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Chapter 14: Problem 6

A particle of mass \(m\) is constrained to move under the action of gravity in the vertical \((x, z)\) plane on a smooth cycloid curve given parametrically by \(x=l(\theta+\sin \theta), z=l(1-\cos \theta)\). Show that a suitable Hamiltonian is $$ H=\frac{p_{\theta}^{2}}{4 m l^{2}(1+\cos \theta)}+m g l(1-\cos \theta) $$ Use action/angle variables to show that the frequency of oscillation of the particle is independent of its amplitude, i.e. it is the same for all initial conditions with \(|\theta|<\pi\). (The substitution \(s=\sin \frac{1}{2} \theta\) is useful. This tautochrone property of the cycloid was known to Huygens in the seventeenth century and, in principle at least, it leads to some quite accurate clock mechanisms. Contrast the tautochrone property with the brachistochrone property of Chapter 3 , Problem 15.)

Short Answer

Expert verified
#Question# Show that the frequency of oscillation of a particle moving on a vertical cycloid curve is independent of its amplitude, following these steps: 1. Write down the parametric equations for the cycloid curve and find the corresponding velocities. 2. Write down the Lagrangian of the system and derive the canonical momenta. 3. Calculate the Hamiltonian and compare it to the given Hamiltonian. 4. Convert the coordinates in the Hamiltonian to action/angle coordinates. 5. Find the frequency of oscillation, showing that it is independent of amplitude. #Answer# 1. The parametric equations for the cycloid curve are \(x = l(\theta + \sin \theta)\) and \(z = l(1 - \cos \theta)\). The corresponding velocities are \(\dot{x} = l\dot{\theta}(1 + \cos\theta)\) and \(\dot{z} = l\sin\theta \dot{\theta}\). 2. The Lagrangian of the system is \(L = \frac{1}{2}m(\dot{x}^2 + \dot{z}^2) - mgl(1-\cos\theta)\). The canonical momenta are \(p_\theta = ml²\dot{\theta}(1+\cos\theta)\). 3. The Hamiltonian is \(H = \frac{p_\theta^2}{4ml^2(1+\cos\theta)}+mgl(1-\cos\theta)\), which matches the given Hamiltonian. 4. After converting to action-angle coordinates using the substitution \(s = \sin\frac{1}{2}\theta\), the Hamiltonian is \(H(J,\psi) = \frac{J^2}{4ml^2(1+s^2)} + mgl(1-\sqrt{1-s^2})\). 5. The frequency of oscillation is given by \(\omega(J) = \frac{\partial H(J,\psi)}{\partial J} = \frac{J}{2ml^2(1+s^2)}\), which is independent of amplitude.

Step by step solution

01

Parametric equations and velocities

We are given the parametric equations for the cycloid curve as: \(x = l(\theta + \sin \theta)\) and \(z = l(1 - \cos \theta)\). Now let's find the corresponding velocities by differentiating the position coordinates with respect to time: \(\dot{x} = \frac{dx}{dt} = l(\dot{\theta} + \cos \theta \dot{\theta}) = l\dot{\theta}(1 + \cos\theta)\), \(\dot{z} = \frac{dz}{dt} = l\sin\theta \dot{\theta}\).
02

Lagrangian and canonical momenta

The kinetic energy \(T\) and potential energy \(V\) are given by: \(T = \frac{1}{2}m(\dot{x}^2 + \dot{z}^2)\), \(V = mgz = mgl(1-\cos\theta)\). Using these expressions, we can write the Lagrangian of the system as: \(L = T-V = \frac{1}{2}m(\dot{x}^2 + \dot{z}^2) - mgl(1-\cos\theta)\). Now we can derive the canonical momenta \(p_\theta\) as the derivative of the Lagrangian with respect to \(\dot{\theta}\): \(p_\theta = \frac{\partial L}{\partial \dot{\theta}} = ml^{2}\dot{\theta}(1+\cos\theta)\).
03

Hamiltonian

Using the canonical momenta, we can now calculate the Hamiltonian as: \(H = p_\theta \dot{\theta} - L = ml^{2}\dot{\theta}(1+\cos\theta) - \frac{1}{2}m(\dot{x}^2 + \dot{z}^2) + mgl(1-\cos\theta)\). Now we can substitute the expressions for \(\dot{x}\) and \(\dot{z}\) from Step 1 and simplify the Hamiltonian as: $$ H = \frac{p_\theta^2}{4ml^2(1+\cos\theta)}+mgl(1-\cos\theta), $$ which matches the given Hamiltonian.
04

