PHYS 601 HW3 Solution
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1 3.1 Norl force using Lgrnge ultiplier Using the center of the hoop s origin, we will describe the position of the prticle with conventionl polr coordintes. The Lgrngin is therefore L = 1 2 ṙ r2 θ2 gr cos θ. (1) However, since the prticle is fixed to the hoop (until it flls off), r is lwys equl to the rdius of the hoop (R), nd the first ter in the kinetic energy drops out. Our constrint eqution is then F = r R = 0. Using Euler-Lgrnge: ( ) d L dt θ = L (2) θ r 2 θ = gr sin θ (3) θ = g sin θ (4) r ( ) d L = L dt ṙ r 0 = r θ 2 g cos θ + λ (6) θ 2 = g cos θ λ/r (7) r To solve for λ, we plug the first eqution of otion into the derivtive of the second eqution: 2 θ θ = g r sin θ θ λ/r (8) 2 θ g r sin θ = g r sin θ θ λ/r (9) 3 θ g r sin θ = λ/r (10) λ = 3g sin θ θ (11) λ = 3g cos θ + C, (12) where C is n integrtion constnt. Plugging λ bck into our eqution of otion, we find C = 2g. Thus, when the prticle flls of the hoop (when F r = λ F r = 0): λ = 3g cos θ 2g = 0 cos θ = 2/3, (13) nd the height is given by r cos θ + R = 5R/3 esured fro the botto of the loop. (5) 3.2 Cnonicl oentu electrognetic forces Assue the Lgrngin tkes the for L = T ( q; t) U( q, q; t), (14) where T is the regulr velocity dependent kinetic energy, but the potentil energy U is now lso dependent on velocity. 1
2 The cnonicl oentu is p j = L q j, (15) given by Goldstein Eq. (2.44). Applying our Lgrngin to this eqution: p θ = L T U = θ θ θ = L θ U, (16) θ where L θ is the echnicl ngulr oentu T, in the cse the potentil ws not dependent θ on velocity. We then expnd the second ter using prtils of generl coordinte vectors (chin rule): U θ = i U ṙ i. (17) ṙ i θ Fro Goldstein Eq. (2.50), we get ṙ i θ = r i θ = n r i. (18) With Eqution (17) nd Eqution (18), we rewrite the cnonicl oentu s p θ = L θ U θ = L θ i = L θ i U ṙ i n r i (19) vi U n r i (20) = L θ i n r i vi U. (21) For n electrognetic force, the totl potentil is U = eφ ea v = eφ j e j A j v j Goldstein Eq. (1.62), (22) where φ is the sclr potentil, A the vector potentil, nd e the electric chrge; j denotes r j, the se generl coordinte syste s used in the first prt. Applying this to Eqution (21): p θ = L θ i n r i vi U = L θ i n r i vi eφ j e j A j v j (23) (24) = L θ i n r i ṙ i j e j A j v j (25) = L θ + i n r i e i A i ṙ i ṙ i, (26) since i nd j re fro the se coordinte syste, r i nd r j (v j ) re orthogonl unless i = j. is the unit vector in the ṙ i direction. We rewrite this with few ore chnges, i.e., chnging e i to ṙ q i /c for Gussin units, nd cobining A i i ṙ i into one vector A i : ṙ i ṙ i p θ = L θ + i n r i q i c A i. (27) 2
3 3.3 Constnt of otion for syste with constrint In 3D sphericl coordintes, the kinetic energy of prticle is T = 1 2 ṙ r2 θ r2 sin 2 θ φ 2. (28) However, since both r (= ) nd φ (= ω) re constnt for the given setup, our kinetic energy reduces to T = θ sin 2 θω 2, (29) where would be the ss of the point ss. And since the potentil energy is only due to grvity, it is U = g( + cos θ), (30) if we chose the ss being t the botto of the hoop s zero potentil energy. Thus, our Lgrngin is L = T U = θ sin 2 θω 2 g( + cos θ). (31) This Lgrngin does not explicitly depend on tie, so Hiltonin is conserved quntity. p θ = L θ = 2 θ (32) 1 H = p θ θ L = 2 2 θ sin 2 θω 2 + g( + cos θ). (33) Note tht this conserved quntity H is not the se s T + U. Applying Euler-Lgrnge s eqution to obtin the eqution of otion: ( ) d L dt θ = L θ (34) 2 θ = 2 sin θ cos θω 2 + g sin θ (35) θ = ω 2 sin θ cos θ + g sin θ (36) θ ( = sin θ ω 2 cos θ + g ). (37) Fro this eqution of otion, we see tht there re two sets of equtions for equilibriu ( θ = 0). The ss will be sttionry on the hoop if 1) sin θ = 0 (38) 2) ω 2 cos θ + g = 0. (39) The first solution is for when the ss is either on the very top or the botto of the hoop. Since cos(θ) cn only hve vlue between 1 nd 1, ω 2 hs to be greter tht g/ for there to be solution. Thus the criticl vlue of ω is given by: ω 2 > g for there to be solution (40) ω 0 = g. (41) 3
