Classical Mechanics. Jung Hoon Han

Transcription

1 Classcal Mechancs Jung Hoon Han May 18, 2015

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3 Contents 1 Coordnate Systems Orthogonal Rotaton n Cartesan Coordnates Curved Coordnates Problems Lagrangan and Hamltonan Lagrangan Hamltonan Problems Symmetres and Conservaton Laws Vew from Newtonan Mechancs Vew from Lagrangan Mechancs Vew from Hamltonan Mechancs Problems One-dmensonal Moton Frst-ntegral Approach Effectve One-Dmensonal Moton Problems Harmonc Oscllaton Coupled Harmonc Oscllaton Generalty One-dmensonal Chan wth Identcal Mass One-dmensonal Chan wth Two Masses Anharmonc Oscllaton Moton n a Smooth Potental and Rapdly Oscllatng Feld Problem

4 4 CONTENTS 6 Rotaton of Rgd Bodes Knematcs Euler s equaton Problem Frenkel-Kontorova Model Nearly dentcal perods Commensurate perods Nearly commensurate perods Fnals Project Problems Oscllaton Others

5 Chapter 1 Coordnate Systems 1.1 Orthogonal Rotaton n Cartesan Coordnates Analytc descrpton of classcal moton rests on the defnton of coordnate system wthn whch the moton s descrbed. The most frequently used one s the Cartesan system, labeled by the three real numbers ndcatng the component of a vector along the x, y, and z axes. Mathematcally, we wrte a vector n terms of ts components as v = v xˆx + v y ŷ + v z ẑ. (1.1) The meanng of the above notaton s that a vector v s characterzed by three numbers (v x, v y, v z ) each expressng the amount of steps taken from the orgn along a partcular axs to form that vector. The three axes, {ˆx, ŷ, ẑ} are assumed to be orthogonal, ˆx ŷ = ŷ ẑ = ẑ ˆx = 0. (1.2) The drecton vectors themselves are of course vectors, whose lengths are taken to be unty: ˆx ˆx = ŷ ŷ = ẑ ẑ = 1. In ths way, we arrve at an orthonormal system of bass vectors, ê α, where α = 1, 2, 3 run over the three unt vectors wth the property ê α ê β = δ αβ. (1.3) We use the Kronecker delta symbol δ αβ for a functon +1 when α = β, and 0 when α β. One can easly magne the coordnate frame rotated as a rgd body, to arrve at a new set of bass vectors {ê α}. In the new frame, the same vector v must be assgned dfferent set of coordnates v α, v = α v α ê α = α v αê α. (1.4) 5

6 6 CHAPTER 1. COORDINATE SYSTEMS Furthermore, the two set of components {v α } and {v α} must be related to each other. To see ther relatonshp, take the nner product of both sdes wth ê β, one of the unt vectors n the orgnal coordnate frame. By utlzng the Kronecker relaton, we see that the l.h.s. only gves v β, whch the r.h.s. gves a set of numbers added together α v α(ê α ê β ). The nner product of unt vectors n the new and the old coordnate frames are called drecton cosnes: ê α ê β = λ βα. Now the relaton s neatly summarzed as v β = α λ βα v α. (1.5) If we arrange the components v β as a column vector v = v 1 v 2 v 3 and λ βα as a λ 11 λ 12 λ 13 matrx, λ = λ 21 λ 22 λ 23, Eq. (1.5) s nothng but a matrx equaton λ 31 λ 32 λ 33 v = λv. (1.6) Components of the same vector n the new and old coordnate frames are related by a lnear matrx relatonshp lke the one shown above. If we take the nner product of Eq. (1.4) wth respect to ê β, we would arrve nstead at v α (ê α ê β) = v β. (1.7) α Invokng the defnton of drecton cosnes, t becomes α v αλ αβ = v β, or v = λ T v. (1.8) The uppercase T denotes the transpose of a matrx. On the other hand, we can obtan the same relatonshp by multplyng Eq. (1.6) by the nverse matrx of λ: v = λ 1 v. (1.9) By comparng the two equatons, (1.8) and (1.9) we are forced to conclude that λ obeys a specal property λ T = λ 1. (1.10) The transpose of the drecton cosne matrx s ts nverse. Such matrces are known as orthogonal matrces. The condton λ T λ T = 1 mples λ j λ k = δ jk. (1.11)

7 1.1. ORTHOGONAL ROTATION IN CARTESIAN COORDINATES 7 It looks somewhat odd, but n fact there s a well-defned geometrc meanng to ths dentty. Note that λ j = ê ê j. Substtutng, we realze that the dentty above means (ê ê j)(ê ê k) = δ jk. (1.12) Where does ths relaton come from? Snce ê j = (ê j ê )ê, and ê k = (ê k ê )ê, ther nner product, ê j ê k, becomes (ê j ê )(ê k ê ). On the other hand, ê j ê k = δ jk, so we obtan Eq. (1.11). Once we know the drecton cosne matrx λ, we can map the coordnates from one frame to any other by the lnear multplcaton of the matrx. In two dmensons the drecton cosne matrx s 2 2 gven by ) (ê1 ê λ = 1 ê 1 ê 2 ê 2 ê 1 ê 2 ê = 2 We arrve at the famlar results ( cos θ sn θ sn θ cos θ ). (1.13) x = x cos θ y sn θ y = y cos θ + x sn θ. (1.14) Determnant of a transpose λ T s the same as the determnant tself, Det(λ) = Det(λ T ). On the other hand, determnant of an nverse matrx s the nverse of the determnant: Det(λ 1 ) = 1/Det(λ). Snce λ T = λ 1, we must conclude Det(λ) = 1/Det(λ), or Det(λ) = ±1. Those frame changes wth Det(λ) = +1 are called proper rotatons, whle those wth Det(λ) = 1 are mproper. A notable example of mproper frame change s an nverson, whch takes one of the unt vectors to ts mnus. For nstance, ˆx ˆx, ŷ ŷ, ẑ ẑ s an nverson. The drecton cosne matrx s λ = Dag( 1, 1, 1) and ts determnant s 1. Smlarly, reversng all three axes (ˆx ˆx, ŷ ŷ, ẑ ẑ) s mproper, but reversng only two of the axes s proper (ˆx ˆx, ŷ ŷ, ẑ ẑ). Indeed, the last example s smply the rotaton about the ẑ-axs by the angle π. Successve changes of the coordnate frames are represented as a product of two orthogonal matrces λ 1 λ 2. If we mplement the rotatons n the opposte order, we generally get a dfferent result: λ 2 λ 1 λ 1 λ 2. Objects whch do not coommute are sometmes called non-abelan. Matrces are examples of non- Abelan objects. If we restrct ourselves to rotatons about the same axs, say the ẑ-axs, then the matrces begn to commute agan. Wrte λ(θ) for a rotaton about the ẑ-axs by angle θ, then one can easly show by explct matrx product calculaton that λ(θ 1 )λ(θ 2 ) = λ(θ 1 + θ 2 ) = λ(θ 2 )λ(θ 1 ). (1.15) The subset of rotaton matrces about the common axs s Abelan. Two vectors A and B can be combned to produce a scalar (by takng nner product c = A B) or another vector (by takng an outer product C = A B). Let s examne how the two quanttes transform under the orthogonal rotaton.

