MSc (Sem I) Classical Mechanics A. R.

Pawar

UNIT -I
d dp
Conservation of linear Momentum: If the total force F is zero the i.e. F= ( mv ) = =0and linear
dt dt
momentum is conserved.
Conservation of angular momentum: Let particle of mass m performing rotational motion about an origin
‘O’ with a linear momentum ‘p’ at position vector ‘r’.
Angular momentum of the system is,
∴⃗ L= ( ⃗r × ⃗p )
The torque as moment of force about origin ‘o’
τ⃗ =⃗r × ⃗F
d ⃗p d
∵⃗ F= = ( m ⃗v )
dt dt
d O
above equation c an be written as⃗τ =⃗r × ( ⃗p )
dt
d d r⃗ d
now let ( ⃗r × ⃗p )= × ⃗p + r⃗ × ( ⃗p )
dt dt dt
d d ⃗r
∴ τ= ( ⃗r × ⃗p )− × ⃗p
dt dt

d d ⃗r
¿ ( ⃗r × ⃗p )− v⃗ × m v⃗ Since =v⃗ ∧v⃗ × v⃗ =0
dt dt
d
∴ τ= ( ⃗r × ⃗p )
dt
d⃗ L d⃗L
∴ τ= If the total torque equal to zero L= constant. Thus the angular momentum is
=0 and ⃗
dt dt
conserved.
Conservation of energy: If the particle is acted upon by the forces which are conservative, i.e. if the forces
are driveable from a scalar potential energy function. F = - 𝛻 V ------------1
A conservative force is a force with the property that the work done in moving a particle between two points is independent
of the taken path. Equivalently, if a particle travels in a closed loop, the net work done (the sum of the force acting along the
path multiplied by the distance travelled) by a conservative force is zero.

The work done by the particle is

ṙ∧dr
2 2
dp
2
d
2
∵ p=m = ṙ ⇒ dr=ṙ dt
W 12=∫ F . dr =¿∫ . dr =∫ ( m ṙ ) . dr=¿ ∫ ( m ṙ ) . ṙ dt ¿ ¿ dt
1 1 dt 1 dt 1

2 2
d 1
=∫
1 dt 2
( )
m ṙ 2 dt = ∫ dT= T2−T1 -------------------2
1

Where T1, T2 are denoted Kinetic energy of the particle at position 1 and 2 respectively.

Form equation 2
2 2 2 2
−dV
W 12=∫ F . dr =¿∫ −∇ V . dr =∫ dr =¿∫ −dV =V 1−V 2 ¿ ¿ --------------3
1 1 1 dr 1

From equation 2 and 3,

T2−T1=V 1−V 2∨ T 2+V 2 =T 1 +V 1=Consatant (Since T+ V is total energy)

This says that, the total energy of particle is conserved.

1
MSc (Sem I) Classical Mechanics A. R. Pawar

MECHANICS OF A SYSTEM OF PARTICLES

Let us consider a mechanical system of N particles. From Newton’s second law equation of motion of the
system is,
mi ai= ṗi=F (ie )+ ∑ F ij Where I = 1,2,3,4------N ------------------------(1)
j≠i

In above equation F (ie ) is external force acting on ithparticle.F ij is internal force exerted by jth particle on ith
particles, since all the particles exert forces on one another total internal force is the sum of forces
∑ F ij =0 for i=j.
j=1

Newton’s third law is valid for internal forces

F ij=−F ijSays that internal forces occur in pair and act along the line of joining the two particles and
j i
combination of these forces must be zero. ∴ ∑ F i + F j=0
i, j

Equation (1) can be written as

∑ mi ai =∑ ṗ i=∑ F ei ------------------------------(2)
i i i

Now let the centre of mass of the system is at ⃗ R and define as the average radii vectors of particles,
weighted in proportion to their mass.
∑ mi r i ∑ mi r i
R= i
⃗ = i --------------------------------(3)
∑ mi M
i

Therefore, equation (2) can be written as

d2 R e (e)
M 2 =∑ Fi =F 90090909099090909
dt i

And the total momentum of the system is

d
p= ∑ mi r i=M Ṙ
dt i
From above equation
ṗ=M R̈=F (e)
Says that,
i) Centre of mass moves as if the total external force F (e) acting on entire mass of the system were
concentrated at the centre of mass.
ii) If the total external force vanishes, the total momentum is conserved.

