Chapter 3: Problem 8
A particle of unit mass is moving in the \(x, y\) plane under the influence of the potential $$ U=-\frac{1}{r_{1}}-\frac{1}{r_{2}}, $$ where \(r_{1}\) and \(r_{2}\) are the respective distances from the points \((1,0)\) and \((-1,0)\). Show that if new coordinates are introduced by putting $$ x=\cosh \varphi \cos \theta, \quad y=\sinh \varphi \sin \theta, $$ then the Lagrangian governing the motion becomes $$ L=\frac{1}{2} \Omega\left(\dot{\theta}^{2}+\dot{\varphi}^{2}\right)+2 \Omega^{-1} \cosh \varphi, $$ where \(\Omega=\cosh ^{2} \varphi-\cos ^{2} \theta\). Write down the equations of motion and show that if the particle is set in motion with \(T+U=0\), then $$ \frac{1}{2}\left(\theta^{2}+\dot{\varphi}^{2}\right)-\frac{2 \cosh \varphi}{\Omega^{2}}=0 $$ Deduce that $$ \frac{\mathrm{d}^{2} \theta}{\mathrm{d} \tau^{2}}=0, \quad \frac{\mathrm{d}^{2} \varphi}{\mathrm{d} \tau^{2}}=2 \sinh \varphi $$ where \(\mathrm{d} t / \mathrm{d} \tau=\Omega\). Hence find the path of the particle.
Short Answer
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Key Concepts
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