The Lagrangian and the Classical Hamiltonian of a two-particle system | Quantum Science Philippines

## The Lagrangian and the Classical Hamiltonian of a two-particle system

Vanessa V. Destura, MSU-IIT

##### Figure 1. A two-particle system

We consider two particles in a system, that is, $m_{1}$ and $m_{2}$. The relative coordinate $\vec{r}$, is given by

$\begin{array}{lclr} \vec{r} &=& \vec{r_{1}} - \vec{r_{2}} & \textbf{(1)} \end{array}$

We replace the positions $\vec{r_{1}}$ and $\vec{r_{2}}$ by the center of gravity,

$\vec{r_{G}} = \dfrac{m_{1}\vec{r_1} + m_{2}\vec{r_2}}{m_{1} + m_{2}}$

$\begin{array}{rclr} \vec{r_{G}} (m_{1} + m_{2}) &=& m_{1}\vec{r_1} + m_{2}\vec{r_2} & \textbf{(2)} \end{array}$

Now, using equations $\textbf{(1)}$ and $\textbf{(2)}$ to solve for the expressions of $\vec{r_{1}}$ and $\vec{r_{2}}$, we have

$\begin{array}{rclr} \vec{r_{G}} (m_{1} + m_{2}) - m_{1}\vec{r_{1}} &=& m_{2}\vec{r_{2}}; & \vec{r_{1}} = \vec{r} + \vec{r_{2}} \\ \vec{r_{G}} (m_{1} + m_{2}) - m_{2}\vec{r_{2}} &=& m_{1}\vec{r_{1}}; & \vec{r_{2}} = \vec{r_{1}} - \vec{r} \end{array}$

So,

$\begin{array}{rclr} \vec{r_{G}} (m_{1} + m_{2}) - m_{1} (\vec{r} + \vec{r_{2}}) &=& m_{2}\vec{r_{2}} \\ \vec{r_{G}} (m_{1} + m_{2}) - m_{1}\vec{r} &=& (m_{1} + m_{2})\vec{r_{2}} \end{array}$

$\begin{array}{rclr} \vec{r_G} - \dfrac{m_1}{m_{1} + m_{2}}\vec{r} &=& \vec{r_2} & \textbf{(3)} \end{array}$

$\begin{array}{rclr} \vec{r_{G}} (m_{1} + m_{2}) - m_{2} (\vec{r_{1}} - \vec{r}) &=& m_{1}\vec{r_{1}} \\ \vec{r_{G}} (m_{1} + m_{2}) + m_{2}\vec{r} &=& (m_{1} + m_{2})\vec{r_{1}} \end{array}$

$\begin{array}{rclr} \vec{r_G} + \dfrac{m_2}{m_{1} + m_{2}}\vec{r} &=& \vec{r_1} & \textbf{(4)} \end{array}$

In Classical Mechanics, the two-particle system is described by Lagrangian,

$\begin{array}{rcl} L (\vec{r_{1}}, \dot{\vec{r_{1}}}; \vec{r_{2}}, \dot{\vec{r_{2}}}) &=& T - V \\ &=& \dfrac{1}{2} m_{1}{\dot{\vec{r_{1}}}}^{2} + \dfrac{1}{2} m_{2}{\dot{\vec{r_{2}}}}^{2} - V({\vec{r}})\end{array}$

From equations $\textbf{(3)}$ and $\textbf{(4)}$, we have

$\begin{array}{rcl} \dot{\vec{r_{1}}} &=& \dot{\vec{r_{G}}} + \dfrac{m_{2}}{m_{1} + m_{2}} \dot{\vec{r}} \\ \dot{\vec{r_{2}}} &=& \dot{\vec{r_{G}}} - \dfrac{m_{1}}{m_{1} + m_{2}} \dot{\vec{r}} \end{array}$

Hence, the Lagrangian in terms of $\vec{r_{G}}$ and $\vec{r}$ is given by

$\begin{array}{rclr} L (\vec{r_{G}}, \dot{\vec{r_{G}}}; \vec{r}, \dot{\vec{r}}) &=& \dfrac{1}{2} m_{1} {\left(\dot{\vec{r_{G}}} + \dfrac{m_{2}}{m_{1} + m_{2}}\dot{\vec{r}}\right)}^{2} + \dfrac{1}{2} m_{2} {\left(\dot{\vec{r_{G}}} - \dfrac{m_{1}}{m_{1} + m_{2}}\dot{\vec{r}}\right)}^{2} - V(\vec{r}) \\ \\ L (\vec{r_{G}}, \dot{\vec{r_{G}}}; \vec{r}, \dot{\vec{r}}) &=& \dfrac{1}{2} M{\dot{\vec{r_{G}}}}^{2} + \dfrac{1}{2}\mu {\dot{\vec{r}}}^{2} - V(\vec{r}) = 0 & \textbf{(5)} \end{array}$

where $M = m_{1} + m_{2}$ (total mass) and $\mu = \dfrac{m_{1} m_{2}}{m_{1} + m_{2}}$ (reduced mass).