CONSERVATION OF ENERGY
“If the external forces are derivable from a scalar potential function and if the internal forces are central,
then the total energy of the system is conserved.”
Consider a system of N particles. Suppose a force acting on i th particle is Fi which movesthe particle
through a distance of dri. Amount of work done by this force is,
m i r̈ i dr i =Fi dr i
¿ F (ie ) + ∑ F ij dr i =F (ie ) dr i + ∑ F ij dr i mi r̈ i ṙ i dt=F(ie ) dr i+ ∑ Fij dr i where dr =ṙ dt
( j
) j j

∴ mi r̈ i ∙ ṙ i dt=Fie ∙ dr i+ ∑ F ij ∙ dr i
j

2
MSc (Sem I) Classical Mechanics A. R. Pawar

d 1
dt 2 ( )
m i ṙ 2i dt =F ei ∙ dr i+ ∑ Fij ∙ dr i
j

For system of particles

d 1
( )
m ṙ 2 dt =F ei ∙ dr i+ ∑ Fij ∙ dr i -----------------------1
dt 2 i i i, j

j 1
But net internal forces are∑ F i ∙ dr i= ∑ ( F ij ∙ dr i + Fij ∙ dr j )----------------2
i, j 2 i, j
From Newton’s third lawF =−F ij j
i

Also, we know that, conservative internal forces are derivable from scalar potential function.
∂ j
∂ ri , j
j j −∂
F i =−∇i V i =
∂ ri
( V
j
i) =− V
∂r i , j i ∂ r i ( )
∂ ∂ r 1∧∂ r j
¿−∇ i V ij, j ∙
( ∂r i )
( r i−r j ) Since i =
∂ ri ∂ ri
=0

∴ F ij=−∇ i, j V ij ---------------------------------3
Similarly
∴ F ij=∇i , j V ij ------------------------------------4
Putting equations (3) and (4) in equation (2)
1 −1
∑ F ij ∙ dr i= 2 ∑ (−∇ i, j V ij ∙ dr i +∇ i , j V ij ∙ dr j )= 2 ∑ ∇ i , j V ij d ( r i −r j )
i, j i, j i, j
1
¿− ∑ ∇ i , j V ij dr i , j-----------------------------5
2 i,j
For external conservative force
F ei =−∇ i V ei ------------------------------6
From equations 1,5 and 6,
d 1 1 1 1
dt 2 ( ) ( ) (
m i ṙ 2i dt =−∇i V ei ∙ dr i± ∑ ∇ i , j V ij ∙ dr i , j ∴∫ d m i ṙ 2i =∫ −∇ i V ie ∙ dr i± ∑ ∇ i , j V ij ∙ dr i , j
2 i, j 2 2 i, j )
1 1
∑ 2 mi ṙ 2i =−∑ ∇i V ei − 2 ∑ ❑ V ij +constant
i i i, j

OR
1 1
∑ 2 mi ṙ 2i +∑ ∇ i V ei + 2 ∑ ❑ V ij=Constant
i i i, j

T + V = Constan This proves that, “ if the external forces are derivable from a scalar
t
potentialfunction and if the internal forces are central, then the total energy of the system is conserved”

CONSTRAINTS
A constrained motion is a motion which cannot proceed arbitrary in any manner. Particle motion is
restricted to occur only.
For example
 Motion of particles in one dimension required only one coordinates (x) to describe its motion.
 If a particle moves in space than it required three coordinates (x, y, z) to describe its motion.
 In case of a rigid body, constrains is that the distance between any two particles must remain
same.

