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This 163 page Class Notes was uploaded by Jamaal McGlynn on Tuesday October 20, 2015. The Class Notes belongs to PHYS608 at San Diego State University taught by M.Bromley in Fall. Since its upload, it has received 19 views. For similar materials see /class/225318/phys608-san-diego-state-university in Physics 2 at San Diego State University.
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Date Created: 10/20/15
Lecture 22 Outline Canon transforms o Modi ed Variational Principle Section 85 o brie y Principle of Least Action Section 86 o Canonical Transforms Section 91 o and Cyclic Coordinates Section 93 annuva variational principle 0 This is way cool With Action integral 6 6 f Salt 0 which immediately gives modi ed Hamilton s principle t2 152 616 qq39tgtdt6 piqi Hltqptgtidto t1 t1 0 ie integrand has fq 1121525 Where in the spirit of the Hamiltonian formulation independently vary q and p Effectively have 2n Euler Lagrange eqns d a f a f 8H 8H thus dt q q easy hey we have varied paths in both q and p Least action principle I at end points in phase space all 6ql O and 6191 0 Old 6 Variati0n only had 771051 771052 0 Introduce A Variation with 771051 7 0 771052 7 0 t2 152At2 152 A light adt Odt t1 t1At1 151 then if a H is conserved on both paths and b at the end points we require all Aqi O we get the principle of least action 152 A Pith 0 151 Least action principle II Least action A fit pigl 0 note sum over non relativistic mechanics example with quadratic qi s 1 T Z EMjkqQiji jk If V not velocity dependent H is conserved 33 3T 1 1 iiQi Z EMinj Z EMiij 375139 375139 aE iqz 152 A may 151 Pi but then pigl 2T ie So of all conserving energy path s the chosen one minimises time Fermat s principle of geometrical optics also Hertz s principle particle path has least curvature Cyclic coords are not trivial Solutions to Hamilton s equns simple if all q are cyclic since all the conjugate momenta are constant ie 8H 8 Now if Hqp t Hp H ie doesn t have q s or t p and p 0 thus gal04 then the other set of Hamilton s equations are then 3H 8H qz 306i as the 04 w are constant in time q cal15 too trivial Well consider planar motion MUM q1xorq2yORq1rorq26 ie for central forces 9 3 not cyclic but 6 is gt easy soln Point transformations Where possible want to use cyclic coords 11 with transform equations from 11 to Q are Qi Qlq t con guration space Point transformations see Goldstein Derivation 110 o Lagrange s eqns are invariant under point transforms o Hamiltonian formulation simultaneous point transforms Qi Qzqm t and P1 13101719 t phase space 0 since both q and p are indep variables Kamiltonian 0 But we also want new coordinates Q and P o to also be canonical for some ICQ P 15 ie 31C 31C and o where IC is the Hamiltonian in new coords with both 1 t2 t2 6 Pia memdt and 6 piqi Hqptgtdt0 t1 t1 0 does not mean that both integrands are equal instead dF dt 0 since we are considering zero variation at both ends MIMI H 1C 0 but what does this rnean and how does it help us Scale transforms A M19192 H 1C 71 o multiplicative constant A just a scale transformation eg Q uqz P sz means CQ P WHOM 0 since our guess for IC can show that our 31C LLV 8H I Qi y MQZ o GENERALLY can nd a transform with A 1 anyway ie given qp gt Q P with A 7 1 but then Q7 P gt Q 7 P with A then we have qp gt Q P with A 1 0 so we have a canonical transform when piQiHPiQiKC l I Generating functions The Cil J contribs to variation of action only at end points 0 if F is a function of q p Q Rt it vanishes at end points ie F is used as a canonical variables bridge between half of the old set and half of the new set Simplest general example F F1q Q t then dF1 8F1 8F1 8F1 iiHPi iIC Pi iIC i i m Q dt Q 8t8qu8QiQ o collecting the like terms since independent variables 8F1 3F 3F pi aql DP 862 I 0 ie F generates the eqns of canonical transforms SHO Gen Funcs SHO Canonical Transform SHO Solutions SHO Phase Space Lecture 24 Outline HJ examples 0 more on Canonical Transforms Section 92 o Hamilton Jacobi Examples Section 102 o Hamilton Jacobi onetwo ways Section 103 o if time Action Angle variables Section 106 Problem 922 point transforms 0 Two degrees of freedom given point transforms Q1 9 81101622 qiQ2 i Find most general canonical transform eqn s for 131132 ii Show that with a choice for 31132 that 2 p p H 192 C11 922 91 0 can be transformed such that Q1 and Q2 are ignorable iii then solve the problem nding ql 12191192 as functions of time Point transforms Example Point transforms Example Point transforms Example HJ Example 1 contd From last lecture with H E g m2w2q2 then we used both Harnilton s Principal S and Characteristic functions to show q 732 sinwt 5 and p V2ma coswt 5 as a