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Lecture 9 Hamilton’s Equations

conjugate momentum

cyclic coordinates

Informal derivation

Applications/examples

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€

ddt

∂L∂˙ q i ⎛ ⎝ ⎜

⎞ ⎠ ⎟− ∂L

∂qi = λ jC1j + Qi

Define the conjugate momentum

€

pi = ∂L∂˙ q i

Start with the Euler-Lagrange equations

€

˙ p i = ∂L∂qi + λ jC1

j + Qi

The Euler-Lagrange equations can be rewritten as

derivation

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If we are to set up a pair of sets of odes, we need to eliminate

€

˙ q i

We can write the Lagrangian

€

L = 12

˙ q iM ij ˙ q j −V qk( )

M is positive definite (coming from the kinetic energy), so

€

˙ q j = M ji pi

derivation

€

pi = M ij ˙ q jfrom which

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I have the pair of sets of odes

€

˙ q j = M ji pi

€

˙ p i = ∂L∂qi + λ jC1

j + Qi

where all the q dots in the second set have been replaced by their expressions in terms of p

derivation

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We can write this out in detail, although it looks pretty awful

€

L = 12

˙ q iM ij ˙ q j −V qk( )

derivation

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∂L∂qi = 1

2˙ q m ∂Mmn

∂qi ˙ q n − ∂V∂qiright hand side

€

˙ q m = M mr pr, ˙ q n = M ns pssubstitute

€

∂L∂qi = 1

2M mr pr

∂Mmn

∂qi M ns ps − ∂V∂qi

right hand side

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€

˙ p i = ∂L∂qi + λ jC1

j + Qi

€

∂L∂qi = 1

2M mr pr

∂Mmn

∂qi M ns ps − ∂V∂qi

We will never do it this way!€

˙ p i = 12

M mr pr∂Mmn

∂qi M ns ps − ∂V∂qi + λ jC1

j + Qi

derivation

I will give you a recipe as soon as we know what cyclic coordinates are

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??

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€

˙ p i = ∂L∂qi + λ jC1

j + Qi

If Qi and the constraints are both zero (free falling brick, say) we have

€

˙ p i = ∂L∂qi

and if L does not depend on qi, then pi is constant!

Conservation of conjugate momentum

qi is a cyclic coordinate

Given

cyclic coordinates

€

˙ p i = 0

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YOU NEED TO REMEMBER THAT EXTERNAL FORCES

AND/OR CONSTRAINTSCAN MAKE CYCLIC COORDINATES

NON-CYCLIC!

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??

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What can we say about an unforced single link in general?

€

L = 12

A ˙ θ cosψ + ˙ φ sinθ sinψ( )2

+ B − ˙ θ sinψ + ˙ φ sinθ cosψ( )2

+ C ˙ ψ + ˙ φ cosθ( )2 ⎛

⎝ ⎜ ⎞ ⎠ ⎟

+ 12

m ˙ x 2 + ˙ y 2 + ˙ z 2( ) − mgz

We see that x and y are cyclic (no explicit x or y in L)

€

px = p1 = m˙ x , py = p2 = m˙ y

Conservation of linear momentum in x and y directions

We see that f is also cyclic (no explicit f in L)

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€

pφ = p4 = Asin2ψ + Bcos2ψ( )sin2 θ + Ccos2 θ( ) ˙ φ + A − B( )sinθ sinψ cosψ ˙ θ + Ccosθ ˙ ψ

Does it mean anything physically?!

The conserved conjugate momentum is

the angular momentum about the k axisas I will now show you

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The angular momentum in body coordinates is

€

l = A ˙ θ cosψ + ˙ φ sinθ sinψ( )I3 + B − ˙ θ sinψ + ˙ φ sinθ cosψ( )J3 + C ˙ ψ + ˙ φ cosθ( )IZZK 3

€

I3 = cosψ cosφI0 + sinφJ0( ) + sinψ cosθ −sinφI0 + cosφJ0( ) + sinθK 0( )

J3 = −sinψ cosφI0 + sinφJ0( ) + cosψ cosθ −sinφI0 + cosφJ0( ) + sinθK 0( )

K 3 = −sinθ −sinφI0 + cosφJ0( ) + cosθK 0

The body axes are (from Lecture 3)

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

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The only force in the problem is gravity, which acts in the k direction

Any gravitational torques will be normal to k, so the angular momentum in the k direction must be conserved.