3
MSc (Sem I) Classical Mechanics A. R. Pawar

 For rotating rigid body about a fixed axis, then in addition to the distance of each particle from the
axis fixed. This is the reason it single coordinate (θ) is sufficient to determine the position of each
particle in the body.
In classical mechanics, a constraint is a relation between the coordinates and momenta. In other words, a
constraint is a restriction on the freedom of movement of a system of particles.
TYPES OF CONSTRAINTS
Time constraints
 If constraints, relationsdoes not explicitly depend on time. Constraintsare known as Scleronomic.
For example, rigid body.
 If constraints, relationsdependexplicitly depends on time. Constraintsare known as Rheonomic. For
example a bead sliding on moving wire.
Velocity constraints
 If constraints, relations are or can be made independent of velocity. Constraintsare known as
Holonomic. For example a cylinder rolling without sliding down on inclined plane.
 If constraints, relations are an irreducible function of the velocities.Constraintsare known as Non-
Holonomic. For example a sphere rolling without sliding down on inclined plane.
Equation constraints
 If constraints, relations are in the form of relation. Constraintsare known as Bilateral. For example,
rigid body.
 If constraints, relations are expressed in the form of inequalities. Constraintsare known as
Unilateral. For example motion of a particle in a gas container.
Force Constraints
 If the forces of constraint do not do any work and total mechanical energy of the system is
conserved while performing the constant motion. The constraint is known as a conservative. For
example simple pendulum with rigid support.
 If the forces of constraint do work and total mechanical energy of the system is not conserved
while performing the constant motion. The constraint is known as a conservative. For example
simple pendulum with variable length.

DEGREE OF FREEDOM
The minimum number of independent variables required to fix the position and the configuration of
amoving (dynamical) system which are compatible with the given constraints is called degree of freedom.
These independent variables must be sufficient in number to describe all positions and configurations of
the system consist with the given constraints.
For example
 The motion of a particle in one dimension requires only one variable, therefore the degree of
freedom of such a system is one.
 Particle in three dimensions require three independent (x, y, z) variables to describe its motion
therefore degree of freedom is three.
 Double pendulum in vertical plane requires two coordinates (θ 1, θ2) describe the system
completely.
GENERALISED CO-ORDINATES To describe the configuration of a system, it required the smallest
possible number of variables. These variables are called the generalized coordinates and denoted by letter
q with a numerical subscript.
A set of ‘n’ generalized coordinates would be written as q1, q2, . . . , qn.
 Thus a particle moving in a plane may be described by two coordinates cartesian coordinates x, y,
or the polar coordinates r, θ, or any other suitable pair of coordinates. In generalized coordinate it
can be written as q1, q2

