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MECHANICS
UNIT I Mechanical Systems : The Mechanical System - Gen-
eralized Coordinates - Constraints - Virtual work - Energy and Mo-
mentum. Chapter 1 : Sections 1.1 to 1.5.
UNIT II Lagrange’s Equations : Derivation of Lagrange’s equa-
tions - Examples - Integrals of motion. Chapter 2 : Sections 2.1 to 2.3
(omit sec 2.4)
UNIT III Hamilton’s Equations : Hamilton’s Principle - Hamil-
ton’s Equation - Other variational Principle. Chapter 4: Sections 4.1
to 4.3 omit Section 4.4.
UNIT IV Hamilton - Jacobi Theory : Hamilton Principle func-
tion - Hamilton - Jacobi Equation - Separability. Chapter 5 : Sections
5.1 to 5.3
UNIT V Canonical Transformation : Differential forms and gen-
erating functions - Special Transformations - Lagrange and Poisson
brackets. Chapter 6 : Sections 6.1, 6.2 and 6.3 (omit sections 6.4, 6.5
and 6.6)
Contents and Treatment as in : D. Greenwood. classical Dynam-
ics. Prentices, Prentice Hall of India, New Delhi, 1985.
Reference :
1. H. Goldstein, Classical Mechanics, (2nd Editin) narosa Publish-
ing House, New Delhi.
2. N.C. Rane and P.S.C. Joag, Classical Mechanics, tata McGraw
Hill, 1991.
3. J.L. Synge and B.A. Griffth, Principles of Mechanics (3rd Edi-
tion) MCGraw Hill Book Co New yuk, 1970.
1
UNIT I
Introduction :
Mechanics forms the basis of physics it contributes to our knowl-
edge of people working of nature and universe classical mechanics deals
with the description of actual or possible motion of points like as well
as extended bodies.
The motion of celestial bodies as well as man made objects such as
space problems. Satellites etc., are fields. The various classical dynam-
ics, Hamilton mechanics, provides most modern research in frontial ar-
eas. Particularly relation between symmetric property or conservation
law.
Application :
Mechanics is the study of motion of physical bodies. The motion
of celestial bodies planets, stars etc., path of an artillery shell or of a
space satellites send from a earth to a planet are among its problems.
Classical mechanics denotes part of Mechanics. Classical dynamics
will also be interpreted to include the type of mechanical assising to
include the type of Mechanical assising out of the special theory of
relativity. Mechanics of a system of particles :
When a Mechanical system consists of two (or) more particles we
must distinguish between the external forces acted upon the particles
of the system forces not belongs to the system. The internal forces
assising on account of interaction between themselves.
Mechanics of a particles :
Let r be a radius of the particle from some given origin and v its
the vector velocity then
v =dr
dt.
The linear momentum p of the particle is defined as the product
of particle mass and its velocity. In consequence of interaction with
external objects and field the particle may experience forces of various
types.
2
For example, gravitation or electro dynamics.
The Mechanic of particle is contained in Newton second law mo-
tion.
Equation of Motion :
The motion of particle is described by the differential equation
F ∝ a−→F = m−→a
= m
(
d−→vdt
)
=d
dt(m−→v ) =
d
dt(p)
−→F =
dp
dt
F =d
dt(mv)
= md
dt
(
dr
dt
)
= md2r
dt2
∴ F = md2r
dt2
Thus the equation of motion is a differential equation of second
order.
Equation of motion of System :
The equation of motion of a system of N particles is written as
mi
··−→r =−→Fi +
−→Ri (i = 1, 2, 3, · · · , N)
where miis the mass of the ith particle,−→Fi is the applied force
and−→Ri the constraint force.
Generalized coordinates :
To described the configuration of the system we select the smallest
possible number of variables which define configuraiton of the system.
These are called the generalized co-ordinates of the system.
3
Configuration :
A set of generalized co-ordinates is any set of co-oirdinates which
described the configuration.
Notation for generalized Co-ordinates :
Generalized co-ordinates are q1, q2, · · · , qn(qi, i = 1, 2, · · · , n) the
symbols q1, q2, · · · , qn corresponding to the co-ordinates that we choose
to describe the motion.
1. When the particle move in a plane it described by cartesian co-
ordinates (x, y) or polar co-ordinates and soon
q1 = x
q2 = yor
q1 = r
q2 = θ
2. Some spherical symmetry the sherical co-ordinates is
q1 = r =√
x2 + y2
q2 = θ = cot−1
(
z√
x2 + y2
)
q3 = φ = tan−1
(y
x
)
Definitions :
1. Newton fundamental equation
2. System of particle
3. Degree of freedom
4. Generalized Co-ordinates
5. Configuration space
6. Holonomic constraints
7. Non polynomic constraints
4
8. Scleronomic constraints
9. Rheonomic constraints
10. Clarnomic system
11. Realnomic system
12. Polyateral constraints
13. Unilateral constraints
Degree of freedom :
The number of degree of freedom is equal to the number of co-
ordinates minus the number of independent equation of constraints.
(or)
If the configuration of a system of a N particles described using de-
gree N cartesian relation these co-ordinates then the degrees of fredom
equals 3N − ℓ.
Example :
Let 3 particles connected by the rigid rods to form a triangular
body with the particles are at the corners. There particles are fixed
by a cartesian co-ordinates each rigid rod is represonted mathemati-
cally by indepenent constraints equation. All together there are three
constraints. Hence degree of freedom equals 9 − 3 = 6.
Generalized co-ordinates :
A set of parameters used to represent the configuration of a system
without ambiguity are called Generalized co-ordinates.
Configuration Space :
Let n be the degree of freedom of a system of N - particles. Let
its configurations be specified by n - generalized co-ordinates.
The n - dimensional space representing these Co-ordinates as a
single point q (q1, q2, · · · , qn) is called a configuation space.
Holonomic constraints :
Let the configuration of a system be specified by n - generalized
Co-ordinates (q1, q2, · · · , qn).
5
Let us assume that ethere are k independent equation of constraints
of the form
φj (q1, q2, · · · , qn, t) = 0, j = 1, 2, 3, · · · , k.
A constraints which can be expressed in this fashion is known as a
Holonomic constraints.
Holonomic system :
A system is said to be Holonomic if its constraints are always Holo-
momic.
Example :
Consider the motion of 2 particles connected by a rigid rod of length
ℓ in xy - plane constraint equation is given by
(x2 − x1)2 + (y2 − y1)
2 = ℓ2.
Scleronomic constraints :
The constraints is said to be scleronomic if time t does not appear
explicity.
Example:
Consider the motion of two particle in the xy plane they are con-
nectd by a rigid rod of length ℓ. The corresponding constraint equation
is
ℓ2 = (x2 − x1)2 + (y2 − y1)
2.
where x1, x2, y1, y2 are cartesian co-ordinates of 2 particles.
The equation does not contain the time t explicity.
Scleronomic system :
A system is said to be scleronomic if its constraints are scleronomic.
Rhenomic system :
A system is said to be Rhenomic if its constraints are Rhenomic.
Rhenomic constraints :
If the length ℓ have been generalized as an explicity function of
time then the constraint is Rhenomic. OR Constraint relation depend
6
explicity on time.
Non - Holonomic constraints :
Let a system has m contraints which are non - integrable and dif-
ferential expression is of the form
n∑
i=1
ajidqi + ajtdt = 0, j = 1, 2, 3, · · · ,m
where a′s is a function of the q′s and t.
This type of constraints are known as non- Holom Holkonomic
constraints.
Bilateral constraints :
If any configuration of the system for every small allowable dis-
placement in any fixed time. The negative displacement is also al-
lowable then the constraints are called bilateral constraint. They are
expressed as equalities.
Unilateral constraint :
In any configuration of the system for every small allowable dis-
placement in any fixed time either a positive or negative displacement
is allowable then the constraints are called unilateral constraints. They
are experessed as in equlities.
f (q1, q2, · · · , qn, t) ≤ 0 (unilateralconstraints)
f (q1, q2, · · · , qn, t) = 0 (Bilateral)
Example :
Let a free from particle be contained in a fixed hollowsphere of
radius r if its centre at the origin of a cartesian co-ordinates system.
If (x, y, z) are the co-ordinates of the particle then x2 + y2 + z2 ≤ 0
where r is trhe radius.
Virtural displacement :
Let a system of N particles is given by 3N cartesian co-ordinates
x1, x2, · · · , x3N with respect to some intertial frame adn may be subject
7
to the contraints. At any given time let us assume that the co-ordinates
have small infinite displacement δx1, δx2, · · · , δx3N .
They are assumed to occure without passing of tiem and donot
necessary conform to the constraints. These small change Sk in the
configuration is known as Virtual displacement.
Virtual work :
Consider a system of N particles. Let the configuration to repre-
sented by the cartesian co-ordinates x1, x2, · · · , x3N .
Let f1, f2, · · · , f3N are the component forces applied at the corre-
sponding
Co-ordinates in a positive sense then the virtual workdone (δW )
of these forces for a virtual displacement δx is given by
δW =3N∑
j=1
−→Fj
−−→δxj
Note :
In a vector form δW =N∑
i=1
−→Fi· δ−→ri where
−→Fi is the applied force of
ith particle −→ri is the positive vector.
The forces are assumed to be constant through out the virtual
displacement.
Workless constraints :
A workless constraints is any bilateral constraint such that the
virtual work done by the corresponding constraint force is zero for any
virtual displacement, which is consistent with the constraint.
Examples :
1. Sliding motion on a frictionless susface.
2. Rolling contact without slipping.
Book work : The virtual displacement is not ingeneral a possible
real displacement.
Proof : In the virtual displacement, the moving constraints are
8
assumen to be slopped.
Let the system have k holonomic constraints.
φj (q1, q2, · · · , q3N , t) = 0, j = 1, 2, 3, · · · , k. (1)
Let us take the totla differential of φ and obtain
dφj =∂φj
∂q1dq1 +
∂φj
∂q2dq2 + · · · +
∂φj
∂q3N
dq3N +∂φj
∂tdt = 0.
Suppose the system is subject to k holonomic constraints
φj (x1, x2, · · · , x3N , t) = 0, j = 1, 2, 3, · · · , k.
dφj =∂φj
∂x1
dx1 +∂φj
∂x2
dx+ · · · +∂φj
∂x3N
dx3N +∂φj
∂tdt = 0.
Virtual displacement
dxi = δxi, dt = 0 (2)
dφj =3N∑
i=1
∂φj
∂xi
dxi +∂φj
∂tdt = 0, j = 1, 2, 3, · · · , k.
Case (i) :
Suppose te system has non-holonomic constraint. Then
3N∑
i=1
ajidxi + ajtdt = 0, j = 1, 2, 3, · · · , k.
Virtual displacement dt = 0, dxi = δxi
n∑
i=1
ajiδxi = 0 (j = 1, 2, 3, · · · ,m)
Form (1) and (2) shows that any holonomic constraint must also
9
be scleronomic constraint
i.e.,∂φj
∂t= 0, j = 1, 2, 3, · · · , k
Suppose non-holonomic constraint aj = 0 . These constraint are
not ment in the general case.
Hence a virtual displacement is not possible as a real displacement.
State and prove the principle of virtual work :
Statement:
The necessary and sufficient condition for the static equilibrium of an
initially mottionless sceleronomic which is subject of the system in to
the virtual workdone for an arbitrary virtual displacement satisfying
the constraint is zero.
Proof :
Case (i)
Let us assume that we are given an sclernomic system of N particles.
Let us assume that their system is in static equilibirium. Then for
each particle−→Fi +
−→Ri = 0
where−→Fi is the external force.
−→Ri is the constraint force acting on ith particle is
(−→Fi +
−→Ri
)
δ−→ri = δW, i = 1, 2, 3, · · · , n
where −→ri is the position vector of ith particle.
∴ Virtual workdone by all the force in moving through an arbitrary
vitrual displacement consistant with the constraint is zero.
N∑
i=1
(−→Fi +
−→Ri
)
δ−→ri = 0
N∑
i=1
−→Fiδ
−→ri +N∑
i=1
−→Riδ
−→ri = 0
10
constraints are workless, thenN∑
i=1
−→Riδ
−→ri = 0
N∑
i=1
−→Fiδ
−→ri = 0
i.e., δW = 0
⇒ workdone = 0, i.e., the virual workdone by the applied force is zero.
Case (ii) :
Let us assume that the same system of particles are initially mo-
tionless. But it is not in static equlibirium.
The motion is s to C we can always choose a virtual displacement
in the direction of actual motion at each particle.
In this case, the virtual work is positive
N∑
i=1
−→Fiδ
−→ri +N∑
i=1
−→Riδ
−→ri > 0
δW =N∑
i=1
−→Fiδ
−→ri > 0,−→Fis 6= 0
δW > 0
If the system is not in equlibrium then the virtual workdone is not
zero.
1. Show that virtual workdowne by the constraint forces
on a system of two particles connected by the rigid massless
rod is zero.
Solution :−→Ri,
−→Ri are the constraint force along the rod on the 2 particles.
They are opoosite in sense with equal magnitude.
−→er is the unit vector along the rod. δ−→r1 , δ−→r2 are the virtual dis-
placement at the ends virtual workdone by a constraint force is given
11
by
δWC =−→R1δ
−→r1 +−→R2δr2
−→R1 = R1 (−−→er ) = −R−→er
−→R2 = R2 (−→er ) = −R1
virtual displacement along the rod all equal
δ−→r1er = δ−→r2er
δWC = −R1−→er δ
−→r1 +R2−→er δ
−→r2= (R2 −R1)
−→er δr1
δWC = 0 (∵ R2 = R1 equal magnitude) .
Example for bilateral constraints :
2. Given an equation of scleronomic with workless constraints
(bilateral) to which the principle of the virtual work can be
applied (or)
Two frictionless blocks of equal mass m all connected by
the massless rigid rod using x1 and x2 as co-ordinates. Solve
for the force F2. If the system is in static equlibirium.
Solution :
R1 and R2 are the external constraint acting at the wall and floor
repectively.
The applied forces are the gravitational force acting in the blocks
and the external forces F2 system is in static equlibirium.
Virtual workdone by external force (F ) = 0
Workdone by constraint force (R) = 0
We have
mg(δx1) + F2(δx2) = 0. (1)
The virtual displacement along the rod at ends are equal.
∴ δx1 sin θ = δx2 cos θ (2)
12
Multiply (1) by sin θ,
⇒ mg sin θδx1 + F2 sin θδx2 = 0
mgδx2 cos θ + F2 sin θδx2 = 0
(mg cos θ + F2 sin θ) δx2 = 0.
But displacement δx2 6= 0.
A frictionless system which is constrained to move in the vertical
plane.
mg cos θ + F2 sin θ = 0
F2 sin θ = −mg cos θ
F2 = −mg cot θ
3. Using a suitable example, show how the concept of
virtual work can be applied to system with unilateral con-
straints.
Solution :
Let us assume a cube of mass is placed between two frictionless
mutually perpenticular plane.
Assume that the motion is an vertical plane. Le x1 = x2 = 0 at the
equilibirium position then unilateral constraints are x1 ≥ 0, x2 ≥ 0.
The only applied force is the weight mg. It has components F1, F2 in
x1 and x2 direction.
m
450
450
g
x1
R 1 R2
x 2
When in equilibuim R1 = −F1 and R2 = −F2
δWC =mg√
2(δx1 + δx2) > 0
13
F1 = F2 = −mg√2
Virtual workdone by applied forces is
δW =−mg√
2(δx1 + δ2) ≤ 0
Thus the virtual δW ≤ 0 for any virtual displacement consistent
with unilateral constraints.
To find workdone by constraint forces
R1 = R2 =mg√
2
Virtual workdone by constraint forces
δWC = R1δx1 +R2δx2 =mg√
2(δx1 + δx2) ≥ 0.
Hence for a system with unilateral constraints
1. Workdone by external force is −ve.
2. Workdone by constraint force is + ve.
D’ Alembert’s Principle :
Let the system has N particles. Consider on ith particle to the
system mi is the mass of the particle.−→ri is the position vector of particle.−→Fi is applied force of particle.
Let−→Ri be constraint force of particle.
Let··−→ri be the acceleration of ith particle relative to an inertial
frame. −→mi
··−→ri is the intertial force acting on the ith particle.
Now we have the equation
−→Fi +
−→Ri −−→mi
··−→ri = 0, i = 1, 2, 3, · · · , N. (1)
(The sum of all the force real and inertial acting on each particle of a
system is zero)
This result is known as D’ Alembert’s principle.
14
Note :−→Fi and
−→Ri together are called as real forces.
Lagrange’s form of D’ Alembrt’s Principle :
Let us assume that the system has N particles. Let mi be the mass
of the ith particle frame.−→Fi is the applied forced and
−→Ri is the constraint force on the ith
particle.
−−→mi
··−→ri is inertial force acting on the particle. Now we have the
equation
−→Fi +
−→Ri −−→mi
··−→ri = 0 (i = 1, 2, 3, · · · , N) .
Now the virtual workdone by the system is
N∑
i=1
(−→Fi +
−→Ri −−→mi
··−→ri
)
δ−→r i = 0
−→Ri is the workless constraint forces
N∑
i=1
−→Riδ
−→ri = 0
N∑
i=1
(−→Fi −−→mi
··−→ri
)
δ−→ri = 0.
This equation is the Lagrange’s form of D’ Alemberts principle.
1. Obtain the differential equation of the given pendulum
(sphericla pendulum) or A particle of mass m is subspended
by a massless wire of length r = a + b cos ωt(a > b > 0) to form
a sphorical pendulum. Find the equation of motion.
Solution :
Let us assume that the spherical co-ordinates θ and φ.θ is mea-
sured from the upward vertical. φ is the angle between a verticla
plane through a supporting point and a vertical plane containing the
pendulum.
15
O
r
m
efeq er
g
q
f
The acceleration of a particle whose spherical co-ordinates are
(r, θ, φ) is as follows
··−→r =
(
··
r − r·
θ2
− r·
φ sin2 θ
)
−→er +
(
r··
θ + 2·
r·
θ − r·
φ2 sin θ cos θ
)
−→eθ
+
(
r··
φ sin θ + 2·
r·
φ sin θ + 2r·
θ·
φ cos θ
)
−→eφ
where −→eθ ,−→er ,
−→eφ are unit vectors forming an orthogonal triad.
A virtual displacement constraint with the constraint is,
δ−→r = rδθ−→eθ + r sin δφ−→eφ .
The applied gravitational force is,
−→F = mg cos
(
1800 − θ)−→er +mg sin
(
1800 − θ)−→eθ
−→F = −mg cos θ−→er +mg sin θ−→eθ
we have(−→F −m
··
r)
δ−→r = 0 (By D’ Alemberts principle)
16
(−mg cos θ−→er +mg sin θ−→eθ ) −m
((
··
r − r·
θ2
− r·
φ sin2 θ
)
−→er
)
+
(
r··
θ + 2·
r·
θ − r·
φ2 sin θ cos θ
)
−→eθ +
((
r··
φ sin θ + 2·
r·
φ sin θ + 2r·
θ·
φ cos θ
)
−→eφ
)
× rδθ−→eθ + r sin θδφ−→eφ = 0
Taking dot product,
mgr sin θ −mr
(
r··
θ + 2·
r·
θ − r·
φ2 sin θ cos θ
)
δθ
−m(
r··
φ sin θ + 2·
r·
φ sin θ + 2r·
θ·
φ cos θ
)
r sin θδφ = 0
δφ and δθ are independent Virtual displacements.
∴ co-efficeient are equal to zero respectively
mr
[
g sin θ −(
r··
θ + 2·
r·
θ − r·
φ sin θ cos θ
)]
= 0 (1)
Since mr sin θ 6= 0
− sin θmr
(
r··
φ sin θ + 2·
r·
φ sin θ + 2r·
θ·
φ cos θ
)
= 0 (2)
r = a+ b cosωt
r = −bω sinωt
(1) ⇒ (a+ b cosωt)··
φ sin θ−2bω sinωt·
θ−(a+ b cosωt)·
φ2
sin θ cos θ−g sin θ = 0
(2) ⇒ (a+ b cosωt)··
φ sin θ−2bω sinωt sin θ·
φ+2 (a+ b cosωt)·
θ·
φ cos θ = 0.
These are differential equation of the motion.
Generalized forces :
Let us assume that a given set of force F1, F2, · · · , , F3N is applied
to system of N particles.
17
The virtual workdone by these force is
δW =3N∑
j=1
Fjδxj (1)
Let us suppose that the 3N ordinary constraint co-ordinates are
related with the n - generalized co-ordinates q1, q2, · · · , qn.Let these are related by the relations
xj = xj(q1, q2, · · · , qn, t), j = 1, 2, 3, · · · , 3N.
Using displacement is virtual workdone we get,
dxj =∂xj
∂q1dq1 +
∂xj
∂q2dq2 + · · · + ∂xj
∂qndqn +
∂xj
∂tdt
δxj =∂xj
∂q1δq1 +
∂xj
∂q2δq2 + · · · + ∂xj
∂qnδqn
δxj =n∑
i=1
∂xj
∂qiδqi (j = 1, 2, · · · , 3N)
(1) ⇒ δW =3N∑
j=1
Fj
(
n∑
i=1
∂xj
∂qiδqi
)
=n∑
i=1
(
3N∑
j=1
Fj
∂xj
∂qi
)
δqi
3N∑
j=1
Fj
∂xj
∂qiis called the generalized force.
δW =n∑
i=1
Qiδqi.
Principle of work and kinetic evergy :
Statement:
The increase in K.E. of a particle as its moves from one arbitary point
to another point is equal to the workdone by the force acting on the
particle during the given intervals (or) chang in K.E. = workdone.
Proof:
Kinetic Energy is given by T =1
2mv2 v is the velocity of particle and
18
m is the mass of the particle.
Let W be the workdone by the force−→F in moving particle from A
to B is
W =
B∫
A
−→F · d−→r
=
B∫
A
m··−→r d−→r , ∵
−→F = m
··−→r
= m
B∫
A
··−→r d−→r
W =1
2m
B∫
A
d
dt
(
·−→r·−→r)
dt
=1
2m
B∫
A
d
(
·−→r)2
=m
2
B∫
A
d
(
·−→r)2
=m
2
B∫
A
d(
v2)
=m
2
(
v2)B
A
=m
2
(
v2B − v2
A
)
=1
2mv2
B − 1
2mv2
A
W = TB − TA
Book work : State and prove the principle of conservation of
energy.
Statement: Let the only force acting on the given particle be conser-
vative.
Proof : Let−→F be the conservative force acting on the particle at the
19
point (x, y, z).
m is the mass of the particle.
T is the K.E. of the particle.
W is the workdone of the particle, from A to B
W = VA − VB = TA − TB
Since the points A and B are arbitrary.
TA +VA = TB +TB = E of the energy is conservative. We conclude
that the total mechanical evergy E remains constant during the motion
of the particle.
This is the principle of conservation of evergy.
Book work :
An equilibrium configuration at a conservative holonomic system
with workless fixed constraints must occur at a position where the P.E.
has a stationary value.
Proof :
consider a system of N particles whose applied forces are conser-
vative and derived from the P.E. function V (x1, x2, · · · , x3N).
The virtual workdone by the forces
δW =3N∑
j=1
Fxjδxj
But Fxj =−∂V∂xj
Workdone δW =3N∑
j=1
(−∂V∂qi
)
δxj = −δV.
This is the first variation of the P.E.
By principle of virtual work the necessary condition for the static
equilibrium of the system is δW = 0, δV = 0.
∂V
∂xj
= 0, j = 1, 2, 3, · · · , 3N.
20
Let us assume the P.E is interms of generalized co-ordinates
V = V (q1, q2, · · · , qn)
δV =n∑
i=1
∂V
∂xi
(δqi) , q′s are independent
We conclude that an equilibrium configuration of a conservative
holonomic system with workless fixed constraints must occur at a po-
sition.
Where the P.E has a stationary value. Consider the stability of
this system at position of static equilibrium.
∴ δV = 0
⇒ ∂V
∂qi= 0, i = 1, 2, 3, · · · , n.
These condition imply that the P.E. is at a stationary value system.
Expanding the P.E. function V (q1, q2, · · · , qn) about the reference
value V0, we have
V = V0 +
(
∂V
∂q1
)
0
δq1 +
(
∂V
∂q2
)
0
δq2 + · · ·
+1
2
(
∂2V
∂q21
)
(δq1)2 + · · · +
(
∂2V
∂q1∂q2
)
(δq1δq2) + · · ·
∴ ∆V is the change in P.E. from its value of equation position.
∆V = V − V0
=1
2
(
∂2V
∂q21
)
(
δ2q1)
+
(
∂2V
∂q1∂q2
)
(δq1δq2) +1
2
(
∂2V
∂q22
)
(
δ2q2)
+ · · ·
where a zero subscript on a function implices that it is to be evalu-
ated at the reference value of the q’s. The δq’s represent infinitesimal
change from this reference configuration.
(i) If ∆V > 0 ⇒ The equilibrium is stable
(ii) If ∆V < 0 ⇒ The equilibrium is unstable
21
(iii) If ∆V = 0 ⇒ The equilibrium is nutral.
K..onig’s Theorem :
The total K.E. of a system is equal to the sum of
(i) The K.E. due to a particle having a mass equal to the total mass
of the system and moving with the velocity of the centre of mass
and
(ii) the K.E. due to the motion the system relative to its centre of
mass.
Find the K.E. of the system of particle.
Proof :
Consider a system of N particles.
Let −→ri be the position vector of the ith particle (whose mass is no)
relative to a point O fixed in an inertial frame.
jm
im
yO
irr
im
.c m
Crr
Crr
x
z
Let −→rC be the position vector of the mass centre om.
Let mi be the mass of the system.
Let T be the K.E. of the system.
T =N∑
i=1
1
2mi
·−→r2
·−→r =·−→r C +
·−→ρi
22
·−→r2
=·−→r
2
C + −→ρi2 + 2
·−→rC
·−→ρi
T =1
2
(
N∑
i=1
mi
)
·−→r2
C +1
2
N∑
i=1
mi
·−→ρ i
2
+N∑
i=1
mi
(
·−→r C
·−→ρ i
)
=1
2
·−→r2
C
N∑
i=1
mi +1
2
N∑
i=1
mi
·−→ρ i
2
+·−→r C
N∑
i=1
mi
·−→ρ i
=1
2
·−→r2
C
N∑
i=1
mi +1
2
N∑
i=1
mi
·−→ρ i
2[
∵
N∑
i=1
mi
·−→ρ i = 0
]
T =1
2m
·−→r2
C +1
2
N∑
i=1
mi
·−→ρ i
2
T = K.E of mass centre + K.E. relative to the centre of mass.
K.E. of a particle of mass m with vector of c.m.(centre of mass)
Book work :
Kinetic Energy of a system of a particle using an arbitrary reference
point.
