Legendre transformation of the CR3BP equations

AI Thread Summary
The discussion revolves around the implementation of a symplectic integrator for the closed restricted three-body problem, focusing on deriving Hamiltonian equations. The user expresses confusion regarding the notation and the relationship between the Hamiltonian and Lagrangian formulations, particularly how to derive the Hamiltonian from the equations of motion. They initially misidentify the Lagrangian as simply \(\Omega(x,y)\) and seek clarification on the correct approach to derive the Hamiltonian, questioning the origin of the kinetic energy term in the Hamiltonian expression. The user ultimately seeks guidance on the correct interpretation of the equations and the derivation process, emphasizing the complexity of the notation involved.
Deadstar
Messages
99
Reaction score
0
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?
 
Physics news on Phys.org
Deadstar said:
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?
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...
Back
Top