How to prove that the Kronecker delta is the unique isotropic tensor of order 2?












1














Is there a way to prove that the Kronecker delta $delta_{ij}$ is indeed the only isotropic second order tensor (i.e. invariant under rotation), i.e. so we can write $T_{ij} = lambda delta_{ij}$ for some constant $lambda$?



By rotational invariance I mean:
$$
T_{ij} = T^prime_{ij} = R_{ip} R_{jq} T_{pq}text{,}
$$
where the matrices $R_{ij}$ are orthogonal.



It is very straightforward to show that $delta_{ij}$ is invariant, but how can I show that it is unique?






rotations tensors invariance







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  • If any of the answeres below were useful to you, then you should upvote all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this.
    – gebruiker
    Mar 30 '16 at 13:28
















1














Is there a way to prove that the Kronecker delta $delta_{ij}$ is indeed the only isotropic second order tensor (i.e. invariant under rotation), i.e. so we can write $T_{ij} = lambda delta_{ij}$ for some constant $lambda$?



By rotational invariance I mean:
$$
T_{ij} = T^prime_{ij} = R_{ip} R_{jq} T_{pq}text{,}
$$
where the matrices $R_{ij}$ are orthogonal.



It is very straightforward to show that $delta_{ij}$ is invariant, but how can I show that it is unique?






rotations tensors invariance







share|cite|improve this question






















  • If any of the answeres below were useful to you, then you should upvote all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this.
    – gebruiker
    Mar 30 '16 at 13:28














1












1








1


1





Is there a way to prove that the Kronecker delta $delta_{ij}$ is indeed the only isotropic second order tensor (i.e. invariant under rotation), i.e. so we can write $T_{ij} = lambda delta_{ij}$ for some constant $lambda$?



By rotational invariance I mean:
$$
T_{ij} = T^prime_{ij} = R_{ip} R_{jq} T_{pq}text{,}
$$
where the matrices $R_{ij}$ are orthogonal.



It is very straightforward to show that $delta_{ij}$ is invariant, but how can I show that it is unique?






rotations tensors invariance







share|cite|improve this question













Is there a way to prove that the Kronecker delta $delta_{ij}$ is indeed the only isotropic second order tensor (i.e. invariant under rotation), i.e. so we can write $T_{ij} = lambda delta_{ij}$ for some constant $lambda$?



By rotational invariance I mean:
$$
T_{ij} = T^prime_{ij} = R_{ip} R_{jq} T_{pq}text{,}
$$
where the matrices $R_{ij}$ are orthogonal.



It is very straightforward to show that $delta_{ij}$ is invariant, but how can I show that it is unique?






rotations tensors invariance




rotations tensors invariance






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jul 29 '14 at 14:07









mSSM

14617




14617












  • If any of the answeres below were useful to you, then you should upvote all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this.
    – gebruiker
    Mar 30 '16 at 13:28


















  • If any of the answeres below were useful to you, then you should upvote all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this.
    – gebruiker
    Mar 30 '16 at 13:28
















If any of the answeres below were useful to you, then you should upvote all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this.
– gebruiker
Mar 30 '16 at 13:28




If any of the answeres below were useful to you, then you should upvote all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this.
– gebruiker
Mar 30 '16 at 13:28










2 Answers
2






active

oldest

votes


















1














Step 1: Realize $T$ is diagonal. Let
$$R_{kl}=begin{cases}
-1 & text{if }k=l=j\
delta_{kl} & text{otherwise}\
end{cases}$$
therefore $R$ is the reflection w.r.t to the hyperplane perpendicular to the j-th vector in the standard ordered basis.
$$R=R^Tland R^2=IRightarrow R^TR=RR^T=I$$
Therefore:
$$T_{ij}=sum_{p,q}R_{ip}R_{jq}T_{pq}=R_{ii}R_{jj}T_{ij}\
ineq jRightarrow T_{ij}=-T_{ij}Rightarrow T_{ij}=0$$




