= Mean logarithmic stretching in an isotropic renovating shear
{title2=$\mu=\log(1+a^2/4)/(2T)$}
For independent uniform orientations in a <renovating shear flow>, the incident angle relative to each new shear is uniform even after earlier deformations. One interval stretches a unit <material line> by $\ell(\theta)^2=1+a\sin2\theta+a^2\sin^2\theta$. Write this as $A+B\cos(2\theta-\delta)$, where $A=1+a^2/2$ and $A^2-B^2=1$. The angular logarithmic integral gives $\langle\log\ell\rangle=\tfrac12\log[(A+1)/2]$. Add these independent logarithmic increments to get the displayed rate. At fixed shear rate, small $T$ gives $\mu\sim\Lambda^2T/8$ and large $T$ gives $\mu\sim\log(|\Lambda|T/2)/T$. The unique positive maximum is at $|\Lambda|T\approx3.9605826$, obtained from $\log z=2(1-z^{-1})$ with $z=1+a^2/4$.
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