Question

A system has the Lagrangian \(\mathcal{L}\left(t, y, \frac{d y}{d t}\right)=\left(\frac{d y}{d t}\right)^{3}+e^{3 y}\). Find an equation for the path \(y(t)\) that minimizes the action \(\int_{t_{1}}^{t_{2}} \mathcal{L}\left(t, y, \frac{d y}{d t}\right) d t\).

Short answer

The equation is \( \frac{d^2 y}{dt^2} = \frac{e^{3y}}{2 \cdot \left(\frac{dy}{dt}\right)} \).

Step-by-step solution

Write Down the Euler-Lagrange Equation

The path \(y(t)\) that minimizes the action can be found by using the Euler-Lagrange equation, which is given by: \[ \frac{\partial \mathcal{L}}{\partial y} - \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \left( \frac{dy}{dt} \right)} \right) = 0 \].

Compute the Partial Derivative of \(\mathcal{L}\) with Respect to \(y\)

In our Lagrangian \(\mathcal{L} = \left(\frac{dy}{dt}\right)^{3} + e^{3y}\), the partial derivative with respect to \(y\) is calculated as: \[ \frac{\partial \mathcal{L}}{\partial y} = 3e^{3y} \].

Compute the Partial Derivative of \(\mathcal{L}\) with Respect to \(\frac{dy}{dt}\)

The partial derivative of \(\mathcal{L}\) with respect to \(\frac{dy}{dt}\) is: \[ \frac{\partial \mathcal{L}}{\partial \left( \frac{dy}{dt} \right)} = 3 \left( \frac{dy}{dt} \right)^{2} \].

Differentiate with Respect to \(t\)

Now, apply the time derivative to the expression obtained in Step 3: \[ \frac{d}{dt} \left( 3 \left( \frac{dy}{dt} \right)^{2} \right) = 6 \left( \frac{dy}{dt} \right) \left( \frac{d^2 y}{dt^2} \right) \].

Formulate the Euler-Lagrange Equation

Substitute the results from Steps 2 and 4 back into the Euler-Lagrange equation: \[ 3e^{3y} - 6\left(\frac{dy}{dt}\right)\left(\frac{d^2 y}{dt^2}\right) = 0 \].

Simplify the Equation

Simplify the equation from Step 5: \[ e^{3y} - 2\left(\frac{dy}{dt}\right)\left(\frac{d^2 y}{dt^2}\right) = 0 \].

Solve for the Equation of Motion

To find the path \(y(t)\), solve the differential equation: \[ \frac{d^2 y}{dt^2} = \frac{e^{3y}}{2 \cdot \left(\frac{dy}{dt}\right)} \].

Key concepts

Euler-Lagrange Equation

The Euler-Lagrange equation is a fundamental equation in Lagrangian mechanics. It helps determine the path a system will take that minimizes the action, which is a core concept in physics.

• Imagine you are planning a road trip and want the path with the least tolls. The Euler-Lagrange equation is like a GPS that calculates this optimal path for you.
• In the Lagrangian framework, the equation is represented as \[\frac{\partial \mathcal{L}}{\partial y} - \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \left( \frac{dy}{dt} \right)} \right) = 0\]where \( \mathcal{L} \) is the Lagrangian, a function that contains all the information about the system's dynamics.
• Our task is to calculate the derivatives of the Lagrangian in relation to the variables \( y \) and \( \frac{dy}{dt} \). We then substitute these derivatives into the Euler-Lagrange equation to identify the optimal path \( y(t) \) for the system.

This method provides a general way to find the equations of motion in various physical systems, simplifying the process of solving complex problems.

Action Minimization

In Lagrangian mechanics, action minimization is about finding a path that makes the action as small as possible.

• Action is a special quantity defined as the integral of the Lagrangian over a period of time.
• The action \( S \) is mathematically expressed as \[S = \int_{t_1}^{t_2} \mathcal{L} \left( t, y, \frac{dy}{dt} \right) dt\]where \( t_1 \) and \( t_2 \) are the start and end points in time.
• By minimizing this integral, we identify the most efficient path in terms of dynamic behavior.

For this reason, it’s comparable to selecting a route that takes the least fuel for your car trip; the car's path results in the least action possible given the constraints. The minimized path leads to the natural motion of the system.

Differential Equations

Differential equations are a key tool in describing physical systems using Lagrangian mechanics.

• These equations involve derivatives of functions and are used to define how a function changes over time.
• The equation we derived, \[\frac{d^2 y}{dt^2} = \frac{e^{3y}}{2 \cdot \left(\frac{dy}{dt}\right)}\]is a second-order differential equation, indicating the system's acceleration \( \frac{d^2 y}{dt^2} \) in relation to time.
• Such equations typically require initial conditions or boundary conditions to solve, marking where and how the problem starts.

Solving these equations gives us the specific path or function \( y(t) \) that describes the system’s motion. Differential equations are central to predicting how a system evolves and understanding the intricate dance of mechanics at play.