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) m1 m2 |F12 | = G 2 r 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 Why do we study applied mathematics? Aside from the intellectual challenge, it is reason4 to Kepler’s andanHamiltonian dynamics able argue that weproblem do so to obtain understanding of physical 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 applied mathematics? Aside from thedesire intellectual challenge, is reasonof Why muchdo of we thestudy mathematics we do, came from Newton’s to understand theit motion to arguewhich thatdowe so applied to obtain understanding ofintellectual physical challenge, phenomena, and to be Why wedo study Aside from the it is reasonof able the planets, were known to mathematics? obeyanKepler’s laws. to argue that we do so to obtain an understanding physical phenomena, and the to beorigin able to make able predictions about them. Possibly the greatestof example of this, and able to make predictions about them.from Possibly the greatest example of this, and thethe origin of much of the mathematics we do, came Newton’s desire to understand motion 4.1 Kepler’soflaws motion much ofoftheplanetary mathematics we do, came from Newton’s desire to understand the motion of the planets,of which were known to obey Kepler’s laws. the planets, which were known to obey Kepler’s laws. In the early seventeenth century (1609-1619) Kepler proposed three laws of planetary motion 4.1TheKepler’s laws of planetary motion (i) orbits of the planets are ellipses, with the Sun’s centre of mass at one focus of In the early seventeenth century (1609-1619) Kepler proposed three laws of planetary motion theearly ellipse. In the seventeenth century (1609-1619) Kepler proposed three laws of planetary motion 4.1 Kepler’s laws of planetary motion (i) The orbits of the planets are ellipses, with the Sun’s centre of mass at one focus of (ii) (i) The lineorbits joining planet andare theellipses, Sun describes equal areas in equal intervals of time. The oftheathe planets with the Sun’s centre of mass at one focus of ellipse. the ellipse. The line joining a planet and the Sun describes equal areas in equal intervals of time. (iii) The squares(ii) of the periods of the planets are proportional to the cubes of their semimajor The squares of the periods of thedescribes planets are equal proportional cubesintervals of their semi(ii) The axes. line (iii) joining a planet and the Sun areastointhe equal of time. major axes. These based on periods detailedofobservations by Tycho Brahe, and put to rest (iii)laws Thewere squares of the the planets made are proportional to the cubes of their semiThese laws were based on detailed observations made by Tycho Brahe, and put to rest any notion thataxes. planets move in perfectly circular orbits. However, it wasn’t until Newton major any notion that planets move in perfectly circular orbits. However, it wasn’t until Newton proposed his lawproposed of gravitation in 1687 that thethat origins of this motion understood. his law of gravitation in 1687 the origins of this motionwere were understood. These proposed laws were based on detailed observations made by Tycho Brahe, and put to rest Newton that Newton proposed that any notion that planets move in perfectly circular orbits. However, it wasn’t until Newton “Every object in the Universe attracts every other object with a force directed along a line “Every object the of Universe attracts1687 everythat other object with force directed along a line proposed hisinlaw gravitation the origins of athis motion of centres for the two in objects that is proportional to the product of theirwere massesunderstood. and of centres for the two objects that is proportional to the product of their masses and Newton proposed that inversely proportional to the square of the separation of the two objects.” inversely proportional to the square of the separation of the two objects.” on Universe this one statement, it is possible to derive Kepler’s “Every objectBased in the attracts every other object withlaws. a force directed along a line Basedofon this one statement, it is possible to derive Kepler’s centres for the two objects that is proportional to thelaws. 