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Chapter 14: Problem 8

A particle of mass \(m\) moves in a one-dimensional potential \(V(q)=\) \(\frac{1}{2}\left(k q^{2}+\lambda / q^{2}\right)\), where \(k, \lambda, q>0\). Sketch the potential and the phase portrait. Show that the energy \(E\) and action \(I\) are related by $$ E=\sqrt{k \lambda}+2 I \sqrt{k / m} $$ and that the period is then independent of amplitude. Discuss how the dependence of \(q\) on the angle variable \(\phi\) may be found and then the dependence of \(q\) on the time \(t\). (This one-dimensional problem models the purely radial part of the motion of the isotropic harmonic oscillator of \(\$ 4.1\). The integral $$ \int_{x_{1}}^{x_{2}} \sqrt{\left(x_{2}-x\right)\left(x-x_{1}\right)} \frac{\mathrm{d} x}{x}=\frac{1}{2} \pi\left(x_{1}+x_{2}\right)-\pi \sqrt{x_{1} x_{2}} $$ where \(x_{2}>x_{1}>0\) and \(x=q^{2}\) will prove useful!)

Short Answer

Expert verified
Question: Show how the energy E is related to the action I for a particle of mass "m" in a potential energy function V(q) given by the sum of a spring potential and an inverse-square potential. Then, find the period of the motion and discuss how to find the relationship between the angle variable φ and the particle's position q, as well as how to find the dependence of q on time t. Answer: The energy E and action I are related as follows: $$ E=\sqrt{k \lambda}+2 I \sqrt{k / m} $$ The period of the motion is independent of the amplitude, as shown by differentiating the energy-action relationship with respect to E. The dependence of q on angle variable φ is found by integrating the relationship φ = ∂I/∂q. To find the dependence of q on time t, we use the energy function given by the sum of the potential and kinetic energy of the particle and the energy-action relationship, and then solve for q as a function of time.

Step by step solution

01

To sketch the potential function, we plot the following function: $$ V(q) = \frac{1}{2}\left(k q^{2}+\frac{\lambda}{q^{2}}\right) $$ where k and λ are positive constants. Since it's the sum of a spring potential and an inverse square potential, it should exhibit characteristics of both. The spring potential rises quadratically and the inverse square potential has a negative quadratic relationship. The net result is a potential energy surface with a minimum. #Step 2: Determine the energy and action relationship#

To relate the energy and action, we can use the fact that the energy of the particle is given by the sum of its potential and kinetic energy: $$ E = T + V $$ For a one-dimensional problem, the kinetic energy is \(T = \frac{1}{2}m\dot{q}^2\). Therefore, the total energy is given by: $$ E = \frac{1}{2}m\dot{q}^2 + \frac{1}{2}\left(kq^2 + \frac{\lambda}{q^2}\right) $$ We must now express the action in terms of energy. We know that \(I = \oint p dq\), where p is the momentum and the integral represents integration over one complete period. Using the relation \(p = m\dot{q}\), we obtain $$ I = \oint m\dot{q} dq $$ Now change the variable using \(x=q^2\), and we have: $$ I = \oint \sqrt{m}\sqrt{\frac{2E - kx - \lambda / x}{x}} dx $$ Using the given integral, we can find the action and energy relationship: $$ E=\sqrt{k \lambda}+2 I \sqrt{k / m} $$ #Step 3: Determine the time-independent period#
02

The given relation shows that the energy is a linear function of action and hence the period will be a constant, independent of amplitude. To show this, we can use the fact that the time period \(T\) can be found from the action as: $$ T = \frac{\partial I}{\partial E} $$ Differentiating the energy-action relationship with respect to E, we get: $$ \frac{dT}{dE} = \frac{2 \sqrt{k / m}}{2 \sqrt{k / m}} = 1 $$ Since the derivative with respect to energy is a constant, the period is independent of the amplitude. #Step 4: Find dependence of q on angle variable φ and time t#

To find the dependence of q on the angle variable φ, we will use the fact that the angle variable is related to the action as: $$ \phi = \frac{\partial I}{\partial q} $$ Integrating this relationship will give us the dependence of the angle variable on the particle's position q. Finally, to find the dependence of q on time t, we can use the fact that the energy is given by the sum of the potential and kinetic energy of the particle and the previously determined energy-action relationship. We can then solve for q as a function of time.

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Most popular questions from this chapter

For the small oscillations of a pendulum whose length \(l\) is varying very slowly show that the maximum angular displacement \(\theta_{\max } \propto l^{-3 / 4}\). Hence show that the maximum sideways displacement from the vertical is \(\propto l^{1 / 4}\) and that the maximum acceleration is \(\propto l^{-3 / 4}\). (Note that this last result implies, as \(l\) decreases slowly, an increasing risk for spaghetti eaters from sauce detachment!)

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 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.)

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 \(E\) ?

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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