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Capita SelectaHamiltonian Mechanics

in company of Arnold’s MMCMHenk Broer

Johann Bernoulli Institute voor Wiskunde en Informatica

Rijksuniversiteit Groningen

Miscellanea

• Exercises and examples

• Miscellanea from Calculus on Manifoldsintroducing the language of MathematicalPhysics

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)

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

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)

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))

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

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

2π

REMARK: Can also substitute τ = sin2 ϕ

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=

2π

ν= T (E) as should be . . .

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

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)

Gauß as Stokes ISTOKES:

∮

∂c

ω =

∫

c

dω

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

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

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

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

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?

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 . . .

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 ⇔ (∗)

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 . . .

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♯

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

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 . . .

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

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)

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

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

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

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

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

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

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

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 (∗)

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?

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

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

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

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

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

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

Hamiltoniana

• Towards a symplectic formulationof Hamiltonian Mechanics

• A summary of Arnold’s MMCM

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

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

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

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 . . .

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

∫

Jγ

ω =

∫

gτγ

dH

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∫

Jγ

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 . . .

Phase flow IIIFinish proof by Stokes:

0 =

∫

Jc

dω∗=

∫

∂Jc

ω

=

(∫

gτ c

−∫

c

±∫

J∂c

)

ω

=

∫

gτ c

ω −∫

c

ω

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

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

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

dω

EXAMPLE: g : R2n −→ R2n canonical

⇒ 1–form ω =∑

j pjd qj = p d q relative ii

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

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

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

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

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

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

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 . . .

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

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 !!

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

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

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

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

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

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

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?

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

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)

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

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

2π

∮

H−1(E)

p d q

ϕ =2π

T (E)t

then d I ∧ d ϕ = d p ∧ d q

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

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

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

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

Action-angle variables VI• . . . elliptic integral

• Vq(x) = 12ν2q + aq3 (Duffing)

t = ±∫ q d q

√

2(E − 12ν2q − aq3)

‘immediately’ elliptic

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

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!

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

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

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

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

dϕ

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

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

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

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

dϕ

d t= ω(f) and ϕ(t) = ϕ(0) + t ω

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

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)

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)

Adiabatic invariance IISUMMARY:For oscillations with nowhere zero frequency

I(E, ν) =1

2π

∮

H−1(E)

p d q ai

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

We are going to prove this . . .

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 . . .

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

∫ 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)

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

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

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

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

2π

∮

H−1(E)

p d q ai

• For harmonic oscillator

H = 12p2 + 1

2ν2q2, I = E/ν

ratio of energy and frequency ai

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