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 a particle moving it under a central force, such as gravity, islaws. constant. By definition, the Based on thisofone statement, is possible to derive Kepler’s angular momentum L to of aderive, particle and with is mass m and velocity u isthe angular momentum Keplers second law is the simplest a statement that of 4.1.1 a particle movinglaw under a central force, such as gravity, is constant. By definition, the dr Second L=r^m , (53) dt angular momentum L of a particle with mass m and velocity u is Keplers second lawrisisthe to derive, and is The a statement thatof the angular momentum where the simplest vector position of the particle. rate of change angular momentum is dr given by under a central force, such as gravity, is constant. By definition, the of a particle moving L= (53) dLr ^ m dt , = r ^ fm= and r ^ f velocity (r)r̂ = 0, u is (54) angular momentum L of a particle with mass dt where r is the vector the force, particle. The only rateonofthechange angular momentum where fposition (r)r̂ is the of central depending distanceof r= |r| and pointing in the is “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 L=r^m dr , dt (53) 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)r̂ = 0, (54) dt where f (r)r̂ is the central force, depending only on the distance r = |r| and pointing in the direction r̂ = 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 joining 13 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, 2 2 dt 2m where L = |L| is a constant. The area swept out is therefore also constant. 4.1.2 First law (55) 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 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 the planets it. The The area equation motion for a planet is where L = in |L|orbit is a around constant. sweptofout is therefore also constant. 4.1.2 4.1.2 First law First law Toplane provepolar Keplers first law consider In coordinates the planets in orbit around it. The dr = dt d2 r = 2 dt In plane polar coordinates d2 r m 2 = f (r)r̂. (56) dt the sun as being stationary (i.e., infinitely heavy), and equation of motion for a planet is ˙ˆ ṙr̂ + (57a) d2rr✓✓, m 2 = f (r)r̂. (56) dt ˙2 ˙ ✓. ˆ (r̈ r✓ )r̂ + (r✓¨ + 2ṙ✓) (57b) In component form, equation (56) dr therefore becomes ˆ = ṙr̂ + r✓˙✓, dt m(r̈ r✓˙2 ) = f (r), d2 r ¨(r̈ ˙✓˙2 )r̂ ˙ ✓. ˆ m(r ✓ + 2 ṙ ✓) =+ 0. = r (r✓¨ + 2ṙ✓) 2 dt Putting (57a) into (53) gives (56) therefore becomes In component form, equation dr 2 2˙ f (r), L =m(r̈ r ^ mr✓˙ ) ==|mr ✓|. dt ˙ = 0. m(r✓¨ + 2ṙ✓) Thus Putting (57a) into (53) gives r2 ✓˙ = l, (57a) (58a) (58b) (57b) (58a) (59) (58b) (60) where l = L/m is the angular momentum perdrunit mass. ˙ Given a radial force f (r), equaL= r^m = |mr2 ✓|. (59) dt r and ✓ as functions of t. A more practical tions (58a) and (58b) can now be solved to obtain 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), equaRewriting equations ofnow motion in terms of the rnew requiresofthe tions (58a)the and (58b) can be solved to obtain andvariable ✓ as functions t. Aidentities more practical result is to solve for r(✓), however,1 and this1 requires definition of a new variable d✓ du the du ṙ = u̇ = = l , (62a) u2 u2 dt1 d✓ d✓ (61) 2 d du r =˙ u d2.u 2 2d u r̈ = l = l✓ 2 = l u . (62b) dt d✓ d✓ d✓2 In component form, equation (56) therefore becomes dr 2˙ Putting (57a) into (53) gives Lm(r̈ = r ^r✓m ✓|. (59) 2 ˙ )dt == f|mr (r), (58a) ˙ = 0. 