{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Morin_Lagrangians

# Morin_Lagrangians - Chapter 6 The Lagrangian Method...

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

Chapter 6 The Lagrangian Method Copyright 2007 by David Morin, [email protected] (draft version) In this chapter, we’re 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 doesn’t 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) Don’t worry, we’ll 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( ωt + φ ) in this problem, which is obtained by integrating the equation of motion twice. VI-1

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

View Full Document
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 θ ). At this point, you may be thinking, “That was a nice little trick, but we just got lucky in the spring problem. The procedure won’t work in a more general situation.” Well, let’s see. How about if we consider the more general problem of a particle moving in an arbitrary potential V ( x ) (we’ll 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), gives m ¨ x = - dV dx . (6.6) But - dV/dx is the force on the particle. So we see that eqs. (6.1) and (6.3) together say exactly the same thing that F = ma says, when using a Cartesian coordinate in one dimension (but this result is in fact quite general, as we’ll see in Section 6.4). Note that
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### 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
Ask a homework question - tutors are online