Morin_Lagrangians - Chapter 6 The Lagrangian Method...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 6 The Lagrangian Method Copyright 2007 by David Morin, (draft version) In this chapter, were going to learn about a whole new way of looking at things. Consider the system of a mass on the end of a spring. We can analyze this, of course, by using F = ma to write down m x =- kx . The solutions to this equation are sinusoidal functions, as we well know. We can, however, figure things out by using another method which doesnt explicitly use F = ma . In many (in fact, probably most) physical situations, this new method is far superior to using F = ma . You will soon discover this for yourself when you tackle the problems and exercises for this chapter. We will present our new method by first stating its rules (without any justification) and showing that they somehow end up magically giving the correct answer. We will then give the method proper justification. 6.1 The Euler-Lagrange equations Here is the procedure. Consider the following seemingly silly combination of the kinetic and potential energies ( T and V , respectively), L T- V. (6.1) This is called the Lagrangian . Yes, there is a minus sign in the definition (a plus sign would simply give the total energy). In the problem of a mass on the end of a spring, T = m x 2 / 2 and V = kx 2 / 2, so we have L = 1 2 m x 2- 1 2 kx 2 . (6.2) Now write d dt L x = L x . (6.3) Dont worry, well show you in Section 6.2 where this comes from. This equation is called the Euler-Lagrange (E-L) equation . For the problem at hand, we have L/ x = m x and L/x =- kx (see Appendix B for the definition of a partial derivative), so eq. (6.3) gives m x =- kx, (6.4) which is exactly the result obtained by using F = ma . An equation such as eq. (6.4), which is derived from the Euler-Lagrange equation, is called an equation of motion . 1 If the 1 The term equation of motion is a little ambiguous. It is understood to refer to the second-order differential equation satisfied by x , and not the actual equation for x as a function of t , namely x ( t ) = A cos( t + ) in this problem, which is obtained by integrating the equation of motion twice. VI-1 VI-2 CHAPTER 6. THE LAGRANGIAN METHOD problem involves more than one coordinate, as most problems do, we just have to apply eq. (6.3) to each coordinate. We will obtain as many equations as there are coordinates. Each equation may very well involve many of the coordinates (see the example below, where both equations involve both x and ). At this point, you may be thinking, That was a nice little trick, but we just got lucky in the spring problem. The procedure wont work in a more general situation. Well, lets see. How about if we consider the more general problem of a particle moving in an arbitrary potential V ( x ) (well stick to one dimension for now). The Lagrangian is then L = 1 2 m x 2- V ( x ) , (6.5) and the Euler-Lagrange equation, eq. (6.3), givesand the Euler-Lagrange equation, eq....
View Full Document

This note was uploaded on 11/24/2011 for the course PHYS 3113 taught by Professor Kennefick during the Fall '11 term at Arkansas.

Page1 / 55

Morin_Lagrangians - Chapter 6 The Lagrangian Method...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online