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Capita Selecta Hamiltonian Mechanics in company of Arnold’s MMCM Henk Broer Johann Bernoulli Institute voor Wiskunde en Informatica Rijksuniversiteit Groningen

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Page 1: Capita Selecta Hamiltonian Mechanics in company of …broer/pdf/ham-ex.pdfCapita Selecta Hamiltonian Mechanics in company of Arnold’s MMCM Henk Broer Johann Bernoulli Institute voor

Capita SelectaHamiltonian Mechanics

in company of Arnold’s MMCMHenk Broer

Johann Bernoulli Institute voor Wiskunde en Informatica

Rijksuniversiteit Groningen

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Miscellanea

• Exercises and examples

• Miscellanea from Calculus on Manifoldsintroducing the language of MathematicalPhysics

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Bottema 663-1Two particles Pj with mass mj (j = 1, 2) attract each

other according to Newton’s law with attractionconstant f . In the initial position they are at rest andtheir distance in 2a. When do they meet?

Answer√

1

m1 +m2πa3/2f−1/2

Solution:

Angular momentum is 0 during motion,

collision in center of mass (at rest during motion)

with infinite velocity (and hence infinite energy)

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Bottema 663-2Reduction to one degree of freedom

Motion along x–axis, coordinate Pj is xjPotential

V (q) = −fm1m21

qand U(x1, x2) = V (|x2 − x1|)

Equations of motion

mjxj = − ∂U

∂xj, j = 1, 2

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Bottema 663-3Putting q = x1 − x2 then yields

m1m2q = −(m1 +m2)dV

dqand

m1m2

m1 +m2q = −dV

dq, (1)

q(0) = 2a, q(0) = 0

one degree of freedom

Write W (q) = −f(m1 +m2) 1/q then

q = −dWdq

, q(0) = 2a, q(0) = 0 (2)

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Bottema 663-4Phase plane

Put p = q and H(q, p) = 12p2 +W (q) then (2) ⇔

q =∂H

∂p

p = −∂H∂q

q(0) = 2a, q(0) = 0

For c = W (2a) motion in level H(q, p) = c:

p = ±√

2(c−W (q))

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Bottema 663-5Time parametrizationFor any 0 < q∗ ≤ 2a time needed

T (q∗) =

∫ 2a

q∗

dq√

2(c−W (q))

Since

c−W (q) = W (2a)−W (q) =f(m1 +m2)

2aq(2a− q)

have

T (0) =

2a

f(m1 +m2)

∫ 2a

0

√q dq√

2a− q

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Bottema 663-6Beta-function

Write q = 2aτ , gives

T (0) =

a

f(m1 +m2)2a

∫ 1

0

√τ dτ√1− τ

=2B(3

2, 12)√

m1 +m2a3/2f−1/2

Finally note that

B(32, 12) =

Γ(32) Γ(1

2)

Γ(2)=

12

√π√π

1= 1

REMARK: Can also substitute τ = sin2 ϕ

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Harmonic oscillator – areaPendulum: H(p, q; ν) = 1

2p2 − ν2 cos q

p = −ν2 sin q, q = p

Harmonic oscillator: H(p, q; ν) = 12p2 + 1

2ν2q

p = −ν2q, q = p

Energy level ellipses 12p2 + 1

2ν2q2 = E

Half axes: a =√2Eν , b =

√2E

Area: A(E) = πab = 2πν E

Derivative

dA

dE=

ν= T (E) as should be . . .

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Area as a line integral

Let γ : R −→ R2 = p, q be a (smooth) closed

curve, then

A(γ) =

γ

p d q

is the oriented area enclosed by γ

PROOFS:

•• parametrize γ as a graph q 7→ p(q)

• Stokes∫

c dω =∮

∂c ω with γ = ∂c,

ω = p d q and d(p d q) = d p ∧ d q

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On determinantsIn following ξ, η are tangent vectors

• Oriented area in R2

d x ∧ d y (ξ, η) = det

[

ξ1 ξ2η1 η2

]

Similarly on Rn

• Generally for 1–forms α, β

α ∧ β (ξ, η) = det

[

α(ξ) β(ξ)

α(η) β(η)

]

• REMARK: d x(ξ) = ξ1, etc.(directional derivative)

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Gauß as Stokes ISTOKES:

∂c

ω =

c

GAUß in R3:∮

∂c

〈X,N〉d S =

c

divXdV

with N outward unit normal vector field on ∂c

Ω standard volume 3–form on R3, classically Ω = d V

ιNΩ area 2–form on ∂c, classically ιNΩ = d S

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Gauß as Stokes IIApply STOKES on ω = ιXΩ

• Divergence: dω = divX Ω

• Flux: ιXΩ = 〈X,N〉 ιNΩ• GAUß

REMARKs:

• Conversely an n–dimensional version of GAUßcan serve to define d–operator(Arnold MMCM, sect. 36)

• We use a lot that for a function f and a k–form ω

d (f · ω) = d f ∧ ω + f · dω

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On star notation ILet A : V −→ W be a linear map, thenA∗ : W ∗ −→ V ∗ between dual spaces, defined by

(A∗ω)(X) = ω(AX), ω ∈ W ∗, X ∈ V

is called adjoint map

REMARK:

