2.0 CLASSICAL MECHANICS

2.1 FUNDAMENTAL TECHNIQUES

2.1.1 The Virial Theorem

The equation of motion of a system can be written in the form

r

r

F pi i− =& 0. (2.1)

We are interested in the quantity

G p ri ii

= r r

⋅∑ . (2.2)

Differentiating this expression gives

dG

dtr p p m r r pi i

ii ri

ii i i

ii ri

i

= + = +⋅ ⋅ ⋅ ⋅∑ ∑ ∑ ∑r r r r r rr r

& & & & & . (2.3)

From Equation 2.1 this reduces to

dG

dtT Fi ri

i

= + ⋅∑2r

r . (2.4)

Averaging over a period of time τ, we obtain

1 10 2

0τ τ

ττ

dG

dtdt G G T F ri i

iz ∑= − = + ⋅( ) ( )

r

r

. (2.5)

If the motion is periodic, with τ = period, then then left hand side of Equation2.5 is zero and we see that

T F ri ii

= − ⋅∑1

2

r

r

. (2.6)

2.1.2 D'Alembert's Principle

From Equation 2.1 it follows that the virtual work done by this system is alsozero,

r

r r

F p ri ii

i− ⋅ =∑ &d i δ 0. (2.7)

The total force will be a combination of externally applied forces and internalconstraints,

r r r

F F fi ia

i= ′ + , (2.8)

so that Equation 2.8 reduces to

′ − ⋅ − ⋅ =∑ ∑r

r r

r

r

F p r f ria

ii

i ii

i&e j δ δ 0. (2.9)

If we restrict our attention to rigid bodies and other systems for which theforces of constraint do no work then we conclude that the condition forequilibrium of a system is given by D'Alemberts principle which states

r

r r

F p ria

ii

i− ⋅ =∑ &e j δ 0. (2.10)

2.1.3 Lagrange's Equation

If r

ri is a function of independent variables qi, then

δ∂∂

δr

r

r =r

qii

jj∑ q j . (2.11)

Dropping the superscript "a" for convenience, the first term from Equation 2.10is

r

r

r

r

F r Fr

qq Q qi i

ii

i j

i

jj j

jj⋅ = ⋅ =∑ ∑ ∑δ

∂∂

δ δ,

, (2.12)

where Qj is the generalized force. The second term in Equation 2.10 is

r r r r r

r

& && &&

,

p r m r r m rr

qqi i

ii i i

ii i

i j

i

jj⋅ = ⋅ = ⋅∑ ∑ ∑δ δ ∂

∂δ . (2.13)

By definition

m rr

q

d

dtm r

r

qm r

d

dt

r

qi ii

jii i

i

ji i

i

ji

r

r

r

r

r

r

&& & &⋅ = −F

HG

I

KJ

F

HG

I

KJ

F

HGG

I

KJJ∑ ∑ ⋅ ⋅

∂∂

∂∂

∂∂

, (2.14)

and

d

dt

r

q

r

q

v

qi

j

i

j

i

j

∂∂

∂∂

∂∂

r r rF

HG

I

KJ = =

&

. (2.15)

Similarly

r

vik

= ∂∂

∂∂

r

qq

r

ti

kk

i& +∑ (2.16)

so it follows that

∂∂

∂∂

r r

v

q

r

qi

j

i

j&

= . (2.17)

Using Equations 2.15 and 2.17 in Equation 2.14 we find that

m rr

q

d

dtm v

v

qm v

v

qi ii

jii i

i

ji i

i

ji

r

r

r

r

r

r

&&

&⋅ ⋅ ⋅∑ ∑=

F

HG

I

KJ −

F

HG

I

KJ

∂∂

∂∂

∂∂

. (2.18)

Equations 2.12 and 2.18 are combined to give

d

dt qm v

qm v Q q

ji i

i ji i

ij

jj

∂∂

∂∂

δ&

1

2

1

202 2∑ ∑∑

F

HGG

I

KJJ

F

HGG

I

KJJ

−F

HGG

I

KJJ −

F

HGG

I

KJJ

= , (2.19)

or

d

dt

T

q

T

qQ q

j jj j

∂∂

∂∂

δ&

F

HG

I

KJ −

F

HG

I

KJ −

F

HGG

I

KJJ

=∑j

0. (2.20)

