Lagrange Multipliers. A Somewhat Silly Example. Monday, 25 September 2013


 Phillip Wilkerson
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1 Lagrange Multplers Monday, 5 September 013 Sometmes t s convenent to use redundant coordnates, and to effect the varaton of the acton consstent wth the constrants va the method of Lagrange undetermned multplers. As a bonus, we obtan the generalzed forces of constrant. Physcs 111 A Somewhat Slly Example To help us wrap our mnds around the challenge of ncorporatng constrants nto the Lagrangan formalsm, let s start wth the slly use of redundant coordnates llustrated at the rght. The mass sldes down the nclned plane, whch provdes a normal force suffcent to make t do so. Of course, t would be more straghtforward to use a coordnate algned wth the surface of the nclned plane, but we have happened to decde to use the x and y coordnates llustrated n the fgure. Along the plane, therefore, y = x tanα. We may wrte the lagrangan n terms of these generalzed coordnates, L = 1 m(ẋ + ẏ ) mg y y α m x and from t produce the acton ntegral whose varaton we set to zero: {[ L δs = x d ] [ L L δx + dt ẋ y d ] } L δy dt = 0 dt ẏ If we blthely pretend that the varatons δx and δy are uncorrelated, then we conclude that for the acton to be statonary we must have each term n square brackets vansh, whch leads to mẍ = 0 and mÿ = g. Of course, these are the correct equatons of moton for a partcle n free fall, not sldng down an nclne. To stay on the nclne, the vrtual dsplacements n x and y must be coordnated: d y = tanα dx = δy = δx tanα Substtutng nto the expresson of statonary acton yelds {[ L δs = x d ] [ L L + dt ẋ y d ] } L tanα δx dt = 0 dt ẏ where we now may take an arbtrary varaton δx. Evaluatng the term n braces and settng t to zero gves mẍ + ( mg mÿ ) tanα = 0 ẍ + ÿ tanα = g tanα Physcs of 6 Peter N. Saeta
2 Dfferentatng the constrant equaton wth respect to tme twce, ÿ = ẍ tanα, allows us to elmnate y from the equaton: ẍ + ẍ tan α = g tanα ẍ sec α = g tanα ẍ = g snαcosα whch s ndeed the correct equaton of moton for x. The General Case Let s generalze to the case of a mechancal system wth N P degrees of freedom, whch we descrbe wth N generalzed coordnates and P equatons of constrant. We wll assume that dsspaton may be neglected so that the system may be descrbed by a Lagrangan. Let us suppose that the equatons of constrant may be wrtten n the form G j (q, t) = 0 = dg j = dq + dt = 0 q t for j = 1,,..., P. For notatonal convenence, defne Translaton: just to confuse you. q = a j and t = a j t Then the j th constrant equaton may be wrtten a j dq + a j t dt = 0 or a j q + a j t = 0 (P equatons) Note that the coeffcents a j and a j t may be functons of the generalzed coordnates q and the tme t, but not the generalzed veloctes q. Hamlton s prncple says that of all possble paths, the one the system follows s that whch mnmzes the acton, whch s the tme ntegral of the lagrangan: tb δs = δ L(q, q, t)dt = 0 (1) t a Expandng the varaton n the lagrangan and ntegratng by parts, we obtan tb [ L δs = d ] L δq dt q dt q t a Remember that we are usng the summaton conventon; we sum over. whch must vansh on the mnmum path. When the coordnates q formed a mnmal complete set, we argued that the vrtual dsplacements δq were arbtrary and that the only way to ensure that the acton be mnmum s for each term n the square brackets to vansh. Now, however, an arbtrary varaton n the coordnates wll send us off the constrant surface, leadng to an mpossble soluton. One way to trck Prof. Hamlton nto fndng the rght soluton (the soluton consstent wth the constrant equatons) would be to sneak somethng nsde the brackets that would make the varaton n the acton zero Physcs 111 of 6 Peter N. Saeta
3 for dsplacements that take us off the surface of constrant. That s hardly cheatng; t just takes away the ncentve to cheat! If we managed to fnd such terms, then t wouldn t matter whch way we vared the path; we d get zero change n the acton, to frst order n the varaton. We could then treat all the varatons δq as ndependent. Before runnng that lttle operaton, consder what t mght mean for those terms to represent the generalzed forces of constrant. Snce the allowed vrtual dsplacements of the system are all orthogonal to the constrant forces, those forces do no work and they don t change the value of the acton ntegral. In fact, they are just the necessary forces to ensure that the moton follows the constraned path. So, f we can fgure out what terms we need to add to the lagrangan to make the llegal varatons vansh, we wll have also found the forces of constrant. Each of the constrant equatons s of the form G j (q, t) = 0, so f we were to add a multple of each constrant equaton to