the Lagrangian of the planar circular restricted three-body problem in the rotating coordinate frame is:
$$\mathcal{L}(x,y,\dot{x},\dot{y})=\frac{1}{2}(\dot{x}-\Omega y)^2 + \frac{1}{2}(\dot{y}+\Omega x)^2- \frac{Gm_1}{r_1}-\frac{Gm_2}{r_2}$$
where: $$r_1=\sqrt{(x-d_1)^2+y^2}, \quad r_2=\sqrt{(x-d_2)^2+y^2}$$
and $d_1,d_2$ being the designated positions of the primaries on $x$-axis.
By means of the Legendre transformation: $$p_x= \dot{x}-\Omega y , \quad p_y= \dot{y}+\Omega x$$
we can transition to the Hamiltonian:
$$H(x,y,p_x,p_y)= \frac{(p_x+\Omega y)^2}{2}+\frac{(p_y-\Omega x)^2}{2} -\frac{1}{2}\Omega^2(x^2+y^2)-\frac{Gm_1}{r_1}-\frac{Gm_2}{r_2}.$$
For utilizing bounded motion, we prefer to work on the "state space" $(x,y,\dot{x},\dot{y})$ rather than phase space $(x,y,p_x,p_y)$. We can show that the transformation:
$$x\rightarrow x, y\rightarrow y,\quad p_x\rightarrow \dot{x}-\Omega y, \quad p_y\rightarrow \dot{y}+\Omega x\tag{1}$$
has unitary Jacobian, and thus preserves the volume while transitioning from phase space to state space. However, this is not a canonical transformation, since it does not maintain the symplectic structure of Hamilton's equations. Βeyond that, according to Liouville's theorem, the volume in phase space is preserved under time evolution.
I have the following question:
Is it a valid claim, that the volume in state space is still preserved? My train of thought is the following:
If we begin with a "bulk" of initial conditions on state space for which we want to track its volume while time evolves, we can firstly transform into phase space, then time propagate, and then "go back" to state space by utilizing transformation (1). All transformation implemented in the above process maintain phase space volume, so i would expect that the volume remains unchanged in state space.
I hope my argument is clear.
