Lagrangian Mechanics Made Simple
Part 3
Central Force Motion

Ron Steinke <rsteinke@w-link.net>

This month we examine the mechanics of a system of two particles which are attracted by a force which depends on the distance between them. This, when combined with our discussion of rotating coordinate frames last month, will allow us to begin our treatment of orbital mechanics in the next installment.

Two Particles in a Central Potential

Consider two particles at positions ${\bf r}_1$ and ${\bf r}_2$, attracted by a force that depends on the distance

\begin{displaymath} r = \left\vert{\bf r}_1 - {\bf r}_2\right\vert \end{displaymath} (1)

between them. The kinetic energy of the system is
\begin{displaymath} T = \frac{1}{2} m_1 \dot{\bf r}_1^2 + \frac{1}{2} m_2 \dot{\bf r}_2^2 \,, \end{displaymath} (2)

and the potential energy $U$ is a function of the distance $r$. The Lagrangian is therefore
\begin{displaymath} L = \frac{1}{2} m_1 \dot{\bf r}_1^2 + \frac{1}{2} m_2 \dot{\bf r}_2^2 - U(r) \,. \end{displaymath} (3)

We want to simplify the problem by removing the motion of the center of mass of the system. In a system where there are no external forces acting on the particles, the center of mass of the system,

\begin{displaymath} {\bf R} = \frac{m_1 {\bf r}_1 + m_2 {\bf r}_2}{m_1 + m_2} \,, \end{displaymath} (4)

moves with a constant velocity, independent of the motions of the particles. Instead of writing the Lagrangian as a function of the positions ${\bf r}_1$ and ${\bf r}_2$ of the particles, we will write it as a function of the position of the center of mass ${\bf R}$ and the particle separation
\begin{displaymath} {\bf r} = {\bf r}_1 - {\bf r}_2 \,. \end{displaymath} (5)

In terms of these variables, the positions of the particles are
$\displaystyle {\bf r}_1$ $\textstyle =$ $\displaystyle {\bf R} + \frac{m_2}{m_1 + m_2} {\bf r}$  
$\displaystyle {\bf r}_2$ $\textstyle =$ $\displaystyle {\bf R} - \frac{m_1}{m_1 + m_2} {\bf r} \,,$ (6)

and the Lagrangian is
\begin{displaymath} L = \frac{1}{2} \left(m_1 + m_2\right) \dot{\bf R}^2 + \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} \dot{\bf r}^2 - U(r) \,. \end{displaymath} (7)

The Lagrangian is a sum of terms containing only ${\bf R}$ and terms containing only ${\bf r}$. This means that ${\bf R}$ and ${\bf r}$ don't appear in each others' equations of motion. A Lagrangian with this property is said to be separable, and can be written as a sum of two Lagrangians each governing part of the system. From (7), we see that the equation of motion for ${\bf R}$ is
\begin{displaymath} \ddot{\bf R} = 0 \,. \end{displaymath} (8)

The center of motion is not accelerating, which confirms that momentum is conserved in this system. We will drop ${\bf R}$ for the remainder of this article.

The Central Potential

The Lagrangian for the separation ${\bf r}$ between the particles is

\begin{displaymath} L = \frac{1}{2} \mu \dot{\bf r}^2 - U(r) \,, \end{displaymath} (9)

where the quantity
\begin{displaymath} \mu = \frac{m_1 m_2}{m_1 + m_2} \end{displaymath} (10)

is called the reduced mass of the system. Solving the equation of motion for the particle separation is equivalent to solving the equation of motion for a particle of mass $\mu$ which is attracted to the point ${\bf r} = 0$ by an external force. Notice that if the mass of the first particle is much larger than that of the second (e.g., if the particles are the Earth and a satellite orbiting it), the denominator of (10) is approximately equal to $m_1$, and the reduced mass $\mu$ is approximately equal to $m_2$. This reflects the fact that, in the case of a large mass difference, the heavier particle is "fixed" and the lighter one orbits around it.

We can deduce several things about the particle motion without knowing the exact form of $U(r)$. The equation of motion derived from (9) is

\begin{displaymath} \mu \ddot{\bf r} = - \frac{\bf r}{r} \frac{dU}{dr} \,. \end{displaymath} (11)

The force is along the direction of the displacement ${\bf r}$. This means that the particle will always lie in the plane defined by its position vector ${\bf r}$ and its velocity $\dot{\bf r}$.

