Classical Mechanics II - Padakshep · Preprint typeset in JHEP style - HYPER VERSION Classical...

Preprint typeset in JHEP style - HYPER VERSION

Classical Mechanics II

Ashoke Sen

Harish-Chandra Research Institute Chhatnag Road,

Jhusi, Allahabad 211019, India

Abstract: This note is based on a handwritten note by Prof Sen. This note is typed by

Sampath Mukherjee and Arnab Rudra under Padakshep Open Teaching Project. Padakshep

is an NGO which is working to in education sector in West Bengal, India. Padakshep Open

Teaching Project is an initiative to provide free education for all. More information can be found

at www.padakshep.org

Classical Mechanics II: Ashoke Sen Padakshep Open Teaching Project

Contents

1. Lagrangian dynamics: 1

2. Hamiltonian System (II) 14

1. Lagrangian dynamics:

1.1 Definition: Legendre transformation of the Hamiltonian

uα ≡(

∂H∂pα

)p,q

=⇒ Gives uα as a function of pα & qα

Assume that these relation are invertible, So that this allows us to solve pi as function of

uα & qα

L(q, u) =n∑

α=1

pαuα −H(p, q) (1.1)

pα express as function of qα and uα

dL =

N∑α=1

dpαuα −

N∑α=1

(∂H

∂pα)p,qdpα −

N∑α=1

(∂H

∂qα

)p,q

dqα (1.2)

=

N∑α=1

(pαduα − ∂H

∂qαdqα) (1.3)

⇒(∂L

∂uα

)q,u = pα,

(∂L

∂qα

)q,u = −

(∂H

∂qα

)p,q

(1.4)

Equation of motion:

∂qα

∂t=

(∂H

∂pα

)p,q

= uα (1.5)

d

dt

(∂L

∂uα

)q,u = −

(∂H

∂qα

)p,q

=

(∂L

∂qα

)p,u

(1.6)

Under what condition is a transformation qα&uα************text missing ***** of the

equation of motion.

1


1.2 Lagrangian : Alternate formulation

Dynamical Variables.

qα, uα 1 ≤ α ≤ N (1.7)

Equations of motion:

dqα

dt= uα (1.8)

d

dt

(∂L

∂uα

)q,u =

(∂L

∂qα

)q,u

(1.9)

Unlike the Hamilton’s equation, these equations do not have nice tensorial form,and hence

it is not clear what condition one must impose on the Lagrangian so that equation of motion

are preserved under a transformation(q, u) −→ (˜q, ˜u)

First take a brute force approach

If you take Lagrangian equation as 2N dimensional vector equation then qα & uα are not

in equal footing. So its covariance cannot be demanded so easily **********missing text******

First take

qα −→ qα = fα (q) (1.10)

uα −→ uα =dfα (q)

dt=∂fα (q)

∂qαuβ =

∂qα

∂qβuβ (1.11)

This givesdqα

dt= uα (1.12)

Define L(˜q, ˜u) ≡ L (q, u) (Justified later in Problem -3)

d

dt(∂L

∂uα) =

d

dt(∂L

∂uβ∂uβ

∂uα+∂L

∂qβ∂qβ

∂uα) (1.13)

=d

dt(∂L

∂uβ∂qβ(q)

∂qα) (1.14)

=d

dt(∂L

∂uβ)∂qβ

∂qα+

∂L

∂uβ∂2qβ

∂qr∂qαur (1.15)

=⇒ ∂L

∂qα=

∂L

∂qβ∂qβ

∂qα+

∂L

∂uβ∂uβ

∂qα(1.16)

=∂L

∂qβ∂qβ

∂qα+

∂L

∂uβ∂2qβ

∂qα∂qrur (1.17)

2


1.3 Equation of motion:

d

dt(∂L

∂uβ)∂qβ

∂qα+

∂L

∂uβ∂2qβ

∂qr∂qαur =

∂L

∂qβ∂qβ

∂qα+

∂L

∂uβ∂2qβ

∂qα∂qrur (1.18)

=⇒ d

dt(∂L

∂uβ)∂qβ

∂qα∂qα

∂qr=

∂L

∂qβ∂qβ

∂qα∂qα

∂qr(1.19)

=⇒ ∂qβ

∂qα∂qα

∂qr+∂qβ

∂uα∂uα

∂qr=∂qβ

∂qr= δβr (1.20)

=⇒ d

dt(∂L

∂uβ)δβr =

∂L

∂qβδβr (1.21)

=⇒ d

dt(∂L

∂ur)− ∂L

∂qr= 0 (1.22)

Thus the equation of motion have the same form in the new coordinate system if

L(˜q, ˜u) ≡ L(q, u) (1.23)

i.e. if

L(˜q, ˜u) ≡ L (q, u) (1.24)

Associated with such a continuous symmetry transformation there is a conserved quantity.

