A novel solution to Kepler s problem

Transcription

1 INSTITUTE OF PHYSICSPUBLISHING Eu. J. Phys. 4 (003) EUROPEANJOURNAL OF PHYSICS PII: S (03) A novel solution to Keple s poblem Jn Vbik Deptment of Mthemtics, Bock Univesity, 500 Glenidge Avenue, St Cthines, LS 3A1, Cnd Received 18 June 003 Published 5 Septembe 003 Online t stcks.iop.og/ejp/4/575 Abstct The stndd (unpetubed) Keple poblem is expessed in Kustnheimo Stiefel fom nd solved utilizing the lgeb of qutenions. This povides the necessy bckgound to undestnd some of the new techniques of celestil mechnics. 1. Intoduction The im of this ppe is to intoduce students to moden nd inteesting ppoch to celestil mechnics. In the pocess, they len fundmentls of qutenion lgeb elted to the geomety of thee-dimensionl spce (ottions in pticul). With this bckgound, they cn follow the subsequent pesenttion of solution to Keple s poblem, nd undestnd its bsic fetues. This constitutes significnt stepping stone fo studying nd undestnding dynmics-elted fetues of single-st sol system (such s ou own).. Keple poblem The bsic lw of gvity sttes tht two spheicl bodies ttct ech othe with foce popotionl to the poduct of thei msses nd invesely popotionl to the sque of thei distnce (with the diection long the line connecting thei centes). Usully, one of the two bodies is lge (nd clled the pimy), nd the othe smll (the stellite). Bsed on the pevious lw, one cn wite the following diffeentil eqution fo the loction of the stellite (expessed s vecto in non-otting Ctesin coodintes, hving the oigin t the pimy s cente): + µ 3 = 0 (1) whee µ is the sum of the two msses (pimy nd stellite), multiplied by the gvittionl constnt. Double dot implies the second deivtive with espect to time t, is shothnd fo vecto with thee components, nmely x(t), y(t) nd z(t),nd. Solving these (effectively thee) equtions (collectively known s the Keple poblem) ws ccomplished by Newton moe thn thee centuies go ( monumentl chievement, consideing tht he fist hd to invent the concept of diffeentil eqution itself), esulting in the following well known nswe: the solutions e ellipses, with the pimy s cente t /03/ $ IOP Publishing Ltd Pinted in the UK 575

2 576 JVbik one of the foci. The initil conditions (stipultingthe stellite s loction nd velocity t time t = 0) will of couse pick out only one such ellipse s the stellite s obit. The fct tht plnets obits e ellipticl hd been discoveed empiiclly bsed on Tycho Bhe s dt bykeple, nd it is nowclled the fist Keple lw. The time dependence of the stellite s pssge though its obit is bsed on the emining two Keple s lws. Sevel decdes lte, Newton s theoy confimed Keple s empiicl lws one of the fist scientific tiumphs of this kind. The objective of this ppe is to demonstte new nd inteesting wy of solving the sme poblem using qutenions (n option not vilble to Newton qutenions becme pt of mthemtics only duing the 19th centuy). 3. Algeb of qutenions The lgeb hs thee imginy units (insted of the usul one) denoted i, j nd k, ech of them squing to 1. Any two of these nticommute (e.g. ij = ji), futhemoe, ij = k, jk = i nd ki = j. Togethe with the odiny (el) unit 1,which commutes with ech of them, they constitute the lgeb s bsis. An element of this lgeb cn thus be witten s A + 3 i + j + 1 k A + A () nd intepeted geometiclly s vecto ( 1, nd 3 epesenting its x, y nd z coodintes, espectively note the delibete ssocition of i with the z coodinte),ppended to scl A. Adding nd subtcting qutenions is done in the usul component-wise mnne: (A + ) ± (B + b) = A ± B + ± b. (3) Qutenion multipliction is bit moe ticky, it mounts to: (A + )(B + b) = AB b + Ab + B b. (4) Execise. Veify the coectness of this fomul. Also pove tht this multipliction is ssocitive (it is clely non-commuttive). Most of the usul functions llow qutenion guments, yielding qutenion in etun, fo exmple: exp(a + ) = exp(a) exp(â) = exp(a)(cos + â sin ) (5) whee nd â e the mgnitude nd unit diection, espectively, of. Anothe elted opetion is tht of (qutenionic) conjugte,nmely evesing the sign of i, j nd k,thus: A + = A. (6) Note tht AB = B A. (7) One cn lso esily show tht AA = AA (8) is el nd non-negtive. We cn thus define qutenion s mgnitude by A AA (9) nd lso constuct qutenion s invese by A 1 = A AA. (10)

