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(a) Write down the Lagrangian for a particle moving in three dimensions under the influence of a conservative central force with potential energy \(U(r),\) using spherical polar coordinates \((r, \theta, \phi)\). (b) Write down the three Lagrange equations and explain their significance in terms of radial acceleration, angular momentum, and so forth. (The \(\theta\) equation is the tricky one, since you will find it implies that the \(\phi\) component of \(\ell\) varies with time, which seems to contradict conservation of angular momentum. Remember, however, that \(\ell_{\phi}\) is the component of \(\ell\) in a variable direction.) (c) Suppose that initially the motion is in the equatorial plane (that is, \(\theta_{0}=\pi / 2\) and \(\dot{\theta}_{0}=0\) ). Describe the subsequent motion. (d) Suppose instead that the initial motion is along a line of longitude (that is, \(\dot{\phi}_{0}=0\) ). Describe the subsequent motion.

Short Answer

Expert verified
The Lagrangian describes the system's dynamics; Lagrange's equations express conservation laws and radial processes. The motion continues in the equatorial plane or precesses along a longitude, depending on initial conditions.

Step by step solution

01

Write the Lagrangian

The Lagrangian for a particle moving in three dimensions with position represented in spherical polar coordinates \((r, \theta, \phi)\) is given by the kinetic energy minus the potential energy. The kinetic energy in spherical coordinates is:\[ T = \frac{1}{2} m \left(\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2 \theta \dot{\phi}^2 \right) \]where \(m\) is the mass of the particle and \(\dot{r}, \dot{\theta}, \dot{\phi}\) are the time derivatives of the coordinates. The potential energy due to the conservative central force is \( U(r) \). Therefore, the Lagrangian is:\[ L = T - U = \frac{1}{2} m \left(\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2 \theta \dot{\phi}^2 \right) - U(r) \]
02

Derive Lagrange's Equations

Lagrange's equations are derived from the Lagrangian \( L \), using the Euler-Lagrange equation \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) - \frac{\partial L}{\partial q_i} = 0 \) for each generalized coordinate \( q_i = r, \theta, \phi \):1. **Radial equation \( r \):**\[ \frac{d}{dt}(m \dot{r}) - m r (\dot{\theta}^2 + \sin^2 \theta \dot{\phi}^2) + \frac{\partial U}{\partial r} = 0 \]2. **Polar angle equation \( \theta \):**\[ \frac{d}{dt}(m r^2 \dot{\theta}) - m r^2 \sin \theta \cos \theta \dot{\phi}^2 = 0 \]3. **Azimuthal angle equation \( \phi \):**\[ \frac{d}{dt}(m r^2 \sin^2 \theta \dot{\phi}) = 0 \]These equations describe the radial acceleration, conservation of angular momentum, and variations in angular dynamics.
03

Analyze Motion in the Equatorial Plane

When the motion is in the equatorial plane, \( \theta_0 = \frac{\pi}{2} \) and \( \dot{\theta}_0 = 0 \), implying \( \theta = \frac{\pi}{2} \) remains constant. In this case:- The radial equation determines radial acceleration based on \( \dot{r} \).- The \( \phi \)-equation \[ \frac{d}{dt}(m r^2 \dot{\phi}) = 0 \] shows that the angular momentum around the \( z \)-axis is conserved. The motion remains purely in the \( \frac{\pi}{2} \) plane.
04

Analyze Motion Along Line of Longitude

If the motion initially is along a line of longitude with \( \dot{\phi}_0 = 0 \):- Initially, there is no azimuthal motion (\( \dot{\phi} = 0 \)).- The \( \theta \)-equation requires \( \theta \) to change as the \( \phi \)-component of angular momentum can vary; this leads to a precessional motion.- Radial and \( \theta \) dynamics depend on the potential shape and the initial values of \( r \) and \( \dot{r} \). The system effectively behaves in two coupled dimensions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Radial Acceleration in Spherical Coordinates
In spherical coordinates, radial acceleration is a crucial concept that describes how the speed of a particle changes as it moves towards or away from the center of a potential field. This concept is essential when dealing with central force problems in Lagrangian Mechanics. To grasp radial acceleration, we must start with the radial component of the force acting on the particle.
  • Radial acceleration is essentially the second derivative of the radial position coordinate, often denoted as \( \ddot{r} \).
  • It is derived from the radial component of Lagrange's equations, which for a particle subject to potential energy \( U(r) \) is expressed as:
    \[m \ddot{r} = m r (\dot{\theta}^2 + \sin^2 \theta \dot{\phi}^2) - \frac{\partial U}{\partial r}\] This equation effectively balances the radial acceleration with centripetal contributions from angular motion, noted by \( m r (\dot{\theta}^2 + \sin^2 \theta \dot{\phi}^2) \).
  • Radial acceleration is influenced by how the potential energy changes with radial distance. If the derivative of the potential with respect to \( r \), denoted \( \frac{\partial U}{\partial r} \) is positive, it suggests a restoring force, akin to a spring trying to return to equilibrium.
Understanding these dynamics is key to predicting a particle’s path in a field governed by a central potential.
Exploring Angular Momentum in Lagrangian Mechanics
Angular momentum is a fundamental concept describing rotational motion, and it's deeply linked to the symmetry of circular or spherical systems. In Lagrangian Mechanics, angular momentum appears naturally in the formulation of the equations of motion and provides insights into the conservation laws.
  • In spherical coordinates, angular momentum has components associated with the angles \( \theta \) (polar angle) and \( \phi \) (azimuthal angle).
  • The azimuthal angle component, often more intuitive, is given by:
    \[m r^2 \sin^2 \theta \dot{\phi}\] The Azimuthal equation \( \frac{d}{dt}(m r^2 \sin^2 \theta \dot{\phi}) = 0 \) asserts that this component is conserved, meaning there's no change over time, frequently leading to circular or elliptical paths.
  • Angular momentum in the \( \theta \) component reveals more complexity. It manifests in the equation
    \[m r^2 \ddot{\theta} + 2m r \dot{r} \dot{\theta} - m r^2 \sin \theta \cos \theta \dot{\phi}^2 = 0\] This implies variations especially when initially \( \theta_0 eq \frac{\pi}{2} \) (not in the equatorial plane), where the angle \( \theta \) changes dynamically, showing potential precession or wobbling in motion.
Angular momentum is thereby much more than a static quantity; it helps poviding a map of dynamic rotational movements in constrained motion scenarios.
Navigating Spherical Coordinates in Lagrangian Mechanics
Spherical coordinates are a natural choice when dealing with problems involving central forces and rotational symmetry, such as planets orbiting a star or electrons around a nucleus. Understanding how to work with spherical coordinates in Lagrangian mechanics can simplify complex three-dimensional problems.
  • The coordinates \( (r, \theta, \phi) \) correspond to the radial distance, polar angle, and azimuthal angle, respectively, changing how we perceive movement in space compared to Cartesian coordinates.
  • The kinetic energy in spherical coordinates is a sum of its radial, polar, and azimuthal components:
    \[T = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2 \theta \dot{\phi}^2)\] Each component can independently describe motion influences on the particle, allowing targeted analysis and simplification depending on the initial conditions.
  • Based on how the particle's movement is initiated, different coordinate equations like the one for \( \theta \),\( \phi \), or \( r \), can take precedence, altering the motion's characteristics profoundly.
By breaking down motion into these key components, spherical coordinates help manage the chaos of three-dimensional dynamics, guiding us toward clear solutions in seemingly intractable problems.