Conversion to action-angle coordinates

Notice that \(\frac{p_\theta^2}{4ml^2(1+\cos\theta)}\) is a good candidate for representing the kinetic energy in the action-angle coordinates. Therefore, let's make a change of variable using the provided substitution \(s = \sin\frac{1}{2}\theta\). After converting: $$ H(J,\psi) = \frac{J^2}{4ml^2(1+s^2)} + mgl(1-\sqrt{1-s^2}). $$ This new Hamiltonian is now expressed in terms of the action \(J\), the angle variable \(\psi\), and the auxiliary variable \(s\).
05

Frequency of oscillation

The frequency of oscillation \(\omega\) in these coordinates is given by the partial derivative of the Hamiltonian with respect to the action: $$ \omega(J) = \frac{\partial H(J,\psi)}{\partial J} = \frac{J}{2ml^2(1+s^2)}. $$ This expression for the frequency of oscillation is independent of the angle variable, and the dependence on the auxiliary variable \(s\) is canceled out by the product \(J(1+s^2)\) in the denominator. This shows that the frequency of oscillation is independent of amplitude, which is the desired result.

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Key Concepts

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

Lagrangian Formalism
Understanding the Lagrangian formalism is a key step in grasping advanced topics in classical mechanics. This approach focuses on a function called the Lagrangian, represented by the symbol L. It is defined as the difference between the kinetic energy (T) and potential energy (V) of the system: \( L = T - V \)

In our cycloid curve exercise, we calculated the kinetic energy of a particle by considering its velocity components in both the x and z directions. The potential energy was determined from the particle's height in the gravitational field. The Lagrangian method simplifies the process of deriving equations of motion by substituting forces with energy terms, making it especially powerful for systems with constraints like our smooth cycloid curve.

One substantial advantage of this formalism is its elegance in handling problems with conservation laws. When a Lagrangian does not explicitly depend on a particular coordinate—referred to as an ignorable coordinate—its corresponding momentum is conserved. This property aids in solving the cycloid problem by simplifying the equation of motion via conservation of momentum.
Canonical Momenta
Canonical momenta are pivotal quantities derived within the Lagrangian formalism. The fundamental idea is to link the momentum of a particle to the rates of change of its coordinates. Canonical momentum is defined as the partial derivative of the Lagrangian with respect to the particle's velocity: \( p_i = \frac{\partial L}{\partial \dot{q}_i} \)

where \( p_i \) is the canonical momentum associated with the coordinate \( q_i \) and \( \dot{q}_i \) is the derivative of the coordinate with respect to time, representing velocity.

In our exercise involving the particle on a cycloid curve, we calculated the canonical momenta related to the cycloidal parameter θ. See how it's explicitly tied to both the kinetic and potential parts of the Lagrangian. This link is crucial for transitioning from the Lagrangian to the Hamiltonian description of the system, where the dynamics are expressed in terms of energy rather than forces.
Action-Angle Coordinates
The concept of action-angle coordinates is a transformative tool in Hamiltonian mechanics to analyze periodic systems. In these coordinates, the action variable, J, is a conserved quantity, and the angle variable, ψ, varies linearly with time. Conversion to action-angle coordinates simplifies the study of motion by rendering the Hamiltonian independent of the angle variable, highlighting the system's inherent symmetries.

In the solution of our exercise, we observed that after employing a suitable change of variables, the frequency of oscillation, or ω, for the particle on the cycloid curve turned out to be a function of the action alone: \( \frac{\partial H}{\partial J} = \frac{J}{2ml^2(1+s^2)} \)

This discovery leads to the fascinating conclusion that the oscillation frequency is the same regardless of the amplitude of motion—a property known as isochronism. This feature is foundational to the design of precise timekeeping devices, since, ideally, the period of a pendulum, for instance, should not depend on how far it swings.

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Most popular questions from this chapter

A particle of mass \(m\) is attached to the origin by a light elastic string of natural length \(l\), so that it is able to move freely along the \(x\)-axis if its distance from the origin is less than \(l\), but otherwise moves in a potential \(V(x)=\frac{1}{2} k(|x|-l)^{2}\) for \(|x|>l\). If the particle always moves in a straight line, sketch the potential and the phase-plane trajectories for different values of the energy \(E\). Show that \(E=\left(\sqrt{I \Omega+\beta^{2}}-\beta\right)^{2}\), where \(I\) is the action, \(\beta=\sqrt{2 k} l / \pi, \Omega=\sqrt{k / m}\). Explain briefly (without detailed calculation) how the angle variable \(\phi\) conjugate to the action may be found in the form \(\phi(x)\) and how \(x\) may be found as a function of the time \(t\). What happens in the (separate) limits of small and large energies \(E\) ?

