L03: Kepler problem & Hamiltonian dynamics 18.354 Ptolemy circa.85 (Egypt) -165 (Alexandria) Greek geocentric view of the universe

Wednesday, February 5, 14 Tycho Brahe 1546 (Denmark) - 1601 (Prague)

"geo-heliocentric" system

last of the major naked eye astronomers

Wednesday, February 5, 14 Johannes Kepler 1571 - 1630 (Germany)

Kepler’s 3 laws

Wednesday, February 5, 14 Isaac Newton 1643 - 1727 (England) m m F = G 1 2 | 12| r2

Wednesday, February 5, 14 Friedrich Bessel 1784 (Germany) - 1846 (Prussia)

Cygni 61 d~10.3 ly gravitational many-body problem: prediction of Sirius B, Bessel functions

Wednesday, February 5, 14

Friedrich Bessel 1784 (Germany) - 1846 (Prussia)

Cygni 61 d~10.3 ly gravitational many-body problem: prediction of Sirius B, Bessel functions

Wednesday, February 5, 14 4 Kepler’s problem and Hamiltonian dynamics

Why do we study applied mathematics? Aside from the intellectual challenge, it is reason- able4 to Kepler’sargue that we problem do so to obtain and an Hamiltonian understanding of physical dynamics phenomena, and to be able to make predictions about them. Possibly the greatest example of this, and the origin 4 Kepler’s problem and Hamiltonian dynamics ofWhy much do of we the study mathematics applied mathematics? we do, came from Aside Newton’s from the desire intellectual to understand challenge, the it motion is reason- ofable the planets, to argue whichWhy that do wewere we do study known so applied to obtain to mathematics?obey an Kepler’s understanding Aside laws. from the of intellectual physical challenge, phenomena, it is reason- and to be able to makeable predictions to argue that about we do them. so to obtain Possibly an understanding the greatest of example physical phenomena, of this, and and theto be origin of much of theable mathematics to make predictions we do, about came them. from Possibly Newton’s the greatest desire example to understand of this, and the the origin motion 4.1 Kepler’sof laws much of of the planetary mathematics we motion do, came from Newton’s desire to understand the motion of the planets,of which the planets, were which known were to known obey to Kepler’s obey Kepler’s laws. laws. In the early seventeenth century (1609-1619) Kepler proposed three laws of planetary motion 4.1 Kepler’s laws of planetary motion (i)4.1 The Kepler’s orbits of the laws planets of planetary are ellipses, motion with the Sun’s centre of mass at one focus of In thethe early ellipse. seventeenthIn the early seventeenth century (1609-1619) century (1609-1619) Kepler Kepler proposed proposed three three laws laws of of planetary planetary motion motion (i) The orbits of the planets are ellipses, with the Sun’s centre of mass at one focus of (ii)(i) The The line orbits joining ofthe a the ellipse. planet planets and are the ellipses, Sun describes with the equal Sun’s areas centre in equal of mass intervals at one of time.focus of the ellipse. (iii) The squares(ii) of The the line periods joining of a planet the planets and the Sun are describes proportional equal areas to the in equal cubes intervals of their of time. semi- (ii)major The axes. line(iii) joining The squares a planet of the and periods the Sun of the describes planets are equal proportional areas to in the equal cubes intervals of their semi- of time. major axes. These(iii) laws The were squares based of the on periods detailed of observations the planets madeare proportional by Tycho Brahe,to the cubes and put of their to rest semi- any notion thatThese planets laws move were based in perfectly on detailed circular observations orbits. made However, by Tycho it Brahe, wasn’t and until put to Newton rest major axes.any notion that planets move in perfectly circular orbits. However, it wasn’t until Newton proposed his lawproposed of gravitation his law of gravitation in 1687 that in 1687 the that origins the origins of this of this motion motion were were understood. understood. NewtonThese proposed laws wereNewton that based proposed on detailedthat observations made by Tycho Brahe, and put to rest any notion that planets move in perfectly circular orbits. However, it wasn’t until Newton “Every object in“Every the Universe object in the attracts Universe every attracts other every object other object with with a force a force directed directed along along a linea line proposed his lawof centres of gravitation for the two in objects 1687 that that is proportional the origins to the of thisproduct motion of their were masses understood. and Newtonof centres proposed for the two thatinversely objects proportional that is proportional to the square of to the the separation product of the of two their objects.” masses and inversely proportional to the square of the separation of the two objects.” “Every objectBased in the on Universe this one statement, attracts it isevery possible other to derive object Kepler’s with laws. a force directed along a line Basedof on centres this one for statement, the two objects it is possible that is proportional to derive Kepler’s to the laws. product of their masses and 4.1.1 Second law inversely proportional to the square of the separation of the two objects.” 4.1.1 SecondKeplers law second law is the simplest to derive, and is a statement that the angular momentum Based on thisof one a particle statement, moving it under is possible a central force,to derive such as Kepler’s gravity, is laws. constant. By definition, the Keplers second lawangular is the momentum simplestL toof a derive, particle and with ismass a statementm and velocity thatu is the angular momentum of a particle moving under a central force, such as gravity,dr is constant. By definition, the 4.1.1 Second law L = r m , (53) angular momentum L of a particle with mass m and^ velocitydt u is Keplers secondwhere lawr isis the the simplestvector position to derive, of the particle. and is The a statement rate of change that of the angular angular momentum momentum is of a particle movinggiven by under a central force, suchdr as gravity, is constant. By definition, the L =dLr m , (53) angular momentum L of a particle with mass=^r fmdt=andr f velocity(r)rˆ =0, u is (54) dt ^ ^ where r is the vectorwhere f position(r)rˆ is the of central the force, particle. depending The only rate on of the change distance ofr = angularr and pointing momentum in the is dr | | given by direction rˆ = r/r. It can thereforeL = ber seenm that the, angular momentum of a particle moving (53) under a central forcedL is constant, a consequence^ dt of this being that motion takes place in a = r f = r f(r)rˆ =0, (54) where r is the vector positiondt of the^ particle.^ The rate of change of angular momentum is wheregivenf( byr)rˆ is the central force, depending only on the distance r = r and pointing in the dL | | direction rˆ = r/r. It can therefore be= seenr thatf = ther angular13f(r)rˆ =0 momentum, of a particle moving(54) under a central force is constant, adt consequence^ of^ this being that motion takes place in a where f(r)rˆ is the central force, depending only on the distance r = r and pointing in the | | direction rˆ = r/r. It can therefore be seen that the angular momentum of a particle moving under a central force is constant, a consequence of this being that motion takes place in a 13