4
MSc (Sem I) Classical Mechanics A. R. Pawar

 A particle moving in space is located by three coordinates, which may be cartesian coordinates x,
y, z, or spherical coordinates r, θ, φ or cylindrical coordinates ρ, z, φ, or, in generalized q 1 q2, q3.
 The configuration of a system of N particles may be specified by the 3N cartesian coordinates viz.
X1 Y1, Z1 X2, Y2, Z2, . . ., XN, YN, ZN of its particles, or by any set of 3N generalized coordinates q l
q2, . . . , q3N.
 Since for each configuration of the system the generalized coordinates must have some definite
set of values, the coordinates q l q2, . . . , q 3N will be functions of the cartesian coordinates, and
possibly also of the time in the case of moving coordinate systems:
q1 = q1 (X1 Y1, Z1 X2, Y2, Z2,. . ., XN, YN, ZN, t),
q2 = q2 (X1 Y1, Z1 X2, Y2, Z2,. . ., XN, YN, ZN, t), ………………… 1
.
.
.
q3 N = q3 N (X1 Y1, Z1 X2, Y2, Z2,. . ., XN, YN, ZN, t),
Since the coordinates q l q2, . . . , q3N specify the configuration of the system, it must also be possible to
express the cartesian coordinates in terms of the generalized coordinates:
Xl = X 1 (q 1 q2, . . . , q3N, t),
Y1 = Y 1 (q 1 q2, . . . , q3N, t), …………….. (2)
:
:
Z N = Z N (q 1 q2, . . . , q3N, t).
If Eqs. (1) are given, they may be solved for X1 , Y1, . . . , ZN to obtain Eqs. (2), and vice versa.
EQUATIONS IN TERM OF GENERALIZED COORDINATES
Let us consider the motion of a particle in one dimension. Thus, it requires only one coordinate x (t) to
describe its motion.
Let q be the generalized coordinate. Then x (t) = x [q (t), t] says that x is a function of generalized
coordinates q and time t.
 Velocity: Since x (t) = x [q (t), t] says that x is a function of generalized coordinates q and time t
∂x ∂x
dx= dq+ dt
∂q ∂t
dx ∂ x dq ∂ x
= +
dt ∂q dt ∂ t
∂x ∂x dx dq
ẋ= q̇+ -------------------------(1) Where ẋ= And q̇=
∂q ∂t dt dt
Thus the velocity of a particle is a function of q , q̇ ,t i.e. velocity is a function of generalized
position q generalized velocityq̇ and timet.
 Kinetic energy: kinetic energy of a particle in one dimension is a function of velocity.
1
T = m x˙2
2
1 ∂x ∂x 2
T= m
2 ( )
∂q
q̇+
∂t
…………………………….(2)

1 ∂x 2 ∂x ∂x 1 ∂x 2
T= m
2 ( ) ( )( )
∂q
q̇ + m
∂q
q̇ + m
∂t 2 ∂t ( )
The above equation can be written as

T = T2 + T1 + T0
Where

5
MSc (Sem I) Classical Mechanics A. R. Pawar

1 ∂x 2
T2=
2
m
∂q ( )
q̇ is a quadratic equation in generalized velocity.
∂x ∂x
T1 = m
∂q ( )( )
q̇
∂t
is linear terms.

1 ∂x 2
T0 = m
2 ∂t ( ) is independent of velocity and equal to zero for the fixed coordinate system.

 Momentum: Linear momentum of a particle in terms of cartesian coordinates is

∂T
pc =m ẋ= From equation (2)
∂ ẋ
Let us define the generalized linear momentum P associated with the generalized coordinate q by
∂ T ( q , q˙,t )
p=
∂ q̇
∂ T ∂ ẋ ∂ ẋ
p= = pc
∂ ẋ ∂ q̇ ∂ q̇
∂ ẋ ∂ x ∂t ∂x
p= p c = pc ⇒ p= pc
∂ q̇ ∂t ∂q ∂q
D’ALEMBERT PRINCIPAL
According to principal of virtual work the system is subjected to an infinitesimal displacement consistent
with the forces and constraints are imposed on the system at the given instant t. This change in
configuration is not associated with a change in time, i.e. there is no actual displacement during which
forces and constraint may change and hence displacement is termed as virtual displacement.
Suppose the system is at equilibrium says that total force acting on the system (on every particle) is zero.
This implies the small virtual displacement δri is also zero.
For the system of N particle
N

∑ F i ∙ δ r i=0 ⟹ ∑ ( F ai +f i ) δ r i=0whereF ai applied force and f i forces of constraint.

i=1 i

Consider a system for which forces of constraint are equal to zero i.e. f i=0.
Thus the above equation can be written as
∑ ( Fia) δ r i=0
i

“The condition for equilibrium of a system that the virtual work of applied forces vanishes”.The equation is
termed as principle virtual work.
A system will remain in equilibrium under the action of a force equal to the actual force Fiplus reversed
effective force of inertia on ith particle is ṗi.
F i− ṗi=0
Thus the principle of virtual work can be written as
∑ ( F i− ṗi ) ∙ δ r i =0This is D’Alembert’s Principal.
i