2 dr m(r✓¨ + 2ṙ✓) (58b) ˙ L = r ^ m = |mr ✓|. (59) Thus dt r2 ✓˙ = l, (60) Putting (57a) into (53) gives Thus where l = L/m is the angular momentum2 per Given a radial force f (r), equadr unit mass. ˙ 2˙ r ✓ = l, (60) = r ^tomobtain = r|mr ✓|. (59) tions (58a) and (58b) can now beLsolved and ✓ as functions of t. A more practical dt resultl = is to solve for r(✓), however, and this the definition of a new variable where L/m is the angular momentum perrequires unit mass. Given a radial force f (r), equaThus tions (58a) and (58b) can now be solved to obtain 2˙ 1l, r and ✓ as functions of t. A more practical r ✓ = = . the definition of a new variable (60) (61) result is to solve for r(✓), however, and thisr requires u where l = L/m is the angular momentum per unit mass. Given a radial force f (r), equa1 Rewriting motion in terms new✓ variable requires the identities tions (58a) the andequations (58b) can of now be solved to robtain as functions of t. A more practical =of the . r and (61) u result is to solve for r(✓), however, and of a new variable 1 this requires 1 d✓ duthe definition du = in 2terms u̇ = of 2the new=variable l , requires the identities (62a) Rewriting the equations of ṙmotion u u 1 dt d✓ d✓ . d2 u (61) 2 d du r = 1 1 d✓ du du u 2 2d u ˙ l u̇ = = l✓ 2 = = ll u , 2 . (62b) ṙ r̈ == (62a) 2 2 dt d✓ d✓ d✓ d✓ requires the identities Rewriting the equations of motionuin terms uof dt thed✓ new variable 2u 2u d du d d Equation (58a) becomes 2 2 ˙ du =✓ l ◆ r̈ = l1 = 1 l✓d✓ u . (62b) du 2 2 = 1l 2 ṙ = d u , (62a) 1 dt2 u̇d✓= d✓ d✓ u2 + u = u2 dt2d✓2 f d✓. (63) d✓ ml u u 2 Equation (58a) becomes d du d2 u ✓ ◆ 2 2d u ˙ r̈ = d2 ul = l✓1 2 = 1l u . (62b) 2 dt d✓ d✓ d✓ + u = 142 2 f . (63) 2 d✓ ml u u Equation (58a) becomes ✓ ◆ 2 d u 1 1 + u = f . (63) 14 d✓2 ml2 u2 u 14 ṙ = r̈ = 1 1 d✓ du du u̇ = = l , u2 u2 dt d✓ d✓ 2 d du d2 u 2 2d u ˙ l = l✓ 2 = l u . dt d✓ d✓ d✓2 Equation (58a) becomes d2 u +u= d✓2 1 f ml2 u2 ✓ ◆ 1 . u (62a) (62b) (63) 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 governing of a particle under a central force. d2 u the motion k + uof=the orbit . r = r(✓), the force function can (64) Conversely, if one is given the polar equation be d✓2 ml2 derived by di↵erentiating and putting the result into the di↵erential equation. According to Newtons of gravitation This has thelaw general solution f (r) = k/r2 , so that k d2 u k u = Acos(✓ ✓ ) + + u = 0 2. 2, 2 d✓ ml ml (65) (64) where A the andgeneral ✓0 are solution constants of integration that encode information about the initial This has 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) 2 ml 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, which is the equation of a conic section with the origin at the focus. This can be rewritten ✓ ◆ 1 in standard form k 1+✏ r =r = Acos✓ (66) r0 + ml2 , (67) 1 + ✏cos✓ where which is the equation of a conic section with the origin at the focus. This can be rewritten 2 2 Aml ml in standard form ✏= , r = . (68) 0 k r = r 1 + ✏ , k(1 + ✏) (67) 0 1 + ✏cos✓ ✏ is called the eccentricity of the orbit: where 2 2 u = Acos(✓2 ✓0 ) + , (65) 2 mlk d u + u = . (64) 2 2 d✓ ml where A and ✓0 are constants of integration that encode information about the initial conditions. Choosing ✓0 =0solution and replacing u by the original radial coordinate r = 1/u, This has the general ✓ ◆ 1 k k = Acos(✓ ✓ ) + , r = u Acos✓ + (66) (65) 0 2 2 ml ml and ✓0 ofare constants of with integration that information about the initial whichwhere is theAequation a conic section the origin at encode the focus. This can be rewritten conditions. in standard formChoosing ✓0 =0 and replacing u by the original radial coordinate r = 1/u, 1+✏ ◆ 1 r = r0 ✓ , (67) k 1 + ✏cos✓ r = Acos✓ + (66) 2 ml where 2 Aml ml2 , r0 = origin at . (68) which is the equation of✏ a=conic k section with thek(1 + ✏) the focus. This can be rewritten in standard form ✏ is called the eccentricity of the orbit: 1+✏ r = r0 , (67) 1 + ✏cos✓ • ✏ = 0 is a circle, where 2 2 Aml ml • ✏ < 1 is an ellipse, ✏= , r0 = . (68) k k(1 + ✏) • ✏ = 1 is a parabola and ✏ is called the eccentricity of the orbit: • ✏ > 1 is a hyperbola. • ✏ = 0 is a circle, For an elliptical orbit, r0 is the distance of closest approach to the sun, and is called the • ✏ < 1 is an ellipse, perihelion. Similarly, (69) • ✏ = 1 is a parabola and r1 = r0 (1 + ✏)/(1 ✏) is the furthest from the sun and is called the aphelion. Orbital eccentricities are • ✏ > 1distance is a hyperbola. small for planets, whereas comets have parabolic or hyperbolic orbits. Interestingly though, Halley’s For comet a veryorbit, eccentric but, according toapproach the definition