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Write the centre position as $\mathbf{x}$ and the bead positions as ${\mathbf{r}}_{1,2}=\mathbf{x}\mp \mathbf{R}/2$. In the affine velocity field, $\mathbf{u}({\mathbf{r}}_{1,2})=\mathbf{u}(\mathbf{x})\mp K\mathbf{R}/2$. The midpoint follows the flow, so its two slip velocities areEach bead loses mechanical power $\zeta$ times its squared slip speed. Adding the two powers proves the drag dissipation of a bead-spring pair:The factor one half comes from two equal-and-opposite bead slips, not from assigning only one bead a drag force.

If the relative bead velocity $\mathbf{v}$ is independent of its separation and has a centered Maxwell-Boltzmann distribution with $⟨v_i{v}_j⟩=(2k_{B}T/m){\delta }_{ij}$, averaging the drag dissipation of a bead-spring pair gives the displayed gross loss. The cross term vanishes by the zero conditional mean. At zero velocity gradient, this remains $3n\zeta k_{B}T/m>0$, although mechanical stress power is zero. A thermal bath replenishes microscopic kinetic energy at equilibrium. Equating this gross loss to polymer stress power omits heat-bath input and cannot follow from the independently specified Maxwellian velocity distribution. The overdamped free-energy balance of a Hookean dumbbell uses a probability current velocity instead.