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Chapter 6: Problem 8

(Harmonic oscillator) For a simple harmonic oscillator of mass \(m\), spring constant \(k\), displacement \(x\), and momentum \(p\), the Hamiltonian is \(H=\frac{p^{2}}{2 m}+\frac{k x^{2}}{2}\) Write out Hamilton's equations explicitly. Show that one equation gives the usual definition of momentum and the other is equivalent to \(F=m a\). Verify that \(H\) is the total energy.

Short Answer

Expert verified
The Hamiltonian for a simple harmonic oscillator is given by \(H(x, p) = \frac{p^2}{2m} + \frac{kx^2}{2}\). Applying Hamilton's equations, we find that \(\frac{dx}{dt} = \frac{p}{m}\), which is the usual definition of momentum, and \(\frac{dp}{dt} = -kx\), which is equivalent to \(F = ma\) with \(F = -kx\). The Hamiltonian indeed represents the total energy of the system as \(H(x, p) = \frac{p^2}{2m} + \frac{kx^2}{2} = T + U\), where T is kinetic energy and U is potential energy.

Step by step solution

01

Identify the Generalized Coordinates and Momenta

For a simple harmonic oscillator, the displacement x(t) and momentum p(t) of the mass m are the generalized coordinates and momenta of the system. The Hamiltonian H(x, p) is given as: \(H(x, p) = \frac{p^2}{2m} + \frac{kx^2}{2}\)
02

Write Hamilton's Equations

Hamilton's equations consist of two sets of equations - one for the time evolution of the generalized coordinates, and the other one for the time evolution of the generalized momenta. These equations are given by: \(\frac{dx}{dt} = \frac{\partial H}{\partial p}\) and \(\frac{dp}{dt} = -\frac{\partial H}{\partial x}\) Using these equations, we will find the expressions for \(\frac{dx}{dt}\) and \(\frac{dp}{dt}\) using the given Hamiltonian H(x, p).
03

Calculate \(\frac{dx}{dt}\) Using Hamilton's Equation

\(\frac{dx}{dt} = \frac{\partial H}{\partial p} = \frac{\partial (\frac{p^2}{2m} + \frac{kx^2}{2})}{\partial p} = \frac{2p}{2m} = \frac{p}{m}\) Thus, the partial derivative of the Hamiltonian with respect to p gives the velocity of the particle: \(\frac{dx}{dt} = \frac{p}{m}\) This is the usual definition of momentum.
04

Calculate \(\frac{dp}{dt}\) Using Hamilton's Equation

\(\frac{dp}{dt} = -\frac{\partial H}{\partial x} = -\frac{\partial (\frac{p^2}{2m} + \frac{kx^2}{2})}{\partial x} = -\frac{kx}{1}\) Thus, the partial derivative of the Hamiltonian with respect to x gives: \(\frac{dp}{dt} = -kx\) This is equivalent to the equation \(F = ma\), where the force acting on the particle is \(F = -kx\) (the restoring force in Hooke's law), and the acceleration is given by \(a = \frac{dp}{dt}\).
05

Verify that Hamiltonian Represents the Total Energy of the System

The given Hamiltonian H(x, p) is: \(H(x, p) = \frac{p^2}{2m} + \frac{kx^2}{2}\) We can identify the first term as the kinetic energy, \(T = \frac{p^2}{2m}\), and the second term as the potential energy, \(U = \frac{kx^2}{2}\). The total energy of the system, E, is given by the sum of the kinetic and potential energies: \(E = T + U = \frac{p^2}{2m} + \frac{kx^2}{2} = H(x, p)\) Thus, the given Hamiltonian, H(x, p), represents the total energy of the simple harmonic oscillator.