Proof :
Consider a system of N - particle. Let −→ri be the position vector
of an ith particle (mass m) relative to a point O fixed in an inertial
frame.
Let P be an arbitrary point. −→rC is the position vector of mass
centre c.m.
km
y
O
irr
im .c m
P
Crr
irr
Crr
z
23
−→ρc is the position vector of mass centre relative to the point P .
Let T be the kinetic energy of the system.
T =N∑
i=1
1
2mi
·−→r2
·−→r =·−→r ρ +
·−→ρi
·−→r2
i =
(
·−→r ρ +·−→ρi
)2
=·−→r
2
ρ +
·
−→ρi2 + 2
·−→r ρ
·−→ρi
T =1
2
N∑
i=1
mi
·−→r2
ρ +1
2
N∑
i=1
mi
·−→ρ i
2
+N∑
i=1
mi
·−→r ρ
·−→ρ i
=1
2
N∑
i=1
mi
·−→ρ i
2
+1
2
·−→r2
ρm+·−→r ρ
N∑
i=1
mi
·−→ρ i
T =1
2m
·−→r2
ρ +1
2
N∑
i=1
mi
·−→ρ i
2
+·−→r ρm
·−→ρ C
[
ρC =1
m
∑
mi
·−→ρ C
]
Hence the total K.E. is the sum of this 3 parts.
1. The K.E. due to the particle having a mass and moving with
reference point P .
2. The K.E. of the system due to its motion relative to P .
3. The sclar product of the velocity of the reference point and the
linear momentum of the system relative to a reference point.
Note :
ρi = ρC + ρj
·
ρi =·
ρC +·
ρj∑
m·
ρi =∑
m·
ρC +∑
m·
ρj
But∑
mρj = 0,∑
m·
ρi =∑
m·
ρC
24
Kinetic Energy of rigid body :
Take CM as the origin of vectors. Let dV be a small volume element
of a rigid body with density.
Each element of the rigid body is having in general translational
and rotational velocity.
We can assume that the rigid body is having instantaneous axis of
rotaion. The typical volume element dV can be chosen so small.
ie, rotational K.E is negligibe compared with its translational K.E
Each element of a rigid body can be considered as a particle of
infinite small mass.
m = V × ρ
Hence the limiting case of K.E of system of N - particles.
Then the K.E of the system
T =1
2m
·−→r2
C +1
2
N∑
i=1
mi
·−→ρ2
i
=1
2m
·−→r2
C +1
2
∫
V
ρ·−→ρ
2
i dV
−→ρ is the position vector of volume element dV relative to mass
centre c.m.
m is the mass of the body. The first term is called translational
K.E. and second term is called rotational K.E.
Let assume that the body be rotating with angular velocity −→ω .
To find
·−→ρ = −→ω ×−→ρ·−→ρ
2
= (−→ω ×−→ρ )2
= −→ω · −→ρ ×·−→ρ
= (−→ω ×−→ρ ) ··−→ρ = −→ω ·
[
−→ρ(
−→ω ×·−→ρ)]
·−→ρ2
= −→ω · [(−→ρ · −→ρ )−→ω − (−→ρ · −→ω )−→ρ ]
25
= −→ω ·[−→ρ 2−→ω − (−→ρ · −→ω )−→ρ
]
·−→ρ2
= −→ω ·[−→ρ 2−→ω − (−→ρ · −→ω )−→ρ
]
Let ~ρ = x~i+ y~j + z~k ~ω = ωx~i+ ωy
~j + ωz~k
~ω ·[
~ρ2~ω − (~ρ · ~ω)~ρ]
=∑
[
(
x2 + y2 + z2)
ωx~i− (xωx + yωy + zωz)
·
~xi
]
ω
=∑
[(
y2 + z2)
ωx − xyωy − xzωz
]
=
[(
y2 + z2)
ωx − (xyωy + xzωz)]
i+ [ ]j + [ ]k
ω
T =1
2
∫
V
−→ρ.−→ρ 2dV
=1
2~ω
∫
V
−→ρ([(
y2 + z2)
ωx − (xyωy + xzωz)]
i
+ [ ]j + [ ]k
)
ω
dV
Define m.I and product of inertial as follows. Let
Ixx =
∫
V
(
y2 + z2)−→ρ dV
Iyy =
∫
V
(
z2 + x2)−→ρ dV
Izz =
∫
V
(
x2 + x2)−→ρ dV
Ixy = Iyx = −∫
V
(xy)−→ρ dV
Iyz = Izy = −∫
V
(yz)−→ρ dV
26
Izx = Ixz = −∫
V
(zx)−→ρ dV
Trot =1
2Ixxω
2x +
1
2Iyyω
2y +
1
2Izzω
2z + Ixyωxωy + Iyyωyωz + Izxωzωx
=3∑
i=1
3∑
j=1
1
2Iijωiωj
=1
2−→ω T I−→ω
=1
2
ωx
ωy
ωz
T
Ixx Ixy Ixz
Iyx Iyy Iyz
Izx Izy Izz
ωx
ωy
ωz
=1
2(−→ω )
T −→I (−→ω )
Kinetic Enery of rigid body = 1
2m
·−→rc + −→ω T I−→ωAngular Momentum of a system :
The angular momentum of a system of N particles and the total
mass ‘M ’ about a fixed point O is equal to the angular momentum
about θ of a single particle of mass m which is moving with centre of
mass plus the angular momentum of the system about the centre of
mass m.
Find the angular momentum of a system of particles about a fixed
point θ.
moment of momentum = Angular momentum.
Proof :
Let us consider a system of N - particles Let mi is the mass of the
ith particle whole position vectors given by −→ri .
The angular momentum of the ithparticle = r × ρ
= r ×mv
27
= m·−→r × r
AM = −→ri ×m·−→r
Angular momentum of the system H =N∑
i=1
−→ri ×m·−→r .
ButN∑
i=1
mi−→ρi = 0 ⇒
N∑
i=1
m·
i−→ρi = 0
Angular momentum of the system
=
(
×·−→rC
)
m+ 0 + 0 +N∑
i=1
−→ρi ×m·−→ρ
= −→rC ×m·−→rC +
N∑
i=1
−→ρi ×m·−→ρ
H =
(
−→rC ×m·−→rC
)
+HC
wher HC is the angular momentum of the system relative to a
centre of mass.
Note :
Tra =1
2−→ω∫
η−→ρ × (−→ω ×−→ρ ) dV
=1
2−→ω−→HC
Angular momenum of a rigid body :
Le dV be small volume element of a rigid body having density η.
Each element of the rigid body is having translation and relation.
We can assume that the rigid body is having instantaneous axis of
rotation.
The typical volume element dV can be chosen so small.
28
Its rotational K.E is negligible compared with its translational K.E
each element of rigid body can be considered as particle of infinite
decimal mass.
The angular momentum of a systme of particle−→H = −→rC ×m
·−→rC +N∑
i=1
−→ρi ×m·−→ρ C .
In the case of rigid body
−→H = −→rC ×m
·−→rC +
∫
V
η
(
−→ρ ×·−→ρ)
dV
But·−→ρ = −→ω ×−→ρ , where −→ω is the angular velocity of the rigid body.
∴−→ρ ×
·−→ρ = −→ρ × (−→ω ×−→ρ ) = −→ρ 2−→ω − (−→ρ · −→ω )−→ρ
Let us consider a cartesian system with its origin at the centre of mass.
−→ρ i = x−→i + y
−→j + z
−→k , −→ω = ωx
−→i + ωy
−→j + ωz
−→k
~ρ2~ω· (~ω)~ρ =∑
(
x2 + y2 + z2)
ωx − (xωx + yωy + zωz)−→xI
=∑
(
y2 + z2)
ωx− (xyωy + xzωz)−→i
=[
∑
(
y2 + z2)
ωx − (xyωy + xzωz)]
i
Ixx =
∫
V
(
y2 + z2)
ηdV
Iyy =
∫
V
(
z2 + x2)
ηdV
Izz =
∫
V
(
x2 + y2)
ηdV
Ixy = Iyx = −∫
V
(xy) ηdV
Iyz = Izy = −∫
V
(yz) ηdV
Izx = Ixz = −∫
V
(zx) ηdV
29
∴
∫
V
η
(
−→ρ ×··−→ρ)
dV
=
∫
V
η[−→ρ 2−→ω − (−→ρ · −→ω )−→ρ dV
]
=
∫
V
[(
ηy2 + z2)
ωx − ηxyωy − ηxzωz
]−→i
+[
η(
z2 + x2)
ωy−ηyzωx − ηyxωx
]−→j
+[
η(
x2 + y2)
ωx−ηzxωy − ηzyωy
]−→k
dV
= Ixxωx
−→i + Iyyωy
−→j + Izzωz
−→k + Ixyωy
−→i
+Iyzωz
−→j + Izxωx
−→k + Ixzωz
−→i + Iyxωx
−→j + Izyωy
−→k
−→HC = (Ixxωx + Ixzωz + Ixyωy)
−→i + (Iyyωy + Iyzωz + Iyxωx)
−→j
+ (Izzωz + Izxωx + Izyωy)−→k
−→HC = I−→ω where I =
Ixx Ixy Ixz
Iyx Iyy Iyz
Izx Izy Izz
∴
−→H = −→rC ×m
·−→rC +−→HC where
−→HC = I−→ω .
Problem:
Prove that Trot =1
2−→ω · −→HC .
Proof:
The rotational K.E of a rigid body M given by
T =1
2
∫
V
η·−→ρ dV (1)
η is the density of the rigid body, −→ρ is the position of the rigid
body is the position vector of elementary volume dv with respect to
centre of mass.
Let −→ω be the angular velocity of the body. Let −→rC be the position
30
vector of centre of mass w.r. to origin ‘O’. Then−→H =
(
−→rC ×m·−→rC
)
+
−→HC
where−→HC =
∫
V
η (−→ρ × (−→ω ×−→ρ )) dV (2)
But
·−→ρ2
=·−→ρ
·
·−→ρ =·−→ρ · (−→ω ×−→ρ )
= (−→ω ×−→ρ ) ··−→ρ
= −→ω ·(
−→ρ ×·−→ρ)
= −→ω ·[
−→ρ ×(
·−→ρ ×−→ω)]
Trot =1
2
∫
V
·−→ρ2
ηdV
=1
2
∫
V
η−→ω [−→ρ × (−→ω ×−→ρ )] dV
=1
2−→ω · −→HC ( using (2))
Angular momentum with respect to an arbitrary point :
Let P be arbitrary point, −→rP is the position vector of P w.r. to O,
mi is the position of the ith particle with position vector −→ri , c.m. is
the centre of mass of the system, −→ρi is the position vector of the ith
particle w.r. to ‘P ’.
Angular momentum of the ith particle w.r.to p = ρimi
·−→ρi .
Angular moment of the system about the position P is
−→HP =
N∑
i=1
−→ρi ×mi
·−→ρi
−→ρi = −→ri −−→rP
31
But
−→rP = −→rC−−→ρ
C
−→ρi = −→ri −(−→r
C−−→ρC
)
= −→ri −−→rC
+ −→ρC
·−→ρi =·−→ri −
·−→rC
+·−→ρC
−→HP =
N∑
i=1
(−→ri −−→rC
+ −→ρC
)
×mi
(
·−→ri −·−→rC
+·−→ρC
)
=N∑
i=1
mi
(
−→ri ×·−→ri −−→ri ×
·−→rC
+ −→ri ×·−→ρC−−→r
C×
·−→ri +−→rC ×
·−→rC −−→rC ×·−→ρC
+ −→ρC×
·−→ri −−→ρC×
·−→rC
+ −→ρC×
·−→ρC
)
AM of the system about the arbitrary point
=N∑
i=1
−→ri ×mi
·−→ri −N∑
i=1
mi−→ri ×
·−→rC
+N∑
i=1
mi−→ri ×
.−→ρC
−−→rC×
N∑
i=1
mi
·−→ri + −→rC ×m·−→rC −−→rC ×m
·−→ρC
+−→ρC×
N∑
i=1
mi
·−→ρi −−→ρC×m
·−→rC
+ −→ρC×m
·−→ρC
But
−→r C =
N∑
i=1
mi−→ri
N∑
i=1
mi
⇒N∑
i=1
mi−→r C =
N∑
i=1
mi−→ri
·−→r C =N∑
i=1
mi
·−→ri
32
AM of the system about the arbitrary point
−→HP =
−→H ·m−→r C ×
·−→r C +m−→r C ×·−→ρ C −−→r C ×m
·−→r C + −→r C ×m·−→r C
−−→r C ×m·−→ρ C + −→ρ C ×m
·−→r C −−→ρ C ×m·−→r C +
·−→ρ C ×m·−→ρ C
=−→H −−→r C ×m
·−→r C + −→ρ C ×m·−→ρ C
=N∑
i=1
−→r i ×mi
·−→r i −−→r C ×m·−→r C + −→ρ C ×m
·−→ρ C
1. Considering the K.E of a system of N particles show with usual
notation that
T =1
2m
·−→r 2C +
1
2
N∑
i=1
mi
·−→ρ 2i
2. Prove with usual notation that
Trot =1
2−→ω−→HC
3. Obtain the expression for the rational K.E. (Rigid body on sys-
tem of particle)
Generalized Momentum :
Consider a system with n - generalized co-ordinates q1, q2, · · · , qn.Let T be K.E of the system. V is the potential energy of the
system. Let us defined the Lagrangian function L(
q,·
q, t)
as follows
L = T − V .
The generalized momentum pi associated with qi is defined as
pi =∂L
∂·
qi
=∂T
∂·
qi
[
∵
[
∂V
∂·
qi
]
= 0
]
.
Note : i) L is almost a function of quadratic in q′s. Hence pi is a
linear function of q′s. If V = V (q, t), then∂L
∂qi=∂T
∂qi− ∂V
∂qi.
ii) Consider a particle of mass in with cortesian co-ordinates (x, y, z).
33
The K.E of the particle is given by
T =1
2m(
·
x2
+·
y2
+·
z2)
px =∂T
∂·
x=
1
2m2
·
x = m·
x
Similarly py = m·
y and
pz = m·
z
px is the x component of the linear momentum. If the position of the
particle is given by spherical polar co-ordinates (r, θ, φ) ,then
T =1
2m
(
·
r2
+ r2·
θ2
+ r2·
φ2
sin2 θ
)
pr =∂T
∂·
r=
1
2m2
·
r = m·
r
pθ =∂T
∂·
θ=
1
2mr22
·
θ = mr2·
θ
pφ =∂T
∂·
φ=
1
2mr22
·
φ sin2 θ = mr2 sin2 θ·
φ
pris called linerar momentum component radial direction.
pθ is called horizandal component of hte angular momentum.
pφ is called vertical component of angular momentum.
1. Three particles are conneeted by 2 rigid rods having a joint be-
tween them to form a system verticla force F and a momentum
M are applied as shown in figure. The configuration of the sys-
tem is given by the ordinary co-ordinates (x1, x2, x3) or by the
generalized co-ordinates (q1, q2, q3) where x1 = q1 +q2 + 1
2q3;x2 =
q1 − q3;x3 = q1 − q2 + 1
2q3. Find the expression for K.E and gen-
eralized momentum.
Solution :
Let T be the K.E. of the system.
∴ T =1
2m(
·
x2
1 +·
x2
2 +·
x2
3
)
34
x1 = q1 + q2 +1
2q3
x2 = q1 − q3
x3 = q1 − q2 +1
2q3
·
x1 =·
q1 +·
q2 +1
2
·
q3
·
x2 =·
q1 −·
q2
·
x3 =·
q1 −·
q2 +1
2
·
q3
∴ T =1
2m
(
3·
q2
1 + 2·
q2
2 +3
2
·
q2
3
)
p1 =∂T
∂·
q1
=1
2m3(
2·
q1
)
= 3m·
q1
p2 =∂T
∂·
q2
=1
2m4
·
q2 = 2m·
q2
p3 =∂T
∂·
q3
=1
2m
3
22·
q3 =3m
2
·
q3
p1, p2, p3 are called the generalized moment equal are inertial co-efficeient.
2 Three particles are connected by two rigid rods having a joint
between them to form the system shown in the figure. A virtual
force F and momentum m are applied to the system as shown the
configuration of the system is given by the ordinary co-ordinates
(x1, x
2, x
3) or by the gneralized co-ordinates (q
1, q
2, q
3), where
x1 = q1 + q2 + 1
2q3;x2 = q1 − q3;x3 = q1 − q2 + 1
2q3. Find the
generalized force Q1 ,Q2,Q
3
l l
mm
m M3
4
l
2x 3
x1
x
Solution :
Assume small motions the applied force on the system F1 =3F
4, F2 =
F
4− m
ℓ, F3 =
m
ℓ.
35
Let Q1, Q2, Q3 be the corresponding generalized forces on the sys-
tem.
Qi =3N∑
j=1
Fj
∂xj
∂qi⇒
3N∑
j=1
Fj
∂xj
∂qi
Q1 = F1
∂x1
∂q1+ F2
∂x2
∂q1+ F2
∂x3
∂q1
=3F
4(1) +
(
F
4− m
ℓ
)
+m
ℓ
=3F
4+F
4= F
Q2 = F1
∂x1
∂q2+ F2
∂x2
∂q2+ F2
∂x3
∂q2
=3F
4(1) +
(
F
4− m
ℓ
)
(0) +m
ℓ(−1)
=3F
4− m
ℓ
Q1 = F1
∂x1
∂q1+ F2
∂x2
∂q1+ F2
∂x3
∂q1
=3F
4
(
1
2
)
+
(
F
4− m
ℓ
)
(−1) +m
ℓ
(
1
2
)
=3F
8− F
4+m
ℓ+m
2ℓ
=F
8+
3m
2ℓ
Thus Q1 = F,Q2 = 3F4− m
ℓand Q3 = F
8+ 3m
2ℓ.
Energy of momentum conservative systems :
Let (x, y, z) be the position of a particle. Let−→F be the total force
acting on the partical with components
Fx = −∂V∂x
, Fy = −∂V∂y
, Fz = −∂V∂τ
where V (x, y, z) is the potential energy function depending on position
only. In this case F is called conservative force.
Book work : Prove that for any constervative force
∮ −→F × d−→r = 0
Proof : Let−→F be the conservative force acting on the particele at
36
(x, y, z). Let V (x, y, z) be the potential energy of the particle.
Fx = −∂V∂x
, Fy = −∂V∂y
, Fz = −∂V∂y
Workdone =(
Fx
−→i + Fy
−→j + Fz
−→k)
· d(
x−→i + y
−→j + z
−→k)
W =
∫
Fdr
dW =−→F · dr = Fxdx+ Fydy + Fzdz
=
(
−∂V∂x
)
dx+
(
−∂V∂y
)
dy +
(
−∂V∂z
)
dz
= −(
∂V
∂xdx+
∂V
∂ydy +
∂V
∂zdz
)
dW = −dV (x, y, z)
Thus dW is an exact differential equation
−→F · d−→r = −dV
Let W be the workdone by the force−→F is the moving particle from
A to B.
Total workdone = W =
B∫
A
− dV = −B∫
A
dV = − [V ]BA = VA − VB
∴ Workdone on the particle is depending on the initial and final
positions.
Workdone is moving around a closed curve is
∮ −→F d−→r = 0
[
∵
√A =
√B]
37
UNIT II
Lagrange’s Equation :
Let us consider a system of N particles whose positions relative to
an inertial frame are given by the cartesian co-ordinates x1, x2, · · · , x3N .
Kinetic energy of the system
T =1
2
3N∑
k=1
mk
·
x2
k (1)
where m1 = m2 = m3,m4 = m5 = m6 is the mass of the first particle,
second particle respectively.
Let q1, q2, · · · , qn be the generalized co-ordinates of the system.xk =
xk (q, t) , k = 1, 2, 3, · · · , 3N where we assume that tehse functions are
twice differentiable with respect to the q′s and t. We find that
·
xk
(
q,·
q, t)
=n∑
i=1
∂xk
∂qi
·
qi +∂xk
∂t
·
xk =∂xk
∂q1
·
q1 +∂xk
∂q2
·
q2 + · · · + ∂xk
∂qn
·
qn +∂xk
∂t
·
xk =n∑
i=1
∂xk
∂qi
·
qi +∂xk
∂t
Put in (1), we get
∴ T =1
2
3N∑
k=1
mk
[
n∑
i=1
∂xk
∂qi
·
qi +∂xk
∂t
]2
=1
2
3N∑
k=1
mk
[
n∑
i=1
n∑
j=1
∂xk
∂qi
∂xk
∂qj
·
qi
·
qj + 2n∑
i=1
∂xk
∂qi
·
qi
∂xk
∂t+
(
∂xk
∂t
)2]
=1
2
n∑
i=1
n∑
j=1
[
3N∑
k=1
mk
∂xk
∂qi
∂xk
∂qj
]
·
qi
·
qj +n∑
i=1
[
3N∑
k=1
mk
∂xk
∂qi
∂xk
∂t
]
·
qi
+1
2
3N∑
k=1
mk
(
∂xk
∂t
)2
.
38
Let3N∑
k=1
mk
∂xk
∂qi
∂xk
∂qj= mij and
3N∑
k=1
mk
∂xk
∂qi
∂xk
∂t= ai
T =1
2
n∑
i=1
n∑
j=1
mij
·
qi
·
qj +n∑
i=1
ai
·
qi +1
2
3N∑
k=1
mk
(
∂xk
∂t
)2
T(
q,·
q, t)
= T2 + T1 + T0.
where T2 is a homogeneous quadratic in q′s, T1 is homogeneous linear
function in·
qi, T0 is a function of q′s and t′s in general.
Note:
(i) T is a functin of(
q,·
q, t)
.
∴ T = T(
q,·
q, t)
(ii) For a sceleronomic system T0 = 0
T1 = 0
since∂xk
∂t= 0
In this case T = T2.
2. Obtain Lagrange’s equatin from D’ Alembert’s principle interms
of generalized co-ordinates.
Proof :
Let us assume that the system has N particles with 3N cartesian
co-ordinates from D’ alembert’s principle, we have
3N∑
k=1
(
Fk −mk
··
xk
)
δxk = 0 (1)
where Fk is the applied force component associated with xk.
39
Let us assume that the system is determined by n-generalized co-
ordinates q1, q2, · · · , qn.
xk = xk (q, t)
xk = xk(q1, q2, · · · , qn, t)
δxk =n∑
i=1
∂xk
∂qiδqi +
∂xk
∂tδt
(1) ⇒3N∑
k=1
(
Fk −mk
··
xk
)
n∑
i=1
∂xk
∂qiδqi = 0
n∑
i=1
3N∑
k=1
(
Fk −mk
··
xk
) ∂xk
∂qiδqi = 0 (2)
·
xk =n∑
i=1
(
∂xk
∂qi
·
qi
)
Differentiating w.r.to·
qi
∂·
xk
∂·
qi
=∂xk
∂qi(3)
d
dt
(
∂·
xk
∂·
qi
)
=d
dt
(
∂xk
∂qi
)
=∂
·
xk
∂qi
T =1
2
3N∑
k=1
mk
·
x2
k
pi =∂T
∂·
qi
=1
2
3N∑
k=1
mk 2·
xk
∂·
xk
∂qi
=3N∑
k=1
mk
·
xk
∂·
xk
∂qi
d
dt
(
∂T
∂·
qi
)
=3N∑
k=1
[
mk
··
xk
∂·
xk
∂qi+mk
·
xk
d
dt
(
∂·
xk
∂qi
)]
40
=3N∑
k=1
mk
∂xk
∂qi
··
xk +3N∑
k=1
mk
·
xk
∂·
xk
∂qi(4)
∂T
∂qi=
1
2
3N∑
k=1
mk 2·
xk
∂·
xk
∂qi(5)
d
dt
(
∂T
∂qi
)
− ∂T
∂qi=
3N∑
k=1
mk
··
xk
∂xk
∂qi(6)
Let us define the generalized force Qi asq
Qi =3N∑
k=1
Fk
∂xk
∂qi
By (2) we have
n∑
i=1
[
3N∑
k=1
Fk
∂xk
∂qi−
3N∑
k=1
mk
··
xk
∂xk
∂qi
]
δqi = 0
n∑
i=1
[
Qi −d
dt
(
∂T
∂·
qi
)
+∂T
∂qi
]
δqi = 0 (7)
This is the Lagrange’s form of D’ Alemberts principle interms of gen-
eralized co-ordinates.
The above discussion the system is under instantaneous constraints.
Let us assume the system be holonomic and described by n- inde-
pendet generalized co-ordinates q1, q2, · · · , qn.The δq′is are independent.
Hence the co-efficeient of δq’s in equation (7) must be zero. q’s are
independent.
∴ co-efficient of δq = 0
d
dt
(
∂T
∂·
qi
)
− ∂T
∂qi= Qi, i = 1, 2, 3, · · · , n.
These equations are Lagrange equations from holonomic system.
Book work : Obtain the standard form of Lagrange’s equation for a
41
holonomic system.
Proof : Let us assume that all the generalized forces are derivable form
a potential function V (q, t) as let the system is holonomic conservative.
Qi = −dVdqi
=d
dt
(
∂T
∂·
qi
)
− ∂T
∂qi= −∂V
∂qi
d
dt
(
∂T
∂·
qi
)
− ∂T
∂qi+∂V
∂qi= 0
d
dt
(
∂T
∂·
qi
)
− ∂
∂qi(T − V ) = 0
Let L = T − V (Lagarangian function) T = L+ V.
∂T
∂·
qi
=∂L
∂·
qi
[
∵ L(
q,·
q, t)]
d
dt
(
∂T
∂·
qi
)
− ∂L
∂·
qi
= 0, i = 1, 2, 3, · · · , N.
Note :
1. The standard form of Lagrange’s equations
d
dt
(
∂T
∂·
qi
)
− ∂L
∂qi= 0, i = 1, 2, 3, · · · , N.
since pi =∂L
∂·
qi
∴
d
dt(Pi) −
∂L
∂qi= 0
where pi =∂L
∂qi, i = 1, 2, 3, · · · , n.
2. Let us assume the generalized forces Qi is not completly deriva-
42
tive from a potential function
Qi = −∂V∂qi
+Q′
i
we haved
dt
(
∂T
∂·
qi
)
− ∂T
∂qi= Qi
d
dt
(
∂T
∂·
qi
)
− ∂T
∂qi= −∂V
∂·
qi
+Q′
i
d
dt
(
∂T
∂·
qi
)
− ∂L
∂·
qi
+∂T
∂qi= Q′
i
Let L = T − V.
d
dt
(
∂T
∂qi
)
− ∂L
∂·
qi
= Q′
i, i = 1, 2, 3, · · · , n.
Book work : Obtain the Lagrange’s equation for a non-holonomic
system.