4 This gives us three possible sttionry points, θ 1 = 0, θ 2 = π, nd θ 3 = cos 1 ( g/ω 2 ), with θ 3 only existing if ω > ω 0. To deterine the stbility of ech solution, we will consider the Tylor expnsion of our eqution of otion with respect to sll chnge of δ in θ. δ = ω 2 sin(θ + δ) cos(θ + δ) + g sin(θ + δ) ω 2 (sin θ + δ cos θ) (cos θ δ sin θ) + g (sin θ + δ cos θ). We first consider θ 1 = 0, the point t the top of the hoop, nd find δ = ω 2 δ + g ( δ = ω 2 + g ) δ. (42) Thus, ny chnge in the ngle θ 1 will cuse n ccelertion wy fro equilibriu, indicting θ 1 is unstble. Next, we consider θ 2 = π, the botto of the hoop, δ = ω 2 δ g ( δ = ω 2 g ) δ. (43) If the coefficient is positive, which occurs when ω > ω 0, we hve the se sitution s θ 1 nd hve n unstble equilibriu. If ω < ω 0, then the coefficient is negtive nd we hve restoring force, ening θ 2 is stble. Finlly, consider θ 3 = cos 1 ( ω0 2/ω2 ), which fter siplifiction (using Mthetic) nd throwing wy higher order ters hs the expnsion ( ) ω δ = ω ω 4 1 δ. (44) This solution only exists when ω > ω 0, ening the coefficient will be negtive. This indictes restoring force, nd the solution is stble. To surize, when ω < ω 0, the sttionry points re t the top of the hoop (unstble) nd the botto of the hoop (stble). When ω > ω 0, the sttionry points re now the botto of the hoop (unstble) nd θ = cos 1 ( ω 2 0 /ω2 ) (stble). 3.4 Conserved quntities for two-diensionl hronic oscilltor, including one not ssocited with Noether s theore (i) If we perfored Legendre trnsfortion on L, we would get the Hiltonin: H = 1 2 (p2 x + p 2 y) ω2 (x 2 + αy 2 ) = E. (45) Since we know tht dh dt = H t, if H t = 0, dh dt = 0, which is the cse in this proble. Therefore, since the Hiltonin is not explicitly dependent on tie, the totl energy does not chnge over tie, nd is conserved. Since tie trnsltion is the syetry of the lgrngin, the corresponding conserved quntity is energy. 4
5 (ii) Liner oentu is not conserved, becuse x nd y both show up in the Lgrngin, indicting tht they re not ignorble coordintes. To verify, we cn use Lgrnge s equtions to find d dt p x = L x = ω2 x, d dt p y = L y = αω2 y. The liner oent re therefore only conserved in the trivil cses of no otion. This lso indictes tht x nd y trnsltion is NOT the syetry of the Lgrngin. (iii) To better represent ngulr oentu, we ke the trnsfortion into polr coordintes, resulting in: H = 1 2 (p2 r + p 2 θ ) ω2 (r 2 sin 2 θ + αr 2 cos 2 θ). (46) Angulr oentu is conserved if H θ ṗ θ = H θ = 0. For this syste, = r2 2 ω2 (sin θ cos θ α sin θ cos θ), (47) which would be zero only if α = 1. This lso indictes tht rottion is NOT the syetry of the Lgrngin unless α = 1. (iv) We hve the equtions of otion ẍ + ω 2 x = 0, ÿ + αω 2 y = 0. If we ultiple the first eqution by ẋ nd the second by ẏ nd then dd the two together, we hve d dt ẍẋ ÿẏ + ω 2 xẋ αω 2 yẏ = 0 ( 1 2 ( ẋ 2 ẏ 2) + 1 ( 2 ω2 x 2 αy 2) ) = 0 The quntity = 1 2 ( ẋ 2 ẏ 2) ω2 ( x 2 αy 2), is therefore conserved. Noether Theore tells us tht every continuous syetry will correspond conserved quntity. However, not ll conserved quntities cn coe fro continuous syetry. This proble shows us such n exple. Other thn continuous syetries, there re soe other discrete syetries of the Lgrngin such s x x nd y y. 3.5 Two sses connected to three springs 5
6 The Lgrngin is given by L = 1 2 ẋ ẋ kx k(x 1 x 2 ) kx2 2, where x 1/2 re displceents wy fro the equilibriu points. Set η = (x 1, x 2 ) T, we got fro eqution of otion tht: The eigenfrequencies cn be solves s η = 4k 3k 3k 4k η ω 2 1 = k, ω2 2 = 7k Norl ode 1: ω 2 1 = k ( ) ( ) k 4 3 A = k 3 4 B = 1 ( ) 1 B 2 1 B Norl ode 2: ω 2 1 = 7k ( ) ( ) k 4 3 A = 7k 3 4 B = 1 ( ) 1 B 2 1 B 6
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