# The mathematical model of the Galilean transformation in Hamiltonian mechanics

In my previous question, I asked about the Galilean invariance of the Hamiltonian. I've got already two answers, probably good but I have difficulties interpreting them. Both answers write the Hamiltonian with the coordinates of the moving observer. I cannot interpret this. The Hamiltonian is a function defined on the phase space. Since phase space doesn't contain the time, I see no sense of such thing as "Hamiltonian from the viewpoint of a moving observer", or such expression as $$H(x+vt,p+mv)$$. From a physical viewpoint, of course, I understand this but I don't see the clear mathematical model behind it. I can express the Galilean invariance of the Hamiltonian function only so that I require that for each solution $$(x(t),p(t))$$ of the Hamiltonian equations $$(x(t)+vt,p'(t))$$ is also a solution with some function $$t\mapsto p'(t)$$. This is a different thing than "transforming" the Hamiltonian. So, my question is, how can be modeled the Galilean invariance of the Hamiltonian mathematically correctly?

• Motion is a canonical transformation: the quoin of classical mechanics. – Cosmas Zachos Mar 17 at 19:39
• The Hamiltonian is allowed to change; it doesn't have to be invariant. In fact, if you think about it, the kinetic energy is frame-dependent, so it'd better change. Initial conditions are also expected to change. It's the eqs. of motion that don't change. The fact that eqs. of motion don't change is contained in, $$\left\{ X,P\right\} =\left\{ x+Vt,p+mV\right\} =\left\{ x,p\right\}$$ as @CosmasZachos said. – joigus Mar 17 at 21:43
• @joigus Could you explain this in detail, or give a reference which explains? – mma Mar 20 at 17:11
• en.wikipedia.org/wiki/Canonical_transformation – joigus Mar 20 at 18:38
• @joigus Thank you, I know what canonical transformation is, but I don’t know what it has to do with Galilean invariance. – mma Mar 20 at 20:57