3 A novel solution to Keple s poblem 577 Simil to mtix invese, we lso hve (AB) 1 = B 1 A 1. (11) Finlly, we define specil clss of qutenions, clled (in this ppe) ottions, which meet the following condition RR = 1. (1) It is quite obvious tht exp(w), wheew is ny vecto, is ottion. The evese is lso tue, i.e.nysuchr cn be expessed in the exp(w) fom. Poof. Clely, ny such R cn be witten s ± 1 + =± 1 + â, whee is vecto whose mgnitude is less thn o equl to 1, nd ± implies specific choice of eithe sign. Now, we find w such tht cos(w) =± 1 nd sin(w) =,ndtkew wâ. 4. Thee-dimensionl ottions Suppose tht we wnt to otte points of thee-dimensionl spce (these will now be epesented by pue qutenions,i.e. qutenions without the el component, thus: = zi+ yj+xk)ound n xis pssing though the oigin. Let the diection of vecto u (lso epesented by pue qutenion) specify this xis, nd its mgnitude u be the ngle of ottion (in dins). One cn show tht such ottion is fcilitted by ( exp u ) ( ) u exp RR. (13) Execise. Supply the poof. It is lso possible to show tht ny such R = exp( u ) cn be witten in the seemingly moe complicted (but geometiclly the meningful) mnne of ( exp i ψ ) ( exp k θ ) ( exp i φ ) (14) whee ψ, θ nd φ e clled Eule ngles. Poof. Expnding nd simplifying (14) yields cos θ ( exp i ψ + φ ) + k sin θ ( exp i ψ φ ). (15) On the othe hnd, ny qutenion cn be evidently witten s P exp(iα) + kq exp(iβ) (16) whee P, Q, α nd β e el. All we hve to show is tht, in the cse of R, P + Q = 1; fte tht, the ctul pmete mtching of (15) nd (16) is tivil. This cn be done by 1 = RR = [P exp(iα) + kq exp(iβ)][p exp( iα) Q exp( iβ)k] = P + kqpexp[i(β α)] PQexp[i(α β)]k + Q = P + Q (17) since e iξ k = ke iξ (18) fo ny el ξ.

4 578 JVbik 5. Refomulting the Keple poblem Let us etun to (1), whee is now seen s vecto in its qutenion epesenttion. This by itself does not chnge much, until we ty to simplify the eqution (elly shothnd fo set of equtions) by convenient tnsfomtion of both (the dependent vible) nd t (the independent vible). As we ll know, diffeentil eqution often simplifies when new dependent vible is intoduced, usully s some simple combintion of the old dependent nd independent vibles(e.g.v = x y). Less fequently, we simplify equtions by intoducing new independent vible, usully s function of the old independent vible (e.g. z = x ). But, one should elize tht it is quite legitimte fo new independent vible to be lso function of both (independent nd dependent) oiginl vibles, nd tht is exctly how we will poceed. We intoduce new dependent vible U (this time full qutenion) by UkU (19) nd new independent vible s by dt ds = (0) µ whee is positive constnt, chosen (t this point) bitily, nd is the mgnitude of. Note tht, in tems of the new U, = UU. Poof. = = UkUUkU = (UU) = UU. Also note tht insted of n explicit fomul fo t (in tems of s nd ), we hve n expession fo its deivtive. This mkes the tnsfomtion bit moe complicted (o pehps just unusul), but tht is wht is needed in this cse. In tems of the new vibles, (1) hs the following fom (known s the Kustnheimo Stiefel eqution [1, ]): U (U U 4) U +ku Ɣ + ku ( Ɣ ) = 0 (1) whee denotes diffeentition with espect to s, is shothnd fo UU,ndƔ fo UkU U ku = µ (Uk U UkU) () (clely el,sinceit is equl to its own conjugte). Poof. µ Ɣ ṙ = Uk U (3) pemultiplied by µ ku, implies µ kuṙ = U kuɣ. (4) Applying d µ dt d ds to the lst eqution yields: ( ) ( ) kuɣ µ µ kuṙ = U. (5)

