2

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?

mSSM
  • 231
  • 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 Answers2

3

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.

DrWill
  • 81
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. $$R=R^T\land R^2=I\Rightarrow R^TR=RR^T=I$$ Therefore: $$T_{ij}=\sum_{p,q}R_{ip}R_{jq}T_{pq}=R_{ii}R_{jj}T_{ij}\\ i\neq j\Rightarrow T_{ij}=-T_{ij}\Rightarrow T_{ij}=0$$

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. $$(P^TP)_{kl}=\sum_{m}P^T_{km}P_{ml}=\sum_{m}P_{mk}P_{ml}=\sum_{m\neq1,j}P_{mk}P_{ml}+\sum_{m=1,j}P_{mk}P_{ml}\\ =\sum_{m\neq1,j}\delta_{mk}\delta_{ml}+\delta_{jk}\delta_{jl}+\delta_{1k}\delta_{1l}=\sum_{m}\delta_{mk}\delta_{ml}=\delta_{kl}$$ Therefore: $$T_{jj}=\sum_{p,q}P_{jp}P_{jq}T_{pq}=\sum_{q}P_{jq}^2T_{qq}=\sum_{q}\delta_{1q}^2T_{qq}=\sum_{q}\delta_{1q}T_{qq}=T_{11}$$

Jack's wasted life
  • 8,989
  • 1
    wow, whole answer in math mode :D – MyUserIsThis Jun 06 '15 at 08:16
  • 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
  • @Jack'swastedlife Why do we assume T is diagonal? – f_garcia May 20 '21 at 01:58
  • @FernandoGarciaCortez Step 1 proves T is diagonal, we don't have to assume anything. – Jack's wasted life May 27 '21 at 19:58