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Lagrangian of two body problem with spin

  1. Jul 15, 2015 #1
    I know how to solve "typical" Kepler problem but I'm interested in a global view to "binary" systems. For example Earth - Moon. If I set lagrangian of system as ##L=\frac{1}{2}(m_1\dot{r}_1^2 + m_2\dot{r}_2^2)-V(|r_2-r_1|)## there isn't included a spin.
    My questions are:
    1) If it is solved as two body (I guess, two points) problem. Is possible to put there a term describing a spin?
    2) Why the spin isn't in general solution of Kepler problem?
    3) Whether it is possible. How? What term could describe the spins? ##L_s=\frac{1}{2}(J_1\dot{\phi}_1^2 + J_2\dot{\phi}_2^2)##? ; J - moment inertia
    4) Anyway, if or not would be possible to create lagrangian with spin. Does it change some characteristic of motion? (shape of orbit, period, etc.)?

    Thank you for your replies.
    Last edited: Jul 15, 2015
  2. jcsd
  3. Jul 15, 2015 #2
    Spin is easy to include, but the dynamics only change if there is an interaction with one or both spins.

    Including the interaction (for example to account for tides) is the hard part.
  4. Jul 15, 2015 #3
    Thank you very much for you reply.
    So if I understand in good way, it is answer to my question 1),2) and 4). Could you please comment the third?

    I totally hope, it change something. I know, changes are arising from tidal effects. But why it doesn't change anything? Lagrangian is changed and I believe that extra term is not a total derivative of some function...or? Because this is a only one case, which I know, when the equations of motion (Lagrange eq.), are not changed.

    If I may, I have last question: How could the term, I mean the easiest one (not real), which describes for example tidal effect (total toy model)?
  5. Jul 15, 2015 #4
    When I'm thinking about tidal force, honestly I have to say, I don't know results of it. What do tidal forces cause on system Earth-Moon? Which way? Changes in velocity of orbiting, spinning or...? I'm not asking for exact mechanisms, but their results on system.
  6. Jul 15, 2015 #5
    Your term for spin is right, but if there is only a kinetic energy term in a variable (no potential energy), the dynamics are trivial.
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