Mechanics – A Brief Introduction to Alternative Formulations Concepts of primary interest: Newtonian Mechanics Lagrangian Mechanics Generalized Coordinates qi L Generalized Momenta pi qi Hamiltonian Mechanics Poisson Brackets Sample calculations: Box on an online Object rolling down an incline 1D particle motion Tools of the Trade Total Derivatives Newton’s ‘overdot’ notation Alternative Formulations of Mechanics – a very quick look The goal of this handout is to introduce two alternative formulations of mechanics that apply well to problems with conservative forces and normal constraint forces. Energy methods proved so fruitful in introductory mechanics that extensions must exist that essentially allow an energy viewpoint to replace the Newtonian force viewpoint. This energy viewpoint is a prerequisite for quantum mechanics so we will just bite the bullet, push the ‘I believe’ button and proceed. In some sense, the energy-view alternatives, the Lagrangian and the Hamiltonian formulations, can be derived or deduced from Newton’s Laws. No such logical development is to be presented here. In fact, the prescriptions in this handout have restricted validity. The whole truth is to found in your junior year mechanics course. Our point of departure is Newton’s Second Law: ma Fi V() r Fj. [MechAlt.1] i,, forceson j forces do the particle nowork The forces have been separated into those derived from potentials and those do to normal constraints, the classes of forces handled most easily in the Lagrangian formalism. The normal force on a box sliding down an incline and the tension force exerted by a string on a particle tethered to move around a circular path are examples of normal constraint forces. As they are directed at right Contact: [email protected] angles to the displacement, normal constraint forces due no work. Our example problems are to be a box sliding down a frictionless incline and an object rolling down an incline without slipping. The sliding box could be treated using the Cartesian coordinates x and y, but, as the box moves along a line, using two coordinates is over-kill. Instead a coordinate q is chosen – basically the distance from the top of the incline to the box as measured along a parallel to the inclined surface. The restriction to motion along a line means that the motion has a single degree of freedom. In the developments that follow care is to be taken to restrict the number of coordinates chosen to be equal to the degrees of freedom for the motion. A bead might be constrained to move along a curved wire requiring a generalized or unusual coordinate to describe its single-degree-of-freedom position. All coordinates are to be called generalized coordinates are to be represented by qi where the index i assumes j values, the number of degrees of freedom. The total time derivatives of the coordinates dq are called coordinate velocities: q i . i dt q y x Example I: A problem with a conservative force, the gravitational attraction due to the earth, and a normal constraint force that does no work and serves to prevent the box from moving into the space occupied by the incline. As the mass of the earth is huge compared to that of the box and the problem is restricted to altitudes near the earth’s surface, the potential energy can be represented as mgy. As illustrated, q is the distance that the box has moved parallel to the incline. It is the generalized coordinate. The kinetic energy for uniform translation can always be computed from its Cartesian form: 1122ˆˆ1 ˆˆˆˆ 22 T(xyz 22,mvm,x2iy )jzk()xiy jzk) m()xyz ( 9/3/2009 Physics Handout Series.Tank: Alternative Forms of Mechanics MechAlt- 2 Given the choice of coordinates, it follows that: (, the incline angle is a constant.) x = xo + q cos; y = yo - q sin xq cosy q sin Note that zand z are not needed here. Substituting, 1 2 Tqq(,) 2 mq and Vqq( , ) mgyo mgq sin 122212 Exercise: Show that 2mx() y z 2 mq for the example above. The Lagrangian process: The description that follows is specialized to simple cases. The term usually means that it is true for our applications, but may need modification for more general applications. I.) Choose a set of generalized coordinates (q1, …, qj) where j is the number of degrees of freedom. II.) Find expressions for the total kinetic energy T and potential energy V in terms of the generalized coordinates and there corresponding coordinate velocities, the qi . The natural variables for a Lagrangian description are the q’s, the qs ' and the time t. jj LLL dL dqik dq dt qqi k t