The Hamiltonian method

1 chapter 15 The Hamiltonian methodCopyright 2008 by David Morin, Version 2, October 2008)This chapter is to be read in conjunction withIntroduction to Classical Mechanics, With Problemsand Solutionsc 2007, by David Morin, Cambridge University text in this version is the same as in Version 1, but some new problems and exercises havebeen information on the book can be found at: present, we have at our disposal two basic ways of solving mechanics problems. InChapter 3 we discussed the familiar method involving Newton s laws, in particularthe second law,F=ma.

2 And in chapter 6 we learned about the Lagrangianmethod. These two strategies always yield the same results for a given problem, ofcourse, but they are based on vastly different principles. Depending on the specificsof the problem at hand, one method might lead to a simpler solution than the this chapter , we ll learn about a third way of solving problems, theHamil-tonianmethod. This method is quite similar to the Lagrangian method , so it sdebateable as to whether it should actually count as a third one. Like the La-grangian method , it contains the principle of stationary action as an it also contains many additional features that are extremely useful in otherbranches of physics, in particular statistical mechanics and quantum the Hamiltonian method generally has no advantage over (and in fact isinvariably much more cumbersome than)

3 The Lagrangian method when it comes tostandard mechanics problems involving a small number of particles, its superioritybecomes evident when dealing with systems at the opposite ends of the spectrumcompared with a small number of particles, namely systems with an intractablylarge number of particles (as in a statistical-mechanics system involving a gas), orsystems with no particles at all (as in quantum mechanics, where everything is awave).We won t be getting into these topics here, so you ll have to take it on faithhow useful the Hamiltonian formalism is.

4 Furthermore, since much of this book isbased on problem solving, this chapter probably won t be the most rewarding one,because there is rarely any benefit from using a Hamiltonian instead of a Lagrangianto solve a standard mechanics problem. Indeed, many of the examples and problemsin this chapter might seem a bit silly, considering that they can be solved much morequickly using the Lagrangian method . But rest assured, this silliness has a purpose;the techniques you learn here will be very valuable in your future physics outline of this chapter is as follows.

5 In Section we ll look at the sim-XV-1XV-2 chapter 15. THE Hamiltonian method ilarities between the Hamiltonian and the energy, and then in Section we llrigorously define the Hamiltonian and derive hamilton s equations , which are theequations that take the place of Newton s laws and the Euler- lagrange Section we ll discuss the Legendre transform, which is what connects theHamiltonian to the Lagrangian. In Section we ll give three more derivations ofHamilton s equations , just for the fun of it. Finally, in Section we ll introducethe concept of phase space and then derive Liouville s theorem, which has countlessapplications in statistical mechanics, chaos, and other EnergyIn Eq.

6 ( ) in chapter 6 we defined the quantity,E (N i=1 L qi qi) L,( )which under many circumstances is the energy of the system, as we will see then showed in Claim thatdE/dt= L/ t. This implies that if L/ t= 0(that is, iftdoesn t explicitly appear inL), thenEis constant in time. In thepresent chapter , we will examine many other properties of this quantityE, or moreprecisely, the quantityH(theHamiltonian) that arises whenEis rewritten in acertain way explained in Section before getting into a detailed discussion of the actual Hamiltonian , let s firstlook at the relation betweenEand the energy of the system.

7 We chose the letterEin Eq. ( ) because the quantity on the right-hand side often turns out tobe the total energy of the system. For example, consider a particle undergoing 1-Dmotion under the influence of a potentialV(x), wherexis a standard Cartesiancoordinate. ThenL T V=m x2/2 V(x), which yieldsE L x x L= (m x) x L= 2T (T V) =T+V,( )which is simply the total energy. By performing the analogous calculation, it like-wise follows thatEis the total energy in the case of Cartesian coordinates inNdimensions:L=(12m x21+ +12m x2N) V(x1.)

8 , xN)= E=((m x1) x1+ + (m xN) xN) L= 2T (T V)=T+V.( )In view of this, a reasonable question to ask is: DoesEalways turn out to be thetotal energy, no matter what coordinates are used to describe the system? Alas,the answer is no. However, when the coordinates satisfy a certain condition,Eisindeed the total energy. Let s see what this condition a slight modification to the above 1-D setup. We ll change variablesfrom the nice Cartesian coordinatexto another coordinateqdefined by, say,x(q) =Kq5, or equivalentlyq(x) = (x/K)1/5.

9 Since x= 5Kq4 q, we can rewrite theLagrangianL(x, x) =m x2/2 V(x) in terms ofqand qasL(q, q) =(25K2mq82) q2 V(x(q)) F(q) q2 Vu(q),( ) ENERGYXV-3whereF(q) 25K2mq8/2 (so the kinetic energy isT=F(q) q2). The quantityEis thenE L q q L=(2F(q) q) q L= 2T (T V) =T+V,( )which again is the total energy. So apparently it is possible for (at least some)non-Cartesian coordinates to yield anEequaling the total can easily demonstrate that in 1-D,Eequals the total energy if the newcoordinateqis related to the old Cartesian coordinatexby any general functionaldependence of the form,x=x(q).

10 The reason is that since x= (dx/dq) qby thechain rule, the kinetic energy always takes the form of q2times some function is,T=F(q) q2, whereF(q) happens to be (m/2)(dx/dq)2. This functionF(q)just goes along for the ride in the calculation ofE, so the result ofT+Varises inexactly the same way as in Eq. ( ).What if instead of the simple relationx=x(q) (or equivalentlyq=q(x)) wealso have time dependence? That is,x=x(q, t) (or equivalentlyq=q(x, t))? Thetask of Problem is to show thatL(q, q, t) yields anEthat takes the form,E=T+V m(( x q)( x t) q+( x t)2),( )which isnotthe total energy,T+V, due to the x(q, t)/ t6= 0 assumption.