Infinitesimal transformation:

qα −→ qα = qα + ϵϕα(q) (1.25)

uα −→ uα = uα + ϵ∂ϕα

∂qβuβ (1.26)

L(˜q, ˜u) ≡ L (q, u) + ϵ

∂L

∂qαϕα + ϵ

∂L

∂uα∂ϕα

∂qβuβ (1.27)

=⇒ ∂L

∂qαϕα +

∂L

∂uα∂ϕα

∂qβuβ = 0 (1.28)

Conserved quantity:

F =∂L

∂uαϕα (1.29)

Proof :

dF

dt=

d

dt(∂L

∂uαϕα) =

d

dt

∂L

∂uαϕα +

∂L

∂uα∂ϕα

∂qβuβ (1.30)

=∂L

∂qαϕα +

∂L

∂uα∂ϕα

∂qβuβ = 0 (1.31)

=⇒F remains constant as q, u evolve according to the equations of motion.

3


Example:

H =

3∑α=1

(pα)2

2m+ V (r) (1.32)

uα = (∂H

∂pα)p,q =

pαm

(1.33)

L =∑α

uαpα −H = m∑α

uαuα − 1

2m∑α

uαuα − V (r) (1.34)

=1

2m∑α

uαuα − V (r) (1.35)

Invariant under:

qα −→ qα = Rαβqβ RRT = I (1.36)

uα −→ uα = Rαβuβ (1.37)

Infinitesimal Transformations:

Rαβ = δαβ + ϵωαβ (1.38)

qα = qα + ϵωαβqβ (1.39)

where ϕα(q)

=⇒ F =∂L

∂uαϕα(q) = muαωαβq

β (1.40)

ω = a1T1 + a2T2 + a3T3 (1.41)

F = a1F1 + a2F2 + a3F3 (1.42)

F1 = m(u2q3−u3q2), F2 = m(u3q1−u1q3), F3 = m(u1q2−u2q1) −→ Angular Momenta (1.43)

Transformation Laws of uαare determined from the beginning in terms of those of qα.

We do not need to check if the transformations are canonical.

More general transformations :

qα −→ qα = fα(q, u) (1.44)

uα −→ uα =dqα

dt=∂fα

∂qβuβ +

∂fα

∂uβuβ (1.45)

4


Figure 1: Type I Dualities.

−→Cannot be viewed as a transformation of phase space coordinates:

P is not determined in terms of P .

Passive view point seems to be lost.

Way out: Equation of motion can be used to express duβ

dt in terms of u& q

=⇒ uα = hα(q, u) (1.46)

The new transformation Laws agree with the old ones while acting on solutions the equation

of motion.

Must do this in order to write the transformation laws of p, q in hamiltonian formalism.

We can stick to the original transformations in the Lagrangian formulation.

Active point of view still holds.

A solution of equation of motion: q0(t) =⇒all derivatives of q0(t)

qα0 (t)& uα0 (t) are defined.

The transformation (qα, uα) −→ (qα, uα) is a symmetry if it takes all solutions of the

equation of motion to solutions.

We would like to determine what kind of transformations satisfy this property and call them

symmetry transformations.

A related problem

L(q, u

)= L(q, u) (1.47)

=⇒ qα = fα(q, u), uα =∂fα

∂qβuβ +

∂fα

∂uβuβ (1.48)

L(q, u

)= function of qα, uα& u (1.49)

5


L(q, u

)= L(q, u) : Function of qα, uα (1.50)

=⇒L cannot be equal to L in a generic Case.Change point of view.

Fundamental Dynamical Variables.

Functions q(t) & not q, u

Actually here fundamental variable is path which is varying.

Instead of using the Lagrangian L, use

S(q(t)) =t2∫

t1

L(q(t), ˙q(t))dt (1.51)

where q(t1) = q1, q(t2) = q2 =Boundary conditions.

S: Functions of function q(t)=⇒Functionals.

For given q(t),we have a definite rule for computing S.

If q(t) is a classical solution then S always increases under a local deformation of the path

away from the boundaries.

6


δS =

t2∫t1

(∂L

∂qαδqα +

∂L

∂uαδqα) (1.52)

=

t2∫t1

(∂L

∂qαδqα − d

dt(∂L

∂uα)δqα) = 0 (1.53)

Noticed this fact S be extremum for classical path we will use it get SYMMETRY CONDI-

TION

H(q, p) −→ L(q, u) (1.54)

A change of coordinate

qα −→ qα = fα (q, u) (1.55)

uα −→ uα =dfα (q, u)

dt(1.56)

makes sense not on phase space coordinates but on functions qα(t)

Paths:

Reverse is also true.