13 4 Kepler’s problem and Hamiltonian dynamics

Why do we study applied mathematics? Aside from the intellectual challenge, it is reason- able to argue that we do so to obtain an understanding of physical phenomena, and to be able to make predictions about them. Possibly the greatest example of this, and the origin of much of the mathematics we do, came from Newton’s desire to understand the motion of the planets, which were known to obey Kepler’s laws.

4.1 Kepler’s laws of planetary motion In the early seventeenth century (1609-1619) Kepler proposed three laws of planetary motion

(i) The orbits of the planets are ellipses, with the Sun’s centre of mass at one focus of the ellipse.

(ii) The line joining a planet and the Sun describes equal areas in equal intervals of time.

(iii) The squares of the periods of the planets are proportional to the cubes of their semi- major axes.

These laws were based on detailed observations made by Tycho Brahe, and put to rest any notion that planets move in perfectly circular orbits. However, it wasn’t until Newton proposed his law of gravitation in 1687 that the origins of this motion were understood. Newton proposed that

“Every object in the Universe attracts every other object with a force directed along a line of centres for the two objects that is proportional to the product of their masses and inversely proportional to the square of the separation of the two objects.”

Based on this one statement, it is possible to derive Kepler’s laws.

4.1.1 Second law Keplers second law is the simplest to derive, and is a statement that the angular momentum of a particle moving under a central force, such as gravity, is constant. By definition, the angular momentum L of a particle with mass m and velocity u is dr L = r m , (53) ^ dt where r is the vector position of the particle. The rate of change of angular momentum is given by dL = r f = r f(r)rˆ =0, (54) dt ^ ^ where f(r)rˆ is the central force, depending only on the distance r = r and pointing in the | | direction rˆ = r/r. It can therefore be seen that the angular momentum of a particle moving under a central force is constant, a consequence of this being that motion takes place in a

plane. The area swept out by the line joining13 a planet and the sun is half the area of the parallelogram formed by r and dr.Thus 1 1 dr L dA = r dr = r dt = dt, (55) 2| ^ | 2 ^ dt 2m where L = L is a constant. The area swept out is therefore also constant. | | 4.1.2 First law To prove Keplers first law consider the sun as being stationary (i.e., infinitely heavy), and the planets in orbit around it. The equation of motion for a planet is d2r m = f(r)rˆ. (56) dt2 In plane polar coordinates dr =˙rrˆ + r✓˙✓ˆ, (57a) dt d2r =(¨r r✓˙2)ˆr +(r✓¨ +2˙r✓˙)✓ˆ. (57b) dt2 In component form, equation (56) therefore becomes

m(¨r r✓˙2)=f(r), (58a) m(r✓¨ +2˙r✓˙)=0. (58b)

Putting (57a) into (53) gives dr L = r m = mr2✓˙ . (59) ^ dt | | Thus r2✓˙ = l, (60) where l = L/m is the angular momentum per unit mass. Given a radial force f(r), equa- tions (58a) and (58b) can now be solved to obtain r and ✓ as functions of t. A more practical result is to solve for r(✓), however, and this requires the definition of a new variable 1 r = . (61) u Rewriting the equations of motion in terms of the new variable requires the identities 1 1 d✓ du du r˙ = u˙ = = l , (62a) u2 u2 dt d✓ d✓ d du d2u d2u r¨ = l = l✓˙ = l2u2 . (62b) dt d✓ d✓2 d✓2 Equation (58a) becomes d2u 1 1 + u = f . (63) d✓2 ml2u2 u ✓ ◆ 14 plane. The area swept out by the line joining a planet and the sun is half the area of the parallelogram formed by r and dr.Thus 1 1 dr L dA = r dr = r dt = dt, (55) 2| ^ | 2 ^ dt 2m plane. The area swept out by the line joining a planet and the sun is half the area of the where L = L is a constant. The area swept out is therefore also constant. parallelogram| | formed by r and dr.Thus 4.1.2 First law 1 1 dr L dA = r dr = r dt = dt, (55) 2| ^ | 2 ^ dt 2m To prove Keplers first law consider the sun as being stationary (i.e., infinitely heavy), and thewhere planetsL = inL orbitis a aroundconstant. it. The The area equation swept of out motion is therefore for a planet also constant. is | | d2r 4.1.2 First law m = f(r)rˆ. (56) dt2 InTo plane prove polar Keplers coordinates first law consider the sun as being stationary (i.e., infinitely heavy), and the planets in orbit around it. The equation of motion for a planet is dr ˙ ˆ =˙rrˆ +d2rr✓✓, (57a) dt m = f(r)rˆ. (56) d2r dt2 =(¨r r✓˙2)ˆr +(r✓¨ +2˙r✓˙)✓ˆ. (57b) In plane polar coordinates dt2 In component form, equation (56)dr therefore becomes =˙rrˆ + r✓˙✓ˆ, (57a) dt m(¨r r✓˙2)=f(r), (58a) d2r m=(¨(r✓¨ +2˙r r✓r˙)=0✓˙2)ˆr +(r. ✓¨ +2˙r✓˙)✓ˆ. (58b)(57b) dt2 PuttingIn component (57a) into form, (53) equation gives (56) therefore becomes

