PX285 - C6c - two particles in 1D

basic case

consider two particles with the same mass, $m$ , in positions, $x_{1}$ and $x_{2}$ , so $d = 4$
the lagrangian:

L = - = \frac{1}{2} m ({\dot{x}}_{1}^{2} + {\dot{x}}_{2}^{2}) - V (x, y)

the euler-lagrange equations:

\begin{matrix} - \frac{\partial V}{\partial x_{1}} = \frac{d}{d t} m {\dot{x}}_{1} = \frac{d}{d t} p_{1} = F_{1} \\ - \frac{\partial V}{\partial x_{2}} = \frac{d}{d t} m {\dot{x}}_{2} = \frac{d}{d t} p_{2} = F_{2} \end{matrix}

define $\vec{p} = (\begin{matrix} p_{x} \\ p_{y} \end{matrix})$ , and $\vec{F} = (\begin{matrix} F_{1} \\ F_{2} \end{matrix})$
therefore, newton's law is obtained again:

\vec{F} = \frac{d}{d t} \vec{p}

special case

now, consider $V (x_{1}, x_{2}) = U (x_{1} - x_{2}) = U (z)$

\begin{matrix} - \frac{\partial V}{\partial x_{1}} = - \frac{\partial U}{\partial z} \frac{\partial z}{\partial x_{1}} = - U^{'} (z) \\ - \frac{\partial V}{\partial x_{1}} = F_{1} = - U^{'} (z) \end{matrix}

\begin{matrix} - \frac{\partial V}{\partial x_{2}} = - \frac{\partial U}{\partial z} \frac{\partial z}{\partial x_{2}} = + U^{'} (z) \\ - \frac{\partial V}{\partial x_{2}} = F_{2} = + U^{'} (z) \end{matrix}

newton's third law can be seen:

F_{1} = - F_{2}

there is also no time-dependence in momentum:

\frac{d}{d t} (m ({\dot{x}}_{1}^{2} + {\dot{x}}_{1}^{2})) = \frac{d}{d t} (p_{1} + p_{2}) = F_{1} + F_{2} = 0