If S always increases under a local deformation of a path away from the boundary, then the

path describes a solution of the classical equation of motion.

0 = δS =

t2∫t1

(∂L

∂qαδqα +

∂L

∂uαδqα) (1.57)

=

t2∫t1

δqα(t)(∂L

∂qα− d

dt(∂L

∂qα)dt (1.58)

δqαarbitrary. Take δqα = δαβδ(t− t′)(say)at t′ we have

7


=⇒ ∂L

∂qβ− d

dt(∂L

∂qβ) = 0 (1.59)

−→valid for allβand all t′

=⇒A Classical solution

q, u −→ q, u (1.60)

=⇒ qα = fα(q, u), uα =dqα

dt(1.61)

L(q, u) ≡ L(q, u, ˙u, ü.....) (1.62)

=

t2∫t1

L(q, u,˙u,

ü.....) (1.63)

=

t2∫t1

L(q, u) = S(q(t)) (1.64)

If S is minimized for the path q(t), then S is minimized for the path q(t)

=⇒Equation of motion of q is obtained by minimizing S.

We want to ask under what condition is this equation has the same form as the equation of

motion for q(t).

Ans. : If L(q, u,˙u,

ü.....) = L(q, u) + d

dtK (q, u,˙u,

ü.....) for some K

Proof : In this case

S =

t2∫t1

L(q, u) +K( q2, u2, ...)−K (q1,u1, ...) (1.65)

Consider a variation δq(t) of the path away from the boundaries.

8


δS =

t2∫t1

dt[∂L

∂qαδqα +

∂L

∂uα∂t(δq

α)] (1.66)

t2∫t1

dtδqα[∂L

∂qα− d

dt

∂L

∂uα] (1.67)

Physical significance: Suppose q(t) = q0(t) is a solution of the equations of motion.

Then qα(t) = fα(q0(t), u0(t) =dq0(t)dt ) is also a solution of the equations of motion.

=⇒A Symmetry in the active senses.

Conservation Law

If under an infinitesimal transformation

qα −→ qα = qα + ϵϕα(q, u) (1.68)

L(q, u) = L(q, u) + ϵd

dtN(q, u) (1.69)

Then

Q =∂L(q, u)

∂uαϕα(q, u) + N (1.70)

is conserved

Proof :

L(q, u) ≡ L(q, u) (1.71)

L(q, u) = L(q, u) + ϵ∂L

∂qαϕα + ϵ

∂L

∂uαdϕα

dt+ ϵ

dN

dt(1.72)

=⇒ dQ

dt=

d

dt(∂L

∂uαϕα) +

dN

dt(1.73)

=⇒ ∂L

∂qαϕα +

∂L

∂uαdϕα

dt+dN

dt= 0 (1.74)

=⇒Q is conserved along the motion

Example

Time translation Symmetry

Given a solution qα(t), qα(t+ a) is also a solution.

For small a = ϵ, qα(t+ ϵ) = qα(t) + ϵuα(t) where uα(t) ≡ ϕα(q, u)

uα(t) −→ uα(t) + ϵuα(t) (1.75)

L(q, u) ≡ L(q, u) = L(qα + ϵuα, uα + ϵuα) (1.76)

9


= L(q, u)− ϵ ∂L∂qα

uα − ϵ ∂L∂uα

duα

dt(1.77)

= L(q, u)− ϵdLdt

(1.78)

=⇒ N = −L (1.79)

Q =∂L

∂uαϕα +N (1.80)

=∂L

∂uαuα − L (1.81)

∂L

∂uα= pα =⇒ Q = pαu

α − L = H (1.82)

We are checking the mathematical consistency of two formalism.

Compare with the transformations generated by the conserved charge H in the Hamiltonian

formalism :

δxi = ϵxi,H

(1.83)

δqα = ϵ qα,H = ϵ∂H

∂pα= ϵuα (1.84)

δpα = ϵ pα,H = −ϵ ∂H∂qα

(1.85)

=⇒ δuαH =

(∂uα

∂pβ

)p,q

δpβ +

(∂uα

∂qβ

)p,q

δqβ (1.86)

Compare with

δuαL = ϵduα

dt(1.87)

uα =∂H

∂pα(1.88)

=⇒ duα

dt=

∂2H

∂pα∂pβdqβ

dt+

∂2H

∂pα∂pβdpβ

dt(1.89)

=∂2H

∂pα∂qβ− ∂2H

∂pα∂pβ∂H

∂qβ(1.90)

-

=⇒ ϵduα

dt=

(∂uα∂pβ

)δpβ +

(∂uα∂qβ

)δqβ (1.91)

10


=⇒The transformation agree after using equations of motion

L0(q, p): Some given Lagrangian.