8 8 CHAPTER 1. COORDINATE SYSTEMS In the new frame, we have c = A B. Accordng to the transformaton rule A = λa and B = λb, the new nner product becomes For explct proof, c = (λa) (λb) = (A B) = c. (1.16) c = A B = (R j A j )(R k B k ) = (R j R k )A j B k = δ jk A j B k = A j B j = A B. (1.17) The nner product remans unchanged under the coordnate rotaton. Such quanttes are called scalars. On the other hand, C = A B s gven by C = (λa) (λb) = λ(a B) = λc. (1.18) Gven the transformaton property, C s a vector, whch of course t s. To prove t, frst work out both λ(a B) and (λa) (λb): [λ(a B)] = λ j (A B) j = λ j ε jmn A m B n, [(λa) (λb)] = ε jk (λa) j (λb) k = ε jk λ jm λ kn A m B n. (1.19) To prove the equalty we need to show λ j ε jmn = ε jk λ jm λ kn. Multplyng both sdes by λ p and summng over gves ε pmn = ε jk λ p λ jm λ kn. (1.20) The r.h.s. s the determnant of a matrx consstng of three columns p, m, n of the λ matrx. If any two columns concde, the determnant s zero. For all three columns non-overlappng, we get ether +1 or -1 dependng on the order the column vectors are arranged. Proof s complete. Two useful vector denttes are lsted below. A (B C) = B(A C) C(A B), (A B) (C D) = (A C)(B D) (A D)(B C). (1.21) The frst one can be proven wth the dentty ε jk ε lm = δ jl δ km δ jm δ kl. Usng smlar deas, one can also show (B C) = B( C) C( B) + (C )B (B )C, ( C) = ( C) 2 C. (1.22)

9 1.2. CURVED COORDINATES Curved Coordnates Sometmes we choose to descrbe the moton of a partcle n cylndrcal, or sphercal coordnates. Instead of three orthonormal bass vectors whch are fxed, the new coordnate frames employ bass vectors that are co-movng wth the partcle s trajectory tself. In cylndrcal coordnates, the partcle poston r s expressed as r = rˆr + zẑ, (1.23) where the radal unt vector ˆr follows from the defnton ˆr = (cos θ, sn θ). (1.24) The relaton (x, y) = r(cos θ, sn θ), where r = x 2 + y 2 s the radal dstance of the partcle measured from the orgn, easly establshes the dentty of the cylndrcal coordnate representaton of the poston vector r wth that of the Cartesan system. Snce the moton along the z-axs s trval, we wll confne ourselves to the two-dmensonal moton n whch r = rˆr. The translaton of the partcle s coordnate from Cartesan to cylndrcal system s purely knetc. In other words they are just two mathematcally equvalent ways of wrtng the same poston vector r. When we come to dscuss Newtonan dynamcs t s mportant to know how to express both the velocty and the acceleraton vectors, gven as frst and second dervatves of the coordnate r respectvely, n cylndrcal coordnates. Velocty and acceleraton vectors n the Cartesan coordnate system are trval to obtan as ṙ = ẋˆx + ẏŷ + żẑ, r = ẍˆx + ÿŷ + zẑ, (1.25) due to the fact that bases vectors ˆx, ŷ, ẑ are ndependent of tme. It s no longer the case that the bass vectors would reman tme-dependent n other coordnate systems. When we take the tme dervatve of r to obtan the velocty, not only do we need to worry about ṙ but also the tme dervatve of ˆr. So let s calculate t. d dt ˆr = d r dt r = ṙ r ṙr r 2. (1.26) Ths looks awkward. In fact there s a better way to express the tme dervatve of ˆr, f we frst recall the dentty d(ˆr ˆr)/dt = d(1)/dt = 0 = 2ˆr ˆr. So whatever ˆr s, t ought to be orthogonal to the orgnal vector ˆr. In two dmensons, there s only one such drecton. We call t ˆθ = ( sn θ, cos θ). (1.27)

10 10 CHAPTER 1. COORDINATE SYSTEMS But stll t s not very easy to prove, based on the above expresson for dˆr/dt, that t s orthogonal to ˆr and parallel to ˆθ. So we choose a dfferent strategy and go back to the coordnate expressons of ˆr and ˆθ, whch were ˆr = (cos θ, sn θ), ˆθ = ( sn θ, cos θ). (1.28) Now t s qute obvous by drect dfferentaton that dˆr dt = θˆθ, dˆθ dt = θˆr. (1.29) Now the velocty s calculated easly: v = ṙˆr + r dˆr dt = ṙˆr + r θˆθ. (1.30) Repeatng the same steps, one can derve the acceleraton vector n the cylndrcal coordnates as well: a = ( r r θ 2 )ˆr + (r θ + 2ṙ θ)ˆθ. (1.31) In the case of central forces, the acceleraton has only the radal component whle the tangental component must vansh. The vanshng quantty can be wrtten as r 2 θ + 2rṙ θ = d dt (r2 θ) = 0. (1.32) Ths s just a re-statement of Kepler s second law of planetary moton, or the conservaton of angular momentum. Another popular coordnate system used n the physcal descrpton s the sphercal coordnates. Here all three bass vectors gven below are co-movng wth the partcle s locaton. ˆr = (sn θ cos φ, sn θ sn φ, cos θ) ˆθ = (cos θ cos φ, cos θ sn φ, sn θ) ˆφ = ( sn φ, cos φ, 0). (1.33) The second unt vector ˆθ s obtaned from the frst by dfferentaton, θˆr = ˆθ. The thrd one s obtaned as the cross product of the frst two: ˆφ = ˆr ˆθ. Tme dervatve of each unt vector can be expressed as the lnear combnaton of the other two snce t has no component along ts own bass vector. Startng from r = rˆr, the velocty vector and the acceleraton vector can be derved as v = ṙˆr + r θˆθ + r sn θ φ ˆφ, a = ( r r φ 2 sn 2 θ r θ 2 )ˆr + (r θ + 2ṙ θ r φ 2 sn θ cos θ)ˆθ + (r φ sn θ + 2ṙ φ sn θ + 2r θ φ cos θ) ˆφ. (1.34)

11 1.3. PROBLEMS Problems 1. Prove Eq. (1.15). 2. Prove Eq. (1.16). 3. Prove Eq. (1.18). 4. Prove Eq. (1.21). 5. Prove Eq. (1.22). 6. Prove Eq. (1.31). 7. Prove Eq. (1.34). You wll frst have to show how to express ˆr, ˆθ, and ˆφ n terms of ˆr, ˆθ, ˆφ. 8. An earth revolves around the sun n a crcle of radus R s wth angular velocty ω s. On ts own, the earth of radus R e rotates about ts own axs wth angular speed ω e. Fnd the acceleraton at arbtrary pont on the surface of the earth, usng the sun as the orgn of the coordnate system. For smplcty take the axs of earth s rotaton to be perpendcular to the plane of earth s moton about the sun, nstead of beng tlted as t s n realty.

