Take the 2-minute tour ×
Astronomy Stack Exchange is a question and answer site for astronomers and astrophysicists. It's 100% free, no registration required.

i try to extend my analysis from planar restricted three body problem to the general one.

Is there anyone that can help me to find out the Hamiltonian for this prolem in cartesian coordinates?

I need the hamiltonian of 3D case somethink like eq.(1) in this article or in this one paper again eq.1 . Because my symplectic integrator can be used only if I have the fixed cartesian coordinates.

Thanks

share|improve this question
    
This should probably be migrated over to Physics –  Rory Alsop Apr 29 at 18:41
    
Could you please edit your question to include more information on what exactly you have problems with? The way you currently specified it, it's too broad and rather unclear what exactly it even pertains to (tags you used help a bit, but not too much). If I get you correctly, you're after Hamiltonian form of the three-body problem? If that's the case then you might find this article useful. –  TildalWave Apr 29 at 19:00
    
Yes, something like that but, in the wiki example the Hamiltonian is for the planar case. I need the hamiltonian of 3D case somethink like eq.(1) in this article link or in this one link again eq.1 –  Panichi Pattumeros PapaCastoro Apr 29 at 19:15

1 Answer 1

The condition $$\sum_{i=1}^{3}m_iq_i'=0$$ of this paper generalizes to $$\sum_{i=1}^{n}m_iq_i'=0,$$ meaning the barycenter should be at the origin of the frame.

Equation (1) generalizes to $$H=\frac{1}{2}\sum\frac{|p_i'|^2}{m_i}-\sum_{i=1}^n\sum_{j=1, j\neq i}^n\frac{m_im_j}{|q_i'-q_j'|}.$$ The first part is the total kinetic energy of the system relative to the barycenter, written in terms of momenta: $$E=\frac{1}{2}mv^2=\frac{1}{2}m\left(\frac{p}{m}\right)^2=\frac{1}{2}\frac{p^2}{m}.$$ The second part is the sum of all self-potential energy between each pair of (spherical) objects, with Newton's gravitational constant $G$ set to 1.

See also n-body problem on Wikipedia.

The 3-dimensional 3-body (or n-body) problem is like the planar version. Just take 3-dimensional vectors for $p_i'$ and $q_i'$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.