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14.4: C.4 3rd-Order Isotropic Tensors

  • Page ID
    18107
  • Let a 3rd-order isotropic tensor \(\underset{\sim}{A}\) be subjected to an infinitesimal rotation \(\underset{\sim}{\delta} +\underset{\sim}{r}\). Then

    \[A_{i j k}^{\prime}=A_{l m n} C_{l i} C_{m j} C_{n k}=A_{i j k}.\]

    Reasoning as in the last section (see derivation of Equation 14.3.1), we obtain

    \[A_{i l k} r_{l j}+A_{l j k} r_{l i}+A_{i j l} r_{l k}=0.\label{eqn:1}\]

    Again, we have changed all dummy indices to \(l\) for tidiness. (And again, this is safe only because they appear in separate terms!) Equation \(\ref{eqn:1}\) represents 27 equations, one for each combination of the index values 1, 2 and 3. It will simplify things if we classify those 27 combinations as follows:

    • all values equal (111, 222 and 333)
    • two values equal and one different (e.g., 223)
    • all values different (123, 231, 312, 213, 321 and 132).

    Case 1: \(i\) = 1, \(j\) = 1, \(k\) = 2

    With {\(i, j, k\)} = {1,1,2}, Equation \(\ref{eqn:1}\) becomes

    \[\begin{aligned}
    A_{112} r_{11} &+A_{122} r_{21}+A_{132} r_{31} \\
    +A_{112} r_{11} &+A_{212} r_{21}+A_{312} r_{31} \\
    +A_{111} r_{12} &+A_{112} r_{22}+A_{113} r_{32}=0.
    \end{aligned}\]

    Using the antisymmetry of \(\underset{\sim}{r}\), we can write this as

    \[\left(A_{122}+A_{212}-A_{111}\right) r_{21}+\left(A_{132}+A_{312}\right) r_{31}+A_{113} r_{32}=0.\]

    Since the coefficients must vanish separately, we have three equations:

    \[\begin{align}
    A_{122}+A_{212} &=A_{111} \label{eqn:2}\\
    A_{132} &=-A_{312} \label{eqn:3}\\
    A_{113} &=0 \label{eqn:4}
    \end{align}\]

    The third equation tells us that an element with two equal indices is zero. Let us guess that this is true for all such elements. If we have guessed right, then the first equation tells us that an element with all three indices equal is zero. Finally, in the second equation, interchanging two indices changes the sign.

    Are these patterns generally true? Let us try another case to check.

    Case 2: \(i\) = 2, \(j\) = 3, \(k\) = 2

    Repeating the previous case with {\(i, j, k\)} changed to {2,3,2}, we have

    \[\begin{aligned}
    A_{212} r_{13} &+A_{222} r_{23}+A_{232} r_{33} \\
    +A_{132} r_{12} &+A_{232} r_{22}+A_{332} r_{32} \\
    +A_{231} r_{12} &+A_{232} r_{22}+A_{233} r_{32}=0
    \end{aligned}\]

    or

    \[\begin{align}
    A_{332}+A_{233} &=A_{222}\label{eqn:5} \\
    A_{132} &=-A_{231}\label{eqn:6} \\
    A_{212} &=0.\label{eqn:7}
    \end{align}\]

    The pattern is the same as in case 1. The third equation tells us that an element with two equal indices is zero. If this is generally true, then the first equation tells us that an element with three equal indices is zero. Finally, interchanging two indices changes the sign.

    You can check as many cases as you like; the results are always the same. Note that the only “rule” that matters here is the second one: interchanging two indices changes the sign. The other two rules follow from this one, because they both involve elements with two or more equal indices. Interchanging two equal indices makes no difference, but it also changes the sign. That only works if the value is zero.

    We conclude that a 3rd-order tensor can be isotropic only if it is completely antisymmetric, i.e., interchanging any two indices changes the sign. In section D.1, we show that the only completely antisymmetric 3rd-order tensor is, to within a multiplicative constant, the Levi-Civita alternating tensor \(\underset{\sim}{\varepsilon}\). The most general isotropic 3rd-order tensor is therefore

    \[\underset{\sim}{A}=a \underset{\sim}{\varepsilon}\]

    where \(a\) is any scalar.