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Find the Hamiltonian \(\mathscr{H}\) for a mass \(m\) confined to the \(x\) axis and subject to a force \(F_{x}=-k x^{3}\) where \(k>0 .\) Sketch and describe the phase-space orbits.

Short Answer

Expert verified
The Hamiltonian is \( \mathscr{H} = \frac{p^2}{2m} - \frac{k}{4}x^4 \). Phase-space orbits are closed loops around the origin.

Step by step solution

01

Understand the Hamiltonian

The Hamiltonian represents the total energy of the system and is typically given by \( \mathscr{H} = T + V \), where \( T \) is the kinetic energy and \( V \) is the potential energy. For a mass \( m \) moving along the x-axis, the kinetic energy \( T \) is \( \frac{p^2}{2m} \), where \( p \) is the momentum. The potential energy \( V \) due to a force \( F_x = -kx^3 \) can be found by integrating the force with respect to \( x \).
02

Compute the Potential Energy

Integrate the force \( F_x = -kx^3 \) with respect to \( x \) to find the potential energy \( V(x) \):\[ V(x) = \int -kx^3 \, dx = -\frac{k}{4}x^4 + C \]Assuming the integration constant \( C = 0 \) for simplicity, we have:\[ V(x) = -\frac{k}{4}x^4 \].
03

Write the Hamiltonian Expression

Combine the kinetic and potential energy expressions to formulate the Hamiltonian:\[ \mathscr{H}(x,p) = \frac{p^2}{2m} - \frac{k}{4}x^4 \].This expression represents the energy of the system in terms of the position \( x \) and momentum \( p \).
04

Analyze the Phase Space

Phase space is a graphical representation of all possible states of a system with position and momentum as coordinates. In this case, \( x \) is the position and \( p \) is the momentum. The phase space orbits can be visualized by holding the Hamiltonian constant and plotting \( p \) versus \( x \).
05

Sketch the Phase-Space Orbits

For a fixed energy level, the phase space orbits correspond to curves derived from the equation:\[ \frac{p^2}{2m} - \frac{k}{4}x^4 = E \] Rearranging gives:\[ p^2 = 2m \left(E + \frac{k}{4}x^4 \right) \].Plotting \( p \) versus \( x \) results in closed loops, resembling orbits around the origin, demonstrating how the state of the system evolves over time.

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Key Concepts

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

Potential Energy
Potential energy is a key concept in understanding how forces influence movement and interaction. It is stored energy that has the potential to do work, depending only on the position in space rather than the specific trajectory taken to get there.
Imagine potential energy like a stretched rubber band that stores energy because of its shape or position. When released, the stored energy causes movement.In the context of our exercise, we consider a particle experiencing a force in one dimension according to the equation \( F_x = -kx^3 \). The negative sign indicates this is a restoring force, pulling back towards equilibrium. To find the potential energy \( V(x) \), we integrate the force across distance:
  • Potential Energy Formula: \( V(x) = -\frac{k}{4}x^4 \)
This function shows that potential energy depends on \( x^4 \). As \( x \) increases or decreases from zero, the potential energy becomes more negative, indicating that the system can release this energy to perform work.
Kinetic Energy
Kinetic energy is the energy of motion. Think about objects rolling, flying, or simply moving over time - they all possess kinetic energy. It's proportional to mass and the square of velocity, showcasing how both mass and speed influence the energy a system can have.
For a particle confined along the x-axis, its kinetic energy can be described using momentum \( p \), where momentum links speed with mass. The formula for kinetic energy \( T \) becomes:
  • Kinetic Energy Formula: \( T = \frac{p^2}{2m} \)
Here, \( p^2/2m \) signifies that kinetic energy increases with the square of momentum. This relationship is crucial because it allows us to express energy in terms of variables (momentum and position) that align with Hamiltonian mechanics.Combining kinetic and potential energies helps us understand the full dynamics of a system, calculated through the Hamiltonian. Both energies are parts of the total energy a system possesses, illustrating how an object can convert from potential to kinetic states and vice versa, influencing its present and future states.
Phase Space
Phase space is a fundamental concept in Hamiltonian mechanics, serving as an abstract representation of all possible states of a system. Each point in phase space is defined by a set of coordinates, typically position \( x \) and momentum \( p \) here, showing the system's state at any given time.
Imagine phase space as a landscape of mountains and valleys - where the position marks the landscape points and momentum represents movement across this landscape.For mechanical systems, like a particle on the x-axis, analyzing phase space involves plotting \( p \) against \( x \). By keeping the Hamiltonian \( \mathscr{H}(x, p) \) constant, we visualize phase space orbits:
  • These orbits reflect system states over time related to specific energy levels.
  • Creating curves or loops shows the dynamic trajectory, akin to tracing a journey through this abstract landscape.
From our exercise, orbits emerge from the energy equation \( \frac{p^2}{2m} - \frac{k}{4}x^4 = E \), showing closed loops that describe how a system evolves with time. These phase space orbits help us understand not only the state transitions but also stability and periodicity in the system behaviors.

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Most popular questions from this chapter

Consider a mass \(m\) confined to the \(x\) axis and subject to a force \(F_{x}=k x\) where \(k>0 .\) (a) Write down and sketch the potential energy \(U(x)\) and describe the possible motions of the mass. (Distinguish between the cases that \(E>0 \text { and } E<0 .)\) (b) Write down the Hamiltonian \(\mathcal{H}(x, p)\), and describe the possible phase-space orbits for the two cases \(E>0\) and \(E<0\). (Remember that the function \(\mathcal{H}(x, p)\) must equal the constant energy \(E .\) ) Explain your answers to part (b) in terms of those to part (a).

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.

(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 beam of particles is moving along an accelerator pipe in the \(z\) direction. The particles are uniformly distributed in a cylindrical volume of length \(L_{\mathrm{o}}\) (in the \(z\) direction) and radius \(R_{\mathrm{o}} .\) The particles have momenta uniformly distributed with \(p_{z}\) in an interval \(p_{\mathrm{o}} \pm \Delta p_{z}\) and the transverse momentum \(p_{\perp}\) inside a circle of radius \(\Delta p_{\perp} .\) To increase the particles' spatial density, the beam is focused by electric and magnetic fields, so that the radius shrinks to a smaller value \(R\). What does Liouville's theorem tell you about the spread in the transverse momentum \(p_{\perp}\) and the subsequent behavior of the radius \(R ?\) (Assume that the focusing does not affect either \(L_{\mathrm{o}}\) or \(\Delta p_{z}\).)

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.]

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