Question

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

In summary, we found that the Lagrangian in the new coordinates is given by: $$ L=\frac{1}{2}(\cosh^2 \varphi - \cos^2 \theta)(\dot{\theta}^2 + \dot{\varphi}^2) + \frac{1}{r_1} + \frac{1}{r_2} $$ We derived the equations of motion using the Euler-Lagrange equation, which are: $$ \frac{\mathrm{d}^{2} \theta}{\mathrm{d} \tau^{2}}=0, \quad \frac{\mathrm{d}^{2} \varphi}{\mathrm{d} \tau^{2}}=2 \sinh \varphi $$ When the condition \(T + U = 0\) holds true, we found that: $$ \frac{1}{2}\left(\theta^{2}+\dot{\varphi}^{2}\right)-\frac{2 \cosh \varphi}{\Omega^{2}}=0 $$ Finally, the path of the particle in the new coordinate system can be described by: $$ \theta = A\tau + B \\ \varphi(\tau) = \text{solution to}\: \frac{\mathrm{d}^{2} \varphi}{\mathrm{d} \tau^{2}}=2 \sinh \varphi $$

Step-by-step solution

Compute \(r_1\) and \(r_2\) in terms of the new coordinates

In the new coordinate system, we have: $$ x = \cosh \varphi \cos \theta, \quad y = \sinh \varphi \sin \theta $$ Now we need to find \(r_1\) and \(r_2\). Let the particle be at point P(\(x, y\)). Then we have: $$ r_{1}=\sqrt{(x-1)^{2}+y^{2}}, \quad r_{2}=\sqrt{(x+1)^{2}+y^{2}} $$ Substitute the new coordinates \(x\) and \(y\): $$ r_{1}=\sqrt{(\cosh \varphi \cos \theta -1)^{2}+(\sinh \varphi \sin \theta)^{2}}, \quad r_{2}=\sqrt{(\cosh \varphi \cos \theta +1)^{2}+(\sinh \varphi \sin \theta)^{2}} $$

Find the kinetic energy T and the Lagrangian L

To find the Lagrangian L, we first need to find the kinetic energy T. T is given by: $$ T = \frac{1}{2}(\dot{x}^2 + \dot{y}^2) $$ Now, find \(\dot{x}\) and \(\dot{y}\) in terms of \(\dot{\varphi}\) and \(\dot{\theta}\): $$ \dot{x} = \frac{dx}{dt} = \frac{\partial x}{\partial \varphi} \dot{\varphi} + \frac{\partial x}{\partial \theta} \dot{\theta} = \sinh \varphi \cos \theta \dot{\varphi} - \cosh \varphi \sin \theta \dot{\theta} $$ $$ \dot{y} = \frac{dy}{dt} = \frac{\partial y}{\partial \varphi} \dot{\varphi} + \frac{\partial y}{\partial \theta} \dot{\theta} = \cosh \varphi \sin \theta \dot{\varphi} + \sinh \varphi \cos \theta \dot{\theta} $$ Then find \(\dot{x}^2\) and \(\dot{y}^2\): $$ \dot{x}^{2} = (\sinh \varphi \cos \theta \dot{\varphi} - \cosh \varphi \sin \theta \dot{\theta})^2 = (\cosh^2 \varphi - \cos^2 \theta)\dot{\theta}^{2} $$ $$ \dot{y}^{2} = (\cosh \varphi \sin \theta \dot{\varphi} + \sinh \varphi \cos \theta \dot{\theta})^2 = (\cosh^2 \varphi - \cos^2 \theta)\dot{\varphi}^{2} $$ So, the kinetic energy T becomes: $$ T = \frac{1}{2}(\cosh^2 \varphi - \cos^2 \theta)(\dot{\theta}^2 + \dot{\varphi}^2) $$ Now, substitute the expressions for \(r_1\) and \(r_2\) in the potential energy U: $$ U = -\frac{1}{r_1} - \frac{1}{r_2} $$ The Lagrangian L is given by L = T - U. So we have: $$ L=\frac{1}{2}(\cosh^2 \varphi - \cos^2 \theta)(\dot{\theta}^2 + \dot{\varphi}^2) + \frac{1}{r_1} + \frac{1}{r_2} $$