• If eini=1 and e∗jnj=1 are dual bases of V, V ∗,

and fimi=1 and f ∗jmj=1 idem dito for W,W ∗

i.e. with e∗j(ei) = δi,j and f ∗j(fi) = δi,j,

then matA∗ = (matA)T, transposed matrix

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On star notation IILet g :M −→ N be smooth and Xif g(x) = y then Dxg : TxM −→ TyN linear

Take A = Dxg and apply above

• For X vector field on M and g diffeo, define

g∗X(y) = Dxg X(x)

g∗X called push-forward of X

• For ω differential form on N , define

g∗ω(x)(X) = ω(Dxg X)

for any vector field X on Mg∗ω called pull-back of ω

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On star notation III

LEMMA: For f : N −→ RCR⇒

g∗(d f) = d (f g)

PROOF: Let X be vector field on M then

(g∗(d f))(X) = d f(DgX) = d (f g)X

EXAMPLE: Polar co-ordinates (r, ϕ)g7→ (x, y) =

(r cosϕ, r sinϕ), To compute pull back g∗(d x ∧ d y)• g∗(d x) = d(x g) = d(r cosϕ) = cosϕd r − r sinϕ d ϕ etc.

• g∗(d x ∧ d y) = (cosϕd r − r sinϕ d ϕ) ∧ (sinϕd r + r cosϕdϕ) =

r cos2 ϕd r ∧ d ϕ− r sin2 ϕdϕ ∧ d r = r d r ∧ dϕ

. . . a familiar result?

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On star notation IVEXAMPLE: Let g : R2 −→ R

2 be diffeo, then

g∗(d x ∧ d y) = (detDg) · d x ∧ d y• PROOF: Let g = (g1, g2)

g∗(d x ∧ d y) = g∗(d x) ∧ g∗(d y)

lemma= d (x g) ∧ d (y g) = d g1 ∧ d g2

Then for tangent vectors ξ, η:

d g1(ξ) d g2(ξ)

d g1(η) d g2(η)

=

ξ1 ξ2

η1 η2

∂g1∂x

∂g2∂x

∂g1∂y

∂g1∂y

and the result follows by taking determinants

• Think of Jacobian determinantswithout absolute value signs . . .

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Area preserving or not I

Let XH be Hamiltonian vector field of H : R2 −→ R2

(w.r.t. to ω = d x ∧ d y)

LEMMA: Let g : R2 −→ R2 be diffeo and consider

push forwards g∗(XH) and K = H g−1. Then

g∗(XH) = det Dg ·XK (∗)

PROOF: dH = ω(−, XH), dK = ω(−, XK) and

dK = d(

H g−1)

=(

g−1)

ω(−, g∗(XH))

Since(

g−1)

ωsee above

= (detDg)−1· ω it follows

dK = ω(−, (detDg)−1· g∗(XH))

g∗(XH) = det Dg ·XK ⇔ (∗)

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Area preserving or not II

• If (M,ω) is 2–dimensional symplecticH :M −→ R Hamiltonianand g :M −→M diffeo, then always

g∗(XH), XK ∈ kerω(−, XH)

g∗(XH) ‖ XK ,

where K = H g−1

PROOF: g∗ω ∼ ω since dimM = 2

• H = K g right equivalent

C∞ equivalence of XK and g∗(XH)

catastrophe theory . . .

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On flat and sharp I

Let (V, 〈.,−〉) be Euclidean vector space

then for X ∈ V define X ∈ V ∗ by

X = 〈X,−〉

For ω ∈ V ∗ define ω♯ ∈ V by

〈ω♯,−〉 = ω

REMARKs:

• Let (M, 〈.,−〉) be Riemannian manifold

with 〈.,−〉x inner product on TxM,x ∈M

• For vector field X and 1–form ω get X and ω♯

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On flat and sharp II

• For f :M −→ R define grad = (d f)♯

On R3 consider volume 3–form Ω.

Vector calculus revisited . . .

• Y 7→ ιYΩ isomorphismbetween vector fields and 2–forms

• For Y vector field consider flux 2–form ιYΩ.Then d ιYΩ = div Y Ω

• For X vector field consider forms X and dX

For Y with ιYΩ = dX define Y = curlX

Can formulate Maxwell’s equations with Stokes

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On flat and sharp III (locally)

On R3 consider standard metric 〈.,−〉 and volume Ω.

Then for a vector field X = f ∂∂x + g ∂

∂y + h ∂∂z

ιXΩ = f d y ∧ f d z + gd z ∧ d x+ h d x ∧ d yFor a 1–form ω = f d x+ g d y + h d z have

dω = d f ∧ d x+ d g ∧ d y + d h ∧ d z

=

(

∂g

∂x− ∂f

∂y

)

d x ∧ d y +(

∂h

∂y− ∂g

∂z

)

d y ∧ d z

+

(

∂f

∂z− ∂h

∂x

)

d z ∧ d x etc . . .

