Gradient of higher rank tensor

[chowdhury]

TL;DR Summary

Algebraic manipulation

How to write following equation in index notation?

$$\nabla \cdot \left( \mathbf{e} : \nabla_{s} \mathbf{u} \right)$$
where $e$ is a third rank tensor, $u$ is a vector, $\nabla_{s}$ is the symmetric part of the gradient operator, : is the double dot product.

The way I approached is
$$\mathbf{e} : \nabla_{s} \mathbf{u} = e_{iJ} \nabla_{Jk} u_{k}$$, where i = x,y,z, and J = 1,2,3,4,5,6
$$\nabla \cdot \left( \mathbf{e} : \nabla_{s} \mathbf{u} \right) = \nabla_{i} e_{iJ} \nabla_{Jk} u_{k} = \frac{\partial e_{iJ}}{\partial r_{i}} \nabla_{Jk} u_{k} = e_{iJ} \nabla_{i} \nabla_{Jk} u_{k} = e_{iJ} \nabla_{i} u_{k, \color{red}{kJ}} = e_{iJ} u_{k, \color{red}{ikJ}} $$

But in the book it is written as

$$ \nabla_{i} e_{iL} \nabla_{Lj} u_{j} = \color{red}{e_{ikj} u_{j, ik}}$$

I am from the book, B. A. Auld, Acoustic fields and waves in solids, vol. 1, 1973.

 

[Orodruin]

The notation used by your book is troublesome at best. In particular, using two indices on the single derivative is highly confusing and leads you to obtain three indices after a comma in the $u_{k,ikJ}$. It is not at all clear what this would mean. I suggest you throw the notation with the capital indices out of the window or use a more transparent way of writing the translation from symmetric rank 2 tensors to capital indices. For example, you could introduce an object $S_{Jik}$ symmetric in $ik$ whose only purpose is translating between the 6-index and two symmetric 3-index notations although personally I would avoid this.

Anyway, for your case, the easiest way forward is (assuming $e$ is symmetric in the last two indices:
$$
\nabla\cdot {\bf e}:\nabla_s {\bf u} =
\partial_i e_{ijk} \partial_j u_k =
e_{ijk} u_{k,ji}
$$
If $e$ is not symmetric, only the symmetric part will contribute, but it can be helpful to explicitly note the symmetrization:
$$
\nabla\cdot {\bf e}:\nabla_s {\bf u} =
\partial_i e_{ijk} \partial_{\{j} u_{k\}}
= e_{ijk} u_{\{k,j\}i}.
$$

Summary:: Algebraic manipulation

Note: If you do not know the answer, that is fine, but please be respectful to address the query, not telling me, it is so obvious, why din't the you read the whole book, etc. etc. It is most likely that I am missing something that I do not realize.

This is not a good way to start a thread where you ask to be helped for free. It comes across as ungrateful from the beginning and is more likely to make people who know the answer simply not answer.

 

[chowdhury]

The notation used by your book is troublesome at best. In particular, using two indices on the single derivative is highly confusing and leads you to obtain three indices after a comma in the $u_{k,ikJ}$. It is not at all clear what this would mean. I suggest you throw the notation with the capital indices out of the window or use a more transparent way of writing the translation from symmetric rank 2 tensors to capital indices. For example, you could introduce an object $S_{Jik}$ symmetric in $ik$ whose only purpose is translating between the 6-index and two symmetric 3-index notations although personally I would avoid this.

Anyway, for your case, the easiest way forward is (assuming $e$ is symmetric in the last two indices:
$$
\nabla\cdot {\bf e}:\nabla_s {\bf u} =
\partial_i e_{ijk} \partial_j u_k =
e_{ijk} u_{k,ji}
$$
If $e$ is not symmetric, only the symmetric part will contribute, but it can be helpful to explicitly note the symmetrization:
$$
\nabla\cdot {\bf e}:\nabla_s {\bf u} =
\partial_i e_{ijk} \partial_{\{j} u_{k\}}
= e_{ijk} u_{\{k,j\}i}.
$$This is not a good way to start a thread where you ask to be helped for free. It comes across as ungrateful from the beginning and is more likely to make people who know the answer simply not answer.

@Orodruin : Thanks for the help. I was seeking for the answer for quite a while, until your response. This clarified my query.