not the an has elliptical r0 is orbit the distance of closest to thejust sun,given, and isis called a comet! perihelion. Similarly, r1 = r0 (1 + ✏)/(1 ✏) (69) is the furthest distance from the sun and is called the aphelion. Orbital eccentricities are • ✏ = 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, r1 = r0 (1 + ✏)/(1 ✏) (69) 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! P 15 A 4.1.3 Third law 4.1.3 Third lawthird law go back to equation (55). Integrating this area law over time To prove Keplers gives To prove Keplers third law go back to equation Z ⌧ (55). Integrating this area law over time l⌧ 4.1.3 Third law gives A(⌧ ) = Z dA = , (70) ⌧ 2 0 l⌧ To prove Keplers third law go back equation A(⌧to) = dA (55). = ,Integrating this area law over time (70) where ⌧ is the period of the orbit. The area0A of an ellipse can also be written as A = ⇡ab 2 gives Z ⌧ where a and b are the semi-major and semi-minor axes, l⌧ yielding where ⌧ is the period of the orbit. The area A of an ellipse can also be written as A = ⇡ab A(⌧ ) = dA = , (70) 2 where a and b are the semi-major and semi-minor axes, yielding l⌧ 0 = ⇡ab (71) where ⌧ is the period of the orbit. The area A of an ellipse can also be written as A = ⇡ab 2 l⌧ = ⇡ab axes, yielding (71) where a and b are the semi-major and semi-minor The ratio of a and b can be expressed in 2terms of the eccentricity l⌧ p The ratio of a and b can be expressed in of the eccentricity = ⇡ab (71) b terms =2 1 ✏2 . (72) a b p 2 . eccentricity The ratio of a and b can be expressed in=terms 1 of✏the (72) Using this expression to substitute b ina (71) b p = 2 1p ✏2 . (72) Using this expression to substitute b in (71) 2⇡a a 2 ⌧= 1 ✏ (73) l 2p 2⇡a Using this expression to substitute b in (71) ⌧= 1 ✏2 (73) The length of the major axis is l 2⇡a2 p ⌧= 1 2ml ✏22 (73) The length of the major axis is l 2a = r0 + r1 = . (74) k(1 2✏2 ) 2ml The length of the major axis is 2a = r0 + r1 = . (74) k(1 2✏2 ) Squaring (71) and replacing gives 2 m 2ml 4⇡ 2a = ⌧r02 + r = (74) = 1 k(1 a3 , ✏ 2 ) . (75) Squaring (71) and replacing gives k2 4⇡ m 3 2 confirming Kepler’s 3rd law. ⌧ = a , (75) Squaring (71) and replacing gives k2 4⇡ m 3 ⌧2 = a , (75) confirming Kepler’s 3rd law. 4.2 Hamiltonian dynamics of many-body systems k confirming law. The Kepler Kepler’s problem 3rd is essentially a two-body problem. In the remainder of this course, we 4.2 Hamiltonian dynamics of many-body systems will be interested in classical (non-quantum) systems that consist of N 2 particles. The Squaring (71) and replacing gives 4⇡ 2 m 3 ⌧ = a , k 2 (75) 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 X p2n H= + U (x1 , . . . , xN ), 2mn (76a) n=1 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 sufficient to assume that we can decompose (76a) into a sum of pair interactions 1 X U (x1 , . . . , xN ) = 2 (xn , xk ). (76b) n,k:n6=k 16 Given H, Newton’s equations can be compactly rewritten as ẋn = rpncan H ,be compactly ṗn = rrewritten xn H. Given H, Newton’s equations as (77) That this so-called Hamiltonian dynamics ẋn = rpn H , is indeed ṗn = equivalent rxn H. to Newton’s laws of motion (77) can be seen by direct insertion, which yields That this so-called Hamiltonian dynamics is indeed equivalent to Newton’s laws of motion pn can be seen by direct ẋinsertion, ,whichṗyields (78) n = n = mn ẍn = rxn U. mn pn ẋ = , ṗ = mn ẍn = rxn U. certain conservation laws. (78) n An important observation is that mn many nphysical systems obey 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. X ⇥ (76a) itself remains conserved ⇤ under the time-evolution (77) For instance, dthe Hamiltonian H = (rpn H) · ṗn + (rxn H) · ẋn dt ⇤ nX ⇥ d ⇤ H =X ⇥ (rpn H) · ṗn + (rxn H) · ẋn dt = (79) n (rpn H) · ( rxn H) + (rxn H) · rpn H ⌘ 0. X ⇥ ⇤ n Given H, Newton’s equations can be compactly rewritten as Given H, Newton’s equations can be compactly rewritten as ẋn = rpn H , ṗn = rxn H. (77) Conservation laws ẋn = r rxn H. (77) pn H , n = equivalent That this so-called Hamiltonian dynamics is ṗ indeed to Newton’s laws of motion can be seen by direct