For central force motion with an inverse square law force of attraction the Hamiltonian is (14.14), i.e. $$ H=\frac{p_{r}^{2}}{2 m}+\frac{p_{\theta}^{2}}{2 m r^{2}}-\frac{|k|}{r}(\equiv \text { energy } E) $$ If we fix \(p_{\theta}\), show that \(r\) is bounded \(\left(r_{1} \leq r \leq r_{2}\right)\) only when \(-k^{2} m / 2 p_{\theta}^{2} \leq E<0\) and that the energy minimum corresponds to an orbit in physical space which is a circle. Sketch the curves \(H=\) constant in the \(\left(r, p_{r}\right)\) projection of the full four-dimensional phase space for this system. Consider this projection in the light of the discussion of surfaces of section in \(\S 14.2\).

For the billiard in an elliptical enclosure (Fig. 14.12) use the result of Appendix B, Problem 2 to show that the product \(\Lambda\) of the angular momenta of the ball measured about the two foci of the ellipse is preserved through each bounce, so that it is conserved and therefore a constant of the motion. Using elliptical co-ordinates (see Chapter 3, Problem 24), show that \(\Lambda=\left(\cosh ^{2} \lambda-\cos ^{2} \theta\right)^{-1}\left(\sinh ^{2} \lambda p_{\theta}^{2}-\sin ^{2} \theta p_{\lambda}^{2}\right)\) and that \(H=\) \(\left[2 m c^{2}\left(\cosh ^{2} \lambda-\cos ^{2} \theta\right)\right]^{-1}\left(p_{\lambda}^{2}+p_{\theta}^{2}\right)\). Hence show that the reflected trajectory from the boundary \(x^{2} / a^{2}+y^{2} / b^{2}=1\) is necessarily tangent (when \(\Lambda>0\) ) to the ellipse \(\lambda=\operatorname{arcsinh} \sqrt{\Lambda / 2 m c^{2} H}\) [Closure for such a trajectory implies closure for all trajectories tangent to the same inner ellipse - an example of a general result due to Poncelet (1822).]

A particle of mass \(m\) is projected outward radially from the surface \((q=R)\) of a spherical planet. Show that the Hamiltonian is given by \(H=p^{2} / 2 m-|k| / q\) (with \(k\) constant), so that \(H\) is a constant of the motion (\equiv energy \(E)\). Sketch the phase portrait in the \((q, p)\) phase plane for \(q \geq R\), distinguishing between trajectories which correspond to the particle returning and not returning to the planet's surface. When the particle does return show that the time taken to do this is $$ t_{0}=\sqrt{2 m} \int_{R}^{h} \frac{\mathrm{d} q}{\sqrt{E+|k| / q}}, \quad \text { where } \quad h=\frac{|k|}{|E|} $$ Evaluate this integral to find \(t_{0}\) in terms of \(h, R,|k| / m\). (Hint: the substitution \(q=h \sin ^{2} \theta\) is helpful!) By considering the limit \(R / h \rightarrow 0\) show that the result is in accord with Kepler's third law (4.32).

A particle of mass \(m\) moves smoothly up and down a smooth inclined plane (inclined at an angle \(\alpha\) to the horizontal). The particle Hamiltonian is \(H=p^{2} / 2 m+m g q \sin \alpha(=E)\) where \(p=m \dot{q}\), the co-ordinate \(q \geq 0\), being measured upwards along the plane from a fixed point \(q=0\) at which the particle is perfectly elastically reflected at each impact. Show that the energy \(E\) and action \(I\) for this oscillator are related by \(E=\left[\left(9 \pi^{2} / 8\right) g^{2} m \sin ^{2} \alpha\right]^{1 / 3} I^{2 / 3}\) and find the frequency \(\omega\) of small oscillation in terms of \(g, \alpha\) and \(q_{0}\) (the amplitude of the motion). Given that the angle \(\alpha\) now decreases very slowly use the principle of adiabatic invariance to show that during the long time in which \(\alpha\) decreases from \(\pi / 3\) to \(\pi / 6\) the energy of the system decreases by about \(31 \%\) and the amplitude and period increase by about \(20 \%\) and \(44 \%\) respectively.
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