= Solution
The <principal strain rates> $a,b,c$ are the <eigenvalues> of the symmetric <strain-rate tensor>. <Incompressibility> makes its <trace> zero, so $a+b+c=0$. The polynomial identity
$$
a^3+b^3+c^3-3abc
=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
$$
therefore gives $a^3+b^3+c^3=3abc$. Since the <trace> of $S^3$ is the sum of the cubes of its <eigenvalues>, the <Betchov relation> becomes
$$
3\langle abc\rangle=-\frac34\langle\omega_i\omega_jS_{ij}\rangle,
\qquad
\boxed{\langle abc\rangle=-\frac14\langle\omega_i\omega_jS_{ij}\rangle}.
$$
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