Summation Convention – Substitution Rule

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.

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

Homework Helper
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?
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

FluidStu said:
Many thanks.

FluidStu
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
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.

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

FluidStu
What do you get when multiplying an identity matrix with another identity matrix?
I * I = I

Homework Helper
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##?

FluidStu
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##?
cik = aiibik + aijbjk + aikbkk

Homework Helper
cik = aiibik + aijbjk + aikbkk
Almost, just that the index to be summed must be written as numbers, not as indices as you did there.

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

cik = a11b13 + a12b23 + a13b33 ?

Homework Helper
Do you mean:

cik = a11b13 + a12b23 + a13b33 ?
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##.

Homework Helper
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.
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. FluidStu FluidStu 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##. 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: Then δijδjk=δi1δ1k+δi2δ2j+δi3δ3jδijδjk=δi1δ1k+δi2δ2j+δi3δ3j\delta_{ij}\delta_{jk}= \delta_{i1}\delta_{1k}+ \delta_{i2}\delta_{2j}+ \delta_{i3}\delta_{3j} 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. Science Advisor Homework Helper However, I still don't see how this leads to the proof that δijδjk= δik. 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? FluidStu 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? 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: δijδjk= δik ? Science Advisor Homework Helper \delta_{ij}\delta_{jk} = \delta_{i1}\delta_{1k} + \delta_{i2}\delta_{2k} + \delta_{i3}\delta_{3k} = \delta_{ik} Now you already have cik = ai1b1k + ai2b2k + ai3b3k 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. FluidStu FluidStu \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. 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? 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. FluidStu Science Advisor Homework Helper 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? 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. Science Advisor Homework Helper Gold Member 2022 Award 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)##) FluidStu 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 $j=i$.

FluidStu
FluidStu
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.
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:

cik = ai1b1k + ai2b2k + ai3b3k
δijδjk=δi1δ1k+δi2δ2k+δi3δ3k=δik

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