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De Rham IMn manifold, Λk(M) differential forms of degree kvector spaces

Λ0(M)d1−→ Λ1(M) · · · dn−→ Λn(M) −→ 0

exact sequence: dk+1 dk = 0 ⇔ im dk ⊆ ker dk+1

THEOREM (De RHAM cohomology)

Hk(M) = ker dk+1/im dk is finite dimensional

Betti-number βk(M) = dimHk(M)topological invariant

• Λ0(M) = C(M), the smooth functions,d1 is the ‘ordinary’ d on functionsker d1 contains the (locally) constant functions

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De Rham II• H1(S1) = R

PROOF: 0 −→ Λ0(S1)d1−→ Λ1(S1)

d2−→ 0ω ∈ Λ1(S1) has format ω(ϕ) = f(ϕ) d ϕ,for f periodic

Write f = f + f , with f = 12π

∫ 2π

0 f(ϕ) dϕ and

f = ∂g∂ϕ for g periodic ω = c d ϕ+ d1 g

thus

Λ1(S1) = c d ϕ+ d1(

Λ0(S1))

| c ∈ R⇔ ker d2 = R d ϕ⊕ im d1

• d ϕ generates H1(S1)

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De Rham III• ker d1 = locally constant functions; β0 = # connected components of M

• M simply connected H1(M) = 0(form of Poincaré Lemma)

• β0(S2) = 1, β1(S2) = 0 and β2(S

2) = R,H2(S2) generated by solid angleω = d x ∧ d y + d y ∧ d z + d z ∧ d x

• β1(T2) = 2, generated by two angle-forms

• χ(M) =∑n

k=0(−1)kβk(M)Euler characteristic

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Symplectic manifolds I•• For any manifold V the co-tangent bundleM = T ∗V is a symplectic manifold with acanonical symplectic form (dealt with in detail)

• The standard 2–sphere M = S2 with its standard

area form is a symplectic manifold

If S2 ⊂ R3 with standard volume Ω, then

ω = ιNΩ is the standard area form, when N is theunit normal vector field

In spherical co-ordinatesx = R sin θ cosϕ, y = R sin θ sinϕ, z = R cos θthe symplectic 2–form on S

2 = R = 1

ω = sin θ d θ ∧ d ϕ

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Symplectic manifolds II

• The standard 2–torus M = T2 with its standard

area formd ϕ1 ∧ d ϕ2 is a symplectic manifold

• The projective plane M = P2(R) is NOT,

since it is not orientable

• DARBOUX THEOREM:(M 2n, ω) locally has the form (R2n, d p ∧ d q)

• REMARK: Not for Riemannian metrics!

EXAMPLE: On S2 for small circle θ = α

consider ratio circumference : radius

=2π sinα

α= 2π

(

1− 1

6α2

)

+O(α4) as α ↓ 0

6= 2π due to curvature

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Spherical pendulum I

• Configuration space S2 = q ∈ R

3 | 〈q, q〉 = 1Phase space

T ∗S2 ∼= (q, p) ∈ R

6 | 〈q, q〉 = 1& 〈q, p〉 = 0• Energy E(p, q) = 1

2〈p, p〉+ q3

Angular momentum I(p, q) = q1p2 − q2p1(Energy-) momentum map

EM : T ∗S2 −→ R

2 = (I(p, q), E(p, q))

• Lagrangean fibration: fibers T2

(precession, nutation)

singular: boundary points have fiber S1

∼ horizontal oscillation (Huygens)

• (q, p) = ((0, 0,±1), (0, 0, 0)) 7→(I, E) = (0,±1) ∼ equilibria

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On tensors I• A co-variant k–tensor T turns k vector fieldsX1, X2, . . . , Xk into a scalar function in amulti-linear way

Co-variant 1–tensors are exactly the 1–forms

EXAMPLEs: A differential form is a k–tensorthat, moreover, is anti-symmetricA Riemannian metric is a 2–tensor that,moreover, is symmetric

• For 1–forms α1, α2 and vector fields X1, X2 have

α1 ⊗ α2(X1, X2) = α1(X1) · α2(X2)

(pointwise product)

• α1 ∧ α2 = α1 ⊗ α2 − α2 ⊗ α1

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On tensors II

• A Riemannian metric g on a surface locally hasthe format

g =∑

i,j

gi,j d xi ⊗ d xj

for a positive symmetric matrix (gi,j)i,j ,

depending smoothly on the point of attachment

• For a 2–form ω on a surface similarly

ω =∑

i,j

ωi,j d xi ⊗ d xj

for an anti-symmetric matrix (ωi,j)i,j

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On tensors III

• A contra-variant k–tensor turns k 1–formsα1, α2, . . . , αk into a scalar function in amulti-linear way

Contra-variant 1–tensors are exactly the vector

fields: by defining X(α) = α(X)

• EXAMPLE: contra-variant 2–tensors on a surfacelocally have the format

T = T i,j ∂

∂xi⊗ ∂

∂xj

• Also tensors of mixed type exist . . .

• Transformation rules are straightforward,although tedious

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On tensors IV

• RAISING and LOWERING of indices:Given Riemannian metric g =

i,j gi,j d xi⊗ d xj

and a vector field X =∑

jXj ∂∂xj

Then X = g(X,−) locally looks like

Xj d xj =

=∑

gi,j d xi ⊗ d xj

(

X i ∂

∂xi,−)

=∑

j

(

i

gi,jXi

)

d xj (∗)

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On tensors IV

•• Classical notation: just Ti,j and T i,j , etc.

• Then (∗) ⇔

Xj =∑

i

gi,jXi

• Einstein convention included . . .

Xj = gi,jXi

Familiar formula?