d2r L =m(¨rr mr✓˙ )== mrf(2r✓˙)., (59)(58a) ^ dt | | ¨ ˙ m(r✓ +2˙r✓)=0 . (58b) Thus Putting (57a) into (53) gives r2✓˙ = l, (60) where l = L/m is the angular momentum perdr unit mass. Given a radial force f(r), equa- L = r m = mr2✓˙ . (59) tions (58a) and (58b) can now be solved to^ obtaindt r and| ✓ as| functions of t. A more practical result is to solve for r(✓), however, and this requires the definition of a new variable Thus r2✓˙ 1= l, (60) r = . (61) where l = L/m is the angular momentum peru unit mass. Given a radial force f(r), equa- Rewritingtions (58a) the and equations (58b) can of now motion be solved in terms to obtain of ther newand variable✓ as functions requires of thet. A identities more practical result is to solve for r(✓), however,1 and this1 requiresd✓ du thedu definition of a new variable r˙ = u˙ = = l , (62a) u2 u2 dt1 d✓ d✓ r = 2. 2 (61) d du ˙ud u 2 2 d u r¨ = l = l✓ 2 = l u 2 . (62b) Rewriting the equations of motion dt ind terms✓ ofd the✓ new variabled✓ requires the identities Equation (58a) becomes 1 1 d✓ du du r˙ =d2u u˙ = 1 =1 l , (62a) u+2 u = u2 dt df✓ . d✓ (63) d✓2 ml2u2 u 2 ✓ ◆ 2 d du ˙ d u 2 2 d u r¨ = l = l✓ 2 = l u 2 . (62b) dt d✓ 14 d✓ d✓ Equation (58a) becomes d2u 1 1 + u = f . (63) d✓2 ml2u2 u ✓ ◆ 14 plane. The area swept out by the line joining a planet and the sun is half the area of the parallelogram formed by r and dr.Thus plane. The area swept out by the line joining a planet and the sun is half the area of the parallelogram formed by r and dr1.Thus 1 dr L dA = r dr = r dt = dt, (55) 2| ^ | 2 ^ dt 2m 1 1 dr L dA = r dr = r dt = dt, (55) where L = L is a constant. The2| ^ area| swept2 out ^ isdt therefore 2m also constant. | | plane. The area swept out by the line joining a planet and the sun is half the area of the where L = L is a constant. The area swept out is therefore also constant. 4.1.2parallelogram First| | formed law by r and dr.Thus 1 1 dr L 4.1.2To prove First Keplers law first lawdA consider= r thedr sun= asr beingdt stationary= dt, (i.e., infinitely heavy),(55) and 2| ^ | 2 ^ dt 2m the planets in orbit around it. The equation of motion for a planet is To prove Keplers first law consider the sun as being stationary (i.e., infinitely heavy), and where L = L is a constant. The area swept out is therefore also constant. the planets in orbit around it. The equationd2r of motion for a planet is | | m = f(r)rˆ. (56) dt2 4.1.2 First law d2r In plane polar coordinates m 2 = f(r)rˆ. (56) To prove Keplers first law consider the sundt as being stationary (i.e., infinitely heavy), and dr Inthe plane planets polar in coordinates orbit around it. The=˙ equationrrˆ + r of✓˙✓ˆ motion, for a planet is (57a) dt dr2 d2r d r ˙ ˆ2 =˙=(¨rmrˆr+2rr✓=✓✓˙ ,f)ˆr(r+()rˆ.r✓¨ +2˙r✓˙)✓ˆ. (56)(57a)(57b) dtdt2 dt d2r InIn componentplane polar form,coordinates equation (56)=(¨ thereforer r✓˙2 becomes)ˆr +(r✓¨ +2˙r✓˙)✓ˆ. (57b) dt2 dr =˙mr(¨rrˆ +rr✓✓˙˙2✓ˆ)=, f(r), (57a)(58a) In component form, equation (56)dt therefore becomes ¨ ˙ d2r m(r✓ +2˙r✓)=0. (58b) =(¨m(¨rr rr✓˙✓2˙2)=)ˆr +(rf✓¨(+2˙r), r✓˙)✓ˆ. (57b)(58a) dt2 Putting (57a) into (53) gives m(r✓¨ +2˙r✓˙)=0. (58b) In component form, equation (56) therefore becomes dr 2 Putting (57a) into (53) gives L = r m2 = mr ✓˙ . (59) m(¨r ^r✓˙ )=dt f| (r), | (58a) m(r✓¨+2˙r✓d˙)=0r . (58b) Thus L = r m = mr2✓˙ . (59) ^ dt | | r2✓˙ = l, (60) Putting (57a) into (53) gives Thus where l = L/m is the angular momentum 2 perdr unit mass. Given a radial force f(r), equa- L = r rm✓˙ = l,= mr2✓˙ . (59)(60) tions (58a) and (58b) can now be solved^ to obtaindt r| and |✓ as functions of t. A more practical whereresultl = is toL/m solveis the for r angular(✓), however, momentum and this per requires unit mass. the definition Given a radial of a new force variablef(r), equa- tionsThus (58a) and (58b) can now be solved to obtain r and ✓ as functions of t. A more practical 2 ˙ 1 result is to solve for r(✓), however, and thisr r✓ requires== l,. the definition of a new variable (60)(61) u where l = L/m is the angular momentum per unit mass. Given a radial force f(r), equa- 1 Rewritingtions (58a) the and equations (58b) can ofnow motion be solved in terms tor obtain= of the. r and new✓ variableas functions requires of t. A the more identities practical(61) u result is to solve for r(✓), however, and1 this requires1 d✓ du the definitiondu of a new variable Rewriting the equations ofr motion˙ = in termsu˙ = of the new= variablel , requires the identities(62a) u2 u21dt d✓ d✓ r = . 2 2 (61) 1 d du 1 ud✓˙ dduu du2 2 d u r˙ r¨ == l u˙ = = l✓ 2 == ll u , 2 . (62a)(62b) Rewriting the equations of motionu in2dt termsd✓u of2 dt thedd✓ new✓ variable d✓ d✓ requires the identities Equation (58a) becomes d du d2u d2u r¨ = l 1 = 1l✓d˙✓ du = l2duu2 . (62b) r˙ = d2udtud˙ ✓= d1✓2 = 1l ,d✓2 (62a) u2+ u =u2 dt d✓ f d✓. (63) d✓2 ml2u2 u Equation (58a) becomes d du d2u ✓ ◆ d2u r¨ = l = l✓˙ = l2u2 . (62b) d2u 1 2 1 2 dt+ ud✓= d✓ f . d✓ (63) d✓2 ml142u2 u Equation (58a) becomes ✓ ◆ d2u 1 1 + u = f . (63) d✓2 14ml2u2 u ✓ ◆ 14 plane. The area swept out by the line joining a planet and the sun is half the area of the parallelogram formed by r and dr.Thus 1 1 dr L dA = r dr = r dt = dt, (55) 2| ^ | 2 ^ dt 2m where L = L is a constant. The area swept out is therefore also constant. | | 4.1.2 First law To prove Keplers first law consider the sun as being stationary (i.e., infinitely heavy), and the planets in orbit around it. The equation of motion for a planet is d2r m = f(r)rˆ. (56) dt2 In plane polar coordinates dr =˙rrˆ + r✓˙✓ˆ, (57a) dt d2r =(¨r r✓˙2)ˆr +(r✓¨ +2˙r✓˙)✓ˆ. (57b) dt2 In component form, equation (56) therefore becomes