1.) How would I write the transpose of a third rank tensor?
$d_{ijk}^{t} = d_{kji}$, then it is equal to $= d_{kij}$, is it because of symmetry?

Reference:
I found in Wikipedia , strain-charge form
https://en.wikipedia.org/wiki/Piezoelectricity

2.) How to represent indices in double differentiation?
$$\partial_{i} \partial_{j} u_{k} = u_{k, ji} \color{red}{\textup{ or }} u_{k, ij}$$

3.) In the book, I do not understand how the divergence of product of a third rank tensor and gradient of a vector is written as product, meaning taken out of the parentheses ( ),like this.

$$\nabla \cdot (\mathbf{e} : \nabla_{s} \mathbf{u}) = \frac{\partial }{\partial r_{i}} e_{iJ} \nabla_{Jk} u_{k} = \nabla_{i}e_{iJ} \nabla_{Jk}u_{k}$$

4.) Same issue here
$$\nabla \cdot (\mathbf{\epsilon}^{S} \cdot \nabla \mathbf{\phi}) = \nabla \cdot \mathbf{\epsilon}^{S} \cdot \nabla \mathbf{\phi} = \epsilon_{ij}^{S} \frac{\partial }{\partial r_{i}} \frac{\partial \phi}{\partial r_{j}} = \epsilon_{ij}^{S} \phi_{ij} \color{red}{\textup{ or }} \epsilon_{ij}^{S} \phi_{ji} $$

5.) How to I choose the appropriate index in, what I mean $u_{k}$ or $u_{i}$
$$
\nabla\cdot {\bf e}:\nabla_s {\bf u} =
\partial_i e_{ijk} \partial_j u_k =
e_{ijk} u_{k,ji}
$$

 

[Orodruin]

1.) How would I write the transpose of a third rank tensor?

First you would have to define what you mean by ”transpose”. It is not a generally applicable term and is typically only used for matrices. The more useful concept is whether or not a tensor has symmetries (or antisymmetries, or both), implying relations between the components. This can be between any pair of indices.

I strongly suggest looking at a textbook that is not 50 years old for a more modern approach.

2.) How to represent indices in double differentiation?

Partial derivatives commute (at least under conditions applicable in most physics applications).

3 and 4) Parentheses are mainly for keeping track of order of operations and what the derivative acts on. If you can do this without the parenthes, they are not required.

5) It does not matter what dummy indices are named.

 

[chowdhury]

@Orodruin : Thanks for the help. I was seeking for the answer for quite a while, until your response. This clarified my query.

1.) How would I write the transpose of a third rank tensor?
$d_{ijk}^{t} = d_{kji}$, then it is equal to $= d_{kij}$, is it because of symmetry?

Reference:
I found in Wikipedia , strain-charge form
https://en.wikipedia.org/wiki/Piezoelectricity

2.) How to represent indices in double differentiation?
$$\partial_{i} \partial_{j} u_{k} = u_{k, ji} \color{red}{\textup{ or }} u_{k, ij}$$

3.) In the book, I do not understand how the divergence of product of a third rank tensor and gradient of a vector is written as product, meaning taken out of the parentheses ( ),like this.

$$\nabla \cdot (\mathbf{e} : \nabla_{s} \mathbf{u}) = \frac{\partial }{\partial r_{i}} e_{iJ} \nabla_{Jk} u_{k} = \nabla_{i}e_{iJ} \nabla_{Jk}u_{k}$$

4.) Same issue here
$$\nabla \cdot (\mathbf{\epsilon}^{S} \cdot \nabla \mathbf{\phi}) = \nabla \cdot \mathbf{\epsilon}^{S} \cdot \nabla \mathbf{\phi} = \epsilon_{ij}^{S} \frac{\partial }{\partial r_{i}} \frac{\partial \phi}{\partial r_{j}} = \epsilon_{ij}^{S} \phi_{ij} \color{red}{\textup{ or }} \epsilon_{ij}^{S} \phi_{ji} $$

5.) How to I choose the appropriate index in, what I mean $u_{k}$ or $u_{i}$
$$
\nabla\cdot {\bf e}:\nabla_s {\bf u} =
\partial_i e_{ijk} \partial_j u_k =
e_{ijk} u_{k,ji}
$$View attachment 297815

@Orodruin : If there a way to thank you by offering and serving you a good coffee, I would be pleased to do so to you. But we live far apart, so Thank you is all I can say at this moment.