5 A novel solution to Keple s poblem 579 Expnding the left-hnd side esults, with the help of (1), in 4kU + µ ku ṙ = 4 U + ku (U ku + Ɣ) (6) since ku U = U (7) nd µ Ɣ ṙ = UkU +. (8) Substituting (6) fo the left-hnd side of (5) finlly gives: 4 U ( ) + ku Ɣ (U ku + Ɣ) = U ku ku Ɣ (9) (the est is just engement of tems). So,whee is the simplifiction? Eqution (1) looks moe complicted thn eqution (1), nd we hve qutenions to woy bout on top of it! Well, thee is still some hope. We hve eplced the old thee-component by fou-component U. This is not n unusul thing to do; we ll emembe tht to solve y + (x)y + b(x)y = 0 one intoduces two functions, u nd v, inplce of y,nd emoves the esulting edundncy lte by intoducing convenient constint (physicists cll it guge)whicheffectively mkes u function of v. Let us do something simil hee. A convenient guge is clely Ɣ 0(theimpotnt thing is tht it is one-dimensionl), since it emoves two unplesnt tems of (1). The eqution then simplifies to U U U U = 0 (30) which seems lot moe mngeble, especilly if U U poves to be constnt, s it does! Poof. Diffeentiting U U with espect to s yields U U + U U U U (U U + UU ). (31) Replcing U by U U U U U nd U by U U U ( U U is el) esults in U U + UU U U (U U + UU ) = 0. (3) One cn lso show tht U U ( v = µ 1 ) (33) whee v ṙ. This mkes it popotionl both to nd the totl enegy of the system (no wonde it is constnt consevtionof totl enegy is one of the holiest pinciples of physics). Poof. U U = 4 U U µ. (34)

6 580 JVbik Since ṙ = Uk U + UkU = Uk U = UkU (35) (due to ou Ɣ = 0constint), we get v = ṙṙ = 4 UkUUk U = 4 U U. (36) To mke things inteesting, the totlenegymust be negtive (othewise, the two bodies would just flypt, the thn obiting ech othe). To simplify eqution (30) futhe, we set (so f bity but positive) to mke (33) equl to 1. All wehvetosolve in the end is thus the simple looking U + U = 0. (37) 6. Solving the Keple poblem The pevious eqution is still qutenion (fou-component) eqution, but line, fully decoupled nd esy to del with (effectively, it is the good old y + y = 0epeted fou times, once fo ech component). The genel solution is thus U = P sin s + Q cos s (38) whee P nd Q e two constnt, lmost bity (subject to the Ɣ = 0condition, which educes the numbe of fee pmetes fom eight to seven, nd consistent with ou pevious U U = 1choice) qutenions. Mthemticlly, we e done. But whee e the ellipses? And wht is the physicl mening of the seven fee pmetes of ou solution? To nswe these questions equies bit moe wok. Wht we need to do is to expess (38) in moe meningful mnne. Fist we eplce sin s by exp(is) exp( is) (39) i nd cos s by exp(is) +exp( is). (40) Note tht we could hve chosen ny othe unit diection in plce of i. But, i is the most convenient choice (being othogonl to both j nd k), in view of eqution (19). Once tht is done, we cn clely e-wite (38) in n equivlent fom of U = P exp(is) + Q exp( is). (41) To mke the solution tuly meningful, we hve to e-pmetize it futhe: U = exp(kδ){a(1+jγ)exp[i(s s p )]+B exp[ i(s s p )]}R (4) whee δ, γ, A, B nd s p e el constnt pmetes, nd R is (constnt) ottion. Note tht, ltogethe, we still hve eight fee pmetes cle peequisite fo mking (41) nd (4) equivlent. Execise. Pove tht thee is one-to-onecoespondencebetween the ight-hndsides of (41) nd (4). Let us now ty to intepet the individul pmetes of (4).