Step 2: Realize $T_{jj} = T_{11}$. Let
$$P_{kl}=begin{cases}
delta_{jl} & text{if } k=1\
delta_{1l} & text{if } k=j\
delta_{kl} & text{otherwise}
end{cases}$$
therefore $P$ is the permutation matrix that interchanges the 1st and j-th rows on left multiplication.
$$(P^TP)_{kl}=sum_{m}P^T_{km}P_{ml}=sum_{m}P_{mk}P_{ml}=sum_{mneq1,j}P_{mk}P_{ml}+sum_{m=1,j}P_{mk}P_{ml}\
=sum_{mneq1,j}delta_{mk}delta_{ml}+delta_{jk}delta_{jl}+delta_{1k}delta_{1l}=sum_{m}delta_{mk}delta_{ml}=delta_{kl}$$
Therefore:
$$T_{jj}=sum_{p,q}P_{jp}P_{jq}T_{pq}=sum_{q}P_{jq}^2T_{qq}=sum_{q}delta_{1q}^2T_{qq}=sum_{q}delta_{1q}T_{qq}=T_{11}$$






share|cite|improve this answer



















  • 1




    wow, whole answer in math mode :D
    – MyUserIsThis
    Jun 6 '15 at 8:16










  • It is actually a horrible idea to compose your whole answer in math mode. It make it difficult to copy the text for some people (when you right click a formula a different menu pops up), screen readers will not work correctly (for visually-impaired people), on mobile devices the formulas with a big width are a pain, etc. Please consider writing only formulas in math mode.
    – Najib Idrissi
    Dec 17 '15 at 14:13










  • The very first argument: why do you assume that $R_{ij}$ has that form?
    – mSSM
    Mar 31 '16 at 11:05










  • @mSSM I don't assume every orthogonal R has that form. That R is a tensor which meets the criterion of orthogonality and which is designed in that specific way for the purpose of bringing out the diagonal nature of T.
    – Jack's wasted life
    Mar 31 '16 at 12:16



















1














There is a simpler proof at http://www.damtp.cam.ac.uk/user/reh10/lectures/nst-mmii-chapter3.pdf, pg 61, that proceeds as follows



Starting with an isotrophic 2nd order tensor written in matrix form Tij, its coordinates are rotated 90 deg. around the 3 axis to get T'ij, then by comparing T'ij (which equals Tij as T is isotrophic) and Tij and concluding by inspection that T11 = T22, T13 = T23 = -T13 = 0, T31 = T32 = -T31 = 0, and then rotating the coordinates 90% about the 2 axis to conclude that T11 = T33 and T21 = T32 = 0 and T21 = T23 = 0.



That is, the proof is simplicity itself and is given in detail in the reference.






share|cite|improve this answer























  • That is indeed a very nice proof. Thank you very much!
    – mSSM
    Dec 3 '18 at 15:47











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Step 1: Realize $T$ is diagonal. Let
$$R_{kl}=begin{cases}
-1 & text{if }k=l=j\
delta_{kl} & text{otherwise}\
end{cases}$$
therefore $R$ is the reflection w.r.t to the hyperplane perpendicular to the j-th vector in the standard ordered basis.
$$R=R^Tland R^2=IRightarrow R^TR=RR^T=I$$
Therefore:
$$T_{ij}=sum_{p,q}R_{ip}R_{jq}T_{pq}=R_{ii}R_{jj}T_{ij}\
ineq jRightarrow T_{ij}=-T_{ij}Rightarrow T_{ij}=0$$




Step 2: Realize $T_{jj} = T_{11}$. Let
$$P_{kl}=begin{cases}
delta_{jl} & text{if } k=1\
delta_{1l} & text{if } k=j\
delta_{kl} & text{otherwise}
end{cases}$$
therefore $P$ is the permutation matrix that interchanges the 1st and j-th rows on left multiplication.
$$(P^TP)_{kl}=sum_{m}P^T_{km}P_{ml}=sum_{m}P_{mk}P_{ml}=sum_{mneq1,j}P_{mk}P_{ml}+sum_{m=1,j}P_{mk}P_{ml}\
=sum_{mneq1,j}delta_{mk}delta_{ml}+delta_{jk}delta_{jl}+delta_{1k}delta_{1l}=sum_{m}delta_{mk}delta_{ml}=delta_{kl}$$
Therefore:
$$T_{jj}=sum_{p,q}P_{jp}P_{jq}T_{pq}=sum_{q}P_{jq}^2T_{qq}=sum_{q}delta_{1q}^2T_{qq}=sum_{q}delta_{1q}T_{qq}=T_{11}$$






share|cite|improve this answer



















  • 1




    wow, whole answer in math mode :D
    – MyUserIsThis
    Jun 6 '15 at 8:16










  • It is actually a horrible idea to compose your whole answer in math mode. It make it difficult to copy the text for some people (when you right click a formula a different menu pops up), screen readers will not work correctly (for visually-impaired people), on mobile devices the formulas with a big width are a pain, etc. Please consider writing only formulas in math mode.
    – Najib Idrissi
    Dec 17 '15 at 14:13










  • The very first argument: why do you assume that $R_{ij}$ has that form?
    – mSSM
    Mar 31 '16 at 11:05










