Why Do Canonical Transformations Preserve the Form of Hamilton's Equations?

[WiFO215]

1. Whenever we perform canonical transformations in Hamiltonian mechanics, we look for those generating functions which leave the form of the canonical equations unchanged.

Why do we restrict ourselves to those transformations which leave the equations unchanged?

Can I not do some transformation,
$$q_{i} \rightarrow Q_{i}(q, p, t)$$
$$p_{i} \rightarrow P_{i}(q, p, t)$$

Which changes the form of the Hamilton equations, but leaves the physics unchanged?

Is it very obvious that by changing the equations, our physics changes?
(My own answer is yes, but I just want to clarify)

2. Why cannot my generating function depend on more than 2 variables? Why not the old p, q AND the new Q or some such dependence? Can it depend on all four variables p,q, P and Q?

 

[Valerie Park]

Yes, your generating function can depend on more than two variables, including the old p, q, as well as the new Q. However, the transformation must still obey Hamilton's equations, which means that the generating function must be a function of the coordinates and momenta only, without any explicit time dependence. Therefore, the generating function cannot depend on all four variables (p, q, P, and Q), since that would introduce an explicit time dependence, violating Hamilton's equations.