Proof : Let us assume that system has ‘m’ non holonomic constraint
equationn∑
i=1
ajidqi + ajidt = 0 (j = 1, 2, 3, · · · ,m) .
Let us assume a virtual displacement consistent with the con-
straints.
Now we have
n∑
i=1
aijδqi = 0, j = 1, 2, 3, · · · ,m (1)
The constraints are assumed to be workless. Hence we have,
n∑
i=1
ciδqi = 0 (2)
43
where ci is the generalized constraints force
Multiply equation (1) by Lagrange’s multiply λj
λj
n∑
i=1
ajiδqi = 0, j = 1, 2, 3, · · · ,m (3)
Adding these m equations and subtracting from (2) we get
(2) − (3) ⇒n∑
i=1
(
ci −m∑
j=1
λjaji
)
δqi = 0
Choose the λ′s arbitrary such that
ci =m∑
j=1
λjaji, i = 1, 2, 3, · · · ,m
of the generalized forces are not wholly derivable form a potential
function.
QQi = −∂V∂qi
+Q′
i;
we haved
dt
(
∂L
∂ξi
)
− ∂L
∂ξi
= Q′
i, i = 1, 2, 3, · · · , n
In our discussion Q′
i nothing but the ci
d
dt
(
∂L
∂·
qi
)
− ∂L
∂qi=
m∑
j=1
λjaji, i = 1, 2, 3, · · · , n.
This is the standard form of Lagrange’s equation for the non-holonomic
systyem.
1. Find the differential equation of motion for a spherical pendulum
of length ℓ (using Lagrange’s equation).
Solution : Let use the spherical co-ordinates θ and φ. θ is mea-
sured from the upward vertical, φ is the angle between a vertical plane
through supporting point ‘O’ and a vertical plane containing the pen-
44
dulum.
The spherical co-ordinates are (r, θ, φ) for this problem.
r = ℓ (constant) ⇒ r′ = 0
Let T be the K.E. and V is the P.E.
T =1
2m
(
ℓ2·
θ2
+ ℓ2·
φ2
sin2 θ
)
V = −mg (−ℓ cos θ) = mgℓ cos θ
L = T − V
=1
2m
(
ℓ2·
θ2
+ ℓ2·
φ2
sin2 θ
)
−mgℓ cos θ
The Lagrange’s equation is given byd
dt
(
∂L
∂·
qi
)
− ∂L
∂qi= 0. For our
problem, qi = ℓ, q2 = 0, q3 = φ, we have
d
dt
(
∂L
∂·
θ
)
− ∂L
∂θ= 0
d
dt
(
1
2mℓ22
·
θ
)
−[
1
2mℓ2
·
φ22 sin θ cos θ +mgℓ sin θ
]
= 0
mℓ2··
θ −mℓ2·
φ2 sin θ cos θ −mgℓ sin θ = 0
Dividing by mℓ
ℓ··
θ − ℓ·
φ2 sin θ cos θ − g sin θ = 0 (1)
q
f
mg
sinl q
cosl ql
O
180 q-
45
Also
d
dt
(
∂L
∂·
φ
)
− ∂L
∂φ= 0
d
dt
(
1
2mℓ22
·
φ sin2 θ
)
− 0 = 0
d
dt
(
mℓ2·
φ sin2 θ
)
= 0
Differentiating w.r.to t
mℓ2 sin2 θ··
φ+ 2m·
φℓ2 sin θ cos θ·
θ = 0
Dividing by mℓ
cosh l q=cosh
lq=
g
h
m
sin2 θ··
φ+ 2 sin θ cos θ·
φ·
θ = 0 (2)
sin θ··
φ+ 2 cos θ·
φ·
θ = 0
we have
∂L
∂φ= 0
d
dt
(
∂L
∂·
φ
)
= 0
46
∴
∂L
∂φ= pφ = constant
i.e., mℓ2 sin2 θ.·
φ = pφ = constant
where pφ is the angular momentum above a vertical axis flow the sup-
porting point.
Inertial Co-efficeient :
The co-efficient of the q is the expression for the generalized mo-
ments are known an inertial co-efficeient.
In the Book work 2, 3n, 2m,3m
2are inertial co-efficeient.
2 A doubel pendulum consists of particles supported by mass less
rods assuming that all motions take place in a vertical plane.
Find the differential equation of the motion linearise these equa-
tions assuming small motions.
Solution :
Let T be the K.E of the system and V is the potential energy of
the system.
Let L = T − V. Let V be the velocity of the particle.
V = The velocity of upper particle + velocity of the lower particle.
l
f&l
f&l
q&l
f q-
f q-q
O
q
m B
T =1
2mv2
1 +1
2mv2
2
v1 = ℓ·
θ
v2 =
√
ℓ2·
φ2 + ℓ2·
θ2
+ 2ℓ·
φℓ·
θ cos (φ− θ)
T =1
2mℓ2
·
θ2 +1
2m
[
ℓ2·
φ2 + ℓ2·
θ2
+ 2ℓ·
φℓ·
θ cos θ
]
47
=1
2mℓ2
[
2·
θ2 +·
φ2 + 2·
φ·
θ cos (φ− θ)
]
V = −mg [ℓ cos θ + ℓ cosφ+ ℓ cos θ]
= −mg (2 cos θℓ+ ℓ cosφ)
L = T − V
=1
2mℓ2
[
2·
θ2 +·
φ2 + 2·
φ·
θ cos (φ− θ)
]
+mg (2 cos θ + ℓ cosφ)
l cosql
BA
C
std level
cosfl
cosql
( )1p mg
f
f
mg
∴The equation
d
dt
(
∂L
∂·
θ
)
− ∂L
∂θ= 0
d
dt
[
1
2mℓ22.2
·
θ +1
2mℓ2.2
·
φ cos (φ− θ)
]
−1
2mℓ22
·
θ·
φ − sin θ (φ− θ) (−1) + (−2mgℓ sin θ) = 0
48
l
O
q
f
g
lm
mℓ2d
dt
[
2·
θ +·
φ cos (φ− θ)
]
−mℓ2·
θ·
φ sin (φ− θ) + 2mgℓ sin θ = 0
mℓ2[
2··
θ +··
φ cos (φ− θ) +·
φ (− sin (φ− θ))
(
·
φ−·
θ
)
−mℓ2·
θ·
φ sin (φ− θ) + 2mgℓ sin θ
]
= 0
mℓ2[
2··
θ +··
φ cos (φ− θ) −·
φ2 sin (φ− θ)
]
+ 2mgℓ sin θ = 0 (1)
Now the equation
d
dt
(
∂L
∂φ
)
− ∂L
∂φ= 0
d
dt
1
2mℓ2
(
2φ+ 2θ cos (φ− θ))
− 1
2mℓ22θφ(− sin (φ− θ)) −mgℓ sinφ
= 0
mℓ2d
dt
[
φ+ θ cos (φ− θ)]
+mℓ2θφ sin(
φ− θ)
+mgℓ sinφ = 0
mℓ2[
φ+ θ cos (φ− θ) + θ2sin (φ− θ)
]
+mgℓ sinφ = 0 (2)
49
(1) and (2) are the differential equation of motion linearise the
equation. Let us assume the system having small oscillations (θ, φ)
are small.
∴ sin θ = θ, sinφ = φ
cos (φ− θ) = 1, sin (φ− θ) = φ− θ
Neglecting higher order terms equation (1) becomes
mℓ2(
2θ + φ)
+ 2mgℓθ = 0
and equation (2) becomes
mℓ2(
θ + φ)
+mgℓφ = 0
3 A block of mass m2 can slide on another block of mass m1 which
in turn, slides on a horizontal surface, as shown in the figure.
Using x1 and x2 as co-ordinates, obtain the differential equations
of motion. Solve for the accelerations of the two blocks as they
move under the influence of gravity, assuming that all surfaces
are frictionless. Find the fore of interaction between the blocks.
Solution : Let x1 be the displacement of m1, let x2 be the dis-
placement of m2 relative to m1. Let v1 be the velocity of m1 and v2
be the velocity of m2 relative to m1.
T =1
2m1v
21 +
1
2m2v
22
v1 = x1
v2 =√
x21 + x2
2 − 2x1x2 cos 450
=
√
x21 + x2
2 −√
2x1x2
∴ T =1
2m1x
21 +
1
2m2
(
x1 + x2 −√
2x1x2
)
.
50
2m
2x
1x
045
Let V be the P.E. of the system
V −m2gx2 sin 450 = − 1√2m2gx2
∴ L = T − V
=1
2m1x
21 +
1
2m2
(
x21 + x2
2 −√
2x1x2
)
+1√2m2gx2
x1− Lagrange’s equation is
d
dt
(
∂L
∂x1
)
−(
∂L
∂x1
)
= 0
d
dt
(
1
2m12x1 +
1
2m22x1
)
− 1√2m2x2 = 0
m1x1 +m2x1 −1√2m2x2 = 0
(m1 +m2) x1 −1√2m2x2 = 0 (1)
[width=6cm]m12
x2− Lagrange’s equation is
d
dt
(
∂L
∂x2
)
−(
∂L
∂x2
)
= 0
d
dt
(
1
2m12x2 −
1√2m2x1
)
− 1√2m2g = 0
m2x2 −1√2m2x1 =
1√2m2g (2)
51
(1) + (2)1√2m2x2 −
1√2m2x1 =
1√2m2g
(1) ⇒ (m1 +m2) x1 =1√2
[
1√2m2x1 +m2g
]
(
m1 +m2 −m2
2
)
x1 =1√2
m2g√2
x1 =m2g
2m1 +m2
(1) ⇒ (m1 +m2)m2g
2m1 +m2
=1√2m2x2
x2 =√
2(m1 +m2) g
2m1 +m2
4 A particle of mass m can side without friction on the inside of a
small tube which is bent in the form of a circle of radius r, the
tube rotates about a vertical diameter with a constant angular
velocity ω, as shown in figure write the differential equation of
motion.
Solution :
Let ‘O’ be the centre of the tube and r be the radius of the tube.
Let T be the K.E of the particle of mass m,V is the P.E. of the
particle.
T =1
2m(
r2θ2+ r2ω2 sin2 θ
)
=1
2mr2
(
θ2+ ω2 sin2 θ
)
V = mgh = mgr cos θ
L = T − V
=1
2mr2
(
θ2+ ω2 sin2 θ
)
−mgr cos θ, r is constant
52
The Lagrange’s equation is
d
dt
(
∂L
∂θ
)
− ∂L
∂θ= 0
d
dt
(
1
2mr22θ
)
−[
1
2mr2ω22 sin θ cos θ +mgr sin θ
]
= 0
d
dt
(
mr2θ)
−[
1
2mr2ω22 sin θ cos θ +mgr sin θ
]
= 0
mr2θ −mrω2 sin θ cos θ −mgr sin θ = 0
5 Mass spring system : A particle of mass m is connected by a
massless spring of stiffness K and unstressed length r0 to a point
P which is moving along a circubr path of radius a at a uniform
angular rate ω. Assuming that the particle moves without friction
on a horizontal plane, find the differential equations of motion.
Solution :
Resolving r and rθ along aω and rθ .
In this probem, we have dealing with a single particle of mass ‘m’.
Let O taken as origin of refeence length ℓ.
~rP is the position vector of the point P w.r.to the point O.·
~ρ is the velocity of the particle w.r.to P.
Let T be the K.E. of the particle, V is the P.E of th+e particle.
T =1
2m
·
~r·
~rP +1
2m
·
~ρ2
+·
~rPm·
~ρO
·
~ρO is the position vector of the system w.r.to ρ.
In the above case, ρC
= ρ
T =1
2m
·
~r·
~rP +1
2m
·
~ρ2
+·
~rPm·
~ρC
·
~rP = aω·
~ρ2
= r2 + r2θ2
·
~rPm·
~ρ = maω[
r sin (θ − ωt) + rθ cos (θ − ωt)]
53
T =1
2ma2ω2 +
1
2m(
r2 + r2θ2)
+maω[
r sin (θ − ωt) + rθ cos (θ − ωt)]
V =1
2K (r − r0)
2
L = T − V
=1
2ma2ω2 +
1
2m(
r2 + r2θ2)
+maω[
r sin (θ − ωt) + rθ cos (θ − ωt)]
− 1
2K (r − r0)
2
The Lagrange’s r equation is
d
dt
(
∂L
∂r
)
− ∂L
∂r= 0
d
dt
[
1
2m2r +maω sin (θ − ωt)
]
−
1
2m2rθ
2+maω
(
θ cos (θ − ωt))
− 1
2K2 (r − r0)
= 0
mr +maω cos (θ − ωt)(
θ − ω)
−mrθ2 −
maωθ cos (θ − ωt) +K (r − r0) = 0
mr −maω2 cos (θ − ωt) −mrθ2+K (r − r0) = 0
The Lagrange’s θ equation is given by
d
dt
(
∂L
∂θ
)
− ∂L
∂θ= 0
d
dt
[
1
2m2r2θ +maω cos (θ − ωt)
]
−
maωr cos (θ − ωt)(
−maωrθ sin (θ − ωt))
= 0
mr2θ + 2mrθ +maωr cos (θ − ωt) −maωrθ sin (θ − ωt) +maωr2 sin (θ − ωt) −maωr cos (θ − ωt) +maωr sin (θ − ωt) = 0
mr + 2mrθ +maω2r cos (θ − ωt) = 0
54
Knife edge problem :
Example of a non-holonomic rheonomic system.
Two particles are connected by a rigid massless rod of length ℓ
which rotates in a horizontal plane with a constant angular vlocity
ω. Knife - edge supports at the two particles prevent either particle
from having a velocity component along the rod, but the particles can
slide without friction in a direction perpendicular to the rod. Find the
differential equations of motion. Solve for x, y and the constraint force
as functions of time if the center of mass is initially at the origin and
has a velocity v0 in the positive y direction.
Solution :
Let (x, y) be the co-ordinates of the CM at time the system has
translational and rotational K.E
Velocity component along the rod = 0
∴ (x1 + x2) cosωt+ (y1 + y2) sinωt = 0 (1)
T =1
2m(
x21 + y2
1
)
+1
2m(
x22 + y2
2
)
=1
2m(
x21 + x2
2 + y21 + y2
2
)
=1
2m
(
2x2 + 2y2 + ℓ2ω2
2
)
= m(
x2 + y2)
+ +mℓ2ω2
4
∴ V = 0
Lagrange’s equation of for non-haolonomic system is
d
dt
(
∂L
∂qi
)
− ∂L
∂qi=
m∑
i=1
λjaij = λ1a11.
The velocity of centre of mass exsited only in a direction perpen-
dicular to the rod. Hence it has no velocity along the rod.
55
x1 = x− ℓ
2cosωt
y1 = y − ℓ
2sinωt
x2 = x+ℓ
2cosωt
y2 = y +ℓ
2sinωt
x1 = x+ℓω
2sinωt
y1 = y − ℓω
2cosωt
(2)
x2 = x− ℓω
2sinωt
y 2 = y +ℓω
2cosωt
From (1) and (2)
x cosωt+ y sinωt = 0
x cosωt+ y cos(
900 − ωt)
= 0
x cosωt+ y sinωt = 0 (A)
[
x+ℓω
2sinωt+ x− ℓω
2sinωt
]
cosωt+
sinωt
[
y − ℓω
2cosωt+ y +
ℓω
2cosωt
]
= 0
2x cosωt+ 2y sinωt = 0
cosωtdx+ sinωtdy = 0n∑
i=1
ajidqi + ajtdt = 0
a11x+ a12y = 0, q1 = x1, q2 = y2
56
a11 = cosωt
a12 = sinωt
n∑
i=1
ajiaij = 1
d
dt
(
∂L
∂x
)
− ∂L
∂x= λa11
L = T − V
= m(
x2 + y2)
+m
4ℓ2ω2
∂L
∂x= 2mx
∴
d
dt(2mx) − 0 =
n∑
i=1
λjaji = λ1 cosωt
2mx = λ1 cosωt (B)
d
dt
(
∂L
∂y
)
− ∂L
∂y= λa12
d
dt(2my) = λ1a12
2my = λ1 sinωt (C)
(C)
(B)⇒ tanωt =
y
x
x cosωt+ y sinωt = 0
− xy
= tanωt
y
x= − x
y2xx+ 2yy = 0
2xd (x) + 2yd (y) = 0
d
dt
(
x2 + y2)
= 0
d
dt
(
2x2 + 2y2)
= 0
2x2 + 2y2 = constant
57
Initially x = 0, y = 0,
0 + v20 = 0,
x2 + y2 = v20
x cosωt+ y sinωt = 0
Hence the centre of mass moves with constant velocity, satisfying those
equation, we have
x = −v0 sinωt
y = v0 cosωtdx
dt= −v0 sinωt
dx = −v0 sinωtdt∫
dx = −∫
v0 sinωtdt
x =v0
ωcosωt+ C1
x = 0, t = 0
0 =v0
ω+ C1
C1 = −v0
ω
x =v0
ωcosωt− v0
ω
=v0
ω(cosωt− 1)
y0 = 0; t = 0
C2 = 0
∴ y =v0
ωsinωt
Eliminating t by ω, x and y we have circular path. The generalized
58
forces are
Ci =m∑
i=1
λjaji, j = 1, 2, 3, · · · ,m
C1 = λ1a11
C1 = cosωt2mx
cosωt
[
∵ λ1 =2mx
cosωt
]
x = −v0 sinωt
x = −v0ω cosωt
∴ C1 = −2m (v0 cosωt)
C2 = λ2a12
=2my
cosωtsinωt
=−2mv0 sinωt cosωt
cosωt= −2mv0 sinωt
Note:
In the above case,
x =v0
ω(cosωt− 1)
x+v0
ω=
v0
ωcosωt
(
x+v0
ω
)2
= y +v2
0
ω2
Hence the system moves in a circular path with radius v0/ω.
2.3 Integrals of the motion:
Ignorable co-ordinates:
Let the holnomic system be described by n standard Lagrange’s
equationd
dt
(
∂L
∂qi
)
− ∂L
∂qi= 0, i = 1, 2, 3, · · · , n
Let (q, q, t) containing all q1, q2, · · · , qn, but some of the q′s, say q1, q2, · · · , qnare missing from the Lagrangian. These k− co-ordinaties are called ig-
norable co-ordinates. Since∂L
∂qiis zero for each ignorable co-ordinate,
59
it follows that
d
dt
(
∂L
∂qi
)
= 0, (i = 1, 2, 3, · · · , k) ,
or
pi =∂L
∂qi= βi, i = 1, 2, 3, · · · , k
where the β′s are constants evaluated from the initial conditions. Hence
we find that the generalized momentum corresponding to each ignor-
able co-ordinate is constant, that is, it is an integral of the motion.
Example :
Let us consider the Kepler’s problem, that is, the problem of the
motion of a particle of unit mass which is attracted by an inverse square
gravitational force to a fixed point O. Using polar co-ordinates, the
kinetic and potential evergies are
T =1
2
(
r2 + r2θ2)
V = −µr
where µ is a positive constant known as the gravitational coefficeient.
The Lagrangian function is
L = T − V =1
2
(
r2 + r2θ2)
+µ
r
L is a independent co-ordinates
Lagrangian ‘r’ equation is
d
dt
(
∂L
∂r
)
− ∂L
∂r= 0
d
dt(r) − 1
2
(
2rθ2)
+µ
r2= 0
r − rθ2+µ
r2= 0. (1)
Since θ does not appear explicity in the Lagrangian function, it is an
60
ignorable co-ordinate. The θ equation of motion is
d
dt
(
r2θ)
= 0
or∂L
∂θ= constant
1
2r22θ = β ⇒ θ =
β
r2
r2 =β
θ(2)
where β is constant and is equal to the angular momentum of the
particle about the attracting centre O.
r − rβ2
r4+µ
r2= 0 ( using (2) in (1))
r − β
r3+µ
r2= 0
The Routhian function :
Consider a standard holonmic system with q1, q2, · · · ., qk are ignor-
able co-ordinates
L = L (qk+1, · · · , qn, q1, q2, · · · , qn, t) .
Now let us define a Routhian function
R (qk+1, · · · , qn, qk+1, · · · , qn, β1, β2, · · · , βk, t)
as follows,
R = L−k∑
i=1
βiqi
where βi =∂L
∂qi, i = 1, 2, 3, · · · , k.
Obtain the Lagrange’s equation using Routhian funcion.
Solution :
61
Consider the holonomic system with ‘n’ independent generalized
co-ordinates
q1, q2, · · · .., .qn. Letq1, q2, · · · .., qk be the ignorable co-ordinates. The
Legrangian function is
L = L (qk+1, · · · , qn, q1, q2, · · · , qn, t) .
Now∂L
∂qi= βi, i = 1, 2, 3, · · · , k (1)
Now the Routhian funciton is given by
R = L−k∑
i=1
βiqi (2)
Subsitute the values for q1, q2, · · · , qk
Using (1) we have
R = R (qk+1, · · · , qn, qk+1, · · · , qn, β1, β2, · · · , βk, t) .
Differentiating (2)
δR = δL− δk∑
i=1
βiqi (3)
δL =n∑
i=k+1
∂L
∂qiδqi +
k∑
i=1
∂L
∂qiδqi +
n∑
i=k+1
∂L
∂qiδqi +
∂L
∂tδt
δR =n∑
i=k+1
∂R
∂qiδqi +
n∑
i=k+1
∂R
∂qiδqi +
n∑
i=1
∂L
∂βi
δβi +∂R
∂tδt (4)
δ
k∑
i=1
βiqi =k∑
i=1
∂L
∂qiδqi +
k∑
i=1
qiδβi
62
(3)⇒
δ
(
L−k∑
i=1
βiqi
)
=n∑
i=k+1
∂L
∂qiδqi +
n∑
i=k+1
∂L
∂qiδqi−
k∑
i=1
qiδβi +∂L
∂tδt (5)
From (4) and (5)
The corresponding coefficients must be equal.
∂L
∂qi=∂R
∂qi, i = k + 1, k + 2, k + 3, · · · , n
∂L
∂qi=∂R
∂qi, i = k + 1, k + 2, k + 3, · · · , n
(6)
and
qi = − ∂R
∂βi
, i = 1, 2, 3, · · · , k
∂L
∂t=
∂R
∂t.
Now let us substitute from equation (6) in to Lagrangian equations
and obtain
d
dt
(
∂R
∂qi
)
− ∂R
∂qi= 0, i = k + 1, k + 2, k + 3, · · · , n
∂R
∂βi
= −qi, i = 1, 2, 3, · · · , k
Illustrate Kepler’s problem using Routhian method.
Solution :
Let us assume that the particle has unit mass. Let (r, θ) be the
polar co-ordinates of the particle.
Let T be the kinetic energy of the particle. V is the potential evegy
of the particle.
T =1
2
(
r2 + r2θ2)
V = −µr
L = T − V
63
=1
2
(
r2 + r2θ2)
+µ
r∂L
∂θ= 0
∴ θ is ignorable solution
∂L
∂θ= β ⇒ r2θ = β
Now the Routhian function is given by
R = L− βθ
=1
2
(
r2 + r2θ2)
+µ
r− βθ
=1
2r2 +
1
2r2β
2
r4+µ
r− β2
r2
=1
2r2 +
µ
r− 1
2
β2
r2
Routhian equation is given by
d
dt
(
∂R
∂qi
)
−(
∂R
∂qi
)
= 0
d
dt
(
∂R
∂r
)
−(
∂R
∂r
)
= 0 (1)
Here there are two co-ordinates (r, θ)
∂L
∂θ= β ( constant)
∴The Routhian equation is only at r
d
dt(r) −
[
(
− µ
r2
)
− 1
2β2(
−2r−3)
]
= 0
r +µ
r− β2
r3= 0
or
r − β2
r3+µ
r= 0
64
Conservative system :
Write down the 3 condition defining a conservative system
and show that these conditions are sufficeient to ensure the
existence of an energy integral or Jacobi integral.
Proof :
1. The standard form of Lagrange’s equation (holonoic or nonholo-
nomic) applies.
2. The Legrangian function L is not an explicit function of time.
3. Any constraint equatins can be expressed in the differential form
n∑
i=1
ajidqi = 0, j = 1, 2, 3, · · · ,m
ie, all the coefficients are equal to zero.
Consider the standard nonholonomic form of Lagrange’s equation
d
dt
(
∂L
∂qi
)
− ∂L
∂qi=
n∑
i=1
λjaji, i = 1, 2, · · · , n (1)
where L (q, q) is not an explicit function of time.
Now let us consider the total derivative
αjt =∂φ
∂t
dL
dt=
n∑
i=1
∂L
∂qiq +
n∑
i=1
∂L
∂qiqi (2)
(1) ⇒ ∂L
∂qi=
d
dt
(
∂L
∂qi
)
−m∑
j=1
λjaji
(2) ⇒ dL
dt=
n∑
i=1
[
d
dt
(
∂L
∂qi
)
−m∑
j=1
λjaji
]
qi +n∑
i=1
∂L
∂qiqi
65
=n∑
i=1
[
d
dt
(
∂L
∂qi
)
qi +∂L
∂qiqi
]
−n∑
i=1
m∑
j=1
λjajiqi
=d
dt
[
n∑
i=1
(
∂L
∂qi
)
qi
]
−m∑
j=1
(
n∑
i=1
ajiqi
)
λj
dL
dt=
d
dt
[
n∑
i=1
(
∂L
∂qi
)
qi
]
− 0
d
dt
[
L−n∑
i=1
(
∂L
∂qi
)
qi
]
= 0
d
dt
[
n∑
i=1
(
∂L
∂qi
)
qi − L
]
= 0
Integrating w.r.to t
n∑
i=1
(
∂L
∂qi
)
qi − L = h ( constant)
∴ h =n∑
i=1
(
∂L
∂qi
)
qi − L
L =n∑
i=1
(
∂L
∂qi
)
qi − h
h = T2 − T0 + V
=n∑
i=1
(
∂L
∂qi
)
qi − L
This integral is called Jacobi integral or energy integral. This en-
ergy integral exists for all conservative system.
Book work :
Define a natural system. Prove that for a conservative
system, Jacobi integral is not equal to the total evergy.
Proof :
For every conservative system
h = T2 − T0 + V
66
For the conservative system energy integral
h =n∑
i=1
(
∂L
∂qi
)
qi − L (1)
L = T − V
= T2 + T1 + T0 − V
We know that
T2 =1
2
n∑
i=1
m∑
j=1
mjiqiqj
T1 =n∑
i=1
aiqi
T0 =1
2
3N∑
k=1
mk
(
∂xk
∂t
)2
∴
n∑
i=1
∂L
∂qiqi =
n∑
i=1
∂T
∂qiqi
=n∑
i=1
(
∂T2
∂qi+∂T1
∂qi+∂T0
∂qi
)
qi
= 2T2 + T1 + 0
(1) ⇒ h = 2T2 + T1 − T1 − (T − V )
= 2T2 + T1 − T2 − T0 + V
= T2 − T0 + V = T ′ + V ′
∴ The energy T ′ + V ′ is constant for any conservative system but
it is not equal to total energy.
