/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 A system has the Lagrangian \(\m... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 2

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

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

Step by step solution

01

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: \[ \frac{\partial \mathcal{L}}{\partial y} - \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \left( \frac{dy}{dt} \right)} \right) = 0 \].
02

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: \[ \frac{\partial \mathcal{L}}{\partial y} = 3e^{3y} \].
03

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: \[ \frac{\partial \mathcal{L}}{\partial \left( \frac{dy}{dt} \right)} = 3 \left( \frac{dy}{dt} \right)^{2} \].
04

Differentiate with Respect to \(t\)

Now, apply the time derivative to the expression obtained in Step 3: \[ \frac{d}{dt} \left( 3 \left( \frac{dy}{dt} \right)^{2} \right) = 6 \left( \frac{dy}{dt} \right) \left( \frac{d^2 y}{dt^2} \right) \].
05

Formulate the Euler-Lagrange Equation

Substitute the results from Steps 2 and 4 back into the Euler-Lagrange equation: \[ 3e^{3y} - 6\left(\frac{dy}{dt}\right)\left(\frac{d^2 y}{dt^2}\right) = 0 \].
06

Simplify the Equation

Simplify the equation from Step 5: \[ e^{3y} - 2\left(\frac{dy}{dt}\right)\left(\frac{d^2 y}{dt^2}\right) = 0 \].
07

Solve for the Equation of Motion

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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鈥檚 motion. Differential equations are central to predicting how a system evolves and understanding the intricate dance of mechanics at play.

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Most popular questions from this chapter

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.

In the examples below, identify whether \(f\) is a function or a functional. \(\bullet\) A parabola is described by \(f(x)=x^{2}\). \(\bullet\) Given two forms of energy and a path \(y(t), f\) is the Lagrangian of the system \(\mathcal{L}\left(t, y, \frac{d y}{d t}\right)\) \(\bullet\) Given the magnitude of the velocity \(|\vec{v}(t)|\) of an object, \(f\) represents the distance that the object travels from time 0 to time 3600 seconds. \(\bullet\) Given the position \((x, y, z)\) in space, \(f(x, y, z)\) represents the distance from that point to the origin.

The purpose of this problem is to derive the shortest path \(y(x)\) between the points \(\left(x_{0}, y_{0}\right)\) and \(\left(x_{1}, y_{1}\right) .\) Consider an arbitrary path between these points as shown in the figure. We can break the path into differential elements \(d \vec{l}=d x \hat{a}_{x}+d y \hat{a}_{y} .\) The magnitude of each differential element is $$|d \vec{l}|=\sqrt{(d x)^{2}+(d y)^{2}}=d x \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}$$ The distance between the points can be described by the action $$\mathbb{S}=\int_{x_{0}}^{x_{1}} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x .$$ To find the path \(y(x)\) that minimizes the action, we can solve the Euler- Lagrange equation, with \(\mathcal{L}=\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\) as the Lagrangian, for this shortest path \(y(x)\). This approach can be used because we want to minimize the integral of some functional \(\mathcal{L}\) even though this functional does not represent an energy difference \([163,\) p. 33\(]\). Set up the Euler-Lagrange equation, and solve it for the shortest path, \(y(x)\).
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