m(¨r r✓˙2)=f(r), (58a) m(r✓¨ +2˙r✓˙)=0. (58b)

Putting (57a) into (53) gives dr L = r m = mr2✓˙ . (59) ^ dt | | Thus r2✓˙ = l, (60) where l = L/m is the angular momentum per unit mass. Given a radial force f(r), equa- tions (58a) and (58b) can now be solved to obtain r and ✓ as functions of t. A more practical result is to solve for r(✓), however, and this requires the definition of a new variable 1 r = . (61) u Rewriting the equations of motion in terms of the new variable requires the identities 1 1 d✓ du du r˙ = u˙ = = l , (62a) u2 u2 dt d✓ d✓ d du d2u d2u r¨ = l = l✓˙ = l2u2 . (62b) dt d✓ d✓2 d✓2 Equation (58a) becomes d2u 1 1 + u = f . (63) d✓2 ml2u2 u ✓ ◆ 14

This is the di↵erential equation governing the motion of a particle under a central force. Conversely, if one is given the polar equation of the orbit r = r(✓), the force function can be derived by di↵erentiating and putting the result into the di↵erential equation. According to Newtons law of gravitation f(r)= k/r2, so that This is the di↵erential equation governingd2u the motionk of a particle under a central force. Conversely, if one is given the polar equation+ u of= the orbit. r = r(✓), the force function can(64) be d✓2 ml2 derived by di↵erentiating and putting the result into the di↵erential equation. According Thisto Newtons has the law general of gravitation solution f(r)= k/r2, so that d2u k k u = Acos(+✓u =✓0)+. , (64)(65) d✓2 ml2 ml2 whereThis hasA theand general✓0 are solution constants of integration that encode information about the initial conditions. Choosing ✓0=0 and replacing u by the original radial coordinate r =1/u, k u = Acos(✓ ✓0)+ 21, (65) k ml r = Acos✓ + (66) ml2 where A and ✓0 are constants of integration✓ that encode◆ information about the initial conditions. Choosing ✓ =0 and replacing u by the original radial coordinate r =1/u, which is the equation of0 a conic section with the origin at the focus. This can be rewritten 1 in standard form k r = Acos✓ +1+✏ (66) r = r0 ml2 , (67) ✓ 1+✏cos✓◆ wherewhich is the equation of a conic section with the origin at the focus. This can be rewritten 2 2 in standard form Aml ml ✏ = ,r1+✏0 = . (68) k r = r , k(1 + ✏) (67) 0 1+✏cos✓ ✏ is called the eccentricity of the orbit: where Aml2 ml2 ✏ =0isacircle, ✏ = ,r= . (68) • k 0 k(1 + ✏) ✏ < 1 is an ellipse, ✏ is• called the eccentricity of the orbit: ✏ = 1 is a parabola and • ✏ =0isacircle, • ✏ > 1 is a hyperbola. ✏ < 1 is an ellipse, •• For✏ = an 1 elliptical is a parabola orbit, andr0 is the distance of closest approach to the sun, and is called the perihelion• . Similarly, ✏ > 1 is a hyperbola. r = r (1 + ✏)/(1 ✏) (69) • 1 0 is theFor furthest an elliptical distance orbit, fromr0 is the the sun distance and is of called closest the approachaphelion to. the Orbital sun, and eccentricities is called the are smallperihelion for planets,. Similarly, whereas comets have parabolic or hyperbolic orbits. Interestingly though, Halley’s comet has a very eccentricr1 orbit= r0(1 but, + ✏ according)/(1 ✏) to the definition just given, (69)is not a comet! is the furthest distance from the sun and is called the aphelion. Orbital eccentricities are small for planets, whereas comets have parabolic or hyperbolic orbits. Interestingly though, Halley’s comet has a very eccentric orbit but, according to the definition just given, is not a comet!