the lagrangan, t would leave the acton unchanged. So, we form the augmented Lagrangan: L = L + λ j G j () where I m usng the summaton conventon and the λ j are Lagrange s undetermned multplers, one per equaton of constrant; all may be functons of the tme. Then the (augmented) acton s t 1 S = t t 1 L dt (but snce we added zero, t s the same as the unaugmented acton). By Hamlton s prncple, the acton s mnmzed along the true path followed by the system. We may now effect the varaton of the acton and force t to vansh: t [ L δs = δq + L ] δ q + λ j δq dt = 0 q q q The prme on L does not mply dfferentaton, merely that ths s the augmented Lagrangan functon defned n Eq. (). Integrate the mddle term by parts (and remember we re usng the summaton conventon): t [ L δs = d ] L + λ j a j δq dt = 0 q dt q t 1 We now have N varatons δq, only N P of whch are ndependent (snce the system has only N P degrees of freedom). Usng the P ndependent Lagrange multplers, we may ensure that all N terms nsde the square bracket vansh, so that no matter what the varatons δq the value of the ntegral doesn t vary. Thus we have L = λ j a j (N equatons) dt q q a j dq + a j t dt = 0 or a j q + a j t = 0 (P equatons) or, f you prefer, d dt ( L dq + dt = 0 q t q ) L = λ j q q or q + q t (N equatons) = 0 (P equatons) Physcs of 6 Peter N. Saeta
4 Snce there are N generalzed coordnates and P Lagrange multplers, we now have a closed algebrac system. When we derved Lagrangan mechancs startng from Newton s laws, we showed that d T T = F tot = F tot r dt q q q where F s the total force on the partcle and F tot s the generalzed force correspondng to the th generalzed coordnate. If we separate the forces nto those expressble n terms of a scalar potental dependng only on postons (not veloctes), the forces of constrant, and anythng left over, then ths becomes L = F constrant +F noncons (3) dt q q Comparng wth the N Lagrange equatons above, we see that when all the forces are conservatve, Summed over j. F constrant = F constrant r = λ j a j = λ j q q In other words, the sum λ j q s the generalzed constrant force. Example 1 y θ A hoop of mass m and radus R rolls wthout slppng down a plane nclned at angle α wth respect to the horzontal. Solve for the moton, as well as the generalzed constrant forces. Usng the ndcated coordnate system, we have no moton n y, but coordnated moton between x and θ, whch are lnked by the constrant condton R dθ = dx or R dθ dx = 0 Therefore, a 1θ = R, a 1x = 1, a 1t = 0 x α The knetc energy s T = m ẋ + mr θ and the potental energy s V = mg x snα, so the Lagrangan s L = T V = m ẋ + mr θ + mg x snα We wll frst solve by usng the constrant equaton to elmnate θ: R θ = ẋ, so L = m ẋ + m ẋ + mg x snα = mẋ + mg x snα Physcs of 6 Peter N. Saeta
5 Ths Lagrangan has a sngle generalzed coordnate, x, and thus we obtan the equaton of moton L dt ẋ x = 0 mẍ mg snα = 0 ẍ = g snα whch s half as fast as t would accelerate f t sld wthout frcton. If we delay the gratfcaton of nsertng the constrant and nstead use the lagrangan wth two generalzed coordnates, we get L x = λ 1a 1x = mẍ mg snα = λ 1 ( 1) dt ẋ ( dt θ ) L θ = λ 1a 1θ = mr θ 0 = λ1 R From the second equaton, we obtan λ 1 = mr θ = mẍ, where I have used the constrant equaton R θ = ẍ n the last step. Substtutng nto the frst equaton, we agan obtan ẍ = g / snα. What about the constrant forces? The generalzed constrant force n x s F x = λ 1 a 1x = mẍ = mg snα. Ths s the force headng up the slope produced by frcton; t s responsble for the slowed moton of the center of mass. The generalzed constrant force n θ s F θ = λ 1 R = mrẍ = mg R snα. Ths s the torque about the center of mass of the hoop caused by the frctonal force. Summary When you wsh to use redundant coordnates, or when you wsh to determne forces of constrant usng the Lagrangan approach, here s the recpe: 1. Wrte the equatons of constrant, G j (q, t) = 0, n the form where a j = q. a j dq + a j t dt = 0. Wrte down the N Lagrange equatons, L = λ j a j dt q q (summaton conventon) where the λ j (t) are the Lagrange undetermned multplers and F = λ j a j s the generalzed force of constrant n the q drecton. 3. Solve, usng the N Lagrange equatons and the P constrant equatons. 4. Compute the generalzed constrant forces, F, f desred. Physcs of 6 Peter N. Saeta
6 Problem 1 Use the method of Lagrange undetermned multplers to calculate the generalzed constrant forces on our venerable bead, whch s forced to move wthout frcton on a hoop of radus R whose normal s horzontal and forced to rotate at angular velocty ω about a vertcal axs through ts center. Interpret these generalzed forces. What do they correspond to physcally? Physcs of 6 Peter N. Saeta
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