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

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).

Short Answer

Expert verified
The Hamiltonian includes terms with velocity \(V\) and is not simply \(T + U\).

Step by step solution

01

Define the inertial frame coordinates

Consider an inertial frame fixed to the ground where the position of the ball is given by \[(X, Y, Z) = (x + Vt, y, z)\] where \(Vt\) is the displacement of the car due to its varying velocity \(V(t)\).
02

Define the generalized coordinates

The generalized coordinates relative to the moving car are \((x, y, z)\), where the potential energy \(U\) is simply a function of \(z\), such as \(U = mgz\) due to gravity.
03

Calculate the velocities in the inertial frame

The velocity components in the inertial frame are \[ \dot{X} = \dot{x} + V, \dot{Y} = \dot{y}, \dot{Z} = \dot{z} \]These include the effect of the moving train.
04

Express the Lagrangian in the inertial frame

The Lagrangian \(\mathcal{L}\) for the system is given by \[ \mathcal{L} = T - U = \frac{1}{2}m(\dot{X}^2 + \dot{Y}^2 + \dot{Z}^2) - mgz \] Substitute \(\dot{X}, \dot{Y}, \dot{Z}\) to get \[ \mathcal{L} = \frac{1}{2}m((\dot{x} + V)^2 + \dot{y}^2 + \dot{z}^2) - mgz \]
05

Calculate the conjugate momenta

Calculate the momentum \(p_x\) by differentiating the Lagrangian: \[ p_x = \frac{\partial \mathcal{L}}{\partial \dot{x}} = m(\dot{x} + V) \] Similarly, compute \(p_ye=g \dot{y}\) and \(p_z = m\dot{z}\).
06

Write the Hamiltonian using conjugate momenta

The Hamiltonian \(\mathcal{H}\) is defined as \[ \mathcal{H} = \sum p_i \dot{q}_i - \mathcal{L} \] Substitute the generalized coordinates and momenta used: \[ \mathcal{H} = p_x \dot{x} + p_y \dot{y} + p_z \dot{z} - \mathcal{L} \]
07

Express the Hamiltonian explicitly

Substitute \(\dot{x} = \frac{p_x}{m} - V\), \(\dot{y} = \frac{p_y}{m}\), \(\dot{z} = \frac{p_z}{m}\) into the equation for \(\mathcal{H}\) and solve:\[ \mathcal{H} = \frac{m}{2}((\frac{p_x}{m} - V)^2 + (\frac{p_y}{m})^2 + (\frac{p_z}{m})^2) + mgz \]
08

Verify the Hamiltonian is not T + U

Note that the Hamiltonian includes the terms involving \(V\), indicating it cannot be expressed simply as a sum of kinetic and potential energies \(T + U\), due to the velocity \(V(t)\) of the moving frame being time-dependent. Hence, \(\mathcal{H} eq T + U\) in either frame.

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.

Generalized Coordinates
Generalized coordinates are a powerful tool in classical mechanics that allow you to describe a system using the most convenient set of variables. For a ball moving inside a railroad car, the natural choice might be to use the position coordinates relative to the car itself, such as
  • The horizontal position along the length of the car, represented by \(x\).
  • The vertical position upwards from the floor of the car, represented by \(y\).
  • The height above the floor, represented by \(z\).
Generalized coordinates are favored because they simplify the equations of motion. However, when the relationship between these coordinates and the standard Cartesian coordinates depends on time, they are not deemed "natural." In such cases, you need to take into account extra terms when calculating quantities like the Hamiltonian of a system. This consideration becomes crucial when transitioning between different frames of reference, such as a moving train and the ground from which it is observed.
Inertial Frame
An inertial frame of reference is one in which Newton's laws of motion are valid. These frames move at constant velocities and are not accelerating. For example, consider an observer on the ground watching a train pass by. This observer is in an inertial frame because the ground is considered stationary. While analyzing the motion of a ball inside a moving train, it is important to translate the observations to an inertial frame to properly define physical quantities like
  • Velocity: In the inertial frame, total velocity includes the pace of the train in addition to the ball's relative movement inside the train.
  • Position: Incorporates both the ball’s position within the train and the distance the train has moved over time.
By computing them in the inertial frame such as the ground, we ensure our observations and subsequent calculations hold true universally according to classical mechanics principles.
Conjugate Momenta
Conjugate momenta provide insight into the dynamics of a system beyond mere velocities. They are derived from the Lagrangian of a system and are essential components in forming the Hamiltonian mechanics framework. For a particle moving along a path defined by generalized coordinates \(q_i\), the conjugate momentum \(p_i\) is calculated by taking the partial derivative of the Lagrangian \(\mathcal{L}\) with respect to the corresponding generalized velocity \(\dot{q}_i\). For example,
  • The momentum \(p_x = m(\dot{x} + V)\) embodies not only the movement of the ball along the train (\(\dot{x}\)) but also the train’s velocity (\(V\)).
  • Similarly, \(p_y = m \dot{y}\) and \(p_z = m \dot{z}\) define linear elastic momenta in other directions.
Hence, unlike plain velocity considerations, the conjugate momentum comprehensively includes all motivating factors, including movement of the coordinate system itself.
Lagrangian
The Lagrangian \(\mathcal{L}\) is a central concept in analytical mechanics, representing the difference between kinetic energy \(T\) and potential energy \(U\) of the system i.e., \(\mathcal{L} = T - U\). Within an inertial frame, it efficiently describes the dynamics of the system by forming a bridge to derive both the equations of motion and the Hamiltonian.Consider a ball moving within a train; the kinetic energy \(T\) in this scenario must account for combined speeds due to both the moving train and the ball’s movement within it:\[T = \frac{1}{2}m((\dot{x} + V)^2 + \dot{y}^2 + \dot{z}^2)\]Similarly, the gravitational potential energy for a ball at height \(z\) is \(U = mgz\).The resulting Lagrangian allows deriving dynamic properties like
  • Equations of motion: Using the Euler-Lagrange equation based on \(\mathcal{L}\).
  • Hamiltonian: As a transition to formulating Hamiltonian mechanics, which provides a different yet richly informative outlook on the system.
Overall, the Lagrangian provides a simplified yet exhaustive way to characterize physical systems and analyze their behaviors in a variety of conditions.

