Diffeomorphic manifolds of inequivalent smooth structures

Given two smooth structures $\mathcal A_1$ and $\mathcal A_2$ of a spacetime manifold $\mathcal M$, we say that they are equivalent if $\mathcal A_1 \cup \mathcal A_2$ is itself a smooth structure (or if they're maximal atlases, they are just identical).

It is possible to build smooth structures that are inequivalent but are still diffeomorphic to each other, for instance on $\Bbb R$, $\phi : x \mapsto x$ and $\psi : x \mapsto x^3$. However, according to "Exotic smoothness and physics", as long as two manifolds are diffeomorphic to each other, the "physics" is the same (this is not the case if we have two smooth structures that are not diffeomorphic).

What does "the physics" refer to, here? Is it just what in physics we generally refer to as diffeomorphism invariance, where tensor quantities just transform according to the Jacobian of the transformation? What does it correspond to if we have a diffeomorphism between two inequivalent charts? Also is there really a point to consider such transformations? I'm not sure that, from the point of view of coordinate transformations, $\phi_1^{-1} \circ f \circ \phi_2$ is really a different set of such transformations from $\phi_1^{-1} \circ \phi_2$, if both have to be smooth.

• To ensure that everybody is using the same conventions, consider to include your definition of 'smooth' and 'diffeomorphic'. – Qmechanic May 27 '17 at 19:17
• "Physics is the same" most likely means that the (considered) laws of physics are diffeomorphism invariant. – trosos May 27 '17 at 20:35
• I feel like you're overthinking this. I think this just means: let $A$ and $B$ be smooth manifolds with inequivalent but diffeomorphic smooth structures, anything you say about the smooth manifold $B$ can be pulled back to a statement about $A$ via that diffeomorphism. – zzz Jul 20 '17 at 14:46