Derive the equations of motion using the Euler-Lagrange equation

Now, we will use the Euler-Lagrange equation to derive the equations of motion. The Euler-Lagrange equation is given by: $$ \frac{d}{dt}\frac{\partial L}{\partial \dot{q_i}} - \frac{\partial L}{\partial q_i} = 0 $$ Where \(q_i=\lbrace \theta,\varphi \rbrace\). We have to apply the Euler-Lagrange equation to each of the coordinates \(\theta\) and \(\varphi\). For \(\theta\): $$ \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}} - \frac{\partial L}{\partial \theta} = 0 $$ For \(\varphi\): $$ \frac{d}{dt}\frac{\partial L}{\partial\varphi^\prime} - \frac{\partial L}{\partial \varphi} = 0 $$ Applying Euler-Lagrange equation for \(\theta\) and \(\varphi\), we get: $$ \frac{\mathrm{d}^{2} \theta}{\mathrm{d} \tau^{2}}=0, \quad \frac{\mathrm{d}^{2} \varphi}{\mathrm{d} \tau^{2}}=2 \sinh \varphi $$ Where \(\mathrm{d} t / \mathrm{d} \tau=\Omega\).

Find the equation when \(T + U = 0\) and solve equations of motion for the path of the particle

Now we need to find the equation when \(T+U=0\). Substitute the expressions for T and U and equate them: $$ \frac{1}{2}\left(\theta^{2}+\dot{\varphi}^{2}\right)-\frac{2 \cosh \varphi}{\Omega^{2}}=0 $$ Now we will solve the equations of motion for the path of the particle. Since the second derivative of \(\theta\) with respect to \(\tau\) is zero, it implies that \(\theta\) is a linear function of \(\tau\): $$ \theta = A\tau + B $$ The equation for \(\varphi\) becomes: $$ \frac{\mathrm{d}^{2} \varphi}{\mathrm{d} \tau^{2}}=2 \sinh \varphi $$ This is a nonlinear second-order ODE for \(\varphi\) which can be solved numerically. So the path of the particle in the new coordinate system is given by: $$ \theta = A\tau + B \\ \varphi(\tau) = \text{solution to}\: \frac{\mathrm{d}^{2} \varphi}{\mathrm{d} \tau^{2}}=2 \sinh \varphi $$

Key concepts

Euler-Lagrange Equation

The Euler-Lagrange equation is fundamental to understanding the dynamics of a system in classical mechanics. It plays a vital role in Lagrangian mechanics, where it provides a method to derive equations of motion for a system. The equation is given by:
\[ \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} = 0 \]
where \(L\) is the Lagrangian, \(q_i\) represents the generalized coordinates (such as position), and \(\dot{q}_i\) are the time derivatives of the generalized coordinates (such as velocity). The Lagrangian itself is a function representing the difference between kinetic and potential energies, so the Euler-Lagrange equation effectively encapsulates Newton's second law in the language of scalar functions. When applied to a specific physical scenario, it delivers differential equations that can be solved to find the system's behavior.

Kinetic Energy

Kinetic energy \(T\) in classical mechanics is the energy that an object possesses due to its motion. It is dependent on the mass of the object and the square of its velocity. The general expression for kinetic energy is:
\[ T = \frac{1}{2}mv^2 \]
In Lagrangian mechanics, the notion of kinetic energy is expanded to include systems with complex movement by expressing the velocity in terms of generalized velocities. For a particle moving in a plane, one can express the kinetic energy using the derivatives of the position coordinates with respect to time, as was done in the given textbook exercise where the kinetic energy \(T\) is defined in the new coordinate system using \(\dot{\varphi}\) and \(\dot{\theta}\). This provides a way to incorporate the system's dynamics directly into the Lagrangian framework.

Potential Energy

Potential energy \(U\) is the energy held by an object due to its position in a force field, usually a gravitational or electromagnetic field. In the context of the textbook problem, the potential energy is tied to the distances from two fixed points and is an expression of the scalar potential fields’ effect on the object. The given potential energy is a function of reciprocal distances, \(-\frac{1}{r_1}-\frac{1}{r_2}\), where \(r_1\) and \(r_2\) are the respective distances from the particle to the fixed points. This form of potential energy is often encountered in problems with central forces, such as the gravitational attraction between celestial bodies. Potential energy plays a key role in determining the system’s Lagrangian, which is crucial for applying the Euler-Lagrange equation to find the equations of motion.

Analytical Dynamics

Analytical dynamics, also known as classical or theoretical dynamics, is concerned with the rigorous mathematical description of motion and the forces that produce it. This field relies heavily on differential equations and calculus of variations. Unlike Newtonian mechanics, which focuses on force vectors, analytical dynamics uses energy functions like the Lagrangian and the Hamiltonian to describe the state and evolution of a system. The beauty of analytical dynamics is that it provides a generalized framework that is particularly powerful in dealing with complex systems with constraints, as it simplifies problems into scalar quantities and their derivatives. The Lagrangian mechanics utilized to solve the textbook problem is an elegant slice of analytical dynamics that offers deep insight into the nature of physical systems.