# Show that the Kronecker delta retains its form under any transformation

#### blue24

Homework Statement
Show that 2nd order tensor, a*d_ij, where a is an arbitrary constant, retains its form under any transformation, Q_ij. This form is then an isotropic 2nd order tensor.
Homework Equations
a'_ij = Q_ip * Q_jq * a_pq (General transformation relation for 2nd order tensor)
Backstory - I have not been in school for 5ish years, and am returning to take some grad classes in the field of Solid Mechanics. I am freaking out a bit about the math (am rusty). I have not started class yet, but figured I would get my books and start working through problems. This problem is from my Theory of Elasticity Class.

The book provides the general transformation relation for a 2nd order tensor. Applying a rotation to the given tensor, a*d_ij:

a*d'_ij = Q_ip * Q_jq * a*d_pq

I am not sure where to go from here. My understanding is that the repeated indices "p" and "q" imply summation, so then the first term in the matrix a*d'_ij would be:

a*d'_11 = Q_1p*Q_1q*a*d_pq = a*[Q_11*Q_11*d_11 + Q_12*Q_12*d_22 + Q_13*Q_13*d_33]

Since d_11 = d_22 = d_33 = 1, then a*d'_11 = a*[Q_11*Q_11 + Q_12*Q_12 + Q_13*Q_13]

But my guess is I've gone wrong somewhere, because I don't know what I do with this. My apologies if this is a dumb question. Still trying to get my head wrapped around this. Thank you.

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#### strangerep

What does your book give as the definition of $Q_{ij}$ ?

#### Michael Price

It is best to absorb notions of raised indices (contravariant) and lowered indices (covariant) first, before answering questions about kronecker-delta (and levi-civta).

#### blue24

What does your book give as the definition of $Q_{ij}$ ?
Thank you for the response. The book defines $Q_{ij}$ as the angle between the $x^{'}_i$ and $x_j$ axes.

#### blue24

It is best to absorb notions of raised indices (contravariant) and lowered indices (covariant) first, before answering questions about kronecker-delta (and levi-civta).
Thank you for the reply. Unfortunately those terms, "covariant" and "contravariant", are not in my textbook. I have a pdf (legally purchased) and I searched for them. I will do some searching online.

What is throwing me is that the textbook literally says, "[referring to isotropic tensors] It can easily be verified (see Exercise 1.8) that the Kronecker delta $d_{ij}$ has such a property." So I figured, hey I better do Exercise 1.8, it'll be easy. And I'm stuck.

Can anyone give me any hints as to where I am going wrong, what steps I should follow, etc? I want to repeat - I am not even in the class yet, first day of class is September 6. Most likely this will never actually be an assigned homework problem. I am just trying to work ahead because I know I will need to do so in order to get help. So I am not posting on here trying to cut corners and get you guys to do my homework for me or anything like that.

Thanks for any help you can provide!
Andrew

#### strangerep

Thank you for the response. The book defines $Q_{ij}$ as the angle between the $x^{'}_i$ and $x_j$ axes.
Well, either that's not exactly what the book says, or it's being extremely sloppy.
You should probably post a link so I can look at what the book actually says.

I suspect what's meant is that $Q_{ij}$ denotes the transformation coefficients corresponding to a certain 3D rotation. I.e., probably something like
$$Q_{ij} ~=~ \frac{\partial x'_i}{\partial x_j} ~~.$$ You also need to know that for any matrix with components $A_{ij}$, we have $$A_{ij} \delta_{jk} ~=~ A_{ik} ~,$$ where there is an implicit summation over repeated indices, in this case $\sum_j$. The above is true because is just matrix multiplication of $A$ by the identity matrix $1$ (all 1's on the diagonal).

• blue24 and Michael Price

#### Michael Price

Thank you for the reply. Unfortunately those terms, "covariant" and "contravariant", are not in my textbook. I have a pdf (legally purchased) and I searched for them. I will do some searching online.

