The Lagrangian Method - Harvard University

1 chapter 6 The Lagrangian MethodCopyright 2007 by David Morin, version)In this chapter , we re going to learn about a whole new way of looking at things. Considerthe system of a mass on the end of a spring. We can analyze this, of course, by usingF=mato write downm x= kx. The solutions to this equation are sinusoidal functions, as we wellknow. We can, however, figure things out by using another Method which doesn t explicitlyuseF=ma. In many (in fact, probably most) physical situations, this new Method is farsuperior to usingF=ma. You will soon discover this for yourself when you tackle theproblems and exercises for this chapter .

2 We will present our new Method by first stating itsrules (without any justification) and showing that they somehow end up magically givingthe correct answer. We will then give the Method proper The Euler-Lagrange equationsHere is the procedure. Consider the following seemingly silly combination of the kinetic andpotential energies (TandV, respectively),L T V.( )This is called theLagrangian. Yes, there is a minus sign in the definition (a plus sign wouldsimply give the total energy). In the problem of a mass on the end of a spring,T=m x2/2andV=kx2/2, so we haveL=12m x2 12kx2.

3 ( )Now writeddt( L x)= L x.( )Don t worry, we ll show you in Section where this comes from. This equation is calledtheEuler-Lagrange (E-L) equation. For the problem at hand, we have L/ x=m xand L/ x= kx(see Appendix B for the definition of a partial derivative), so eq. ( ) givesm x= kx,( )which is exactly the result obtained by usingF=ma. An equation such as eq. ( ),which is derived from the Euler-Lagrange equation, is called anequation of the1 The term equation of motion is a little ambiguous. It is understood to refer to the second-orderdifferential equation satisfied byx, and not the actual equation forxas a function oft, namelyx(t) =Acos( t+ ) in this problem, which is obtained by integrating the equation of motion 6.

4 THE Lagrangian Method problem involves more than one coordinate, as most problems do, we just have to apply eq.( ) to each coordinate. We will obtain as many equations as there are coordinates. Eachequation may very well involve many of the coordinates (see the example below, where bothequations involve bothxand ).At this point, you may be thinking, That was a nice little trick, but we just got luckyin the spring problem. The procedure won t work in a more general situation. Well, let ssee. How about if we consider the more general problem of a particle moving in an arbitrarypotentialV(x) (we ll stick to one dimension for now).

5 The Lagrangian is thenL=12m x2 V(x),( )and the Euler-Lagrange equation, eq. ( ), givesm x= dVdx.( )But dV/dxis the force on the particle. So we see that eqs. ( ) and ( ) togethersay exactly the same thing thatF=masays, when using a Cartesian coordinate in onedimension (but this result is in fact quite general, as we ll see in Section ). Note thatshifting the potential by a given constant has no effect on the equation of motion, becauseeq. ( ) involves only the derivative ofV. This is equivalent to saying that only differencesin energy are relevant, and not the actual values, as we well a three-dimensional setup written in terms of Cartesian coordinates, the potentialtakes the formV(x, y, z), so the Lagrangian isL=12m( x2+ y2+ z2) V(x, y, z).

6 ( )It then immediately follows that the three Euler-Lagrange equations (obtained by applyingeq. ( ) tox,y, andz) may be combined into the vector statement,m x= V.( )But V=F, so we again arrive at Newton s second law,F=ma, now in three s now do one more example to convince you that there s really something nontrivialgoing on (Spring pendulum):Consider a pendulum made of a spring with a massmonthe end (see Fig. ). The spring is arranged to lie in a straight line (which we can arrange l+xmFigure , say, wrapping the spring around a rigid massless rod). The equilibrium length of thespring is`.

7 Let the spring have length`+x(t), and let its angle with the vertical be (t).Assuming that the motion takes place in a vertical plane, find the equations of motion forxand .Solution:The kinetic energy may be broken up into the radial and tangential parts, so wehaveT=12m( x2+ (`+x)2 2).( )The potential energy comes from both gravity and the spring, so we haveV(x, ) = mg(`+x) cos +12kx2.( )The Lagrangian is thereforeL T V=12m( x2+ (`+x)2 2)+mg(`+x) cos 12kx2.( ) THE EULER-LAGRANGE EQUATIONSVI-3 There are two variables here,xand . As mentioned above, the nice thing about the La-grangian Method is that we can just use eq.

8 ( ) twice, once withxand once with . So thetwo Euler-Lagrange equations areddt( L x)= L x= m x=m(`+x) 2+mgcos kx,( )andddt( L )= L = ddt(m(`+x)2 )= mg(`+x) sin = m(`+x)2 + 2m(`+x) x = mg(`+x) sin .= m(`+x) + 2m x = mgsin .( )Eq. ( ) is simply the radialF=maequation, complete with the centripetal acceleration, (`+x) 2. And the first line of eq. ( ) is the statement that the torque equals the rateof change of the angular momentum (this is one of the subjects of chapter 8). Alternatively,if you want to work in a rotating reference frame, then eq. ( ) is the radialF=maequation, complete with the centrifugal force,m(`+x) 2.

9 And the third line of eq. ( ) isthe tangentialF=maequation, complete with the Coriolis force, 2m x . But never mindabout this now. We ll deal with rotating frames in chapter :After writing down the E-L equations, it is always best to double-check them by tryingto identify them asF=maand/or =dL/dtequations (once we learn about that). Sometimes,however, this identification isn t obvious. And for the times when everything is clear (that is, whenyou look at the E-L equations and say, Oh, of course! ), it is usually clear onlyafteryou ve derivedthe equations. In general, the safest Method for solving a problem is to use the Lagrangian methodand then double-check things withF=maand/or =dL/dtif you can.

10 At this point it seems to be personal preference, and all academic, whether you use theLagrangian Method or theF=mamethod. The two methods produce the same , in problems involving more than one variable, it usually turns out to bemucheasier to write downTandV, as opposed to writing down all the forces. This is becauseTandVare nice and simple scalars. The forces, on the other hand, are vectors, and it iseasy to get confused if they point in various directions. The Lagrangian Method has theadvantage that once you ve written downL T V, you don t have to think anymore. Allyou have to do is blindly take some jumping from high in a tree,Just write down delLby delLbyzdot,Thent-dot what you ve got,And equate the results (but quickly!)