15

15 This is the di↵erential equation governing the motion of a particle under a central force. Conversely, if one is given the polar equation of the orbit r = r(✓), the force function can be derived by di↵erentiating and putting the result into the di↵erential equation. According to Newtons law of gravitation f(r)= k/r2, so that d2u k This is the di↵erential equation governing+ u = the. motion of a particle under a central(64) force. d✓2 ml2 Conversely, if one is given the polar equation of the orbit r = r(✓), the force function can be Thisderived has the bygeneral di↵erentiating solution and putting the result into the di↵erential equation. According to Newtons law of gravitation f(r)= k/r2, so that k u = Acos(✓ ✓0)+ , (65) d2u mlk2 + u = . (64) d✓2 ml2 where A and ✓0 are constants of integration that encode information about the initial conditions.This has Choosing the general✓0=0 solution and replacing u by the original radial coordinate r =1/u,

1 k k r = uA=cosAcos(✓ + ✓ ✓0)+ , (66) (65) ml2 ml2 ✓ ◆ whichwhere is theA equationand ✓0 ofare a conic constants section of with integration the origin that at encodethe focus. information This can be about rewritten the initial in standardconditions. form Choosing ✓0=0 and replacing u by the original radial coordinate r =1/u, 1+✏ r = r0 , 1 (67) 1+✏cos✓ k r = Acos✓ + (66) where ml2 Aml2 ✓ ml◆2 ✏ = ,r= . (68) which is the equation of a conick section with0 thek origin(1 + ✏) at the focus. This can be rewritten in standard form ✏ is called the eccentricity of the orbit: 1+✏ r = r , (67) 0 1+✏cos✓ ✏ =0isacircle, • where Aml2 ml2 ✏ < 1 is an ellipse, ✏ = ,r= . (68) • k 0 k(1 + ✏) ✏ = 1 is a parabola and • ✏ is called the eccentricity of the orbit: ✏ > 1 is a hyperbola. • ✏ =0isacircle, • For an elliptical orbit, r is the distance of closest approach to the sun, and is called the ✏ < 1 is an ellipse,0 perihelion•. Similarly, ✏ = 1 is a parabola and r1 = r0(1 + ✏)/(1 ✏) (69) • is the furthest✏ > 1 distance is a hyperbola. from the sun and is called the aphelion. Orbital eccentricities are small for• planets, whereas comets have parabolic or hyperbolic orbits. Interestingly though, Halley’sFor comet an has elliptical a very orbit, eccentricr0 is orbit the distance but, according of closest to approach the definition to the just sun, given, and is is called not the a comet!perihelion. Similarly, r = r (1 + ✏)/(1 ✏) (69) 1 0 is the furthest distance from the sun and is called the aphelion. Orbital eccentricities are small for planets, whereas comets have parabolic or hyperbolic orbits. Interestingly though, Halley’s comet has a very eccentric orbit but, according to the definition just given, is not a comet!

15

15 This is the di↵erential equation governing the motion of a particle under a central force. Conversely, if one is given the polar equation of the orbit r = r(✓), the force function can be derived by di↵erentiating and putting the result into the di↵erential equation. According to Newtons law of gravitation f(r)= k/r2, so that d2u k + u = . (64) d✓2 ml2 This has the general solution k u = Acos(✓ ✓ )+ , (65) 0 ml2 where A and ✓0 are constants of integration that encode information about the initial conditions. Choosing ✓0=0 and replacing u by the original radial coordinate r =1/u,

1 k r = Acos✓ + (66) ml2 ✓ ◆ which is the equation of a conic section with the origin at the focus. This can be rewritten in standard form 1+✏ r = r , (67) 0 1+✏cos✓ where Aml2 ml2 ✏ = ,r= . (68) k 0 k(1 + ✏) ✏ is called the eccentricity of the orbit:

✏ =0isacircle, • ✏ < 1 is an ellipse, • ✏ = 1 is a parabola and • ✏ > 1 is a hyperbola. • For an elliptical orbit, r0 is the distance of closest approach to the sun, and is called the perihelion. Similarly, r = r (1 + ✏)/(1 ✏) (69) 1 0 is the furthest distance from the sun and is called the aphelion. Orbital eccentricities are small for planets, whereas comets have parabolic or hyperbolic orbits. Interestingly though, Halley’s comet has a very eccentric orbit but, according to the definition just given, is not a comet!

