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Lagrangian for the Kepler Problem

  1. Aug 26, 2013 #1
    In a two-body solution ( Kepler Problem) how do you find the Lagrangian, L, if the position vector is:

    [tex]x = (v_x * t + x_0, v_y * t + y_0, v_z * t + z_0)[/tex]

    Action from Wikipedia is:

    [tex]S = \int_{t_1}^{t_2} L * dt [/tex]
     
    Last edited: Aug 26, 2013
  2. jcsd
  3. Aug 26, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    That is certainly not true for all t, as this would mean linear motion (no acceleration).

    The Lagrangian can be found similar to all other Lagrangians for a particle moving in a potential, the potential is GMm/r where ##r=|x|##.
     
  4. Aug 26, 2013 #3
    If the central mass is large enough, the Lagrangian is

    $$\mathscr{L} = \frac{1}{2}m(\dot{r}+ r \dot{\theta})^{2} - U(r)$$

    Plugging it into the Euler-Lagrange equation:

    $$\frac{\partial \mathscr{L}}{\partial q} = \frac{d}{dt}\frac{\partial \mathscr{L}}{\partial \dot{q}}$$

    For the radial component, we have ##m \ddot{r} = mr \dot{\theta}^{2}-U'(r)##. The ##\theta## component gives us ##0##, because ##\frac{\partial \mathscr{L}}{\partial \theta} = 0 = \frac{d}{dt}(mr^2 \dot{\theta})##, we can call ##mr^2 \dot{\theta} = L##

    At this point, you can pretty much solve the equations of motion with respect to time to get the position.

    I hope that wasn't too convoluted of an answer.

    Edit: Here's a link you might be interested in reading: MIT - Kepler Problem
     
    Last edited: Aug 26, 2013
  5. Aug 26, 2013 #4
    I found a solution for Classical Mechanics

    http://www.wolframalpha.com/input/?i=kepler+problem

    The Kinetic Energy came out to be [tex]\frac {1}{2} (\frac{dx}{dt})^2[/tex]
    The Potential Energy came out to be [tex]\frac {\lambda}{\frac {dx}{dt}}[/tex]

    Not sure what lambda is. Can someone explain?
     
  6. Aug 26, 2013 #5
    None of that is correct. First, Kepler's problem is usually set in at least 2 dimensions. Second, the potential energy does not depend on velocity; it depends on displacement.
     
    Last edited: Aug 26, 2013
  7. Aug 26, 2013 #6
    I've got no idea what you're doing.

    Look, the Lagrangian IS the kinetic minus the potential. ##\mathscr{L} = KE - U##

    In the Kepler Problem, where ##M >> m## $$\mathscr{L} = \frac{1}{2}m(\dot{r}+ r \dot{\theta})^{2} - \frac{GMm}{r}$$

    As Voko says, this can be reduced to a one dimensional problem by playing noting that ---->

    $$\frac{\partial \mathscr{L}}{\partial \theta} = 0 = \frac{d}{dt}(mr^2 \dot{\theta})$$ where we can set ##mr^2 \dot{\theta} = L## and play with the equations from here.

    What EXACTLY are you trying to figure out? State your question CLEARLY, because I don't have a clue what you're asking for.
     
  8. Aug 27, 2013 #7

    vanhees71

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    The potential is spherically symmetric and thus angular momentum is conserved. The trajectory of the planet thus is in a plane perpendicular to the angular momentum and thus in a plane.

    The kinetic energy is
    [tex]T=\frac{m}{2} \dot{\vec{x}}^2.[/tex]
    Using polar coordinates in the plane perpendicular to angular momentum
    [tex]\vec{x}=r \begin{pmatrix} \cos \varphi \\ \sin \varphi \end{pmatrix}, [/tex]
    you get
    [tex]T=\frac{m}{2} (\dot{r}^2+ r^2 \dot{\varphi}^2).[/tex]
    The Lagrangian thus is
    [tex]L=T-V=\frac{m}{2} (\dot{r}^2+ r^2 \dot{\varphi}^2 )+ \frac{G mM}{r}.[/tex]
    Now you see that [itex]\varphi[/itex] is cyclic and thus
    [tex]J=\frac{\partial L}{\partial \dot{\varphi}}=m r^2 \dot{\varphi}=\text{const}.[/tex]
    This is the conservation of angular momentum in this special frame of reference.

    Rewriting the equation of motion for [itex]r[/itex] using this first integral as a function of [itex]\varphi[/itex] and doing the substitution [itex]s=1/r[/itex] you get an easy to solve differential equation, leading to ellipses, parabolas, or hyperbolas as trajectories.

    The case [itex]J=0[/itex] has to be discussed separately.
     
  9. Aug 27, 2013 #8
    Thanx
     
    Last edited: Aug 27, 2013
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