Natual system :
A natual system is a consevative system.
Proof :
A system is said to be natural system if
67
(i) it is described by standard form of holonomic Lagrangian equa-
tion.
(ii) the Lagrangian function L is not an explicit function of time.
(iii) any constraint equation can be expressed in the differential form
n∑
i=1
ajidqi = 0
(iv) the K.E is expressed as homogeneous quadratic functio of q’s.
To prove Jacobi integral = Total energy for the natural system.
T =1
2
n∑
i=1
m∑
j=1
mijqiqj = T2
T0 = 0;T1 = 0
We know that
L = T − V, L = T2 − V∂L
∂qi=
∂T2
∂qin∑
i=1
∂L
∂qiqi =
n∑
i=1
∂T2
∂qiqi = 2T2 [ by the above Book work]
h =n∑
i=1
∂L
∂qiqi − L
= 2T2 − (T2 − V )
= T2 + V
Jacobi integral = Total evergy.
Example : 1 Suppose a mass - spring system is attached to a
frame which is translating with a uniform velocity v0 as shown
in figure. Let ℓ0 be the unstressed spring length and use the
elongation x as the generalized co-ordinate. Find the Jacobi
integral for the system.
68
Solution : Let m be the mass of the body. Let x be the displacement
of the body from the relating postion at time t .
0v
km
0x+l
Velocity of the body = x
Velocity of the system = v0 + x
The K.E of the system =1
2m (v0 + x)2
T =1
2m(
v20 + x2 + 2v0x
)
T2 =1
2mx2
T1 =1
2mv0x
T0 =1
2mv2
0
Let V be the potntial Energy of the system.
∴ V =1
2kx2
L = T − V
=1
2mv2
0 +1
2mx2 +mv0x =
1
2kx2
L is independent of ‘t’
Hence the system is conservative.
Jacobian integral h = T2 − T0 + V
h =∂L
∂xx− L
= (mx+mv0) x−
1
2mv2
0 +1
2mx2 +mv0x
2 − 1
2kx2
69
= mx2 +mv0x−1
2mv2
0 −1
2mx2 −mv0x
2 +1
2kx2
=1
2mx2 − 1
2mv2
0 +1
2kx2
= T2 − T0 + V
Example 2 : A small tabe, bent in the form of a circle of
radius r, rotates about a vertical diameteer with a constant
angular velocity ω. A particle of mass m can slide without
friction inside the tube. At any given time, the configuration
of the system is specified by the angle θ which is measured
from the upward vertical to the line connecting the center O
and the particle. Find the Jacobi integral.
Solution :
w
g
O
q
r
m
Let O be the center of the tube. The transformation equations
relating the generalized co-ordinate θ and the position (x, y, z) of the
particle.
x = r sin θ cosωt
y = r sin θ sinωt
z = r cos θ
70
AP
OP= sin θ ⇒ AP = r sin θ
OA
OP= cos θ ⇒ OA = r cos θ
where r is the radius of the tube.
Let T be the K.E of the particle of mass m, V is the P.E of the
particle
T =1
2m(
r2θ2+ r2ω2 sin2 θ
)
=1
2mr2
(
θ2+ ω2 sin2 θ
)
V = mgr cos θ, T2 =1
2mr2θ
2, T1 = 0, T0 =
1
2mr2ω2 sin2 θ
L = T − V =1
2mr2
(
θ2+ ω2 sin2 θ
)
−mgr cos θ
L is independent of t.
Hence the system is conservative Jacobi method.
h =n∑
i=1
∂L
∂qiqi − L
=∂L
∂θθ − L
=(
mr2θ)
θ −(
1
2mr2θ
2+
1
2r2ω2 sin2 θ −mgℓ cos θ
)
=1
2mr2θ
2 − 1
2r2ω2 sin2 θ +mgℓ cos θ
= T2 − T0 + V.
Example 3 :
Two particles, each of mass m, are connected by a rigid mass-
less rod of length ℓ. The particles are supported by knife
edges placed perpendicular of the rod. Assuming that all
moion is confined to the horizontal xy plane, find the Jacobi
integral.
Solution: Let (x1, y1), (x2, y2) be the co-ordinate of the two ends of
71
the rod at time ‘t’.
Let (x, y) be the co-ordinates of the cm and we have holonomic con-
straint
(x1 − x2)2 + (y1 − y2)
2 = ℓ2 (1)
We know that
tanωt =y2 − y1
x2 − x1
y2 − y1 = (x2 − x1) tanωt (2)
x1 = x− ℓ
2cosωt
x2 = x+ℓ
2cosωt
y1 = y − ℓ
2sinωt
y2 = y +ℓ
2sinωt
x1 + x2 = 2x
y1 + y2 = 2y
Let T be the K.E. of the system. Let V be the P.E. of the system.
V = 0
T =1
2m(
x21 + y2
1
)
+1
2m(
x22 + y2
2
)
x1 = x+ℓ
2ω sinωt
x2 = x− ℓ
2ω sinωt
y1 = y − ℓ
2ω cosωt
y2 = y +λ
2ω cosωt
x21 + x2
2 + y21 + y2
2 = 2(
x2 + y2)
+ℓ2
2ω cos
ω
2
= 2(
x2 + y2)
+ 2ℓ2ω2
4
72
T = m(
x2 + y2)
+mℓ2ω2
4V0 = 0, T2 = mx2 +my2
T0 =mℓ2ω2
4,
h = T2 − T0 + V
L = T − V = T − 0 = T
L is independent of t.
cosωtx+ sinωty = 0
ajit = 0
The system is conservative.
Jacobi integral h =n∑
i=1
∂L
∂qiqi − L
=∂L
∂xx+
∂L
∂yy − L
= (2mx) x+ (2my) y −(
mx2 +my2 +mℓ2ω2
4
)
= mx2 +my2 − 1
4mℓ2ω2.
These result show that the toal mass is M.I about the cente of mℓ2/2.
Orthogonal System :
Orthogonal system is an natural system. A natural system is said
to be orthogonal if
i) It is described by standard holonomic Lagrangian equation.
ii) The Lagrangian function L is not an explicit functio of time.
iii) The K.E contain’s only q2i and no cross products in the q′s.
Liouville’s system :
A system is said to be Liouville’s system if
73
i) It is described by standard holomic Lagrange’s equation.
ii) The Lagrange’s function is not an explicit function of time.
iii) Any constraint equation can be expressed in the differential form
n∑
i=1
ajidqi = 0, j = 1, 2, 3, · · · ,m
iv) The K.E. Tand P.E. V are given by
K.E.T =1
2f
n∑
i=1
mi (qi) (qi)2 where mi (qi) > 0
P.E.V =1
f
n∑
i=1
vi (qi)
f =n∑
i=1
fi (qi > 0)
Liouville’s system special case of orthogonal system.
Book Work : Liouville’s system special case of orthogonal system
(OR)
With usual notation show that the following system is separable (or)
Reduce the orthogonal system to quadrature. (Or)
Explain one of the Quadratic system to be orthogonal system.
T =1
2f
n∑
i=1
q2i
V =1
f
n∑
i=1
vi (qi)
f =n∑
i=1
fi (qi > 0)
Proof : Lagrange’s equation is in the form
d
dt
(
∂L
∂qi
)
− ∂L
∂qi= 0
74
f = f1 (q1) + f2 (q2) + · · · + fi (qi) + · · · + fn (qn)
L = T − Vd
dt
(
∂ (T − V )
∂qi
)
− ∂ (T − V )
∂qi= 0
d
dt
(
∂T
∂qi
)
− ∂T
∂qi+∂V
∂qi= 0
d
dt(f qi) −
1
2
∂fi
∂qi
n∑
i=1
q2i +
[
1
f
∂vi
∂qi− 1
f 2
∂fi
∂qi
n∑
i=1
vi (qi)
]
= 0
d
dt(f qi) −
1
2
∂fi
∂qi
n∑
j=1
q2j +
1
f
∂vi
∂qi− v
f
∂fi
∂qi= 0 (1)
This is a natural system.
∴ T + V = h; qi = qj
1
2f
n∑
j=1
q2j + V = h
1
2
n∑
j=1
q2j =
h− V
f
(1) ⇒ d
dt(f qi) −
h− V
f
(
∂fi
∂qi
)
+1
f
(
∂vi
∂qi
)
− V
f
(
∂vi
∂qi
)
= 0
d
dt(f qi) −
h
f
(
∂fi
∂qi
)
+1
f
(
∂vi
∂qi
)
= 0
Multiply 2f qi on both sides
2f qid
dt(f qi) − 2f qi
(
h
f
)(
∂fi
∂qi
)
+2f qif
(
∂vi
∂qi
)
= 0
2f qi
[
d
dt(f qi)
]
− 2hfqi
(
∂fi
∂qi
)
+ 2qi
(
∂vi
∂qi
)
= 0
d
dt
(
f 2q2i
)
− 2h
(
dqidt
)(
∂fi
∂qi
)
+ 2qi
(
∂vi
∂qi
)
= 0
d
dt(f qi)
2 − 2h
(
dqidt
)(
∂fi
∂qi
)
+ 2qi
(
∂vi
∂qi
)
= 0
75
d
dt(f qi)
2 − 2h
(
dqidt
)(
∂fi
∂qi
)
+ 2
(
∂vi
∂qi
)(
dqidt
)
= 0 (2)
d
dt(f qi)
2 − 2h
(
∂fi
∂t
)
+ 2
(
∂vi
∂t
)
= 0
d
dt(f qi)
2 = 2h
(
∂fi
∂t
)
− 2
(
∂vi
∂t
)
= 2d
dt(hfi − vi)
T + V =1
2f
n∑
i=1
q2i + V = h
1
2f
n∑
i=1
q2i + 2
n∑
i=1
vi (qi) = h
Multifly 2f on both sides,
f 2
n∑
i=1
q2i + 2
n∑
i=1
vi (qi) − 2fh = 0 (3)
(2) ⇒
d
dt
(
f 2i q
2i
)
= 2
[
d
dt(hfi − vi)
]
∫
d
dt
(
f 2q2i
)
= 2
∫ [
d
dt(hfi − vi)
]
f 2qi = 2 (hfi − vi) + 2ci
q2i =
2
f 2(hfi + ci − vi)
Let 2ci = f 2q2i − 2fhi + 2vi.
2n∑
i=1
ci = 0 ⇒n∑
i=1
ci = 0 [by (3)]
Here c′is and h together comprise n - independent constant of mo-
tion.
76
q2i =
2
f 2i
[(hfi − vi) + ci]
qi =
√
2 [(hfi − vi) + ci]
f
dqidt
=
√
2 [(hfi − vi) + ci]
fdqi
√
2 [(hfi − vi) + ci]=
dt
f
dq1√
2 [(hf1 − v1) + c1]=
dq2√
2 [(hf2 − v2) + c2]= · · ·
· · · =dqn
√
2 [(hfn − vn) + cn]=dt
f= dτ (4)
Hence required system reduced to quadratures.
Note:
Replacing dqi by√
mi (qi)dqi
T =1
2f
n∑
i=1
mi (qi) q2i where mi (qi) > 0
V =1
f
n∑
i=1
vi (qi)i ; f =n∑
i=1
fi (qi) > 0
A natural system of this type is Liouville’s system.
The equation (4) ⇒
dq1√
Q1 (q1)=
dq2√
Q2 (q2)= · · · =
dqn√
Qn (qn)=dt
f= dτ
Qi (qi) =2
mi
(hfi − vi + ci) , i = 1, 2, 3, · · · , n
f1dq1√
Q1 (q1)=
f2dq2√
Q2 (q2)= · · · =
fndqn√
Qn (qn)=f · dtf
77
f1dq1√
Q1 (q1)+
f2dq2√
Q2 (q2)+ · · · + fndqn
√
Qn (qn)=
(f1 + f2 + · · · + fn)
fdt
=f
fdt = dt
∴ dt =n∑
i=1
fidqi√
Q1 (qi)= t+ β1,whereβ1is constant.
Show that spherical pendularn problem can be solved com-
pletly by quadratures.
Solution :
T =1
2m(
ℓ2θ2+ ℓ2φ
2sin2 θ
)
V = mgℓ cos θ
where m is the mass of the particle.
h = T + V, L = T − V
The system is conservative and natural.
h =1
2m(
ℓ2θ2+ ℓ2φ
2sin2 θ
)
+mgℓ cos θ
L =1
2m(
ℓ2θ2+ ℓ2φ
2sin2 θ
)
−mgℓ cos θ
L is an independent on φ,∴ φ is ignorable.
∂L
∂φ= a constant = αφ
1
2mℓ22φ sin2 θ = αφ
φ =2φ
mℓ2 sin2 θ
(1) ⇒ h =1
2mℓ2θ
2+
1
2
mℓ2αφ2 sin2 θ
m2ℓ4 sin4 θ+mgℓ cos θ
=1
2mℓ2θ
2+
αφ2
2mℓ2 sin2 θ+mgℓ cos θ
78
Multiply by 2mℓ2 sin2 θ,
m2ℓ4 sin2 θ θ2+ 2φ2 = 2mℓ2 sin2 θ (h−mgℓ cos θ)
θ =
√
2mℓ2 sin2 θ (h−mgℓ cos θ) − 2φ2
mℓ2 sin θ
θ =dθ
dt.
This values apply dφ
θ∫
θ0
mℓ2 sin θ√
2mℓ2 sin2 θ (h−mgℓ cos θ) − 2φ2
dθ =
t∫
t0
dt = t− t0
And
αφ
mℓ2 sin2 θ= φ
dφ
dt=
αφ
mℓ2 sin2 θ
dφ =αφ
mℓ2 sin2 θdt
(1) ⇒ αφ
mℓ2 sin2 θdt =
mℓ2 sin θdθ
2mℓ2 sin2 θ×
αφ
mℓ2 sin θ(h−mgℓ cos θ) − αφ2
=αφdθ
sin θ√
2mℓ2 sin2 θ (h−mgℓ cos θ) − αφ2
φ∫
φ0
dφ =
θ∫
θ0
αφdθ
sin θ√
2mℓ2 sin2 θ (h−mgℓ cos θ) − αφ2
φ− φ0 =
θ∫
θ0
αφdθ
sin θ√
2mℓ2 sin2 θ (h−mgℓ cos θ) − αφ2
where φ0 = φ (t0) .
Now we have obtained the required four constraints of motion
79
namely αφ, h, t0, φ0.
Discuss probem of the spherical pendulum considering it as
a Lioville’s system.
Solution : We know that
T =1
2m(
ℓ2θ2+ ℓ2φ
2sin2 θ
)
V = mgℓ cos θ
For Lioville’s system
T =1
2f
(
n∑
i=1
mi (qi) q2i
)
V =1
f
n∑
i=1
vi (qi)i
1
2(fθ + fφ) = mθθ
2+mφφ
2
=1
2
[
mℓ2θ2+mℓ2 sin2 θφ
2]
(fθ + fφ)Mθ = mℓ2
(fθ + fφ)Mφ = mℓ2 sin2 θ
Mθ
Mφ
=1
sin2 θ,Mφ = 1,Mθ =
1
sin2 θ
(fθ + fφ) v = v1 (θ) + v (φ)(
Mℓ2 sin2 θ)
×mgℓ cos θ = v1 (θ)
vθ = m2gℓ3 sin2 θ cos θ
v1 (θ) = m2gℓ3 sin2 θ cos θ
fθ + fφ = mℓ2 sin2 θ
fθ = mℓ2 sin2 θ [∵ fφ = 0]
We know that
φi (qi) =2
Mi
(hfi − vi + ci) (i = 1, 2, · · · , n)
φθ =2
Mθ
(hfθ − vθ + cθ)
80
=2
Mθ
[
hmℓ2 sin2 θ −m2gℓ3 sin2 θ cos θ]
= 2 sin2 θ[
Mℓ2 sin2 θ (h−mgℓ cos θ) + cθ]
φφ =2
Mφ
[hfφ − vφ + cθ]
= 2 [h (0) − 0 + cφ]
= 2cφ
Using
f 2φ2
i = 2 [hfi (gi) − vi (qi) + ci]n∑
i=1
ci = 0
= 2 [hfφ − vφ + cφ]
∂L
∂φ= mℓ2φ sin2 θ = αφ
(fθ + fφ)2 φ
2= 2 [h (0) − 0 + cφ]
mℓ2φ sin2 θ = 2cφ
2cθ = −2cφ = −2φ2
2cφ = −2cθ =(
mℓ2φ sin2 θ)2
= αφ2
n∑
i=1
∫
fidqi√
φi (qi)= t+ β1
θ∫
θ0
fθdθ√
φθ (θ)= t− t0
θ∫
θ0
mℓ2 sin2 θ√
φθ (θ)dθ = t− t0
We have
∫
dq1√
φ1 (q1)−∫
dqi√
φj (qj)= βj, j = 1, 2, 3, · · · , n
81
θ∫
θ0
dθ√
φ0 (θ)=
φ∫
φ0
dφ√
2cφ
[
∵ φφ = 2cφ]
θ∫
θ0
αφdθ√
φ0 (θ)= φ− φ0
Now we have obtained four required constraint cθ, cφ, φℓ, φ0.
Book work :Small Oscillations
Obtain the equation of the natural system by given a small
motion about its equilibrium position. OR
Obtain the equation of natural system in matrices form as
mq + kq = 0
Proof :
Let the natural system be given by ‘n’ independent generalized co-
ordinates q1, q2, q3, · · · · · · ., qn . Let us assume all the q′s all measurable
from a position of equilibrium.
Consider small motion about this equilibrium position.
Let v0 be the reference point is zero. The P.E. can be written in
the form
∴ V =1
2
n∑
i=1
n∑
j=1
(
∂2v
∂qi∂qj
)
0
qiqj = 0.
Neglecting terms of higher p order terms the second in the q′s we obtain
V =1
2
n∑
i=1
n∑
j=1
(
∂2v
∂qi∂qj
)
0
qiqj
V =1
2
n∑
i=1
n∑
j=1
kijqiqj where kij =∂2v
∂qi∂qj
is the stiffness co-efficient.
V is the homogenous as quoctratic of the q′s for small motion near.
Small motion near a positon of equilibrium.
Let us assume the system consists N - particles with caresian Co-
82
ordinates x1, x2, · · · .., x3N then the K.E of the system is given by
T =1
2
n∑
i=1
n∑
j=1
mij qiqj
where mij = mji =3N∑
k=1
mk
(
∂xk
∂qi
)(
∂xk
∂qj
)
L = T − V
=1
2
n∑
i=1
n∑
j=1
mij qiqj −1
2
n∑
i=1
n∑
j=1
kijqiqj (1)
The standard Lagrange’s equation for a holomic system is
d
dt
(
∂L
∂qi
)
− ∂L
∂qi= 0, i = 1, 2, 3, · · · , n
(1) ⇒n∑
j=1
mij qj +n∑
j=1
kijqj = 0
mj qj + kijqj = 0
∴ mq + kq = 0
This is the equation of natural system
Memory point : i) For a spherical pendulum
L = T − V
T =1
2m(
ℓ2θ2+ ℓ2φ
2sin2 θ
)
V = mgℓ sin θ
ii) For double pendulum
v1 = ℓθ, v2 =
√
ℓ2θ2+ ℓ2φ
2+ 2ℓθφ cos (φ− θ)
T =1
2m(
v21 + v2
2
)
83
=1
2mℓ2θ
2+
1
2m[
ℓ2θ2+ ℓ2φ
2+ 2ℓθφ cos (φ− θ)
]
V = −mgℓ (2 cos θ + cosφ)
iii) For a small tube bent is
T =1
2m(
r2θ2+ r2ω2 sin2 θ
)
L = mgr cos θ
iv) For mass storing system
T =1
2mr2ρ2 +
1
2m
·
~ρ+·
~rPm··
~ρC
=1
2m
·
~r2
P +1
2m
·
~ρ2
+·
~rP
·
~rP = aω,·
~ρ = r2 + r2θ2
·
~rPm··
~ρ = maω[
r sin (θ − ωt) + rθ cos (θ − ωt)]
T =1
2ma2ω2 +
1
2m(
r2 + r2θ2)
+maω[
r sin (θ − ωt) + rθ cos (θ − ωt)]
V =1
2k (r − r0)
2
v) For Kepler’s Probem
T =1
2
(
r2 + r2θ2)
V = −µr, L = T − V
R = L− βθ
vi) Two particles each of masses m find Jjacobi integral.
(x1 − x2)2 + (y1 − y2)
2 = ℓ2
tanωt =y2 − y1
x2 − x1
(y2 − y1) = (x2 − x1) tanωt
84
UNIT III
Hamilton’s Equations
Stationary Value :
To find the stationary values of the function f 6= z subject to
constraints
φ1 = x2 + y2 + z2 − 4 = 0
φ2 = xy − 1 = 0
combined force F = f + λ1φ1 + λ2φ2.
Solution:
Let augmented function be
F = z + λ1φ1 + λ2φ2
= z + λ1
(
x2 + y2 + z2 − 4)
+ λ2 (xy − 1)
∂F
∂x= 2xλ1 + λ2y = 0 (1)
∂F
∂y= 2yλ1 + λ2x = 0 (2)
∂F
∂z= 1 + 2zλ1 = 0 (3)
∂F
∂λ1
= x2 + y2 + z2 − 4 = 0 (4)
∂F
∂λ2
= xy − 1 = 0 (5)
(2)
(1)⇒ 2y
2x=
−x−y
⇒ x2 = y2 ⇒ x = ±y ⇒ y = ±x
(5) ⇒ xx− 1 = 0
x2 − 1 = 0
x2 = 1
85
x = ±1
If x = 1, y = 1 and if x = −1, y = −1
(4) ⇒ x = 1, y = 1 ⇒ 1 + 1 + z2 − 4 = 0
z2 = 2 ⇒ z = ±√
2
Here solving the equation, we have stationary at(
1, 1,√
2)
,(
1, 1,−√
2)
,(
−1,−1,√
2)
and(
−1,−1,−√
2)
.
(3) ⇒ 1 + 2zλ1 = 0 ⇒ λ1 =−1
2√
2if z =
√2
If z = −√
2, λ1 =1
2√
2, λ1 = ± 1
2√
2
(2) ⇒ 2 (1)1
2√
2+ λ2 = 0
λ2 = − 1√2, λ2 =
1√2
λ2 = ± 1√2
It is seen that(
1, 1,√
2)
,(
−1,−1,√
2)
are the constraints of maxi-
mum oints and the other two points are minimum points.
Euler Lagrage equation or stationary values a definite inte-
gral.
Solution : Suppose we wish to find the stationary of
I =
x1∫
x0
f [y (x) , y′ (x) , x] dx
where y′ (x) =dy
dxand the limits are fixed.
Let us assume that derivatives f (y, y′, x) has two continuous deriva-
tives in each of its arguments. We have to find a function y′ (x) which
86
gives a stationary value for I we’ve
y (x) = y′ (x) + δy (x)
where δy (x) is small variation in y. But small variation
δy = αη (x)
∴ δy (x) = αη (x)
∴ y (x) = y∗ (x) + 2η (x)
where η (x) is a arbitary function having the required smoothness and
α is an parameter which does not independent of x.
Hence for any given η (x) we can consider the varied curve y to be
function of α and x.
i.e., y (α, x) = y∗ (x) + αη (x) .
Let us assume the variation δy = 0 at the end points
2η (x0) = 0 and αη (x1) = 0
∴ y (x0) , y (x1) are fixed.
We see that the integral I is a function of α only for any given
η (x) .
I having a stationary value y∗ (x)
δI =
(
dI
dα
)
α=0
= 0
∴ x0, x1 are independent on α
y (x, α) = y∗ (x) + αη (x)
∂y
∂α= η (x) ;
∂y′
∂α= η′ (x)
87
I =
x∫
x0
f (y, y′, x) dx
dI
dx=
x∫
x0
(
∂f
∂y
∂y
∂α+∂f
∂y′∂y′
∂x
)
dx = 0
x∫
x0
∂f
∂yη (x) dx+
x∫
x0
∂f
∂y′η′ (x) dx = 0
Since η (x0) = 0, η (x1) = 0, we have
x∫
x0
∂f
∂yη (x) dx+
[
η (x)∂f
∂y′
]x
x0
−x∫
x0
η (x)d
dx
(
∂f
∂y′
)
dx = 0
x∫
x0
∂f
∂yη (x) dx+
[
η (x)∂f
∂y′
]x
x0
−x∫
x0
η (x)d
dx
(
∂f
∂y′
)
dx = 0
x∫
x0
∂f
∂yη (x) dx−
x∫
x0
η (x)d
dx
(
∂f
∂y′
)
dx = 0
Since η (x) is arbitary we have
∂f
∂y− d
dx
(
∂f
∂y′
)
= 0
⇒ d
dx
(
∂f
∂y′
)
− ∂f
∂y= 0
This equation is known as Euler- Lagrage’s equation for any curve
y = y∗(x) which result in a stationary value of I.
Branchistochrone Probem :
It is a find a curve y (x) between origin ‘O’ and the point (x1, y1)
such that a particle starting from rest at ‘O’ and sliding down the
curve without friction under the influence of uniform gravity field in a
minimum time to reach the end of the curve.
88
Solution :
The vertical distance described by the particle to corresponding to
the point (x1, y1) is x with velocity.
Let P be the position of the particle at time with velocity v.
From principle of conservation of energy we obtain
mgx =1
2mv2
v = ±√
2gx
Let t be the time taken to reach the point (x1, y1) is found by noting
first that an infinitesimal path element ds is given by from O
ds = ±√
1 + y′2dx
t =
s∫
0
ds
v=
x1∫
0
√
1 + y′2√2gx
dx.
