# 14.3: C.3- 2nd-order isotropic tensors


Let a 2nd-order isotropic tensor $$\underset{\sim}{A}$$ be subjected to an infinitesimal rotation $$\underset{\sim}{\delta} + \underset{\sim}{r}$$. Then

\begin{aligned} A_{i j}^{\prime}=A_{k l} C_{k i} C_{l j} &=A_{k l}\left(\delta_{k i}+r_{k i}\right)\left(\delta_{l j}+r_{l j}\right)=A_{k l}\left(\delta_{k i} \delta_{l j}+\delta_{k i} r_{l j}+r_{k i} \delta_{l j}+r_{k i} r_{l j}\right) \\ &=A_{i j}+A_{i l} r_{l j}+A_{k j} r_{k i}=A_{i j} \end{aligned} \nonumber

where we have neglected the product of the infinitesimal matrices $$\underset{\sim}{r}$$. This leaves us with

$A_{i k} r_{k j}+A_{k j} r_{k i}=0,\label{eqn:1}$

which must be true for all antisymmetric matrices $$\underset{\sim}{r}$$ whose elements are $$\ll 1$$. (Note that we have renamed the dummy index $$l$$ as $$k$$ for tidiness. There is no potential for confusion because the pairs of $$k$$s are in separate terms.)

Equation $$\ref{eqn:1}$$ represents nine equations, one for each combination of the free indices $$i$$ and $$j$$. It will be enough to consider three of these.

Case 1: $$i$$ = 1, $$j$$ = 1

$A_{11} r_{11}+A_{12} r_{21}+A_{13} r_{31}+A_{11} r_{11}+A_{21} r_{21}+A_{31} r_{31}=0. \nonumber$

Remembering that $$r_{11}$$ = 0, we can write this as

$\left(A_{12}+A_{21}\right) r_{21}+\left(A_{13}+A_{31}\right) r_{31}=0. \nonumber$

Because this must be true for all values of $$r_{21}$$ and $$r_{31}$$, the coefficients of those quantities must vanish separately:

$A_{12}+A_{21}=0 ; \quad A_{13}+A_{31}=0.\label{eqn:2}$

Case 2: $$i$$ = 1, $$j$$ = 2

$A_{11} r_{12}+A_{12} r_{22}+A_{13} r_{32}+A_{12} r_{11}+A_{22} r_{21}+A_{32} r_{31}=0. \nonumber$

Because $$r_{11}$$ = $$r_{22}$$ = 0 and $$r_{21}$$ = $$−r_{12}$$, we can write this as

$\left(A_{11}-A_{22}\right) r_{12}+A_{13} r_{32}+A_{32} r_{31}=0. \nonumber$

Because this must be true for all $$\underset{\sim}{r}$$,

$A_{11}=A_{22} ; \quad A_{13}=0 ; \quad A_{32}=0.\label{eqn:3}$

Case 3: $$i$$ = 1, $$j$$ = 3

$A_{11}=A_{33} ; \quad A_{12}=0 ; \quad A_{23}=0.\label{eqn:4}$

Combining Equation $$\ref{eqn:2}$$, $$\ref{eqn:3}$$ and $$\ref{eqn:4}$$, we have

$A_{11}=A_{22}=A_{33} ; \quad A_{12}=A_{21}=A_{32}=A_{23}=A_{13}=A_{31}=0.\label{eqn:5}$

If $$A_{11}$$ has the value $$a$$, then $$\underset{\sim}{A}$$ must therefore be proportional to the identity matrix:

$\underset{\sim}{A}=\left(\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right)=a \underset{\sim}{\delta} \nonumber$

We conclude that the only isotropic 2nd order tensors are those that are proportional to the identity.

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