12 12 CHAPTER 1. COORDINATE SYSTEMS

13 Chapter 2 Lagrangan and Hamltonan 2.1 Lagrangan Newton s force law states m dv dt = F = V, (2.1) assumng a conservatve force F = V that can be derved from the scalar functon (called the potental energy) through the operaton of takng the gradent on t. Multplyng both sdes of Newton s equaton by v gves ( ) d 1 dt 2 mv2 = dr dt whch n turn mples the conservaton of total energy (r) V = dv, (2.2) dt d dt (K + V ) = 0, K = 1 2 mv2. (2.3) Startng from Newton s force law, whch s expressed n terms of vector quanttes, the prncple of energy conservaton s derved followng the steps outlned above. One mght say that n ths approach force s fundamental, whle energy s the concept derved from the underlyng noton of force. Another school of thought takes the energy as the more fundamental quantty, from whch Newton s law wll be derved accordng to some frm prncple of nature. Champons of such thoughts are Lebnz, Bernoull, Lagrange, and Hamlton, among others. To get a grasp of the Lagrangan approach to mechancs, t s essental to understand frst the noton of a functonal - a functon of a functon. Knetc energy, K = (m/2)ṙ 2, for nstance, s a functon of the velocty v. But snce the velocty tself s a functon of the trajectory r(t) whch depends on tme t, we may say more correctly that the knetc energy s a functon of the whole 13

14 14 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN hstory of the trajectory r(t) as a functon of tme t. The same argument can be made wth the potental energy V (r(t)). Let s wrte down the followng quantty, K = t2 t 1 K(ṙ(t))dt, V = t2 t 1 V (r(t))dt, (2.4) whch as we argued s a functonal of the trajectory, r(t). The dfference of these two, S = K V, called the acton, s also a functonal of the trajectory r(t). The ntegrand S = K V s called the Lagrangan. The acton s then the tme ntegral of the Lagrangan for a partcular chosen path r(t). Thnk of a partcular trajectory r(t) as a pece of rubber stretched n some way n space. Next consder a trajectory whch s just a slght dsplacement of the orgnal trajectory, r (t) = r(t) + δr(t). At each tme t, δr(t) s assumed an nfntesmal quantty. What s the dfference n the value of the acton S adopted by the two trajectores? Ths s bascally an enlarged verson of the queston: what s the dfference n the value of the functon f(x) when x s dsplaced to x + δx? Such queston led to the noton of dfferentaton as we know so well. A smlar queston can be posed for the varaton n path, and the dfference t causes n the functonal S. To answer the queston, smply consder the values of K for the two paths and subtract: δk = t2 t 1 [K(r + δr) K(r)]dt t2 t 1 mv d δr. (2.5) dt In gong to the last expresson we gnored terms of order (δr) 2. Integratng by parts, wth the boundary condtons δr(t 1 ) = 0 = δr(t 2 ), we arrve at A smlar exercse gves t2 δk = δr d (mv). (2.6) t 1 dt Varaton n the acton s δv = t2 t 1 δr V. (2.7) t2 ( ) d δs = δk + δv = dtδr (mv) + V. (2.8) t 1 dt The quantty nsde parenthess s exactly Newton s equaton. If the orgnal path we have chosen to vary by a tny amount happened to concde wth the actual, physcal trajectory obeyng Newton s law, the correspondng acton S wll have a vanshng dfference aganst the small varaton n the path. In other words, the acton wll acheve the extremal value for the path actually realzed n nature.

15 2.1. LAGRANGIAN 15 The sprt of the statement can be reversed: If a path s found such that t gves the extremal value of the acton, then such path would be the physcal path obeyng Newton s law. Hence, the entre classcal mechancs can be formulated n terms of the new quantty, the acton S, together wth the varatonal prncple. Ths we mght state as an equally compellng formulaton of the laws of classcal mechancs. We know that some problems n mechancs can be better tackled n curved coordnate frames. Sometmes t s better to descrbe the partcle s trajectores n terms of angles and dstances from the orgn. It s fttng that we enlarge the noton of coordnates to refer to any set of varables beng used to completely specfy the partcle s locaton at any gven tme. We call them q, = 1,, N. The quantty L = T V s also now a functon of the generalzed coordnates q and ther dervatves q : L(q, q ). Applyng the varatonal prncple for the acton wrtten n terms of the generalzed coordnates and ther dervatves results n δs = t2 t 1 ( L δq + L ) t2 ( L δ q dt = d q q t 1 q dt L q ) δq dt. (2.9) The fnal expresson s once agan obtaned by dong ntegraton by parts subject to the boundary condton δq (t 1 ) = δq (t 2 ) = 0. Assumng the varatonal prncple really works, we obtan a set of equatons governng the classcal moton as L d L = 0. (2.10) q dt q The Lagrange s equaton can be used n place of Newton s force equaton as the law of classcal mechancs. As an example of the Lagrangan and the varatonal prncple at work, let us begn wth the smplest example of one-dmensonal smple harmonc oscllator (SHO), K = (m/2)ẋ 2, V = (k/2)x 2. You can derve Newton s equaton mẍ = kx by applyng Lagrange s equaton n no tme. As a second, more challengng example, consder a smple pendulum of length l wth a mass m attached at ts end. The trajectory of a swngng mass spans the two dmensonal plane, but apparently one only needs one coordnate θ (angle from the vertcal poston) to descrbe the moton. Somehow the reducton of the number of coordnates from two to one has to be accounted for. Frst wrte down the Lagrangan of a two-dmensonal moton takng place n the xz plane, subject to a constant gravty F = gẑ. L = m 2 (ẋ2 + ż 2 ) + mgz. (2.11) Ths s not qute the whole story, snce the two coordnates are constraned by the condton x 2 + z 2 = l 2 whch mples, among other thngs, xẋ + zż = 0. One way to deal wth the constrant s to ntroduce a new coordnate system n whch the constrant s naturally removed. Take x = l sn θ, z = l cos θ, and ẋ = θ cos θ, ż = θ sn θ. The Lagrangan becomes

16 16 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN L = ml2 2 θ 2 + mgl cos θ. (2.12) The varatonal prncple then mmedately gves out the equaton of the pendulum moton: ml 2 g θ = mgl sn θ θ = sn θ. (2.13) l Restrcton to the small-ampltude oscllaton ( θ 1) wll recover the SHO: θ = ω 2 θ, ω 2 = g/l. Another way to solve the problem of constraned moton such as that of a pendulum s to use a trck that bear Lagrange s name, called the Lagrange multpler method. For the pendulum problem the constrant was smple enough that we could solve t rght away wthn the sutable new coordnates that dd away wth the constrant. The method wll not work, however, f the constrant got too complcated. In ths case the trck s to supplement the orgnal acton S by a term proportonal to the constrant. S = t2 t 1 dt ( m t2 2 (ẋ2 + ż 2 ) + mgz) + dt λ(x 2 + z 2 l 2 ). (2.14) t 1 The new varable λ, also tme-dependent, s to be regarded as an addtonal coordnate, to be vared n the same manner as x and z. Varaton of λ results n the Lagrange s equaton L λ = x2 + y 2 λ 2 = 0, (2.15) whch s nothng but the constrant. Now the constrant s mplemented as part of an equaton of moton. Other components of Lagrange s equatons are mẍ = 2λx, mÿ = 2λy + mg. (2.16) In physcal terms, λ s proportonal to the tenson exercsed by the strng on the mass. So far we have shown how to make the Lagrangan formulaton of classcal mechancs for constraned and non-constraned systems. The method can be appled even for non-nertal frames. Take a mass n two-dmensonal free space wth the Lagrangan L = m 2 (ẋ2 + ẏ 2 ). (2.17) Another observer sees the moton of ths free partcle n a rotatng frame wth angular velocty ω = (0, 0, ω). New coordnates are related to the old by x = x cos ωt y sn ωt, y = y cos ωt + x sn ωt. (2.18)

