= Thermal bead drag and heat-bath power
{title2=$D_{\rm gross}=3n\zeta k_BT/m+(n\zeta/2)\operatorname{tr}(KCK^T)$}
If the relative bead velocity $\mathbf v$ is independent of its separation and has a centered <Maxwell-Boltzmann distribution> with $\langle v_iv_j\rangle=(2k_BT/m)\delta_{ij}$, 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 $3n\zeta k_BT/m>0$, 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.
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