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

Evaluate the three-dimensional divergence \(\nabla \cdot\) v for each of the following vectors: (a) \(\mathbf{v}=k \mathbf{r}\) (b) \(\mathbf{v}=k(z, x, y),(\mathbf{c}) \mathbf{v}=k(z, y, x),(\mathbf{d}) \mathbf{v}=k(x, y,-2 z),\) where \(\mathbf{r}=(x, y, z)\) is the usual position vector and \(k\) is a constant.

A bead of mass \(m\) is threaded on a frictionless wire that is bent into a helix with cylindrical polar coordinates \((\rho, \phi, z)\) satisfying \(z=c \phi\) and \(\rho=R,\) with \(c\) and \(R\) constants. The \(z\) axis points vertically up and gravity vertically down. Using \(\phi\) as your generalized coordinate, write down the kinetic and potential energies, and hence the Hamiltonian \(\mathcal{H}\) as a function of \(\phi\) and its conjugate momentum \(p\) Write down Hamilton's equations and solve for \(\ddot{\phi}\) and hence \(\ddot{z}\). Explain your result in terms of Newtonian mechanics and discuss the special case that \(R=0\).

Consider a particle of mass \(m\) moving in two dimensions, subject to a force \(\mathbf{F}=-k x \hat{\mathbf{x}}+K \hat{\mathbf{y}}\) where \(k\) and \(K\) are positive constants. Write down the Hamiltonian and Hamilton's equations, using \(x\) and \(y\) as generalized coordinates. Solve the latter and describe the motion.

(a) Evaluate \(\left.\nabla \cdot \mathbf{v} \text { for } \mathbf{v}=k \hat{\mathbf{r}} / r^{2} \text { using rectangular coordinates. (Note that } \hat{\mathbf{r}} / r^{2}=\mathbf{r} / r^{3} .\right)\) (b) Inside the back cover, you will find expressions for the various vector operators (divergence, gradient, etc.) in polar coordinates. Use the expression for the divergence in spherical polar coordinates to confirm your answer to part (a). (Take \(r \neq 0\).)

A roller coaster of mass \(m\) moves along a frictionless track that lies in the \(x y\) plane \((x\) horizontal and \(y\) vertically up). The height of the track above the ground is given by \(y=h(x) .\) (a) Using \(x\) as your generalized coordinate, write down the Lagrangian, the generalized momentum \(p,\) and the Hamiltonian \(\mathcal{H}=p \dot{x}-\mathcal{L}\) (as a function of \(x\) and \(p\) ). (b) Find Hamilton's equations and show that they agree with what you would get from the Newtonian approach. [Hint: You know from Section 4.7 that Newton's second law takes the form \(F_{\text {tang }}=m \ddot{s},\) where \(s\) is the distance measured along the track. Rewrite this as an equation for \(\ddot{x}\) and show that you get the same result from Hamilton's equations.]

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