Consequently,

d

dt

T

q

T

qQ

j jj

∂∂

∂∂&

F

HG

I

KJ −

F

HG

I

KJ = . (2.21)

If F Vi i= −∇ , then

Q Fr

qV

r

q

V

qj ii

jii

i

ji j =

r

r r

⋅∑ ∑= − ∇ ⋅ = −∂∂

∂∂

∂∂

. (2.22)

Equation 2.21 is equivalent to

d

dt

T

q qT V

j j

∂∂

∂∂&

( )F

HG

I

KJ − −

F

HG

I

KJ = 0. (2.23)

We define L=T–V to be the Lagrangian. If V is not a function of time then

d

dt

L

q

L

qj j

∂∂

∂∂&

F

HG

I

KJ −

F

HG

I

KJ = 0, (2.24)

which is known as Lagrange's equation, can be used to determine the equationsof motion. We also define

pL

qjj

= ∂∂ &

, (2.25)

to be the canonical, or conjugate, momentum. If the Lagrangian does notcontain a given coordinate qj then the coordinate is said to be cyclic orignorable.

2.1.3.1 The Two Body Central Force Problem

As an example of the application of Lagrange's equation, consider a system oftwo mass points m1 and m2 subject to an interaction potential V, where V is anyfunction of the vector between the particles. The kinetic energy is

T m r m r= 1

2

1

21 12

2 22

& & + . (2.26)

The kinetic energy can also be written as the kinetic energy of the center ofmass plus the kinetic energy about the center of mass. We define

r

R = positionof the center of mass, and

r r r

r r r= −1 2 = vector between m1 and m2. The kineticenergy of the center of mass is given by

T m m Rcm = 1

2 1 22( + ) & . (2.27)

Relative to the center of mass, the position of m1 and m2 are given by

r r

rm

m mr1

2

1 2

' =−

+, (2.28)

and

r r

rm

m mr2

1

1 2

' =+

. (2.29)

Therefore, the kinetic energy about the center of mass is given by

T m r m rm m

m mr' & & &=

F

HGI

KJ1

2

1

2

1

21 12

2 22 1 2

1 2

2' + ' = +

. (2.30)

Consequently, the Lagrangian is

L m m Rm m

m mr r rV=

F

HGI

KJ−

1

2

1

21 22 1 2

1 2

2( + ) ++

( , ,...) &

& & . (2.31)

We see immediately that because the potential is only a function of the vectorbetween the particles the conjugate momentum of R, (the momentum of thecenter of mass), is constant. That is, the motion of the center of mass has no

effect on the motion about the center of mass. This also implies that there willbe no out of plane motion.

2.1.3.2 The Inverse Square Law of Forces

When V is a function of r only, as is the case for gravitational or electrostaticforces, Equation 2.31 may be expressed in polar coordinates as

L r r V r= −1

22 2 2µ θ&

&+ ( )e j . (2.32)

Note that we have chosen to ignore the term describing the motion of the centerof mass since it has no effect on other parameters and we have introduced thedefinition

µ =+

m m

m m1 2

1 2, (2.33)

where µ is termed the reduced mass. The equations of motion are found fromLagrange's equation, (Equation 2.24). For the variable q = θ we have

d

dt

L L d

dtr

i i

∂∂θ

∂∂θ

µ θ&

( )&F

HGI

KJ− = =2 0, (2.34)

while for q = r we have

d

dt

L

r

L

rr r

V

r

∂∂

∂∂

µ µ θ ∂∂&

&&

&

FHG

IKJ

− = − + =2 0 . (2.35)

For many problems of interest, such as orbital mechanics, m m1 2>> andµ → m2. The physical consequence is that the smaller particle, m2, is subjectedto the largest perturbation in its motion. From this point forward we willfollow the usual convention and replace µ with the symbol m with theunderstanding that it refers to the motion of the smaller of m1 and m2 about thecenter of mass.