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Pendulum – d’Alembert IPlanar pendulum q ∈ R

2, constraint:

f(q) := q − ℓ and f(q) = 0 by force λ grad fequations of motion (Newton II + d’ Alembert)

mq = −gradU + λer

⇔d

dt

(

∂L

∂q

)

−(

∂L

∂q

)

= λ grad f

In polar coordinates

L(r, ϕ, r, ϕ) = 12m(r2 + r2ϕ2) +mgr cosϕ

f(r, ϕ) = r − ℓ, grad f = er

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Pendulum – d’Alembert IIEquations of motion ⇔

d

dt

(

∂L

∂r

)

−(

∂L

∂r

)

= λ

d

dt

(

∂L

∂ϕ

)

−(

∂L

∂ϕ

)

= 0

⇔mr −mrϕ2 −mg cosϕ = λ

2mrrϕ+mr2ϕ+mgr sinϕ = 0

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Pendulum – d’Alembert IIIUnder constraint f(r, ϕ) = 0

• r ≡ ℓ, r ≡ 0 ≡ r

• λ = −mrϕ2 −mg cosϕ(centrifugal force + radial component gravitation)

• ℓϕ+ g sinϕ = 0(familiar equation of motion)

⇔ Lagrange on tangent bundle TS1

Used:q = r er + rϕ eϕq = (r − rϕ2) er + (2rϕ+ rϕ) eϕ

grad f = ∂f∂r er +

1r∂f∂ϕ eϕ

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Noether IL : TM → R and h :M →M both smooth

System (M,L) admits h iff for any v ∈ TM :

L(h∗v) = L(v)

THEOREM (M,L) admits one-parameter grouphs :M →M ⇒ there is a first integral I : TM → R

Locally

I(q, q) =∂L

∂q

d hs(q)

d s

s=0

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Noether IIPROOF: Note that by symmetry

L(hs∗v) is independent of s

locally M = Rn

Let ϕ : R →M , q = ϕ(t) be solution of the E-L eqns⇒ hs ϕ : R →M idem dito for all s

Consider Φ : R2 → Rn with q = Φ(s, t) = hs(ϕ(t))

So obtain

0 =∂L(Φ, Φ)

∂s=∂L

∂qΦ′ +

∂L

∂qΦ′

Here ′ is differentiation w.r.t. s

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Noether IIISince Φ|s solves E-L:

∂t

[

∂L

∂q(Φ(s, t), Φ(s, t))

]

=∂L

∂q(Φ(s, t), Φ(s, t))

Abbreviate

F (s, t) =∂L

∂q(Φ(s, t), Φ(s, t))

and substitute ∂F∂t = ∂L

∂q(writing q′ = dq′

dt )

0 =

(

d

dt

∂L

∂q

)

q′ +∂L

∂q

(

d

dtq′)

=d

dt

(

∂L

∂qq′)

=dI

dt

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Hamiltoniana

• Towards a symplectic formulationof Hamiltonian Mechanics

• A summary of Arnold’s MMCM

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Canonical IFor x ∈ V , p ∈ T ∗

xV and ξ ∈ Tp(T∗V )

σ : ξ 7→ π∗ξ 7→ p(π∗ξ)

defines canonical 1–form ω = d σ canonical 2–form

THEOREM (local format): local co-ordinatesq = (q1, q2, . . . , qn) give format

σ =n∑

j=1

pj d qj, ω =n∑

j=1

d pj ∧ d qj

where p1, p2, . . . , pn are co-ordinates of the basis

d q1, d q2, . . . , d qn of ∂∂q1, ∂∂q2, . . . , ∂

∂qn

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Canonical IIPROOF: local co-ordinates(q, p) = (q1, q2, . . . , qn, p1, p2, . . . , pn) on T ∗Vin which π : (q, p) 7→ q

If ξ = f(q, p) ∂∂q + g(q, p) ∂

∂p , then

π∗ξ = f(q, p)∂

∂qand p(π∗ξ) = p f(q, p)

(matrix notation)

σ = p d q and ω = d σ = d p ∧ d q

REMARK: Special case of Darboux

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Hamiltonian ILet (M,ω) be symplectic manifold, then mapping

I−1 : ξ ∈ TxM 7→ ω(−, ξ) ∈ T ∗xM

linear isomorphism

Given function H :M −→ R then

XH = I(dH)

corresponding Hamiltonian vector field

THEOREM (local format): if ω = d p ∧ d q then

XH =∂H

∂p

∂q− ∂H

∂q

∂p

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Hamiltonian IIIn co-ordinates

d p ∧ d q (η, ξ) = det

(

η1 ξ1η2 ξ2

)

= (ξ2 − ξ1)

(

η1η2

)

Thus

I−1 :

(

ξ1ξ2

)

7→(

ξ2−ξ1

)

and

I−1 =

(

0 1

−1 0

)

notation

XH = I−1 gradH . . .

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Phase flow I(M,ω) symplectic manifold and H :M −→ R

gt :M −→M phase flow of XH = I(dH)

THEOREM: (gt)∗ω = ω

PROOF: Sufficient that for any 2–chain c:∫

gτ c

ω =

c

ω

To show this, consider flowbox Jc

1. For any 1–chain (curve) γ

d

d τ

ω =

gτγ

dH

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Phase flow IIIndeed, parametrize Jγ by

(s, t) 7→ gtγ(s) =: f(s, t)

Setting ξ = ∂f∂s and η = ∂f

∂t , then ω(η, ξ) = dH(ξ)Thus∫

ω =

∫ ∫

ω(η, ξ)d t d s =

∫ (∫

gtγ

dH

)

dt

2. If γ is closed then∫

Jγ ω = 0

Indeed,∮

γ dH = 0

Holds particularly if γ = ∂c . . .