ik11 That is: The natural variables show up nicely in the differential of the function. This introduction does not include cases with time dependent forces so all functions are to be written as forms depending on the q’s and qs ' . When partial derivatives are indicated, each qi is to be considered to be an independent variable even independent of its conjugate qi. III) Form the Lagrangian (function) as: LqqqqTqqqqVqqqq(11 , ;...;j ,jj ) (11 , ;...; ,j ) (11 , ;...;j ,j ) . [MechAlt.2] IV.) For each coordinate, compute the conjugate generalized momentum. L p [MechAlt.3]. qi qi Note that the process calls for a partial derivative with respect to qi . The variables qi and qi are to be considered as independent variables as these partial derivatives are computed. V.) Compute the equations of motion for each coordinate degree of freedom. 9/3/2009 Physics Handout Series.Tank: Alternative Forms of Mechanics MechAlt- 3 dL L d jj where qqi k [MechAlt.4] dt qii q dtik11 q i qk t In this last expression, a total time derivative is to be computed. Note the mixing of the d d conventional /dt and the Newtonian dot to represent total time derivatives. The /dt form is used for emphasis. Example I: Box on the incline: I.) There is one degree of freedom and the single coordinate q, the distance moved down the incline is chosen to be the coordinate. 1 2 II.) It has been shown that: Tqq(,) 2 mq and Vqq( , ) mgyo mgq sin . 1 2 III.) Lqq(,) = Tqq(,) Vqq (,) 2 mq mgyo mgqsin . L IV.) p mq q q dL d L V.) mq mq m gsin q g sin . The box accelerates down dt q dt qi along the incline with a linear acceleration g sin as expected. Example II: An symmetric object of radius R with moment if inertia I rolls down the same incline without slipping. As before there is one degree of freedom, and the net angle of rotation is to be chosen as the one generalized coordinate. In terms of this coordinate, the center of mass coordinates of the rolling object are: X = Xo + q cos; Y = Yo - q sin and its components of velocity are: Xq cos and Yq sin where q = R. The kinetic energy decomposes into the translational energy of the center of mass and the kinetic energy of rotation about the center of mass. 1122 TMXYI2[]2CM . The potential energy depends on the center of mass position only. V = M g Y. Example II: Rotationally symmetric object rolling w/o slipping on the incline the distance that the center of mass has moved parallel to the incline is q = R . I.) There is one degree of freedom; we choose the single coordinate , the net angle of rotation. 1122 2 II.) It follows that: TMRI(,) 22CM andVVMgR(,) 0 sin . 9/3/2009 Physics Handout Series.Tank: Alternative Forms of Mechanics MechAlt- 4 1122 2 III.) LMRIVMgR(,) 220 sin . L IV.) p MR2 I . Note that the momentum conjugate to the angle has the form of an angular momentum, but it is not just I , the angular momentum due to the rotation about the center of mass. The contribution MR2 is the angular momentum associated with the center of mass motion relative to the instantaneous point of contact (or any other point of the surface of the incline). The angular momentum of a rigid rotating body decomposes into the angular momentum associated with the center of mass motion plus that due to the body’s rotation about the center of mass. Adopting a single coordinate to describe the rotation and translation requires that the momentum conjugate to represents the combination of the coupled motions. 2 dL d 22L gMRsin V.) MR I MR I MgRsin . 2 dt dt R MR I 2 2 1 If the object is a uniform cylinder, I = ½ M R and 3 gR sin . Center of Mass Theorems: Exercise: Repeat the object rolling down the incline problem, but choose q, the displacement of the center of mass along a line parallel to the incline. Find expressions for the kinetic energy T(q, q ) and the generalized momentum pq. Discuss the result for pq given that it represents the total mass- motion and that the rotation and translation are coupled. The Hamiltonian Process: The Hamiltonian picture is built on the Lagrangian picture by a process L involving the conjugate momenta computed as: p . The Hamiltonian (function) is defined to qi qi j jj LLL be Hqpt(,ii ,) pq kk Lqqt (,,) ii . As dL dqik dq dt , qqi k t k 1 ik11 jjjj LLL dH(, qii p ,) t d pkk q dL(,,) q ii q t pkk dq q k dp k qqt dqi dq k dt . kki111i k1k L As ppq i , i qi jjj LL dH(, qii p ,) t pkdqkqk dp kqt dqi pm dqm dt ki11i m1 jj LL dH(, qii p ,) t q k dp k qt dqi dt k11i i As they appear as the differentials, qi, pi and t are the natural variables for the hamiltonian.