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With equal rod lengths $\ell$ and absolute downward-vertical angles $\theta ,\varphi$, the endpoint separation is $s^2=2{\ell }^2[1+\cos (\theta -\varphi )]$. A zero-rest-length spring between the fixed support and the final mass contributes $-\kappa {\ell }^2[1+\cos (\theta -\varphi )]$ to the Lagrangian. It favours bending away from maximum extension and can reduce the gravitational bending stiffness.

For identical no-slip spheres of radius $a$ linked by a zero-rest-length spring, with $R\gg a$ and $\zeta =6\pi \mu a$, the leading Oseen tensor interaction reduces their relative spring relaxation rate by $3a/(2R)$. The midpoint follows an imposed incompressible linear velocity field. Add the two bead force balance equations to prove midpoint advection; subtract them and use $\mathsf{G}(\mathbf{R})\mathbf{R}=\mathbf{R}/(4\pi \mu R)$ to obtain the displayed equation. This deterministic far-field model omits bead inertia, thermal forcing, connector drag and near-contact lubrication.

Use the deterministic Hookean dumbbell model, with a zero-rest-length spring and negligible connector drag. For $\mathbf{R}={\mathbf{r}}_1-{\mathbf{r}}_2$, its elastic forces on the two beads areHere $H>0$ is the spring constant. No independent thermal forcing from Brownian motion or bead inertia is included. The unperturbed linear velocity field is $\mathbf{Lr}$, with $\mathrm{tr}\mathbf{L}=0$ by incompressibility.

The fluid velocity used in each bead's drag law is the imposed flow plus the disturbance from the other bead, with its own singular field omitted. At leading order in $a/R$, the evenness of the Oseen tensor, $\mathsf{G}(-\mathbf{R})=\mathsf{G}(\mathbf{R})$, givesThese are the pair hydrodynamic mobility matrix relations. Their signs follow from balancing each spring force with $-\zeta ({\dot{\mathbf{r}}}_i-{\mathbf{u}}_{\mathrm{ext},i})$, the fluid drag on that bead.

Adding the equations cancels both the direct spring slips and the cross-induced velocities, since ${\mathbf{F}}_1+{\mathbf{F}}_2=0$. For the midpoint ${\mathbf{r}}_c=({\mathbf{r}}_1+{\mathbf{r}}_2)/2$,The midpoint therefore follows the undisturbed linear velocity field. Subtracting the equations instead gives the hydrodynamic interaction of a Hookean dumbbell:The dimensionless interaction matrix isIts transverse and longitudinal eigenvalues are $3a/(4R)$ and $3a/(2R)$, respectively. Since the spring force is longitudinal, the separation equation can also be writtenThe relative correction to the free-draining spring relaxation rate is therefore $3a/(2R)$, explicitly proportional to $a/R$. Entrainment by the other bead reduces their relative approach for given opposing spring forces.

This is a far-field hydrodynamic interaction, valid for $R\gg a$; it is not a near-contact law and omits higher reflections and lubrication. The printed target also fixes the spring's rest length implicitly. For an ordinary spring obeying Hooke's law with nonzero rest length $R_0$, the elastic force would be $-H(R-R_0)\hat{\mathbf{R}}$, and every spring term above would acquire the factor $1-R_0/R$. The displayed result uses the zero-rest-length spring convention appropriate to the ideal Hookean dumbbell model.

Measure both angles from the downward vertical, as in the PDF. With the support as origin and downward coordinate $z$,Thus $|AC{|}^2=2{\ell }^2[1+\cos (\theta -\varphi )]$. The requested expression implicitly uses a zero-rest-length spring: its energy is $\kappa |AC{|}^2/2$. The exact Lagrangian for equal masses isExpand the cosines and count angles and velocities as first order. The cosine correction multiplying $\dot{\theta }\dot{\varphi }$ is fourth order and is dropped. After removing constants,The positive spring term in $L_2$ means the spring lowers its energy when the straight assembly bends. If its natural length were $s_0$, this term would instead be $\kappa \ell (2\ell -s_0)(\theta -\varphi {)}^2/4$. An unspecified ordinary natural length cannot simply be ignored; the displayed target fixes the zero-length idealization.

A zero-rest-length spring has quadratic potential energy in its actual length $s$, with no subtracted natural length. It is a useful idealization, not an ordinary unstretched spring of arbitrary length. For an end-to-end spring on a two-link pendulum, changing the natural length changes the quadratic bending stiffness about the straight configuration.