check we do have energy conserved ie H E a p2 mw2 2 q 1 cos2wt 5 sin2wt 5 2mg 2a PUNCHLINE Harnilton s principal function is the generator of a canonical transform to a new coordinate that measures the phase angle of the oscillation The new coordinate is the total energy HJ Example 2 HJ Example 2 contd H J Example 3 setup HJ Example 3 dodge the bullet The Other HJ Deal Whenever the original Hamiltonian is indep of time with the equations of transformation of type II 8W 8W 3W qi 139 63 am then we can use the restricted Hamiltonian Jacobi Eqn 8W Hqyp 1061 Pi ie choose one integration constant as energy IC a1 81C 8T2i O chosen Pl 041 3 1 372 1 other 0 so only one coord not constant of motion Q1 15 51 E 3 While the others Qi 51 8 W Energy H time 80 ie canonical eqns B and also Q1 Setup Summary two ways Periodic Orbits eg Pendulum in phase space Actionangles w J Actionangles example Lecture 7 Outline Multi Pliers Lagrange Multipliers Section 24 Friction Section 15 25 Conservation of Momentum Sections 26 Conservation of Energy Sections 27 Lagrange Multiplier example II take II 0 Particle 0n hemisphere reduX Lagrange undetermined multipliers 0 effective Lagrangian L L 2211 Aafa c which we can expand using the calculus of variations as t2 6L d 6L 6 ldt gt6th0 an d an 0 Thus we now know that 8 a 8 O and so d 813 813 m 8 ZA LPQk dt 3 5 1 5 c which means that we now have to solve n m eqns 0 solving for each of the n coordinates ql gm 0 including m eqn s faq1 qn7 dditl 615 O 0 get the forces of constraint Qk through the solution Spare Lagrange Multiplier example III 0 bead moving on rotating Wire setup problem Velocity Dependent Potentials 15 o if L contains only conservative forces 0 and Qj represents forces not from potential d dt 5 39k 0 Consider frictional forces such as Ffm cxvx Qk 0 These come from Rayleigh s dissipation function 1 a 7 E kyvii where Ff 30m 0 for 239 particles Or we can write Fi Vm7 o The work done against friction is then dwf 41M 13fz7dt m Iggy kzvgmt 25E Fraction too much friction 7 kyvji o Generalised force was originally given by a aa am an a F i Vluif VIUiF Q3 2 f 3 3 3 3 l 0 With the Lagrange Eqn s of motion are d 63 63 37 8 f 37 0 Where A E 0 Where two scalars functions are reqdincl f constraint 0 eg Stokes Law sphere radius a moving at velocity 17 in medium viscosity 77 experiences drag 67T77a17 Practical Example Stokes Law Generalised Momentum I Generalised Momentum II Energy Function SometimeS h T V E Part I Sometimes h T V E Part II Final Note on Dissipation Lecture 18 Outline Arti cial Forces 0 Finite and In nitesimal Rotations Section 4748 o Non inertial reference frames Section 49410 Passive vs Active 0 Can think of transform either on axes or on vector o 18 passive vs active transforms Rotations don t commute 0 Problem is that rotations dont commute AB y BA 0 eg consider two rotations one 900 about 2 o and also one 900 about new y Finite Rotations 0 Consider active rotation of vector clockwise by angle I 0 ie xed co ordinate system gives Rotation formula F Fcosltlgt Fl cosltlgt7quotgtlt sinltlgt 0 thus for a small change dltlgt we have dFF Fm Fl lFgtlt dltlgtFgtlt dltlgt a but from the In nitesimal d F x d so d dCID Rate of Change of vectors During time dz vector 3 moves in space frame but doesn t move in body frame Difference is a rotation d space d body dmt d body d x 6 ie counter clockwise rotation dltIgt of body in space frame needs opposite dF F x d d x F d mt Dividing by the differential in time d 6 d 6 a x 3 dz dz w space body where J is instantaneous angular velocity didt d6 and d is instantaneous velocity during 2 gt t dt Frame Transformation In effect operator 8 r af x Consider the stars as the inertial frame s frame 0 Consider a particle moving on rotating earth r fraine Then the velocityies of the radius vector F are d d an a nnxn w 7 gt vvr w 7 dz 8 dz r S o and the accelerations are found by a 2nd application idz7s dz7s 3 as dts dzr M US dUr d wgtltrwgtltvrwgtltr Effective Rotation Forces In terms of F mag and Feff ma l e 2mwxa mwxcamel According to observer in the rotating system sees Feff Coriolis term is when object moves 6 75 0 Centrifugal term is the J x a x F perpendicular to J Centrifugal Effect EeffF 2mc3gtltz7r mc3gtlt tax a Magnitude of centrifugal term is mw27 sin 6 0 Example Stationary particle on Earth ie 27 0 0 add Sun Earth s rotation wE 27T24 x 3600 w w siderealdaysinyear 27T 36625 N 7 3X10581 Stars E solar daysinyear 24 x 3600 36525 N 39 o centripetal acceleration at equator wgtarsr m 338 377152 a only about 03 of the earths gravitational acceleration Coriolis Effect Coriolis term is when object moves 27715 x 6 75 0 Consider Coriolis De ection ie planar motion on surface in Northern hemisphere veers to right