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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