Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Here is a simple example of a canonical transformation that illustrates how the Hamiltonian formalism lets one mix up the \(q^{\prime}\) s and the \(p\) 's. Consider a system with one degree of freedom and Hamiltonian \(\mathcal{H}=\mathcal{H}(q, p)\). The equations of motion are, of course, the usual Hamiltonian equations \(\dot{q}=\partial \mathcal{H} / \partial p\) and \(\dot{p}=-\partial \mathcal{H} / \partial q .\) Now consider new coordinates in phase space defined as \(Q=p\) and \(P=-q .\) Show that the equations of motion for the new coordinates \(Q\) and \(P\) are \(\dot{Q}=\partial \mathcal{H} / \partial P\) and \(\dot{P}=-\partial \mathscr{H} / \partial Q ;\) that is, the Hamiltonian formalism applies equally to the new choice of coordinates where we have exchanged the roles of position and momentum.

Short Answer

Expert verified
The transformation is canonical, preserving the Hamiltonian structure.

Step by step solution

01

Identify Original Equations

For the original system with coordinates \(q\) and \(p\), the equations of motion are determined by the Hamiltonian \(\mathcal{H}\): \[ \dot{q} = \frac{\partial \mathcal{H}}{\partial p} \] and \[ \dot{p} = -\frac{\partial \mathcal{H}}{\partial q}. \] These are the standard Hamiltonian equations describing evolution in the original coordinates.
02

Define New Coordinates

We introduce a new set of coordinates \(Q = p\) and \(P = -q\). This means that in the new coordinate system, the position \(Q\) is actually the momentum \(p\) of the original system, and \(P\) is the negative of the original position \(q\). The role of \(q\) and \(p\) has been interchanged in terms of the assignment of new coordinates \(Q\) and \(P\).
03

Determine New Equations of Motion

Using the definition of the new coordinates, substitute into the original equations of motion. We have \(q = -P\) and \(p = Q\). The equations become: \[ \dot{Q} = \dot{p} = -\frac{\partial \mathcal{H}}{\partial (-P)} = \frac{\partial \mathcal{H}}{\partial P}, \] and \[ \dot{P} = -\dot{q} = -\frac{\partial \mathcal{H}}{\partial Q}. \] These equations follow the same form as the Hamiltonian equations when interpreted with the new variables.
04

