Lecture 9 Hamilton s Equa4ons

Transcription

1 Lecture 9 Hamilton s Equa4ons Informal deriva5on conjugate momentum cyclic coordinates Applica5ons/examples 1

2 deriva4on Start with the Euler Lagrange equa5ons d dt L q i L = λ q i j C j 1 + Q i Define the conjugate momentum p i = L q i The Euler Lagrange equa5ons can be rewrigen as p i = L + λ q i j C j 1 + Q i 2

3 deriva4on If we are to set up a pair of sets of odes, we need to eliminate q i We can write the Lagrangian L = 1 2 q i M ij q j V ( q k ) from which p i = M ij q j M is posi5ve definite (coming from the kine5c energy), so q j = M ji p i 3

4 deriva4on I have the pair of sets of odes q j = M ji p i p i = L + λ q i j C j 1 + Q i where all the q dots in the second set have been replaced by their expressions in terms of p 4

5 deriva4on We can write this out in detail, although it looks pregy awful L = 1 2 q i M ij q j V ( q k ) right hand side L = 1 q i 2 q M m mn q i q n V q i subs5tute q m = M mr p r, q n = M ns p s right hand side L = 1 q i 2 M M mr p mn r q i M ns p s V q i 5

6 deriva4on p i = L + λ q i j C j 1 + Q i L = 1 q i 2 M M mr p mn r q i M ns p s V q i p i = 1 2 M M mr p mn r q i M ns p s V + λ q i j C j 1 + Q i We will never do it this way! I will give you a recipe as soon as we know what cyclic coordinates are 6

7 ?? 7

8 cyclic coordinates Given p i = L + λ q i j C j 1 + Q i If Q i and the constraints are both zero (free falling brick, say) we have p i = L q i and if L does not depend on q i, then p i is constant! Conserva4on of conjugate momentum q i is a cyclic coordinate p i = 0 8

9 YOU NEED TO REMEMBER THAT EXTERNAL FORCES AND/OR CONSTRAINTS CAN MAKE CYCLIC COORDINATES NON CYCLIC! 9

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11 What can we say about an unforced single link in general? ( ) 2 + B( θ sinψ + φ sinθ cosψ) 2 + C( ψ + φ cosθ) 2 L = 1 2 A θ cosψ + φ sinθ sinψ ( z 2 ) mgz m x 2 + y 2 + We see that x and y are cyclic (no explicit x or y in L) Conserva5on of linear momentum in x and y direc5ons p x = p 1 = m x, p y = p 2 = m y We see that φ is also cyclic (no explicit φ in L) 11

12 The conserved conjugate momentum is p φ = p 4 = (( Asin 2 ψ + Bcos 2 ψ)sin 2 θ + Ccos 2 θ) φ + ( A B)sinθ sinψ cosψθ + Ccosθψ Does it mean anything physically?! the angular momentum about the k axis as I will now show you 12

13 The angular momentum in body coordinates is l = A( θ cosψ + φ sinθ sinψ)i 3 + B( θ sinψ + φ sinθ cosψ)j 3 + C( ψ + φ cosθ)i ZZ K 3 The body axes are (from Lecture 3) ( ) + sinψ cosθ( sinφi 0 + cosφj 0 ) + sinθk 0 I 3 = cosψ cosφi 0 + sinφj 0 J 3 = sinψ cosφi 0 + sinφj 0 ( ) ( ( ) + sinθk 0 ) ( ) + cosψ cosθ sinφi 0 + cosφj 0 K 3 = sinθ( sinφi 0 + cosφj 0 ) + cosθk 0 and the k component of the angular momentum is k l = A( θ cosψ + φ sinθ sinψ)sinψ sinθ + B( θ sinψ + φ sinθ cosψ)cosψ sinθ + C( ψ + φ cosθ)cosθ 13