We want to study system with constrains:ϕi(q) = 0 i = 1, 2, ...,m

L(q, p) = L0(q, p)−1

2η

m∑i=1

(ϕi(q))|2 (1.92)

For η −→ 0 this will constrain the system to move along ϕi(q) = 0

We change coordinates such that ϕ1(q), ....ϕm(q) are the first m coordinates and the rest os

(n-m) coordinates are some functions of q′s.

How do the equation change under a coordinate transformation?

qα = fα(q) (1.93)

uα =∂fα(q)uβ

∂qβ? (1.94)

duα

dt= uα automatically (1.95)

Define L(q, u) = L(q, u)

∂L

∂uβ=

∂L

∂qα∂qα

∂uβ+

∂L

∂uα∂uα

∂uβ(1.96)

=

=∂L

∂uα∂fα(q)

∂qβ(1.97)

d

dt(∂L

∂uβ) =

d

dt(∂L

∂uα∂fα

∂qβ) (1.98)

=d

dt(∂L

∂uα)∂fα

∂qβ+

∂L

∂uα∂2fα

∂qβ∂qrur (1.99)

∂L

∂uβ=

∂L

∂qα∂qα

∂qβ+

∂L

∂uα∂uα

∂qβ(1.100)

=∂L

∂qα∂fα

∂qβ+

∂L

∂uα∂2fα

∂qβ∂qrur (1.101)

=⇒ d

dt(∂L

∂uβ)− ∂L

∂qβ= 0 (1.102)

=⇒ (d

dt(∂L

∂uα)− ∂L

∂qα)∂fα

∂qβ= 0 (1.103)

=⇒ d

dt(∂L

∂uα)− ∂L

∂qα= 0 (1.104)

11


L(q, u) is the new Lagrangian.

Now go back to the problem at hand

f i(q) = ϕi(q) = 0 i = 1, 2, ...,m (1.105)

f i(qarbitrary for i = 1, 2, ...,m

L0(q, u) ≡ L0(q, p)

L(q, u) = L0(q, u)−1

2η

m∑i=1

q2i (1.106)

=⇒ d

dt(∂L(q, u)

∂uα)− ∂L(q, u)

∂qα= 0 (1.107)

=⇒ d

dt(∂L0(q, u)

∂ui)− ∂L0(q, u)

∂qi− 1

ηq = 0 i = 1, ...,m (1.108)

=⇒ d

dt(∂L0(q, u)

∂uα)− ∂L0(q, u)

∂qα= 0 for α = m+ 1, ..., n (1.109)

η −→ 0 qi = 0 =⇒ ui = 0 (1.110)

Define :

Lnew (q, u) = L0(q, u)|qi=0,ui=0 i = 1, ....,m (1.111)

(q1, ..., qn−m) = (qm+1, ..., qn)|qi=0,ui=0 (1.112)

(u1, ..., un−m) = (um+1, ..., un) (1.113)

=⇒ d

dt

∂Lnew (q, u)

∂uk− ∂Lnew

∂qk= 0 (1.114)

1.3.1 Force due to constraint :

−1

ηqi = +(

d

dt

∂L0(q, u)

∂ui− ∂L(q, u)

∂qi) (1.115)

Prescription: Simply eliminate m,qvariables using the constraints & express the Lagrangian

in terms of remaining q, u .

=⇒The New Lagrangian .

Once we have the New Lagrangian we can use Action Principle,Hamiltonian Formalism etc.

What about ϕi(q) = 0 i = 1, 2, ...,m

See M.R.Flannory, Am.J.Physics, Vol.73,p 265.

If the constrain equation contains terms higher than 1storder,then there is a sign ambigu-

ity.One can choose any sign and solve .But one has to check whether the solution approaches

the boundary as it evolves in time.

12


i.e.,suppose constrain equation is x2 + y2 + z2 = 1 ⇒ z = ±√

1− x2 − y2 one can put

z = +√

1− x2 − y2 (upper Hemisphere) and checks whether the solution approaches z = 0 (at

the boundary).It approaches the boundary in finite time then one has to deal with both the

cases.

i.e., locally the sign ambiguity has no meaning.It takes meaning only in global sense.

Symmetry of the equations of motion underqα −→ qα = fα(q, u) requires

L(q, u) = L(q, u) +d

dtk(q, u) (1.116)

for some k

Infinitesimal transformation

qα −→ qα = qα + ϵϕα(q, u) (1.117)

L(q, u) = L(q, u) + ϵd

dtN (q, u) + 0(t2) (1.118)

=⇒ ∂L

∂qαϕα +

∂L

∂uαdϕα

dt+dN

dt= 0 (1.119)

Associated conserved quantity

Q =∂L

∂uαϕα(q, u) +N (1.120)

dQdt = 0 as a result of 1.Symmetry 2.Equation of motion

Caution : While checking a given transformation is a symmetry,never use equations of

motion

Otherwise every transformation will appear to be a symmetry

∂L

∂qαϕα +

∂L

∂uαdϕα

dt=

d

dt

∂L

∂uαϕα +

∂L

∂uαdϕα

dt=

d

dt(∂L

∂uαϕα) (1.121)

Corresponding conserved quantity ∂L∂uαϕα −N = 0

Reason for this “symmetry”

δS = 0to first order in δqαfor any δq for the classical path

S (q) = S(q + ϵϕ(q, u)

)(1.122)

for any ϕfor the classical path.By using equation of motion we are restricting our checking

to classical paths only.