Let f (y, y′, x) =
√
1 + y′2
2gx. The Euler Lagrang equation is
∂F
∂y− d
dx
(
∂F
∂y′
)
= 0 (1)
∂F
∂y= 0
0 − d
dx
(
∂F
∂y′
)
= 0
∂F
∂y′= C where C is constant
∂F
∂y′=
∂
∂y′
√
1 + y′2
2gx
= C
89
⇒ 1
2
√
1 + y′2
2gx
· 2y′
2gx= C
y′√
2gx√
1 + y′2= C
y′2 = C2[
2gx(
1 + y′2)]
= 2C2gx+ 2gxC2y′2
=(
1 − 2gxC2 − 2gxC2)
y′2 − 2gxC2y′2 = 2C2gx⇒ y′2(
1 − 2gxC2)
= 2C2gx
y′2 =2C2gx
(1 − 2gxC2)
y′ =
√
2C2gx
(1 − 2gxC2)
Let x = a (1 − cos θ) where a =1
4gC2.
y′ =
√
√
√
√
√
2gC2 · 1
4gC2
(1 − cos θ)
1 − 2gC21
4gC2(1 − cos θ)
√
√
√
√
√
√
1
2(1 − cos θ)
1 − 1
2(1 − cos θ)
=
√
√
√
√
√
√
1
2(1 − cos θ)
1 − 2gC21
4gC2(1 − cos θ)
√
1 − cos θ
2 − cos θ=
√
1 − cos θ
1 + cos θ=
√
√
√
√
√
√
2 sin2θ
2
2 cos2θ
2
dy
dx=
sinθ
2
cosθ
2
⇒ dy =sin
θ
2
cosθ
2
dx
dy =sin
θ
2
cosθ
2
a sin θdθ
90
=sin
θ
2
cosθ
2
a2 sinθ
2cos
θ
2dθ
= 2a sin2 θ
2dθ
∫
dy =
∫
2a sin2 θ
2dθ = a
∫
(1 − cos θ) dθ
y = −a (θ − sin θ) + c
At x = y = 0, θ = 0 ⇒ c = 0.
Path is y = a(θ − sin θ), x = a (1 − cos θ) which is cycloid. The
constant ‘a’ is choosen δt the poth always through the point (x1, y1).
By comparing other paths through ‘O’ and (x1, y1) we can prove only
the cycloid path has minimum time.
Note:
y′2 − 1 − y′2√α− y
√
1 + y′2= c
−1√α− y
√
1 + y′2= c
−√
1 + y′2√α− y =
1
c
Taking square on both sides
(
1 + y′2)
(α− y) =1
c2= c1
y′2 =c1
α− y− 1 =
c1 − α+ y
α− y
y′ = ±√
c1 − α+ y
α− ytake y′ = −
√
c1 − α+ y
α− y
dy
dx= −
√
c1 − α+ y
α− y√α− y√
c1 − α+ ydy = dx (1)
91
Take α− y = c1 sin2θ
2, dy = −c12 sin
θ
2cos
θ
2dθ
2Geodesic Problem :
Definition :
The Problem of finding the shortest path between two points in a
given space is called geoclesic.
The shortest path between two points is given space is the path
which forms a greaest circle which is the geodesic.
To prove this let us consider the equation of the problem.
Find the geodesic for the spherical surface.
(or)
Find the path of minimum length between two given points on the
two dimensional surface of a sphere of radius ‘r’
Let the spherical co-ordinate, (θ, φ) of the point of spherical surface.
Proof:
Let us use the spherical co-ordinate (θ, φ) as variable since r is a
constant.
We know that
x = r sin θ cosφ
y = r sin θ sinφ
z = r cos θ
The differential elements of length ds is given by
ds2 = r2dθ2 + r2 sin2 θdφ2
= r2dθ2
[
1 + sin2
(
dφ
dθ
)2]
ds = r
√
1 + sin2
(
dφ
dθ
)2
dθ
s = r
θ′∫
θ0
√
1 + φ2
1 sin2 θdθ where φ′ =dφ
dθ.
92
Let f (φ, φ′, θ) =√
1 + φ′2 sin2 θ.We want to find the shortest dis-
tance of the path P and Q.
∴we have to find the stationary value of Euler - Lagrangian equa-
tion is∂f
∂y− d
dx
(
∂f
∂y′
)
= 0.
But here f is a function of (φ, φ′, θ) .
The new Euler’s Lagrangian equation for thus problem is
∂f
∂φ− d
dθ
(
∂f
∂φ′
)
= 0.
But f =√
1 + φ′2 sin2 θ.
∴
∂F
∂φ= 0,
∂f
∂φ′=
1
2√
1 + φ′2 sin2 θ2φ′ sin2 θ
∴ Using (1) Euler’s Lagrangian equation
d
dθ
(
∂f
∂φ′
)
= 0
∂f
∂φ′= c
∂F
∂φ′=
φ′ sin2 θ√
1 + φ′2 sin2 θ= c
φ′ sin2 θ = c
√
1 + φ′2 sin2 θ
Squaring and rearranging we get
sin4 θφ′2 = c2(
1 + φ′2 sin2 θ)
sin2 θφ′2(
sin2 θ − c2)
= c2
φ′2 =c2
sin2 θ(
sin2 θ − c2)
φ′ =c
sin θ√
sin2 θ − c2
φ =
∫
c
sin θ√
sin2 θ − c2dθ
93
=
∫
c
sin2 θ
√
1 − c2
sin2 θ
dθ
φ =
∫
c
sin2 θ√
1 − c2 cosec2 θdθ
=
∫
c cosec2 θ√1 − c2 cosec2 θ
dθ
= c
∫
cosec2 θ√
1 − c2 (1 + cot2 θ)dθ
= −c∫
d (cot θ)√1 − c2 − c2 cot2 θ
= −c∫
d (cot θ)
c
√
(
1 − c2
c2
)
− cot2 θ
φ = cos−1
cot θ√
(
1 − c2
c2
)
+ φ0
= cos−1
cot θ√
(
1 − c2
c2
)
= cos−1
(
c cot θ√1 − c2
)
cos (φ− φ0) =c√
1 − c2cot θ
=c cot θ√1 − c2
cosφ cosφ0 − sinφ sinφ0 =c√
1 − c2cos θ
sin θ
Multiply by r sin θ,
r sin θ cosφ cosφ0 + r sin θ sinφ sinφ0 =c√
1 − c2cos θ
sin θr sin θ
=c√
1 − c2r cos θ
94
The above equation becomes
x cosφ0 + y sinφ0 =c√
1 − c2z
Then the equation of the form ax + by + cz = 0 . This represents
the plane through the origin. This plane intersects the sphere along a
great circle which is geodesic.
Here φ0 and c are two constants which are adjustable such that the
greats circle will pass through P and Q.
General Euler’s Lagrangian equation :
Let
I =
x1∫
x0
f (y1, y2, · · · , yn, y′
1, y′
2, · · · , y′n, x) .
We have found a necessary condition for a stationary value of I.x0, r
are fixed limits.
State and prove Hamilton’s principle :
Variational principle or integration principle[Integrated form
of D’Alemberts principle]
Statement :
The actual path in configuration space followed by a holonomic
dynomical system during the fixed interval t0 to t1 is such that the
integral I =
t1∫
t0
Ldt is the stationary value with respect to path vari-
ation which vanish at the end points δI = 0. Here L is a function
L (q, q′, t) .q is a n dimensional point q1, q2, · · · , qn.
Proof :
Let ~r1, ~r2, · · · , ~rN be the position vectors of the N particles.
By the form of D’ Alembert’s principle
~r
N∑
i=1
(
~Fi −mi
··
~ri
)
δ~ri = 0
95
where ~Fi is the applied force acting on the ith particle.
N∑
i=1
~Fiδri −N∑
i=1
mi
··
~riδ~ri = 0 (1)
Virtual work of A.F.workdone = δW =N∑
i=1
~Fiδ~ri.
Now variation in K.E.
δT = δ
[
1
2
N∑
i=1
mi
·
~ri
]
=1
2
N∑
i=1
mi2·
~riδ·
~ri
δT =N∑
i=1
mi
·
~riδ·
~ri (2)
Butd
dt
[
N∑
i=1
mi
·
~riδ~ri
]
=N∑
i=1
mi
··
~riδ~ri +N∑
i=1
mi
·
~riδ·
~ri (3)
whered
dt(δ~ri) = δ
·
~ri = δW + δT
From (2) and (1)
d
dt
[
N∑
i=1
mi
·
~riδ~ri
]
=N∑
i=1
mi
··
~riδ~ri +N∑
i=1
miδ·
~ri
·
~ri
= δT +N∑
i=1
~Fiδ~ri
⇒ δT + δW =d
dt
[
N∑
i=1
mi
·
~riδ~ri
]
(∵ equation (3))
⇒ (δT + δW ) dt = d
[
N∑
i=1
mi
·
~riδ~ri
]
96
t1∫
t0
(δT + δW ) dt =
t1∫
t0
d
[
N∑
i=1
mi
·
~riδ~ri
]
=
[
N∑
i=1
mi
·
~riδ~ri
]t1
t0
= 0
Now let us assume that the configuration of the system is fixed at the
time t0 to t1 implying the variation δ~ri are zero at these times.
⇒N∑
i=1
[
mi
·
~riδ~ri
]t1
t0
= 0
⇒t1∫
t0
(δT + δW ) dt = 0 (4)
For a given virtual displacement and a time the value of δT and
δW are independent of these Co-ordinates system.
This case of generalized forces equation (4) can be written in the
formt1∫
t0
(
δT +N∑
i=1
Qiδqi
)
dt = 0 (5)
Version of Hamilton’s principle :
Let us assume all the applied forces and derivable from a potentaial
function V (q, t) .
∴ δW = −δV
(4) ⇒t1∫
t0
(δT + δW ) dt =
t1∫
t0
(δT − δV ) dt = 0
=
t1∫
t0
δ (T − V ) dt = 0
=
t1∫
t0
δLdt = 0
97
= δ
t1∫
t0
δLdt
δI = 0 ⇒ δ
t1∫
t0
Ldt = 0 is Hamilton’s principle.
Hamilton principle and lagrangian equation are equiva-
lent.
(Or)
Derive the lagrangian equation from Hamilton principle.
Solution :
By Hamilton’s principle
δI = 0
where I =
t1∫
t0
L (q, q, t) dt.
Hence L (q, q, t) corresponds to f(y, y, x). Assuming δq′s are inde-
pendent and using Euler Lagrangian equation takes the form
∂L
∂qi− d
dt
(
∂L
∂qi
)
= 0
d
dt
(
∂L
∂qi
)
− ∂L
∂qi= 0 for all i = 1 to n.
Hamilton’s equation or Hamilton’s cononical equation of
motion :
Consider a holonomic system which can be described by the stan-
dard form by of lagrangian equation, namely.
d
dt
(
∂L
∂qi
)
− ∂L
∂qi= 0 , i = 1 to n. (1)
The generalized momentum conjugate to qi is given by
pi =∂L
∂qi(2)
98
(1) ⇒ d
dt(pi) −
∂L
∂qi= 0
d
dt(pi) =
∂L
∂qi∂L
∂qi= pi (3)
Now, let us define the Hamilton’s function H (q, p, t) for the system
as
H (q, p, t) =n∑
i=1
piqi − L (q, q, t) (4)
H is an explict function of q′s and p′s and t
H (q, p, t) =N∑
i=1
piqi − L (q, q, t)
δH =n∑
i=1
∂H
∂qiδqi +
n∑
i=1
∂H
∂pi
δpi +∂H
∂tδt
=n∑
i=1
piδqi +n∑
i=1
∂H
∂pi
δpi −n∑
i=1
∂L
∂qiδpi −
n∑
i=1
∂L
∂qiδqi −
∂L
∂tδt
n∑
i=1
∂H
∂qiδqi+
n∑
i=1
∂H
∂pi
δpi+∂H
∂tδt =
n∑
i=1
piδqi+n∑
i=1
qiδpi−n∑
i=1
piδqi−∂L
∂tδt
Equating co-efficients on both sides
δpi ⇒ ∂H
∂pi
= qi
δqi ⇒ ∂H
∂qi= −pi;
∂L
∂t= −∂H
∂t
∂H
∂pi
= qi;∂H
∂qi= −p (5)
∂L
∂t= −∂H
∂t
The 2n first order Lagrangian equations is given the set of equation
(5) is called Hamilton equationn of motion.
Hamiton’s funciton is generalized quadratic in the p’s or
99
the form of Hamiltons’s function.
Proof :
We have
∂T
∂qi= pi =
n∑
j=1
mij (q, t) qj + ai (q, t)
=n∑
j=1
mij qj + ai
n∑
i=1
piqi =n∑
i=1
n∑
j=1
mijqiqj +n∑
i=1
qiqi = 2T2 + T1
We know that
H =n∑
i=1
piqi − L (q, q, t)
=n∑
i=1
piqi − (T − V )
= 2T2 + T1 − T1 − T2 − T0 + V
= T2 − T0 + V
T2 =1
2
n∑
i=1
n∑
j=1
mij qiqj.
This can be put in the matrix form as follows
T2 =1
2mT qq.
But q = b (p− a) bm−1. Both b,m are symmetric matrix.
∴ bT = b,m = mT
∴ T 2 =1
2bT (p− a)T bb−1 (p− a)
=1
2b (p− a) (p− a)T
100
Expanding this
T2 =1
2
n∑
i=1
n∑
j=1
bijpipj −n∑
i=1
n∑
j=1
bijaipi +1
2
n∑
i=1
n∑
j=1
bijaiaj
T0, V are function of q′s and t adding we have
H (q, p, t) =1
2
n∑
i=1
n∑
j=1
bijpipj−n∑
i=1
n∑
j=1
bijaipi+1
2
n∑
i=1
n∑
j=1
bijaiaj−T0+V
Grouping term by their degree
H = H2 +H1 +H0
H2 =1
2
n∑
i=1
n∑
j=1
bijpipj
H1 = −n∑
i=1
n∑
j=1
bijpjai
H0 =1
2
n∑
i=1
n∑
j=1
bijaiaj − T0 + V
Hence the Hamilton’s function is in general quadratic in p’s.
The Hamilton function of scleronomic system is equal to
toal energy.
Let us consider a system of cartesian co-ordinates do not certain
time t explicity.
It follows that a’s all zero and T = T2.
In sceleronomic system T0 = T1 = 0.
H2 = T2
∴ T1 = T2 and H1 = 0;H0 = V
i.e., H = T + V
101
∴The Hamilton’s function of sceleronomic system is equal to T.K.
∴ H (q, p, t) =1
2
n∑
i=1
n∑
j=1
bijpipj + V (q, t)
Non - holonomic system has a constant Hamilton’s func-
tion.
Solution :
Let us have a conservative holonomic system we haveH = H (q, p, t) .
dH
dt= H =
n∑
i=1
([
∂H
∂qi
∂qi∂t
]
+
[
∂H
∂pi
∂pi
∂t
])
+
[
∂H
∂t
∂t
∂t
]
H =n∑
i=1
([
∂H
∂qiqi
]
+
[
∂H
∂pi
pi
])
+∂H
∂t
Using Hamilton’s canonical equation
∂H
∂pi
= qi,∂H
∂qi= −pi (A)
and∂H
∂t= −∂L
∂t
∴ H =n∑
i=1
(−piqi + piqi) −∂L
∂t= −∂L
∂t.
If the system is conservative. L does not contain t explicity.
∴
∂L
∂t= 0 ⇒ H = 0
∴ H = C ( constant)
Hence the proof.
Phase space :
Hamilton’s canonical eqn of motion consisting of a set of 2n first
order equation giving q, p as function q’s, p’s are after considered as
components of a single vector X.
Thus the equation of motion for a standard holonomic system can
102
be written as x = X (x, t) where X is a 2n dimensional vector consist-
ing of nq’s and np’s.
Consider the q’s as x1, x2, · · · , xn and p’s as xn+1, xn+2, · · · , x2n.This
2n dimensional x space is called phase space.
Extended phase space :
In a non-holonomic conservative system, let us consider the variable
qn+1 as choose a parameter q as the new independent variable. This
system consist of (n+ 1)q’s and corresponding (n+ 1)p’s.
This space is called extened phase space.
Autonomous - Non autonomous :
In a conservative system H is not explicit function of time and
configuration of motion of form x = X (x) .
A holonomic system described by equation of this form is called
autonomous where as x = X (x) is called non autonomous.
Liouville’s theorem :
The phase fluid incompressibel.
Proof :
Let a holonomic system be described by n independents q’s.
Let us consider a gropp of phase p’s described fraectorcies in a 2n
dimensional phase space.
Considering the moving points in a small elementry volume dV =
dq1, dq2, · · · , dqn; dp1, dp2, · · · , dpn.
Consisting the moving points of a fluid being called phase fluid.
The phase velocity v of a fluid. Particle is given by the 2n component
(qi, pi) these can be expresed as function of q’s and p’s by canonical
equation
qi =∂H
∂pi
, pi = −∂H∂qi
As a given volume elements of the phase fluid moves it will ingeneral
change in it shape. But the neighbouring particles will remain close to
each other. We can show that volume of each phase fluid is constant
103
during the motion.
∇~v =n∑
i=1
[
∂
∂qiqi +
∂
∂pi
pi
]
=n∑
i=1
[
∂
∂qi
(
∂H
∂pi
)
+∂
∂pi
(
−∂H∂qi
)]
=n∑
i=1
[
∂2H
∂qi∂pi
− ∂2H
∂pi∂qi
]
= 0
∇~v = 0
A geometric interpretation of this results that the phase fluid is
incompressible.
Kepler’s problem :
Use the Jacobi form of the principle of least action. Obtaing the
orbit for the Kepler’s problem.
Solution :
Let a particle of mass m be attracted to a fixed point O by an
inverse square force.
V = −µmr, Fr = −µm
r2
T =1
2mv2 =
1
2m(
r2 + r2θ2)
h = T + V
=1
2m(
r2 + r2θ2)
− µm
r(1)
By Jacobi form of principle of least action
δ
∫
√
2 (h− v)ds = 0
⇒ δ
∫
√
2(
h+µm
r
)
ds = 0 (2)
104
we have ds2 = m(
dr2 + r2dθ2)
(
ds
dθ
)2
= m
[
(
dr
dθ
)2
+ r2
]
= m(
r′2 + r2)
ds
dθ=
√
m (r′2 + r2)
ds =√
m (r′2 + r2)dθ
⇒ δ
∫√
2(
h+µm
r
)
√
m (r′2 + r2)dθ = 0
⇒ δ
∫
√
2m(
h+µm
r
)
(r′2 + r2)dθ = 0
where the end points this is the form of
δ
θ∫
θ0
f (r, r′) dθ = 0
where f (r, r′) =
√
2m(
h+µm
r
)
(r′2 + r2).
Applying the Euler’s Lagrange’s equation for f (r, r′) we have
d
dθ
(
∂f
∂r′
)
− ∂f
∂r= 0.
The corresponding energy integral
∑ ∂L
∂qiqi − L = h
∂f
∂r′r′ − f = C ( constant)
f (r, r′) =
√
2m (r′2 + r2)(
h+µm
r
)
105
∂f
∂r′=
2m(
h+µm
r
)
2r′
2
√
2m (r′2 + r2)(
h+µm
r
)
=2m(
h+µm
r
)
r′
√
2m (r′2 + r2)(
h+µm
r
)
=
√
√
√
√
2m(
h+µm
r
)
(r′2 + r2)r′
r′∂f
∂r′− f = C
√
√
√
√
2m(
h+µm
r
)
(r′2 + r2)r′ · r′ −
√
2m (r′2 + r2)(
h+µm
r
)
= C
√
2m(
h+µm
r
)
[
r′2√
(r′2 + r2)−√
(r′2 + r2)
]
= C
√
2m(
h+µm
r
)
[
r′2 − r′2 − r2
√
(r′2 + r2)
]
= C
−
√
2m(
h+µm
r
)
√
(r′2 + r2)r2 = C (3)
Now
h = T + V
=1
2m(
r2 + r2θ2)
− µm
r
m(
r2 + r2θ2)
= 2(
h+µm
r
)
⇒ m2
(
r2 + r2θ2)
= 2m(
h+µm
r
)
(4)
r =dr
dt=dr
dθ
dθ
dt
106
r =dr
dt= r′θ
θ. =r
r′
r = r′θ
(4) ⇒ m2
(
r′2θ2+ r2θ
2)
= 2m(
h+µm
r
)
m2θ2 (
r′2 + r2)
= 2m(
h+µm
r
)
In (3) use this
−
√
√
√
√
2m(
h+µm
r
)
(r′2 + r2)r2 = C
−
√
m2θ2(r′2 + r2)
(r′2 + r2)r2 = C
−√
m2θ2r2 = C
−mθr2 = C
θ = − C
mr2
Angular momentum is constant
θ2
=C2
m2r4
To find the equation of orbit
(3) ⇒ −
√
√
√
√
2m(
h+µm
r
)
(r′2 + r2)r2 = C
Squaring and rearranging we get
2m(
h+µm
r
)
r4 = C2(
r′2 + r2)
C2r′2 = 2m(
h+µm
r
)
r4 − C2r2
107
r′2 =2mr4
Cdr2
(
h+µm
r
)
− r2
=2mr2
C2
[
hr2 + µmr − C2
2m
]
(
dr
dθ
)2
=2mr2
C2
[
hr2 + µmr − C2
2m
]
dr
dθ=
√
2mr2
C2
[
hr2 + µmr − C2
2m
]
dr√
2mr2
C2
[
hr2 + µmr − C2
2m
]
= dθ
∫
dθ =C√2m
∫
dr√
hr4 + µmr2 − C2r2
2m
θ =C√2m
r∫
r0
dr
r2
√
h+µm
r− C2
2m
=
r∫
r0
d
(
µm2
C2− 1
r
)
√
µ2m4
C4+
2mh
C2−(
µm2
C2− 1
r
)
θ = sin−1
(
µm2
C2− 1
r
)
√
µ2m4
C4+
2mh
C2
r
r0
= sin−1
(
µm2
C2− 1
r
)
√
µ2m4
C4+
2mh
C2
− π
2
sin(
θ +π
2
)
=
µm2
C2− 1
r√
µ2m4
C4+
2mh
C2
108
cos θ
√
µ2m4
C4+
2mh
C2=
µm2
C2− 1
r
1
r=
µm2
C2−√
µ2m4
C4+
2mh
C2cos θ
Multiplying byC2
µm2we get
C
µm2/r= 1 −
√
1 +2hc2
µ2m3cos θ
This is a conic with eccentricity
i.e.,
√
1 +2hc2
µ2m3
(
ℓ
r= 1 + e cos θ
)
To find h :
h =1
2m(
r2 + r2θ2)
− µm
r
At r = r0 = rmin, θ = θ0 = 0
m
2
[
0 + r2θ2
0
]
= h+µm
r0
Also
mr20θ0 − C ⇒ θ0 =
−Cmr2
0
(1) ⇒ m
2
[
r20
C2
m2r40
]
= h+µm
r0C2
2mr20
= h+µm
r0
h =C2
2mr20
− µm
r0
Example Given a holonomic system with Lagrangian form L =1
2m (x2
1 + x22 + x2
3)−mgx0 and a constraint x1 − x2 + x3 = 0. We argumented Lagrangian
function to obtain the differential equation of motion for x1.
109
Solution :
L =1
2m(
x21 + x2
2 + x23
)
−mgx0
constraint:
x1 − x2 + x3 = 0
λ = L+ µg
=1
2m(
x21 + x2
2 + x23
)
−mgx3 + µ (x1 − x2 + x3)
d
dt
(
∂λ
∂x1
)
− ∂λ
∂x1
= 0
we have
d
dt(mx1 + µ) = 0
mx1 +dµ
dt= 0 (1)
d
dt
(
∂λ
∂x2
)
− ∂λ
∂x2
= 0
d
dt(mx2 − µ) = 0
mx2 −dµ
dt= 0 (2)
d
dt
(
∂λ
∂x3
)
− ∂λ
∂x3
= 0
d
dt(mx3 + µ) +mg = 0
mx3 +mg +dµ
dt= 0 (3)
(1), (2) and (3) are the differential equation of motion. Itreating µ
are of the we haved
dt
(
∂λ
∂µ
)
− ∂λ
∂µ= 0
x1 − x2 + x3 = 0 (4)
110
Adding (1) and (2), we get
x1 + x2 = 0 ⇒ x1 = −x2
Adding (2) and (3) we get
x2 + x3 = 0 ⇒ x2 = −x1 (*)
(∗) ⇒ x1 − x3 = g
x3 = −g − x2
= −g + x1
x1 + x1 + x1 − g = 0
3x1 = g
x1 =g
3
(4) ⇒ x1 + x3 = x2
x3 = x2 − x1
= −x1 − x1
= −2x1 = −2g
3
Definition : Action In mechanics action is deficed as
A =
t1∫
t0
n∑
i=1
piqidt.
State and prove Least action or Principle of least action
The actual path of a conservative holonomic system is such that
the action is stationary with respect to varied pathes having the same
energy integral and the same end points in a ‘q’ space.
Proof :
In this case we are considering a more general type of variation. In
111
this variation the point (q + δq, t+ δt) corresponds to (q, t).
Let ‘d’ indicates differential change along individual path.
δ indicates variations in going from actual path to varied path
considering the small quadirlateral in the above figure.
qdt+ δq + dδq = δq + (q + δq) (dt+ dδt)
qdt+ dδq = qdt+ qdδt+ δqdt+ δqdδt
Dividing by dt
dδq = qdδt+ δqdδt+ δqdtd
dt(δqi) = qi
d
dt(δt) + δqi + δqi
d
dt(δt)
Omitting δqid
dt(δt)
d
dt(δqi) = qi
d
dt(δt) + δqi
δqi =d
dt(δqi) − qi
d
dt(δt) (1)
Let I =
t1∫
t0
L (q, q, t) dt.
