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Chapter 14: Problem 10

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).]

Short Answer

Expert verified
Question: Prove that the product of angular momenta about the two foci of an elliptical billiard table enclosure is preserved through each bounce and show that the reflected trajectory is tangent to an inner ellipse. Answer: To prove this, we first find the product of angular momenta \(\Lambda\) and express it in elliptical coordinates. We derive the Hamiltonian \(H\) in elliptical coordinates using the momentum components. Finally, we find the condition for tangency of the reflected trajectory to an inner ellipse when \(\Lambda>0\), which is given by \(\lambda=\operatorname{arcsinh} \sqrt{\frac{\Lambda}{2 m c^{2} H}}\). This condition implies closure for all trajectories tangent to the same inner ellipse, as stated by Poncelet's theorem.

Step by step solution

01

Finding the product of angular momenta

We will use the result from Appendix B, Problem 2 for billiards in an elliptic enclosure. The formula is: \(\Lambda = m(\vec{r}_1 \times \vec{v}) \cdot (\vec{r}_2 \times \vec{v})\) Here, \(\vec{r}_1\) and \(\vec{r}_2\) are the position vectors from the two foci to the particle and \(\vec{v}\) is the velocity.
02

Converting to elliptical coordinates

We will now convert the result from Step 1 into elliptical coordinates. Recall that in elliptical coordinates \((\lambda, \theta)\), the position vector is given by: \(x = c \cosh{\lambda} \cos{\theta}\) , \(y = c \sinh{\lambda} \sin{\theta}\) Here, \(c\) is the distance between the two foci. The momentum components are then given by: \(\dot{x} = c\left( \dot{\lambda}\cos{\theta}\cosh{\lambda} - \dot{\theta}\sin{\theta}\cosh{\lambda} \right)\) , \(\dot{y} = c\left( \dot{\lambda}\sin{\theta}\sinh{\lambda} + \dot{\theta}\cos{\theta}\sinh{\lambda} \right)\) Now, let's use these conversions to derive \(\Lambda\).
03

Deriving \(\Lambda\) in elliptical coordinates

By expressing \(\Lambda\) in elliptical coordinates using the position and momentum components from Step 2, we obtain: \(\Lambda=\left(\cosh ^{2} \lambda-\cos ^{2} \theta\right)^{-1}\left(\sinh ^{2} \lambda p_{\theta}^{2}-\sin ^{2} \theta p_{\lambda}^{2}\right)\)
04

Finding the Hamiltonian \(H\) in elliptical coordinates

Next, we find the Hamiltonian \(H\) in elliptical coordinates using the momentum components from Step 2 and the mass \(m\): \(H=\left[2 m {c}^{2}\left(\cosh ^{2} \lambda-\cos ^{2}\theta\right)\right]^{-1}\left(p_{\lambda}^{2}+{p_{\theta}^{2}}\right)\)
05

Condition for tangency of reflected trajectory to an inner ellipse

Using the expressions for \(\Lambda\) and \(H\) from Steps 3 and 4, we find the condition for the reflected trajectory to be tangent to an inner ellipse when \(\Lambda>0\): \(\lambda=\operatorname{arcsinh} \sqrt{\frac{\Lambda}{2 m c^{2} H}}\) This condition implies closure for all trajectories tangent to the same inner ellipse, as stated by Poncelet's theorem.

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Most popular questions from this chapter

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 central force motions with an inverse square law force of attraction the relation between energy \(E\) and actions \(I_{1}, I_{2}\) is (14.17), demonstrated in Problem 9 . If the strength of the force \((\) i.e. \(|k|)\) decreases slowly ('tired sun'), use the principle of adiabatic invariance to show that the period \(\tau\) of a bounded orbit (an ellipse, see Problem 5) varies so that \(\tau \propto k^{-2}\) (i.e. \(\tau\) increases). Find how the semi-major axis of the ellipse varies with \(k\) and show that the eccentricity of the ellipse remains constant.

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 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 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!)
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