Which coordinate is cyclic in this case

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In a two-particle system with point masses at coordinates x1 and x2, the potential energy depends on their coordinate difference, V = V(x1 - x2). The Lagrangian is expressed as L = ½m(x1')² + ½m(x2')² + V(x1 - x2), and while total momentum is conserved, no cyclic coordinates are evident. Transforming the Lagrangian to new coordinates y1 = x1 + x2 and y2 = x1 - x2 reveals a structure that could potentially allow for cyclic coordinates. However, achieving a situation where all coordinates are cyclic may not be feasible without employing the Hamiltonian formulation, which allows for greater flexibility in variable variation. Ultimately, the transformation to a fully cyclic coordinate system remains uncertain, particularly with a general potential V.
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Consider a simple two particle system with two point masses of mass m at x1 and x2 with a potential energy relative to each other which depends on the difference in their coordinates V = V(x1-x2)

The lagrangian is:

L = ½m(x1')2 + ½m(x2')2 + V(x1-x2)

Obviously their total momentum is conserved d/dt(mx1' + mx2') = 0, which can be verified by plugging into the lagrangian. But there is no cyclic coordinates in the lagrangian. Is it possible to put it in a form where this hidden cyclic coordinate is shown?
 
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Hint: Try to express your lagrangian in terms of
y1=x1+x2
y2=x1-x2
 
ahh nice.
So you get:

1/4my12 + 1/4my22 - V(y2) = 0

Is it possible to transform to a situation with all coordinates cyclic?
 
It seems to be the motion in an inertial frame of reference. Well in that case x1 - x2 must be constant I think, so potential energy can be omitted as a constant. What is clear is that potential energy is a function of generalised coordinates, so it has to be Cartesian coordinates as well.
 
First of all the Lagrangian in the new coordinates
L=\frac{\mu}{2}(\dot{y}_1^2+\dot{y}_2^2)-V(y_2)
with \mu=m/2.

To answer the question, whether you can find a set of coordinates, which all are cyclic, you should read about the Hamilton-Jacobi partial differential equation and action-angle variables.

In your case there is for sure another conserved quantity! Think which that might be!
 
well that's the energy but that has nothing to do with cyclic coordinates.
Also I did read Hamilton-Jacobi theory but that takes its basis in the hamiltonian formulation. So I guess it's not really possible to transform to a frame with all coordinates cyclic UNLESS you use the hamiltonian formulation with more freedom to vary your conjugate variables?
 
To get a second cyclic variable, you would need some parameter which describes the time-evolution of your (y2-)system with fixed energy. If V is quadratic (or at least gives oscillations in some way), this would be the phase of the oscillation, for example.
I doubt that you can do this transformation in an explicit way with a general V.
 
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