Here's an elementary type argument I learned, based on unique splitting $T=S+A$ of linear operators $T: \mathbb{R}^n \to \mathbb{R}^n$. 

(i) every linear operator $T$ can be uniquely decomposed $T=S+A$ where $S$ is symmetric ${}^tS=S$ and $A$ is antisymmetric ${}^tA=-A$. 

The meaning of symmetric and antisymmetric operators has a physical interpretation, where every force $T$ is factored as a radial force along the orthogonal eigenspaces of $S$, and the curl-type rotational force $A$. 

(ii) Now with respect euclidean quadratic cost $c(x,y)=-x\cdot y$, the $c$-optimal transportation plans are those plans which have zero rotational component, i.e. $T=S=\nabla \phi$ for a function $\phi$. It is elementary that $T$ is positive (semi) definite iff $\phi$ is (semi) convex.  

In *physical* terms, I understand Brenier-McCann's polar factorization theorem as saying that energy-optimal optimal transport plans move particles along rotation-free trajectories. Every source particle $dx$ has a target $dy$ and is direction by the energy-optimal transport plan $\pi$ to "Keep straight, and don't twist or rotate until you arrive at your destination."