Legendre transformation of the CR3BP equations

[Deadstar]

I'm not quite sure where to post this but I suppose it should go here given it's about classical mechanics...

Anyhoo. I'm currently on the long road to implementing a symplectic integrator to simulate the closed restricted 3 body problem and I'm in the process of deriving the Hamiltonian equations for it. I'm having a problem with this part as I'm finding the notation slightly confusing/misleading.

So anyway let's crack on...

We have the equations of motion for the three body problem as;

[tex]\ddot{x} - 2\dot{y} = \Omega_x[/tex]

[tex]\ddot{y} + 2\dot{x} = \Omega_y[/tex]

where [tex]\Omega(x,y) = \frac{x^2 + y^2}{2} + \frac{1-\mu}{r_1} + \frac{\mu}{r_2}[/tex].

Now, I'm following the below pdf where the Hamiltonian is found about halfway down page 5.

http://www.cds.caltech.edu/~koon/papers/specialist_final.pdf

It's given as;

[tex]H = \frac{(p_x + y)^2 + (p_y - x)^2}{2} - \Omega(x,y)[/tex] (I'm only dealing with planar case so have dropped the z part.

Now my information on Legendre transformations come from the wikipedia link here http://en.wikipedia.org/wiki/Legendre_transformation#Hamilton-Lagrange_mechanics.

We see that H is defined as...

[tex]H(q_i,p_j,t) = \sum_m \dot{q}_m p_m - L(q_i,\dot q_j(q_h, p_k),t)[/tex].

Now, from the pdf I'm following, it would appear that the L(...) part in the above definition is just [tex]\Omega(x,y)[/tex] and the sum part is;

[tex]\frac{(p_x + y)^2 + (p_y - x)^2}{2}[/tex]

but I can't quite figure out how to get the sum part.

As I understand it...

[tex]p_j=\frac{\partial L}{\partial \dot{q}_j}[/tex]

and so [tex]\sum_m \dot{q}_m p_m = \dot{x} \frac{\partial L}{\partial \dot{x}} + \dot{y} \frac{\partial L}{\partial \dot{y}}[/tex]

with [tex]q_1 = x[/tex], [tex]q_2 = y[/tex] (Perhaps this is what I'm getting wrong...)

But then that would equal zero which is clearly not right! (no [tex]\dot{x}[/tex] or [tex]\dot{y}[/tex] terms in L)

So can someone help me see where I am going wrong here? Do I have to alter or change [tex]\Omega[/tex] at all or does that just stay as it is?

 

[Deadstar]

with [tex]q_1 = x[/tex], [tex]q_2 = y[/tex] (Perhaps this is what I'm getting wrong...)

But then that would equal zero which is clearly not right! (no [tex]\dot{x}[/tex] or [tex]\dot{y}[/tex] terms in L)

Ok this is where I have figured out my misunderstanding is.

I was taking L to be [tex]\Omega[/tex] which is not correct.

Am I right in saying that the set of equations below...

[tex]\ddot{x} - 2\dot{y} = \Omega_x[/tex]

[tex]\ddot{y} + 2\dot{x} = \Omega_y[/tex]

is the Lagrangian of the system?

If so then is below the right approach?

[tex]H(q_1,p_1) = \dot{q}_1 p_1 + \dot{q}_2 p_2 - L(q_1,\dot q_1)[/tex]

=> [tex]H(x,\frac{\partial L}{\partial \dot{x}}) = \dot{x} \frac{\partial L}{\partial \dot{x}} + \dot{y} \frac{\partial L}{\partial \dot{y}} - L(x,\dot{x})[/tex]

[tex] = 2\dot{x} - 2\dot{y} - L(x, \dot{x})[/tex]

Then do the same for [tex]H(q_1, p_2)[/tex] etc...

A quick look tells me this won't work but I can't see what I'm supposed to do instead. Notations very confusing...

Does m sum from 1 to 2 (in this case)?

 

[Deadstar]

Ok this will be my final post on the matter as there doesn't seem to be any interest in this question.

My question now is as follows...

Where does [tex]\frac{(p_x + y)^2 + (p_y - x)^2}{2}[/tex]come from?

Why is it not just [tex]\frac{p_x^2 + p_y^2}{2m}[/tex] which would make sense being kinetic energy would it not?