Classical Mechanics, Galilean Transformations and Hint of Special Relativity?

In (1) a Galilean transformation x’ = x-vt and t’=t is applied to the Schrodinger equation id/dt (partial) W(x,t) = -1/2m d/dx d/dx W(x,t) for a free particle, where W(x,t) is the wavefunction, and it is argued that W’(x’,t’) does not equal W(x(x’t,t’),t(x’,t’)). Instead it is shown that W’(x’,t’) = W(x,t) exp( -i/bar [mvx’+.5mvvt’] with x=x’+vt’ and t=t’. The author’s of (1) suggest this is due to remnants of special relativity creeping into the nonrelativistic situation. They then argue that this issue is already found in nonrelativistic classical mechanics, in particular in the Hamiltonian-Jacobi equation H(x, dS/dx) + dS/dt = 0 where S is the free particle action= t *Lagrangian and H is the Hamiltonian i.e 1/2m dS/dx dS/dx. They show that S’(x’,t’) = S(x,t) - (mvx’ +.5mvvt’) and also argue that this is due to a remnant of special relativity creeping into nonrelativistic classical mechanics. In this note, we consider the free particle action S= -Et+px nonrelativistically. (This form also holds for the relativistic free particle). It may be seen that such a solution satisfies the Hamilton-Jacobi equation, but the catch seems to be that one must transform both (x,t) and (p,E) to find a solution in a moving frame. The Galilean transformation leads to two different transformations, one for (x,t) and another for (p,E). A Lorentz transformation on the other hand treats (x,t) and (p,E) in the same way and -Et+px behaves as a dot product with a (1,-1) type metric. We argue that given the physical example of a particle with rest mass mo at x=0 at t=to, a moving frame transforms this into a new energy (including mo and kinetic energy), a p and x’,t’ such that x’/t’=v where -v is the velocity of the frame. One may suggest a more general transformation than the Galilean i.e. replace t’=t with t’=g(v) t. This then leads to -Et+px written in terms of E’,t’, p’,x’ equalling -E’t’ + p’x’ if g(v) = 1/sqrt(1-vv/bb) where b is a constant to make v/b dimensionless. Applying the transformation, which we assume holds for both (x,t) and (p,E) to light yields b=c. One may then show that -Et+px, written in terms of E’,t’,p’,x’, yields -E’t’ + p’x’