€

k ⋅l = A ˙ θ cosψ + ˙ φ sinθ sinψ( )sinψ sinθ +

B − ˙ θ sinψ + ˙ φ sinθ cosψ( )cosψ sinθ + C ˙ ψ + ˙ φ cosθ( )cosθ

€

p4 = Asin2ψ + Bcos2ψ( )sin2 θ + C cos2 θ( ) ˙ φ + A − B( )sinθ sinψ cosψ ˙ θ + C cosθ ˙ ψ

some algebra

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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 left

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??

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Let’s take a look at some applications of this method

Falling brick

Flipping a coin

The axisymmetric top

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Flipping a coin

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A coin is axisymmetric, and this leads to some simplification

€

L = 12

A ˙ θ 2 + ˙ φ 2 sin2 θ( ) + C ˙ ψ + ˙ φ cosθ( )2

( ) + 12

m ˙ x 2 + ˙ y 2 + ˙ z 2( ) − mgz

We have four cyclic coordinates, adding y to the set and simplifying f

€

pψ = p6 = C ˙ ψ + ˙ φ cosθ( )€

pφ = p4 = Asin2 θ + Ccos2 θ( ) ˙ φ + C cosθ ˙ ψ

We can recognize the new conserved term as the angular momentum about K

€

l = A ˙ θ cosψ + ˙ φ sinθ sinψ( )I3 + B − ˙ θ sinψ + ˙ φ sinθ cosψ( )J3 + C ˙ ψ + ˙ φ cosθ( )K 3

Flipping a coin

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We can use this to simplify the equations of motion

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˙ φ =p4 − p6 cosθ( )

Asin2 θ, ˙ ψ = p6

C−

p4 − p6 cosθ( )cosθAsin2 θ

The conserved momenta are constant; solve for the derivatives

(There are numerical issues when q = 0; all is well if you don’t start there)

Flipping a coin

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€

pψ = p6 = C ˙ ψ + ˙ φ cosθ( ) = 0

€

pφ = p4 = Asin2 θ + Ccos2 θ( ) ˙ φ + C cosθ ˙ ψ = 0€

˙ φ = 0 = ˙ ψ

I can flip it introducing spin about I with no change in f or y

This makes p4 = 0 = p6

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∂L∂θ

= Asinθ cosθ ˙ φ 2 − C sinθ ˙ ψ + ˙ φ cosθ( ) = 0

Then

Flipping a coin

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€

p1 = p10, p2 = p20, p3 = p30 − mgt, p4 = p40 = 0, p5 = p50, p6 = p60 = 0

€

q1 = q10 + p10

m, q2 = q20 + p20

m, q3 = q30 + p30

m− 1

2gt 2

q4 = q40, p5 = p50

A, q6 = q60

from which

€

x = x0 + ut, y = y0 + vt, z = z0 + wt − 12

gt 2

φ = φ0 = 0, θ = θ0 + ωt, ψ =ψ 0 = 0

or

Flipping a coin

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I claim it would be harder to figure this out using the Euler-Lagrange equations

Flipping a coin

I’ve gotten the result for a highly nonlinear problem by clever argumentaugmented by Hamilton’s equations

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??

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Same Lagrangian

but constrained

€

xyz

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪= dK = d

sinφsinθcosφsinθ

cosθ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

symmetric top

How about the symmetric top?

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Treat the top as a cone

€

A = 35

m 14

a2 + h2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟= B, C = 3

10ma2

and apply the holonomic constraints

symmetric top

€

xyz

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪= dK = d

sinφ sinθcosφsinθ

cosθ

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

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After some algebra

€

L = 3320

m 12a2 + 31h2 + 4a2 − 31h2( )cos 2θ( )( ) ˙ φ 2 + 3160

m 4a2 + 31h2( ) ˙ θ 2 + 320

ma2 ˙ ψ 2

310

ma2 cosθ ˙ φ ˙ ψ − 34

mgh cosθ

and we see that f and y are both cyclic here.

f is the rotation rate about the vertical — the precession

y is the rotation rate about K — the spin

symmetric top

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The conjugate momenta are

€

p1 = 3160

m 12a2 + 31h2( ) + 12a2 + 31h2( )cos 2θ( )( ) ˙ φ + 310

ma2 cosθ ˙ ψ

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p2 = 380

m 4a2 + 31h2( ) ˙ θ

€

p3 = 310

ma2 cosθ ˙ φ + ˙ ψ ( )

There are no external forces and no other constraints, so the first and third are conserved

symmetric top

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We are left with two odes

€

˙ θ = 80 p2

3m 4a2 + 31h2( )

