PX285 - C6d - two particles in 3D

basic case

consider two particles with the same mass, $m$ , in positions, $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ , so $d = 6$
the lagrangian:

L = - = \frac{1}{2} m (v_{1}^{2} + v_{2}^{2}) - V (x_{1}, \dots z_{2})

where, $v_{d}^{2} = {\dot{x}}_{d}^{2} + {\dot{y}}_{d}^{2} + {\dot{z}}_{d}^{2}$

\begin{matrix} - \frac{\partial V}{\partial x_{1}} = \frac{d}{d t} m {\dot{x}}_{1} = \frac{d}{d t} p_{x_{1}} \\ ⋮ \end{matrix}

now, consider if the potential was a function of only the separation, ie: $V (r)$ , where $r = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}}$
redefine:
$q_{1} = x_{2} - x_{1}$
$q_{2} = y_{2} - y_{1}$
$q_{3} = y_{2} - y_{1}$
$q_{4} = x_{2} + x_{1}$
$q_{5} = y_{2} + y_{1}$
$q_{6} = z_{1} + z_{2}$
there s a one-to-one mapping: $q_{1} \leftrightarrow x / y / z_{i}$ , eg: $z_{1} = \frac{q_{6} - q_{3}}{2}$
there will be three non-trivial equations of motion with a non-zero $L H S :$

\frac{\partial L}{\partial q_{1}} = \frac{d}{d t} \frac{\partial L}{\partial {\dot{q}}_{1}} = \frac{d}{d t} p_{1}

and so on for $q_{2}$ and $q_{3}$

\frac{\partial L}{\partial q_{4}} = \frac{d}{d t} \frac{\partial L}{\partial {\dot{q}}_{4}} = \frac{d}{d t} p_{4} = 0

and so on for $q_{5}$ and $q_{6}$

this is because $V$ has no dependence on them, so the corresponding momenta are constant, and conserved