LAGRAGIAN EQUATION
Suppose the system is at equilibrium says that total force acting on the system (on every particle) is zero.
This implies the small virtual displacement δri is also zero.
According to D’Alembert principle of the system of N particles.
∑ ( F i− ṗi ) ∙ δ r i =0-----------------------1
i

The coordinate equations are

6
MSc (Sem I) Classical Mechanics A. R. Pawar

r i=r i ( q1 , q2, q 3 … … … … … … q n ,t )
d r i ∂ r i dq 1 ∂ r i dq 2 ∂ r i dt ∂r i ∂ ri
= + +... … … …+ ⟹ v i =∑ q̇ j + -------------2
dt ∂ q 1 dt ∂ q2 dt ∂ t dt j ∂ q1 ∂t
∂ ri ∂ ri
δ r i =∑ δ q j+ δt
i ∂qj ∂t
For virtual displacement last term is zero.
∂r
∴ δ r i =∑ i δ q j
j ∂qj

Therefore equation 1 can be written as,

∂r
∑ ( F i− ṗi ) ∙ ∑ ∂ qi δ q j=0
i j j
∂r ∂r
∑ F i ∙ ∂ qi δ q j −¿ ∑ ṗ i ∙ ∂ qi δ q j=0¿
ij j ij j
∂ ri
∑ Q j ∙ δ q j−¿ ∑ ṗi ∙ ∂ q δ q j =0 ¿ -------------------------3
ij ij j
∂ ri
Where    ∑ F i ∙ =Q j and Q j ∙ δ q j have the dimensions of work.
i ∂qj
Noe let second term of equation,
∂r i ∂ ri d ∂ ri
∂ ri
∑ ṗi ∙ ∂ q δ q j
ij j
d
dt
m
(i ṙ i
∂qj
=m
) i
d
dt
( [
ṙ i ) ∂ qj
+ ṙ i
( )]
dt ∂ q j
∂ ri ∂r ∂r d ∂ ri
¿ ∑ mi r̈ i ∙
ij ∂q j
δ q j-----refere note ∴
d
dt ( )
mi ṙ i i =mi r̈ i i + ṙ i
∂qj ∂qj [
dt ∂ q j ( )]
∂r ∂r
¿∑
{( ) }
ij
d
dt
mi v i i −mi v i i δq j
∂qj ∂q j ∴ m i r̈ i
∂r i d
=
∂ q j dt (
m i ṙ i
∂ ri
∂qj )
−m i ṙ i
d ∂ ri
( )
dt ∂ q j
∂v ∂v And ṙ i=v i
¿ ∑{ ( ) }d
dt
mi v i i −mi v i i δq j
∂ q̇ j ∂q j d ∂r i ∂ ∂ r i d q k ∂ ∂ r i dt
ij

d ∂ 1 ∂ 1
( )
dt ∂ q j
=∑
k ∂ qk ∂ q j
+ ( )
dt ∂ t ∂ q j dt ( )
¿ ∑{ ( [ ]) [ ]}
dt ∂ q̇ j 2
mi v 2i −
∂qj 2
mi v 2i δq j ∂2 r i ∂2 r i
ij ¿∑ q̇ k +
k ∂ qk ∂ q j ∂ qj ∂ t
1
∵ m i v 2i =T ∂ ri ∂ ri
2 ∂ ∂ ∂ ri
d ∂T ∂T
¿
∂qj (∑ ∂ qk
q̇ k +
∂t
= ) ( )
∂q j ∂ t
∴=∑
ij {( ) } −
dt ∂ q̇ j ∂ q j
δq j
¿
∂ vi
∂qj
k

Putting above result in equation 3

Differtiating equation 2 w. r. t. q̇ we have
d ∂T ∂T
∑ Q j ∙ δ q j−¿ ∑
j j {( ) } −
dt ∂ q̇ j ∂ q j
δq j=0 ¿

For holonomic constraints qj is independent to each otherand as j =1-----------N

d ∂T ∂T
{( ) } −
dt ∂ q̇ j ∂ q j
=Q j

CONSERVATIVE SYSTEM
Let us consider a conservative system in which forces Fiarederiveablefrom potential function V.