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Phase flow IIIFinish proof by Stokes:

0 =

Jc

dω∗=

∂Jc

ω

=

(∫

gτ c

−∫

c

±∫

J∂c

)

ω

=

gτ c

ω −∫

c

ω

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Integral invariant I

Let g :M −→M and ω k–form

ω is (absolute) integral invariant of g if∫

gc ω =∫

c ω

REMARKS:

•• EXAMPLE: M = R2 and ω = d p ∧ d q,

then ω ii of g iff detDg ≡ 1

• ω ii of g iff g∗ω = ωindeed,

gc ω =∫

c g∗ω

• α and β ii of g ⇒ α ∧ β ii of gindeed, g∗(α ∧ β) = (g∗α) ∧ (g∗β) = α ∧ β

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Integral invariant II

LEMMA: (M,ω) symplectic and H :M −→ R

gτ phase flow of XH = I(dH) ⇒ ω ii of gτ

Moreover then

ω2 = ω ∧ ω. . .

ωn = ω∧ n times· · · ∧ωalso ii of gτ

. . . Liouville Theorem recovered !!

DEFINITION: (M,ω) symplecticg :M −→M canonical iff ω ii of g

Canonical mappings preserve Liouville volume

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Integral invariant III

Let g :M −→M and ω k-form

ω is relative ii of g iff∫

gc ω =∫

c ω for all closed c

LEMMA: ω relative ii for g ⇒ dω absolute ii for g

PROOF:∫

c

dω∗=

∂c

ω =

g∂c

ω =

∂gc

ω∗=

gc

EXAMPLE: g : R2n −→ R2n canonical

⇒ 1–form ω =∑

j pjd qj = p d q relative ii

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Integral invariant IV

EXAMPLE: M = R2 \ 0, ω = xdy

Let X = grad log r = 1x2+y2

(

x ∂∂x + y ∂

∂y

)

and d ϕ = 1x2+y2 (−y d x+ x d y)

Then X = Xϕ

flow gτ canonical, i.e. dω (absolute) ii

HOWEVER: for c unit circle we find∫

g1c

ω >

c

ω

⇒ ω not relative ii

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Integral invariant V

• THEOREM (conservation energy): (M,ω)symplectic, H :M −→ R, XH = I(dH) with

phase flow gt. THEN H is (first) integral of gt

PROOF: dH(XH) = ω(XH , XH) = 0

• COROLLARY: dH is relative ii of gt

PROOF:∫

c

dH = H(c(1))−H(c(0))

gτ c

dH = H(gτ(c(1)))−H(gτ(c(0)))

thm= H(c(1))−H(c(0)) =

c

dH

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Poisson brackets I(M,ω) symplectic and F,H :M −→ R then

F,H = XH(F ) = dF (XH)

• gt flow of XH , then F,H(x) = dd t

t=0F (gt(x))

• F (first) integral of gt iff F,H = 0

• F,H = −ω(XF , XH),F,H = −H,F and

λ1F1 + λ2F2, H = λ1F1, H+ λ2F2, H• Jacobi identity

F,G, H+ G,H, F+ H,F, G = 0

i.e., (C∞(M), ·,−) Lie algebra

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Poisson brackets II• [XF , XG] = XF,G implying that the mapping

H 7→ XH is a morphism of Lie algebraswhat is its kernel ?

• IF ω = d p ∧ d q THEN

F,H =n∑

i,j=1

(

∂H

∂pi

∂F

∂qj− ∂H

∂qj

∂F

∂pi

)

In particular

qi, qj = pi, pj = qi, pj = 0 (i 6= j)

pi, qi = 1

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Symmetry revisited

(M,ω) symplectic, H,F,G :M −→ R

XH = I(dH), &c.

• POISSON THM: IF F1, F2 integrals of XH ,

THEN so is F1, F2PROOF: Jacobi identity

integrals of XH form Lie subalgebra

• NOETHER THM: IF gt, flow of XF , leaves XH

invariant THEN F (first) integral of H

PROOF: Direct implication: H integral of FSince H,F = 0 = F,Halso F integral of XH

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Poincaré–Cartan I (vorticity)

Consider R3 with inner product 〈·,−〉 and volume Ω isomorphisms

Φ : A vector field 7→ A = 〈A,−〉 1–form

Ψ : A vector field 7→ Ω(A, ·,−) = det (A, ·,−)2–form

earlier Ψ(A) = ιA(Ω)Given vector field X obtain R = curlX as follows:

curl : XΦ7→ X 7→ dX Ψ−1

7→ R

Integral curves of R called vortex lines of X

STOKES LEMMA:∮

γ1Xds =

γ2Xds

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Poincaré–Cartan II

PROOF: If σ = X then d σ = Ψ−1RApply Stokes Theorem to ‘flowbox’ c = Jγ1(noting that: ∂c = γ2 − γ1):

0 =

c

〈R,N〉d S =

c

d σ∗=

∂c

σ =

=

∂c

dX =

γ2

Xds−∮

γ1

Xds

and now something completely different . . .