I understand your suggestion on the recent book, but this is what I was given and for some reason, I have to stick to it for reference purposes.

Regarding the maths operation, my confusion in $\nabla \cdot (\bf{c}^{E} : \nabla_{s} \bf{u})$ is that shall I not take the double sum between $\bf{c}^{E}$ and $\bf{u}$ first, then take the divergence operator? I do not understand how it is written as $\nabla \cdot (\bf{c}^{E} : \nabla_{s} \bf{u}) = \nabla_{iK} c_{KL}^{E} \nabla_{Lj}u_{j}$

 

[chowdhury]

@Orodruin :
I understand the transpose operator appears in matrix operations, however, I have seen the transpose operation in a 3rd rank tensor in Wikipedia, as suggested by one of the moderators, @andrewkirk. What I am lacking how to represent them. Thanks for your clarification.

How to say $d^{transpose}$, where $d$ is a third rank tensor.

Reference :
https://en.wikipedia.org/wiki/Piezoelectricity

 

[Orodruin]

In that particular case the meaning may be subtextual as referring to a particular index reordering which, when represented in matrix form and using the single index for the symmetric two-index space, corresponds to the matrix transpose. I would certainly not call it standard nomenclature in tensor analysis.

Furthermore, I have always abhorred the non-indexed formulas such as the ones given in your book (as presented in the OP etc) as they many times are ambiguous in terms of what indices are actually contracted. Using index or abstract index notation serves a much clearer purpose.

 

[anuttarasammyak]

Symbol $\cdot$ is familiar to show inner product of two vectors. I am afraid that this book use this symbol for contraction which higher rank tensors are involved. Say $A^{ik}=b^ic^k$, is
$$\nabla \cdot A^{ik}$$
$$=(\nabla \cdot c) b^{i}?$$or
$$=(\nabla \cdot b) c^{k}?$$
Writing surviving index and index for contraction prevent such a confusion. I expect in the beginning pages of the book, the definition of symbols $\cdot$ and : are explained.
I have not seen the symbol : before for such a double product or convention meaning.

 

[chowdhury]

Symbol $\cdot$ is familiar to show inner product of two vectors. I am afraid that this book use this symbol for contraction which higher rank tensors are involved. Say $A^{ik}=b^ic^k$, is
$$\nabla \cdot A^{ik}$$
$$=(\nabla \cdot a) b^{k}?$$or
$$=(\nabla \cdot b) c^{i}?$$
Writing surviving index and index for contraction prevent such a confusion. I expect in the beginning pages of the book, the definition of symbols $\cdot$ and : are explained.
I have not seen the symbol : before for such a double product or convention meaning.

@anuttarasammyak Here is the notation for double dot in the book,

 

[anuttarasammyak]

Thanks. So you know the definition of the symbols. Why don't you apply these definitions to your problem? Did you find a difficulty in doing it ?

[EDIT] I take $\cdot$ as contraction for the last or the rightest index and : as contraction for the last or the rightest pair indices.

 

[Orodruin]

Symbol $\cdot$ is familiar to show inner product of two vectors. I am afraid that this book use this symbol for contraction which higher rank tensors are involved. Say $A^{ik}=b^ic^k$, is
$$\nabla \cdot A^{ik}$$
$$=(\nabla \cdot c) b^{i}?$$or
$$=(\nabla \cdot b) c^{k}?$$
Writing surviving index and index for contraction prevent such a confusion. I expect in the beginning pages of the book, the definition of symbols $\cdot$ and : are explained.
I have not seen the symbol : before for such a double product or convention meaning.

Precisely my point in the second paragraph of #7 illustrated. The notation is horribly obscure and not generalisable to arbitrary tensors (it effectively only works when the only tensors involved are rank 1 and symmetric rank 2 tensors (and their linear relationships in terms of rank 2, 3, and 4 tensors).

 

[chowdhury]

In that particular case the meaning may be subtextual as referring to a particular index reordering which, when represented in matrix form and using the single index for the symmetric two-index space, corresponds to the matrix transpose. I would certainly not call it standard nomenclature in tensor analysis.

Furthermore, I have always abhorred the non-indexed formulas such as the ones given in your book (as presented in the OP etc) as they many times are ambiguous in terms of what indices are actually contracted. Using index or abstract index notation serves a much clearer purpose.

