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

Last updated
Save as PDF

Page ID: 18093

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

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

The same reasoning leads to

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

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.