δI = δ
t1∫
t0
Ldt =
t1∫
t0
δLdt+
t1∫
t0
Ld (δt)
=
t1∫
t0
[
n∑
i=1
∂L
∂qiδqi +
∂L
∂qiδqi
+∂L
∂tδt+ L
d
dt(δt)
]
dt
=
t1∫
t0
[
n∑
i=1
∂L
∂qiδqi +
∂L
∂qi
[
d
dt(δqi) − qi
d
dt(δt)
]
+∂L
∂tδt+ L
d
dt(δt)
]
dt
112
=
t1∫
t0
[
n∑
i=1
∂L
∂qiδqi +
[
∂L
∂qi
d
dt(δqi)
]
− ∂L
∂qiqid
dt(δt)
+∂L
∂tδt+ L
d
dt(δt)
]
dt
Consider
d
dt
[
n∑
i=1
∂L
∂qi
d
dt(δqi)
]
=n∑
i=1
d
dt
(
∂L
∂qi
)
d
dt(δqi)
n∑
i=1
(
∂L
∂qi
)
d
dt(δqi) =
d
dt
[
n∑
i=1
(
∂L
∂qi
)
d
dt(δqi)
]
−n∑
i=1
d
dt
(
∂L
∂qi
)
δqi
Using this value in above equation we get,
δI =
t1∫
t0
[
n∑
i=1
∂L
∂qiδqi +
d
dt
[
n∑
i=1
(
∂L
∂qi
)
δqi
]
−n∑
i=1
d
dt
(
∂L
∂qi
)
δqi
−n∑
i=1
∂L
∂qiqid
dt(δt) +
∂L
∂tδt+ L
d
dt(δt)
]
dt
=
t1∫
t0
d
dt
(
n∑
i=1
∂L
∂qiδqi
)
dt−t1∫
t0
n∑
i=1
[
d
dt
(
∂L
∂qi
)
− ∂L
∂qi
]
δqidt
−t1∫
t0
[
n∑
i=1
∂L
∂qiqi − L
]
d
dt(δt) dt+
t1∫
t0
∂L
∂tδtdt (*)
= I1 − I2 − I3 + I4
where
I1 =
t1∫
t0
d
dt
(
n∑
i=1
∂L
∂qiδqi
)
dt
⇒[
n∑
i=1
∂L
∂qiδqi
]t1
t0
= 0
113
I2 =
t1∫
t0
n∑
i=1
[
d
dt
(
∂L
∂qi
)
− ∂L
∂qi
]
δqidt = 0
I4 =
t1∫
t0
∂L
∂tδtdt = 0
I3 = −t1∫
t0
[
n∑
i=1
∂L
∂qiqi − L
]
d (δt) = δI
Every varied path having the same evergy integral h.
n∑
i=1
∂L
∂qiqi − L = h
δI = −t1∫
t0
hd (δt) = −h (δt)t1t0
δI = −h (δt1 − δt0) (a)
Let us define the action
A =
t1∫
t0
n∑
i=1
piqidt
=
t1∫
t0
n∑
i=1
∂L
∂qiqidt
=
t1∫
t0
(L+ h) dt
δA = δ
t1∫
t0
Ldt+ δ
t1∫
t0
hdt
= δI +
t1∫
t0
δhdt+
t1∫
t0
hd (δt)
= δI + δh (t1 − t0) + h (δt1 − δt0)
= δI + δh (t1 − t0) − δI using (a)
114
h is same as in the actual path and varied path δh = 0
∴ δA = 0
∴ The P.L.A. δA = δ
t1∫
t0
n∑
i=1
piqidt = 0
Hence the theorem.
Derive Hamilton principle form of least action
Solution :
Write down upto form the above.
Let us assume that the variation be contemperaneous. In this case
δt = 0.
∴ Second integral vanished.
Let us assume all the applied force are derivable from a potential
function.
Hence the least integral vanishes.
Let us assume all the varied paths have fixed end points in a q
space.
Hence the I, integral vanishes.
∴ δI = 0
This is Hamilton principle.
Obtain Jacobis form of principle of least action
Solution :
Write the bookwork Least action completely let us assume that the
system be natural.
Now
n∑
i=1
piqi − L = H
n∑
i=1
piqi = L+H
= T2 − T0 + V + T − V
115
= T2 − T0 + T
= T2 − T0 + T2 + T0 + T1
= 2T2 + T1 [∵ T + v = h]
= 2T + 0
n∑
i=1
piqi = 2T
For the natural system T2 = T, T1 = 0, T0 = 0.
Least action = δA = δ
t1∫
t0
n∑
i=1
piqidt = 0
A =
t1∫
t0
2Tdt = 0
δA = δ
t1∫
t0
2Tdt = 0
Define ds as follows
ds2 =∑
i
∑
j
mij qiqjdt2
= 2Tdt2
ds =√
2Tdt
δA = δ
t1∫
t0
√2T
√2Tdt
= δ
t1∫
t0
√2Tds
= δ
t1∫
t0
√
2 (h− v)ds = 0
This is the principle of least action in Jacobi’s theorem.
116
Problem :
Given a mass spring system consisting of a mass m and a linear
spring of stiffness of k as shown in gigure. Find the equation of motion
using the Hamilton’s procedure. Assume that the displacement x is
measure from the unstressed position of the spring.
Solution :
First we find the K.E and P.E in the usual form for mass spring
system we obtain
T =1
2mx2
V =1
2x2k
where k is the stiffness.
But we know that
L = T − V =1
2mx2 − 1
2x2k
Angular momentum P =∂L
∂x=
1
2m2x
= mx [ here x is only the co-ordinates q = x]
P = mx
x =P
m
T =1
2mx2 =
P 2
2m
The Hamilton function is
H (x, p) = px− L
= p (pm) −[
1
2m (pm)2 − 1
2kx2
]
=p2
m− p2
2m+
1
2kx2
117
=p2
m+
1
2kx2 (1)
H (q, p, t) = H (x, p)
Using Hamilton canonical equation
qi =∂H
∂p; x =
∂H
∂p=
2
2mp =
p
m
⇒ p = −∂H∂x
= −(
1
2k · 2x
)
= −kx
Using (1)
∴ p = −kx (2)
But
p = mx
p = mx (3)
From (2) and (3)
mx = −kxmx+ kx = 0
This is equation of the mass spring system.
Kepler’s problem using Hamilton canonical function.
A particle of mass m is attached to a fixed point ‘O’ by an inverse
square force that is Fr = −µmr2
where µ is the gravitational co-efficient
using the plane co-ordinates (r, θ) to describe the position of the par-
ticle.
Find the equation of motion
qi =∂H
∂pi
; p i = −∂H∂qi
.
118
Solution:
T =1
2m(
r2 + r2θ2)
V =
∫
Frdr =
∫
−µmr2dr
The Lagrange’s function
L = T − V
H = T + V
=1
2m(
r2 + r2θ2)
− µm
r
=1
2m
(
p2r
m2+ r2 p2
θ
m2r4
)
− µm
r
=p2
r
2m+
p2θ
2mr2− µm
r
Using the Hamilton equation we have
r =∂H
∂pr
=2pr
2m=pr
m(1)
pr = −∂H∂r
= −[
1
2mp2
θ (−2) r3 +µm
r2
]
=p2
θ
mr3− µm
r2(2)
From (1)
r =pr
m⇒ r =
pr
m
pθ = −∂H∂θ
⇒ r − 1
m
[
p2θ
mr3
]
+1
m
[µm
r2
]
= 0
⇒ mr − p2θ
mr3+µm
r2= 0
119
pθ is constant say β
⇒ mr − β2
mr3+µm
r2= 0
pr = mr
pr = mr (3)
From equation (1) and (2)
mr =β2
mr3− µm
r2
mr − β2
mr3+µm
r2= 0
This is the equation of motion.
Problem
Using the Hamilton equation discuss the motion for a charged par-
ticle in an electro magnetic field.
Solution :
The Lagrangian function is
T =1
2mv2
V = ρ(
φ− v · A)
L = T − V
=1
2mv2 − ρ
(
φ− v · A)
Momentum
p = mv + ρA
mv = p− ρA
v =p− ρA
m
Assume ρ is the charge our the particle Hamilton H is given by
H = pv − L
120
= pv − 1
2mv2 + ρ
(
φ− vA)
=(
mv + ρA)
v − 1
2mv2 + ρφ− ρvA
= mv2 + ρAv − 1
2mv2 + ρφ− ρvA
=1
2mv2 + ρφ
=1
2m
(
p− ρA
m
)2
+ ρφ
=1
2m(p− ρA)2 + ρφ
=1
2m
[
(px − ρAx)2 + (py − ρAy)
2 + (pz − ρAz)2]
+ ρφ
x =∂H
∂px
=1
2m[2 (px − ρAx)]
=1
m(px − ρAx)
Similarly
y =1
m(py − ρAy)
z =1
m(pz − ρAz)
∴ ~v =1
m
(
~p− ρ ~A)
The second canonical equation
x =∂H
∂px
px = −∂H∂x
= − 1
2m
[
2 (px − ρAx)
(
−ρ∂Ax
∂x
)
+ 2 (py − ρAy)
(
−ρ∂Ay
∂x
)
+2 (pz − ρAz)
(
−ρ∂Az
∂x
)]
+ ρ∂φ
∂x
121
= −ρ∂φ∂x
+ρ
m
[
(px − ρAx)∂Ax
∂x+ (py − ρAy)
∂Ay
∂x
+ (pz − ρAz)∂Az
∂x
]
(1)
Similarly
py = −ρ∂φ∂y
+ρ
m
[
(px − ρAx)∂Ax
∂y+ (py − ρAy)
∂Ay
∂y+ (pz − ρAz)
∂Az
∂y
]
pz = −ρ∂φ∂z
+ρ
m
[
(px − ρAx)∂Ax
∂z+ (py − ρAy)
∂Ay
∂z+ (pz − ρAz)
∂Az
∂z
]
But
∇(
~v · A)
= ∇ [(xi + yj + zk) · (Axi + Ayj + Azk)]
= ∇ (xAx + yAy + zAz)
=∑
i
[
x∂
∂xAx + y
∂
∂yAy + z
∂
∂zAz
]
Substitute the values of x, y , z, we get
∇(
~v · A)
=∑ 1
mi
[
(px − ρAx)∂Ax
∂x+ (py − ρAy)
∂Ay
∂x
+ (pz − ρAz)∂Az
∂x
]
=∑
i
[
px
ρ+∂φ
∂x
]
=
(
px
ρ+∂φ
∂x
)
i+
(
py
ρ+∂φ
∂y
)
j +
(
pz
ρ+∂φ
∂z
)
k
=p
ρ+ ∇φ
p = ρ∇(
~v · ~A)
− ρ∇φ
Hence the proof.
Lagender’s Transformation :
122
Derive Hamilton’s cononical equation using Legender transforma-
tion and Lagrange’s equation.
Solution : Suppose the funcion F (u1, u2, · · · , un, ω1, ω2, · · · , ωm, t)
is associated with the given system.
Hence u’s are active variables and ω’s and t are passive variable.
Let
vi =∂F
∂ui
(i = 1, 2, · · · , n) . (1)
Let n×m Heissian determinant for the transformation be non-zero.
Then∣
∣
∣
∣
∂2f
∂ui∂uj
∣
∣
∣
∣
=
∣
∣
∣
∣
∂
∂uj
(
∂F
∂ui
)∣
∣
∣
∣
=
∣
∣
∣
∣
∂vi
∂uj
∣
∣
∣
∣
6= 0.
Now let us define a new function G (v1, v2, · · · ., vn, ω1, ω2, · · · , ωn, t)
according to the equation
G =n∑
i=1
uivi − f (u, ω, t) . (2)
Consider a variation ∂G associated with arbitrary variation of the
active variable.So
δG =n∑
i=1
∂G
∂vi
δvi
=n∑
i=1
[
uiδvi + viδui −∂F
∂ui
δui
]
orn∑
i=1
∂G
∂vi
δvi =n∑
i=1
[
uiδvi +
(
viδui −∂F
∂ui
)
δui
]
where δu’s and δv’s are assumed to the arbitrary equation .
Equating the co-efficient on both sides
ui =∂G
∂vi
. (3)
For a holonomic system
123
The Lagrange’s equation is
pi =∂L
∂qi, i = 1, 2, 3, · · · , n (4)
where pi =∂L
∂qi, i = 1, 2, 3, · · · , n.
Let us choose L (q, q, t) corresponding to the function F (u, ω, t) .
Also we have∣
∣
∣
∣
∂2L
∂qi∂qj
∣
∣
∣
∣
= |mij| 6= 0.
Since the inertia matrix is positive definite matrix.
Let G (u, ω, t) corresponding to the Hamilton’s funciton
H (p, q, t) =n∑
i=1
piqi − L
where qi =∂H
∂pi
.
∴
∂G
∂vi
= ui
∂H
∂qi(q, p, t) = −∂L
∂qi(q, q, t) (5)
From (4) and (5)∂H
∂qi= −pi ⇒ pi = −∂H
∂qi
Problem:
Obtain Hamiton cononical eqn from modified Hamilton’s principle.
Solution :
Consider a holonomic system having n independent q’s.
By Hamilton’s principle, δI = 0 where I =
t1∫
t0
Ldt.
δI = δ
t1∫
t0
Ldt
124
we have
h = T + V, L = T − V
H =∑
piqi − L
L =∑
piqi −H (p, q, t)
δ
t1∫
t0
Ldt = δ
t1∫
t0
[
∑
piqi −H (p, q, t)]
dt
0 =
t1∫
t0
[
n∑
i=1
(piδqi + qiδpi) −n∑
i=1
(
∂H
∂pi
δpi +∂H
∂qiδqi −
∂H
∂tδt
)
]
dt
125
UNIT IV
Hamilton -Jacobi theorem
Integrals of Motion :
In a holonomic system consisting of n generalized co-ordinates.
We have n second order non-linear differential equation with time as
independent variables. Any general analytic solution of this differential
equation contains one method of expression the general solution.
ie, To obtain ∂R independent funciton of the form
fi(q, q, t) = γj, j = 1, 2, 3, · · · , 2n
We have the γj’s are arbitrary constant.
These two functions are called the integral.
The value of each γ’s are depending on the initial condition. In
principle these 2n equations can be solved for the q’s as function of γ
and t.
It is possible to find q’s and q’s as a function of γ and t.
∴It is possible to find
qi = qi (γ1, γ2, · · · , γ2n, t)
qi = qi (γ1, γ2, · · · , γ2n, t)
1. Corresponding Hamilton’s Principle function. Solve the Hamil-
ton poblem giving the motion in phase space as function of time.
Solution :
Let
I =
t1∫
t0
Ldt.
Consider the cononical integral which is associating with Hamil-
ton’s principle.
126
We want to evaluate this integral over actual dynamical path of a
holonomic system that obeys standard Lagrange’s equation or Hamil-
ton’s equation. If 2n independent initial conditions as specified at the
time t0 any final time t1.
Let us consider the solution of equations
qi1 = qi1 (q0, q0, t0, t1) , i = 1, 2, 3, · · · , n.
Let us assume Jacobian
∂ (q11, q12, · · · , q1n)
∂ (q10, q20, · · · , qn0)
is non zero. Hence we get
qi0 = ηi (q0, q1, t0, t1) , i = 1, 2, 3, · · · , n
We can evaluate the integral
t1∫
t0
Ldt as a funciton of (q0, q0, t0, t1) .
Then subsititution for qi0 from the above equation we have
s = s (q0, q1, t0, t1) =
t1∫
t0
Ldt.
The function s (q0, q1, t0, t1) obtain from the canonical equation.
Required form to be obtained.
Let us assume twice differentiation in all its arguments. This is
called Hamilton’s principle function.
Let us consider a non contemporaneous system general variation
of the canonical integral with the reference in the actual solution.
We have,
H −H (q1, p, t)
dH
dt=
n∑
i=1
[
∂H
∂qi
dqidt
+∂H
∂pi
dpi
dt
]
+∂H
∂tdt
127
=n∑
i=1
[
∂H
∂qiqi +
∂H
∂pi
pi
]
+∂H
∂t(1)
From Hamilton’s canonical equation we have
qi =∂H
∂pi
, pi = −∂H∂qi
, i = 1, 2, · · · , n
∂L
∂t= −∂H
∂t
(1) ⇒ H =∂H
∂t= −∂L
∂t
We know that from the principle of least action
δI =
t1∫
t0
n∑
i=1
d
dt
(
∂L
∂qi
)
δqidt+
t1∫
t0
∂L
∂tδt−
(
n∑
i=1
∂L
∂qiδqi − L
)
d
dt
dt
−t1∫
t0
[
n∑
i=1
d
dt
(
∂L
∂qi
)
− ∂L
∂qi
]
δqidt
=
t1∫
t0
n∑
i=1
d
dt(piδqi) dt+
t1∫
t0
(
−∂H∂t
δt−Hd
dt(δt)
)
dt
=
t1∫
t0
n∑
i=1
d
dt(piδqi) dt −
t1∫
t0
d
dt(Hδt) dt
As our assumtion δS = δI.
∴ δS =
[
n∑
i=1
piδqi −Hδt
]t1
t0
dS =n∑
i=1
piδqi −t1∫
t0
d
dt(Hδt) dt.
This is difference of two forces.
(By writting the above quation in differential form)
s = s (q0, q1, t0, t1)
128
ds =n∑
i=1
(
∂s
∂qi1δqi1 +
∂s
∂qi0δqi0
)
+∂s
∂t1δt1 +
∂s
∂t0δt0 (2)
The (2n+2) arguments of the principle funcitons are independent.
Hence the corresponding co-efficients in (1) and (2) are equal.
pi1 =∂s
∂qi1, pi0 = − ∂s
∂qi0, i = 1, 2, 3, · · · ., n
H1 = − ∂s
∂t1;H0 =
∂s
∂t0
Now, we get pi0 as function f (qi0, qi1, t0, t1) .
Assuning∣
∣
∣
∣
∂2s
∂qi0∂qj1
∣
∣
∣
∣
6= 0
We can solve for each qi1as a funciton of the initial condition
i.e., qi1 = qi1 (q0, p0, t0, t1) , i = 1, 2, 3, · · · , n.
If this is substituted in pi1 =∂s
∂qi1, we have
pi1 = pi1 (q0, p0, t0, t1) .
This completes the solution of the Hamilton’s problem. Note :
(i)
t1∫
t0
Ldt is a fucntion (qi1, qi1, t0, t1) but qi1, qi1 are the funciton of
q0, q0, t0, t1,
(ii) In the whole discussion first finding the principle funcion without
knowing of solution.
Pfaffian Differential forms :
i) A pfaffian form gΩ is the m variables x1, x2, · · · , xm can be writ-
ten as
Ω = X1 (x) dx1 + · · · +Xm (x) dxm.
129
Let us define Cij =∂Xi
∂xj
− ∂Xj
∂xi
ii) If the pfaffian form is an exact differentiation then all the ds are
zero
we have
ds =n∑
i=1
pi1dqi1 −n∑
i=1
pi0dqi0 −H1dt1 −H0dt0.
The R.H.S consists of twon variables are p’s and q’s each expression
can be written as
p1q1 + p2q2 + · · · + pnqn + 0 · dp1 + · · · + 0 · dpn − 1 + h (q, pt) dt
Now, we have m = 2n+ 1 (odd) .
By pfaffican system these equation are of the form
n∑
i=1
Cijdxi = 0, j = 1, 2, 3, · · · ,m.
Applying this equation to the differential equaiton,we have
dqj −∂H
∂pj
dt = 0 (1)
−dpj −∂H
∂qjdt = 0 (2)
(1) and (2) can be written as
(1) ⇒ qj =∂H
∂pj
(2) ⇒ pj = −∂H∂qj
, j = 1, 2, 3, · · · , n
H = H (q, p, t)
∂H
∂t=
n∑
i=1
[
∂H
∂qiqi +
∂H
∂pi
pi
]
+∂H
∂t
These are the Hamilton’s canonical equation.
130
Also using (2), we have
H =∂H
∂t.
Now let us generalized the differential form
ds =n∑
i=1
pi1dqi1 −n∑
i=1
pi0dqi0 −H1dt1 +H0dt0.
Using the 2n parameter (γ1, γ2, · · · , γ2n) to specify the intial con-
dition in place the q0’s and p0’s in otherwise.
We assume a transformation
qi0 = qi0 (γ11, γ21, · · · , γ2n)
pi0 = pi0 (γ11, γ21, · · · , γ2n)
(3)
such that∂ (q10, q20, · · · , qn0, p10, p20, · · · , pn0)
∂ (γ11, γ21, · · · , γ2n)6= 0
Then using (3) we have
n∑
i=1
pi0dqi0 =2n∑
i=1
pi0
∂qi0∂γj
(4)
By the theorem that the 2n can be replaced by nα’s and nβ’s where
the function
αi = αi (γ11, γ21, · · · , γ2n)
βi = βi (γ11, γ21, · · · , γ2n) , i = 1, 2, 3, · · · , n.
choosen such that
n∑
i=1
βidαi =2n∑
i=1
Γ (γ, dγ)
131
in the same manner as (4) we have
Γj (γ) =n∑
i=1
βi
∂αi
∂γj
, j = 1, 2, 3, · · · , n.
It can be show that α’s are not unique
n∑
i=1
pi0dqi0 =n∑
i=1
βjdαi
where α’s and β’s are another representation shows that (q0, p0) and
(α, β) are connected by a homogeneouss cannonical transformation at
a time t0.
Hamilton - Jacobi equation :
Consider a holonomic system giving 2n independent initial condi-
tions at time to as q0 6= p0.
Now we have differential equation
Assume that
ds =n∑
i=1
pi1dqi1 −n∑
i=1
pi0dqi0 −H1dt1 +H0dt0 (1)
where s is the Hamilton’s principle funtion. It is associated with
cononical transformation relating the initial and final point of a path
in a phase space.
Let the initial condition be specified by
αi = αi (q10, q20, · · · , qn0, p10, p20, · · · , pn0)
βi = βi (q10, q20, · · · , qn0, p10, p20, · · · , pn0)
which satisfied.n∑
i=1
pi0dqi0 =n∑
i=1
βidαi
∴ ds =n∑
i=1
pi1dqi1 −n∑
i=1
βidαi −H1dt1 +H0dt0 (2)
132
Now we consider s number can be associated as a function of
s = s (qi1, αi, t1, t0)
ds =n∑
i=1
ds
dqi1dqi1 +
n∑
i=1
∂s
∂αi
dαi +∂s
∂t1dt1 +
∂s
∂t0dt0 (3)
Hence q’s and α’s are independent equality variable.
∴ Equating the corresponding coefficient in ds we have
−βi =∂s
∂αi
, i = 1, 2, 3, · · · , n.
pi1 =∂s
∂qi1, H1 = − ∂s
∂t1, H0 =
∂s
∂t0.
Let us assume initial time t0 = 0 , take the final time at t.
i.e., t1 = t⇒ dt1 = dt
Hence we can drop the subsorip q’s . Hence (1) becomes
ds =n∑
i=1
pidqi −n∑
i=1
βidαi −Hdt. (I)
From this we see that s is a funciton of (q, α, t)
(p is a function of q, q, r, t )
ds =n∑
i=1
∂s
∂qidqi +
n∑
i=1
∂s
∂αi
dαi +∂s
∂tdt (II)
Once again assuming
∣
∣
∣
∣
∂2s
∂qi∂αj
∣
∣
∣
∣
6= 0.
∴ Equating the coefficient in I and II, we have
−βi =∂s
∂αi
, i = 1, 2, 3, · · · , n (4)
pi =∂s
∂qi, i = 1, 2, 3, · · · , n (5)
133
∂s
∂t= −H (6)
Form (4) we can get q’s as function (α, β, t) using this in equation (5)
we can find p as function of (α, β, t) .
Hence we have the solution for Hamilton’s rpinciple.
H is usually as a function of (q, p, t) . Using (5) and (6) we have
∂s
∂t+H (q, p, t) = 0
∂s
∂t+H
(
q,∂s
∂q, t
)
= 0
This is called Hamilton - Jacobi equaiton.
State and prove Jacobi theorem.
Statement :
If s (q, α, t) is any complete solution of the Hamilton Jacobi equa-
tion
∂s
∂t+H
(
q,∂s
∂q, t
)
= 0
−βi =∂s
∂αi
, i = 1, 2, 3, · · · , n (1)
pi =∂s
∂qi, i = 1, 2, 3, · · · , n
where β’s are arbitrary constant are used to solve for qi (α, β, t) and
pi (α, β, t) .
Then these expression provide the general solution of the canonical
equation associated with Hamilton H (q, p, t) .
OR
Prove that any complete solution of the Hamilton Jacobi
equation leads to a solution of this Hamilton problem.
Proof :∂s
∂t+H
(
q,∂s
∂q, t
)
= 0 (2)
∂s
∂qi= pi, i = 1, 2, 3, · · · , n (3)
134
which is a function of (q, α, t) . Differentiating w.r.to αi
d2s
dαi∂t+
n∑
j=1
∂H
∂pj
∂pj
∂αi
= 0 (4)
where pj is considered as a function of (q, α, t) in (3).
[p (q, x, t) are independent to each otherds
dqi= pi]
∂s
∂αi
is a function of (q, α, t) and α’s and β’s are constants.
Now,∂s
∂αi
= −βi, i = 1, 2, 3, · · · , n
Taking the total time derivation of this w.r.to t
∂2s
∂t∂αi
+n∑
j=1
∂2s
∂qj∂αi
qj = 0 (5)
∂2s
∂t∂αi
+n∑
j=1
∂
∂qj
(
∂qj∂t
)(
∂s
∂αi
)
dt = 0
(5) and (4) ⇒ Using (1),(2) and (3), we have
n∑
j=1
(
qj −∂H
∂qjαi
)
∂2s
∂qj∂αi
= 0, i = 1, 2, 3, · · · , n (6)
But
∣
∣
∣
∣
∂2s
∂qi∂αj
∣
∣
∣
∣
6= 0.
qj =∂H
∂pj
, j = 1, 2, 3, · · · , n (7)
Now, ∂s∂t
+H (q, p, t) = 0.pi is a function of (q, α, t), differentiating (1)
partially w.r.to qj, we have
∂2s
∂qj∂t+
n∑
i=1
∂H
∂pi
∂pi
∂qj+∂H
∂qj= 0 (8)
135
(3) ⇒ pj =∂s (q, α, t)
∂qj, j = 1, 2, 3, ..., n.
Taking the total time derivation of this equation we have to differen-
tiating w.r.to t
pj −∂2s
∂t∂qj−
n∑
i=1
∂2s
∂qj∂qiqi = 0 (9)
(6) + (7) ⇒
pj +∂H
∂qj+
n∑
i=1
∂H
∂pi
∂pi
∂qj−
n∑
i=1
∂2s
∂qj∂qiqi = 0
⇒ pj +∂H
∂qj+
n∑
i=1
qi∂pi
∂qj−
n∑
i=1
∂pi
∂qjqi = 0
⇒ pj +∂H
∂qj= 0, pi = − ∂s
∂qi, j = 1, 2, 3, ..., n
pj = −∂H∂qj
Equations (7) and (10) together give the cononical equation of Hamil-
ton. Thus any complete solution of the Hamilton Jacobi equation leads
to a solution of the Hamilton equation.
Modified Hamilton Jacobi equation :
Consider a conservative holonomic system with n independent for
this system H is not an explicit function of time t for this system
H (q, p) = αn = h where h is the value of the Hamilton Jacobi integral.
∂s
∂t+H
(
q,∂s
∂q, t
)
= 0
∂s
∂t+H (q, p) = 0
∂s
∂t= −H = −αn (1)
136
∴ s can be taken as linera funciton of time t is
s (q, α, t) = −αnt+ ω (q, α) .
Then we have omitted arbitrary additive constant.
The funciton
ω (q, α) = ω (q1, q2, ..., qn, α1, α2, ..., αn) .
does not contain time explicitly and is known as a characteristic func-
tion.
Now,
∂s
∂αi
=∂ω
∂αi
, i = 1, 2, 3, ..., (n− 1) (2)
∂s
∂αn
=∂ω
∂αn
− t (3)
∂s
∂qi=
∂ω
∂qi, i = 1, 2, 3, ..., n (4)
Using (3) and (4), Hamilton jacobi equaiton Induces to
This is the modified Hamilton Jacobi equaiton
Note:
We have
βi =∂s
∂αi
, pi =∂s
∂qi∂s
∂αi
=∂ω
∂αi
, i = 1, 2, 3, ..., (n− 1)
∂s
∂αn
=∂ω
∂αn
− t
i∂s
∂qi=
∂ω
∂qi, i = 1, 2, 3, ..., n
Comparing the equation we have
βi =∂ω
∂qi, i = 1, 2, 3, ..., n
137
where βn is the intial time to t0.