Example : Time translational Symmetry

qα −→ qα = qα(t+ ϵ) = qα(t) + ϵuα(t) (1.123)

uα −→ uα = uα(t) + ϵduα

dt(1.124)

13


Conserved charge Q = H

Compare with the transformation laws in the Hamiltonian framework.

Conserved quantity ψ↔transformation lawδxi = ϵxi, ψ

ψ = H

δqα = qα,H = ϵ∂H

∂pα= ϵuα (1.125)

δpα = pα,H = −ϵ(∂H

∂qα)|q,p (1.126)

uα = (∂H

∂pα)|p,q (1.127)

(δuα)H = −ϵ ∂2H

∂pα∂pβ

∂H

∂qβ+ ϵ

∂2H

∂pα∂qβdqβ

dt+ ϵ

∂2H

∂pα∂pβ

dpβdt

(1.128)

= −ϵ ∂2H

∂pα∂pβ

∂H

∂qβ+ ϵ

∂2H

∂pα∂qβ∂H

∂pβ(1.129)

Thus the two transformations agree after using equations of motion:

−→Do constrained dynamics

1. L −→ L− 1ϵ

∑(ϕi(q)

2

2. Variational Principle

3.ϕi(q, u) =non inertial normalization

Lagrangian and action formalism :

More efficient to discover symmetries & hence conservation laws.

After we discover them,how do we use them to analyze the system?

−→Go back to the Hamiltonian formalism.

2. Hamiltonian System (II)

2.1 Conserved charges: Integrability

I0 ∈ H, I1, I2, ....Ip (2.1)

2.1.1 How many do we need to get a solvable theory of Canonical Transformations.

Canonical Transformations:

xi −→ yi

such that

∂yi

∂xk∂yj

∂xlΩkl = Ωij (2.2)

14


In other words, If we define

Mik =∂yi

∂xk(2.3)

then

MΩMT = Ω (2.4)

=⇒ Ω =M−1Ω(MT )−1 (2.5)

=⇒ ω =MTωM ⇒ ∂yi

∂xkωij

∂yj

∂xl= ωkl (2.6)

Infinitesimal Canonical transformation:

xi −→ yi = xi + ϵxi, ψ(x)

(2.7)

where ψ(x) is the generator of the Canonical Transformation.

Now we shall look for generators of finite canonical transformation instead of the hit and

miss approach.

So far we looked for a transformation such that functional form of Hamiltonian remains

the same and check whether it is Canonical transformation or not.——-This is hit and miss

approach.

Phase Space Coordinates: x1

...

...

xn

where n = 2N

where,

x1 = q1, x2 = p1, x3 = q2, x4 = p2, · · · · · · · · ·

and

Ω =

0 1

−1 0

0 1

−1 0. . .

, (2.8)

ω = Ω−1 =

0 −11 0

0 −1

1 0. . .

(2.9)

Consider a covariant vector Ai

15


A1

A2

...

...

A2N

=

p10

p20

...

...

(2.10)

∂iAj−∂jAi = +ωij (2.11)

Canonical Transformation xi −→ xi Ωij −→ Ωij = Ωij and ωij −→ ωij = ωij

x1 = q1, x2 = p1, ......... (2.12)

Define a covariant vector Bi in the new coordinate system:∂iBj−∂jBi = ωij where ∂i =∂∂xi

B1

...

BN

=

p10

p2...

pN

(2.13)

Bi: Components of the vector B in the xcoordinate system.

Bi =∂xk

∂xiBk (2.14)

∂iBj−∂jBi =∂

∂xi(∂xk

∂xjBk)−

∂

∂xj(∂xk

∂xiBk) (2.15)

=∂xk

∂xj∂Bk

∂xi− ∂xk

∂xi∂Bk

∂xj(2.16)

=∂xk

∂xj∂xl

∂xi∂lBk −

∂xk

∂xi∂xl

∂xj∂lBk (2.17)

=∂xk

∂xi∂xl

∂xj(∂kBl − ∂lBk) (2.18)

= ωij (2.19)

Ci = Ai −Bi (2.20)

∂iCj − ∂jCi = 0 (2.21)

16


=⇒ Ci =∂F1

∂xi(2.22)

In any arbitrary coordinate system xi

Ci =∂xj

∂xi∂F1

∂xj=∂F1

∂xi(2.23)

δF = Ciδxi = Aiδx

i − Biδxi (2.24)

= Ak∂xk

∂xiδxi − Bk

∂xk

∂xiδxi (2.25)

Akδxk − Bkδx

k (2.26)

δF1 = pαδqα − pαδqα (2.27)

F1: Function of 2n independent variables .

