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Kepler's problem in lagrangian formalism

  1. Feb 2, 2007 #1

    quasar987

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    In the statement of the problem, it is said that with an appropriate choice of units, the lagrangian for Kepler's problem can be written

    [tex]L=\frac{1}{2}\mathbf{\dot{q}}^2+q^{-1}[/tex]

    A priori, what q is in terms of cartesian coordinates seems irrelevant because the problem only asks to show that the Runger-Lenz vector, defined by

    [tex]A_k=\mathbf{\dot{q}}^2 q_k-\mathbf{q} \cdot \mathbf{\dot{q}}\dot{q_k}-q_k/q \ \ \ \ \ k=1,2,3[/tex]

    or, in vectorial notation,

    [tex]\mathbf{A}=\mathbf{\dot{q}}\times (\mathbf{q}\times \mathbf{\dot{q}})-\mathbf{q}/q[/tex]

    is a constant of the motion associated (in the sense of Noether's thm) to a certain coordinate transformation.

    But then the question asks, "Discuss the properties of this vector." and I'm kind of at a loss about what to say. I can't say much about its direction and its norm is ugly and uninsightful. So I figured if I knew what those q where in terms of cartesian coordinates, maybe I could make some sense out of the Runge-Lenz vector. So I try to get the lagrangian into the above form, right?

    Ok, in cartesian, it is

    [tex]L=\frac{m_1}{2}(\dot{x}_1^2+\dot{y}_1^2+\dot{z}_1^2)+\frac{m_2}{2}(\dot{x}_2^2+\dot{y}_2^2+\dot{z}_2^2)+\frac{Gm_1m_2}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}[/tex]

    We can choose the units of mass, length and time such that the lagrangian becomes

    [tex]L=\frac{1}{2}(\dot{x}_1^2+\dot{x}_2^2+\dot{y}_1^2+\dot{y}_2^2+\dot{z}_1^2+\dot{z}_2^2)+\frac{1}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}[/tex]

    Compare this with where we want to go:

    [tex]L_q=\frac{1}{2}(\dot{q}_1^2+\dot{q}_2^2+\dot{q}_3^2)+\frac{1}{\sqrt{q_1^2+q_2^2+q_3^3}}[/tex]

    Are there really maps [itex]q_i(x_1,y_1,z_1,x_2,y_2,z_2)[/itex] that transform L into L_q ?!? :confused:
     
    Last edited: Feb 2, 2007
  2. jcsd
  3. Feb 3, 2007 #2
    The q's are the generalized coordinates so if your generalized coordinates are the cartesian coordinates, then you should have the answer?
     
  4. Feb 3, 2007 #3

    quasar987

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    What?

    The book says the lagrangian can be written like this:

    [tex]L=\frac{1}{2}\mathbf{\dot{q}}^2+q^{-1}[/tex]

    Or, expanding the vectors into their components (i.e. the generalized coordinates),

    [tex]L_q=\frac{1}{2}(\dot{q}_1^2+\dot{q}_2^2+\dot{q}_3^ 2)+\frac{1}{\sqrt{q_1^2+q_2^2+q_3^3}}[/tex]

    On the other hand, the cartesian lagrangian is

    [tex]L=\frac{1}{2}(\dot{x}_1^2+\dot{x}_2^2+\dot{y}_1^2+ \dot{y}_2^2+\dot{z}_1^2+\dot{z}_2^2)+\frac{1}{\sqr t{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}[/tex]

    Clearly the obvious [itex]q_1=(x_2-x_1)[/itex], [itex]q_2=(y_2-y_1)[/itex], [itex]q_3=(z_2-z_1)[/itex] inspired by comparison of the "potential term" does not work for if we expand [tex](\dot{q}_1^2+\dot{q}_2^2+\dot{q}_3^ 2)[/tex] in terms of x, y, z, we do not get [tex](\dot{x}_1^2+\dot{x}_2^2+\dot{y}_1^2+ \dot{y}_2^2+\dot{z}_1^2+\dot{z}_2^2)[/tex].

    So I ask in disbelief, is there really a coordinate transformation that allows one to write L as

    [tex]L=\frac{1}{2}\mathbf{\dot{q}}^2+q^{-1}[/tex]

    ???
     
  5. Feb 3, 2007 #4

    quasar987

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    P.S. Ideas about what to say about [itex]\mathbf{A}[/itex] are of course welcome as well ! Besides the direction and the norm, what are interesting things to say about a vector? :grumpy:
     
  6. Feb 3, 2007 #5

    quasar987

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    I found it... q is actually just the ordinary position vector like Logarythmic pointed out (I think). So we are actually solving the Newton problem [particle in a 1/r potential] for a reduced mass µ (but with the units chosen to that µ=1). We can then set q=r1-r2 to get the solution to the actual Kepler problem.
     
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