Question

Find the Lagrangian, the generalized momenta, and the Hamiltonian for a free particle (no forces at all) moving in three dimensions. (Use \(x, y, z\) as your generalized coordinates.) Find and solve Hamilton's equations.

Short answer

The Lagrangian is \(\frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)\), the Hamiltonian is \(\frac{1}{2m}(p_x^2 + p_y^2 + p_z^2)\), and the particle moves linearly with constant velocity.

Step-by-step solution

Define the Kinetic Energy

For a free particle moving in three dimensions, the only energy involved is kinetic energy.The kinetic energy is given by:\[T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)\]where \(m\) is the mass of the particle and \(\dot{x}, \dot{y}, \dot{z}\) are the velocities along the \(x, y, z\) axes respectively.

Write the Lagrangian

The Lagrangian \(L\) for a system is defined as the difference between the kinetic and potential energies. Since there is no potential energy for a free particle:\[L = T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)\]

Calculate the Generalized Momenta

The generalized momenta \(p_x, p_y, p_z\) are defined as the partial derivatives of the Lagrangian with respect to the generalized velocities:\[p_x = \frac{\partial L}{\partial \dot{x}} = m\dot{x}\]\[p_y = \frac{\partial L}{\partial \dot{y}} = m\dot{y}\]\[p_z = \frac{\partial L}{\partial \dot{z}} = m\dot{z}\]

Express the Hamiltonian

The Hamiltonian \(H\) is obtained by performing the Legendre transform:\[H = p_x \dot{x} + p_y \dot{y} + p_z \dot{z} - L\]Substituting the expressions for \(p_x, p_y, p_z\) and \(L\):\[H = m\dot{x}\dot{x} + m\dot{y}\dot{y} + m\dot{z}\dot{z} - \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)\]\[H = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)\]Since \(p_x = m\dot{x}, p_y = m\dot{y}, p_z = m\dot{z}\):\[H = \frac{1}{2m}(p_x^2 + p_y^2 + p_z^2)\]

Hamilton's Equations

Hamilton’s equations are given by:\[\dot{q}_i = \frac{\partial H}{\partial p_i}\quad \text{and}\quad \dot{p}_i = -\frac{\partial H}{\partial q_i}\]For our system:\[\dot{x} = \frac{\partial H}{\partial p_x} = \frac{p_x}{m},\quad \dot{y} = \frac{\partial H}{\partial p_y} = \frac{p_y}{m},\quad \dot{z} = \frac{\partial H}{\partial p_z} = \frac{p_z}{m}\]Since \(H\) does not depend on \(x, y, z\), \(\dot{p}_x = \dot{p}_y = \dot{p}_z = 0\), indicating momentum is conserved.

Solve Hamilton's Equations

Given that \(\dot{p}_x = \dot{p}_y = \dot{p}_z = 0\), the momenta \(p_x, p_y, p_z\) are constant. Thus, using \(p_x = m\dot{x}, p_y = m\dot{y}, p_z = m\dot{z}\), we have:\[x(t) = \frac{p_x}{m}t + x_0\]\[y(t) = \frac{p_y}{m}t + y_0\]\[z(t) = \frac{p_z}{m}t + z_0\]where \(x_0, y_0, z_0\) are the initial positions.

Key concepts

Lagrangian mechanics

Lagrangian mechanics is a powerful formulation in classical mechanics. It focuses on the concept of energy rather than forces. The Lagrangian, denoted as \(L\), is defined as the difference between the kinetic energy \(T\) and the potential energy \(V\) of a system:

\[L = T - V\]
For a free particle moving in three dimensions, the potential energy \(V\) is zero because there are no forces acting on the particle. This simplifies the Lagrangian to just the kinetic energy:

\[L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)\]
Here, \(m\) is the mass of the particle, and \(\dot{x}, \dot{y}, \dot{z}\) are the velocities along the \(x, y, z\) axes respectively. Lagrangian mechanics simplifies many problems by reducing them into equations of motion using generalized coordinates, as opposed to the traditional force-based approach. This makes it particularly useful in more complex systems or when analyzing systems in non-Cartesian coordinates.

Generalized coordinates

In classical mechanics, generalized coordinates are variables that describe a system's configuration. They may not always be standard Cartesian coordinates \((x, y, z)\). Instead, generalized coordinates are chosen based on the problem's symmetry or constraints to simplify calculations. For instance, in a spherical pendulum problem, the angles \(\theta\) and \(\phi\) are more appropriate.

The advantage of using generalized coordinates lies in their flexibility. They allow you to choose the most convenient coordinate system for the problem at hand. This makes solving the equations of motion much easier, especially in systems that exhibit complex behaviors or constraints.

In the context of the original problem, the generalized coordinates are simply the Cartesian coordinates \(x, y, z\), representing the position of the free particle in three-dimensional space. This is because there are no constraints or additional symmetry in this scenario that necessitates a different coordinate system.

Conservation of momentum

Conservation of momentum is a fundamental principle in physics. It states that in a closed system with no external forces, the total momentum remains constant over time. In this exercise, we deal with a free particle, meaning there are no external forces acting on it. Thus, the principle of conservation of momentum holds.

When analyzing the system using Hamiltonian mechanics, the conservation laws are revealed through the constancy of the generalized momenta, \(p_x, p_y, p_z\). These quantities are derived from the derivatives of the Lagrangian with respect to the corresponding velocities and remain constant because the Hamiltonian does not depend explicitly on the coordinates:

\[p_x = m\dot{x}, \quad p_y = m\dot{y}, \quad p_z = m\dot{z}\]
Since \(\dot{p}_i = -\frac{\partial H}{\partial q_i} = 0\), the generalized momenta do not change, signifying momentum is conserved.

This is a clear demonstration of the power of Hamiltonian and Lagrangian formulations, where principles like conservation of momentum arise naturally from the mathematics, providing deep insights into the system’s behavior.