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












1












$begingroup$


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









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – gebruiker
    Mar 30 '16 at 13:28
















1












$begingroup$


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









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – gebruiker
    Mar 30 '16 at 13:28














1












1








1


1



$begingroup$


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









$endgroup$




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









mSSMmSSM

14617




14617












  • $begingroup$
    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.
    $endgroup$
    – gebruiker
    Mar 30 '16 at 13:28


















  • $begingroup$
    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.
    $endgroup$
    – gebruiker
    Mar 30 '16 at 13:28
















$begingroup$
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.
$endgroup$
– gebruiker
Mar 30 '16 at 13:28




$begingroup$
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.
$endgroup$
– gebruiker
Mar 30 '16 at 13:28










2 Answers
2






active

oldest

votes


















1












$begingroup$

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











$endgroup$









  • 1




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










  • $begingroup$
    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.
    $endgroup$
    – Najib Idrissi
    Dec 17 '15 at 14:13










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










  • $begingroup$
    @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.
    $endgroup$
    – Jack's wasted life
    Mar 31 '16 at 12:16



















1












$begingroup$

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











$endgroup$













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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f881571%2fhow-to-prove-that-the-kronecker-delta-is-the-unique-isotropic-tensor-of-order-2%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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











$endgroup$









  • 1




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










  • $begingroup$
    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.
    $endgroup$
    – Najib Idrissi
    Dec 17 '15 at 14:13










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










  • $begingroup$
    @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.
    $endgroup$
    – Jack's wasted life
    Mar 31 '16 at 12:16
















1












$begingroup$

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











$endgroup$









  • 1




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










  • $begingroup$
    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.
    $endgroup$
    – Najib Idrissi
    Dec 17 '15 at 14:13










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










  • $begingroup$
    @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.
    $endgroup$
    – Jack's wasted life
    Mar 31 '16 at 12:16














1












1








1





$begingroup$

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











$endgroup$



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 lifeJack's wasted life

7,42111430




7,42111430








  • 1




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










  • $begingroup$
    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.
    $endgroup$
    – Najib Idrissi
    Dec 17 '15 at 14:13










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










  • $begingroup$
    @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.
    $endgroup$
    – Jack's wasted life
    Mar 31 '16 at 12:16














  • 1




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










  • $begingroup$
    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.
    $endgroup$
    – Najib Idrissi
    Dec 17 '15 at 14:13










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










  • $begingroup$
    @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.
    $endgroup$
    – Jack's wasted life
    Mar 31 '16 at 12:16








1




1




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




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












$begingroup$
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.
$endgroup$
– Najib Idrissi
Dec 17 '15 at 14:13




$begingroup$
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.
$endgroup$
– Najib Idrissi
Dec 17 '15 at 14:13












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




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












$begingroup$
@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.
$endgroup$
– Jack's wasted life
Mar 31 '16 at 12:16




$begingroup$
@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.
$endgroup$
– Jack's wasted life
Mar 31 '16 at 12:16











1












$begingroup$

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











$endgroup$













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
















1












$begingroup$

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











$endgroup$













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














1












1








1





$begingroup$

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











$endgroup$



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









DrWillDrWill

185




185












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


















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
















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




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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f881571%2fhow-to-prove-that-the-kronecker-delta-is-the-unique-isotropic-tensor-of-order-2%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more