# Legendre transformation of the CR3BP equations

## Main Question or Discussion Point

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

Related Classical Physics News on Phys.org
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)?

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