1D two particles system (Lagrangian)

Lagrangian symmetry

Let up consider a one dimensional, two particles system. The two particles  have an attractive force between them of the type $F=-kr^2$. Use the general coordinates $X$ (the position of mass center) and $d$ (the distance between the masses).
A) Please write the Lagrangian of this system as function of the general coordinates, their derivatives and time.
B) From the Lagrangian dependence on the general coordinates, please specify the symmetry of this Lagrangian.
C) Please specify what is the conserved quantity of this system
D) Please find the Hamiltonian of this system.


The figure for the two dimensional case is drawn below (it has been drawn for two dimensional case to be more clear). For the one dimensional case the equations are the same. We have  (all letters are vectors in the notation):

2 particles Lagrangian

$r1 = X – d/2$
$r2 = X + d/2$
The kinetic energy of this system is
$Ek = (1/2)*m*(dr1/dt)^2 + (1/2)*m*(dr2/dt)^2 = m*(dX/dt)^2 =m*(\dot X)^2$
(Observation : we note $\dot X = dX/dt$ )
The potential energy is
$F = -d(U)/d(r) =-d(U)/d(d) =+k/d^2$
and therefore by integration one obtains
$U = -K/d$
Hence the Lagrangian of the two particles system is
$L = T-U = m*(\dot X)^2 + K/d$


Because $L =L(\dot x^2, d)$ does depend only on $\dot X^2$ (not on $X$) it is said that there is symmetry of rotation (it means isotropy of space)
(see on  Lagrangian Symmetry Noether’s theorem)


This Lagrangian remains unchanged under a rotation, which  means
$dL/d(X) =0$
$\dot P= dL/d(X) =0$
because the conjugate of the coordinate X is the total linear momentum $\dot P$.
Thus the total linear momentum of the system $P$ is conserved.
$\dot P = d(P)/dt =0$ and $P = constant$


The Hamiltonian of this system (1D two particles) is
$H = T+U = m(\dot X)^2 – K/d$