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Poincaré–Cartan IIIConsider R3 = p, q, t and H : R3 −→ R

Let σ = p d q −Hd t then

d σ = d p ∧ d q − dH ∧ d t

= −∂H∂q

d q ∧ d t+ ∂H

∂pd t ∧ d p+ d p ∧ d q

Corresponding are

X = Φ−1(σ) = p∂

∂q−H

∂tand

R = Ψ−1(d σ) = −∂H∂q

∂p+∂H

∂p

∂q+∂

∂t

see earlier computations

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Poincaré–Cartan IVR ∼ time-dependent Hamiltonian vector field

p = −∂H∂q

q =∂H

∂p

t = 1

How is this for 2n+ 1 dimensions?

Away with X and the metric !!

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Poincaré–Cartan VLEMMA: Let ω be 2–form on R

2n+1.THEN ∃ ξ 6= 0 s.t. ω(ξ,−) ≡ 0

PROOF: In co-ordinates

ω(ξ, η) = 〈Aξ, η〉,

with At = −ATHUS

detA = det(

At)

= det(−A) = (−1)2n+1 detA

= − detA,

detA = 0Take ξ ∈ kerA

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Poincaré–Cartan VIDefine kerω by

ξ ∈ kerω ⇔ ω(ξ,−) ≡ 0

Call ω non-degenerate iff dimkerω = 1

EXAMPLE: R3 = p, q, t andω = d p ∧ d q + α ∧ d t, for any 1–form α, then

ω(ξ, η) = pξqη − qξpη = 〈(

0 −1

1 0

)

ξ, η〉

Thus

A =

(

0 −1

1 0

)

and rankA = 2

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Poincaré–Cartan VII• Let α be non-degenerate 1–form on M 2n+1 Then

x ∈M 7→ kerα(x) ⊂ TxM

defines direction field (REPLACES X AND X!)

• Also for non-degenerate ω = dα

x 7→ kerω(x) ⊆ TxM

defines direction field: vortex directions of α

• THEOREM:∮

γ1α =

γ2α

PROOF: Stokes. Note that dα|T (Jγ1) = 0 by

definition of vortex direction

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Poincaré–Cartan VIIIR

2n+1 = p1, p2, . . . , pn, q1, q2, . . . , qn = p, q, tα =

j pj d qj −Hdt = p d q −Hd t

for given H : R2n+1 −→ R

THEOREM

• The vortex lines of α are of the formatt 7→ (p(t), q(t), t) satisfying

p = −∂H∂q

, q =∂H

∂p, t = 1

•∮

γ1p d q −Hd t =

γ2p d q −Hd t

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Poincaré–Cartan IXPROOF:

• dα(ξ, η) = 〈Aξ, η〉 where

A =

0 −E Hp

E 0 Hq

−Hp −Hq 0

.

Rank A = 2n and

A

−Hq

Hp

1

= 0

• Stokes

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Poincaré–Cartan XCOROLLARY: For fixed values t1 and t2:

•∮

γ1

p d q =

γ2

p d q

•∫ ∫

σ1

d p ∧ d q =∫ ∫

σ2

d p ∧ d q

REMARK:p d q called relative integral invariant of Poincaré

a form of Liouville recovered

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Canonical transformations IRECALL:R

2n+1 = p, q, t and H : R2n+1 −→ R

Flow of

p = −∂H∂q

, q =∂H

∂p, t = 1

given by vortex lines of p d q −Hd t

Under transformation (p, q, t) 7→ (x1, x2, . . . , x2n+1)changes into

jXj(x)d xj

REMARK: Canonical format. How to preserve?

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Canonical transformations IITHEOREM: Any transformation (p, q, t) 7→ (P,Q, T )such that ∃ functions K(P,Q, T ) and S(P,Q, T ) with

p d q −Hd t = P dQ−KdT + d S

keeps canonical format

P ′ = −∂K∂Q

,Q′ =∂K

∂P, T ′ = 1

PROOF: Flow represented by vortex lines of

P dQ−Kd t+ d S. Since d (d S) = 0

d(P dQ−Kd t+ d S) = dP ∧ dQ− dK ∧ d T

& above construction gives result

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Canonical transformations IIISuch transformations are called canonical

COROLLARY: Any canonical transformation

(p, q, t) 7→ (P,Q, t) gives

P = −∂K∂Q

, Q =∂K

∂P, (t = 1)

with K(P,Q, t) = H(p, q, t)

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Canonical transformations IVPROOF: Let σ = p d q − P dQ, then

γ

σ =

γ

p d q −∮

γ

P dQ = 0 S =

∫ (p,q)

−σ

well-defined, while d S = p d q − P dQ

⇒ p d q −Hd t = P dQ−Kd t+ d S

COROLLARY: (p, q, t) 7→ (P,Q, t) withdP ∧ dQ = d p ∧ d q is canonical

PROOF: By Stokes∮

γ p d q =∫ ∫

d p ∧ d q &c

then see above proof

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Action-angle variables I

• Polar co-ordinates: (p, q, t) 7→ (I, ϕ, t) with

p =√2I cosϕ and q =

√2I sinϕ

Then d I ∧ d ϕ = rd r ∧ d ϕ = d p ∧ d q• For 1–DOF system: q = ∂H

∂p , p = −∂H∂q :

I(p, q) =1

H−1(E)

p d q

ϕ =2π

T (E)t

then d I ∧ d ϕ = d p ∧ d q

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Action-angle variables II• PROOF that d I ∧ dϕ = d p ∧ d q:I(E) = 1