What is throwing me is that the textbook literally says, "[referring to isotropic tensors] It can easily be verified (see Exercise 1.8) that the Kronecker delta $d_{ij}$ has such a property." So I figured, hey I better do Exercise 1.8, it'll be easy. And I'm stuck.

Can anyone give me any hints as to where I am going wrong, what steps I should follow, etc? I want to repeat - I am not even in the class yet, first day of class is September 6. Most likely this will never actually be an assigned homework problem. I am just trying to work ahead because I know I will need to do so in order to get help. So I am not posting on here trying to cut corners and get you guys to do my homework for me or anything like that.

Thanks for any help you can provide!
Andrew
Upon reflection the contravariant/covariant thing might not be relevant, since you are doing 3D tensors. They come into their own for 4D tensor transformations.

#### blue24

Well, either that's not exactly what the book says, or it's being extremely sloppy.
My apologies! The book defines $Q_{ij}$ as the cosine of the angle between the $x^{'}_{i}$ and $x_j$ axes. See attachment.

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#### blue24

Ok, so I'd like to zoom out here and get a better understanding of $Q_{ij}$. I'm trying to do a basic example problem where the solution is given in the book (see attached image).

I can calculate the transformation matrix, $Q_{ij}$ (did it without looking at the solution, woot!), and do the transformation of the vector, $a_{i}$, no problem. However I am stumped by the transformation of the 2nd order tensor, $a_{ij}$. The solution given points you to the equation $a^{'}_{ij} = Q_{ip}Q_{jq}a_{pq}$. I don't understand how this turns into $Q_{ij}a_{ij}Q^{T}_{ij}$.

I guess one of the things I am fundamentally not understanding is what is the difference between $Q_{ip}$ and $Q_{jq}$ in the equation you are supposed to use? And how are they different than $Q_{ij}$?

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#### strangerep

I can calculate the transformation matrix, $Q_{ij}$ (did it without looking at the solution, woot!), and do the transformation of the vector, $a_{i}$, no problem. However I am stumped by the transformation of the 2nd order tensor, $a_{ij}$. The solution given points you to the equation $a^{'}_{ij} = Q_{ip}Q_{jq}a_{pq}$. I don't understand how this turns into $Q_{ij}a_{ij}Q^{T}_{ij}$.
It doesn't. It turns into
$$a^{'}_{ij} ~=~ Q_{ip} a_{pq} Q_{jq} ~=~ Q_{ip} a_{pq} Q^T_{qj} ~=~ \Big(Q a Q^T)_{ij} ~.$$
The $i,j,$ are ordinary indices -- they must match up on both sides of the equality. The $p,q$ indices are dummy summation indices (implicit sum over repeated indices). I.e., there's an implicit $\sum_p \sum_q$. See Einstein summation convention. (That reference is for a more general case where there are both upstairs and downstairs indices, whereas in your case all indices are downstairs. Same principle though.)

I guess one of the things I am fundamentally not understanding is what is the difference between $Q_{ip}$ and $Q_{jq}$ in the equation you are supposed to use? And how are they different than $Q_{ij}$?
Trying sticking those explicit "$\sum$" symbols back into the equation, and then relate it to how one multiplies matrices.

Btw, I sense that you really need to study some books on linear algebra and vector/tensor calculus. The Schaum Outline series are usually good for covering a lot of material quickly, with example and exercises. Otherwise, ask for suggestions in the Academic Guidance forum.

• blue24 and Michael Price

#### Michael Price

Ok, so I'd like to zoom out here and get a better understanding of $Q_{ij}$. I'm trying to do a basic example problem where the solution is given in the book (see attached image).

I can calculate the transformation matrix, $Q_{ij}$ (did it without looking at the solution, woot!), and do the transformation of the vector, $a_{i}$, no problem. However I am stumped by the transformation of the 2nd order tensor, $a_{ij}$. The solution given points you to the equation $a^{'}_{ij} = Q_{ip}Q_{jq}a_{pq}$. I don't understand how this turns into $Q_{ij}a_{ij}Q^{T}_{ij}$.

