Question

Set up the Hamiltonian and Hamilton's equations for a projectile of mass \(m\), moving in a vertical plane and subject to gravity but no air resistance. Use as your coordinates \(x\) measured horizontally and \(y\) measured vertically up. Comment on each of the four equations of motion.

Short answer

The Hamiltonian is \( H = \frac{p_x^2}{2m} + \frac{p_y^2}{2m} + mgy \), and Hamilton's equations determine motion as these.

Step-by-step solution

Define the system

The system is a projectile moving in a vertical plane with no air resistance and subject to gravity. The horizontal position is given by the coordinate \(x\) and the vertical position by \(y\). The mass of the projectile is \(m\).

Write the Lagrangian

The Lagrangian \( \mathcal{L} \) is given by the difference between the kinetic energy \(T\) and potential energy \(V\). The kinetic energy is \( T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) \), and the potential energy is \( V = mgy \). Thus, the Lagrangian is:\[ \mathcal{L} = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) - mgy \]

Obtain the generalized momenta

The generalized momenta \( p_x \) and \( p_y \) are obtained as derivatives of the Lagrangian with respect to the velocities \( \dot{x} \) and \( \dot{y} \) respectively:\[ p_x = \frac{\partial \mathcal{L}}{\partial \dot{x}} = m\dot{x} \]\[ p_y = \frac{\partial \mathcal{L}}{\partial \dot{y}} = m\dot{y} \]

Set up the Hamiltonian

The Hamiltonian \( H \) is given by:\[ H = p_x\dot{x} + p_y\dot{y} - \mathcal{L} \]Substituting \( \dot{x} = \frac{p_x}{m} \), \( \dot{y} = \frac{p_y}{m} \) into the expression and simplifying, we find:\[ H = \frac{p_x^2}{2m} + \frac{p_y^2}{2m} + mgy \]

Derive Hamilton's Equations

Hamilton's equations are given by:1. \( \dot{x} = \frac{\partial H}{\partial p_x} = \frac{p_x}{m} \)2. \( \dot{y} = \frac{\partial H}{\partial p_y} = \frac{p_y}{m} \)3. \( \dot{p}_x = -\frac{\partial H}{\partial x} = 0 \)4. \( \dot{p}_y = -\frac{\partial H}{\partial y} = -mg \)

Comment on the Equations of Motion

1. \( \dot{x} = \frac{p_x}{m} \): This equation relates horizontal velocity to horizontal momentum.2. \( \dot{y} = \frac{p_y}{m} \): This equation relates vertical velocity to vertical momentum.3. \( \dot{p}_x = 0 \): The horizontal momentum is conserved, as there are no forces acting in the horizontal direction.4. \( \dot{p}_y = -mg \): The vertical momentum changes due to the gravitational force acting downward.

Key concepts

Hamilton's equations

Hamilton's equations form a fundamental part of Hamiltonian mechanics, providing a way to describe the evolution of a physical system over time. When we set up Hamilton's equations for a projectile subject to gravity in a vertical plane (as in our exercise), we essentially look at how certain quantities related to the system, like position and momentum, change. Hamilton's equations are:

• \( \dot{x} = \frac{\partial H}{\partial p_x} \)
• \( \dot{y} = \frac{\partial H}{\partial p_y} \)
• \( \dot{p}_x = - \frac{\partial H}{\partial x} \)
• \( \dot{p}_y = - \frac{\partial H}{\partial y} \)

Here, \( \dot{x} \) and \( \dot{y} \) relate the rate of change of position to momentum, while \( \dot{p}_x \) and \( \dot{p}_y \) describe the change in momentum due to forces. In the absence of horizontal forces, \( \dot{p}_x = 0 \) meaning horizontal momentum is conserved. On the other hand, \( \dot{p}_y = -mg \) shows a decrease of vertical momentum due to gravity. This structure elegantly decouples the equations, showcasing conservation laws and symmetries inherent in the system.

Lagrangian mechanics

Lagrangian mechanics provides an alternative formulation of classical mechanics, focusing on energy rather than forces. It's rooted in the principle of least action, which suggests that the path taken by a system is the one for which the action integral is stationary. In simpler terms, a projectile chooses the path where its energy difference, described by the Lagrangian \( \mathcal{L} \), balances kinetic and potential energies in the system: \[ \mathcal{L} = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) - mgy \] Here, the kinetic energy \( T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) \) accounts for motion while the potential energy \( V = mgy \) relates to the height. This difference is the essence of the Lagrangian \( \mathcal{L} \). From \( \mathcal{L} \), we compute generalized momenta by differentiating with respect to velocities \( \dot{x} \) and \( \dot{y} \), laying the groundwork for transitioning to Hamiltonian mechanics, where momentum plays a central role.

Projectile motion

Understanding projectile motion is crucial when considering a projectile's path through a gravitational field. In an ideal scenario devoid of air resistance, the only force acting is gravity, affecting the motion vertically. Horizontally, the projectile moves at a constant velocity due to no additional forces acting along that axis. Projectile motion can be modeled using the equations of motion derived through Lagrangian and Hamiltonian approaches. For example, with our determined system: - The horizontal coordinate \( x \) follows a uniform motion with the momentum \( p_x = m\dot{x} \) remaining constant. - Vertically, \( y \) describes a uniformly accelerated motion due to gravity, with \( p_y = m\dot{y} \) decreasing over time since \( \dot{p}_y = -mg \). This reveals that projectiles travel in parabolic trajectories, woven by fundamental laws of physics encapsulated seamlessly within Hamiltonian mechanics. These insights enrich our understanding of the natural world, further illustrating the beauty and coherence of classical mechanics principles.