@Orodruin :
I have noticed that in the field of piezoelectricity, "they" denote like this, Here is another book,

https://www.amazon.com/dp/3540686800/?tag=pfamazon01-20

What is driving me crazy is this statement, from the end of this book

Acoustic Fields and Waves in Solids. Volume I Hardcover – April 20, 1973​

https://www.amazon.com/dp/0471037001/?tag=pfamazon01-20

I understand and agree to all you said in this thread, but not $e_{iJ} = e_{Ji}$ in page 378 of the book.

 

[chowdhury]

Thanks. So you know the definition of the symbols. Why don't you apply these definitions to your problem? Did you find a difficulty in doing it ?

I applied the defintions, but could not find the reasonable notation, that is the problem.

 

[chowdhury]

Precisely my point in the second paragraph of #7 illustrated. The notation is horribly obscure and not generalisable to arbitrary tensors (it effectively only works when the only tensors involved are rank 1 and symmetric rank 2 tensors (and their linear relationships in terms of rank 2, 3, and 4 tensors).

@Orodruin : I was told this book is the Bible, and there is no mistake in it. Without complaining, which is easy to rant to anybody in the world, I took an introspective view and try to educate myself.

 

[anuttarasammyak]

I applied the defintions, but could not find the reasonable notation, that is the problem.

Let me know for examination, how many ranks or indexes $\nabla$, $\nabla_s$, e and u have ?
Please write them down explicitly with indexes.

 

[chowdhury]

@Orodruin:
@anuttarasammyak:

I derived this quantity, following the book

Acoustic Fields and Waves in Solids. Volume I Hardcover – April 20, 1973​

$$ \bf{c^{eff}} = \bf{c^{E}} + e^{transpose} \cdot (\bf{(\epsilon^{S}})^{-1} \bf{e})$$

The notation my be incorrect, but the true nature is CORRECT in the above expression, then when I wanted to express in index notation, I get into trouble,

$$ c^{eff}_{IJ} = c^{E}_{IJ} + e^{transpose}_{iI} (\epsilon^{S}_{??})^{-1} e_{??})
$$

where I, J are expressed in the book and took a picture here. For $\epsilon$ I cannot used the same symbol, more than twice, so I am forcing the permittivity to be diagonal. What it means is that I cannot or should not (??) use materials whose permittivity has off diagonal elements?

 

[chowdhury]

Let me know for examination, how many ranks or indexes $\nabla$, $\nabla_s$, e and u have ?
Please write them down explicitly with indexes.

@anuttarasammyak : I will provide after the last #16 of mine. Thanks for being so patient.

 

[Orodruin]

@Orodruin :
I have noticed that in the field of piezoelectricity, "they" denote like this, Here is another book,

https://www.amazon.com/dp/3540686800/?tag=pfamazon01-20

What is driving me crazy is this statement, from the end of this book

Acoustic Fields and Waves in Solids. Volume I Hardcover – April 20, 1973​

https://www.amazon.com/dp/0471037001/?tag=pfamazon01-20

I understand and agree to all you said in this thread, but not $e_{iJ} = e_{Ji}$ in page 378 of the book.
View attachment 297825

View attachment 297823View attachment 297824

Those are matrix transposes, not tensor transposes.

@Orodruin : I was told this book is the Bible, and there is no mistake in it. Without complaining, which is easy to rant to anybody in the world, I took an introspective view and try to educate myself.

There is a difference between being a subject bible and being an authority on the general math tools used in the subject. Even if there are no errors this does not mean it is a good reference notationally. I could write a book on GR without using any indices or tensor methods at all. It would be correct but it would still be a horrible book.

I applied the defintions, but could not find the reasonable notation, that is the problem.

The reasonable notation is the index or abstract index notations.

 

[chowdhury]

Let me know for examination, how many ranks or indexes $\nabla$, $\nabla_s$, e and u have ?
Please write them down explicitly with indexes.

• $\nabla$ is the usual operator in maths
• $\nabla_s$ is the symmetric part of the gradient operator, which results in strain of the material
• $e$ is the piezoelectric coefficient and it is a third rank tensor
• $u$ is a vector in space, so 3x1 vector.

FYI, I copy the definitions from the book, which is confusing, when I try to think in terms of maths,

 

[anuttarasammyak]

• ∇ is the usual operator in maths
• ∇s is the symmetric part of the gradient operator, which results in strain of the material
• e is the piezoelectric coefficient and it is a third rank tensor
• u is a vector in space, so 3x1 vector.