15 P A 4.1.3 Third law 4.1.3To prove Third Keplers law third law go back to equation (55). Integrating this area law over time Togives prove Keplers third law go back to equation (55). Integrating this area law over time ⌧ l⌧ gives4.1.3 Third law A(⌧)= dA = , (70) ⌧ 0 l2⌧ To prove Keplers third law go backA( to⌧)= equationZ dA (55).= , Integrating this area law over time(70) where ⌧ is the period of the orbit. The area A of an ellipse2 can also be written as A = ⇡ab gives Z0 wherewhere⌧aisand theb periodare the of semi-major the orbit. andThe semi-minor area A⌧ of an axes, ellipsel⌧ yielding can also be written as A = ⇡ab A(⌧)= dA = , (70) where a and b are the semi-major and semi-minorl⌧ 0 axes,2 yielding Z= ⇡ab (71) where ⌧ is the period of the orbit. The areal2⌧ A of an ellipse can also be written as A = ⇡ab where a and b are the semi-major and semi-minor= ⇡ab axes, yielding (71) The ratio of a and b can be expressed in2 terms of the eccentricity l⌧ The ratio of a and b can be expressed inb terms= ⇡ ofab the eccentricity (71) =2 1 ✏2. (72) ba The ratio of a and b can be expressed in= termsp1 of✏ the2. eccentricity (72) Using this expression to substitute b ina (71) p b 2 Using this expression to substitute b in (71)2=⇡a21 ✏ . (72) ⌧ =a 1 ✏2 (73) lp 2⇡a2 Using this expression to substitute b⌧ in= (71) p1 ✏2 (73) The length of the major axis is l 2 2⇡a p 2 The length of the major axis is ⌧ = 1 2ml✏ 2 (73) 2a = r + rl = . (74) 0 1 pk(1 ✏2) The length of the major axis is 2ml 2 2a = r0 + r1 = 2 . (74) Squaring (71) and replacing gives k(1 2✏ ) 2 2ml 2a = r2 + r4⇡=m 3 . (74) Squaring (71) and replacing gives ⌧0 = 1 ka(1, ✏2) (75) k2 2 4⇡ m 3 confirmingSquaring (71) Kepler’s and replacing 3rd law. gives ⌧ = a , (75) k 4⇡2m confirming Kepler’s 3rd law. ⌧ 2 = a3, (75) 4.2 Hamiltonian dynamics of many-bodyk systems confirming Kepler’s 3rd law. 4.2The Kepler Hamiltonian problem is dynamics essentially a of two-body many-body problem. systems In the remainder of this course, we will be interested in classical (non-quantum) systems that consist of N 2 particles. The Thecomplete4.2 Kepler Hamiltonian microscopic problem is dynamics essentially dynamics of a such oftwo-body many-body systems problem. is encoded systems In the in their remainder Hamiltonian of this course, we willThe be Kepler interested problem in classical is essentially (non-quantum) a two-body systems problem. that In the consist remainder of N of2 this particles. course, The we N 2 completewill be interested microscopic in classicaldynamics (non-quantum) of suchpn systems systems is encoded that consist in their of HamiltonianN 2 particles. The H = + U(x1,...,xN ), (76a) complete microscopic dynamics of such2mn systems is encoded in their Hamiltonian nN=1 2 X pn H = + U(x1,...,xN ), (76a) N 2 where mn, pn(t) and xn(t) denote the2mp n mass, momentum and position of the nth particle. H =n=1 n + U(x ,...,x ), (76a) The first contribution on the rhs.X of Eq.2m (76a) is1 the kineticN energy, and U is the potential n=1 n whereenergy.mn For, pn our(t) purposes, and xn(t) it denote is suXcient the mass, to assume momentum that we and can positiondecompose of (76a) the nth into particle. a sum of pair interactions Thewhere firstm contributionn, pn(t) and onxn the(t) denote rhs. of the Eq. mass, (76a) momentum is the kinetic and energy, position and ofU theisn theth particle.potential energy.The first For contribution our purposes, on it the is surhs.cient of Eq. to (76a) assume is the that kinetic we can energy, decompose and U (76a)is the into potential a sum 1 ofenergy. pair interactions For our purposes,U it(x is,..., sucientx )= to assume that(x we, x can). decompose (76a) into a(76b) sum 1 N 2 n k of pair interactions n,k:n=k 1 X6 U(x ,...,x )= (x , x ). (76b) 1 N 21 n k U(x1,...,xN )= n,k:n=k (xn, xk). (76b) 2 X6 16n,k:n=k X6 16 16 4.1.3 Third law To prove Keplers third law go back to equation (55). Integrating this area law over time gives ⌧ l⌧ A(⌧)= dA = , (70) 2 Z0 where ⌧ is the period of the orbit. The area A of an ellipse can also be written as A = ⇡ab where a and b are the semi-major and semi-minor axes, yielding l⌧ = ⇡ab (71) 2 The ratio of a and b can be expressed in terms of the eccentricity b = 1 ✏2. (72) a p Using this expression to substitute b in (71)

2⇡a2 ⌧ = 1 ✏2 (73) l p The length of the major axis is

2ml2 2a = r + r = . (74) 0 1 k(1 ✏2) Squaring (71) and replacing gives 4⇡2m ⌧ 2 = a3, (75) k confirming Kepler’s 3rd law.

4.2 Hamiltonian dynamics of many-body systems The Kepler problem is essentially a two-body problem. In the remainder of this course, we will be interested in classical (non-quantum) systems that consist of N 2 particles. The complete microscopic dynamics of such systems is encoded in their Hamiltonian

N p2 H = n + U(x ,...,x ), (76a) 2m 1 N n=1 n X

where mn, pn(t) and xn(t) denote the mass, momentum and position of the nth particle. The first contribution on the rhs. of Eq. (76a) is the kinetic energy, and U is the potential energy. For our purposes, it is sucient to assume that we can decompose (76a) into a sum of pair interactions 1 U(x ,...,x )= (x , x ). (76b) 1 N 2 n k n,k:n=k X6

