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by: Dr. Carissa Rowe


Dr. Carissa Rowe
GPA 3.97


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Class Notes
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This 31 page Class Notes was uploaded by Dr. Carissa Rowe on Tuesday October 20, 2015. The Class Notes belongs to PHYS608 at San Diego State University taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/225322/phys608-san-diego-state-university in Physics 2 at San Diego State University.


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Date Created: 10/20/15
Lecture 23 Outline Canon transforms a Hamilton equations of motion Section 81 Cyclic Coordinates Section 82 Modi ed Variational Principle Section 85 brie y Principle of Least Action Section 86 Canonical Transforms Section 91 Hamilton s Canonical eqns of motion a the 2n l canonical equations are aH 8H acaH gi 31 pi8qi WE 0 First n eqns give q i s as a function of q7 p7 it ie we are really nding the inverse of pi Lg a Second set of n gives 13 s as a function of qp t o The recipe follows on from as we ve been doing Form qi QM T V Form Hamiltonian which can have qi q pi t Invert pi 3 to get q i as a function of q 192 Eliminate the q from H leaving q p t Main Simpli cation o Shortcut method if T is diagonal then q q39 t is a so last time i wrote H 615 L 1a a a a H 5paT 119 aLoqt Example Spherical Potential Cyclic coords Conservation 0 Again qj is cyclic if doesn t appear in L a then since the canonical momentum pj 3 J d 83 83 dt 8 8 7 but also H cm 2 cm q cm ie 275 0 ie same conservation laws as for q q39 t 0 eg conserved canonical momentum In general though dH 8H 8H 8H Cit agiqu apipz 875 191 1sz 75 at a so if 3 doesn t have 25 then H will be constant in time annuva variational principle a This is way cool With Action integral 6 6ft 3dr 0 which immediately gives modi ed Hamilton s principle t2 t2 t1 t1 0 ie integrand has fq q p1 t Where in the spirit of o the Hamiltonian formulation independently vary q and p o Effectively have 2n Euler Lagrange eqns d 8 f 8 f 8H 8H p39 thus p d 8 f 8 f 8H 8H thus qj ma aw a a easy hey we have varied paths in both q and p Least action principle I at end points in phase space all 6 0 and 619i 0 Old 6 Variati0n only had 77 t1 77 t2 0 Introduce A Variation With 77 t1 75 0 t2 152At2 A dt t1 t1At1 WW2 7A 0 Ozdt 2 0dt 751 then if H is conserved on both paths and b at the end points we require all Aqi 0 we get the principle of least action 752 A 19102 0 t1 Least action principle II Least action A 12 pg 0 note sum over non relativistic mechanics example with quadratic q i s 1 T I Z Mjk61qjqk jk If V not velocity dependent H is conserved 813 8T l 1 pr Ma qu Z Esz Qj Z EMiij quot quot z z 752 but then pig 2T ie A Tdt 0 t1 So of all conserving energy paths 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 813 8H p a and MZ aqi Now if Hqpt Hp H ie doesn t have q s or t 0 thus p 04 then the other set of Hamilton s equations are then 8H 8H qz all 8062 as the 04 w are constant in time q wit 3 too trivial Well consider planar motion 012061 q1xorq2yORq1rorq26 ie for central forces 51334 not cyclic but 6 is gt easy soln Point transformations Where possible want to use cyclic coords qi with transform equations from qi to Qi are Qz Q q 2 con guration space Point transformations see Goldstein Derivation 110 o Lagrange s eqns are invariant under point transforms Hamiltonian formulation simultaneous point transforms Q Qzlqm 1 and Pi 1346171971 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 CQ 132 ie 81C 81C 1 d e Q an an add a where C is the Hamiltonian in new coords with both 152 t2 6 Pia mo Roch 0 and 6 piqi Hltqptgtdt 0 t1 t1 0 does not mean that both integrands are equal instead dF dz 0 since we are considering zero variation at both endsl pz Qz H 1C a but what does this mean and how does it help us Scale transforms A Mpqu H 1C ii 1 a multiplicative constant A just a scale transformation eg Q ugh P m means lC Q P WH 6171 0 since our guess for C can show that our 81C LLV 8H Qi V MQ L o GENERALLY can nd a transform with A l anyway ie given qp gt Q P with A y l but then Q P gt Q 7 P with A then we have q7p gt Q7 P with A 1 a so we have a canonical transform when o dF piQiH PLQilCE Generating functions a The 71 