17 2.1. LAGRANGIAN 17 Veloctes are related by ẋ = ẋ cos ωt ẏ sn ωt ω(x sn ωt + y cos ωt) = (ẋ ωy ) cos ωt (ẏ + ωx ) sn ωt. ẏ = ẏ cos ωt + ẋ sn ωt + ω( y sn ωt + x cos ωt) = (ẏ + ωx ) cos ωt + (ẋ ωy ) sn ωt. (2.19) Lagrangan n the new frame reads L = m 2 [ (ẋ ωy ) 2 + (ẏ + ωx ) 2]. (2.20) Dong away wth the annoyng prmes, we can also wrte the Lagrangan n nce vector notaton: L = m 2 (ṙ + ω r)2. (2.21) On expandng the square, t appears as though the partcle s subject to a velocty-dependent potental V (r, ṙ) = (mω 2 /2)r 2 mṙ (ω r). From d dt ( L ṙ L r ) = m(ṙ ω ω (ω r)) = m( r + ω ṙ), (2.22) Euler-Lagrange equaton follows as r = ω (ω r) + 2ṙ ω. (2.23) The frst term on the r.h.s. s the centrfugal force, the second s the Corols force. They are both apparent forces caused by the non-nertal nature of the frame of reference. For rotaton about the ẑ axs we may wrte ω = ωẑ, so that the centrfugal force takes on ω (ω r) = +ω 2 r. (2.24) It s tryng to push the partcle outwards. A partcle at rest n an nertal frame appears to be dong a crcular moton r = R(cos ωt, sn ωt) (2.25) n a frame rotatng wth the angular velocty ω = +ωẑ. Such moton gves rse to the equaton of moton r = ω 2 r whle the centrfugal force contrbutes +ω 2 r to the acceleraton. The dscrepancy, equal to 2ω 2 r, s provded by the Corols term whch gves 2ṙ ω = 2ω 2 r n the case of constant crcular

18 18 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN moton. Furthermore, the Corols force resembles the Lorentz force actng on a charged partcle by the magnetc feld. Mathematcally, they are the same knd of forces, lnearly dependent on the partcle s velocty and orthogonal to t. Now that the subject of velocty-dependent force has been brought up, let s dscuss how to wrte down the Lagrangan for such forces when they are really there. Specfcally we wll try to wrte down the Lagrangan for the moton of a charged partcle n electrc and magnetc felds, and confrm that the correspondng Lagrange s equaton s none other than Lorentz equaton. The Lagrangan reads L = m 2 ṙ2 e(a 0 ṙ A), (2.26) where A 0 and A are the scalar and vector potentals, respectvely. To obtan Lagrange s equaton, L r = e A 0 + e (ṙ A) = d (mṙ + ea) = m r + eȧ. (2.27) dt Electrc (E) and magnetc (B) felds are obtaned from the vector potentals accordng to E = A 0 A, B = A. (2.28) t To cast the vector potentals as electrc and magnetc felds, frst express Rearrangng some terms, Ȧ = A t + (ṙ )A. (2.29) m r = ee + e (ṙ A) e(ṙ )A. (2.30) To recognze that the fnal two terms on the r.h.s. n fact equal the Lorentz force, (ẋ j A j ) ẋ j j A = ẋ j ( A j j A ) = ẋ j ε jk B k = (v B). (2.31) We have derved the equaton of moton of a charged partcle (wth charge e) n the presence of electrc and magnetc felds, as well as the Lagrangan whch produces t. 2.2 Hamltonan The Hamltonan approach s yet another way to reproduce the Newton s force law. Lke the Lagrangan formulaton, t can be easly generalzed to study quantum-mechancal laws as well as the laws of classcal physcs. As a physcst, one s requred to be famlar wth both approaches. Actually Hamltonan

19 2.2. HAMILTONIAN 19 formulaton bulds upon the Lagrangan formulaton n that t requres the defnton of a canoncal momentum p as the conjugate varable of the canoncal coordnate q, gven by the partal dfferental p = L q. (2.32) Instead of workng wth canoncal coordnates q and ts tme dervatves, we wll be workng wth q and ther conjugate momentum p. Once p s defned n the above manner, we can ntroduce the quantty, Hamltonan, as H[q, p ] = p q L[q, q ]. (2.33) Summaton over the repeated ndex s assumed. Each q s to be vewed as some combnaton of p s and q s. Let s take the dervatve of H wth respect to q and p and see what we get. H q j = p j L q ( j L = p j L ) qj L. (2.34) q q q q q j q j q q The terms nsde the parenthess become zero due to the defnton of p j, whle the last term accordng to the Euler-Lagrange equaton equals the tme dervatve of p, hence H/ q smply equals ṗ. Dfferentaton for p gves H q j = q + p j L ( q j = q + p j L ) qj = q. (2.35) p p q j p q j p In Hamlton s theory, Euler-Lagrange equatons are replaced by a par of equatons that read q = H p, ṗ = H q. (2.36) Except for the relatve mnus sgn, p and q appear n a symmetrcal fashon. Let s apply the Lagrangan Hamltonan transformaton rule just worked out to a smple problem. Startng from the Lagrangan L = (m/2)ṙ 2 V (r), Hamltonan follows by frst wrtng down the canoncal momentum, p = L/ ṙ = mṙ, and then usng t to wrte q = p/m, ( p H = p m) p2 p2 + V (r) = + V (r). (2.37) 2m 2m For a physcal trajectory, Hamltonan s nothng other than the total energy E. The observaton that H plays the role of the total energy can be generalzed to the case n whch the Lagrangan s gven n terms of a generalzed set of coordnates q and veloctes q. Wth an assumpton that the knetc energy K(q, q ) s the only place where the dependence on the velocty q occurs, we can conclude