So with indices $$\nabla_i, (\nabla_s)_i, e_{ijk}, u_i$$, right ? Did you use these notations in your trial ?

 

[Orodruin]

So with indices $$\nabla_i, (\nabla_s)_i, e_{ijk}, u_i$$, right ? Did you use these notations in your trial ?

No, the way $\nabla_s$ is being used it is an object with two indices, one symmetric index (taking values from 1 through 6 (xx, yy, zz, xy, xz, yz)) and one regular index. It is a linear differential operator from the space of vectors to the space of symmetric tensors. Similarly, the way $e$ is being used, it has two indices one regular and one symmetric.

Of course, if you use standard notation and only use regular indices, $e$ and $\nabla_s$ would have three indices. The symmetries involved would be apparent from the relevant symmetries of these objects when exchanging indices.

 

[anuttarasammyak]

I have no regard to physics of piezoelectricity but math formula. @chowdhury Do you apply $(\nabla_s)_{ijk}$ as Oroduin suggested ?

 

[chowdhury]

@anuttarasammyak : yes in most cases,

Query 1.)

here is what I try to follow
$$\nabla \cdot (\bf{\epsilon}^{S} \cdot \nabla \phi) = \nabla \cdot (\bf{e} : \nabla_{s} \bf{u}) $$
Now
$$\nabla_{i} \epsilon_{ij}^{S} \nabla_{j} \phi = \nabla_{i} \epsilon_{ij}^{S} \phi_{,j} = \epsilon_{ij}^{S}\nabla_{i} \phi_{,j} = \epsilon_{ij}^{S}\phi_{,ij}$$

I don't know how to full the full 3-index notation for e. It was immensely helpful by @Orodruin : With his insight, it turned out to be,

$$ \nabla \cdot (\bf{e} : \nabla_{s} \bf{u}) = \nabla_{i} e_{ijk} \nabla_{j} u_{k} = e_{ijk} u_{k, ji}$$

Hence
$$\epsilon_{ij}^{S}\phi_{,ij} = e_{ijk} u_{k, ji}$$

I do not understand what was the basis for the book to say symmetry condition for the e-coeficient in their index?

Query 2.)
$$ \bf{c^{eff}} = \bf{c^{E}} + e^{transpose} \cdot (\bf{(\epsilon^{S}})^{-1} \bf{e})$$

The notation my be incorrect, but the true nature is CORRECT in the above expression, then when I wanted to express in index notation, I get into trouble,

$$ c^{eff}_{IJ} = c^{E}_{IJ} + e^{transpose}_{iI} (\epsilon^{S}_{??})^{-1} e_{??})
$$

I cannot write in single index notation, meaning not combined capital I,J etc.

Query 3.) For this equation,

$$\nabla \cdot (\bf{c}^{E} : \nabla_{s}\bf{u}) -\rho \frac{\partial^2 \bf{u}}{\partial t^2} = - \nabla \cdot (\bf{e}^{transpose} \cdot \nabla \phi)$$

I derived
$$(c^{E}_{ijkl} u_{k,l})_{,j} - \rho u_{i,tt} = - (e^{transpose}_{ijk} \phi_{,k})_{,j} =- (\color{red}{e}_{kji} \phi_{,jk}) $$
But the book derived as in the picture above. There might be difference in the indices, I cannot reconcile between mine and the book.

 

[chowdhury]

I have no regard to physics of piezoelectricity but math formula. @chowdhury Do you apply $(\nabla_s)_{ijk}$ as Oroduin suggested ?

@anuttarasammyak : I am sorry, I am not clear how to proceed, better, I would seek @Orodruin as the Judge!

 

[anuttarasammyak]

Summary:: Algebraic manipulation

How to write following equation in index notation?

∇⋅(e:∇su)
where e is a third rank tensor, u is a vector, ∇s is the symmetric part of the gradient operator, : is the double dot product.

Say $\nabla_s$ has rank one, it has rank zero,
$$\sum_{a,b,c}\nabla_c e_{cab} (\nabla_s)_au_b$$
Say $\nabla_s$ has rank three , it has rank two,
$$\sum_{a,b,c}\nabla_c e_{iab} (\nabla_s)_{jca}u_b$$
Are they OK?