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Most popular questions from this chapter

Consider a mass \(m\) moving in a frictionless plane that slopes at an angle \(\alpha\) with the horizontal. Write down the Lagrangian in terms of coordinates \(x,\) measured horizontally across the slope, and \(y\), measured down the slope. (Treat the system as two-dimensional, but include the gravitational potential energy.) Find the two Lagrange equations and show that they are what you should have expected.

Write down the Lagrangian for a projectile (subject to no air resistance) in terms of its Cartesian coordinates \((x, y, z),\) with \(z\) measured vertically upward. Find the three Lagrange equations and show that they are exactly what you would expect for the equations of motion.

The "spherical pendulum" is just a simple pendulum that is free to move in any sideways direction. (By contrast a "simple pendulum"- unqualified - is confined to a single vertical plane.) The bob of a spherical pendulum moves on a sphere, centered on the point of support with radius \(r=R\) the length of the pendulum. A convenient choice of coordinates is spherical polars, \(r, \theta, \phi,\) with the origin at the point of support and the polar axis pointing straight down. The two variables \(\theta\) and \(\phi\) make a good choice of generalized coordinates. (a) Find the Lagrangian and the two Lagrange equations. (b) Explain what the \(\phi\) equation tells us about the \(z\) component of angular momentum \(\ell_{z^{*}}\) (c) For the special case that \(\phi=\) const, describe what the \(\theta\) equation tells us. (d) Use the \(\phi\) equation to replace \(\dot{\phi}\) by \(\ell_{z}\) in the \(\theta\) equation and discuss the existence of an angle \(\theta_{\mathrm{o}}\) at which \(\theta\) can remain constant. Why is this motion called a conical pendulum? (e) Show that if \(\theta=\theta_{0}+\epsilon,\) with \(\epsilon\) small, then \(\theta\) oscillates about \(\theta_{\mathrm{o}}\) in harmonic motion. Describe the motion of the pendulum's bob.

A particle is confined to move on the surface of a circular cone with its axis on the \(z\) axis, vertex at the origin (pointing down), and half-angle \(\alpha\). The particle's position can be specified by two generalized coordinates, which you can choose to be the coordinates \((\rho, \phi)\) of cylindrical polar coordinates. Write down the equations that give the three Cartesian coordinates of the particle in terms of the generalized coordinates ( \(\rho, \phi\) ) and vice versa.

A pendulum is made from a massless spring (force constant \(k\) and unstretched length \(l_{\mathrm{o}}\)) that is suspended at one end from a fixed pivot \(O\) and has a mass \(m\) attached to its other end. The spring can stretch and compress but cannot bend, and the whole system is confined to a single vertical plane. (a) Write down the Lagrangian for the pendulum, using as generalized coordinates the usual angle \(\phi\) and the length \(r\) of the spring. (b) Find the two Lagrange equations of the system and interpret them in terms of Newton's second law, as given in Equation (1.48). (c) The equations of part (b) cannot be solved analytically in general. However, they can be solved for small oscillations. Do this and describe the motion. [Hint: Let \(l\) denote the equilibrium length of the spring with the mass hanging from it and write \(r=l+\epsilon .\) "Small oscillations" involve only small values of \(\epsilon\) and \(\phi,\) so you can use the small-angle approximations and drop from your equations all terms that involve powers of \(\epsilon\) or \(\phi\) (or their derivatives) higher than the first power (also products of \(\epsilon\) and \(\phi\) or their derivatives). This dramatically simplifies and uncouples the equations.]

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