20 20 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN p L q = K q. (2.38) Furthermore, we wll assume that K s some quadratc functon of the veloctes, gven by K = j K j q q j. It s straghtforward to show that for arbtrary matrx of coeffcents K j one has q K/ q = 2K, whch gves H = q p L = 2K (K V ) = K + V. (2.39) The fnal expresson equals the total energy. The proof does not work for cases of velocty-dependent potental as occurs n electrodynamcs. From the sngle-partcle Hamltonan (2.37) we derve the Hamlton s equatons ṙ = H p = p m, ṗ = H r = V (r). (2.40) The frst equaton tells us that the momentum p s gven by the famlar product of mass and velocty: mṙ. The second equaton s the Newton s second law. Equaton (2.36) tells us how the two canoncal varables p and q evolve over tme. All other dynamcal varables are n turn functons of p s and q s. Denote such a dynamcal varable as f(p, q, t) where an explct tme dependence s allowed as well. The tme evoluton of f s governed by the equaton df dt = f t + = f t + ( f ṗ + f q p q ) ( f p H q + f q H p ). (2.41) The fnal expresson n the second lne has a neat structure known as the Posson bracket, whch s defned between a par of varables f and g as [f, g] = ( f g f ) g. (2.42) p q q p Usng the Posson bracket we can wrte the tme evoluton of the varable f as df dt = f + [H, f]. (2.43) t Ths s the generalzaton of the Hamlton s equatons for p and q frst found n Eq. (2.36). In partcular Posson brackets for the generalzed coordnates and momenta are [q, q j ] = 0 = [p, p j ], [p, q j ] = δ j. (2.44)

21 2.2. HAMILTONIAN 21 In the prevous secton we ntroduce the Lagrangan of a partcle subject to an electromagnetc potental L = m 2 ṙ2 e(a 0 ṙ A). (2.45) Here the canoncal momentum follows p = mṙ + ea. It s no longer smply mṙ. Takng mṙ = p ea everywhere, we get H = p p ea m = (p ea)2 + ea 0 ea p ea 2m m 1 2m (p ea)2 + ea 0. (2.46) When the magnetc feld s unform B = (0, 0, B), the vector potental can be chosen as A = ( By, 0, 0) gvng us the Hamltonan H = 1 2m (p x + eby) 2 + p2 y 2m. (2.47) There are four, frst-order equatons followng from ths Hamltonan. Two of them are ṗ x = 0, ṗ y = eb m (p x + eby). (2.48) We wrte the constant p x as eby 0, and ntroduce a Larmor frequency ω L = eb/m. Then The other two equatons are ṗ y = mω 2 L(y y 0 ). (2.49) ẋ = 1 m (p x + eby) = ω L (y y 0 ), ẏ = p y m. (2.50) By comparng the above two equatons we also learn ṗ y + mω L ẋ = 0, whch means p y = mω L (x 0 x). The only two ndependent equatons left to solve can be chosen as ẋ = ω L (y y 0 ), ẏ = ω L (x x 0 ). (2.51) They are solved as x x 0 = A cos(ω L t + α), y y 0 = A sn(ω L t + α). In quantum mechancs, the Larmor frequency becomes known as the cyclotron frequency. The Landau level spacng becomes h tmes the cyclotron frequency. The same problem can be solved a lttle more formally by use of the Posson brackets. Defnng Π = p + ea, one can re-wrte the Hamltonan

22 22 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN H = 1 2m (Π2 x + Π 2 y). (2.52) Frst one can show that the Posson bracket for Π x and Π y s [Π x, Π y ] = eb. (2.53) Next one works out the equaton of moton for Π x and Π y, Π x = [H, Π x ] = eb m Π y, Π y = [H, Π y ] = eb m Π x. (2.54) The moton s that of a crcular moton wth the angular frequency ω = eb/m. 2.3 Problems 1. Prove the energy conservaton law for mult-partcle systems governed by m r = F, (2.55) where = 1,, N runs over all the partcles each carryng a mass m. The force F actng on the -th partcle s derved from the par-wse nteracton energy V = <j V (r r j ). 2. Take the acton of a one-dmensonal free partcle L = t 2 t 1 (m/2)ẋ 2 dt. We know the physcal moton s that of constant velocty x p (t) = L t t 1 t 2 t 1 (2.56) consstent wth boundary condtons x(t 1 ) = 0, x(t 2 ) = L. Suppose an alternatve path to the physcal one had been chosen, ( x(t) = x p (t) + η sn π t t ) 1. (2.57) t 2 t 1 Prove that the new path x(t) gves more acton than the physcal path x p (t). 3. A pendulum of length l and mass m has ts pont of support attached to a block of mass M that s free to slde horzontally. Coordnates of each mass are gven by (X, 0) and (x, y), wth the constrant (x X) 2 +y 2 = l 2. Wrte down the Lagrangan for ths system and derve the Euler-Lagrange equaton for X and θ, the angle of the pendulum swng.

23 2.3. PROBLEMS A partcle of mass m moves n the plane subject to a central force feld wth the potental energy V (r). Wrte down the correspondng Lagrangan n the cylndrcal coordnates and derve the Euler-Lagrange equatons of moton. 5. Prove Eq. (2.22). 6. Based on Eq. (2.23), what can an observer n a rotatng frame conclude about the conservaton of knetc energy E = (m/2)ṙ 2? Is t conserved? If not, what s the tme dependence of the energy de/dt? 7. Derve the equaton of moton startng from the Lagrangan for the relatvstc partcle L = mc 2 1 ṙ ṙ/c 2 V (r). Show that t reduces to Newton s law n the lmt ṙ c. 8. Prove Eq. (2.44). 9. A sphercal pendulum of length l has a mass m attached at ts end whch s free to swng n three dmensons. Its poston s descrbed by two angles of the sphercal coordnates (θ, φ). Wrte down the Lagrangan and derve the equaton of moton. There s a constant force actng vertcally, F = mgẑ. We wll get back to the analyss of the derved equaton of moton n a later chapter. 10. Derve the Hamltonan of a smple harmonc oscllator startng from the Lagrangan L = (m/2)ẋ 2 (k/2)x Startng from the Lagrangan for the partcle of mass m n a central force feld V (r) n two dmensons, derve the correspondng Hamltonan H, then derve Hamlton s equaton of moton. 12. Startng from the Lagrangan (2.21) n the rotatng frame, derve the Hamltonan for the free partcle n the rotatng frame of constant angular velocty ω. 13. Angular momentum vector L = (L x, L y, L z ) s defned as L = r p. Derve the Posson bracket relaton among the components: [L x, L y ] = L z and ts cyclc permutatons. Further prove that [L 2, L ] = 0 for = x, y, z. Both relatons have ther analogues n quantum-mechancal commutators. 14. Show that any functon of the radus r = r, φ(r), has a Posson bracket wth L z equal to zero: [φ(r), L z ] = 0.