I guess one of the things I am fundamentally not understanding is what is the difference between $Q_{ip}$ and $Q_{jq}$ in the equation you are supposed to use? And how are they different than $Q_{ij}$?
Have a thorough read of

• blue24

#### blue24

Btw, I sense that you really need to study some books on linear algebra and vector/tensor calculus. The Schaum Outline series are usually good for covering a lot of material quickly, with example and exercises. Otherwise, ask for suggestions in the Academic Guidance forum.
Thank you for the feedback. This is exactly what I am trying to figure out. This class is going to be a heavy lift for me, and so I need to figure out what the gaps are in my understanding and start working on addressing those. I have not taken a linear algebra class (was optional in my undergrad), and so I think that is a definite weakness. I just ordered Schaum's Outline of Linear Algebra.

#### blue24

It doesn't. It turns into
$$a^{'}_{ij} ~=~ Q_{ip} a_{pq} Q_{jq} ~=~ Q_{ip} a_{pq} Q^T_{qj} ~=~ \Big(Q a Q^T)_{ij} ~.$$
I don't understand how you got from $Q_{ip} Q_{jq} a_{pq}$ to $Q_{ip} a_{pq} Q_{jq}$. Why did the order of the terms change?

#### nrqed

Homework Helper
Gold Member
I don't understand how you got from $Q_{ip} Q_{jq} a_{pq}$ to $Q_{ip} a_{pq} Q_{jq}$. Why did the order of the terms change?
It's because of the transpose (you dropped it in your question). What happens here is that $Q_{jq} = (Q^T)_{qj}$ which is just saying that the jq entry of a matrix is equal to the qj entry of the transpose of the matrix.

#### blue24

It's because of the transpose (you dropped it in your question). What happens here is that $Q_{jq} = (Q^T)_{qj}$ which is just saying that the jq entry of a matrix is equal to the qj entry of the transpose of the matrix.
The given equation from the book is: $a^{'}_{ij}=Q_{ip}Q_{jq}a_{pq}$ (Eq 1)

Per strangerrep's post from Yesterday at 1:55PM, this "turns into $a^{'}_{ij}=Q_{ip}a_{pq}Q_{jq}$..." (Eq 2)

He does not introduce the transpose until after shifting the $a_{pq}$ term before the $Q_{jq}$ term. I don't understand what allows for this shift. In other words, how are equations 2 and 3 equal?

I do understand that $Q_{jq} = (Q^T)_{qj}$

#### nrqed

Homework Helper
Gold Member
The given equation from the book is: $a^{'}_{ij}=Q_{ip}Q_{jq}a_{pq}$ (Eq 1)

Per strangerrep's post from Yesterday at 1:55PM, this "turns into $a^{'}_{ij}=Q_{ip}a_{pq}Q_{jq}$..." (Eq 2)

He does not introduce the transpose until after shifting the $a_{pq}$ term before the $Q_{jq}$ term. I don't understand what allows for this shift. In other words, how are equations 2 and 3 equal?

I do understand that $Q_{jq} = (Q^T)_{qj}$
I am not sure what Eqs 2 and 3 are. I Guess you mean Eqs 1 and 2.
These are just numbers so you can move them around and put them in any order you want $Q_{jq} a_{pq}=a_{pq}Q_{jq}$ for example.

#### strangerep

@blue24 : When asking questions out of a textbook like you're doing in this thread, it's wise (even mandatory) to say exactly which textbook it is, because sometimes questions can be answered more effectively by pointing you to another section of the book.

(I haven't responded to your most recent questions above, because nrqed already did.)

#### blue24

I am not sure what Eqs 2 and 3 are. I Guess you mean Eqs 1 and 2.
These are just numbers so you can move them around and put them in any order you want $Q_{jq} a_{pq}=a_{pq}Q_{jq}$ for example.
Sorry, yes, Eqs 1 and 2.