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Key Concepts

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

Harmonic Oscillator
A harmonic oscillator is a system that, once displaced from its equilibrium position, experiences a restoring force proportional to the displacement. This concept is fundamental in physics because it models various systems, ranging from simple pendulums to electrical circuits.
For a simple harmonic oscillator:
  • Mass is denoted as \(m\).
  • Spring constant is \(k\).
  • Displacement is \(x\).
  • Momentum is \(p\).
The harmonic oscillator equation is:\[ H = \frac{p^2}{2m} + \frac{kx^2}{2} \]In this equation:
  • The term \(\frac{p^2}{2m}\) represents the kinetic energy.
  • \(\frac{kx^2}{2}\) represents the potential energy stored in the spring.
Together, these terms describe the harmonic motion of the oscillator, where energy oscillates between kinetic and potential forms.
Hamilton's Equations
Hamilton's Equations are central in Hamiltonian mechanics. They describe the evolution over time of a dynamical system's generalized coordinates (\(x\)) and momenta (\(p\)). These two equations are:
  • \(\frac{dx}{dt} = \frac{\partial H}{\partial p}\)
  • \(\frac{dp}{dt} = -\frac{\partial H}{\partial x}\)
For a harmonic oscillator, the calculation using the Hamiltonian \(H(x, p)\) shows that:
  • \(\frac{dx}{dt} = \frac{p}{m}\), which is the velocity of the particle. This equation aligns with the standard definition of momentum where \(p = mv\).
  • \(\frac{dp}{dt} = -kx\). This represents the force \(F\) as per Hooke's law, with \(F = ma = -kx\).
These equations show the intuitive connection between classical mechanics and the concept of Hamiltonian dynamics. They not only provide a framework for analyzing motion but also link to how energy and forces interact in physical systems.
Conservation of Energy
The principle of conservation of energy is one of the most important concepts in physics. It states that energy cannot be created or destroyed, only transformed from one form to another.
In the context of a harmonic oscillator, we can express this principle as:
  • \(E = T + U\)
  • Where kinetic energy \(T = \frac{p^2}{2m}\)
  • And potential energy \(U = \frac{kx^2}{2}\)
The Hamiltonian \(H(x, p)\) of a harmonic oscillator is the system's total energy: \(H(x, p) = \frac{p^2}{2m} + \frac{kx^2}{2}\). Since the Hamiltonian remains constant over time for an isolated system, it highlights the conservation of energy.
  • This means that energy merely shifts between kinetic and potential forms with no net gain or loss of total energy over time.
  • The harmonic oscillator's energy phases in time, where energy increases in the kinetic form as the speed increases and equivalently decreases as potential energy during displacement.
The elegant nature of energy conservation in a harmonic system illustrates a central tenet of physics that resonates through many disciplines.

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Most popular questions from this chapter

(Two-mode laser) According to Haken (1983, p. 129), a two-mode laser produces two different kinds of photons with numbers \(n_{1}\) and \(n_{2}\). By analogy with the simple laser model discussed in Section \(3.3\), the rate equations are $$ \begin{aligned} &\dot{n}_{1}=G_{1} N n_{1}-k_{1} n_{1} \\ &\dot{n}_{2}=G_{2} N n_{2}-k_{2} n_{2} \end{aligned} $$ where \(N(t)=N_{0}-\alpha_{1} n_{1}-\alpha_{2} n_{2}\) is the number of excited atoms. The parameters \(G_{1}, G_{2}, k_{1}, k_{2}, \alpha_{1}, \alpha_{2}, N_{0}\) are all positive. a) Discuss the stability of the fixed point \(n_{1}{ }^{*}=n_{2}{ }^{*}=0\). b) Find and classify any other fixed points that may exist. c) Depending on the values of the various parameters, how many qualitatively different phase portraits can occur? For each case, what does the model predict about the long-term behavior of the laser?

(Rotational dynamics and a phase portrait on a spherc) The rotational dy. namics of an object in a shear flow are govemed by $$ \tilde{\theta}=\cot \phi \cos \theta, \quad \dot{\phi}=\left(\cos ^{2} \phi+A \sin ^{2} \phi\right) \sin \theta $$ where \(\theta\) and \(\phi\) are spherical coordinates that describe the orientation of the object. Our convention here is that \(-\pi<\theta \leq \pi\) is the "longitude," i.c., the angle around the \(z\)-axis, and \(-\frac{\pi}{2} \leq \phi \leq \frac{\pi}{2}\) is the "latitude," i.e., the angle measured northward from the equator. The parameter \(A\) depends on the shape of the object. a) Show that the equations are reversible in two ways: under \(t \rightarrow-t, \theta \rightarrow-\theta\) and under \(t \rightarrow-t, \phi \rightarrow-\phi\). b) Investigate the phase portraits when \(A\) is positive, zero, and negative. You may sketch the phase portraits as Mercator projections (treating \(\theta\) and \(\phi\) as rectanguIar coordinates), but it's better to visualize the motion on the sphere, if you can. c) Relate your results to the tumbling motion of an object in a shear flow. What happens to the orientation of the object as \(t \rightarrow \infty\) ?

(Why we need to assume isolated minima in Theorem 6.5.1) Consider the system \(\dot{x}=x y, \dot{y}=-x^{2}\). a) Show that \(E=x^{2}+y^{2}\) is conserved. b) Show that the origin is a fixed point, but not an isolated fixed point. c) Since \(E\) has a local minimum at the origin, one might have thought that the origin has to be a center. But that would be a misuse of Theorem \(6.5 .1\) : the theorem does not apply here because the origin is not an isolated fixed point. Show that in fact the origin is not surrounded by closed orbits, and sketch the actual phase portrait.

For each of the following systems, find the fixed points, classify them, sketch the neighboring trajectories, and try to fill in the rest of the phase portrait. $$ \dot{x}=1+y-e^{-x}, \dot{y}=x^{3}-y $$

Consider the following "rabbits vs. sheep" problems, where \(x, y \geq 0\). Find the fixed points, investigate their stability, draw the nullclines, and sketch plausible phase portraits. Indicate the basins of attraction of any stable fixed points. $$ \dot{x}=x(3-2 x-y), \quad y^{\prime}=y(2-x-y) $$
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