Southernleft At equator a is horizontal so no horizontal de ection Max magnitude Coriolis acceleration 2cm x 14 x 10 41 visualise in stars frame cons of angular momentum Effect of Coriolis Effect Winds eg trade Winds gulf stream cyclones air moves from high to low pressure regions ignore vertical motion air pressure balances gravity so Winds are generally in horizontal direction Freefall Coriolis So given that of lies along North South 0 Freefall results in East motion With Coriolis acceleration d2a m dt2 0 of Goldstein Page 179 2m a X 27 2mwvz sin6 More Freefall Coriolis Lecture 15 Outline Rigid Body 0 Rigid Body Coordinates Section 41 Orthogonal Transformations Section 42 o more basic Linear Algebra Section 43 Euler Angles Section 44 Euler s Theorem on Rigid Body Motion Section 46 if time possibly further Rigid Body Coordinates Rigid body xed relative coordinates between particles Requires 6 independant coordinates o 3 to specify location of a reference point in the body 3 to specify orientation of the body 0 Consider N particles 3N degrees of freedom 1 NN 1 eqns cm 0 but i th point only needs distances to 3 other points Assigning Coordinates 3 reference points have 9 degrees of freedom With constraint eqns 7712 12 7723 c23 7713 C13 ie point 1 needs 3 coords point 2 then constrained to sphere 2 coords and point 3 around axis 1 coord How to assign the 6 indep coordinates Can take Cartesian coords at reference point Minimum 3 coords for origin 3 for orientation There are many ways to specify the orientation Direction Cosines 0 Can specify the orientation with 9 direction cosines 0 eg 931 axis speci ed by 3 direction cosines a11a12a13 o with eg all cos611 i icos611 i iEii o iezquotcos611icos612jcos613k 0 eg if 613 7T2 just rotation in 93113 plane Directions from direction cosines o 9 direction cosines specify orientation of axes 0 ie the 931 932 93 axes relative to 91 92 93 axes i cos 6115 cos 612 3 cos 613 E j cos 6215 cos 622 3 cos 623 E I cos 6315 cos 632 3 cos 633 E 0 other vectors 77 x1i 932339 9331 9332 x gj xgk 0 Inner product gives lengths in terms of 93113 3733 931 77 2quot x1x2x3 cos 611cos 612cos 613 cos 611 931 cos 612 932 cos 613 93 Redundant Direction Cosines Out of 9 only need 3 needed for orientation due to Orthogonality E E 2 quot 0 along With o normalised 5 E 1 means orthonormal 0 Use these conditions to nd constraint relations 0 along with inverses 239 cos 611 zquot cos 621 j cos 631 k 3 3 iiE cos26111 and ijE cos611c0s6120 3 o in general 6 eqns E cos 61m cos 6m 6m 1 0 ie direction cosines are not generalised coordinates Transformation Matrix Really all about Linear Algebra with orthogonal transformations such as 77 A77 and consider the direction cosines as matrix elements a cos 6 1 0011 0012 0013 1 2 0021 0022 0023 2 933 0031 0013 933 Warning Goldstein uses Einstein summation notation 0032 l j ie sum is implied Whenever index appears note problem with gt LUZxi ie l lOt Orthogonal Transform vector remains unchanged by change in coordinates 2 r xix 9 aijxjaikxk aijaikxjxk ljkxjxk ie we have siX equations alljam ljk Vj k 1 2 3 such transformations are orthogonal eg planar rotation cos gb sin gb 0 all an O sin gb cos gb O am an O L O 0 1J L O 0 1J four matrix elements three equations with j k 1 2 eg allall algalg cos2 gb sin2 gb 1 leaves one coordinate reqd to describe the actual rotation Passive vs Active 0 Can think of tranform either on axes or on vector 0 ie passive vs active transforms Linear Algebra ultra brief Matrix addition is commutative ie A B B A matrix multiplication is non commutative in general ieCAB7 BAD eg consider B Where as bijj 93 Oilkm aikbijj cijxj and similarly the matrix elements allj bikakj 7 cm But both matrix addition and multiplication are associative eg ABC ABC Inverse and Orthogonality De ne inverse A 1 that takes 77 back to 77 91 alm as akixi ama m j x 6ijj xk 19 J ie since xjlg are independent we need aman 6 This is just the matrix product AA 1 I Defn Orthogonal matrix has A 1 A transpose of A Proof consider the following double sum aklamagj clian 6Man agj Where used the orthogonality condition aklam 6 but using the inverse amongan akldkj akldkj a Which means that agj a ie just the transpose and since aman ama 6 ie AA 1 I Inversion and Determinants o In general we have determinants AB A B o and for an orthogonal matrix A i1 eg cosgb singb O A singb cosgb O has A 1 O O 1 cos gb sin gb singb cosgb 0 Thus for an orthogonal matrix AA I A2 1 0 ie A i1 Consider inversion matrix With A 1 1 0 0 1 0 0 0 0 0 S I O 1 O O 1 O O O O 0 0 1J 0 0 OJ 0 0 1J o Transforms right handed system into left handed Euler Angles 0 so the 9 a are not independent and we 0 want orthogonal transform matrices with A 1 o introduce the three Euler angles