Thanks. I am definitely confused on this point. I thought the terms were matrices, not numbers. If you look at the book solution (my post from yesterday at 12:46), it appears that they are matrices. And I thought that order matters for matrix multiplication.

#### blue24

@blue24 : When asking questions out of a textbook like you're doing in this thread, it's wise (even mandatory) to say exactly which textbook it is, because sometimes questions can be answered more effectively by pointing you to another section of the book.

(I haven't responded to your most recent questions above, because nrqed already did.)
Good idea. The book I am using is Elasticity: Theory, Applications, and Numerics; Third Edition; by Martin Sadd.

#### nrqed

Homework Helper
Gold Member
Sorry, yes, Eqs 1 and 2.

Thanks. I am definitely confused on this point. I thought the terms were matrices, not numbers. If you look at the book solution (my post from yesterday at 12:46), it appears that they are matrices. And I thought that order matters for matrix multiplication.
As soon as we have indices, like $Q_{pq}$, we are talking about the entries of the matrix, which are just numbers. If we write $Q$ , we mean the matrix and the order matters.

#### blue24

As soon as we have indices, like $Q_{pq}$, we are talking about the entries of the matrix, which are just numbers. If we write $Q$ , we mean the matrix and the order matters.
Thanks for the reply. That was my understanding before I got the textbook for this class and started reading through it. However, this textbook seems to use a notation where $a_{ij}$ represents a matrix. See the image below.

Also, in order to calculate the solution, we get to $Q_{ip}a_{pq}Q^{T}_{qj}$. We are transposing the third symbol in the term, $Q_{qj}$. It only makes sense to transpose $Q_{qj}$ if it is a matrix, and not a number, right?

I have no doubt that you are right and I am wrong. Just trying to highlight the things I don't understand, and I'm hoping to see the light.

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#### nrqed

Homework Helper
Gold Member
You have good questions. The problem is the sloppiness of notation that is unfortunately widespread, which makes things confusing indeed.

1) When we write $Q_{qj}^T$, we mean $(Q^T)_{qj}$. The parentheses are important, but people usually drop them. So we mean the qj entry of the matrix $Q^T$.

2) As for the notation used by the book, you are right that it is confusing. I am a physicist often working with mathematicians and I have been scolded by them more than once for writing things this way. Now I realize they they are entirely right, one should never write something like that!! When people write a_{ij} = a matrix, they are badly abusing notation. What they mean is that if we let the indices range over all their values, we span the entire matrix. But it is incorrect to write that a_{ij} is equal to a matrix. Unfortunately, this is an abuse often done by physicists.

To be precise, a_{ij} is always an entry in the matrix, when we work with equations and manipulate terms.

• blue24

#### blue24

Thank you. That is a really helpful answer. So when I am looking at equations which are written in index notation, I should treat the elements of the equation as numbers, not matrices.

For me, that begs a new question. In order to arrive at the final solution (see first attached image), matrices are used. I don't understand the logic that determines what order the matrices are multiplied in or why the one matrix is transposed.

I think I understand the basic concepts of Einstein notation (is explained in the book, and strangerep explained it above as well). In order to convince myself that I understand Einstein notation, I calculated the first term of the transformed matrix, $a^{'}_{11}$ using summation, and then also calculated it using $(QaQ^{T})_{ij}$ and found them to be the same (see second attached image).

So I understand that $(QaQ^{T})_{ij}$ gets you the final answer. I just don't understand the thought process for getting to $(QaQ^{T})_{ij}$.

My guess is that I simply need to shore up my fundamentals, which I am planning on doing. If there is no clear answer to my question, no worries. I will give up on this problem for now and come back when I have the background to understand. Thanks everyone for the help.

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#### strangerep

I don't understand the logic that determines what order the matrices are multiplied in or why the one matrix is transposed.
OK. Consider the column vector in your example, whose components are denoted $a_i$. Do you understand why $$a'_i ~=~ Q_{ij} a_j ~~~?$$ And what is this in matrix language? (Once you're clear on this, we'll move on to 2nd rank tensors.)

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