In order to solve for the classical Hamiltonian, $H$, we alter the expression of the reduced mass $\mu$ from the Lagrangian (equation $\textbf{(5)}$) into $\dfrac{1}{\mu}$. So,

$\begin{array}{rcl} \dfrac{1}{\mu} &=& \dfrac{m_{1} + m_{2}}{m_{1} m_{2}} \\ &=& \dfrac{1}{m_1} + \dfrac{1}{m_2} \end{array}$

Now, we differentiate $\textbf{(5)}$ in terms of $\dot{\vec{r_G}}$ and $\dot{\vec{r}}$ to obtain the conjugate momenta of the variables $\vec{r_G}$ and $\vec{r}$, that is,

$\begin{array}{rclr} \vec{p_G} &= & M\dot{\vec{r_G}} & \textbf{(6)}\\ \vec{p} &=& \mu \dot{\vec{r}} & \textbf{(7)} \end{array}$

We further expand $\textbf{(6)}$ by substituting the reduced mass, $M$ and $\dot{\vec{r_G}} = \dfrac{m_{1} \dot{\vec{r_1}} + m_{2} \dot{\vec{r_2}}}{m_1 + m_2}$. So,

$\begin{array}{rcl} \vec{p_G} = M\dot{\vec{r_G}} &=& (m_1 + m_2) \dfrac{m_{1} \dot{\vec{r_1}} + m_{2} \dot{\vec{r_2}}}{m_1 + m_2} \\ &=& m_1 \dot{\vec{r_1}} +m_2\dot{\vec{r_2}} \\ &=& \vec{p_1} + \vec{p_2} \end{array}$

where $\vec{p_1} = m_1 \dot{\vec{r_1}}$ and $\vec{p_2} = m_2 \dot{\vec{r_2}}$ are the conjugate momenta of the two particles.

Since, $\dot{\vec{r_1}} = \dot{\vec{r}} + \dot{\vec{r_2}}$ and $\dot{\vec{r_2}} = \dot{\vec{r_1}} - \dot{\vec{r}}$, we have

$\begin{array}{rcl} M \dot{\vec{r_G}} &=& m_1 (\dot{\vec{r}} + \dot{\vec{r_2}}) + m_2 (\dot{\vec{r_1}} - \dot{\vec{r}}) \\ &=& (m_1 - m_2) \dot{\vec{r}} + m_1 \dot{\vec{r_2}} + m_2 \dot{\vec{r_1}} \\ M \dot{\vec{r_G}} &=& (m_1 - m_2) \dot{\vec{r}} \\ (m_1 + m_2) \dfrac{\vec{p_1} + \vec{p_2}}{m_1 + m_2} &=& (m_1 - m_2)\dot{\vec{r}} \end{array}$

$\begin{array}{rclr} \dot{\vec{r}} &=& \dfrac{\vec{p_1} + \vec{p_2}}{(m_1 - m_2)} & \textbf{(8)} \end{array}$

So, substituting $\textbf{(8)}$ into $\textbf{(7)}$ we find

$\begin{array}{rcl} \vec{p} = \mu \dot{\vec{r}} &=& \left(\dfrac{m_{1} m_{2}}{m_{1} + m_{2}}\right) \left(\dfrac{\vec{p_1} + \vec{p_2}}{m_1 - m_2}\right) \\ &=& \dfrac{m_2 \vec{p_1} - m_1 \vec{p_2}}{m_1 + m_2}\end{array}$

or

$\dfrac{\vec{p}}{\mu} = \dfrac{\vec{p_1}}{m_1}-\dfrac{\vec{p_2}}{m_2}$

In this case, $\vec{p_G}$ is the total momentumĀ of the system and $\vec{p}$ is theĀ relative momentum of the two particles.

Thus, from the Lagrangian, we arrived to

$H (\vec{r_G}, \vec{p_G}; \vec{r}, \vec{p}) = \dfrac{{\vec{p_G}}^2}{2M} + \dfrac{{\vec{p}}^2}{2 \mu} + V(\vec{r})$