gb 6 gb 0 there are other common conventions headingpitchroll Euler Angle Transformations 0 We have vectors 5 D53 6 7 C5 and Li BS7 0 so the overall transform is 2 Ag BCDf cos gb sin gb O 1 O 0 cos w sin w 0 A sin gb cos gb O 0 cos 6 sin 6 sin w cos w 0 O O 1 O sin 6 cos 6 O O 1 0 Full matrix expansion in Goldstein 446 0 eg all coswcosgb cos6singbsin a33 cos6 o importantly we also have easy inverse Liquot A 1f A52 Euler s Thm of rigid body motion Euler s Theorem The general displacement of a rigid body with one point xed is a rotation about some axis For this to be true there must be one vector lying along rotation axis that is unaffected by the rotation operation R AR R This is just an eigenvalue problem APE AI 6 Where real orthogonal matrix A has eigenvalue A 1 The eigenvector I X Y Z is speci c to A Which is found by solving the A eigenproblem all AX 0Lng 0Lng 0 0Lng G22 MY a23Z O a31X a32Y 0033 NZ 0 an example Lecture 19 Outline Coupled Oscillators 0 end of non inertial reference frames Section 49 410 o Oscillations theory of Section 61 62 a Normal Coordinates Section 63 0 Linear Triatornic Molecule Section 64 Effect of Coriolis Effect Winds eg trade Winds gulf stream cyclones air moves from high to low pressure regions ignore vertical motion air pressure balances gravity so Winds are generally in horizontal direction Freefall Coriolis So given that of lies along North South 0 Freefall results in East motion With Coriolis acceleration d2a m dt2 0 of Goldstein Page 179 2m a X 27 2mwvz sin6 More Freefall Coriolis Oscillations formalism intro interested in motion about equilibrium system is in equilibrium when generalised forces 8V i 0 Q 80 ie extremum at equilibrium q01 q02 qu As per usual equilibrium Classi ed stable unstable in terms of the deviations of system energy from equilibrium Oscillations formalism potential o All about Taylor series fa n a ZOO WW 1 aquot n0 n 0 Choose deviations of generalised coords from equilib qr 901 771 a so that the potential can be expanded about each gm 8V 1 82V V n V 7 7 n 917927 79 901 102 C10aqi0772aqiaqj077773 o implied summation We then set Vq01 q02 qu 0 o and at equilibrium we have 33 0 so 1 0 1 82V 1 VC17 Q27 7 Qn E aqaq 771417 Evain 2 J 0 Example LTlVI Linear Triatomic Molecule 0 Three atoms two of m and one of M both springs k a At equilibrium distance b 5102 5101 5103 5102 apart a potential energy simpli es with 77 xi 330 lt w3 NIET NDIPT NDIP ltltx2 x1 bf g ltltx3 x2 bf 773 7722 772 7712 2 2 Di 2773 772 771772 772771 772773 773772 m M xi 933 933 307 773 7073 2 Oscillations formalism kinetic o In the distant past 27 CZ Z 877139quot a j aqqu 815 a see Goldstein equations 171 and 172 on page 25 0 our generalised coords don t explicitly depend on time 2 1 1 97 1 Tzzgmweaw Zak z z j m o The coeffs mij are functions of qk so Taylor expand 377 MijltQ17Q277QTLgt mijQO17QO277QOn a 13 77k Qk 0 a only include constants since T has quadratic terms in q i a so we have mij E mzjq017 9027 7 Qon Tu and 1 1 T m j77 77j 515711717 Oscillations formalism Lagrangian So we have Lagrangian L a jnmj Vijnmj so if the 77 s are the generalised coords d 83 83 1 1 E877 an 5T 5 0 ie n equations of motion note TU and V17 Further most of the time Tm is diagonal such that 3 I an WWW so equations of motion Tm Vijnj 0 only sum over j eg for LTM kinetic only depends on each motion about equilibrium but potential change in one spring affects potential of the other Eigenvalue equations So differential equations Ej j lgjnj 0 Suggests oscillating solution Where Cat is some possibly complex amplitude With common C just some scale factor V coords But note that actual motion is given by real part so n linear homogeneous equations for each aj 2 6 ij CU O This is just an eigenproblem VEL ATEL requiring V11 ATll V12 AT12 V21 AT21 V22 AT22 0 LTM Building T and V tensors V I g 77 277 77 771772 772771 772773 7737722 T 07 77 773 a so we have the tensor forms k k 0 m 0 0 V k 2k k and T 0 M 0 0 k k 0 0 m a so we need to nd the eigenvalues w Via k w2m k 0 IV w2T k 2k MM k 0 0 k k w2m o ie cubic eqn w2k w2mkM 2777 w2Mm 0 LTM Eigenproblemo 0 16 cubic eqn w2k w2mkM 2m w2Mm 0 Frequencies Equations of motion satisfied by 77 Cam m with n free Vibration frequencies cal wn Since the differential equations are satisfied by 77 then the linear combination of 77 s also is and since actual motion only dependent on real part of 77 77 t Oa kewkt fka k cosw t 6k initial co ordinate conds set amplitude fk and phase 67 Problem is that 77 23 won t necessarily repeat unless the frequencies wk are related by fractions Lecture 3 Outline D Alembert and co 0 Constraints and Generalised Coordinates Section 13 a some examples a Bernoulli and D Alembert s principles Section 14 a if time start on