7 A novel solution to Keple s poblem 581 Fist, we hve to elize tht the ctul physicl solution is, not U. Whenwesubstitute the ight-hnd side of(4) into (19), the δ-pmetecncels out. It hs, theefoe, no physicl mening t ll, nd cn be set to 0 (we e thus fixing ou guge) without ffecting. Similly, substituting the ight-hnd side of (4) into () esults in Ɣ = 4γ A. Poof. Fist we evlute UkU = R{A exp[ i(s s p )](1 jγ)+ B exp[i(s s p )]} k{a(1+jγ)exp[i(s s p )] B exp[ i(s s p )]}ir = R{A (1 γ ) exp[ i(s s p )] B exp[i(s s p )] γ AB cos[(s s p )]}jr +γ A. (43) Adding the coesponding conjugte yields zeo fo the fist tem, nd 4γ A fo the second one. To ensue tht Ɣ = 0, we hve to set γ = 0(theothe possible choice, nmely A = 0, would effectively eliminte both A nd γ the emining solution would no longe be fully genel). Finlly, we wnt to mkesuetht U U = 1. Since U U is constnt of motion, mking it equl to 1tone vlue of s mkes it so fo ll times. Wewill thus fix it t s = s p, whee the vlue of U U equls (A B). (44) (A + B) Poof. Diffeentiting U ={Aexp[i(s s p )]+Bexp[ i(s s p )]}R (45) with espect to s yields U ={iaexp[i(s s p )] ib exp[ i(s s p )]}R (46) implying tht U U,evluted t s = s p,equls (A B). Similly = UU, evluted t s = s p,equls (A + B). Expession (44) will thus equl to 1 only if A + B = (showing this is quite tivil). It is now convenient to tde off the A nd B pmetes fo nd β B,ndwite the A finl vesion of ou solution s U = 1+β [ei(s s p) + β e i(s s p) ]R (47) which esily yields (see (19)) = R[e i(s s p) + βe i(s s p) k ] 1+β [ei(s s p) + βe i(s s p) ]R = R k 1+β [ei(s s p) + βe i(s s p) ] R = R k 1+β [ei(s s p) + β e i(s s p) +β]r (48) due to (18). The lst expession is clely cuve in the x y plne (it hs only k nd j components), otted inther 0 R mnne of (13). So, wht kind of cuve is it?

8 58 JVbik e Its x nd y components (befoe the ottion, which of couse does not chnge its shpe) 1+β [(1+β ) cos ω +β] = cos ω + β 1+β (49) nd 1+β (1 β ) sin ω (50) espectively, whee ω (s s p ) is the so-clled eccentic nomly. Thecuvethus meets the following eqution ( x β 1 β which is n ellipse with eccenticity e = β 1+β. ) ( 1+β ) + 1 β y = (51) Poof. Fistly, 1 e = 1 4β (1+β ) = ( 1 β ).The 1+β pevious eqution cn thus be witten in moe ecognizble fom of (x e) y + = 1. (5) (1 e ) This is n ellipse with (semi) mjo nd mino xes of length nd 1 e,espectively, nd n eccenticity of (1 e ) = e. The distnce of ech focus fom the ellipse s cente is (1 e ) = e (the left of the foci is thus locted t the oigin). As = UU = UU = 1+β (1+β +β cos ω) = (1+ecos ω) (53) we cn esily integte the ight-hnd side of (0) with espect to s w + s p,toobtin the following eltionship between time t nd the eccentic nomly ω: t t 0 = 3/ µ (ω + e sin ω). (54) This is vesion ofthe so-clledkeple eqution,which seves to estblish the vying speed of the stellite though its ellipticl obit. 7. Futhe chllenges Suppose now tht, in ddition to being ttcted by the pimy, dditionl (smll) foces e cting on the stellite, due to: the gvittionl pull of othe stellites, the pimy nd/o the stellite not being pefectly spheicl, tmospheic dg, tidl foces, etc. The bsic eqution (1) then chnges ccodingly, by cquiing n exttem fo ech such effect, thus becoming the so-clled petubed Keple poblem. One of the most chllenging of these is the lun poblem: the moon, petubed by n ext pull of the (the distnt, but huge) sun cusing 0.6% distotion of the egul foce. Anothe inticte sitution ises when the petubing foce hs peiod commensuble (in n intege tio, such s :1) to tht of the stellite phenomenon clled esonnce. This hppens quite often in ou sol system, the steoid belt (petubed by Jupite) being pime exmple. The technique just descibed cn be extended to solve, itetively (i.e. the solution is expnded in tems of smll pmete), ll of these possible cses. It is pticully well

9 A novel solution to Keple s poblem 583 suited to del with esonnces, nd its utiliztion of qutenions then becomes indispensble. The inteested ede should efe to [3] fo mthemticl desciption of the technique, nd to [4] nd [5] fo some of its pplictions (undestnding the cuent ppe is necessy peequisite). Refeences [1] Stiefel E L nd Scheifele G 1971 Line nd Regul Celestil Mechnics (Belin: Spinge) [] Hestenes D 1986 New Foundtions of Clssicl Mechnics (Dodecht: Kluwe) [3] Vbik J 1995 Petubed Keple poblem in qutenionic fom J. Phys. A: Mth. Gen [4] Vbik J 000 Petubtive solution of the motion of n steoid in esonnce with Jupite Mon. Not. R. Aston. Soc [5] Vbik J 1997 Oblteness petubtions to fouth ode Mon. Not. R. Aston. Soc