Let $(\rho^{(i)}_t,{\bf v}^{(i)}_t)$ for $i = 1,2$ be two solutions of the continuity equation $$\partial \rho^{(i)}_t + \nabla\cdot \left({\bf v}^{(i)}_t \rho^{(i)}_t\right) = 0 \label{1}\tag{1}$$ on a compact domain $\Omega$ (with no-flux boundary conditions), and suppose that $(\rho^{(i)}_t,{\bf v}^{(i)}_t)$ are absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}^d$ for every $t$ and that $\rho^{(i)}$ are absolutely continuous curves in $\mathbb{W}_p(\Omega)$. Then Corollary 5.25 in the monograph Optimal Transport for Applied Mathematicians stated that $$\frac{d}{dt}\left(\frac{1}{p} W^p_p\left(\rho^{(1)}_t,\rho^{(2)}_t\right)\right) = \int_{\Omega} \left(x-T_t(x)\right)\cdot \left({\bf v}^{(1)}_t(x) - {\bf v}^{(2)}_t(T_t(x))\right)\rho^{(1)}_t ~dx \label{2}\tag{2}$$ where $T_t$ is the optimal transport map from $\rho^{(1)}_t$ to $\rho^{(2)}_t$ for the cost $\frac{1}{p}|x-y|^p$.

On the other hand, in the paper Convergence to equilibrium in Wasserstein distance for Fokker-Planck equations the continuity equation \eqref{1} with the specific velocity field $${\bf v}_t = -(\nabla \log \rho_t + A) \label{3}\tag{3}$$ is considered. Here the vector field $A$ takes the form $A = \nabla V + F$ where $F$ satisfies $\nabla \cdot (\mathrm{e}^{-V} F) = 0$, so that the equilibrium solution/distribution is of the form $\nu = \mathrm{e}^{-V}$. Now Theorem 2.1 in the aforementioned paper claimed that $$\frac{d}{dt}\left(\frac{1}{2} W^2_2\left(\rho_t,\nu\right)\right) = -\int (x-\nabla \psi_t)\cdot \left(\nabla \log \rho_t + A\right) ~ d\rho_t\label{4}\tag{4}$$ where $\nabla \psi_t$ push forwards $\rho_t$ to $\nu$ for every $t \geq 0$.

Of course, we can take $T_t(x) = \nabla \psi_t(x)$ due to property of $W^2_2$. But I am failing to see the equivalence between \eqref{2} and \eqref{4} (when $p=2$). Specially, I think the formulation \eqref{2} contains the term $$\int (x-\nabla \psi_t)\cdot {\bf v}_t(T_t(x))~d\rho_t$$ while such term is missing in the formulation \eqref{4}. May I know what is going on here?