Ignorable co-ordinates :
Consider a holonomic system described by the standard form
d
dt
(
∂L
∂qi
)
− ∂L
∂qi= 0,
qi is said to be an ignorable co-ordinate if∂L
∂qi= 0.
Also pi =∂L
∂qi= βi only for the ignorable co-ordinate.
Case (i)
Consider a system with ignorable co-ordinates q1, q2, ..., qn .Let us
assume the system is not consecutive. We know that
pi = αi, i = 1, 2, 3, ..., k.
We can assume principle solution is of the form
s (q, α, t) =k∑
i=1
αiqi + s′ (qk+1, ..., qn, α1, α2, ..., αn, t)
Hamilton Jacobi equation leads to
∂s
∂t+H
(
q,∂s
∂q, t
)
= 0
∂s′
∂t+H
(
qk+1, ..., qn, α1, α2, ..., αk,∂s′
∂q′k+1
, ...,∂s′
∂q′n, t
)
= 0.
The complete solution of these equation involve (n − k) non additive
constants exclusive of the known solution of this equation is obtained
form.
−βi = qi +∂s′
∂αi
, i = 1, 2, 3, ..., k
pi = αi, i = 1, 2, 3, ..., k
pi =∂s′
∂qi, i = k + 1, k + 2, k + 3, ..., n
pi = −qi0
138
−βi =∂s′
∂αi
, i = k + 1, k + 2, k + 3, ..., n
Case (ii)
Consider a system with ignorable co-ordinate q1, q2, ..., , qk and con-
servative with the help of the previous result we have
s (q, α, t) =k∑
i=1
αiqi − αnt+ ω′ (qk+1, ..., qn, α1, α2, αn, t)
The modified Hamilton Jacobi equation becomes
H
(
qk+1, ..., qn, α1, α2, ..., αk,∂ω′
∂q′k+1
, ...,∂ω′
∂q′n
)
= qn
The complete solution for ω′ in this case involve (n− k − 1) non addi-
tive constants αk+1, αk+2, ..., αn−1 the energy constent αn and constant
moments α1, α2, ..., αk.
The motion of the system is given by
−βi = qi +∂ω′
∂αi
, i = 1, 2, 3, ..., k
−βi =∂ω′
∂αi
, i = 1, 2, 3, ..., (k − 1)
t− βn =∂ω′
∂αn
pi = αi (i = 1, 2, 3, ..., k) ⇒ pi =∂ω′
∂qi, (i = k + 1, k + 2, ..., n)
Ordinary Conservation of linear momentum
This follows from Newton’s second law of motion
d
dt(mv) =
dp
dt= F.
If the total force is zero, thendp
dt= 0 at the linear momentum is
conserved.
Note :
The word conservation applied in the of connectedness.
139
Conservation of angular momentum :
Angular momentum is the angular of linear momentum in the case
of relational motion.
Consider a particle at mass m and the linear momentum p at a
positin vector ~r relative to the origin O at an inertial frame.
Angular momentum l at the particle with respect to the origin O
is L = r × ρ
Moment of a force about a origin is N = r × F then F =dp
dt.
N = r × F ⇒ N = r × dp
dt(1)
⇒ N =d
dt(r × ρi) −
dr
dt× ρ
⇒ N =d
dt(r × ρ) − v ×mv (2)
⇒ N =d
dt(r × ρ) − 0 (3)
∴ N =d
dt(r × ρ) (4)
∴The second term is zero both vectors are parallel now.
N =d
dt(r × ρ)
=dL
dt.
Thus the rate of change of angular momentum of a particle is equal
to the force acting on it is the total force N = 0.
ThendL
dt= 0,angular momentum is considered in the absence of
an external force.
Principle of conservation of evergy :
Let the only force acting on the given particle be conservation is
of the force are derivable from scalar potential evergy function.
F = −∇V, then the T.E.=K.E.+P.E.
Suppose under the action of such force F particles moves from the
140
position 1 to 2. Then the workdone
ω
2=
2∫
1
Fds
ω =
∫
Fds =
2∫
1
Fds =
2∫
1
dF
dtds
The particle moves ds distance in the time dt with velocity r, so
p = mv = mr.
dr
dt= r, dr = rdt
ω
2=
2∫
1
d
dt(mr) rdt
=
2∫
1
d
dt
(
mr2)
dt
=
2∫
1
2d
dt
(
1
2mr2
)
dt
= T2 − T1
T2, T1 denote the K.E of the particle of the K.E of the position 2 and
1 respectively.
F = −∇V
ω12 =
2∫
1
−∇V dr =
2∫
1
− dV
dxdr
−2∫
1
dV = V1 − V2
141
From (1) and (2),
T2 − T1 = V1 − V2
T2 + V2 = V1 + T1 = constant
∴ T + V = constant
⇒ T − E of a particle is conserved (constant)
Problem :
For a simple spring system using Hamilton Jacobi method to solve
it.
The K.E and P.E are given by
T =1
2mx2, V =
1
2kx2
P =∂T
∂x=
1
2m2x = mx
x =ρ
m⇒ x2 =
P 2
m2
H = T + V
=P 2
2m+kx2
2
We are considering a conservative systm.
∴ H = T + V = α
The modified homilton Jacobi equation is
1
2m
(
∂ω
∂x
)2
+1
2kx2 = α
(
∂ω
∂x
)2
= 2m
(
α− 1
2kx2
)
142
where α is energy constant.
∂ω
∂x=
√
2m
(
α− 1
2kx2
)
=
√
2m
(
2α− kx2
2
)
=√mk
√
(
2α
k− x2
)
Taking
a =
√
2α
mω2and ω =
√
k
m
∂ω
∂x=
x∫
x0
√mxdx
x∫
x0
dt√
a2 − ξ2
ω (x, α) = mω
x∫
x0
√
a2 − ξ2dt, ω =
√
x
m
∂ω
∂α= mω
∫
1
2
2a√a2 − ω2
1
2
√
2α
ω2m
2
mω2dω
= t− β
=1
ω
x∫
x0
dt√a2 − t2
= − 1
ω
x∫
x0
dt√t2 − a2
= − 1
ω
[
cos−1 t
a
]x1
x0
=1
ω
[
cos−1 x0
a− cos−1 x1
a
]
ρ is initial time t0. i.e., β = t0.
t− t0 =1
ω
[
φ− cos−1
(x
a
)]
ω (t− t0) = φ− cos−1
(x
a
)
143
φ− ω (t− t0) = cos−1
(x
a
)
− [ω (t− t0) − φ] = cos−1
(x
a
)
x = a cos [ω (t− t0) − φ]
=
√
2α
mω2cos [ω (t− t0) − φ] .
Let x (t0) = x0. Then x (t0) = v0. We know that
α = T + V
=1
2mx2 +
1
2kx2
=1
2mv2
0 +1
2kω2x2
0
=mω2
2
[
v20
ω2+ x2
0
]
(1)
sinφ =√
1 − cos2 φ
=
√
a2 − x20
a2
=1
a
√
a2 − x20
=1
a
√
2α
mω2− x2
0 using (1)
=1
a
√
v20
ω2+ x2
0 − x20
=v0
aω
Now,
x = a cos [ω (t− t0) − φ]
= a [cosω (t− t0) cosφ+ sinω (t− t0) sinφ]
=x0
cosφ[cosω (t− t0) cosφ+ sinω (t− t0) sinφ]
= x0 cosω (t− t0) +x0
cosφsinω (t− t0)
v0
aω
= x0 cosω (t− t0) + a sinω (t− t0)v0
aω
144
= x0 cosω (t− t0) +v0
ωsinω (t− t0)
Differentiating,
x = 0
Amplitute oscillation is given by
a =
√
v20
ω2+ x2
0
Kepler’s Problem :
Use Hamilton Jacoi equation to analysis the keplers problem or
modified Jacobi method.
Solution :
Suppose a particle of unit mass attacted by an inverse square grav-
itational force at a fixed point ‘O’. The position of a given problem
is given interms of the polar co-ordinates (r, θ) in the plane of the
orbits.
The K.E and P.E. are
T =1
2m(
r2 + r2θ2)
V = −µr
where µ is gravitation coefficient.
Lagrangian function is
L = T − V
=1
2m(
r2 + r2θ2)
+µ
r
We find that the generalized momentum are given by
pr =∂L
∂r; θ =
pθ
r2
αθ = pθ =∂L
∂θ= r2θ
145
θ =αθ
r2
The system is natural.
H = T + V
H =1
2
(
p2r +
pθ
r2
)
− µ
r= αt (1)
where αt represents constants values of the total energy (1) does not
appear in H.L.
∴ It is ignorable.
∂L
∂θ= 0 ⇒ ∂L
∂θ= p0 = constant = α0
The modified Hamilton Jacobi equation is
1
2
(
∂ω′
∂r
)2
+α2
θ
2r2− µ
r= αt
r is ignorable.
1
2
(
∂ω′
∂r
)2
= αt +µ
r− α2
θ
2r2
(
∂ω′
∂r
)2
= 2
(
αt +µ
r− α2
θ
2r2
)
∂ω′
∂r=
√
2αt + 2µ
r− α2
θ
r2
ω′ =
r∫
r0
√
2αt + 2µ
r− α2
θ
r2dr
We have
βi = qi +∂ω′
∂α
θ − θ0 = − ∂ω′
∂αθ
146
= −r∫
r0
(
− 1
r2
)
2αθ
2
√
2αt + 2µ
r− α2
θ
r2
dr
= −r∫
r0
αθ
r2
√
2αt + 2µ
r− α2
θ
r2
dr
= −r∫
r0
dr
r2αθ
√
2αt
αθ
+ 2µ
rαθ
− 1
r2
= −r∫
r0
dr
r2
√
2αt
αθ
+2µ
rαθ
− 1
r2
= −r∫
r0
dr
r2
√
2αt
α2θ
+µ2
α4θ
−(
1
r− µ
αθ
)2
= −r∫
r0
d
(
1
r− µ
αθ
)
r2
√
2αt + α2θ + µα
α4θ
−(
1
r− µ
αθ
)2
θ − θ0 = cos−1
1
r− µ
αθ√
2αt
α2θ + µ2
α4θ
r
r0
Let θ = θ0 where r = r0.
θ − θ0 = cos−1
[
α2θ − µr
r√
2αt + α2θ + µ2
]
− cos−1
[
−µr0ro
√
µ2
]
= cos−1
[
α2θ − µr
r√
2αt + α2θ + µ2
]
− cos−1 (−1)
cos θ =−µr + α2
θ
r√
2αt + α2θ + µ2
147
⇒ θ0 = 0 where r0 = 0√
2αtα2θ + 1 cos θ = −µr + αθ2
√
2αtα2θ + 1
µ2cos θ =
−µ+αθ2
µ
µ
αθ2
µ
r= 1 +
√
1 + 2αtα2θ
µ2cos θ
This is the forml
r= 1 + e cos θ.
This the cone whose eccentricity is
e =
√
µ2 + 2αtα2θ
µ2
This is the equation of a conic section having eccentricity.
r =αθ2
µ
1 +
√
µ2 + 2αtα2θ
µ2 cos θ
Hence the Kepler’s problem.
Seperability :
The index of seperability associated with the solution of P.D.E by
a solution to that is by expressing the solution interms of integrals
each involving only one variable.
Orthogonal system :
It is conservative holonomic system whose K.E function contains
only squared forms and no produed tems in this variable. If a system
is seperable that is possible to find a charatorstic function ω such that
ω =n∑
i=1
ωi (qi) where each contain only one of the q’s.
Note :
A particularly simple example of a seperable system occure in the
ω all but of the co-ordinates all ignorable.
148
Liouville’s System :
It is an orthogonal system whose K.E. and P.E. are of the form
T =1
2
[
r∑
i=1
pi (qi)
]
n∑
i=r
[
(qi)2
Ri (qi)
]
(1)
=R1P
21 +R2P
22 + ...+RnP
2n
2 (f1 + f2 + ...+ fn)
=v1 (q1) + v2 (q2) + ...+ vn (qn)
f1 (q1) + f2 (q2) + ...+ fn (qn)
where fi, qi and vi are each function of qi.We assume thatn∑
i=1
fi (qi) > 0
and Ri (qi) > 0.
1
2a = φ2 − ψ
a = 2φ2 − 2ψ
This is identical with (a).
Hence the system is seperable.
Using the pfaffian equation, derive the Hamilton canonical equation
from Hamilton funciton
ds =n∑
i=1
pi1dqi1 −n∑
i=1
pi0dqi0 −H1dt1 +H0dt0
Solution :
Book Work 5: pff equation.
Discuss the Kepler’s probelm using seperability.
Solution :
Let us consider a particle of unit mass p attached towards ‘O’ by
an inverse square gravitational force. Let T be the K,E, of the system.
Let V be the P.E. of the system.
T =1
2
[
r2 + r2θ2+ r2φ
2sin2 θ
]
dr = −µr
149
L = T − V =1
2
[
r2 + r2θ2+ r2φ
2sin2 θ
]
+µ
r
Now
pr =∂L
∂r=
1
22r = r
pθ =∂L
∂θ=
1
2r22θ = r2θ
pφ =∂L
∂φ=
1
2r2 sin2 θ2φ = r2 sin2 θφ
T =1
2
[
p2r +
p2θ
r2+
p2φ
r2 sin2 θ
]
where pr, pθ and pφ are the generalized moments. This system is an
orthogonal system.
∴ we have H = T + V = αt.
The modified Hamilton Jacobi equation is
1
2
(
∂ωr
∂r
)2
+1
2r2
(
∂ωθ
∂θ
)2
+1
2r2 sin2 θ
(
∂ωφ
∂φ
)1
− µ
r= αt
where
ω = ωr (r) + ωθ (θ) + ωφ (φ) (1)
∵ φ is missing from the above equation since φ is ignorable co-
ordinates.
∵ [rφ (φ) = αφ (φ)] ∴ pφ = αφ =∂ωφ
∂φ
∴ (1) × 2r2
⇒ r2
(
∂ωr
∂r
)2
− 2r2
(µ
r+ αt
)
+
(
∂ωφ
∂φ
)2
+1
sin2 θ(αφ)
2 = 0 (2)
The first two terms are functions of r only and the last terms are
fuctions of θ.
Hence they are each to seperable constant.
Stackel’s thoorem
150
Statement : Consider an orthogonal system whose K.E is given by
T =1
2
n∑
i=1
miq2i =
1
2
n∑
i=1
cip2i
where ci (q1, q2, ..., qn) > 0. Stackel’s theorem asserts that this is a
seperable system iff there exists a non-singular matrix n × n [Φij (qi)]
and a column matrix [Ψ (qi)] exists such that
i) CT Φ = (1, 0, ..., 0)
ii) CT Ψ = V, where V (q1, q2, ..., qn) is the P.E. and C is a column
matrix composed of the nc’s.
Proof :
Necesary part :
Let us assume that the general orthogonal system be seperable.
Hence it has a characterstic function ω(q, α) which consist of the
sum of terms of the form ωi (qi, α1, α2, ..., αn) .
This characteristic function is the complete integral of the modified
H.J equation
1
2
n∑
i=1
ci
(
∂ωi
∂qi
)2
+ V = α1 (1)
Here α1 is chosen as the total energy because of the system is assumed
to be separability.
∴
(
∂ωi
∂qi
)2
is a function of (qi, α1, α2, ..., αn) .
We can choose the seperation constants such that the d’s appear
linearly.
The most general form involving the single co-ordinate qi is
(
∂ωi
∂qi
)2
= −2Ψi (qi) + 2n∑
j=1
Φij (qj)αj (I)
where the mumerical co-efficients are chosen for our convenient for our
convenient
151
Then subustitute (I) in (1)
1
2
n∑
i=1
ci
[
−2Ψi (qi) + 2n∑
j=1
Φij (qj)αj
]
+ V = α1 (2)
−n∑
i=1
ciΨi (qi) +n∑
i=1
n∑
j=1
ciΦij (qj)αj + V = α1
⇒ CT Φ + CT Φα + V = α1.
Comparing the terms containing α’s we find that CT Φ = (1, 0, 0, ..., 0) ,which
we recognise as the first Stackel’s conditions. Similary the term in (3)
which do not involve α’s must sum to zero reading to CT Ψ = V,which
is Stackel’s condition.
Sufficient part :
Define a column matrix a =
(
∂ωi
∂qi
)2
.
The modified Hamilton Jacobi equation (1) can be written as
CT (a) + V = α.
Using second Stacicel’s condition
1
2CTa+ CT Ψ = (1, 0, 0, ..., 0)α (4)
⇒ CT Φ = (1, 0, 0, ..., 0)(
CT Ψ = V)
CT = (1, 0, 0, ..., 0) Φ−1 (5)
Sustituting (5) in (4)
1
2(1, 0, 0, ..., 0) + CT Φ = α1
1
2(1, 0, 0, ..., 0) Φ−1a+ (1, 0, 0, ..., 0) Φ−1Ψ = (1, 0, 0, ..., 0)α
152
(1, 0, 0, ..., 0)
[
1
2Φ−1a+ Φ−1Ψ
]
= (1, 0, 0, ..., 0)α
1
2Φ−1a+ Φ−1Ψ = α
1
2Φ−1a = α− Φ−1Ψ
Φ (α− Ψ) =1
2a
a = 2 (Φα− Ψ)
This is identical with equation (I).
∴ The system is separable.
Note:
(
∂ωθ
∂θ
)2
+dφ2
sin2 θ= α2
θ
⇒ r2
(
∂ωθ
∂θ
)2
+∂r2
(µαr
r+) = α2
θ
We all now assume that the system is separable.
∂ωr
∂r=
√
2(µ
r+ dt
)
− α2θ
r2
∂ωθ
∂θ=
√
α2θ −
dφ2
sin2 θ
which is immediatery integrable and find ωr, ωθ are already found out.
Hence the system is separable and hence reduces of quadratare
ωr + ωθ + ωφ = 0.
Discuss the kepler’s problem using separability.
Let us consider a particle of unit mass p attached towards O by an
iverse square gravitational force.
Let T be the K.E. of the system.
Let V be the P.E. of the system.
153
T =1
2
[
r2 + r2θ2+ r2φ
2sin2 θ
]
dr = −µr
L = T − V =1
2
[
r2 + r2θ2+ r2φ
2sin2 θ
]
+µ
r
Now, we can find generalized moments
pr =∂L
∂r=
1
22r = r
pθ =∂L
∂θ=
1
2r22θ = r2θ
pφ =∂L
∂φ=
1
2r2 sin2 θ2φ = r2 sin2 θφ⇒ φ =
pφ
r2 sin2 θ
T =1
2
[
p2r +
p2θ
r2+
p2φ
r2 sin2 θ
]
where pr, pθ and pφ are the generalized moments.
This system is an orthogonal system, ∴ we have H = T + V = dt.
The modified Hamilton Jacobi equation is
1
2
(
∂ωr
∂r
)2
+1
2r2
(
∂ωθ
∂θ
)2
+1
2r2 sin2 θ
(
∂ωφ
∂φ
)1
− µ
r= dt (1)
where ω = ωr (r) + ωθ (θ) + ωφ (φ) .
∵ φ is missing from the above equation since φ is ignorable co-
ordinates..
∵ [rφ (φ) = dφ (φ)] ∴ pφ = αφ =∂ωφ
∂φ
∴ (1) × 2r2
⇒ r2
(
∂ωr
∂r
)2
− 2r2
(µ
r+ αt
)
+
(
∂ωθ
∂θ
)2
+1
sin2 θ
(
∂ωφ
∂φ
)2
= 0
The first two terms are functions of r only and the last terms are
fuctions of θ.
154
Hence they are each to seperable constant.
r2
(
∂ωr
∂r
)2
− 2r2
(µ
r+ αt
)
= −αθ2 (2)
(
∂ωθ
∂θ
)2
+1
sin2 θ
(
∂ωφ
∂φ
)2
= αθ2 (3)
(2) ⇒ r2
(
∂ωr
∂r
)2
= −α2θ + 2r2
(µ
r+ αt
)
= 2r2
(µ
r+ αt
)
− α2θ
∂ωr
∂r=
√
2(µ
r+ αt
)
− α2θ
r2
(3) ⇒(
∂ωθ
∂θ
)2
= α2θ −
(
∂ωφ
∂φ
)21
sin2 θ
∂ωθ
∂θ=
√
α2θ −
α2φ
sin2 θ
which is immeadiately integrable and find ωr, ωθ are already found out.
Hence the system is separable and hence reduces to quadrative
ωr + ωθ + ωφ = 0.
Jacobi Theorem :
If s (q, α, t) is any complete solution of the Hamilton Jacobi equa-
tion
∂s
∂t+H
(
q,∂s
∂q, t
)
= 0, βi =∂s
∂αi
, pi =∂s
∂qi, i = 1, 2, 3, ..., n
where β’s are arbitrary constants are used to solve qi (α, β, t) and pi
(α, β, t) . Then these expression provide the general solution of the
canonical equation associated with Hamilton H(q, p, t).
155
Proof :∂s
∂t+H
(
q,∂s
∂q, t
)
= 0 (1)
Differentiating (1) w.r.to αi, by considering pj as a function of (q, α, t)
∂2s
∂αi∂t+
n∑
j=1
∂H
∂pj
∂pj
∂αi
= 0 (2)
Given∂s
∂αi
= −βi
Differentiating both sides w.r.to t
∂2s
∂t∂αi
+n∑
j=1
∂2s
∂qj∂αi
qj = 0 (3)
(3) − (2) ⇒
n∑
j=1
qj∂2s
∂qj∂αi
− ∂H
∂pi
∂pi
∂αi
= 0
pi =∂s
∂qj
Differentiating w.r.to αi,
∂pi
∂αi
=∂2s
∂αi∂qj
Using this into second term (4) becomes
n∑
j=1
qj∂2s
∂qj∂αi
− ∂H
∂pi
∂2s
∂αi∂qj= 0
n∑
j=1
(
qj −∂H
∂pi
)
∂2s
∂αi∂qj= 0.
Now∣
∣
∣
∣
∂2s
∂αi∂qj
∣
∣
∣
∣
6= 0
156
∴ Each co-efficeient must be in this
∴ q =∂H
∂pj
.
This is Hamilton first equation.
In order to derive the Hamilton second equation we again differen-
tiate (1) partially w.r.to qj and associated pi = f (q, α, t)
∂2s
∂qj∂t+
n∑
i=1
∂H
∂pi
∂pi
∂qj+∂H
∂qj= 0 (5)
pj =∂s
∂qj⇒ pj =
∂2s
∂qj∂t
pj −∂2s
∂qj∂t−
n∑
i=1
∂2s
∂qj∂qiqi = 0 (6)
(5) + (6) ⇒
n∑
i=1
∂H
∂pi
∂pi
∂qj+∂H
∂qj+ pj −
n∑
i=1
∂2s
∂qj∂qiqi = 0 (7)
But∂H
∂qj= qj and
∂pi
∂qj=
∂2s
∂qi∂qj
∴ First and last terms get called in (8) reduces to
∂H
∂qj+ pj = 0
pj = −∂H∂qj
This is second kind equation.
157
Unit-V
Canonical transformation
Let us consider only a holonomic system which can be described by
the standard form of either Hamilton’s (or) Lagrange’s equation.
Hamilton’s principle applies to these system the generalized coordi-
nates are q1, q2, . . . , qn . We know that the Hamilton principle is
δI = 0 where I =
t1∫
t0
Ldt
∴ δ
t1∫
t0
L (q, q, t) dt = 0
where the endpoints of the varied paths are fixed in both configuration
space and time.
Let L = T − V.
Now if a new set of coordinates Q1, Q2, . . . , Qn is Q related to the
formal set by a point transformation. The Larangian function is given
by,
L∗
(
Q, Q, t)
= L (q, q, t) = T − V
i.e., L and L∗are equal. Now the Hamilton’s principle function
δI = 0 .
δ
t∫
t0
L∗
(
Q, Q, t)
dt = 0.
Now we consider the Lagrangian equation is
L∗
(
Q, Q, t)
= L (q, q, t) − d
dtφ (q,Q, t)
158
where φ (q,Q, t) is twice differential. Then,
δI = δ
t∫
t0
L∗
(
Q, Q, t)
dt
= δ
t∫
t0
[
L (q, q, t) − d
dtφ (q,Q, t)
]
dt
= δ
t∫
t0
L (q, q, t) dt− δ
t∫
t0
dϕ (q,Q, t)
= δ
t∫
t0
L (q, q, t) dt− δ [φ (q,Q, t)]tt0
δI = δ
t∫
t0
L (q, q, t) dt
i.e., δq ’s and δQ ’s are zero at the fixed time t0 and t .
⇒ δ
t∫
t0
L∗
(
Q, Q, t)
= 0.
Now Hamilton are described by the equation
H (q, p, t) =n∑
i=1
piqi − L (q, q, t)
K (Q,P, t) =n∑
i=1
piQi − L∗
(
Q, Q, t)
where the generalized moment are given by,
pi =∂L
∂qi, Pi =
∂L∗
∂Qi
qi =∂H
∂pi
, pi = −∂H∂qi
, i = 1, 2, . . . , n.
159
Here the function L∗
(
Q, Q, t)
.
∴ The Hamilton canonical function
Qi =∂K
∂Pi
, Pi = − ∂H
∂Qi
, i = 1, 2, . . . , n
A transformation form (q, p) to (Q,P ) which preserves the canon-
ical form of the equation of motion is known as canonical transforma-
tion.
Differential forms of Pfaffian differential equation:
Let us consider a Hamilton function H (q, p, t) suppose the trans-
formation equation is of the form,
Qi = Qi (q, p, t) , Pi = Pi (q, p, t) i = 1, 2, . . . , n
where each function is atleast twice differentiable. Now the Lagrangian
equation is given by,
L∗
(
Q, Q, t)
= L (q, q, t) − d
dt[φ (q,Q, t)] (1)
Hamilton function is defied by,
H (q, p, t) =∑
piqi − L (q, q, t) (2)
K (Q,P, t) =∑
PiQi − L∗
(
Q, Q, t)
. (3)
Using (2) and (3) in (1) we get
∑
PiQi −K (Q,P, t) =∑
piqi −H (q, p, t) − d
dt[φ (q,Q, t)]
⇒∑
Pi
d
dt(Qi) −K =
∑
pi
d
dt(qi) −H − d
dt(φ)
⇒ dφ =∑
pidqi −Hdt−∑
PidQi +Kdt (4)
where φ is the generalized function. Equation (4) is called the differ-
ence of two pfaffian differential equation and (4) is an exact differential
equation
160
Note: The function φ (q,Q, t) is called the generating function for
the transformation.
