Classic description of the problem two bodies under the action of a central force (Hydrogen
Atom)
Consider two particles of mass m1 and m2, whose position (in relation to an inertial frame) is
represented by the vectors r1 and r2, respectively, which interact through a central force
F(|r1 - r2|) = -∇V(|r1 - r2|),
where V is the potential energy associated with the conservative force F. The Lagrangian of the
system is
ℒ = 1/2 m1 |ṙ1|^2 + 1/2 m2 |ṙ2|^2 - V(|r1 - r2|). with r = r1 - r2, R = (m1 r1 + m2 r2) / (m1 + m2),
Changing to the referential in the center of mass, where R = 0, the Lagrangian can be written in
the form
ℒ = 1/2 μ |ṙ|^2 - V(r), with r = |r|, μ = (m1 m2) / (m1 + m2).
Note that equation (1) describes the Lagrangian of a particle of mass μ (known as reduced
mass of the system) and position r, with potential energy V(r).
a) Show that the fact that
L = r × p, with P = μ |ṙ|.
implies that the trajectory of the system is restricted to the plane formed by the vectors r and
p. So, the movement has two degrees of freedom, which can be described by means of the
polar coordinates (r; ϕ) and, therefore, the Lagrangian can be expressed in the form
ℒ = 1/2 μ (ṙ^2 + r^2 ṡ^2) - V(r).
b) According to the information in the previous item, show that
L = μ r^2 ṡ
it is a motion constant that is equivalent to the total angular momentum of the system.
c) Show that the Hamiltonian of the two-body system can be written in the form
ℋ = P_r^2 / 2μ + L^2 / 2μr^2 + V(r),
where
P_r = μ ṙ
is the generalized moment associated with the coordinate r.