Lagrange s Equations Section 14 Constraints To solve mechanics problems just equation of motion d277 mi F Z j dt2 holonomic constraints are those that can be written f7 17 2t 0 eg the rigid body has Ti T jl Cz jZO First problem Constraints couple 72 s and diff eqn s but usually there are non holonomic constraints eg from last lecture a Z 0 Second problem when Forces of constraints are unknowns eg act Via some effect on motion 33 Generalised Coordinates With no constraints N particles 3N Cartesian co ords With constraints k eqns f7quot17quot2 t 0 leaves 3N k degrees of freedom Introduce indep variables q1q2 q3Nk eg 2 D double pendulum has 4 co ords 2 variables 0 Useful in systems with Without constraints eg 7 6 gb Constraints Hemisphere Constraints Pendulum Constraints Pendulum contd Constraints Vertical Disk Constraints Vertical Disk contd Virtual Displacement Consider a Virtual in nitesimally small displacement 67 of each particle in the system consistent with all of the system s constraints at time 23 Suppose that the system is at equilibrium ie 6 so during 67 the Virtual work7 llVi 0 and the total Virtual work for all particles is 0 213572 212057 2182573 given applied forces Ff and constraint forces fi Bernoulli s Principle of Virtual Work a Virtual displacement 677 with 17 19677 0 0 Assume constraint forces 12677 0 do no net Virtual work eg 0 then the applied forces 17 19677 0 must also do no net Virtual work even though possibly PE 75 6 a note the 67 may be connected by constraints D Alembert s Principle 0 Rewrite the equation of motion dgi d a d Fi d pt O thus Fi Z6m O 0 ie the system will always be under equilibrium with a force equal to the reversed e eetive force 117139 dt 0 again split into applied and constrained forces 0212am m 62 02Ea d m 0 ie this is D Alembert s Principle for the Lost Force 672 o for systems where forces of constraint do no work a at this point let 17 2 example of D Alembert s Principle 0 Consider free particle motion 0 24 672 back into Generalised Coordinates 0 Now rewrite in terms of generalised coordinates 77 q1 12 mg 13 by the vector chain rule an 8 8t 8t d77 87an 1 dz k 0 Thus a Virtual displacement 67 displaces the set of 6 8F 57 quot5 Virtual displacement only on coordinates so 62 0 2155572 Z ltqu 62ij z j m 9 note Qj are the components of generalised Force Qj6qj has units of work eg if Qj E Nj then qu E 667 Lecture 12 Outline In and Out Burger o The Kepler Problem Section 38 o Hohmann Transfer Parabolic motion 6 1 Hyperbolic motion 6 gt 1 Intro to Scattering Section 310 Hohmann Transfer 0 Requires two velocity Changes for planet transfer 0 Which is the most energy ef cient way to do so Hohmann Transfer II o For both circles and ellipses we have E k2a o For circular path of the planets we get simple velocities k 1 k k k E mv2 thus 211 and v2 2T0 2 7 0 m r l 77173 a Transfer ellipse has at 7 1 7 2 With conserved energy k k k l k 2 mv2 k 2T2 6 271 Up gt 01 and va lt 02 mm 7 1 7 2 mrg 7 1 7 2 0 Whilst most energy ef cient also the slowest transit o and return must wait a While for the planets to be re aligned Parabolic Motion in time for V7 c7 Parabolic Motion in time for F7 c7 2 t e3 f0 d0 WW2 90 1 l ecos9 0 2 0 eg For Parabola s e 1 use 1 3086 2 C082 3 0 4 6 3 tang 1 2 t 4mk20 sec 2 d6 2mk20 51 da 0 Above used a tan so da sec2 d6 0 and also 8602 1 tar o Sowehave 7Tlt6lt7T o Where 6 7T gt t oo7 oo o Where 6 0 gt t 07 apsidal Stealing Gravity from a kid planet o But for interplanetary travel steal energy from a planet 0 Let U and v be the velocities relative to P inertial frame a then we must have from an attractive force 27f gt a Conservation of energy says that Planet s z7f lt Intergalactic Planetary o Voyager 2 path departing 1977 a this procedure is also known as gravity assist Central Force Scattering 0 Consider effective one body scattering 0 Where problem has rotational symmetry about axis a De ne cross section 09 of units of area for scattering in given direction as 0Qd 2 0Q27T sin dG the of particles scattered into solid angle d9 per unit time DIVIDED by the incident intensity I which is of particles crossing area per unit time Impact Parameter Given the impact parameter s and initial speed 210 imagine in the lab frame particle passing by with no central force Then angular momentum must be 6 mvos 5V 2mE Given known E7 5 then we know 6 and thus we also know the scattering angle 9 as well Solid Angle Element d9 0 The 7 of particles scattered into d9 27T sin dG lying between 9 and 9 d is equal to the of particles with s to s d5 ie 27TIsds 27w sin 9d9 0 Given that can consider s E differential cross section E d 09 sin 9 0 eg for repulsive scattering Solid Angle Expression 0 Want to nd an expression for 95 take a so we go back