Question

A particle of unit mass is subject to an inverse-square-law central force $$ \boldsymbol{F}=-\frac{\boldsymbol{r}}{r^{3}} $$ where \(r=|\boldsymbol{r}|\) and \(\boldsymbol{r}\) is the position vector from the origin of an inertial frame. Show that the motion is governed by the Lagrangian $$ L=\frac{1}{2} \dot{r} \cdot \dot{r}+\frac{1}{r} $$ Write down the equations of motion in spherical polar coordinates and show that there are solutions with \(\theta=\pi / 2\) throughout the motion.

Short answer

In conclusion, we showed that the given Lagrangian describes the motion of a particle subject to an inverse-square law force correctly. We also found the equation of motion in spherical polar coordinates and showed that there exist solutions with \(\theta = \frac{\pi}{2}\) throughout the motion. This result indicates that the particle's trajectory is restricted to a plane with constant \(\theta\), satisfying the given condition.

Step-by-step solution

Lagrangian Formalism Basics

In classical mechanics, a system's motion can be described by the Lagrangian \(L\) which is defined as the difference between the kinetic (\(T\)) and potential (\(V\)) energies: $$ L = T - V $$ To find the equations of motion, we need to apply the Euler-Lagrange equation: $$ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right) - \frac{\partial L}{\partial q_i} = 0 $$ where \(q_i\) denotes the generalized coordinates and dots represent derivatives with respect to time.

Writing the Lagrangian for our problem

The problem is described by a central force \(\boldsymbol{F}\). Since a central force has no component along the angular directions, we can see that the potential energy \(V\) is only a function of the radial coordinate \(r\). Therefore, the Lagrangian for this problem can be separated into radial and angular parts. The radial part depends on \(\dot{r}\) and the potential energy, while the angular part depends on \(\dot{\theta}\) and \(\dot{\phi}\). Let's find the potential energy \(V\) from the given force \(\boldsymbol{F}\). We have: $$ \boldsymbol{F} = -\nabla V $$ Since \(\boldsymbol{F}\) only depends on the radial coordinate \(-\frac{\boldsymbol{r}}{r^{3}}\), we can write: $$ V = -\int \frac{1}{r^2} dr = \frac{1}{r} + C $$ where \(C\) is an integration constant. Since we are only interested in relative potentials, we can set \(C=0\). Now, let's write the Lagrangian as: $$ L = T - V = \frac{1}{2}\left(\dot{r}^2 + r^2\dot{\theta}^2 + r^2\sin^2\theta\dot{\phi}^2\right) - \frac{1}{r} $$ The given Lagrangian includes only radial components, so for our case, it simplifies to: $$ L = \frac{1}{2}\dot{r}^2 + \frac{1}{r} $$

Equations of motion in spherical polar coordinates

Now, we need to derive the equations of motion from our Lagrangian. Since our Lagrangian is time-independent, and for simplicity, let's use only one Euler-Lagrange equation for \(r\): $$ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{r}}\right) - \frac{\partial L}{\partial r} = 0 $$ Calculating the relevant derivatives: $$ \frac{\partial L}{\partial r} = -\frac{1}{r^2}, \quad \frac{\partial L}{\partial \dot{r}} = \dot{r} $$ and taking their time derivative: $$ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{r}}\right) = \ddot{r} $$ Plug these derivatives into the Euler-Lagrange equation and we get: $$ \ddot{r} = \frac{1}{r^2} $$ This differential equation describes the motion of the particle in spherical polar coordinates.

Solutions with \(\theta = \frac{\pi}{2}\) throughout the motion

Since the force \(\boldsymbol{F}\) is central, the angular momentum of the particle is conserved. In spherical polar coordinates, the angular momentum of a particle with unit mass is given by $$ \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{\dot{r}} = r^2\dot{\theta}\boldsymbol{e}_\phi - r^2\sin\theta\dot{\phi}\boldsymbol{e}_\theta $$ For a trajectory that lies in the plane \(\theta = \frac{\pi}{2}\), we have that the components of the angular momentum are constant throughout the motion, $$ L_\theta = -r^2\sin\theta\dot{\phi}=0 \Rightarrow \dot{\phi} = 0 \text{ or } \sin\theta = 0 $$ Since the problem admits non-trivial solutions where the particle moves along the trajectory, we must have \(\sin\theta = 0\). Thus, the solutions for which the condition \(\theta = \frac{\pi}{2}\) holds throughout the motion can be found.

Key concepts

Lagrangian Mechanics

Considered the cornerstone of analytical dynamics, Lagrangian mechanics offers a powerful framework for analyzing the motion of particles and systems. At its heart, it involves the concept of the Lagrangian, denoted as \( L \), which is the difference between the kinetic energy (\( T \)) and potential energy (\( V \)) of the system. Mathematically, \( L = T - V \).

Unlike Newtonian mechanics, which focuses on forces and accelerations, Lagrangian mechanics utilizes the principle of least action. This principle states that the true path taken by a system is the one that minimizes the action, which is the integral of the Lagrangian over time. To determine the equations that govern a system's dynamics, one applies the Euler-Lagrange equations to the Lagrangian. These equations are highly versatile, allowing for the analysis of complex systems and the inclusion of constraints in a straightforward manner. Moreover, the Lagrangian formulation is particularly handy when dealing with coordinates that are more complex than the standard Cartesian ones, such as spherical polar coordinates.

Equations of Motion

The equations of motion are the mathematical descriptions that characterize the behavior of a physical system. In the context of Lagrangian mechanics, these equations are derived from the Lagrangian using the Euler-Lagrange equation. The generic Euler-Lagrange equation is \[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right) - \frac{\partial L}{\partial q_i} = 0 \] where \( \dot{q}_i \) represents the time derivative of generalized coordinates \( q_i \).

In practice, these equations encapsulate all the information about a system's dynamics, such as its potential and kinetic energies. They show how the system evolves over time, responding to various forces, and they can accommodate both conserved quantities, like energy or momentum, and dissipative forces, such as friction. The beauty of using these equations is that they can be applied in any coordinate system that best suits the problem, including spherical polar coordinates.

Inverse-square-law Force

One of the fundamental forces in physics, an inverse-square-law force is characterized by its intensity diminishing as a function of the square of the distance from the source of the force. A key example is the gravitational force, described by Newton's law of universal gravitation.

In the context of our problem, the inverse-square-law force is represented by \( \boldsymbol{F} = -\frac{\boldsymbol{r}}{r^3} \), where \( \boldsymbol{r} \) is the position vector, and \( r = |\boldsymbol{r}| \) is the scalar distance. This force implies a potential energy that varies as \( V = \frac{1}{r} \), a characteristic feature of central forces that govern the motion of celestial bodies, as well as dynamics in atomistic and subatomic scales. These forces are central to understanding orbital motions and play a pivotal role in the conservation of angular momentum.

Spherical Polar Coordinates

Spherical polar coordinates are an ideal choice for problems with symmetry around a point, such as those involving central forces. This coordinate system is defined by three parameters: radial distance \( r \), polar angle \( \theta \), and azimuthal angle \( \phi \).

In these coordinates, the position of a particle is specified by the distance from the origin (\( r \)), the angle from the positive z-axis (\( \theta \)), and the angle from the positive x-axis in the xy-plane (\( \phi \)). These coordinates have the advantage of simplifying calculations involving spherical objects or motion around a sphere. Appropriately, using spherical polar coordinates to describe motion affected by an inverse-square-law force allows for a more natural and thus simplified mathematical representation of the system's dynamics.