  • @mSSM I don't assume every orthogonal R has that form. That R is a tensor which meets the criterion of orthogonality and which is designed in that specific way for the purpose of bringing out the diagonal nature of T.
    – Jack's wasted life
    Mar 31 '16 at 12:16
















1














Step 1: Realize $T$ is diagonal. Let
$$R_{kl}=begin{cases}
-1 & text{if }k=l=j\
delta_{kl} & text{otherwise}\
end{cases}$$
therefore $R$ is the reflection w.r.t to the hyperplane perpendicular to the j-th vector in the standard ordered basis.
$$R=R^Tland R^2=IRightarrow R^TR=RR^T=I$$
Therefore:
$$T_{ij}=sum_{p,q}R_{ip}R_{jq}T_{pq}=R_{ii}R_{jj}T_{ij}\
ineq jRightarrow T_{ij}=-T_{ij}Rightarrow T_{ij}=0$$




Step 2: Realize $T_{jj} = T_{11}$. Let
$$P_{kl}=begin{cases}
delta_{jl} & text{if } k=1\
delta_{1l} & text{if } k=j\
delta_{kl} & text{otherwise}
end{cases}$$
therefore $P$ is the permutation matrix that interchanges the 1st and j-th rows on left multiplication.
$$(P^TP)_{kl}=sum_{m}P^T_{km}P_{ml}=sum_{m}P_{mk}P_{ml}=sum_{mneq1,j}P_{mk}P_{ml}+sum_{m=1,j}P_{mk}P_{ml}\
=sum_{mneq1,j}delta_{mk}delta_{ml}+delta_{jk}delta_{jl}+delta_{1k}delta_{1l}=sum_{m}delta_{mk}delta_{ml}=delta_{kl}$$
Therefore:
$$T_{jj}=sum_{p,q}P_{jp}P_{jq}T_{pq}=sum_{q}P_{jq}^2T_{qq}=sum_{q}delta_{1q}^2T_{qq}=sum_{q}delta_{1q}T_{qq}=T_{11}$$






share|cite|improve this answer



















  • 1




    wow, whole answer in math mode :D
    – MyUserIsThis
    Jun 6 '15 at 8:16










  • It is actually a horrible idea to compose your whole answer in math mode. It make it difficult to copy the text for some people (when you right click a formula a different menu pops up), screen readers will not work correctly (for visually-impaired people), on mobile devices the formulas with a big width are a pain, etc. Please consider writing only formulas in math mode.
    – Najib Idrissi
    Dec 17 '15 at 14:13










  • The very first argument: why do you assume that $R_{ij}$ has that form?
    – mSSM
    Mar 31 '16 at 11:05










  • @mSSM I don't assume every orthogonal R has that form. That R is a tensor which meets the criterion of orthogonality and which is designed in that specific way for the purpose of bringing out the diagonal nature of T.
    – Jack's wasted life
    Mar 31 '16 at 12:16














1












1








1






Step 1: Realize $T$ is diagonal. Let
$$R_{kl}=begin{cases}
-1 & text{if }k=l=j\
delta_{kl} & text{otherwise}\
end{cases}$$
therefore $R$ is the reflection w.r.t to the hyperplane perpendicular to the j-th vector in the standard ordered basis.
$$R=R^Tland R^2=IRightarrow R^TR=RR^T=I$$
Therefore:
$$T_{ij}=sum_{p,q}R_{ip}R_{jq}T_{pq}=R_{ii}R_{jj}T_{ij}\
ineq jRightarrow T_{ij}=-T_{ij}Rightarrow T_{ij}=0$$




Step 2: Realize $T_{jj} = T_{11}$. Let
$$P_{kl}=begin{cases}
delta_{jl} & text{if } k=1\
delta_{1l} & text{if } k=j\
delta_{kl} & text{otherwise}
end{cases}$$
therefore $P$ is the permutation matrix that interchanges the 1st and j-th rows on left multiplication.
$$(P^TP)_{kl}=sum_{m}P^T_{km}P_{ml}=sum_{m}P_{mk}P_{ml}=sum_{mneq1,j}P_{mk}P_{ml}+sum_{m=1,j}P_{mk}P_{ml}\
=sum_{mneq1,j}delta_{mk}delta_{ml}+delta_{jk}delta_{jl}+delta_{1k}delta_{1l}=sum_{m}delta_{mk}delta_{ml}=delta_{kl}$$
Therefore:
$$T_{jj}=sum_{p,q}P_{jp}P_{jq}T_{pq}=sum_{q}P_{jq}^2T_{qq}=sum_{q}delta_{1q}^2T_{qq}=sum_{q}delta_{1q}T_{qq}=T_{11}$$