to the equation of orbital motion 7 d 7 67 60 2mE 2mV 1 T0 7 2 T2 T2 72 o with 7 0 007 60 7T and for 67 mm 6min 7T III 00 dr II 7T 6min 7a 2 2mE2mVi mzn 7quot E2 62 T2 a rewrite using 5 5V 2mE AND 27T II II 9 s7r 2mm 5d 0 1 M 2 E Hyperbolic Orbit Equation 0 Coulomb Force f7 ZZ e27 2 thus k ZZ e2 o For E gt 0 we have hyperbolic motion eccentricity 12E 2 1 2E6 1 2E5 2 6 Mk2 771ZZ 822 ZZ e2 o Eqn of orbit Choose 6 7T ie 6 0 at periapsis 1 mk m 6 2 COSlt6 1 ZZ m 2 e ecos6 1 rhyperbola g 0 Now nd 11 from when rhyp gt 00 with 6 7T II 1 6cos6 1 0 thus cosIIl e Lecture 16 Outline Arti cial Forces 0 Finite and In nitesirnal Rotations Section 4748 o Non inertial reference frames Section 49 410 0 Exam Detox o if time start on Section 51 Finite Rotations Consider active rotation of vector clockwise by angle 1 ie the co ordinate system is xed gt gt gt gt gt want 77 ONNVVQ and NP NQ Fgtlt gt gt gt also ON F and NPF ONF Fz gt gt so we have NVNPcosltIgt F Fcoslt13 which With V Q 77 X 7 sin I gives Rotation formula 77 FcosCI F1 COSI77X misinq Rotations don t commute 0 Problem is that rotations dont commute AB BA 0 eg consider two rotations one 900 about z o and also one 900 about new 3 In nitesimal Rotations do commute 0 Consider in nitesimal rotation of some vector 77 931 931 6115171 612L 2 61313 so 6m eijxj c which in matrix form is 2 As I 552quot 0 But two such in nitesimal rotations do commute 161I62 161 62 6162 I 1 62 with A 1 I 6 since I AA 1 I2 e2 I 0 eg for in nitesimal rotation with respect to Euler angles 1 do 0 O A dgb dip 1 d6 0 d6 1 In nitesimal vector changes In general consider three quantities d91d92d93 which specify the in nitesimal rotation 77 I 6 0 dog d92 6 d03 O d l dag d01 0 Consider how 77 changes d7 77 77 6 ie dzrl 9293 9392 1552 9391 x193 dF F x dc dzrg 9192 93391 but 77 FcostID 7707 F1 costID 77X sindlt1gt or 77 77 77 X mold thus the vector d6 dCID Rate of Change of vectors During time dt vector 3 moves in space frame but doesn t move in body frame Difference is a rotation a hma m wmw mhwmx ie counter clockwise rotation I of body in space frame needs an opposite d7 77 X d6 d6 X 77 d ot Dividing by the differential in time dc dc a X dt dt w space body where c is instantaneous angular velocity Q dt d6 and d6 is instantaneous velocity during 15 gt t dt Frame Transformation d In effect operator Cws f ix Consider the stars as the inertial frame s frarne Consider a particle moving on rotating earth r frarne Then the velocityies of the radius vector 77 are d7 d7 EX Egtrwxr gt Usvrwgtltr and the accelerations are found by a 2nd application dUs d6 as a a w x vs d6 i Jr2c3XUrc3Xc3XF d gtwamwxwrwxm Effective Rotation Forces In terms of F mas and Fear moi Fe F 2maxz7r mwxxF According to observer in the rotating system sees 138g Centrifugal term is the a X a X F Coriolis term is when object moves 6 7 O Centrifugal Effect EegF 2mc3gtlt17r mc3gtlt af x o Centrifugal term is perpendicular to J o Magnitude of centrifugal term is mw2r sin 6 0 Example Stationary particle on Earth ie 17 0 o with Earth s rotation wE 27T24 X 3600 the overall sidereal daysinyear 27T 36625 5 1 Wstars WE Q 8 solar daysmyear 24 X 3600 36525 c max centripetal acceleration at equator w2r 338 cms 0 only about 03 of the earths gravitational acceleration Coriolis Effect Coriolis term is when object moves 27716 X 6 7 0 Consider Coriolis Deflection ie planar motion on surface in Northern hemisphere veers to right Southernleft At equator c is horizontal so no horizontal deflection Max magnitude Coriolis acceleration 2M x 14 X 10 411 Problem mainly for weapons other precision bits Effect of Coriolis Effect Winds eg trade Winds gulf stream cyclones air moves from high to low pressure regions ignore vertical motion air pressure balances gravity so Winds are generally in horizontal direction Freefall Coriolis 0 So given that of lies along North South 0 a freefall results in either East or West motion 0 so the Coriolis acceleration in East direction m 2m c X 17 2mwv2 sin6 More Freefall Coriolis Lecture 21 Outline Hamilton is da man 0 Hamilton equations of motion Section 81 0 Cyclic Coordinates Section 82 o Modi ed Variational Principle Section 85 Hamilton formulation Hamilton formulation is far more reaching than Lagrangian procedure Lagrange s eqns are of second order ie so far n eqns of motion for n indep coords init conds n qi s n q39i s or n qi s at 151 n qi s at 152 INSTEAD we want 2n indep rst order eqns of motion choose qi s rst half of generalised coords and choose the other n coords are pl qj 13 15 ie q and p are the canonical variables ILegendre39Irans nmn 0 allows us to transform from q q t to 11 t 0 consider function of two variables f x y with 6 df fdx dy udx only 3m 3y 0 But we want this in terms