Book work: Consider Q =√
2qet cos p, P =√
2qe−t sin p .
Show that the above transformation is canonical.
Solution: Given that
Q =√
2qet cos p,
P =√
2qe−t sin p.
We obtained
pδq − PδQ = δψ (1)
δQ = et
[
2δq cos p
2√
2qδq +
√
2q (− sin p) δp
]
= et
[
cos p√2qδq − sin p
(
√
2q)
δp
]
(1)⇒
pδq − PδQ = pδq −√
2qe−t sin p
[
et
(
cos p√2qδq − sin p
√
2qδp
)]
= pδq − sin p cos pδq + 2q sin2 pδp
δψ = [p− sin p cos p] δq + 2q sin2 pδp (2)
which is the differential form,
pdx+ qdy = f (x)
and
∂P
∂y=
∂Q
∂x⇒ ∂P
∂p=∂Q
∂q∂
∂p(p− sin p cos p) = 1 −
[
sin p (− sin p) + cos2 p]
= 1 −[
− sin2 p+ cos2 p]
= 1 − cos 2p
161
∂
∂p[p− sin p cos p] = 2 sin2 p
∂
∂p[p− sin p cos p] = 2 sin2 p
Similarly,
∂
∂q
[
2q sin2 p]
= 2 sin2 p
∴
∂
∂p[p− sin p cos p] =
∂
∂q
[
2q sin2 p]
∴ The transformation is caonical.
Example: Let us consider the transformation
Q =1
2
(
q2 + p2)
, P = − tan−1
(
q
p
)
Show that the transformation is canonical. Find the new Hamilton
canonical equation.
Solution: Given
P = − tan−1
(
q
p
)
;
Q =1
2
(
q2 + p2)
.
We obtained
pδq − PδQ = dψ (1)
δQ =1
22qδq +
1
22pδp
δQ = qδq + pδp
(1) ⇒
pδq − pδQ = pδq + tan−1
(
q
p
)
(qδq + pδp)
162
= pδq + tan−1
(
q
p
)
qδq + tan−1
(
q
p
)
pδp
δψ =
[
p+ q tan−1
(
q
p
)]
δq + p tan−1
(
q
p
)
δp
∂
∂p
[
p+ q tan−1
(
q
p
)]
= 1 + q
1
1 +q2
p2
(
− q
p2
)
= 1 + q
(
p2
p2 + q2
)(
− q
p2
)
= 1 − q2
p2 + q2
=p2 + q2 − q2
p2 + q2
=p2
p2 + q2
∂
∂p
[
p+ tan−1
(
q
p
)]
=p2
p2 + q2
∂
∂q
[
p+ tan−1
(
q
p
)]
= p
p
1
1 +q2
p2
(
1
p
)
= p
(
p2
p2 + q2
)(
1
p
)
=p2
p2 + q2
∴
∂
∂p
[
p+ tan−1
(
q
p
)]
=∂
∂Q
[
p+ tan−1
(
q
p
)]
∴The transformation is canonical.
Principle forms of generating functions:
Let us consider a generating function φ (q,Q, t) . The other types
of generating functions namely F2 (q, P, t) , F3 (P,Q, t) and F4 (p, P, t)
163
. Now we show the relationship of F2 and F1 and again F3 and F4 and
F1 and F4.
The relationship of the generating function F1 , F2 , F3 and F4 are
called the principle of generating function.
Proof: Given that the generating function φ (q,Q, t) . We know
that the differential form
∑
piδqi −∑
PiδQi −Hdt+Kdt = dF1 (q,Q, t) . (1)
Let us consider
∑
QidPi −∑
PidQi = d(
∑
QiPi
)
. (2)
Adding (1) and (2) we get
∑
Pidqi +∑
QidPi −Hdt+Kdt = dF2 (q, P, t) (3)
where dF2 (q, P, t) = dF1 (q,Q, t) + d
(
n∑
i=1
QiPi
)
.
F2 (q, P, t) = F1 (q,Q, t) +∑
QiPi. (4)
Equation (4) gives the relationship in between F1 and F2 . Now the
total differential form of F2 is the form
dF2 =∑ ∂F2
∂qidqi +
∑ ∂F2
∂Pi
dPi +∂F2
∂tdt. (5)
Equating the coefficient in (3) and (5)
Pi =∂F2
∂qi, Qi =
∂F2
∂pi
, K =∂F2
∂t+H. (6)
Equation (6) is called the canonical transformation of the function F2
.
Similarly we can derive the following equation associated with
F3 (p,Q, t) .
164
Consider
F3 (p,Q, t) = F1 (q,Q, t) −∑
qipi (7)
(7) ⇒F3 = dF1 (q,Q, t) −
∑
qidpi −∑
piqi
Using (1),
dF3 =∑
pidqi −∑
PidQi −Hdt+Kdt−∑
qidpi −∑
pidqi
dF3 ( p,Q, t) = −∑
PidQi −Hdt+Kdt−∑
qidpi (8)
Then the total differentiating dF3 (p,Q, t)
dF3 (p,Q, t) =∑ ∂F3
∂Pi
dpi +∑ ∂F3
∂Qi
dQi +∂F3
∂tdt. (9)
Equating the coefficients in (8) and (9),
−qi =∂F3
∂Pi
(i = 1, 2, . . . , n)
−Pi =∂F3
∂Qi
(i = 1, 2, . . . , n)
K = H +∂F3
∂t
(10)
which is the canonical transformation of equation (8).
Similarly for next generating function F4 (p, P, t) we derived as fol-
lows.
F4 (p, P, t) = F2 (q, P, t) −∑
qiPi (11)
dF4 = dF2 −∑
pidqi −∑
qidpi
Using (3),
dF4 =∑
pidqi +∑
QidPi −Hdt+Kdt−∑
pidqi −∑
qidpi (12)
dF4 =∑
QidPi −Hdt+Kdt−∑
qidpi.
165
Taking the total differential of F4 ,
dF4 =∑ ∂F4
∂pi
dpi +∑ ∂F4
∂Pi
dPi +∂F4
∂tdt (13)
Equating the coefficient in (12) and (13),
Qi =∂F4
∂Pi
(i = 1, 2, . . . , n)
qi = −∂F4
∂pi
(i = 1, 2, . . . , n)
K = H +∂F4
∂t
(14)
which is the canonical transformation of F4 (p, P, t) .
Equations (4), (7) and (11) gives the relation between the generat-
ing function which is the principle function.
Book Work: Consider the transformation
Q = logsin p
q, P = q cot p
Obtain the four major types of generating function associated with
the transformation.
Proof: Given that,
Q = logsin p
q, P = q cot p
We know that,
pδq − PδQ = δψ (1)
δQ =1
sin p
q
[
1
qcos pδP + sin p
(
− 1
q2
)
δq
]
=q
sin p
[
cos p
qδP − sin p
q2δq
]
= cot pδp− 1
qδq
166
pδq − q cot p
[
cot pδp− 1
qδq
]
= δψ
pδq − q cot2 pδp+ cot pδq = δψ
(p+ cot p) δq − q cot2 pδp = δψ
∂
∂p(p+ cot p) = 1 +
(
− cosec2 p)
= 1 − cosec c2p = − cot2 p
∂
∂p(p+ cot p) = − cot2 p
∂
∂q
(
−q cot2p)
= − cot2 p
∂
∂p(p+ cot p) =
∂
∂q
(
−q cot2p)
= δψ
∴ The transformation is canonical.
Let ψ = pq + q cot p (from (1)).
The geometrical representation of the transformation
sin p =q
e−Q= qeQ
cos p =p
e−Q= peQ
cot p =cos p
sin p=peQ
qeQ=p
qq cot p = p
Now,
p2 + q2 =(
e−Q)2
= e−2Q
p2 = e−2Q − q2
p =√
e−2Q − q2
167
=1
eQ
[
√
1 − q2e2Q
]
We know that
cos p = peQ
=1
eQ
[
√
1 − q2e2Q
]
eQ
cos p =√
1 − q2e2Q
p = cos−1
(
√
1 − q2e2Q
)
From the principle of generating function
ψ (p, q) = F1 (q,Q) (2)
F1 (q,Q) = q cos−1
(
√
1 − q2e2Q
)
+√
e−2Q − q2 (3)
∂F1
∂q= cos−1
(
√
1 − q2e2Q
)
= p
∂F1
∂Q=√
e−2Q − q2 = −p
Comparing (2) and (3) equating the coefficient we get
p = cos−1
(
√
1 − q2e2Q
)
,
p =√
e−2Q − q2.
The above canonical can be written as
∂F1
∂q= cos−1
(
√
1 − q2e2Q
)
= p,
∂F1
∂q= −
(
√
e−2Q − q2
)
= −p.
168
To find the generating function F2 (q, P ) . From the principle of gen-
erating function
F2 (q, P ) = F1 (q,Q, t) +∑
QiPi
F2 = pq + q cotP +QP (4)
F2 is the generations function of the variable (q, P ) .
We know that
q cot p = P
cot p =P
q
tan p =q
P⇒ p = tan−1
( q
P
)
From the triangle
P 2 = e−2Q − q2
e−Q =√
p2 + q2
eQ =1
√
p2 + q2
Q = log
(
1√
p2 + q2
)
= log 1 − log(
√
p2 + q2
)
= − log(
√
p2 + q2
)
Substitute p,Q in equation (4) we get
F2 (q, P ) = q tan−1
( q
P
)
+ p+(
− log(
√
p2 + q2
))
p
= q tan−1
( q
P
)
+ p(
1 − log(
√
p2 + q2
))
(5)
From (4) and (5) we get
169
∂F2
∂q= tan−1
( q
P
)
= p
∂F2
∂p= 0 − log
√
q2 + p2 = Q
which are the canonical equation of the function F2 .
To find the function F3 ,
F3 = F3 (p,Q)
We know that,
F3 (p,Q) = F1 (q,Q) −∑
qipi
= qp+ q cotP − qp
F3 = q cot p (6)
q cot p = qcos p
sin p= q · Pe
Q
qeQ= p
Next to find the function F4 (p, P ) .
Similarly
F4 (p, P ) = F2 − qp
= qp+ q cot p+QP − qp
= q cot p+QP
F4 (p, P ) = P +QP
cos p = p = PeQ
eQ =cosP
p
Q = log
(
cosP
p
)
F4 = p+ log
(
cosP
p
)
P
170
F4 = p+ P log
(
cos p
p
)
∂F4
∂p= q (− tan p cot p) = −p tan p = −q
∂F4
∂p= log
(
cosP
p
)
= Q
Hence the four generating function.
Special transformation
1.Identity transformation:
Let us consider the identity transformation which is an obvious
example of a canonical transformation. It is generated by a function
is of the form,
Tn =n∑
i=1
qipi (1)
Then
Pi =∂F2
∂qi= pi (i = 1, 2, . . . , n)
Qi =∂F2
∂Pi
= qi (i = 1, 2, . . . , n) . (2)
The identity transformation is generated by
F3 =n∑
i=1
PiQi.
2. Orthogonal transformation:
This is the simple canonical transformation. The orthogonal trans-
formation of q ’s and p ’s generated by
F2 =n∑
i=1
n∑
j=1
aijpiqj,
where ai ’s are constant meeting the orthogonality conditions
n∑
i=1
aijajk = δjk.
171
δjk is the kroneckal delta the orthogonality can be written in the matrix
form
a · a−1 = a−1 · a = 1.
The transformation equation are
Pj =∂F2
∂qi=∑
aij, Qi =∂F2
∂pi
=∑
aijqj.
These equation can be solved for pi ’s in terms of Pi ’s and Qi ’s in
terms of qi’s then the transformation equations are
Qi =n∑
j=1
aijqi
Pi =n∑
j=1
aijPj (i = 1, 2, . . . , n)
These equations represents the equal rotation is q -space and p -space.
3. Translation:
We know that F2 =∑
qiPi, replace qi by qi + ci and replace Pi by
Pi − di
F2 =∑
qiPi
=∑
(qi + ci) (Pi−di)
=∑
qipi + ciPi − di
We have to omit cidi . Now,
pi =∂F2
∂qi= Pi − di
Qi =∂F2
∂Pi
= qi + ci
Qi = qi + ci,
Pi = Pi + di (i = 1, 2, . . . , n)
172
which represents the required translation.
Homogeneous Canonical transformation:
We know that the differential form is
n∑
i=1
piδqi −n∑
i=1
PiQi = δψ (1)
is an exact differential. Consider the function φ and ψ are identically
zero. Thenn∑
i=1
(Piδqi − PiδQi) = 0
and the corresponding transformation is called a homogeneous canon-
ical transformation. This transformation is also known as Mathieu
transformation (or) contact transformation.
Point transformation:
We consider the class of homogeneous canonical transformation for
which full set of n independent function Ωj (q,Q, t) exist and equal to
zero.
This implies that the two n× n coefficient matrices of
n∑
i=1
∂Ωj
∂qjδqi +
n∑
i=1
∂Ωj
∂Qi
δQi = 0 (j = 1, 2, . . . ,m)
are non singular and the corresponding determinants are non zero,
that is,∣
∣
∣
∣
∂Ωj
∂qi
∣
∣
∣
∣
6= 0,
∣
∣
∣
∣
∂Ωj
∂Qi
∣
∣
∣
∣
6= 0.
Hence we can solve for the Q’s in terms of (q, t) on for the q ’s as
functions of (Q, t) .
We have n equation is of the form
Qi = fi (q, t) i = 1, 2, . . . , n
where the fi ’s are twice differentiable. These equations represent
a point transformation. i.e., they represent a mapping of points in
173
configuration space. If one also includes the equations for the P ’s,
the entire set represents a transformation in phase space and is known
as extended point transformation.
Momentum transformation:
The momentum transformation equation are obtained from
pi =n∑
j=1
Pj
∂Qi
∂qi(i = 1, 2, . . . , n)
which in this case takes the form
pi =n∑
j=1
∂fi
∂qi(i = 1, 2, . . . , n) .
Thus the p ’s are homogeneous linear functions of the P ’s and vice
versa. If we define nΩ ’s of the form
Ωj = Qj − fj (q, t) (j = 1, 2, . . . , n)
If the represents the transformation is called the momentum transfor-
mation.
Non homogeneous point transformation:
Let the basic differential form be
n∑
i=1
piδqi −n∑
i=1
PiδQi = δψ. (1)
Consider
Qi = fi (q, t) ,
Qj = fj (q, t)
⇒ δQj =n∑
j=1
∂fi
∂qiδqi. (2)
174
Substitute the equation (2) in (1) implies
n∑
i=1
piδqi −n∑
i=1
Pi
[
n∑
j=1
∂fi
∂qiδqi
]
= δψ. (3)
In general,
δψ =n∑
i=1
∂ψ
∂qiδqi +
n∑
i=1
∂ψ
∂Pi
δpi (4)
Here the δq ’s and δp ’s are independent.Equation (3) and (4) we get
∂ψ
∂qi=
n∑
i=1
pi −n∑
i=1
Pi
n∑
j=1
∂fj
∂qi
=n∑
i=1
[
pi − Pi
n∑
j=1
∂fj
∂qi
]
pi =∂ψ
∂qi+ Pi
n∑
j=1
∂fi
∂qi(i = 1, 2, . . . , n)
⇒ ∂ψ
∂pi
= 0
Equation (4) implies ψ (q, t) is not an explicit function of the p ’s.
Furthermore it has the proper form for a generating function of the
type of fi . Hence we can choose F1 to be identically equal to ψ for a
point transformation.
Example: Illustrate the theory for the nonhomogeneous case, sup-
pose we are given sceleronomic extended point transformation
Q = tan q, P = (p−mν0) cos2 q,
where m and ν0 are constants.
Solution: Given that Q = tan q, P = (p−mν0) cos2 q , we wish
to show that the transformation is canonical.
δQ = sec2 qδq
= Pδq =[
(p−mν0) sec2 q]
sec2 qδq
175
= δψ
∴ pδq = pδq +mν0δq = δψ
⇒ mν0δq = δψ (1)
which is an exact differential integrating
F1 = ψ = mν0q. (2)
Hence this is a nonhomogeneous canonical transformation. Now the
point transformation is,
Qi (q, t) = fi (q, t)
Q = tan q − f (q) .
From the non homogeneous point transformation
Ωj = Qi − fi (q, t)
Ω = Q− tan q
which yields K = H . Thus the two Hamilton functions are equal
F2 (q, P ) = F1 +QP
= mv0q +QP
= mv0q + tan qP
p =∂F2
∂q= mv0 + P sec2 q
Q =∂F2
∂P= tan q.
Here the canonical equations for the generating functionF2 .
To find F3 :
F3 (P,Q) = F1 − qP
176
= mv0q − qP∂F3
∂P= −q, ∂F3
∂Q=
mv0
1 +Q2.
To find F4 :
F4 (p, P ) = F2 − qp = F1 +QP − qp
= mv0q + tan qP − qp∂F4
∂p= −q, ∂F4
∂P= tan q.
Now let us apply this transformation to a specific physical situation.
Consider a mass spring system which is attached to a plane that is
translating with a constant velocity v0 as shown in the following figure.
0v
km
0q+l
The unstressed length of the spring. Hence we have a conservative
holonomic system (or) Hamilton’s equations.
L = T − V
L =1
2m (v0 + q)2 − 1
2Kq2
p =∂L
∂q=
1
2m2 (v0 + q) = m (v0 + q)
The Hamilton function is H = T2 − T0 + V from the mass spring
system.
H =p2
2m− v0p+
1
2Kq2
177
Hence the canonical equation of the motion for (q, p) variables are,
q =∂H
∂p=
p
m− v0,
p = −∂H∂q
= −1
2K (2q) = −Kq
Now let us obtain L∗
(
Q, Q)
relative to the moving frame. Here,
T =1
2mq2
V =1
2Kq2
Given that Q = tan q ⇒ q = tan−1Q.
q =1
1 +Q2Q
L∗ = T − V
=1
2m
(
Q2
(1 +Q2)2
)
− 1
2K(
tan−1Q)2
=1
2
mQ2
(1 +Q2)2− 1
2K(
tan−1Q)2
P =∂Lα
∂Q=
m2Q
2 (1 +Q2)2=
mQ
(1 +Q2)2
Given that
P = (p−mv0) cos2 q
cos2 q =P
p−mv0
=mQ
(1 +Q2)2mq=
Q
q (1 +Q2)2
cos2 q =Qq
(1 +Q2)2
=1 +Q2
(1 +Q2)2=
1
1 +Q2
[
SinceQ
q= 1 +Q2
]
178
cos q =1
√
1 +Q2
In this case in K.E consist of T2 terms only Hamiltonian function.
H = T2 + V
=p2
2m+
1
2Kq2
K = T + V
=1
2mq2 +
1
2Kq2
=1
2m
[
Q2
(1 +Q2)2
]
+1
2K(
tan−1Q)2
We know that
P =mQ
(1 +Q2)2
mQ = P(
1 +Q2)2
Q =P (1 +Q2)
2
m
K =1
2mQ
Q
(1 +Q2)2+
1
2K(
tan−1Q)2
=1
2p(1 +Q2)
2
(1 +Q2)2p(1 +Q2)
2
m+
1
2K(
tan−1Q)2
=p2
2m
(
1 +Q2)2
+1
2K(
tan−1Q)2
Here the canonical equations are
Q =∂K
∂p=
1
2m2p(
1 +Q2)2
=p
m
(
1 +Q2)2
P = −∂K∂Q
= −[
p2
2m
[(
1 +Q2)
2θ]
+1
2K2 tan−1Q
(
1
1 +Q2
)]
179
= −2p2Q
m
(
1 +Q2)
− K tan−1Q
1 +Q2.
Hence the canonical equation.
Lagrangian and Poisson brackets:
Lagrangian brackets:
Suppose the given transformation equation
Qi = Qi(q, P, t)
Pi = Pi(q, p, t)(i = 1, 2, · · · , n)
Let us consider the canonical differential form
n∑
i=1
piδqi −n∑
i=1
PiδQi = δψ (1)
δQi =n∑
i=1
∂Qi
∂qiδqi +
n∑
i=1
∂Qi
∂Pj
δPj (2)
Using (2) in (1)
n∑
i=1
piδqi −n∑
i=1
Pi
[
n∑
j=1
∂Qi
∂qjδqj +
n∑
j=1
∂Qi
∂Pj
δPj
]
= δψ
n∑
j=1
[
pj −n∑
i=1
Pi
∂Qi
∂qj
]
δqj −n∑
i=1
n∑
j=1
Pi
∂Qi
∂Pj
δPj = δψ
Now,
∂
∂qk
(
pj −n∑
i=1
Pi
∂Qi
∂qi
)
=∂
∂qj
(
pk −n∑
i=1
Pi
∂Qi
∂qk
)
∂
∂pk
(
n∑
i=1
Pi
∂Qi
∂pj
)
=∂
∂pj
(
n∑
i=1
Pi
∂Qi
∂Pk
)
∂
∂pk
(
pj −n∑
i=1
Pi
∂Qi
∂qj
)
=∂
∂qj
(
−n∑
i=1
Pi
∂Qi
∂pk
)
180
where j, k = 1, 2, . . . .n.These equation can be written as
n∑
i=1
(
∂Qi
∂qi
∂Pi
∂qk− ∂Pi
∂qi
∂Qi
∂qk
)
= 0
n∑
i=1
(
∂Qi
∂pj
∂Pi
∂pK
− ∂Pi
∂pj
∂Qi
∂pK
)
= 0
n∑
i=1
(
∂Qi
∂qj
∂Pi
∂pK
− ∂Pi
∂qj
∂Qi
∂pK
)
= δjk
where δjkis the kronecker delta.
Definition: Lagrangian brackets expression of two variable (u, v)
by using the notation
[u, v] =n∑
i=1
(
∂Qi
∂u
∂Pi
∂v− ∂Pi
∂u
∂Qi
∂v
)
where u andv are any two variables q1, q2, . . . , qn, p1, p2, . . . , pn .
Skew symmetry: As a consequence of a two skew symmetry of
the Lagrangian bracket is
[u, v] = − [v, u]
and
[u, u] = [v, v] = 0.
Invariant: A general characteristic of the Lagrangian bracket is
invariant under a canonical transformation.
Example: If the 2n variables (q, p) and in variable (Q,P ) are
related by a canonical transformation. The either set may be used in
equating a given bracket evaluation. Thus we have
[u, v] =n∑
i=1
(
∂qi∂u
∂pi
∂v− ∂pi
∂u
∂qi∂v
)
=n∑
i=1
(
∂Qi
∂u
∂Pi
∂v− ∂Pi
∂u
∂Qi
∂v
)
181
where each set of dynamical variable in assume to be a function of
(u, v) as well as other variable.
Poisson brackets: Suppose that the function of the dynamical
variable and time namely u = u (q, p, t) and v = v (q, p, t). The poisson
bracket expression for function is
(u, v) =n∑
i=1
(
∂u
∂qi
∂v
∂pi
− ∂u
∂pi
∂v
∂qi
)
.
In the case of Lagrangian brackets, we have
(u, v) = − (v, u)
(u, u) = (v, v) = 0.
The poisson brackets is useful testing whether the transformation is
canonical (or) not.
Poisson theorem:
Statement: If u (q, p) and v (q, p) are integrals of a Hamiltonian
system, then the Poisson bracket (u, v) is also an integral, that is (u, v)
is constant of the motion.
Proof: Let u and v be integrals of the motion. We know that,
df
dt= (f,H) +
∂f
∂t,
(u,H) +∂u
∂t= 0, (1)
(v,H) +∂v
∂t= 0. (2)
Also,
d
dt(u, v) = [(u, v) , H] +
∂
∂t(u, v)
d
dt(u, v) = [(u, v) , H] +
(
∂u
∂t+ v
)
+
(
u,∂v
∂t
)
(3)
182
From (1) and (3)
d
dt(u, v) = [(u, v) , H] − [(u,H) , v] − [u, (v,H)]
= [(u, 0) , H] + [(H, u) , v] + [(v,H) , u]
By Jacobi identity,
[u, (v, w)] + [v, (w, u)] + [w, (u, v)] = 0
∴
d
dt(u, v) = 0.
Hence the proof.
Example: Consider the transformation
Q =√
e−2q − p2, P = cos−1 (peq) .
Use the poisson bracket to show that it is canonical.
Solution: We know that the poisson bracket
(u, v) =n∑
i=1
(
∂u
∂qi
∂v
∂pi
− ∂u
∂pi
∂v
∂qi
)
. (1)
Given that
Q =√
e−2q − p2,
P = cos−1 (peq) .
Here u = Q, v = P.
(Q,P ) =∂Q
∂q
∂P
∂p− ∂Q
∂q
∂P
∂p(2)
Q =√
e−2q − p2
∂Q
∂q=
1
2
(
e−2q (−2)√
e−2q − p2
)
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= − e−2q
√
e−2q − p2
=−p
√
e−2q − p2
P = cos−1 (peq)
∂P
∂q=
(−1) peq
√
1 − p2e2q
=−p
√
e−2q − p2
∂P
∂p=
−e2q
√
1 − p2e2q
=−1
√
e−2q − p2.
Using these values in (2) we get
(Q,P ) =e−2q
(
√
e−2q − p2
)2− p2
(
√
e−2q − p2
)2=e−2q − p2
e−2q − p2= 1
We know that (u, u) = (v, v) = 0
⇒ (Q,Q) = 0, (P, P ) = 0
Hence the given transformation is canonical.
Bilinear co-variant:
This is the another method which may be used in testing whether
a given transformation is canonical involves the bilinear co-variant.
Suppose we consider the Pfaffian differential
Ω =n∑
i=1
Xi (x) dxi (1)
where the δx ’s are an independent set of infinitesimal displacements
from the same reference point. Now we can write,
δΩ =n∑
i=1
(δXidxi +Xiδdxi)
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dQ =n∑
j=1
(dXjδxj +Xjdδxj)
where
δXi =n∑
j=1
∂Xj
∂xj
δxj,
dXj =n∑
i=1
∂Xj
∂xi
dxi.
Then we obtain
δΩ − dQ =n∑
i=1
n∑
j=1
(
∂Xi
∂xj
− ∂Xj
∂xi
)
dxiδxj
=n∑
i=1
n∑
j=1
[(δXjdxi − dxjδxj) + (Xjδdxi −Xδxj)] (2)
note that δdxi = dδxi
Now, considering contemporaneous variants where
τ ij =∂Xi
∂xj
− ∂Xj
∂xi
.
We find that
δΩdθ =n∑
i=1
n∑
j=1
τ ijdXiδxj
This is the bilinear co-variant associated with the differential equation
(1).
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