contribs to variation of action only at end points a if F is a function of q p Q 132 it vanishes at end points 0 ie F is used as a canonical variables bridge between half of the old set and half of the new set 0 Simplest general example F F1q7 Q t then o dF1 o 8F1 8F1 8F1 iiHP i ilC P ilC i 1 pg Q dt Q ataqiqaQiQ o collecting the like terms since independent variables an an 8F 8 DP 862 a 0 ie F generates the eqns of canonical transforms Pr SHO Gen Funcs SHO Canonical Transform SHO Solutions Lecture 8 Outline Endo Z Chapter 2 outtro Energy function Section 252627 Friction Section 15 25 Conservation of momentum energy Sections 2627 if time Electrical circuits Section 25 example III Lagrange Multiplier 0 Particle falling down hemisphere spare spare Sometimes h T V E Part I 0 Sometimes hq q39 t is the total energy of the system 0 Recall that T 6H2 TO T1 T2 in gencoords 2 TZ M0ZMijZQij z j jk 0 eg T2 is quadratic in gencoords 26a j 8q 8t o if constraints indep of time then only have T2 a consider 3 0q t 1q q39 t 2q q39 t o most problems do conform to then ac ac2 acl h gqjaqj Lqjaqjqjaqj 2 1 0 Sometimes h T V E Part II Mg q t 27 qjg rig g 2 1 0 Euleris theorem states that for function f of degree n 8f xi i am 1quot then since 2 is quadratic and 1 linear in q39j hqq t2 2 1 2 1 0 2 0 so if transform eqns are indep of time then 2 T and potential indep of q39j then 0 V then energy function h T V E is total energy further if V indep of time total energy conserved dE 0 dt dt Velocity Dependent Potentials 15 o if 3 contains only conservative forces a and Q7 represents forces not from potential d 31 31 Qk 0 Consider frictional forces such as fo kxvx 0 These come from Rayleigh s dissipation function 37 1 7 E k905i where Ffm L o for 239 particles Or we can write Ffz sz7 o The work done against friction is then dwf 441M 13fz7dt m Iggy kzvjmt 23F Fraction too much friction 7 k905i o Generalised force was originally given by a am am an a 39 vvifo Q3 2 f 3 3 3 3 L 0 With the Lagrange Eqn s of motion are i3 5 5 i0 note aiz d258ij aqj aqj 39 aq j aq k a so two scalars functions are reqd 7 73 0 eg Stokes Law sphere radius a moving at velocity 27 in medium viscosity 77 experiences drag Fj 67T77az7 Practical Example Stokes Law Final Note on Dissipation If Frictional Forces derivable from dissipation 7 ie the Lagrange s Equation s with dissipation d 33 33 87 I 0 dt we can then show that the energy function relation db 8 d 8 83 97 39 39 dt at dz 1quotan gt at aqjqj 7 dt s degree 2 since 7 is a homogeneous function of h ac d J dz 5 If 3 not explicitly depend on t and h E then we have 21 27 ie the rate of energy loss Electrical Circuits o are a simple dissipative system Coupled Electrical Circuits o we ll come back to this later chapter for now apply Lagrange s EofMo Generalised Momentum o That s energy conservation What about momentum 0 Consider a system With forces from potentials Vqthen acaT avaT a Zmi 813 813 813 ajg 333 7 912 19er L Z Z Z L i 33 a wrt qi generalised canonical momentum pj 11 Note pj not always has units of linear momentum if VQ then pj not equal to mechanical momentum If 3 indep of qj ie qj is a cyclic coordinate iai 3i Kai dzan aqj dzan dz ie the generalised momentum pj is conserved Translation and Rotation of System I m skipping section 26 on symmetry read for HW Which looks at how conservation works given a system Translation or Rotation Consider a cyclic coordinate qj Where Lagrangian qj Translation of cyclic coordinate qj the generalised force which is really just conservation of linear momentum if change cyclic coordinate qj means system rotation then the generalised force Qj is the component of Torque about rotation axis and the generalised momentum pj is the angular momentum along the same axis Chapter 2 outtro Once equations of motion obtained n diffeqn s second order in time to solve with 2 integrations7 ij t S with 2n constants of integration ans and nd Constraints rst order diffeqn s inc conservation laws Generalised canonical momentum defn pj 37 J gm gmx motlvated by 0 dtag j midi dt dt Newton s law Generalised momentum conserved when qj d if qjqjt then f where h Zj 01 33qu3 L If transform F Fqj q and Vqj then b T V E and if V explicitly indep of time then 0


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