Kepler problem in parabolic coordinates

Click For Summary
SUMMARY

The discussion centers on solving the Kepler problem using parabolic coordinates, specifically addressing the separation constants required for the solution. The equation presented is E=2/(m*(a+b)) * (a*(dWa/da)^2+b*(dWb/db)^2-k)+l^2/(2mab), where l is a constant related to the cyclic variable c. Participants seek clarity on the appropriate separation constants and request a detailed Hamiltonian (H) for the first solution provided.

PREREQUISITES
  • Understanding of classical mechanics and orbital dynamics
  • Familiarity with parabolic coordinates in physics
  • Knowledge of Hamiltonian mechanics
  • Proficiency in calculus, particularly differentiation
NEXT STEPS
  • Research the derivation of separation constants in parabolic coordinates
  • Study Hamiltonian mechanics and its application to the Kepler problem
  • Explore the concept of cyclic variables in classical mechanics
  • Learn about the role of energy equations in orbital dynamics
USEFUL FOR

Students and researchers in physics, particularly those focusing on classical mechanics, orbital dynamics, and mathematical methods in physics.

Kate_12
Messages
7
Reaction score
2
Homework Statement
kepler problem
H=1/2m(px^2+py^2+pz^2)-k/(x^2+y^2+z^2)^1/2
with parabolic coordinates (a,b,c)
x=sqrt(ab)cos c
y=sqrt(ab)sin c
z=(a-b)/2
1) rewrite H as a function of new canonical variables (a,b,c, pa,pb,pc)
2) Hamilton-Jacobi equation in this coordinate system turns out to be completely separable. Using the Ansatz S=Wa(a)+Wb(b)+Wc(c)-Et, write the partial differential equation for each Wa, Wb, Wc with suitable separation constants.
Relevant Equations
Hamilton Jacobi equation
I solve (1).
But to solve (2), What should be the suitable separation constants?
I am so confused...

E=2/(m*(a+b)) * (a*(dWa/da)^2+b*(dWb/db)^2-k)+l^2/(2mab)
where l(constant) is pc since c is cyclic.

What should I do to solve the problem?
 
Physics news on Phys.org
if you do not mind could you write down H of your solution 1 ?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 46 ·
2
Replies
46
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
22
Views
4K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K