Kepler's problem in lagrangian formalism

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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

L=\frac{1}{2}\mathbf{\dot{q}}^2+q^{-1}

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

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

or, in vectorial notation,

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

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

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}}

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

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}}

Compare this with where we want to go:

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}}

Are there really maps q_i(x_1,y_1,z_1,x_2,y_2,z_2) that transform L into L_q ?!? :confused:
 
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The q's are the generalized coordinates so if your generalized coordinates are the cartesian coordinates, then you should have the answer?
 
What?

The book says the lagrangian can be written like this:

L=\frac{1}{2}\mathbf{\dot{q}}^2+q^{-1}

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

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}}

On the other hand, the cartesian lagrangian is

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}}

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

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

L=\frac{1}{2}\mathbf{\dot{q}}^2+q^{-1}

?
 
P.S. Ideas about what to say about \mathbf{A} are of course welcome as well ! Besides the direction and the norm, what are interesting things to say about a vector?
 
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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