14 p 4 = (( Asin 2 ψ + Bcos 2 ψ)sin 2 θ + Ccos 2 θ) φ + ( A B)sinθ sinψ cosψθ + Ccosθψ some algebra k l = A( θ cosψ + φ sinθ sinψ)sinψ sinθ + B( θ sinψ + φ sinθ cosψ)cosψ sinθ + C( ψ + φ cosθ)cosθ The only force in the problem is gravity, which acts in the k direc5on Any gravita5onal torques will be normal to k, so the angular momentum in the k direc5on must be conserved. 14

15 Summarize our procedure so far Write the Lagrangian Apply holonomic constraints, if any Assign the generalized coordinates Find the conjugate momenta Eliminate conserved variables (cyclic coordinates) Set up numerical methods to integrate what is le\ 15

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17 Let s take a look at some applica5ons of this method Flipping a coin The axisymmetric top Falling brick 17

18 Flipping a coin 18

19 Flipping a coin A coin is axisymmetric, and this leads to some simplifica5on ( ) + C( ψ + φ cosθ ) 2 ( ) m L = 1 2 A θ 2 + φ 2 sin 2 θ ( x 2 + y 2 + z 2 ) mgz We have four cyclic coordinates, adding ψ to the set and simplifying φ p φ = p 4 = ( Asin 2 θ + Ccos 2 θ) φ + Ccosθψ ( ) p ψ = p 6 = C ψ + φ cosθ We can recognize the new conserved term as the angular momentum about K l = A( θ cosψ + φ sinθ sinψ)i 3 + B( θ sinψ + φ sinθ cosψ)j 3 + C( ψ + φ cosθ)k 3 19

20 Flipping a coin We can use this to simplify the equa5ons of mo5on The conserved momenta are constant; solve for the deriva5ves ( ) φ = p p cosθ 4 6 Asin 2 θ, ψ = p 6 C ( p p cosθ 4 6 )cosθ Asin 2 θ (There are numerical issues when θ = 0; all is well if you don t start there) 20

21 Flipping a coin I can flip it introducing spin about I with no change in φ or ψ φ = 0 = ψ This makes p 4 = 0 = p 6 Then p φ = p 4 = ( Asin 2 θ + Ccos 2 θ) φ + Ccosθψ = 0 ( ) = 0 p ψ = p 6 = C ψ + φ cosθ L θ = Asinθ cosθ φ 2 Csinθ ψ + φ cosθ ( ) = 0 21

22 Flipping a coin p 1 = p 10, p 2 = p 20, p 3 = p 30 mgt, p 4 = p 40 = 0, p 5 = p 50, p 6 = p 60 = 0 from which q 1 = q 10 + p 10 m, q 2 = q 20 + p 20 m, q 3 = q 30 + p 30 m 1 2 gt 2 q 4 = q 40, p 5 = p 50 A, q 6 = q 60 or x = x 0 + ut, y = y 0 + vt, z = z 0 + wt 1 2 gt 2 φ = φ 0 = 0, θ = θ 0 + ωt, ψ =ψ 0 = 0 22

23 Flipping a coin I claim it would be harder to figure this out using the Euler Lagrange equa5ons I ve gogen the result for a highly nonlinear problem by clever argument augmented by Hamilton s equa5ons 23

24 ?? 24

25 symmetric top How about the symmetric top? Same Lagrangian but constrained x y z sinφ sinθ = dk = d cosφ sinθ cosθ 25

26 symmetric top Treat the top as a cone A = 3 5 m 1 4 a2 + h 2 = B, C = 3 10 ma2 and apply the holonomic constraints x y z sinφ sinθ = dk = d cosφ sinθ cosθ 26

27 symmetric top A\er some algebra ( ( )) L = m 12a2 + 31h 2 + ( 4a 2 31h 2 )cos 2θ φ m 4a2 + 31h ma2 cosθ φ ψ 3 mgh cosθ 4 ( ) θ ma2 ψ 2 and we see that φ and ψ are both cyclic here. φ is the rota5on rate about the ver5cal the precession ψ is the rota5on rate about K the spin 27

28 symmetric top The conjugate momenta are (( ) + ( 12a h 2 )cos( 2θ) ) p 1 = m 12a2 + 31h 2 p 2 = 3 80 m ( 4a2 + 31h 2 ) θ ( ) p 3 = 3 10 ma2 cosθ φ + ψ φ ma2 cosθψ There are no external forces and no other constraints, so the first and third are conserved 28