of du and dy So let gf ux gt dgdf udx xduvdy xdu o and this means that we can associate as gy u or 89 89 89 89 d g 3U du 3y dy so a u and 1 3y 0 see Goldstein for examples of this in Thermodynamics Hamiltonian Hqp t 0 Want Legendre transform q q39 t to 11 If so rstly 813 so piz ql 0 so we can rewrite the Lagrangian differential as 83 83 83 813 GM d 1 d39l dt 39id 1 id39i dt dqi 1in 16 p q Q8t o inspired by the energy function hq1 qn Q1 q t iazaa az0 d158qu 8q dtpl 8 813 h Qipi quyt HULPJ 0 ie Legendre transform dH qidpl pidq39l d which 8t 8t 8H 8H 8H 0 but we also know that dH 8in amelpi Edi 3 3 dH Qidpi i piin 15 in 19 in dtgt QidPi Piin dt Hamilton s Canonical eqns of motion 0 so we have the 2n 1 canonical equations aH 8H acaH qi am pF aqi w w 0 First n eqns give qi s as a function of 11 15 ie we are really nding the inverse of pi W 0 Second set of n gives 15 s as a function of 11 t o The recipe follows on from as we ve been doing 1 Form qi qi t T V 2 Form Hamiltonian which can have qi qi pi t 3 Invert pl 3 5 to get q39l as a function of qpt 4 Eliminate the Q from H leaving q 1915 Example Spherical Pendulum Example contd Possible Simpli cations Hamiltonian Hqp t qipi q q39 t which can often be rewritten as seen before as H Qipi L0Cz 7 15 L1Qi7 13 L2Qi7 UQkal If the generalised coords don t explicitly depend on time then T qukq39m And if the forces are from a conservative potential L0 V ie work indep of path Then the Hamiltonian H T V is the total energy Another Simpli cation 0 Another simpli cation if T is diagonal then q 125 is Loq t dialq t 131101 t Loq t 50 0 ie in Cartesian with 11 9 y z then m 0 0 Li 1771 1222 EQTCIIEGC y z 0 m 0 y mx y z O O m 2 SO we can write ol 0 also Q55axj5ayyazz395 3 A T H qp Qp a qTq Lo p aT 119 00 Loqnf D k 0 Where 13 T640780 cf T1 Jalsog 5T1 Two Shortcut Examples Cyclic coords Conservation Again qj is cyclic if doesn t appear in L then we know that the canonical momentum pl 35 d 63 63 but 8180 Hqpt Jin 01931 16 fj ie same conservation laws as for q q39 15 eg conserved canonical momentum dH 6H 3H 6H 3H 3 alt aqqu apjpl 8t 8t 815 so if L doesn t have t then H will be constant in time annuva variational principle 0 This is way cool With Action integral 6 6 f Salt 0 which immediately gives modi ed Hamilton s principle t2 152 616 qq39tgtdt6 piqi Hltqptgtidto t1 t1 0 ie integrand has fq 1121525 Where in the spirit of the Hamiltonian formulation independently vary q and p Effectively have 2n Euler Lagrange eqns d a f a f 8H 8H thus dt q q easy hey we have varied paths in both q and p Lecture 10 Outline Central Forces 0 The central force problem Sections 3132 0 Effective force problem Section 33 o Bertrand s Theorem Section 36 0 Diff Eqn for the orbit Section 35 Conservation Energy l i a o 2 l2 i 2 dt 2 7 2W2 V7 0 dt 7717 6 0 o Quadratures E evaluating integrals a Two generalised variables 7 6 four integrations needed 0 First two already done converted to 1st order eqns 0 for the third we rewrite the E equation 2 2 d g E V7 2 so dt T dt M 2m 2 l2 u E 2lur2 a which we can integrate from to 0 where 7 7 0 Evaluating the Fourth Integral o Inverting this can give the solution for 7 at time t 0 ie given 7 t we can nd 6t l2 2 t dt d6 dt thus 6 60 l H72 to 7320 a so Four constants of motion E7 l 7 0 60 we can nd 6 o are not the only ones eg 7 0 60 7 0 60 l o in Quantum Mechanics El are the most important It s Centrifugal dear Watson 0 4 constants of integration motion E l 7 0 60 0 ie we know radial and angular init conds 7 0 6071 60 d a Last time also given central 1D f0rce f m dt2 12 dz 2ltrgt 2 2m 2 mm 0 and dt 771729 0 9V 87quot 2 f7 771792 f7 fe r o the fe cU is an effective force partly cticiousl a centrifugal force outwards acts a repulsive potential V7 2 277172 with conserved E 79 Veff7 InverseSquare Law 0 Consider attractive force f 7 kr2 ie V7 E r a plus the cticious potential energy gives a Veffr o For E gt 0 we have unbounded motion 0 Classical particle cannot have Veff gt E Kinetic Energies a kinetic energy radial ET 7 2 angular is E9 a total kinetic energy Ek7 EAT EMT E V o tells us the entire kinematics Bounded States Emm g E g 0 a turning points 717 2 are apsidal distances 0 Circular orbits need 7quot 0 at some radius 7 0 2 ave 2 8 thus f7 0 m 7 7717 0 7 00 0 fefEU a GPS cover is 7 attractive 74 repulsive no stable Circular orbits but bound non Closed orbits Unable to be there 2 277772 a if the potential falls off faster than 0 ie if 7 quot with 77 gt 2 eg V7 a7 3 o with E in hump 7 1 lt 7 lt 7 2 not physically possible a initially 7 lt 7 1 or 7 gt 7 2 transition between impossible Simple Harmonic Oscillator o Isotropic 3D force f7 k7 thus V7 kT Z o For l 0 usual SHO motion V7 7 2 o For l gt 0 motion is bounded never passes through 