share|cite|improve this answer














Step 1: Realize $T$ is diagonal. Let
$$R_{kl}=begin{cases}
-1 & text{if }k=l=j\
delta_{kl} & text{otherwise}\
end{cases}$$
therefore $R$ is the reflection w.r.t to the hyperplane perpendicular to the j-th vector in the standard ordered basis.
$$R=R^Tland R^2=IRightarrow R^TR=RR^T=I$$
Therefore:
$$T_{ij}=sum_{p,q}R_{ip}R_{jq}T_{pq}=R_{ii}R_{jj}T_{ij}\
ineq jRightarrow T_{ij}=-T_{ij}Rightarrow T_{ij}=0$$




Step 2: Realize $T_{jj} = T_{11}$. Let
$$P_{kl}=begin{cases}
delta_{jl} & text{if } k=1\
delta_{1l} & text{if } k=j\
delta_{kl} & text{otherwise}
end{cases}$$
therefore $P$ is the permutation matrix that interchanges the 1st and j-th rows on left multiplication.
$$(P^TP)_{kl}=sum_{m}P^T_{km}P_{ml}=sum_{m}P_{mk}P_{ml}=sum_{mneq1,j}P_{mk}P_{ml}+sum_{m=1,j}P_{mk}P_{ml}\
=sum_{mneq1,j}delta_{mk}delta_{ml}+delta_{jk}delta_{jl}+delta_{1k}delta_{1l}=sum_{m}delta_{mk}delta_{ml}=delta_{kl}$$
Therefore:
$$T_{jj}=sum_{p,q}P_{jp}P_{jq}T_{pq}=sum_{q}P_{jq}^2T_{qq}=sum_{q}delta_{1q}^2T_{qq}=sum_{q}delta_{1q}T_{qq}=T_{11}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 31 '16 at 12:24

























answered Jun 6 '15 at 8:06









Jack's wasted life

7,37111430




7,37111430








  • 1




    wow, whole answer in math mode :D
    – MyUserIsThis
    Jun 6 '15 at 8:16










  • It is actually a horrible idea to compose your whole answer in math mode. It make it difficult to copy the text for some people (when you right click a formula a different menu pops up), screen readers will not work correctly (for visually-impaired people), on mobile devices the formulas with a big width are a pain, etc. Please consider writing only formulas in math mode.
    – Najib Idrissi
    Dec 17 '15 at 14:13










  • The very first argument: why do you assume that $R_{ij}$ has that form?
    – mSSM
    Mar 31 '16 at 11:05










  • @mSSM I don't assume every orthogonal R has that form. That R is a tensor which meets the criterion of orthogonality and which is designed in that specific way for the purpose of bringing out the diagonal nature of T.
    – Jack's wasted life
    Mar 31 '16 at 12:16














  • 1




    wow, whole answer in math mode :D
    – MyUserIsThis
    Jun 6 '15 at 8:16










  • It is actually a horrible idea to compose your whole answer in math mode. It make it difficult to copy the text for some people (when you right click a formula a different menu pops up), screen readers will not work correctly (for visually-impaired people), on mobile devices the formulas with a big width are a pain, etc. Please consider writing only formulas in math mode.
    – Najib Idrissi
    Dec 17 '15 at 14:13










  • The very first argument: why do you assume that $R_{ij}$ has that form?
    – mSSM
    Mar 31 '16 at 11:05










  • @mSSM I don't assume every orthogonal R has that form. That R is a tensor which meets the criterion of orthogonality and which is designed in that specific way for the purpose of bringing out the diagonal nature of T.
    – Jack's wasted life
    Mar 31 '16 at 12:16








1




1




wow, whole answer in math mode :D
– MyUserIsThis
Jun 6 '15 at 8:16




wow, whole answer in math mode :D
– MyUserIsThis
Jun 6 '15 at 8:16












It is actually a horrible idea to compose your whole answer in math mode. It make it difficult to copy the text for some people (when you right click a formula a different menu pops up), screen readers will not work correctly (for visually-impaired people), on mobile devices the formulas with a big width are a pain, etc. Please consider writing only formulas in math mode.
– Najib Idrissi
Dec 17 '15 at 14:13




It is actually a horrible idea to compose your whole answer in math mode. It make it difficult to copy the text for some people (when you right click a formula a different menu pops up), screen readers will not work correctly (for visually-impaired people), on mobile devices the formulas with a big width are a pain, etc. Please consider writing only formulas in math mode.
– Najib Idrissi
Dec 17 '15 at 14:13