Equation 2.34 is the statement of conservation of angular momentum. That is,

l mr= 2&θ , (2.36)

is a constant. Equation 2.36 can be rewritten in the form

ldt mr d= 2 θ , (2.37)

which implies

d

dt

l

mr

d

d=

2 θ, (2.38)

and

d

dt

l

mr

d

d

l

mr

d

d

2

2 2 2= F

HGIKJθ θ

. (2.39)

The area swept out by a moving body is given by

A r r= 1

2θb g . (2.40)

It follows that

dA

dt

l

mr

d

dr r

l= L

NMO

QP=

2

1

2 2θθb g . (2.41)

Because angular momentum is conserved dA/dt is also constant. This isKepler's Second Law which states that the planets sweep out equal areas inequal times.

From the definition of l, equation 2.35 becomes

mrl

mr

V

r&&− = −

2

3

∂∂

, (2.42)

or, because V is only a function of r,

mrd

drV

l

mr&& = − +

F

HG

I

KJ

1

2

2

2. (2.43)

The particle moves in an effective potential given by

V Vl

mreff = +

F

HG

I

KJ

2

22. (2.44)

Equation 2.43 reduces to

mrdV

dreff

&& = − , (2.45)

thus

mrrd

dtmr

dr

dt

dV

dr

dV

dteff eff

&&& & = 1

22F

HGIKJ

= − = − . (2.46)

Consequently,

d

dtmr Veff

1

202

& +FHG

IKJ

= , (2.47)

which is the statement that energy is conserved. From this equation, we alsofind that

&rm

E Vm

E Vl

mreff= ± −F

HGIKJ

= ± − −F

HG

I

KJ

F

HG

I

KJ

2 2

2

2

2d i , (2.48)

or simply

dtdr

mE V

l

mr

=

− −F

HG

I

KJ

2

2

2

2

. (2.49)

Substituting the relation between dt and dθ, Equation 2.37, and we find that

dmE

l

mV

l r

dr

rθ =

− −

1

2 2 12 2 2

2. (2.50)

Consider the case when the potential is of the form Vk

rku= − = − . Equation

2.50 becomes

dmE

l

mk

lu u

duθ = −

+ −

1

2 22 2

2. (2.51)

Integrating this expression gives

θ θf i

umk

l

mE

l

mk

l

l

mk r

El

mk

− =− +

+ FHG

IKJ

=

−F

HG

I

KJ

+

arcsin arcsin2

2

8 2

11

21

2

2 2

2

2

2

2

. (2.52)

Inverting this expression gives

11 1

22

2

2r

mk

l

El

mkf i= − +

L

N

MM

O

Q

PP

−R

S|

T|

U

V|

W|sin θ θd i . (2.53)

This is usually written in the form

11

2r

mk

le f i= − −sin θ θd i , (2.54)

where

eEl

mk= +1

2 2

2. (2.55)

This is the equation for a conic section having several classes of solutions asshown below.

1.) If e > 1, and E > 0, the orbit is a hyperbola.

2.) If e = 1, and E = 0, the orbit is a parabola.

3.) If e < 1, and E < 0, the orbit is an ellipse.

4.) If e = 0, and Emk

l= −

2

22, the orbit is a circle.

2.2 VARIATIONAL TECHNIQUES

2.2.1 The Calculus of Variations

Consider a function f y y x( , & , ) defined on a path y=y(x) between x1 and x2where & /y dy dx= . We wish to find a particular path y(x) such that the integral

J f y y x dx

x

x

= ( , & , )

1

2

z , (2.56)

has a stationary value relative to paths differing infinitesimally from the correctfunction y(x). Since J must have a stationary value for the correct path relativeto any neighboring path, the variation must be zero relative to some particularset of neighboring paths. Such a set of paths can be denoted by

y x y x x( , ) = ( , ) + ( )α αη0 , (2.57)

where y(x, 0) is the correct path and η(x1) – η(x2) = 0. Explicitly,

J f y x y x x dx

x

x

( ) =α α α( ( , ), &( , ), )

