 Hookean dumbbell in planar extension

Represent one polymer by two beads joined by an ideal linear entropic spring, and stipulate a dimensionless conformation tensor $C$ with equilibrium value $I$. The displayed model uses velocity gradient $A=\mathrm{diag}(\dot{ϵ},-\dot{ϵ},0)$ and positive polymer relaxation time $\lambda$. It omits finite chain extensibility, interactions with other chains, and feedback from the polymer onto the flow. It is a teaching model, not a fitted description of Keller's apparatus.

Define the Weissenberg number for this model as $\mathrm{Wi}=\lambda \dot{ϵ}$ and dimensionless time $u=t/\lambda$. With an initially isotropic $C=I$, the diagonal entries obeyBelow $\mathrm{Wi}=1/2$, both transverse and longitudinal responses approach positive finite steady values:For a chosen $\mathrm{Wi}=0.4$, $C_{xx}=5$ and $C_{yy}=5/9$: the longitudinal mean-square separation is five times its quiescent value, while the transverse separation contracts. A root-mean-square length grows as the square root of a variance, so the longitudinal length factor is $\sqrt{5}$, not five. The anisotropy $C_{xx}-C_{yy}$ is an orientational indicator; converting it to measured birefringence requires an additional calibrated optical relation.

At $\mathrm{Wi}=1/2$, $C_{xx}=1+u$. Above that value, the growing homogeneous solution prevents a positive finite steady state. For instance, $\mathrm{Wi}=0.6$ gives $C_{xx}=6e^{u/5}-5$. Real chains have finite contour lengths, so this growth identifies the failure of the linear spring model. It does not establish a universal threshold for arbitrary flow histories, finite chains or other definitions of polymer relaxation time. Combining the growth equation with residence time in planar extension also explains why a large instantaneous Weissenberg number alone does not guarantee a strongly stretched exiting chain.