Take them to be qα and qα

δF1 =∂F1

∂qαδqα +

∂F1

∂qαδqα (2.28)

=⇒ pα =∂F1(q, q)

∂qα, pα = −∂F1(q, q)

∂qα(2.29)

The 1st equation gives qas functions of q, pand 2nd equation gives pas functions of q, p

Given F1(q, q) this gives pαand qαas functions of q, p

Generates a Canonical Transformation:

This freedom can be carried not for any Ai & Bi satisfying

∂iAj−∂jAi=ωij ∂iBj − ∂jBi = ωij (2.30)

Other examples

A1.........

A2N

= −

0

q1

0

q2

...

B1.........

B2N

=

p10

p20...

(2.31)


Generating functions of Canonical Transformation

xi −→ xi

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(qα, pα) −→ (qα, pα)

If (qα, qα) can be taken to be independent phase space coordinates, than there is a function

F1(q, q) such that,pα = ∂F1(q,q)∂qα pα = −∂F1(q,q)

∂qα

The 1st equation gives qαas functions of q, pand 2nd equation gives pαas functions of q, p

Example :

F1 = qαqα (2.33)

pα = qα, pα = −qα (2.34)

In this case (qα, qα) can certainly be taken to be independent phase space coordinate.

Starting point of the Analysis :

A =

p10

p20...

B =

p10

p20...

(2.35)


We can try to carry out the analysis with other choices of A,B satisfying these requirements.

Example :

A = −

0

q10

q2...

B =

p10

p20...

(2.37)

∂iAj−∂jAi = ωij ∂iBj − ∂jBi = ωij (2.38)

Follow exactly the same steps .

Ci = Ai −Bi satisfies (2.39)

∂iCj−∂jCi = 0 =⇒ Ci = −∂F2

∂xi(2.40)

Ci = Ai − Bi =∂F2

∂xi(2.41)

in any coordinate system.

−δF2 = Ciδxi = Aiδx

i − Biδxi (2.42)

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= Aiδxi − Biδx

i = −qαδpα − pαδqα (2.43)

Take F2 as a function ofpα and qα

δF2 =∂F2

∂pαδpα +

∂F2

∂qαδqα (2.44)

=⇒ qα =∂F2(p, q)

∂pα, pα = +

∂F2(p, q)

∂qα

1st equation gives qαas functions of q, pand 2nd equation gives gives pαas functions of q, p

F2 = +pαqα (2.45)

=⇒ qα = qα, pα = pα (2.46)

F2 = +pαqα − ϵψ(p, q) (2.47)

qα = qα − ϵ ∂ψ∂pα

, pα = pα − ϵ∂ψ

∂pα(2.48)

qα = qα + ϵ∂ψ(p, q)

∂pα≃ qα + ϵ

∂ψ(p, q)

∂pα= qα + ϵ qα, ψ (2.49)

pα = pα − ϵ∂ψ(p, q)

∂pα≃ pα − ϵ

∂ψ(p, q)

∂pα= pα + ϵ pα, ψ (2.50)

−→Standard form of infinitesimal transformations.

Other choices:

A =

p10

p20...

B =

0

−q10

−q2...

(2.51)

qα =∂F3(p, q)

∂pα−→ gives qα as a function of (p, q) (2.52)

pα =∂F3(p, q)

∂qα−→ gives pα as a function of (p, q) (2.53)

A =

0

−q1

0

−q2...

B =

0

−q10

−q2...

(2.54)

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qα = −∂F4(p, p)

∂pα, qα =

∂F4(p,p)

∂pα(2.55)

Mixed choices

A =

p10

0

−q2...

(2.56)

In each of the N slots,we can choose

(pα0

)or

(0

−qα

)=⇒2N choices for A

=⇒2N choices for B

Total number of choices 2N × 2N = 22N

The generic Canonical transformations can be represented by any of the 22Ngenerating

functions.

Special clan of Canonical transformations may not admit 1 or more of these

22N representation.

Depending on the specific problem, we may choose to use one or the other kind of generating

functions.1

2.2 Hamilton Jacobi method for solving equations of motion

H(p, q) Original Hamiltonian.

