such that AB = C.​ Then you know from the definition of matrix multiplication that cij = Σ aik bkj for any i, j in the range 1 ≤ i, j ≤ n, where the sum is over k = 1,...,n. Specializing to the case of n = 3, for the case of A = B = C = I3 (the 3D identity matrix), and then using the repeated-index summation convention, you get δik δkj = δij for any i, j in the range 1 ≤ i, j ≤ 3. 19. Feb 24, 2016 blue_leaf77 Note that the last sentence in post #16 is a conditional statement, look what the condition is in order for the conclusion to be true. 20. Feb 24, 2016 PeroK Here's an alternative proof (for which I'll abandon the summation convention): $\Sigma_{j} \delta_{ij} \delta_{jk} = \delta_{ii} \delta_{ik} = \delta_{ik} \ \$ (as $\delta_{ij} = 0 \ (i \ne j)$) Or, more laboriously: $\Sigma_{j} \delta_{ij} \delta_{jk} = \delta_{ii} \delta_{ik} + \Sigma_{j \ne i} \delta_{ij} \delta_{jk} = \delta_{ik} + 0 \ \$ (as $\delta_{ii} = 1, \ \delta_{ij} = 0 \ (i \ne j)$) 21. Feb 25, 2016 Edgardo Try to get an intuitive feeling for what the Kronecker-Delta [itex]\delta_{ij}$ does.
Consider this sum:
$\sum_{j=1}^{n} c_j = c_1 + c_2 + \dots + c_n$

Now observe what happens if we multiply with the Kronecker-Delta:
$\sum_{j=1}^{n} \delta_{ij} c_j = \delta_{i1} c_1 + \delta_{i2}c_2 + \dots + \delta_{ii}c_i + \dots + \delta_{in}c_n$

There will be exactly one summand with $j=i$, that is $\delta_{ii} c_i = 1 \cdot c_i = c_i$.
For the remaining summands with $j \neq i$ we have $\delta_{ij} c_j = 0 \cdot c_j = 0$
(by the definition of the Kronecker-Delta).

We get:
$\sum_{j=1}^{n} \delta_{ij} c_j$
$= \delta_{i1} c_1 + \delta_{i2}c_2 + \dots + \delta_{ii}c_i + \dots + \delta_{in}c_n$
$= 0 \cdot c_1 + 0 \cdot c_2 + \dots + 1 \cdot c_i + \dots + 0 \cdot c_n$
$= c_i$

With the Einstein summation convention we have $\delta_{ij} c_j = c_i$.
Intuitively, $\delta_{ij}$ evaluates the expression $c_j$ at $j=i$.

Now, similarly, $\delta_{ij} \delta_{jk} = \delta_{ik}$ since
intuitively $\delta_{ij}$ evaluates the expression $\delta_{jk}$ at $j=i$.

22. Feb 25, 2016

FluidStu

The condition being that A=B=C=I≡ δij? Then we may replace the a, b and c with δ and the following two are equivalent:

Having this then makes it easy to see how both post #12 and #21 explain the required proof. Thanks.