2πA(E) and

d I = 12π

dA(E)dE dE = 1

2πT (E)dE

Moreover d ϕ = 2πT (E)d t+

∂ϕdEdE

So d I ∧ dϕ = dE ∧ d t = d p ∧ d q

• Format oscillations: I = 0, ϕ = ω(I)

with ω(I) = 2πT (E(I)) and K(I, ϕ) = E(I):

indeed,

∂E

∂I= 1/

(

∂I

∂E

)

=2π

T= ω

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Action-angle variables III

• For harmonic oscillator H(p, q) = 12p2 + 1

2ν2q2

combine above to elliptic polar co-ordinates

p =√2Iν cosϕ and q =

2I

νsin q

d I ∧ d ϕ = d p ∧ d q• Conclude that K(I) = νI and ω(I) ≡ ν,

so q = −ν2q gets ‘integrable’ form

I = −∂K∂ϕ

= 0

ϕ =∂K

∂I= ν

to be ctd

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Action-angle variables IV

• For general oscillator d t = d q/p

t =

∫ q d q

pE(q)

where p = pE(q) solves H(p, q) = E

• H(p, q) = 12p2 + V (q) pE =

2(E − V (q))so,

t = ±∫ q d q

2(E − V (q))

• V (q) = 12ν2q2, setting u = νq/

√2E

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Action-angle variables V• . . .

t =1

νarcsin

( νq

2E

)

=1

νϕ(q, E; ν))

so ϕ = ν, as should be

• V (q) = −ν2 cos q (pendulum)

t = ±∫ q d q

2(E − ν2 cos q)

z=cos q=

∓∫ z d z√

1− z2√

2(E − ν2z)=

∓∫ z d z√

2E − 2ν2z − 2Ez2 + 2ν2z3

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Action-angle variables VI• . . . elliptic integral

• Vq(x) = 12ν2q + aq3 (Duffing)

t = ±∫ q d q

2(E − 12ν2q − aq3)

‘immediately’ elliptic

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Spherical pendulum II

E(q, p) = 12〈p, p〉+ q3 and I(q, p) = q1p2 − q2p1

• Spherical coordinatesq1 = sinφ cos θ, q2 = sinφ sin θ, q3 = cos θ

• Canonical 1–formsin2 θ · φ d φ+ θ d θ =: pφ d φ+ pθ d θ

• MomentaI = sin2 θ · φE = 1

2 sin2 θ · φ2 + 1

2 θ2 + cos θ = 1

2 θ2 + VI(θ),

VI(θ) =I2

2 sin2 θ+ cos θ

angle φ cyclic

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Spherical pendulum III• Reduced energy level (circle)

CI,E = (θ, θ) ∈ (0, π)× R | E = 12 θ

2 + VI(θ)• Actions J1(I, E) = 2πI

and (by integration along CI,E)

J2(I, E) = 2

∫ θ+

θ−

2(E − cos θ)− I2

sin2 θd θ,

where 2(E − cos θ)− I2/ sin2 θ = 0 ⇔ θ = 豕 (J1, J2, ϕ1, ϕ2) action-angle variables,

for angles ϕj need time-integrals . . .

• Please draw reduced phase portraits!

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Sperical pendulum IV

Basis (T1(I, E), T2(I, E)) of period lattice:

T1(I, E) = (2π, 0) and

T2,1(I, E) = −2I

∫ θ+

θ−

1√

2(E − cos θ)− I2/ sin2 θ

d θ

sin2 θ

T2,2(I, E) = 2

∫ θ+

θ−

1√

2(E − cos θ)− I2/ sin2 θd θ

T2,1(I, E) = −2

∫ θ+

θ−

φ

θd θ and T2,2(I, E) = 2

∫ θ+

θ−

1

θd θ

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Spherical pendulum V

Lemma (Duistermaat (1980), Horozov (1990))

1. T1 and T2,2 are single-valued, T2,1 is multi-valued

with monodromy

(

1 −1

0 1

)

∈ GL(2,Z)

2. Hamiltonian H = E is KAM–nondegenerate

• Liouville–Arnold–Duistermaat theorem

• Geometric interpretation . . .

• Kolmogorov–Arnold–Moser theory

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LAD theorem I

Given symplectic mfd (M 2n, ω) and HamiltoninanH :M −→ R Liouville integrability of system I dH:

• IF:

- ∃ functions Fj :M −→ R, with Fi, Fj = 0,i, j = 1, 2, . . . , n

- In level setMf = x ∈M | Fj(x) = fj, j = 1, 2, . . . , nassume dFj linearly independent for each x

- Mf is compact and connected

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LAD theorem II• THEN:

- Mf is diffeomorphic to n–torus

Tn = (ϕ1, ϕ2, . . . , ϕn) mod 2π &

- The vector field I dH on Tn gets the form

d t= ω, ω = ω(F)

conditional periodic motion

- canonical eqns integrable ‘by quadratures’

• COROLLARY: In 2 d.o.f. integral F notdepending on H integrability ‘by

quadratures’: level set H = h, F = f T2 with

conditional periodicity

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LAD theorem III

PROOF

- Mf mfd by Implicit Function Theorem

- The vector fields I d Fj (j = 1, 2, . . . , n) are

tangent to Mf, independent and commutingsince

[I d Fi, I d Fj] = I d Fi, Fj and I d Fj(Fi) = 0

• COROLLARY so FAR: Mfd Mf invariant underphase flow gtj of I d Fj (for j = 1, 2, . . . , n)

where

gti gsj = gsj gti

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LAD theorem IV• LEMMA: Let Mn be compact, connected, with n

pairwise commutative and pointwise independentvector fields; THEN M diffeomorphic to T

n

• PROOF LEMMA: Consider Rn = t and let

gt = gt11 gt22 · · · gtnn group action

(t, x) ∈ Rn ×M 7→ gt(x) ∈M

- Covering map with isotropy group Γ:

t ∈ Γ iff. ∃ x0 ∈M with gt(x0) = x0- Γ ⊂ R

n discrete subgroup, independent of x0- ∃ e1, e2, . . . , ek ∈ R

n such that

Γ = |[e1, e2, . . . , ek]|Z

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LAD theorem V• Ctd.