Verify Hamiltonian Structure

In the new coordinate system, the transformed equations \(\dot{Q} = \frac{\partial \mathcal{H}}{\partial P}\) and \(\dot{P} = -\frac{\partial \mathcal{H}}{\partial Q}\) have the same structure as the original Hamilton's equations. This confirms that the Hamiltonian formalism is preserved under the change of variables, demonstrating the canonical nature of the transformation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Hamiltonian Mechanics
Hamiltonian Mechanics is a powerful framework used in classical mechanics to describe the evolution of a physical system. It provides a different perspective compared to Newtonian mechanics by focusing on energy rather than force. The core idea revolves around the Hamiltonian function, denoted as \( \mathcal{H}(q, p) \), which represents the total energy of the system, combining both kinetic and potential energy.
In Hamiltonian mechanics:
  • We exploit the position \(q\) and momentum \(p\) of the system as variables.
  • The equations of motion are derived based on partial derivatives of the Hamiltonian function.
These equations of motion are expressed as:\[\dot{q} = \frac{\partial \mathcal{H}}{\partial p}, \quad \dot{p} = -\frac{\partial \mathcal{H}}{\partial q}. \]The beauty of Hamiltonian mechanics lies in its ability to simplify complex problems, especially when dealing with higher dimensions and when transformations, such as canonical transformations, are involved. This approach lays the foundation for more advanced theories like Quantum Mechanics, making it an essential concept in the study of dynamics.
Phase Space
Phase Space is an essential concept in Hamiltonian mechanics and a powerful tool used to visualize and analyze the state of a mechanical system. Phase space is a multidimensional space where each state of the system is represented as a point.
  • Each point in phase space corresponds to a unique combination of position \(q\) and momentum \(p\).
  • The evolution of the system is depicted by the trajectory of these points as time progresses.
Phase space provides a complete description of the system's dynamics, as knowing both the position and momentum of each degree of freedom equips us with the information needed to determine its future and past states.
In practical applications, analyzing phase space helps in:
  • Understanding the behavior and stability of a system.
  • Identifying conserved quantities and symmetries.
  • Facilitating the analysis of complex systems where other methods would be cumbersome.
Phase space is also integral to statistical mechanics where it aids in describing systems with a large number of particles, forming the backbone for concepts like the Gibbs ensemble.
Equations of Motion
In the context of Hamiltonian mechanics, the Equations of Motion are fundamental as they describe how the state of a physical system evolves over time. These are expressed as Hamilton's Equations, which link the change in system coordinates to its Hamiltonian. They play a crucial role in understanding the dynamics of the system in phase space.
The standard form of Hamilton's Equations is:\[\begin{align*}\dot{q} &= \frac{\partial \mathcal{H}}{\partial p}, \\dot{p} &= -\frac{\partial \mathcal{H}}{\partial q}.\end{align*}\]These equations reveal how the position \(q\) and momentum \(p\) are influenced by the Hamiltonian function:
  • \( \dot{q} \) represents the rate at which position changes, determined by the partial derivative of the Hamiltonian with respect to momentum.
  • \( \dot{p} \) indicates the rate of change in momentum, corresponding to the negative derivative of the Hamiltonian with respect to position.
Understanding these equations provide insights into the conservation of energy and symplectic structure—key components in classical mechanics. Moreover, when systems undergo transformations, such as the canonical transformations addressed in the exercise, Hamilton's Equations ensure that the integrity and dynamics of the system remain intact.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the Lagrangian formalism, a coordinate \(q_{i}\) is ignorable if \(\partial \mathcal{L} / \partial q_{i}=0 ;\) that is, if \(\mathcal{L}\) is independent of \(q_{i}\). This guarantees that the momentum \(p_{i}\) is constant. In the Hamiltonian approach, we say that \(q_{i}\) is ignorable if \(\mathcal{H}\) is independent of \(q_{i}\), and this too guarantees \(p_{i}\) is constant. These two conditions must be the same, since the result " \(p_{i}=\) const" is the same either way. Prove directly that this is so, as follows: (a) For a system with one degree of freedom, prove that \(\partial \mathcal{H} / \partial q=-\partial \mathcal{L} / \partial q\) starting from the expression (13.14) for the Hamiltonian. This establishes that \(\partial \mathcal{H} / \partial q=0\) if and only if \(\partial \mathcal{L} / \partial q=0 .\) (b) For a system with \(n\) degrees of freedom, prove that \(\partial \mathcal{H} / \partial q_{i}=-\partial \mathcal{L} / \partial q_{i}\) starting from the expression (13.24).

13.13 \star\star Consider a particle of mass \(m\) constrained to move on a frictionless cylinder of radius \(R\), given by the equation \(\rho=R\) in cylindrical polar coordinates \((\rho, \phi, z) .\) The mass is subject to just one external force, \(\mathbf{F}=-k r \hat{\mathbf{r}},\) where \(k\) is a positive constant, \(r\) is its distance from the origin, and \(\hat{\mathbf{r}}\) is the unit vector pointing away from the origin, as usual. Using \(z\) and \(\phi\) as generalized coordinates, find the Hamiltonian \(\mathcal{H}\). Write down and solve Hamilton's equations and describe the motion.

Consider a mass \(m\) constrained to move in a vertical line under the influence of gravity. Using the coordinate \(x\) measured vertically down from a convenient origin \(O\), write down the Lagrangian \(\mathcal{L}\) and find the generalized momentum \(p=\partial \mathcal{L} / \partial \dot{x}\). Find the Hamiltonian \(\mathcal{H}\) as a function of \(x\) and \(p\) and write down Hamilton's equations of motion. (It is too much to hope with a system this simple that you would learn anything new by using the Hamiltonian approach, but do check that the equations of motion make sense.)

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.

The simple form \(\mathcal{H}=T+U\) is true only if your generalized coordinates are "natural" (relation betweeen generalized and underlying Cartesian coordinates is independent of time). If the generalized coordinates are not "natural," you must use the definition \(\mathcal{H}=\sum p_{i} \dot{q}_{i}-\mathcal{L} .\) To illustrate this point, consider the following: Two children are playing catch inside a railroad car that is moving with varying speed \(V\) along a straight horizontal track. For generalized coordinates you can use the position \((x, y, z)\) of the ball relative to a point fixed in the car, but in setting up the Hamiltonian you must use coordinates in an inertial frame \(-\) a frame fixed to the ground. Find the Hamiltonian for the ball and show that it is not equal to \(T+U\) (neither as measured in the car, nor as measured in the ground-based frame).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free