7
MSc (Sem I) Classical Mechanics A. R. Pawar

−∂V
F i=−∇ i V =
∂ ri
The generalised force can be written as,
∂ ri ∂ r −∂ V
Q j =∑ F i ∙ =−∑ ∇i V i ∙ i =
i ∂qj i ∂qj ∂qj
Similarly we know that, equation of for holonomic constraints.
d ∂T ∂T
[ ( ) −
dt ∂ q̇ j ∂ q j ]
=Q j from these two results

d ∂T ∂ T ∂V
( ) − +
dt ∂ q̇ j ∂ q j ∂ q j
=0

d ∂ ( T −V ) ∂ ( T −V )
dt ( ∂ q̇ )
− =0
j ∂q j

Since V is a function of q̇ j
d ∂L ∂L
( ) −
dt ∂ q̇ j ∂ q j
=0Where (T – V ) = L define as Lagrangianfor the conservative system.

VELOCITY DEPENDENT FUNCTION

Ifpotentias are velocity dependent, then L= T - U equation of motion can be written as,
d ∂ ( T −U ) ∂ ( T −U )
dt ( ) ∂ q̇ j
−
∂qj
=0where U is called generalised potential U=U(q, q̇ j )

d ∂L ∂L
( ) −
dt ∂ q̇ j ∂ q j
=0Where (T – U ) = L define as Lagrangianfor the non-conservative system.

HAMILTONIAN PRINCIPLE
Many interesting physics systems describe systems of particles on which many forces areacting. Some of
these forces are immediately externally applied. Other forces are not immediately but they are applied by
theexternal constraints imposed on the system. These forces are often difficult to quantify, but theeffect
of these forces is easy to describe. Trying to describe such a system in terms of Newton'sequations of
motion is often difficult since it requires us to specify the total force. Such a system describing by applying
Hamilton's principle, which allow usto determine the equation of motion for system for which we would
not be able to derive theseequations easily on the basis of Newton's laws. Hamilton'sprinciple does not
provide us with a new physical theory, but it allows us to describe the existingtheories in a new and
elegant framework.
The minimization approach to physics was formalized in detail by Hamilton, and resulted in
Hamilton's Principle which states:
" Of all the possible paths along which a dynamical system may more from one point toanother within a
specified time interval (consistent with any constraints), the actual pathfollowed is that which
minimizes the time integral of the difference between the kineticand potential energies. "
OR
“The motion of system from t1 to t2is such that the line integral”
t2

I =∫ ( T −V ) dt=0
t1

We can express this principle in terms of the calculus of variations:

8
MSc (Sem I) Classical Mechanics A. R. Pawar
t2

I =δ ∫ ( T −V ) dt=0
t1

The quantity T - V is called the Lagrangian L.

CALCULUS OF VARIATION
The main interest of dynamical problem is to locate the position of particle at particular instant of time
and path adopted. The exremum of path can be calculated by diffential calculus by putting ẏ ( x ) =0ect.
If we want information of whole path as longest or shortest for this purpose we rquiredintergralcalculus
of variation.
We have a function f ( y , ẏ , x )defined on a path y = y(x) between two values of x1 and x2. Thus we wish to
find a particular path y(x) such that the line integral J of the function fbetweenx1 and x2
x2

J=∫ ( y , ẏ , x ) dx ----------------------------1
x1

to have a stationary values. The intergrandf taken to be a function of dependent variable y, independent
dy
variable x and ẏ= .
dx
x2