24 24 CHAPTER 2. LAGRANGIAN AND HAMILTONIAN

25 Chapter 3 Symmetres and Conservaton Laws 3.1 Vew from Newtonan Mechancs A partcle n moton subject to a conservatve force, F = V has ts total energy, (m/2)ṙ 2 + V conserved throughout the moton. We showed ths n the prevous chapter. In turn, t says that any non-conservatve system does not have the luxury of mantanng the constant energy. The smplest way to ntroduce non-conservatve forces s to allow explct tme dependence n the force. For nstance, take the gravtatonal force exerted by one object on another, V (r) = k/r. Suppose the source of gravtatonal force at the orgn s shaken n a perodc manner by some external means. The relatve poston vector r wll also pck up the oscllatory dependence and becomes r r 0 (t). The potental V (r r 0 (t)) s no longer energy-conservng even though one stll has To see why, multply both sdes by ṙ, m r = V (r r 0 (t)). (3.1) d ( m dt 2 ṙ2) = dr dt V (r r 0(t)) = dv dt + V t. (3.2) Now the total energy E = K + V obeys de dt = V t = ṙ 0 V, (3.3) whch clearly shows the lack of energy conservaton for tme-dependent potental. The examples shows that energy conservaton s closely ted to the tme(t)-ndependence of the system, or the sotropy of tme. Angular momentum of a partcle s gven by L = r p. Whether ths quantty s conserved or not depends on dl dt = r ṗ = r V. (3.4) 25

26 26 CHAPTER 3. SYMMETRIES AND CONSERVATION LAWS Generally, the r.h.s. (torque) would be non-zero, but n the case of central potental V = V (r), the gradent becomes V = ˆrV (r), and r V = 0. For the central potental actng on the partcle, L s conserved. Central potental V (r) depends only on the dstance from the orgn, makng t nvarant under an arbtrary orthogonal rotaton. We can nfer that angular momentum conservaton has to do wth the sotropy of space. We can also talk about partal angular momentum conservaton. If the potental s gven n the form V (r, z), r = (x 2 +y 2 ) 1/2, we fnd V = ẑ V V z +ˆr r. For the z-component of L, dl z dt = xf y yf x = 0. (3.5) The conserved component s related to the axs about whch rotaton leaves potental form preserved. Lnear momentum conservaton rests on dp dt = V. (3.6) It s conserved when V = V 0, for a free partcle. As n the angular momentum case, we can talk about partal momentum conservaton. If V (x) s only a functon of x, p y and p z are conserved. The conserved drectons are those along whch the potental functon remans nvarant. 3.2 Vew from Lagrangan Mechancs The case of tme-ndependent force corresponds to the Lagrangan L whch does not contan t as an explct varable. In such a case the total tme dervatve of the Lagrangan L s dl dt = L q + L q q q = d ( ) L dt q q + L q q = d dt ( ) L q, (3.7) q the second lne followng from the Euler-Lagrange equaton tself. Takng the dfference of the two sdes, we fnd an nvarant of tme ( ) d L q L = 0. (3.8) dt q In other words, the quantty we prevously defned as the Hamltonan H, when vewed as a functon of the generalzed coordnates q and generalzed veloctes q, does not change over tme. For physcally allows paths (obeyng the Euler- Lagrange equaton), H s a constant of moton provded L/ t = 0. We know very well that H s the total energy.

27 3.2. VIEW FROM LAGRANGIAN MECHANICS 27 In general, f some components of the generalzed coordnates were mssng from the Lagrangan such that L/ q = 0, there s a correspondng conserved quantty p = L/ q. ṗ = d dt ( ) L q = L q = 0. (3.9) Ths s the generalzaton of the noton of partal lnear and angular momentum conservaton dscussed n the prevous secton. From symmetry perspectve the absence of a partcular coordnate q means that the physcal system remans nvarant under the dsplacement of that coordnates by an arbtrary amount q q + a. If q were a rectlnear coordnate the nvarance s a consequence of the homogenety of space along that drecton. If q were an angular coordnates the nvarance s a consequence of the sotropy of space about the axs assocated wth the angle. As a smple applcaton of the above-mentoned prncple relatng nvarance of the Lagrangan to a conservaton law, consder a homogeneous system consstng of N-partcles whose Lagrangan s L = 1 m ṙ 2 V ( r r j ). (3.10) 2 <j The central force between a par of partcles s medated by the potental V ( r r j ). Ths Lagrangan s nvarant under the translaton of all the coordnates by a unform amount r r + u. From the nvarance, follows the concluson L[r + u, ṙ ] = L[r, ṙ ] (3.11) u L = u ( ) ṗ = u d p = 0. (3.12) r dt The relaton has to hold for arbtrary u, hence the total lnear momentum P = p must be conserved. When specalzed to the two-body problem we also conclude =1,2 L = r =1,2 V = r =1,2 F = 0. (3.13) That s, the force actng on 1 by 2 (F 1 ) must be equal and opposte to the force actng on 2 by 1 (F 2 ). The thrd law of Newton, acton-reacton prncple, follows naturally from the momentum conservaton of the two-body problem, whch n turn follows from the homogenety of the Lagrangan under the unform translaton. If a system s nvarant under rotaton, the Lagrangan keeps the same form under

28 28 CHAPTER 3. SYMMETRIES AND CONSERVATION LAWS r r + sˆn r, ṙ ṙ + sˆn ṙ, (3.14) where ˆn serves as the rotaton axs and s s the (small) amount of rotaton about the axs. It follows that ( ˆn r L + ˆn ṙ L ) = ˆn (r ṗ + ṙ p ) r ṙ ( ) = ˆn d r p = d (ˆn L) = 0. (3.15) dt dt The total angular momentum L projected onto the drecton of rotaton ˆn s thus conserved. The two examples dscussed here, one of conservaton of total lnear momentum and the other one of total angular momentum, are nstances of what s known as Noether s theorem. The theorem states that f a Lagrangan remans nvarant under a change change of the coordnates parameterzed by the one constant α, then there exsts one conserved quantty assocated wth such a contnuous change. It also tells us that conservaton laws are assocated wth the contnuous symmetres of the Lagrangan. 3.3 Vew from Hamltonan Mechancs For tme-ndependent Lagrangan, we saw that the energy,.e. the Hamltonan H s ndependent of tme, so that dh/dt = 0. To see ths drectly n the Hamltonan formulaton, calculate dh dt = H q q + H p ṗ = ṗ q + q ṗ = 0. (3.16) Take some quantty A(p, q ) whch depends on the coordnates q and ther canoncal momenta p. To see whether the quantty s conserved, one should calculate da dt = A q + A ṗ = A H A H = [H, A]. (3.17) q p q p p q Ths quantty s the Posson bracket already ntroduced n the prevous chapter. If a physcal quantty A s a constant moton, t must commute wth the Hamltonan H n the sense that ts Posson bracket wth H s zero. As a trval consequence, any observable A whch s a functon of H and not of the p and q ndvdually s a constant of moton. da dt = A H A H = A H H A H H q p p q H q p H p q = A [H, H] = 0. (3.18) H

29 3.4. PROBLEMS 29 A famous example of such an observable quantty s the Botlzmann dstrbuton functon ρ = exp( H/k B T ). Conservaton of lnear momentum would follow from dp dt = [H, p]. (3.19) If the Hamltonan s that of a free partcle, H = p 2 /2m, t follows mmedately that [H, p] = 0, hence ṗ = 0. Conservaton of the angular momentum L = r p for the Hamltonan H = p 2 /2m+V ( r ) can be shown usng the smlar Posson bracket algebra. 3.4 Problems 1. Prove the conservaton of angular momentum L = r p for the Hamltonan H = p 2 /2m + V ( r ) by workng out the Posson bracket [H, L]. 2. A partcle movng n the x < 0 regon of space wth potental energy V = V 1 s ncdent on the x = 0 plane at an angle θ 1 wth the normal and emerges on the x > 0 space at an angle θ 2 where the potental s V = V 2. Calculate the rato of the ncdent and emergent angles by calculatng sn θ 1 / sn θ 2. The answer must be expressed n terms of the potental energy dfference V 1 V 2, and the knetc energy of the ncdent partcle mv1/2. 2 V1 Potental V2 θ 1 θ 2 v 1 X<0 X=0 X>0 Fgure 3.1: Problem 3.2