1

2

z . (2.58)

A necessary condition for a stationary point is

dJ

dα α

FHG

IKJ =

=0

0 . (2.59)

From Equation 2.58

dJ

d

f

y

y f

y

ydx

x

x

α∂∂

∂∂α

∂∂

∂∂α

= +FHG

IKJz

&

&

1

2

. (2.60)

It is easily seen that

∂∂

∂∂α

∂∂

∂∂ ∂α

f

y

y f

y

y

x&

&

&

=2

, (2.61)

so that

∂∂

∂∂α

∂∂

∂∂ ∂α

∂∂

∂∂α

∂∂

∂∂α

f

y

ydx

f

y

y

xdx

f

y

y d

dx

f

y

ydx

x

x

x

x

x

x

x

x

&

&

& &

&

&

1

2

1

2

1

2

1

22

z z z= = −FHG

IKJ

. (2.62)

From the boundary conditions, the first term on the right hand side vanishesand Equation 2.60 reduces to

dJ

d

f

y

d

dx

f

y

ydx

x

x

α∂∂

∂∂

∂∂α

= −FHG

IKJz

&

1

2

. (2.63)

The fundamental lemma of the calculus of variations states that if

M x x dx

x

x

( ) ( )η

1

2

0z = , (2.64)

for all η(x) continuous through the second derivative then M(x) must beidentically zero on the interval. Thus J is stationary only if

∂∂

∂∂

f

y

d

dx

f

y−

F

HGI

KJ=

&

0. (2.65)

If f is a function of many independent variables then

∂∂

∂∂

f

y

d

dx

f

yi i−

F

HGI

KJ=

&

0. (2.66)

2.2.2 Hamilton's Principle

Hamilton's principle states that the motion, in configuration space, of a systemwhere all non-constraining forces are derivable from a generalized scalarpotential that may be a function of coordinates, velocities, and time is such thatthe integral

I Ldt

t

t

=

1

2

z , (2.67)

has a stationary value for the correct path of the motion. That is,

δ δI Ldt

t

t

=

1

2

0z = . (2.68)

The integral I is termed the action, and Hamilton's principle states that thevariation in I is zero. In other words, the action is minimized. By comparisonwith Equation 2.65 it follows that

∂∂

∂∂

L

q

d

dt

L

qi i− =

&

0, (2.69)

which is Lagrange's equation.

2.2.3 Lagrange Multipliers

D'Alemberts principle, and the resulting form of Lagrange's equation, assumeno constraint forces. Consider a treatment when the equations of constraint canbe put in the form

a dq alk k ltk

+ =∑ 0 , (2.70)

where the alk and alt's may be functions of a, t. For virtual displacements itfollows that

a qlk kk

δ =∑ 0 . (2.71)

If this is true, then it must also follow that

λ δl lk kk

a q =∑ 0 , (2.72)

where the λl are undetermined coefficients called Lagrange multipliers. FromEquations 2.63, 2.68, and 2.69 it is seen that Hamilton's principle is equivalentto

∂∂

∂∂

δL

q

d

dt

L

qq

k kkk−

F

HGI

KJ=∑z

&

0. (2.73)

By the same process, Equation 2.72 is equivalent to

λ δl lk kk l

a q dt,

∑z = 0. (2.74)

We combine these two relations to obtain

∂∂

∂∂

λ δL

q

d

dt

L

qa q dt

k kl lk

lk

n

k− +F

HGG

I

KJJ =∑∑z

=&

1

0. (2.75)

The δqk 's are not necessarily independent, but because the values of the λl's areundetermined we may choose them such that

d

dt

L

q

L

qa

k kl lk

l

∂∂

∂∂

λ&

F

HGI

KJ− = ∑ . (2.76)

Thes equations, together with Equation 2.70, can be used to determine theequations of motions for systems with constraining forces.

Example 2.1

Consider the case of a ladder oflength L that is inclined against africtionless wall and floor asshown at right. Find the equationsof motion.