Given H, Newton’s equations can be compactly16 rewritten as

x˙ = H, p˙ = H. (77) Given H, Newton’s equationsn rpn can be compactlyn r rewrittenxn as That this so-called Hamiltonian dynamics is indeed equivalent to Newton’s laws of motion x˙ n = pn H, p˙ n = xn H. (77) can be seen by direct insertion, whichr yields r That this so-called Hamiltonian dynamics is indeed equivalent to Newton’s laws of motion p can be seen by direct insertion,n which yields x˙ n = , p˙ n = mnx¨n = xn U. (78) mn r pn x˙ n = , p˙ n = mnx¨n = xn U. (78) An important observation is thatmn many physical systemsr obey certain conservation laws. For instance, the Hamiltonian (76a) itself remains conserved under the time-evolution (77) An important observation is that many physical systems obey certain conservation laws. For instance,d the Hamiltonian (76a) itself remains conserved under the time-evolution (77) H = ( H) p˙ +( H) x˙ dt rpn · n rxn · n d n H =X ⇥ ( p H) p˙ +( x H) x˙ ⇤n dt = ( rHn ) (· n Hr)+(n · H) H 0. (79) n rpn · rxn rxn · rpn ⌘ nX ⇥ ⇤ =X ⇥ ( p H) ( x H)+( x H) p H⇤ 0. (79) r n · r n r n · r n ⌘ which is just the statement ofn energy conservation. Other important examples of conserved X ⇥ ⇤ quantitieswhich is are just total the statement linear momentum of energy and conservation. angular momentum, Other important examples of conserved quantities are total linear momentum and angular momentum, P = p , L = x p (80) n n ^ n P =n p , L =n x p (80) X n X n ^ n n n if the pair potentials only dependX on the distanceX between particles. ifThere the pair exists potentials a deep mathematical only depend connection on the distance between between such invariants particles. and symmetries of the underlyingThere exists Hamiltonian, a deep mathematical known as Noether’s connection theorem. between For such example, invariants energy and symmetries conservation of is athe consequence underlying ofHamiltonian, the fact that known the as Hamiltonian Noether’s theorem. (76a) is For note example, explicitly energy time-dependent conservation and,is hence, a consequence invariant of under the fact time that translations. the Hamiltonian Similarly, (76a) conservation is note explicitly of linear time-dependent momentum is linkedand, to hence, spatial invariant translation under invariance time translations. and conservation Similarly, of conservation angular momentum of linear momentum to rotational is invariance.linked to spatial translation invariance and conservation of angular momentum to rotational invariance.For the remainder of this course, it will be important to keep in mind that microscopic symmetriesFor the and remainder conservation of this laws course, must it be will preserved be important in coarse-grained to keep in mind macroscopic that microscopic contin- uumsymmetries descriptions. and conservation laws must be preserved in coarse-grained macroscopic contin- uum descriptions. 4.3 Practical limitations 4.3 Practical limitations Deriving Kepler’s laws required us to solve a second-order linear ordinary di↵erential equa- tion,Deriving which was Kepler’s obtained laws by required considering us to solvethe idealised a second-order case in linearwhich ordinary a single planet di↵erential is orbiting equa- thetion, sun. which If we consider was obtained a more by realisticconsidering problem the idealised in which case several in which planets a single orbit planet the sun, is orbiting all in- teractingthe sun. with If we each consider other a via more gravity, realistic the problem problem in becomes which several analytically planets intractable. orbit the sun, Indeed, all in- forteracting just two planets with each orbiting other via the gravity, sun one the encounters problem thebecomes celebrated analytically ‘three-body intractable. problem’, Indeed, for for just two planets orbiting the sun one encounters the celebrated ‘three-body problem’, for which there is no general analytical solution. Lagrange showed that there are some solutions which there is no general analytical solution. Lagrange showed that there are some solutions to this problem if we restrict the planets to move in the same plane, and assume that the to this problem if we restrict the planets to move in the same plane, and assume that the mass of one of them is so small as to be negligible. In the absence of an explicit solution mass of one of them is so small as to be negligible. In the absence of an explicit solution to the ‘three body problem’ one must use ideas from 18.03 to calculate fixed points of the to the ‘three body problem’ one must use ideas from 18.03 to calculate fixed points of the equationsequations and and investigate investigate their their stability. stability. However, However, now now you you see see the the problem. problem. The The critical critical numbernumber of equations of equations for for complicated complicated things things to to happen happen is is three three (for (for ODE’s), ODE’s), and and yet yet any

1717 Given H, Newton’s equations can be compactly rewritten as Given H, Newton’s equations can be compactly rewritten as x˙ = H, p˙ = H. (77) n rpn n rxn x˙ = H, p˙ = H. (77) That this so-called Hamiltoniann r dynamicspn is indeedn r equivalentxn to Newton’s laws of motion can be seen by direct insertion,Conservation which yields laws That this so-called Hamiltonian dynamics is indeed equivalent to Newton’s laws of motion can be seen by direct insertion,p whichn yields x˙ n = , p˙ n = mnx¨n = xn U. (78) mn r pn x˙ n = , p˙ n = mnx¨n = xn U. (78) An important observationm isn that many physical systemsr obey certain conservation laws. For instance, the Hamiltonian (76a) itself remains conserved under the time-evolution (77) An important observation is that many physical systems obey certain conservation laws. For instance, thed Hamiltonian (76a) itself remains conserved under the time-evolution (77) H = ( H) p˙ +( H) x˙ dt rpn · n rxn · n d n H = X ⇥( p H) p˙ n +( xn H) x˙ n ⇤ dt = (r n H)· ( rH)+(· H) H 0. (79) n rpn · rxn rxn · rpn ⌘ Xn ⇥ ⇤ = X ⇥( H) ( H)+( H) H⇤ 0. (79) rpn · rxn rxn · rpn ⌘ which is just the statementn of energy conservation. Other important examples of conserved X quantities are total linear momentum⇥ and angular momentum, ⇤ which is just the statement of energy conservation. Other important examples of conserved quantities are total linear momentumP = p and, angularL = momentum,x p (80) n n ^ n n n P = Xp , L = Xx p (80) n n ^ n if the pair potentials only dependn on the distancen between particles. X X There exists a deep mathematical connection between such invariants and symmetries of if the pair potentials only depend on the distance between particles. the underlying Hamiltonian, known as Noether’s theorem. For example, energy conservation There exists a deep mathematical connection between such invariants and symmetries of is a consequence of the fact that the Hamiltonian (76a) is note explicitly time-dependent the underlying Hamiltonian, known as Noether’s theorem. For example, energy conservation and, hence, invariant under time translations. Similarly, conservation of linear momentum is islinked a consequence to spatial translation of the fact invariance that the Hamiltonian and conservation (76a) of is angular note explicitly momentum time-dependent to rotational and,invariance.Reflection hence, invariant under time of translations. underlying Similarly, conservation symmetry of linear momentum ! is linkedFor to the spatial remainder translation of this invariance course, it and will conservation be important of to angular keep in momentum mind that to microscopic rotational invariance.symmetries and conservation laws must be preserved in coarse-grained macroscopic contin- uumFor descriptions. the remainder(Noether’s of this course, it will be importanttheorem) to keep in mind that microscopic symmetries and conservation laws must be preserved in coarse-grained macroscopic contin- uum descriptions. 4.3 Practical limitations 4.3Deriving Practical Kepler’s limitations laws required us to solve a second-order linear ordinary di↵erential equa- tion, which was obtained by considering the idealised case in which a single planet is orbiting Derivingthe sun. Kepler’s If we consider laws required a more realistic us to solve problem a second-order in which several linear planets ordinary orbit di↵ theerential sun, all equa- in- tion,teracting which with was each obtained other by via considering gravity, the the problem idealised becomes case in analytically which a single intractable. planet is orbiting Indeed, thefor sun. just two If we planets consider orbiting a more the realistic sun one problem encounters in which the celebrated several planets ‘three-body orbit the problem’, sun, all forin- teractingwhich there with is each no general other analyticalvia gravity, solution. the problem Lagrange becomes showed analytically that there intractable. are some solutions Indeed, forto just this two problem planets if we orbiting restrict the the sun planets one encounters to move in the the celebrated same plane, ‘three-body and assume problem’, that the for whichmass thereof one is of no them general is so analytical small as solution. to be negligible. Lagrange In showed the absence that there of an are explicit some solutions solution toto this the problem‘three body if we problem’ restrict one the must planets use to ideas move from in the 18.03 same to calculate plane, and fixed assume points that of the massequations of one and of theminvestigate is so small their stability. as to be negligible. However, now In the you absence see the of problem. an explicit The solution critical tonumber the ‘three of equations body problem’ for complicated one must thingsuse ideas to happenfrom 18.03 is three to calculate (for ODE’s), fixed points and yet of any the equations and investigate their stability. However, now you see the problem. The critical number of equations for complicated things to happen is three (for ODE’s), and yet any 17