€

˙ p 2 = 402 p1

2 + p32( )cosθ − p1p3 3 + cos 2θ( )( )

3m 4a2 + 31h2( )sin3 θ+ 3

4mgh sinθ

The response depends on the two conserved quantitiesand these depend on the initial spin and precession rates

symmetric top

We can go look at this in Mathematica

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Let’s go back and look at how the fall of a single brick will go in this method

Falling brick

I will do the whole thing in Mathematica

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That’s All Folks

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€

pφ = p4 = Asin2ψ + Bcos2ψ( )sin2 θ + Ccos2 θ( ) ˙ φ + A − B( )sinθ sinψ cosψ ˙ θ + Ccosθ ˙ ψ

What is this?!

If A and B are equal (axisymmetric body) , this is much simpler

€

pφ = p4 = Asin2 θ + Ccos2 θ( ) ˙ φ + Ccosθ ˙ ψ

Falling brick

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We know what happens to all the position coordinates

All that’s left is the single equation for the evolution of p5.

After some algebra we obtain

€

˙ p 5 = −p6 − p4 cosθ( ) p4 − p6 cosθ( )

Asin3 θ

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˙ θ = p5

A

Falling brick

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We can restrict our attention to axisymmetric wheelsand we can choose K to be parallel to the axle

without loss of generality

€

L =12

A ˙ θ cosψ + ˙ φ sinθ sinψ( )2

+ B − ˙ θ sinψ + ˙ φ sinθ cosψ( )2

+ C ˙ ψ + ˙ φ cosθ( )2

( )

+12

m ˙ x 2 + ˙ y 2 + ˙ z 2( ) − mg⋅R

€

L =12

A ˙ θ 2 + ˙ φ 2 sin2 θ( ) + C ˙ ψ + ˙ φ cosθ( )2

( ) +12

m ˙ x 2 + ˙ y 2 + ˙ z 2( ) − mg⋅R

mgz

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If we don’t put in any simple holonomic constraint (which we often can do)

€

q =

xyzφθψ

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

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We know v and w in terms of qany difficulty will arise from r

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

€

v =ω × r

€

r = −aJ2

Actually, it’s something of a question as to where the difficulties will arise in general

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€

ddt

∂L∂˙ x i ⎛ ⎝ ⎜

⎞ ⎠ ⎟+ mgx = λ jC1

j

€

ddt

∂L∂ ˙ φ i ⎛ ⎝ ⎜

⎞ ⎠ ⎟= λ jC4

j

€

ddt

∂L∂ ˙ θ ⎛ ⎝ ⎜

⎞ ⎠ ⎟−

∂L∂θ

= λ jC5j

€

ddt

∂L∂ ˙ ψ ⎛ ⎝ ⎜

⎞ ⎠ ⎟= λ jC6

j

€

ddt

∂L∂˙ y i ⎛ ⎝ ⎜

⎞ ⎠ ⎟+ mgy = λ jC2

j

€

ddt

∂L∂˙ z i ⎛ ⎝ ⎜

⎞ ⎠ ⎟+ mgz = λ jC3

j

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We have the usual Euler-Lagrange equations

€

ddt

∂L∂˙ q i ⎛ ⎝ ⎜

⎞ ⎠ ⎟−

∂L∂qi = λ jCi

j

and we can write out the six equations

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The angular momentum in body coordinates is

€

l = A ˙ θ cosψ + ˙ φ sinθ sinψ( )I3 + B − ˙ θ sinψ + ˙ φ sinθ cosψ( )J3 + C ˙ ψ + ˙ φ cosθ( )IZZK 3

€

I3 = cosψ cosφI0 + sinφJ0( ) + sinψ cosθ −sinφI0 + cosφJ0( ) + sinθK 0( )

J3 = −sinψ cosφI0 + sinφJ0( ) + cosψ cosθ −sinφI0 + cosφJ0( ) + sinθK 0( )

K 3 = −sinθ −sinφI0 + cosφJ0( ) + cosθK 0

The body axes are (from Lecture 3)

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

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We will have a useful generalization fairly soonThe idea of separating the second order equations

into first order equations

€

˙ q i = M ir pr

€

˙ p i = 12

M mr pr∂Mmn

∂qi M ns ps − ∂V∂qi + λ jC1

j + Qi

More generically

€

˙ q i = A ji qm( )u j

€

˙ u i = f i qm,un( )