insertion, which yields That this so-called Hamiltonian dynamics is indeed equivalent to Newton’s laws of motion can be seen by direct insertion,pwhich yields ẋn = n , ṗn = mn ẍn = rxn U. (78) m n p ẋn = n , ṗn = mn ẍn = rxn U. (78) m An important observation is nthat many physical systems 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. X ⇥(76a) itself remains conserved For instance, the dHamiltonian ⇤ under the time-evolution (77) H = (rpn H) · ṗn + (rxn H) · ẋn X ⇤ n ⇥ ddt H = X ⇥(rpn H) · ṗn + (rxn H) · ẋn ⇤ dt = n (rpn H) · ( rxn H) + (rxn H) · rpn H ⌘ 0. (79) X ⇤ n ⇥ = (rpn H) · ( rxn H) + (rxn H) · rpn H ⌘ 0. (79) which is just the statementnof energy conservation. Other important examples of conserved quantities are total linear momentum and angular momentum, which is just the statement of energy conservation. Other important examples of conserved X X quantities are total linear momentum and angular P = pn , L = momentum, xn ^ pn (80) X X n n P = pn , L= xn ^ pn (80) if the pair potentials only depend on the distance n n between particles. 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 of the fact invariance that the Hamiltonian (76a)ofisangular note explicitly time-dependent to spatial translation and conservation momentum to rotational and, hence, invariant under time translations. Similarly, conservation of linear momentum is invariance. linked spatial translation invariance of to angular momentum rotational Fortothe remainder of this course, itand willconservation be important keep in mind thatto microscopic invariance. symmetries and conservation laws must be preserved in coarse-grained macroscopic continFordescriptions. the remainder of this course, it will be important to keep in mind that microscopic uum symmetries and conservation laws must be preserved in coarse-grained macroscopic continuum descriptions. Reflection of underlying symmetry ! (Noether’s theorem) 4.3 Practical limitations Deriving Kepler’s limitations laws required us to solve a second-order linear ordinary di↵erential equa4.3 Practical tion, which was obtained by considering the idealised case in which a single planet is orbiting Deriving Kepler’s laws required us to solve a second-order linear ordinary di↵erential equa- uum descriptions. 4.3 Practical limitations Deriving Kepler’s laws required us to solve a second-ord relevant problem in the world contains many more than three degrees of freedom. Indeed a tion,contains which was obtained by systems considering the idealised ca physical problem … typically 1023 interacting particles (Avogadro’s number), which too many particles in interesting theitsun. If we consider a more realistic problem is so great a number that is unclear if the mathematical techniques described above arein which of any use. The central aim of this course is to make theoretical progress towards underteracting with each other via gravity, the problem becom standing systems with many degrees of freedom. To do so we shall invoke the “continuum two planets orbiting theofsun one encounters the hypothesis”, imaginingfor thatjust the discrete variable (e.g. the velocity a particular molecule Solution: continuum theory … leads to(e.g. new challenges of fluid) can be replaced with a continuum the velocity field v(x, t)).solution. There are many which there is no general analytical Lagrange s subtleties that arise in trying to implement this idea, among them; to this problem if we restrict the planets to move in th (i) How does one write down macroscopic descriptions in terms of microscopic constants mass of one of them is so small as to be negligible. In in a systematic way? It would be terrible to have to solve 1023 coupled di↵erential to the ‘three body problem’ one must use ideas from 1 equations! equations investigate their stability. However, now (ii) Forces and e↵ects that a prioriand appear to be small are not always negligible. This turns out to be of fundamental was complicated not recognised universally number of importance, equationsbutfor things until to happen 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