[example: Constant-Speed Motion of a Density]
Let $\mu_0=\rho_0\mathcal L^n$, let $T_\#\mu_0=\mu_1$, and set $T_t(x)=(1-t)x+tT(x)$. By the definition of pushforward, for every bounded test function $\varphi$,
\begin{align*}
\int_{\mathbb R^n}\varphi(y)\,d\mu_t(y)=\int_{\mathbb R^n}\varphi(T_t(x))\,d\mu_0(x).
\end{align*}
Since $\partial_tT_t(x)=T(x)-x$, differentiating the identity for smooth compactly supported $\varphi$ gives
\begin{align*}
\frac{d}{dt}\int_{\mathbb R^n}\varphi(y)\,d\mu_t(y)=\int_{\mathbb R^n}\nabla\varphi(T_t(x))\cdot (T(x)-x)\,d\mu_0(x).
\end{align*}
Define the velocity along transported particles by $v_t(T_t(x))=T(x)-x$. Then the previous identity becomes
\begin{align*}
\frac{d}{dt}\int_{\mathbb R^n}\varphi(y)\,d\mu_t(y)=\int_{\mathbb R^n}\nabla\varphi(y)\cdot v_t(y)\,d\mu_t(y),
\end{align*}
which is exactly the weak form of $\partial_t\mu_t+\nabla\cdot(v_t\mu_t)=0$.
The kinetic action of this curve is computed by changing variables through the pushforward $\mu_t=(T_t)_\#\mu_0$:
\begin{align*}
\int_0^1\int_{\mathbb R^n}|v_t(y)|^2\,d\mu_t(y)\,dt=\int_0^1\int_{\mathbb R^n}|v_t(T_t(x))|^2\,d\mu_0(x)\,dt.
\end{align*}
Using $v_t(T_t(x))=T(x)-x$ gives
\begin{align*}
\int_0^1\int_{\mathbb R^n}|v_t(T_t(x))|^2\,d\mu_0(x)\,dt=\int_0^1\int_{\mathbb R^n}|T(x)-x|^2\,d\mu_0(x)\,dt.
\end{align*}
The integrand is independent of $t$, so
\begin{align*}
\int_0^1\int_{\mathbb R^n}|T(x)-x|^2\,d\mu_0(x)\,dt=\int_{\mathbb R^n}|T(x)-x|^2\,d\mu_0(x).
\end{align*}
Thus the displacement interpolation moves each particle with constant velocity from $x$ to $T(x)$, and its kinetic action equals the quadratic transport cost induced by the optimal map $T$.
[/example]