θ

The position of the center of mass of the ladder, and its orientation, can bedescribed with the variables x, y, θ. The motion of the ladder is constrained bythe wall and floor. We have the two constraints

xL=2

sinθ , (2.77)

and

yL=2

cosθ . (2.78)

From Equation 2.60 it follows that these give the constraining relations

λ θ θ1 20dx

Ld−L

NMO

QP=cos , (2.79)

and

λ θ θ2 20dy

Ld+L

NMO

QP=sin , (2.80)

respectively. By inspection, the kinetic energy is

T m x y I= + +1

2

1

22 2 2

& &

&e j θ , (2.81)

where I mL= 1

122. Similarly, the potential energy is

V mgy= , (2.82)

so that the Lagrangian is

L m x ymL

mgy= + + −1

2 242 2

22

& &

&e j θ . (2.83)

From Equations 2.76, 2.79, and 2.80 the equations of motion are

mx&& = λ1, (2.84)

my mg&&+ = λ2 , (2.85)

and

mL L L2

1 224 2 2&& cos sinθ λ θ λ θ= − F

HGIKJ

+ FHG

IKJ, (2.86)

respectively. From Equations 2.77 and 2.78 we see that

&& sin & cos &&xL L= − +2 2

2θθ θθ , (2.87)

and

&& cos & sin &&yL L= −2 2

2θθ θθ . (2.88)

From Equations 2.84 and 2.87 we see that

λ θθ θθ12

2 2= − +F

HGIKJ

mL L

sin & cos && , (2.89)

while from Equation 2.85 and 2.88 we see that

λ θθ θθ22

2 2= − +F

HGIKJ

mL L

gcos & sin && . (2.90)

When these are combined with Equation 2.86 and simplified we obtain

1

62 2

&& sin & cos && cos cos & sin && sinθ θθ θθ θ θθ θθ θ= − + − − +e jb g e jb gg , (2.91)

which is equivalent to

− = − −5

62 2

&& sin cos & sinθ θ θθ θg . (2.92)

2.2 RIGID BODY MOTION

2.2.1 Rotations

A rigid body in space needs 6 independent generalized coordinates to specify itsconfiguration. For example, 3 coordinates are needed to specify the location ofthe center of mass relative to some external axes and 3 other coordinates areneeded to specify the orientation of the body relative to a coordinate systemparallel to the external axes. The orientation is specified by stating thedirection cosine of the body axes relative to the external axes. That is, if theprime denotes body axes then

$' $ ' $ $ $' $ $ $' $ $i i i i i j j i k k= ⋅ + ⋅ + ⋅d i d i e j , (2.93)

or

$' $ $$i i j k= + +α α α1 2 3 , (2.94)

or

$' cos $' , $ $ cos $' , $ $ cos $' , $ $i i i i i j j i k k= + +d i d i e j . (2.95)

Similarly,

r

t

r

x Rx'= , (2.96)

where

t

R =F

H

GG

I

K

JJ

α α αβ β βγ γ γ

1 2 3

1 2 3

1 2 3

. (2.97)

Because

r r

x x' = , (2.98)

we have

t t t

RR It = , (2.99)

or

α β γ12

12

12 1+ + = , (2.100)

for l = 1, 2, 3. That is,

x a xi ij j'= , (2.101)

and

x x a a x x x xi i ij ik j k i i' '= = . (2.102)

Therefore αijαik = 1, if j = k, αijαik = 0 otherwise. In two dimensions

t

R =−

FHG

IKJ

cos sin

sin cos

φ φφ φ

. (2.103)

2.2.1.1 General Properties of Rotations

If t t t

G = AF and we transform to a new coordinate system thent t t t t t t t t t

BG BAF BAB BF= = -1 we say that t t t

BAB-1 is the form of t

A in the new

coordinate system. t t t t

A BAB'= -1 defines a similarity transformation. In somecoordinate system

t

A'

cos sin

sin cos= −F

H

GG

I

K

JJ

φ φφ φ

0

0

0 0 1

, (2.104)

so that TrAt

= +1 2cosφ . This property holds true in all coordinate systems.