17 Given H, Newton’s equations can be compactly rewritten as

x˙ = H, p˙ = H. (77) n rpn n rxn That this so-called Hamiltonian dynamics is indeed equivalent to Newton’s laws of motion can be seen by direct insertion, which yields

pn x˙ n = , p˙ n = mnx¨n = xn U. (78) mn r An important observation is that many physical systems obey certain conservation laws. For instance, the Hamiltonian (76a) itself remains conserved under the time-evolution (77)

d H = ( H) p˙ +( H) x˙ dt rpn · n rxn · n n X ⇥ ⇤ = ( H) ( H)+( H) H 0. (79) rpn · rxn rxn · rpn ⌘ n X ⇥ ⇤ which is just the statement of energy conservation. Other important examples of conserved quantities are total linear momentum and angular momentum,

P = p , L = x p (80) n n ^ n n n X X if the pair potentials only depend on the distance between particles. There exists a deep mathematical connection between such invariants and symmetries of the underlying Hamiltonian, known as Noether’s theorem. For example, energy conservation is a consequence of the fact that the Hamiltonian (76a) is note explicitly time-dependent and, hence, invariant under time translations. Similarly, conservation of linear momentum is linked to spatial translation invariance and conservation of angular momentum to rotational invariance. For the remainder of this course, it will be important to keep in mind that microscopic symmetries and conservation laws must be preserved in coarse-grained macroscopic contin- uum descriptions.

4.3 Practical limitations Deriving Kepler’s laws required us to solve a second-order linear ordinary di↵erential equa- relevant problem in thetion, world which contains was many obtained more than three by degrees considering of freedom. the Indeed idealised a case in which a single planet is orbiting physical problem… typically too many contains particles 1023 interacting in interesting particles (Avogadro’ssystems number), which is so great a number thatthe it sun. is unclear If we if the consider mathematical a more techniques realistic described problem above are in which several planets orbit the sun, all in- of any use. The centralteracting aim of this with course each is to make other theoretical via gravity, progress towards the problem under- becomes analytically intractable. Indeed, standing systems with many degrees of freedom. To do so we shall invoke the “continuum hypothesis”, imaginingfor that just the discrete two planets variable (e.g. orbiting the velocity the of sun a particular one encounters molecule the celebrated ‘three-body problem’, for Solution: continuum theory … leads to new challenges of fluid) can be replacedwhich with a there continuum is no (e.g. general the velocity analytical field v(x,t)). solution. There are many Lagrange showed that there are some solutions subtleties that arise in trying to implement this idea, among them; to this problem if we restrict the planets to move in the same plane, and assume that the (i) How does one writemass down of macroscopic one of them descriptions is so in terms small of microscopic as to be constants negligible. In the absence of an explicit solution in a systematic way? It would be terrible to have to solve 1023 coupled di↵erential equations! to the ‘three body problem’ one must use ideas from 18.03 to calculate fixed points of the

(ii) Forces and e↵ectsequations that apriori andappear investigate to be small are their not always stability. negligible. However, This now you see the problem. The critical turns out to be ofnumber fundamental of importance, equations but for was complicated not recognised universally things until to happen is three (for ODE’s), and yet any the 1920’s.

(iii) The mathematics of how to solve ‘macroscopic equations’, which are nonlinear partial di↵erential equations, is non-trivial. We will need to introduce many new ideas.17

In tackling these problems we will spend a lot of time doing fluid mechanics, the reason being that it is by far the most developed field for the study of these questions. Experiments are readily available and the equations of motion are very well known (and not really debated!). Furthermore, fluid dynamics is an important subject in its own right, being relevant to many di↵erent scientific disciplines (e.g. aerospace engineering, meteorology, co↵ee cups). We will also introduce other examples (e.g. elasticity) to show the generality of the ideas.

4.4 Suggestions For more details on Kepler’s laws, and a java applet to let you play with them, go to

http://csep10.phys.utk.edu/astr161/lect/history/kepler.html

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