Look for Canonical transformations

pα =∂F3(p, q)

∂qα, qα =

∂F3(p, q)

∂pα(2.57)

Such that H (p, q) ≡ H(p, q) is independent of q

=⇒ H(∂F3(p, q)

∂qα, qα) (2.58)

is independent of qα

qα dependence of H(∂F3(p,q)∂qα , q) comes through qα dependence.

=⇒We need H(∂F3(p,q)∂qα , q) = H(p) independent of q

=⇒a set of partial differential equations of F3.

=⇒Solve to get F3

Suppose we can find such an F3.

Equation of motion in (p, q) coordinate system:

1Non holonomic constrains MR Flannery,Am J. Phys. Vol. 73 Page 265

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dpαdt

= − ∂H∂qα

= 0 =⇒ pα = p(0)α = constant. (2.59)

dqα

dt=∂H(p)

∂qα= Constant (2.60)

=⇒ qα =∂H(p)

∂pα|p(0)t+ qα(0) (2.61)

Once pα &qαare known, pαand qαare obtained by solving the set of equations

pα =∂F3(p, q)

∂qα, qα =

∂F3(p, q)

∂pα(2.62)

This system has N independent conservation laws:

pα (1 ≤ α ≤ N) (2.63)

H : Function of pα −→not an independent conserved quantity.

2.3 Action angle variable

pα, qα qα ≡ qα + 2π (2.64)

H(p) (2.65)

dpαdt

= 0dqα

dt=∂H

∂pα≡ ωα(p) (2.66)

pα = p(0)α (2.67)

qα = ωα(p(0))t+ qα(0) (2.68)

Periodic Orbit :

ω(p(0)) = 0 for all α except α = β (2.69)

qα = qα(0) for α = β (2.70)

qβ = ωβt+ qβ(0) (2.71)

Under t → t+ 2πωβ ,q

β ≡ qβ + 2π

→Same point

⇒The Motion is Periodic.

For generic p, all ωαare non zero

For Periodic motion we need

21


ωα(p)T = 2πnα for some T (2.72)

⇒ ωα

ωβ=nβ

nβ= Rational Number (2.73)

In Generic Case ωα

ωβ is not rational

⇒The Moton is not Periodic

Example : 2-D Harmonic Oscillator

Summary

1. A Hamiltonian System with N degrees of freedom is solvable if it has N conserved

quantities.

2. Conserved quantities ←→Infinitesimal Canonical Transformations that leave the Hamil-

tonian invarient

If we find such aCanonica Transformation, we can construct the corresponding conserved

quantities & vice versa.

Sometimes Symmetries are easily visible.

Sometimes they are not →hidden symmetries.

Example:

Motion with 2 centers of attraction :

H =p2x2m

+p2y2m− k( 1

r1+

1

r2) (2.74)

→Solvable by Hamiltonian - Jacobi method.

⇒Two Conserved quantities:

p1 = C1, p2 = C2 (2.75)

C2,H = 0 (2.76)

⇒ δpα = ϵ pα, C2(p, q) (2.77)

δqα = ϵ qα, C2(p, q) (2.78)

⇒ δH = 0 (2.79)

=⇒ A new symmetry of H

Not visible →Hidden Symmetry.

Method for finding conserved quantities :

1. Use visible symmetries of H/L

2.Hamilton - Jacobi Method

3.Construction of Lax pairs.

22


Method

dpαdt

= − ∂H∂qα

,dqα

dt=∂H

∂pα1 ≤ α ≤ N (2.80)

Suppose we can construct NxN matrices

Aαβ(p, q) and Bαβ(p, q) such that

Aαβ = Aβα, Bαβ = −Bβα (2.81)

with the property

dA

dt= BA−AB (2.82)

as a consequence of equations of motion.

Then we can construct N conserved quantities.

A,B →Lax pairs.

d

dt(An) =

dA

dtAn−1 +A

dA

dtAn−2 + .......... (2.83)

=n−1∑p=0

ApdA

dtAn−p−1 (2.84)

=

n−1∑p=0

Ap(BA−AB)An−p−1 (2.85)

=

n−1∑p=0

(ApBAn−p −Ap+1BAn−p−1) (2.86)

=

n−1∑p=0

ApBAn−p −n−1∑p′=1

Ap′BAn−p′ (2.87)

= (BAn −AnB) (2.88)

d

dtTr(An) = Tr(BAn)− Tr(AnB) = 0 (2.89)

=⇒ Tr(An) is conserved for all n

How many of them are independent?

A: NxN symmetric matrix

Eigenvalues : λ1......λN

A = S

λ1. . .

λN

S−1 (2.90)

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An = SAαS−1.SAαA

−1 · · · · · · · · · n times (2.91)

= S(Aα)nS−1 = S

(λ1)n

. . .

(λN )n

S−1 (2.92)

=⇒ Tr(An) =

N∑α=1

(λα)n (2.93)

where the term in the bracket has N independent variables.