Since the vector ${\bf r}$ lies in a fixed plane, it can be described by its magnitude $r$ and an angular displacement $\theta$. In terms of these variables, the Lagrangian is

\begin{displaymath} L = \frac{1}{2} \mu \left(\dot r^2 + r^2 \dot\theta^2\right) - U(r) \,. \end{displaymath} (12)

Since the Lagrangian depends on $\dot\theta$ but not on $\theta$, this gives an equation of motion
\begin{displaymath} \frac{d}{dt} \frac{\partial L}{\partial \dot\theta} = 0 \,, \end{displaymath} (13)

or
\begin{displaymath} \frac{\partial L}{\partial \dot\theta} = \mu r^2 \dot\theta = l \,, \end{displaymath} (14)

where $l$ is a constant. The expression $\mu r^2 \dot\theta$ is, in fact the angular momentum of the system, and this equation of motion is an expression of angular momentum conservation.

We can obtain a similar equation using energy conservation. The energy of the system is

\begin{displaymath} E = T + U = 2T - L \,. \end{displaymath} (15)

If the energy is constant, its time derivative must be zero. We will demonstrate this by utilizing the Lagrangian equations of motion. The time derivative of the energy is
\begin{displaymath} \frac{dE}{dt} = 2 \frac{dT}{dt} - \frac{\partial L}{\partial... ...t r - \frac{\partial L}{\partial \dot \theta} \ddot \theta \,. \end{displaymath} (16)

We integrate by parts in the $\ddot r$ and $\ddot \theta$ terms and rearrange to get
\begin{displaymath} \frac{dE}{dt} = \frac{d}{dt} \left[ 2T - \frac{\partial L}{\... ...rac{d}{dt} \frac{\partial L}{\partial \dot \theta} \right] \,. \end{displaymath} (17)

Lagrange's equations tell us that
$\displaystyle \frac{\partial L}{\partial r}$ $\textstyle =$ $\displaystyle \frac{d}{dt} \frac{\partial L}{\partial \dot r}$  
$\displaystyle 0$ $\textstyle =$ $\displaystyle \frac{d}{dt} \frac{\partial L}{\partial \dot \theta} \,,$ (18)

so the terms in the second set of brackets in (17) cancel. This leaves
$\displaystyle \frac{dE}{dt}$ $\textstyle =$ $\displaystyle \frac{d}{dt} \left[ 2T - \frac{\partial L}{\partial \dot r} \dot r \frac{\partial L}{\partial \dot \theta} \dot\theta \right]$  
  $\textstyle =$ $\displaystyle \frac{d}{dt} \mu \left[ \dot r^2 + r^2 \dot\theta^2 - \dot r^2 - r^2 \dot\theta^2 \right] \,,$ (19)

which is zero. This result is true in general when the kinetic energy $T$ depends quadratically on the velocities of the particles.

The Equations Of Motion

The equations of motion for ${\bf r}$ have been reduced to conservation of energy,

\begin{displaymath} E = \frac{1}{2} \mu \left( \dot r^2 + r^2 \dot\theta^2 \right) + U(r) \,, \end{displaymath} (20)

and conservation of angular momentum,
\begin{displaymath} l = \mu r^2 \dot\theta \,. \end{displaymath} (21)

It is straightforward to solve the second of these for $\dot\theta$,
\begin{displaymath} \dot\theta = \frac{l}{\mu r^2} \,, \end{displaymath} (22)

and substitute this into the first to obtain an equation of motion for $r$,
\begin{displaymath} \dot r = \sqrt{2\frac{E - U(r)}{\mu} - \frac{l^2}{\mu^2 r^2}} \,. \end{displaymath} (23)

Note how the angular momentum acts like a contribution to the potential $U(r)$, making the area near $r = 0$ "high potential."

In addition to finding $r$ and $\theta$ as functions of the time $t$, it is also possible to find and equation relating the two,

\begin{displaymath} \frac{dr}{d\theta} = \frac{\dot r}{\dot\theta} = r \sqrt{ 2 \left(E - U(r)\right) \frac{\mu r^2}{l^2} - 1} \,. \end{displaymath} (24)

This equation will be the starting point for our discussion of orbital motion next month.

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