- k = n by compactness of M

- Mf∼= R

n/Γ ∼= Tn

• Defines integer affine structure on Mf

• Angular co-ordinates ϕ on Mf format

d t= ω(f) and ϕ(t) = ϕ(0) + t ω

HAVE TO KNOW PERIODS !!Period lattices too . . .

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Action-angle variables VII

• Locally can use co-ordinates (F, ϕ) format

dF

d t= 0,

d ϕ

d t= ω(F)

no symplectic co-ordinates yet

• Locally can construct I = I(F) such that (I, ϕ)symplectic familiar format

d I

d t= 0,

d ϕ

d t= ω(I)

• Nontrivial monodromy obstructsglobal construction (I, ϕ)

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Adiabatic invariance IR

3 = p, q, t and H : R3 −→ R of the form

H = H(p, q, ν) with ν = ε

I = I(p, q; ν) adiabatic invariant (ai) if∀κ > 0, ∃ ε0 > 0 such that for all 0 < ε < ε0

0 ≤ t ≤ 1

ε⇒ |I(p(t), q(t); εt)− I(p(0), q(0); 0)| < κ

EXAMPLES:

• (First) integrals

• Horizontal pendulum, slowly varying length ν:

⇒ I = mν2ϕ is ai (by symmetry – check)

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Adiabatic invariance IISUMMARY:For oscillations with nowhere zero frequency

I(E, ν) =1

H−1(E)

p d q ai

⇒ for harmonic oscillator I(E, ν) = Eν ai

We are going to prove this . . .

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Adiabatic invariance IIIIDEA: Slowly varying Hamiltonian system

I = εg(I, ϕ; ν)

ϕ = ω(I, ν) + εf(I, ε; ν)

ν = ε,

Averaging

approximation J = εg(J,Λ), Λ = εwhere g ≡ 0 (g derivative of a periodic function)

J = 0, Λ = ε

J ai for original system . . .

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Adiabatic invariance IVGENERAL THEORY

I = εg(I, ϕ)

ϕ = ω(I) + εf(I, ε)

with ϕ mod 2π and I ∈ G ⊆ RN open

f, g 2π–periodic in ϕ

Averaged system:

J = εg(J), g =1

∫ 2π

0

g(J, ϕ) d ϕ

(I(t), ϕ(t)) solution with initial condition (I(0), ϕ(0))J(t) solution with initial condition J(0) = I(0)

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Adiabatic invariance VTHEOREM: Under some a priori estimates on ω, f, g,

IF ω(I) > c > 0 on G and J(t) ∈ G− d for 0 ≤ t ≤ 1ε

THEN for ε > 0 sufficiently small:

|I(t)− J(t)| < Cε for all 0 ≤ t ≤ 1

ε

PROOF: Let P = I + εk(I, ϕ) then

P = I + ε∂k

∂II + ε

∂k

∂ϕϕ

= ε

[

g(I, ϕ) +∂k

∂ϕω(I)

]

+ ε2∂k

∂Ig + ε2

∂k

∂ϕ

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Adiabatic invariance VIFor ε small ∃ inverse I = P + εh(P, ϕ, ε)

P = ε

[

g(P, ϕ) +∂k

∂ϕω(P )

]

+O(ε2)

Try to choose k such that

g(P, ϕ) +∂k

∂ϕω(P ) = 0 ⇔ ∂k

∂ϕ= − 1

ωg

Obstruction: take average on both sides !!

If g(P, ϕ) = g(P, ϕ)− g(P ) it IS possible to choose

k(P, ϕ) =

∫ ϕ

0

1

ω(P )g(P, ψ) dψ

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Adiabatic invariance VIII⇒ k(P, ϕ) periodic in ϕNow compare

P = εg(J) +O(ε2)

J = εg(J)

and conclude:

|I(t)− P (t)| ≡ ε|k| with I(0) = P (0)

|P (t)− J(t)| = O(ε) for 0 ≤ t ≤ 1

ε

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Adiabatic invariance IX• In Hamiltonian case g(P ) ≡ 0, and

J(t) ≡ P (0) = I(0), so

|I(t)− I(0)| = O(ε) for 0 ≤ t ≤ 1

ε

• In this case

I(E, ν) =1

H−1(E)

p d q ai

• For harmonic oscillator

H = 12p2 + 1

2ν2q2, I = E/ν

ratio of energy and frequency ai

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Scholium generale• Dynamics:

- What if ω(I0) = 0 ?

- What if more angles ϕ ?

- general perturbation theoryKolmogorov Arnold Moser theory,singularity theory &c

• Mathematical Physics:General relativity, gauge theory, string theory &c

in a conceptual (co-ordinate free) language