δJ =δ ∫ ( y , ẏ , x ) dx=0whereδ is called variation and represents the increase in the quantity to which it is
x1

applied in the switching from stationary path to the comparison path at a fixed value of x.
Since J must have a stationary value for correct path relative to neighbouring path. The variation must be
zero for other path denoted by an infinitesimal parameter α and the set of such path denoted by
y(x,α),with y(x,0) reprent correct path.
y(x,α) = y(x,0) + αη(x) where η(x) is any arbitrary function of x which vanishes at x1 and x2 since it is
variation with fixed ends.
Thus integral J can be written as,
x2

J ( α )=∫ f [ y ( x , α ) , ẏ ( x , α ) , x ] dx
x1

The condition for obtaining a stationary point is,

x2
dJ ∂ f ∂ y ∂ f ∂ ẏ ∂ x ∂x
( )
dα α =0
=∫ (
x1
+ +
∂ y ∂ α ∂ ẏ ∂ α ∂ α )
dx=0since x is not function of α ⇒
∂α
=0
x2

( dαdJ ) α =0
=∫ (
x1
∂f ∂ y ∂ f ∂ ∂ y
∂ y ∂ α ∂ ẏ ∂ α ∂ x )
+ dx
x2 x2

( dαdJ ) α =0
=∫ (
x1
∂f ∂ y
∂ y ∂α )
dx+∫ (
∂f ∂ ∂ y
∂ ẏ ∂ α ∂ x )
x1
dx

By parts

x2 x2 x2
∂f ∂ ∂ y ∂f ∂ y d ∂f ∂ y
∫(
x1
∂ ẏ ∂α ∂ x
dx= ) [
∂ ẏ ∂ α ] x1
−∫
x1
( )
dx ∂ ẏ ∂ α
dx

x2
∂ f ∂ x2 ∂ x1
¿ [ ( −
∂ ẏ ∂ α ∂ α
−∫
x
d ∂f ∂ y
dx ∂ ẏ ∂ α
dx)] 1
( )
since x is not function of α therefore 1st term on RHS = 0
x2 x2
∂f ∂ ∂ y d ∂f ∂ y 9
∴∫
x1
( ∂ ẏ ∂ α ∂ x)dx=−∫
x
dx ∂ ẏ ∂ α
dx
1
( )
MSc (Sem I) Classical Mechanics A. R. Pawar

x2 x2
dJ ∂f ∂ y d ∂f ∂ y
∴
dα( ) α=0
=∫
x1
( ∂ y ∂α
dx−∫
x
)
dx ∂ ẏ ∂ α
dx
1
( )
x2
dJ ∂f d ∂f
∴
dα( ) α=0
=∫
x1
[ −
∂ y dx ∂ ẏ ( )] ∂∂ αy dx
x2
dJ ∂f d ∂f
∴
dα( ) α=0
dα =∫
x1
[ −
∂ y dx ∂ ẏ ( )]( ∂∂ αy ) dα dx
0

∴ ( dJdα ) α=0
dα=increses∈theintegral I as we pass ¿ theextremum path ¿ the

comparision pathat the same value of x.

∴ ( dJdα ) α=0
( ∂∂ αy ) dα =δy
dα =δJ ∧
0
x2
∂f d ∂f
∴ δJ =∫
[ ∂ y dx ∂ ẏ )] δy dx=o
− (
x1

∂f d ∂f
∴ −
∂ y dx ∂ ẏ
=0 ( )
DERIVATION OF LARGANGE’S EQUATION FROM HAMILTON’S PRINCIPLE
The basic problem of the calculus of variation is easily generalized to the case where f function of many
independent variables yi and there dserivatives.
The variation of integral J,
x2