29 symmetric top We are le\ with two odes θ = 80p 2 3m 4a h 2 ( ) p 2 = 40 2 p 2 2 ( 1 + p 3 )cosθ p 1 p 3 3+ cos 2θ 3m( 4a h 2 )sin 3 θ ( ( )) mghsinθ The response depends on the two conserved quan55es and these depend on the ini5al spin and precession rates We can go look at this in Mathema5ca 29

30 Falling brick Let s go back and look at how the fall of a single brick will go in this method I will do the whole thing in Mathema5ca 30

31 That s All Folks 31

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33 Falling brick p φ = p 4 = (( Asin 2 ψ + Bcos 2 ψ)sin 2 θ + Ccos 2 θ) φ + ( A B)sinθ sinψ cosψθ + Ccosθψ What is this?! If A and B are equal (axisymmetric body), this is much simpler p φ = p 4 = ( Asin 2 θ + Ccos 2 θ) φ + Ccosθψ 33

34 Falling brick We know what happens to all the posi5on coordinates All that s le\ is the single equa5on for the evolu5on of p 5. A\er some algebra we obtain ( )( p 4 p 6 cosθ) p 5 = p 6 p 4 cosθ Asin 3 θ θ = p 5 A 34

35 We can restrict our agen5on to axisymmetric wheels and we can choose K to be parallel to the axle without loss of generality ( ( ) 2 + B( θ sinψ + φ sinθ cosψ ) 2 + C( ψ + φ cosθ ) 2 ) L = 1 2 A θ cosψ + φ sinθ sinψ ( z 2 ) mg R m x 2 + y 2 + ( ( ) + C( ψ + φ cosθ ) 2 ) m ( x 2 + y 2 + z 2 ) mg R L = 1 2 A θ 2 + φ 2 sin 2 θ mgz 35

36 If we don t put in any simple holonomic constraint (which we o\en can do) x y z q = φ θ ψ 36

37 We know v and ω in terms of q any difficulty will arise from r v = ω r r = aj 2 Actually, it s something of a ques5on as to where the difficul5es will arise in general This will depend on the surface flat, horizontal surface we ve been doing this flat surface we can do this today general surface: z = f(x, y) this can be done for a rolling sphere 37

38 d dt d dt d dt d dt L j + mg x i x = λ j C 1 L j y i + mg y = λ j C 2 L j + mg z i z = λ j C 3 d dt L j φ i = λ j C 4 L θ L θ = λ C j j 5 d dt L j ψ = λ j C 6 38

39 We have the usual Euler Lagrange equa5ons d dt L q i L = λ q i j C i j and we can write out the six equa5ons 39

40 The angular momentum in body coordinates is l = A( θ cosψ + φ sinθ sinψ)i 3 + B( θ sinψ + φ sinθ cosψ)j 3 + C( ψ + φ cosθ)i ZZ K 3 The body axes are (from Lecture 3) ( ) + sinψ cosθ( sinφi 0 + cosφj 0 ) + sinθk 0 I 3 = cosψ cosφi 0 + sinφj 0 J 3 = sinψ cosφi 0 + sinφj 0 ( ) ( ( ) + sinθk 0 ) ( ) + cosψ cosθ sinφi 0 + cosφj 0 K 3 = sinθ( sinφi 0 + cosφj 0 ) + cosθk 0 and the k component of the angular momentum is k l = A( θ cosψ + φ sinθ sinψ)sinψ sinθ + B( θ sinψ + φ sinθ cosψ)cosψ sinθ + C( ψ + φ cosθ)cosθ 40

41 We will have a useful generaliza5on fairly soon The idea of separa5ng the second order equa5ons into first order equa5ons q i = M ir p r p i = 1 2 M M mr p mn r q i M ns p s V + λ q i j C j 1 + Q i More generically q i i = A j ( q m )u j u i = f i ( q m,u n ) 41