30 30 CHAPTER 3. SYMMETRIES AND CONSERVATION LAWS

31 Chapter 4 One-dmensonal Moton 4.1 Frst-ntegral Approach One-dmensonal moton n classcal mechancs refers to a set of stuatons wheren the dynamcs can be reduced to that of a sngle varable. All other varables can be elmnated by use of conservaton laws. A prototypcal Lagrangan of a one-dmensonal moton s L = 1 2 mẋ2 V (x). (4.1) Conservaton of energy ensures that E = 1 2 mẋ2 + V (x) wll reman constant throughout the moton. We nfer that dx dt = ± E V (x). (4.2) For convenence take the postve branch, and re-wrte the frst-order dfferental equaton n the form Upon ntegratng, we obtan dt = dx E V (x). (4.3) t 2 t 1 = m 2 x2 x 1 dx E V (x). (4.4) Obvously the moton s only allowed n the regon where V (x) < E, whch n turn defnes the turnng ponts E = V (x). In the case when the moton s confned at both ends, wth two turnng ponts x l and x r, one should be able to derve the perod of the moton as T = xr 2m x l dx E V (x). (4.5) 31

32 32 CHAPTER 4. ONE-DIMENSIONAL MOTION Appled to the problem of smple harmonc oscllator where we can take E = (k/2)x 2, where X s the maxmum dsplacement, we have T = X 2m X dx k/2(x2 x 2 ) = 2 1 ω 0 1 dy 1 y 2 = 2π ω 0. (4.6) The desred perod T = 2π/ω 0 s obtaned. It s ndependent of the ampltude of harmonc oscllaton X. θ l m Fgure 4.1: A smple pendulum. Wth the more general moton of a swngng pendulum we have the total energy gven by 1 2 ml2 θ2 mgl cos θ = mgl cos θ 0. (4.7) We used θ 0 to ndcate the maxmum angle of swng. For the angular velocty θ we get dθ 2g dt = l (cos θ cos θ 0) t 2 t 1 = 1 2ω0 θ2 θ 1 dθ cos θ cos θ0, ω 0 = g l. (4.8) For small oscllatons we may approxmate the cosnes by ther Taylor expanson, t 2 t 1 1 ω 0 θ2 θ 1 dθ θ 2 0 θ 2, (4.9) whch would reproduce the harmonc oscllator result. For large swngs the Taylor expanson no longer works, and we have to evaluate the ntegral numercally. The perod of a large oscllaton s gven by ω 0 T = 2 θ0 dθ 2. (4.10) 0 cos θ cos θ0

33 4.2. EFFECTIVE ONE-DIMENSIONAL MOTION 33 Because the ntegral s mpossble to evaluate exactly, t s nstead gven a grand name Jacob complete ellptc ntegral of the frst knd. For small angle θ 0 the perod s gven by ( ω 0 T = 2π ) 16 θ (4.11) Even for a farly large angle of swng the frst correcton term proves qute small compared to 1, and presumably the second correcton term of order θ 4 0 even smaller. In practcal calculatons retanng the frst correcton wll be adequate. 4.2 Effectve One-Dmensonal Moton A system of two pont partcles nteractng by the central force can always be reduced to the sngle-partcle moton. Startng from L = 1 2 m 1ṙ m 2ṙ 2 2 V ( r 1 r 2 ), (4.12) we ntroduce the reduced mass m and the relatve coordnates r = r 1 r 2. In ths case, the Lagrangan can be re-wrtten as L = 1 2 MṘ mṙ2 V (r), (4.13) M = m 1 + m 2, MṘ = m 1ṙ 1 + m 2 ṙ 2. The total momentum s a constant of moton, and can be dropped from the Lagrangan. The relevant problem s that of the relatve moton descrbed by r. L = 1 2 mṙ2 V (r). (4.14) As dscussed n the prevous chapter, ths Lagrangan also happens to have the conserved angular momentum L = r p, p = mṙ. Choosng the conserved L drecton to be the z-drecton, t means that the moton wll only take place n the xy plane. Employng the cylndrcal coordnates, Lagrangan becomes Lagrange s equatons are L = 1 2 m(ṙ2 + r 2 θ2 ) V (r). (4.15) m r = mr θ 2 V (r), d dt (mr2 θ) = 0. (4.16) The conserved quantty mr 2 θ s the angular momentum about the z-axs, L z. Another way to see the conservaton s to note that the coordnate θ does not appear explctly n the Lagrangan. For all such coordnates, we saw n the prevous chapter, there s a correspondng conserved quantty. Denotng the

34 34 CHAPTER 4. ONE-DIMENSIONAL MOTION conserved angular momentum mr 2 θ = L z allows us to re-wrte the remanng equaton as m r = L2 z mr 3 V (r). (4.17) Indeed, the two-body problem nteractng by the central force s reduced to that of a sngle mass executng a one-dmensonal moton subject to the effectve potental V eff (r) = V (r)+l 2 z/2mr 2. Emergence of the effectve potental nstead of the bare one V (r) s a consequence of projectng the problem from hgher dmensons to one. Such phenomena,.e. renormalzaton of the bare nteracton whle elmnatng the degrees of freedom, occurs n all branches of many-body physcs. Soluton of the dfferental equaton for V (r) = k/r wll lead to Kepler s ellptc orbts and all that. Although not qute as powerful as the full soluton, one can extract some relevant nformaton about the moton by consderng the constant of moton E = (m/2)(ṙ 2 + r 2 θ 2 ) + V (r) = (m/2)ṙ 2 + L 2 z/2mr 2 + V (r), nvokng the constancy of angular momentum to wrte down the latter. So, dr 2 dt = m [E V eff(r)]. (4.18) Substtuton of the central potental V (r) = k/r leads to the pcture of the moton for r oscllatng between the two extremum rad set by E + k r We obtan the two roots as L2 z 2mr 2 = 0 r2 + k E r L2 z 2mE = 0. (4.19) ( r 1,2 = k ) 2 k 2E ± + L2 z 2E 2mE. (4.20) For some choce of E < 0 the term nsde the square root vanshes: E c = mk2 2L 2. (4.21) z Only for ths partcular energy s the moton perfectly crcular. For other energes, the oscllaton of the radus between r 1 and r 2 results n the ellptc orbts. Usng the constant of moton mr 2 dθ/dt = L z one can convert Eq. (4.18) to dr dθ = dr/dt dθ/dt = mr2 dr L z dt = r2 2m[E Veff (r)]. (4.22) L z Integraton yelds the relaton