7 0 Circular Orbit Stability 0 Which orbits retrace their steps 0 Obviously circles 0 at radius Where Veff is at minima given 6 and E Vefo O V7 0 62277171 8V 2 7 0 thus 7 feff 0 87 To f 0 mfg 0 TWO cases depending on Whether minima 0r maxima Power Law Stability 0 ie Circular Orbit Stability determined by 2nd derivative 0 Assuming attractive force f To r0 lt 0 then 82V 8 352 3 Olt f 4f 3f7 0 87 m 87 re mro 8r r0 r0 g lt 3f7 0 87 7 0 0 Further assuming power law force f g7n near r0 mrn1 BMW 1 ie n gt 3 a Power law potential must vary slower than Vr 17 2 to have stable Circular orbits Deviate Circles o Deviating slightly from circular motion gives rise to harmonic motion 7 7 01 OR 7 7 0 ozoos 6 0 Here 04 depends on deviation of energy E o and from rst order Taylor expansion of f 7 at 7 0 Bertrand s Theorem 0 Anyways can show that for 7 7 0 04 cos 6 Condition for a closed orbital depends on m 8f k g Vim Em can also show f7 W If rational number pq with integers p q After q revolutions cosp6q cos27rp 0 o Second order can show that only l7 2 are stable f r34 2 either OR k7 2 7 o Bertrand s Theorem 1873 only central forces that have closed orbitals for all bound particles are inverse square and Hooke s law Orbit Diff Eqn 0 So we can sometimes nd 7 03 6t given E7 l 7 0 60 a we want to nd the orbit 7 6 ie eliminate t d l d ldt 2d6 th W us dt W2 d6 0 converts the radial EofM into the orbit d27 2 l d l dr 2 dV7 r m mr3 mr2 mr3 dr 4 7 r du dr a now u thus du Ti2dr hence E Ti2 1 2712 d2u 2713 2 d 1 V f m d62 77173 u du o RHS since 7 thus dr ui2du Thus d2u m d 1 m 1 1 V F d62 u l2 du l2 a Lecture 10 Outline Kepler 0 Diff Eqn for the orbit Section 35 o Bertrand s Theorem Section 36 o The Kepler Problem Sections 37 38 Circular Orbit Stability 0 Which orbits retrace their steps 0 Obviously Circles O at radius Where Veff is at minima given E and E Veff7 0 V040 322771748 3V 32 feff7quot0 E To 0 thus f7quot0 m rg o T W0 cases depending on Whether minima 0r maxima Power Law Stability ie Circular Orbit Stability determined by 2nd derivative Assuming attractive force f To r0 lt 0 then 32V 8 332 3 Olt f 4f 3f7 0 3T2 m 37 To mro 37 r0 r0 g lt 3f7 0 874 T0 Further assuming power law force f crquot near r0 mr 1 Slam 1 ie n gt 3 Power law potential rnust vary slower than V0 17 2 to have stable Circular orbits Deviate Circles o Deviating slightly from Circular motion gives rise to harmonic motion to uo oacos 17 OR 7 7 0 oacos 0 Here 04 depends on deviation of energy E o and from rst order Taylor expansion of f 7 at 7 0 Bertrand s Theorem 0 Anyways can somehow show that for 7 7 0 04 cos 0 Condition for a closed orbital depends on 5 r0 8f k 5 VW Em andcanalsoshow f 7 W o If rational number 5 pq with integers p q 0 After q revolutions cosp6q cos27rp O o Second order can show that only 5 1 2 are stable 1674 7434 g either OR k74 742 o Bertrand s Theorem 1873 only central forces that have closed orbitals for all bound particles are inverse square and Hooke s law Power Laws d 2 2 E V 2mr2 0 Given any Force use Edt mr2d6 to eliminate time Edr WV E V 0 Thus integrating we again simplify with u 17 d6 7quot d u 7 du 6 60 60 2mE 2mV 1 2mE 2mV T0 T2 e2 e2 72 W e2 e2 2 o For Power law potentials V7 WW1 du u 6 60 uO 2mE 2maun1 u2 2 2 0 Integrable n 1 2 3 and n 5 3 O 4 5 7 Inverse Square Law 0 fr and thus V0 c7 0 Several ways to integrate the eqn for orbit eg 16 E2 d E2 d7 2 f7 gt 72 72 W 0 however Goldstein knows that the inde nite integral dx L COS1 5 27x W Y 52 4047 2u 1 oos1 mk d 99 u 9 m mu 2 2 2Eu2 1 0 Note 6 is a constant of integration Thus rearranging Equation of 74 Orbit Contains E E 6 three constants of integration Note 6 is also the turning angle 2E32 1 1 lt mk2 1 mic 2 cos6 9 The missing fourth speci es the location on orbit 60 which we need to nd 7 05 615 again we can use mr2d6 Edt Conic Cosmetic Cosmic Comic Cynic Cyclic o The general equation for a conic is l o C1e cos6 6 0 Johann Bernoulli proved all orbits in potential are conic sections with their focus at the centre of mass Semimajor Axis of Ellipse Relation 0 SO the eccentricity e 1 252 Motion in time for Fr kr2 o Fairly sirnple using conservation Edt mr2d6 again 0 where we throw in TEL f 1 e cos6 9 t m T dr 3 9 d6 2 To m 77162 90 1 ecos6 92 7quot 2mquot2 0 BUT either integrals to give 7 05 615 are tough Parabolic Motion in time for Fr k7 2 t e3 f9 d6 WW2 90 1 l ecost9 t9 2 0 eg For Parabola s e 1 use 1 0086 2 308262 3 6 4 3 tan02 2 2 1 t 4mk20 sec 6 d6 2mk20 x dx 3 W tan62 tan362gt 0 Above used a tan6t2 so dx sec26t2d6 o and also SGC2 1 tar o Sowehave 7Tlt6lt7T o WhereQ 7T gtt ooroo 0 Where 6 O gt t 07 apsidal Elliptical Motion in time for 7 2 Only approximately true Kepler s Equation
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