=⇒There are N independent functions

Tr(An) (2.94)

Take them for 1 ≤ n ≤ N=⇒N conserved charge

or we can take the conserved quantities to be λα 1 ≤ α ≤ NExample 3 body Toda lattice

H =m∑i=1

p2i2

+ V (q) (Take m = 1) (2.95)

V (q) = e−(q1−q2) + e−(q2−q3) + e−(q3−q1) (2.96)

dp1dt

= −e−(q3−q1) + e−(q1−q2) (2.97)

dp2dt

= −e−(q2−q3) − e−(q1−q2) (2.98)

dp3dt

= e−(q3−q1) − e−(q2−q3) (2.99)

dqα

dt= pα (2.100)

a1 ≡1

2e−

12(q1−q2), a2 ≡

1

2e−

12(q2−q3), a3 ≡

1

2e−

12(q3−q1), bα =

pα2

(2.101)

A ≡

b1 a1 a3a1 b2 a2a3 a2 b3

, B =

0 a1 −a3−a1 0 a2a3 −a2 0

(2.102)

Example : Check that

dA

dt= (BA−AB) (2.103)

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=⇒Lax pair =⇒3 Conserved quantities

Tr(A) = b1 + b2 + b3 =1

2

∑pα (2.104)

→Generates

δqα = ϵ, δpα = 0 (2.105)

A2 =

b21 + a21 + a23 b1a1 + a1b2 + a2a3 b1a3 + a1a2 + a3b2a1b1 + b2a1 + a2a3 a21 + b22 + a22 a1a3 + b2a2 + a2b3a3b1 + a2a1 + b3a3 a3a1 + a2b2 + b3a2 a23 + a22 + b23

(2.106)

=⇒ Tr(A2) = b21 + b22 + b23 + 2(a21 + a22 + a23) (2.107)

=1

2H (2.108)

Tr(A3) = (b21 + a21 + a23)2 + 2(a1b1 + b2a1 + a2a3)

2 + 2(a3b1 + a2a1 + b3a3)2 (2.109)

+(a21 + b22 + a22)2 + (a23 + a22 + b23)

2 + 2(a1a3 + b2a2 + a2b3)2 (2.110)

−→Generates a hidden symmetry

The System is solvable by virtue of the three conserved charges.

Introduce Routhian

Some comments on symmetries and conservation Laws :

1.The Symmetry generated by conserved quantity take simple form in the action angle

variable.

Conserved quantities Qα = pαϵβQβ generates

δpα = pα, ϵβQβ = pα,ϵβ pβ = 0 (2.111)

δqα = qα, ϵβQβ = qα,ϵβ pβ = ϵα (2.112)

=⇒The conserved quantities generate translation symmetry of qα

2.If Q1 and Q2are conserved and Q1, Q2 = 0 then the infinitesimal symmetries generated

by Q1 and Q2 commutes .

Define :

F (x) = F (x), δF = F (x)− F (x) (2.113)

δϵ1F = F, ϵ1Q1 (2.114)

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δϵ2G =G, ϵ2Q2

(2.115)

First apply δϵ1 : F −→ F + ϵ1 F,Q1 = G

Then apply δϵ2 : G −→ G+ ϵ2 G,Q2 = F + ϵ2 F,Q2+ ϵ1(F,Q1+ ϵ2 F,Q1 , Q2

)⇒ (1 + δϵ2) (1 + δϵ1)F = F + ϵ2 F,Q2+ ϵ1 F,Q1+ ϵ1ϵ2 F,Q1 , Q2 (2.116)

Reverse Transformation

⇒ (1 + δϵ1) (1 + δϵ2)F = F + ϵ1 F,Q1+ ϵ2 F,Q2+ ϵ1ϵ2 F,Q2 , Q1 (2.117)

⇒ δϵ2δϵ1F − δϵ1δϵ2F = ϵ1ϵ2 (F,Q1 , Q2 − F,Q2 , Q1) (2.118)

Jacobi Identity

Example :

A,B , C+ B,C , A+ C,A , B = 0 (2.119)

F,Q1 , Q2 − F,Q2 , Q1 = −Q1, F , Q2 − F,Q2 , Q1 = Q2, Q1 , F (2.120)

Thus if Q2, Q1 = 0 then δϵ2δϵ1F − δϵ1δϵ2F = 0

=⇒The two symmetries commute.

[Lx, Ly] = 0can be traced to non-commtate of Rotation about X and Y axis.

Rx (θ) =

1 0 0

0 cos θ sin θ

0 − sin θ cos θ

Ry (ϕ) =

cosϕ 0 sinϕ

0 1 0

sinϕ 0 cosϕ

(2.121)

Rx (θ)Ry (θ) = Ry (ϕ)Rx (θ) (2.122)

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