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:
δ_{i j} δ_{j k} = δ_{i 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:
δ_{i j} δ_{j k} = δ_{3 1} δ_{1 1} + δ_{3 2} δ_{2 1} + δ_{3 3} δ_{3 1} = ?

Is my expansion correct?

Many thanks.

 

[blue_leaf77]

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.

 

[HallsofIvy]

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:
δ_{i j} δ_{j k} = δ_{i 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:
δ_{i j} δ_{j k} = δ_{3 1} δ_{1 1} + δ_{3 2} δ_{2 1} + δ_{3 3} δ_{3 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:
δ_ijδ_jk= δ_i1δ_1k+ δ_i2δ_2k+ δ_i3δ_3k

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:
δ_ijδ_jk= δ_i1δ_1k+ δ_i2δ_2k+ δ_i3δ_3k

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 δ_ik. This was really the point in my question – I would like to see step-wise how δ_ijδ_jk = δ_ik.

 

[blue_leaf77]

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

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

 

[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 $B$?

 

[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 $B$?

c_ik = a_iib_ik + a_ijb_jk + a_ikb_kk

 

[blue_leaf77]

c_ik = a_iib_ik + a_ijb_jk + a_ikb_kk

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:

c_ik = a₁₁b₁₃ + a₁₂b₂₃ + a₁₃b₃₃ ?

 

[blue_leaf77]

Do you mean:

c_ik = 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 $A=B=C=I$.

 

[HallsofIvy]

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 δ_ik. This was really the point in my question – I would like to see step-wise how δ_ijδ_jk = δ_ik.

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].<br /> If i= 1, k= 2, then at least one of the factors in every term is 0: \delta_{12}= 0+ 0+ 0= 0[/quote].&lt;br /&gt; If i= 1, k= 3, then at least one of the factors in every term is 0: \delta_{13}= 0+ 0+ 0= 0.&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; If i= 2, k= 2, then only the second term is non-zero and it is 1: \delta_{22}= 0+ 1+ 0= 1.&amp;amp;lt;br /&amp;amp;gt; etc.

 

[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 c_ik = a_i1b_1k + a_i2b_2k + a_i3b_3k?

In this way I can see that A=B=C=I. However, I still don't see how this leads to the proof that δ_ijδ_jk= δ_ik. 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 δ_ik. Very good, thank you both for your help.

 

[blue_leaf77]

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

?

 

[blue_leaf77]

$$\delta_{ij}\delta_{jk} = \delta_{i1}\delta_{1k} + \delta_{i2}\delta_{2k} + \delta_{i3}\delta_{3k} = \delta_{ik}$$
Now you already have

c_ik = a_i1b_1k + a_i2b_2k + a_i3b_3k

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]

$$\delta_{ij}\delta_{jk} = \delta_{i1}\delta_{1k} + \delta_{i2}\delta_{2k} + \delta_{i3}\delta_{3k} = \delta_{ik}$$
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 = (a_ij), B = (b_ij), and C = (c_ij) be n-dimensional matrices such that

AB = C.​

Then you know from the definition of matrix multiplication that

c_ij = Σ a_ik b_kj​

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

δ_ik δ_kj = δ_ij

for any i, j in the range 1 ≤ i, j ≤ 3.

 

[blue_leaf77]

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.

 

[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 $\delta_{ii} = 1, \ \delta_{ij} = 0 \ (i \ne j)$)

 

[Edgardo]

Try to get an intuitive feeling for what the Kronecker-Delta \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]

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.