The very first argument: why do you assume that $R_{ij}$ has that form?
– mSSM
Mar 31 '16 at 11:05




The very first argument: why do you assume that $R_{ij}$ has that form?
– mSSM
Mar 31 '16 at 11:05












@mSSM I don't assume every orthogonal R has that form. That R is a tensor which meets the criterion of orthogonality and which is designed in that specific way for the purpose of bringing out the diagonal nature of T.
– Jack's wasted life
Mar 31 '16 at 12:16




@mSSM I don't assume every orthogonal R has that form. That R is a tensor which meets the criterion of orthogonality and which is designed in that specific way for the purpose of bringing out the diagonal nature of T.
– Jack's wasted life
Mar 31 '16 at 12:16











1














There is a simpler proof at http://www.damtp.cam.ac.uk/user/reh10/lectures/nst-mmii-chapter3.pdf, pg 61, that proceeds as follows



Starting with an isotrophic 2nd order tensor written in matrix form Tij, its coordinates are rotated 90 deg. around the 3 axis to get T'ij, then by comparing T'ij (which equals Tij as T is isotrophic) and Tij and concluding by inspection that T11 = T22, T13 = T23 = -T13 = 0, T31 = T32 = -T31 = 0, and then rotating the coordinates 90% about the 2 axis to conclude that T11 = T33 and T21 = T32 = 0 and T21 = T23 = 0.



That is, the proof is simplicity itself and is given in detail in the reference.






share|cite|improve this answer























  • That is indeed a very nice proof. Thank you very much!
    – mSSM
    Dec 3 '18 at 15:47
















1














There is a simpler proof at http://www.damtp.cam.ac.uk/user/reh10/lectures/nst-mmii-chapter3.pdf, pg 61, that proceeds as follows



Starting with an isotrophic 2nd order tensor written in matrix form Tij, its coordinates are rotated 90 deg. around the 3 axis to get T'ij, then by comparing T'ij (which equals Tij as T is isotrophic) and Tij and concluding by inspection that T11 = T22, T13 = T23 = -T13 = 0, T31 = T32 = -T31 = 0, and then rotating the coordinates 90% about the 2 axis to conclude that T11 = T33 and T21 = T32 = 0 and T21 = T23 = 0.



That is, the proof is simplicity itself and is given in detail in the reference.






share|cite|improve this answer























  • That is indeed a very nice proof. Thank you very much!
    – mSSM
    Dec 3 '18 at 15:47














1












1








1






There is a simpler proof at http://www.damtp.cam.ac.uk/user/reh10/lectures/nst-mmii-chapter3.pdf, pg 61, that proceeds as follows



Starting with an isotrophic 2nd order tensor written in matrix form Tij, its coordinates are rotated 90 deg. around the 3 axis to get T'ij, then by comparing T'ij (which equals Tij as T is isotrophic) and Tij and concluding by inspection that T11 = T22, T13 = T23 = -T13 = 0, T31 = T32 = -T31 = 0, and then rotating the coordinates 90% about the 2 axis to conclude that T11 = T33 and T21 = T32 = 0 and T21 = T23 = 0.



That is, the proof is simplicity itself and is given in detail in the reference.






share|cite|improve this answer














There is a simpler proof at http://www.damtp.cam.ac.uk/user/reh10/lectures/nst-mmii-chapter3.pdf, pg 61, that proceeds as follows



Starting with an isotrophic 2nd order tensor written in matrix form Tij, its coordinates are rotated 90 deg. around the 3 axis to get T'ij, then by comparing T'ij (which equals Tij as T is isotrophic) and Tij and concluding by inspection that T11 = T22, T13 = T23 = -T13 = 0, T31 = T32 = -T31 = 0, and then rotating the coordinates 90% about the 2 axis to conclude that T11 = T33 and T21 = T32 = 0 and T21 = T23 = 0.



That is, the proof is simplicity itself and is given in detail in the reference.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 17:56

























answered Dec 3 '18 at 15:23









DrWill

185




185












  • That is indeed a very nice proof. Thank you very much!
    – mSSM
    Dec 3 '18 at 15:47


















  • That is indeed a very nice proof. Thank you very much!
    – mSSM
    Dec 3 '18 at 15:47
















That is indeed a very nice proof. Thank you very much!
– mSSM
Dec 3 '18 at 15:47




That is indeed a very nice proof. Thank you very much!
– mSSM
Dec 3 '18 at 15:47


















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