35 4.2. EFFECTIVE ONE-DIMENSIONAL MOTION 35 θ θ 0 = r r 0 L z r 2 2m[E V eff (r )] dr. (4.23) In partcular the angle pcked up durng one cycle of r = r 2 to r = r 1 and back, wrtten θ, θ = 2L z r1 dr r 2 r 2 2m[E V eff (r )], (4.24) has to be a ratonal multple of 2π f the orbts were to repeat tself after a fnte tme. Another seemngly two-dmensonal problem reducble to a one-dmensonal one s the moton of a sphercal pendulum. The cartesan coordnates of a sphercal pendulum are gven by x = l sn θ cos φ, y = l sn θ sn φ, z = l cos θ. (4.25) The Lagrangan n terms of θ and φ reads L = m 2 (ẋ2 + ẏ 2 + ż 2 ) mgz = 1 2 ml2 ( θ 2 + φ 2 sn 2 θ) + mgl cos θ. (4.26) Note that the coordnate θ does not appear, mplyng the conservaton of the assocated momentum L/ θ = ml 2 φ sn 2 θ = L z. Ths s the angular momentum component along the drecton whch also serves as the axs of rotaton for angle φ. The equaton of moton for θ follows from the Lagrangan ml 2 θ = ml 2 φ2 sn θ cos θ mgl sn θ, (4.27) but φ can be elmnated usng the constant of moton just derved, φ = L z /ml 2 sn 2 θ. The whole thng can be summarzed as where the effectve potental V (θ) s V (θ) θ = θ (4.28) V (θ) = g l cos θ + L2 z 1 2m 2 l 4 sn 2 θ. (4.29) It appears to be a one-dmensonal problem. The reason for ntroducng a term new V (θ) s that t s now rather see to express the energy conservaton as E = 1 2 θ 2 + V (θ). (4.30)

36 36 CHAPTER 4. ONE-DIMENSIONAL MOTION Solvng t n terms of θ proceeds as t 2 t 1 = 1 θ2 dθ. (4.31) 2 E V (θ) θ 1 For θ values correspondng to the mnmum of V (θ) we obtan θ = 0. The sphercal pendulum stays at a constant angle θ 0. For values of θ devatng slghtly from θ 0 we can expand V (θ) to quadratc order and obtan a harmonc oscllator equaton θ = ω 2 θ. (4.32) 4.3 Problems 1. Prove that Eq. (4.10) s equvalent to 4K(sn θ 0 /2) where K(x) π/2 0 dy 1 x2 sn 2 y (4.33) s known as the complete ellptc ntegral of the frst knd. 2. Evaluate Eq. (4.10) numercally and plot the perod as a functon of θ 0 rangng from 0 to π. Compare the plot wth the approxmate result ω 0 T 2π(1 + θ 2 0/16). 3. Prove Eq. (4.13). 4. Derve the oscllaton frequency ω n Eq. (4.32).

37 Chapter 5 Harmonc Oscllaton 5.1 Coupled Harmonc Oscllaton Generalty For a coupled harmonc oscllator the Lagrangan s gven by L = 1 2 T j q q j 1 2 V jq q j. (5.1) The sum runs over both ndces 1, j N. One can always choose T j and V j to be symmetrc, T j = T j, V j = V j. Lagrange s equatons follow as T j q j = V j q j T q = V q. (5.2) Assume that the matrx T can be nverted, and we would have q = T 1 V q = F q. (5.3) In a normal mode, all components of q oscllate wth the same frequency ω and we may wrte q(t) = e ωt q wth the tme-ndependent q-vector. Then we arrve at an egenvalue equaton ω 2 q = F q, q = q 1.. (5.4) q N If F were symmetrc, we would know for sure that ω 2 are all real-valued. Even though now F s not guaranteed to be a symmetrc matrx, we can stll show that ω 2 must be real. To see ths, we go back a step and wrte the egenvalue problem as ω 2 T q = V q. (5.5) 37

38 38 CHAPTER 5. HARMONIC OSCILLATION Takng the nner product on the left wth the conjugate of q gves ω 2 = q V q q T q, q = (q 1,, q N). (5.6) Due to the fact that both T and V are real and symmetrc, t follows that both the denomnator and the numerators are real. For nstance,j q V j q j =,j V j q q j = real. (5.7) Hence ω 2 must be real, too. It s stll not guaranteed that ω 2 would be postve, though. Postvvty of ω 2 should follow from the dynamcal stablty of the Lagrangan tself. To see ths pont wrte q generally n terms of ts real and magnary parts, q = a + b. Then q V j q j =,j,j a V j a j +,j b V j b j. (5.8) Each term on the rght sde represents the change n the potental energy due to an arbtrary dsplacement vector a and b. Expandng around an stable mnmum, by defnton, mples that a small arbtrary dsplacement leads to an ncrease n energy. Hence V j must be that the two terms on the rght are both postve for arbtrary real-valued vectors a and b. A smlar argument can be made regardng the postvty of q T q, sayng that moton can only ncrease the knetc energy and not the other way around. The structure of the normal mode equaton (5.5) s such that one cannot make a dstncton between solutons of postve ω and negatve ω. For convenence we may always choose ω > 0. For each egenmode ω α we have the correspondng egenvector q α satsfyng ω 2 αt q α = V q α, (5.9) for 1 α N. For a lnear problem, an arbtrary lnear superposton of such solutons s also a soluton, hence the most general soluton to the coupled oscllator problem s q(t) = N A α q α e ωαt, (5.10) α=1 wth an arbtrary set of complex coeffcents A α. Of course the physcal soluton s recovered by takng the real part of the above soluton. At ths pont t s useful to pont out the analogy of the normal-mode solutons to the coupled oscllator problem to the quantum mechancal soluton. One mght regard q α as an egenvector representng the α-th egenstate. Then

39 5.1. COUPLED HARMONIC OSCILLATION 39 Eq. (5.9) has a nce nterpretaton as the average of the potental energy operator V dvded by the average of the knetc energy operator T for the n-th egenstate n, as formally gven by ω 2 = n V n n T n. (5.11) Closely related to the normal modes q α just found above s the noton of normal coordnates, wrtten θ α, 1 α N. By defnton these are the lnear combnatons of the orgnal coordnates θ α = X α q, θα = X α q, (5.12) such that the Lagrangan becomes normal, L = 1 2 T j q q j 1 2 V jq q j = α 1 ( 2 m α θ2 α ωαθα) 2 2. (5.13) In effect we are requrng that the two condtons be met smultaneously: T j q q j = ( ) m α θ2 α = m α X α X αj q q j α α V j q q j = ( ) m α ωαθ 2 α 2 = m α ωαx 2 α X αj q q j. (5.14) α α The two relatons can be summarzed compactly as matrx relatons, T = X T MX, V = X T MΩ 2 X. (5.15) Two dagonal matrces M = dag(m 1,, m N ), and Ω 2 = (ω1, 2, ωn 2 ), have been defned. On the other hand, the normal mode equatons (5.9) for all the 1 α N can be summarzed as a matrx equaton: T QΩ 2 = V Q, (5.16) where Q = ( q 1 q N ) s a collecton of normal mode solutons arranged as a matrx. Insertng Eq. (5.15) nto Eq. (5.16) gves the relaton X T MXQΩ 2 = X T MΩ 2 XQ, (5.17) from whch we deduce XQΩ 2 = Ω 2 XQ. Ths relaton s satsfed provded we choose XQ = 1, an dentty matrx. In other words the desred transformaton matrx X for the normal coordnates are nothng other than the nverse of the matrx correspondng to the normal mode solutons: X = Q 1. The dagonal mass matrx s found from