 Free-energy balance of a Hookean dumbbell

For constant number density $n$, constant temperature, a homogeneous incompressible linear velocity field and decaying configuration tails, defineThis free energy is a relative entropy times $n{k}_{B}T$. Differentiate it using the configuration Fokker-Planck probability current. One integration by parts gives $\dot{\mathcal{F}}=n\int \psi (H\mathbf{R}+k_{B}T\nabla \log \psi )\cdot \mathbf{V}d\mathbf{R}$. Substituting the probability current velocity gives the displayed balance: the diffusive affine contribution is $-n{k}_{B}T\mathrm{tr}K=0$. The polymer stress tensor ${\sigma }_p=n(HC-k_{B}TI)$ has mechanical power ${\sigma }_p:K=nHC:K$. Its instantaneous power need not be nonnegative, because it includes changes of stored free energy. The nonnegative quantity is $\mathcal{D}$, and equality of stress power and dissipation requires a stationary free energy.