# Legendre transformation of the CR3BP equations

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;

$$\ddot{x} - 2\dot{y} = \Omega_x$$

$$\ddot{y} + 2\dot{x} = \Omega_y$$

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

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;

$$H = \frac{(p_x + y)^2 + (p_y - x)^2}{2} - \Omega(x,y)$$ (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...

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

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

$$\frac{(p_x + y)^2 + (p_y - x)^2}{2}$$

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

As I understand it...

$$p_j=\frac{\partial L}{\partial \dot{q}_j}$$

and so $$\sum_m \dot{q}_m p_m = \dot{x} \frac{\partial L}{\partial \dot{x}} + \dot{y} \frac{\partial L}{\partial \dot{y}}$$

with $$q_1 = x$$, $$q_2 = y$$ (Perhaps this is what I'm getting wrong...)

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

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

## Answers and Replies

with $$q_1 = x$$, $$q_2 = y$$ (Perhaps this is what I'm getting wrong...)

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

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

I was taking L to be $$\Omega$$ which is not correct.

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

$$\ddot{x} - 2\dot{y} = \Omega_x$$

$$\ddot{y} + 2\dot{x} = \Omega_y$$

is the Lagrangian of the system?

If so then is below the right approach?

$$H(q_1,p_1) = \dot{q}_1 p_1 + \dot{q}_2 p_2 - L(q_1,\dot q_1)$$

=> $$H(x,\frac{\partial L}{\partial \dot{x}}) = \dot{x} \frac{\partial L}{\partial \dot{x}} + \dot{y} \frac{\partial L}{\partial \dot{y}} - L(x,\dot{x})$$

$$= 2\dot{x} - 2\dot{y} - L(x, \dot{x})$$

Then do the same for $$H(q_1, p_2)$$ 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)?

Where does $$\frac{(p_x + y)^2 + (p_y - x)^2}{2}$$come from?
Why is it not just $$\frac{p_x^2 + p_y^2}{2m}$$ which would make sense being kinetic energy would it not?