Question

A system has Lagrangian \(\mathcal{L}\left(t, y, \frac{d y}{d t}\right)=\frac{1}{2}\left(\frac{d y}{d t}\right)^{2}+\frac{1}{2} \cdot y^{-2}\). Find the corresponding equation of motion.

Short answer

The equation of motion is \( \ddot{y} + \frac{1}{2} y^{-3} = 0 \).

Step-by-step solution

Formulate the Lagrangian

First, write down the given Lagrangian for the system. The Lagrangian is: \( \mathcal{L}(t, y, \frac{dy}{dt}) = \frac{1}{2}\left(\frac{dy}{dt}\right)^2 + \frac{1}{2} \cdot y^{-2} \). This expression contains the kinetic and potential energy terms of the system.

Apply the Euler-Lagrange Equation

To derive the equation of motion, use the Euler-Lagrange equation: \( \frac{d}{dt}\left( \frac{\partial \mathcal{L}}{\partial \dot{y}} \right) - \frac{\partial \mathcal{L}}{\partial y} = 0 \), where \( \dot{y} = \frac{dy}{dt} \).

Compute Partial Derivative with Respect to \( \dot{y} \)

Find \( \frac{\partial \mathcal{L}}{\partial \dot{y}} \). Substitute the Lagrangian components: \( \frac{\partial \mathcal{L}}{\partial \dot{y}} = \frac{\partial}{\partial \dot{y}}\left( \frac{1}{2}\dot{y}^2 + \frac{1}{2}y^{-2} \right) = \dot{y} \).

Compute Total Derivative with Respect to Time for \( \dot{y} \) Term

Calculate \( \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{y}} \right) = \frac{d}{dt} (\dot{y}) = \ddot{y} \), where \( \ddot{y} \) is the second derivative of \( y \) with respect to \( t \).

Compute Partial Derivative with Respect to \( y \)

Find \( \frac{\partial \mathcal{L}}{\partial y} \): \( \frac{\partial \mathcal{L}}{\partial y} = \frac{\partial}{\partial y}\left( \frac{1}{2}\dot{y}^2 + \frac{1}{2}y^{-2} \right) = -\frac{y^{-3}}{2} \).

Substitute into Euler-Lagrange Equation

Substitute the computed derivatives back into the Euler-Lagrange equation: \( \ddot{y} - \left(-\frac{1}{2} y^{-3}\right) = 0 \). This leads to: \( \ddot{y} + \frac{1}{2} y^{-3} = 0 \).

Final Equation of Motion

Thus, the equation of motion for the system is: \( \ddot{y} + \frac{1}{2} y^{-3} = 0 \). This describes how the system behaves dynamically.

Key concepts

Euler-Lagrange Equation

The Euler-Lagrange equation is a fundamental component of Lagrangian mechanics. It helps us derive the equations of motion for a system based on its Lagrangian function. The Lagrangian is usually expressed in terms of the system's coordinates and their time derivatives.
For a Lagrangian function \( \mathcal{L}(t, y, \dot{y}) \), the Euler-Lagrange equation is given by:

• \( \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{y}} \right) - \frac{\partial \mathcal{L}}{\partial y} = 0 \)

Here, \( y \) is a generalized coordinate, and \( \dot{y} \) is its time derivative.
The equation states that the change over time of the partial derivative of the Lagrangian concerning \( \dot{y} \), minus the partial derivative of the Lagrangian concerning \( y \), must equal zero.
This principle is powerful because it relates the dynamics of the system to its energy components, allowing us to explore the "path" that minimizes the action within Lagrangian mechanics.

Equation of Motion

The equation of motion describes the behavior of a physical system over time. It provides insights into how the system will evolve from any initial state, given forces or constraints acting on it. When we use the Lagrangian approach combined with the Euler-Lagrange equation, we can derive these equations elegantly.
The Lagrangian \( \mathcal{L}(t, y, \frac{dy}{dt}) = \frac{1}{2}\left(\frac{dy}{dt}\right)^2 + \frac{1}{2} y^{-2} \) shows us the kinetic and potential energy terms of the system. Utilizing the Euler-Lagrange equation, as shown in the steps of the solved problem, leads us to the derived equation of motion:

• \( \ddot{y} + \frac{1}{2} y^{-3} = 0 \)

This equation tells us how the system's acceleration \( \ddot{y} \), which is the second derivative of \( y \), changes over time based on the current state of \( y \). By solving this equation with appropriate initial conditions, we could predict the future states of the system.

Kinetic and Potential Energy

Lagrangian mechanics often involves expressing a system's motion in terms of its kinetic and potential energies. These energies are integral in formulating the Lagrangian, which is the foundation for deriving the equation of motion.
- **Kinetic Energy (T):** This represents the energy of motion. For a particle with velocity \( \dot{y} \), the kinetic energy is given by

• \( T = \frac{1}{2} \left( \dot{y} \right)^2 \)

indicating how fast the particle is moving.
- **Potential Energy (V):** This represents the stored energy due to the position or configuration of the particle. In the given problem, the potential energy is

• \( V = \frac{1}{2} y^{-2} \)

indicating energy due to the position of \( y \).
The Lagrangian is defined as the difference between kinetic and potential energies:

• \( \mathcal{L} = T - V \)

By defining the Lagrangian this way, we can use the powerful tools of calculus to find the path that the system naturally follows, providing a bridge between energy principles and motion.