 

[chowdhury]

Say $\nabla_s$ has rank one, it has rank zero,
$$\sum_{a,b,c}\nabla_c e_{cab} (\nabla_s)_au_b$$

I understand up to this, I cannot say further below.

Say $\nabla_s$ has rank three , it has rank two,
$$\sum_{a,b,c}\nabla_c e_{iab} (\nabla_s)_{jca}u_b$$
Are they OK?

 

[anuttarasammyak]

Physically how do you expect the rank of result, scalar zero or vector one or tensor two ?

 

[chowdhury]

Physically how do you expect the rank of result, scalar zero or vector one or tensor two ?

In this particular equation, I expect the result to be of rank ZERO.

$$\nabla \cdot (\bf{\epsilon}^{S} \cdot \nabla \phi) = \nabla \cdot (\bf{e} : \nabla_{s} \bf{u}) = \textup{rank zero} $$

 

[Orodruin]

The notation my be incorrect, but the true nature is CORRECT in the above expression, then when I wanted to express in index notation, I get into trouble,

Notation by itself is not correct or incorrect (unless it really is incorrect). However, it may be more or less illuminating.

Say $\nabla_s$ has rank one, it has rank zero,
$$\sum_{a,b,c}\nabla_c e_{cab} (\nabla_s)_au_b$$
Say $\nabla_s$ has rank three , it has rank two,
$$\sum_{a,b,c}\nabla_c e_{iab} (\nabla_s)_{jca}u_b$$
Are they OK?

No. It has rank three but you are missing one contraction.

 

[anuttarasammyak]

No. It has rank three but you are missing one contraction.

I thought for rank three $\nabla_s$ the number of all the indices are 1+3+3+1=8, the number of contraction of $\cdot$ and : are 3 so the number of surviving indices after contraction is 8 - 3*2 = 2 for which I wrote i,j. Further contraction of i and j makes the quantity scalar. I do not find the further contraction in the symbol definition.

 

[chowdhury]

@anuttarasammyak : yes in most cases,

Query 1.)

here is what I try to follow
$$\nabla \cdot (\bf{\epsilon}^{S} \cdot \nabla \phi) = \nabla \cdot (\bf{e} : \nabla_{s} \bf{u}) $$
Now
$$\nabla_{i} \epsilon_{ij}^{S} \nabla_{j} \phi = \nabla_{i} \epsilon_{ij}^{S} \phi_{,j} = \epsilon_{ij}^{S}\nabla_{i} \phi_{,j} = \epsilon_{ij}^{S}\phi_{,ij}$$

I don't know how to full the full 3-index notation for e. It was immensely helpful by @Orodruin : With his insight, it turned out to be,

$$ \nabla \cdot (\bf{e} : \nabla_{s} \bf{u}) = \nabla_{i} e_{ijk} \nabla_{j} u_{k} = e_{ijk} u_{k, ji}$$

Hence
$$\epsilon_{ij}^{S}\phi_{,ij} = e_{ijk} u_{k, ji}$$

I do not understand what was the basis for the book to say symmetry condition for the e-coeficient in their index?
View attachment 297832

Query 2.)
$$ \bf{c^{eff}} = \bf{c^{E}} + e^{transpose} \cdot (\bf{(\epsilon^{S}})^{-1} \bf{e})$$

The notation my be incorrect, but the true nature is CORRECT in the above expression, then when I wanted to express in index notation, I get into trouble,

$$ c^{eff}_{IJ} = c^{E}_{IJ} + e^{transpose}_{iI} (\epsilon^{S}_{??})^{-1} e_{??})
$$

I cannot write in single index notation, meaning not combined capital I,J etc.

Query 3.) For this equation,

$$\nabla \cdot (\bf{c}^{E} : \nabla_{s}\bf{u}) -\rho \frac{\partial^2 \bf{u}}{\partial t^2} = - \nabla \cdot (\bf{e}^{transpose} \cdot \nabla \phi)$$

I derived
$$(c^{E}_{ijkl} u_{k,l})_{,j} - \rho u_{i,tt} = - (e^{transpose}_{ijk} \phi_{,k})_{,j} =- (\color{red}{e}_{kji} \phi_{,jk}) $$
But the book derived as in the picture above. There might be difference in the indices, I cannot reconcile between mine and the book.

Does anybody know how to derivemy above three queries? Thanks.