δJ =∫ f [ y ( x , α ) , ẏ ( x , α ) , x ] dx
x1

The variation must be zero for other path denoted by an infinitesimal parameter α and the set of such
path denoted by y1 (x,α),with y1 (x,0) reprent correct path.
yi (x,α) = yi (x,0) + αηi (x)
where η(x) is any arbitrary function of x which vanishes at 1 and 2 since it is variation with fixed ends.
Thus integral J can be written as,
2
J ( α )=∫ f [ y i ( x , α ) , ẏ i ( x , α ) , x ] dx
1

The condition for obtaining a stationary point is,

2
dJ ∂ f ∂ yi ∂ f ∂ ẏ i ∂x ∂x
( ) (
dα
dα=∫
1 ∂ yi ∂ α
dα +
)
∂ ẏi ∂ α
dα +
∂α
dα dx=0since x is not function of α ⇒
∂α
=0
2
dJ ∂f ∂y ∂ f ∂ ẏ
( dα ) ( ∂ y ∂ α ∂ ẏ ∂ α dα ) dx
dα=∫ dα +
1 i
i

i
i

10
MSc (Sem I) Classical Mechanics A. R. Pawar
2 2
dJ ∂f ∂ y ∂ f ∂ ∂ yi
( )
dα α =0
=∫
1
( ∂ y ∂α
dx+∫ )
1 ∂ ẏ i ∂ α ∂ x
(
dx
)
By parts

x2 x2 x2 x2
∂ f ∂ ∂ yi ∂ f ∂ yi d ∂ f ∂ yi ∂ f ∂ x2 ∂ x1 d ∂ f ∂ yi
∫
x1
( ∂ ẏi ∂ α ∂ x
dx = ) [
−∫
∂ ẏ i ∂ α x x dx ∂ ẏ i ∂ α
dx=
] −
∂ ẏ i ∂ α ∂ α
1
−∫
x dx ∂ ẏ i ∂ α
1
( )
dx
[ ( )] 1
( )
since x is not function of α therefore 1st term on RHS = 0
x2 x2
∂ f ∂ ∂ yi d ∂ f ∂ yi
∴∫
x1
( ( )
∂ ẏ i ∂ α ∂ x
dx=−∫
x
)
dx ∂ ẏi ∂α
dx
1

2 2
dJ ∂f ∂ y d ∂f ∂y
∴ ( ) =∫
dα ( ∂ y ∂ α dx ( ∂ ẏ ) ∂ α dx
dx−∫
1 i
i
) 1 i
i

2
dJ ∂f d ∂f ∂ yi
∴ ( )
dα
=∫
1 ∂ y
−
dx ∂ ẏ i [ ( )] ∂α
dx

2
dJ ∂ f d ∂f ∂ yi
∴ ( )
dα
dα =∫
1 ∂
−
y dx ∂ ẏ i [ ( ) ]( ) ∂α 0
dα dx

∴ ( dJdα ) dα =increses∈theintegral I as we pass ¿ the extremum path ¿the

comparision pathat the same value of x.

∂ yi
∴ ( dJdα ) dα =δJ ∧( ∂ α ) dα =δ y 0
i

x2
∂f d ∂f
∴ δJ =∫
x1
[ −
∂ y i dx ∂ ẏ i ( )]
δ y i dx =o

Since the y i variables are independent ∴ δJ =0

∂f d ∂f
∴ −
∂ y i dx ∂ ẏ i
=0 ( )
Above equation appropriate generstion to the several variables and are known as the Euler- Lagrange
equations.

11
MSc (Sem I) Classical Mechanics A. R. Pawar

UNIT II
CONSERVATION THEROREMS
An important property of the Lagrangian is that conservation laws can easily be read off from it. For
example, if the Lagrangian   does not depend on   itself, then the generalized momentum ( ), given by:

is a conserved quantity, because of Lagrange's equations:

It doesn't matter if   depends on the time derivative   of that generalized coordinate, since the
Lagrangian independence of the coordinate always makes the above partial derivative zero. Such
coordinates are called "cyclic" or "ignorable".

12