Example 2.2

Find the axis of rotation and compute the angle of rotation for

t

R = −

−

F

H

GGGGGG

I

K

JJJJJJ

3

4

6

4

1

46

4

1

2

6

41

4

6

4

3

4

.

It can easily be verified that t t t

RR IT = , so t

R satisfies the requirements of arotation matrix. The axis of rotation can be defined as the direction that anyvector which remains unchanged by

t

R points. That is, if r

t

r

x Rx= , then r

x is theaxis of rotation. We must solve

3

4

6

4

1

46

4

1

2

6

41

4

6

4

3

4

1

2

3

1 2 3−

−

F

H

GGGGGG

I

K

JJJJJJ

F

H

GG

I

K

JJ

=x

x

x

x x xb g, (2.105)

which gives the equations

3 6 41 2 3 1x x x x+ + = , (2.106)

− + + =6 2 6 41 2 3 2x x x x , (2.107)

x x x x1 2 3 36 3 4− + = . (2.108)

These three equations can be solved to show that x1 = x3 and x2 = 0. The

normalized axis of rotation is therefore 1

21 0 1, ,b g . By examination, Tr

t

R = 1

+ 2 cos θ = 2, so that θ = 60º.

2.2.2.2 The Euler Angles

Rather than specify the 9 independent elements of the rotation matrix we maydescribe the orientation in terms of 3 Euler angles. For example, rotate theinitial system of axes, xyz, by an angle ψ counterclockwise about z. Thisdefines the ξηζ axes. Next rotate about the ξ axis by an angle θ in thecounterclockwise direction. This defines the ξ'η'ζ' axes. Finally rotatecounterclockwise by an angle φ about the ζ' axes. This defines the ξ'ψ'ζ' axes.In matrix form

r

t

r

x Ax'= , (2.109)

where

t t t t

A BCD= , (2.110)

and

t

B = −F

H

GG

I

K

JJ

cos sin

sin cos

ψ ψψ ψ

0

0

0 0 1

, (2.111)

t

C =−

F

H

GG

I

K

JJ

1 0 0

0

0

cos sin

sin cos

θ θθ θ

, (2.112)

t

D = −F

H

GG

I

K

JJ

cos sin

sin cos

φ φφ φ

0

0

0 0 1

. (2.113)

2.2.2.3 The Cayley-Klein Parameters

Consider a general linear transformation in 2 dimensional space

r r r

u u v'= +α β , (2.114)

and

r r r

v'= u + vγ δ , (2.115)

where the transformation matrix is

t

Q =FHG

IKJ

α βγ δ

. (2.116)

Note that α, β, γ, δ may be complex. If we require t t t

QQ It = and t

Q = +1 we

find that β = −γ* and δ = α*, that is

t

Q =−

FHG

IKJ

α ββ α* *

. (2.117)

Consider a matrix of the form

t

Pz x iy

x iy z=

−+ −

FHG

IKJ, (2.118)

such that

t t t t

P Q P Qt'= . (2.119)

The hermitian property and the trace of a matrix are unaffected by similaritytransformations. Consequently,

t

P' is of the form

t

Pz x iy

x iy z'

' ' '

' ' '=

−+ −

FHG

IKJ

. (2.120)

If we let x x iy+ = + and x x iy− −= , then

t

Pz x

x z

z x

x z'

' '

' '=

FHG

IKJ

=FHG

IKJ −FHG

IKJ

−−

FHG

IKJ

−

+

−

+

α βγ δ

δ βγ α

. (2.121)

In this way, we may define a 9 element rotation matrix in terms of 4 Cayley-Klein parameters.

2.2.2 The Rate of Change of a Vector

The rate of change of a vector r

r as seen by an observer in the body system ofaxes will differ from the corresponding change as seen by an observer fixed inspace. If the body axes are rotating with angular velocity ω the generalsolution is

dr

dt

dr

dtr

space body

r r

r rFHG

IKJ

= FHG

IKJ

+ ×ω . (2.122)

We have

r r r r

v v rs b= +ω × , (2.123)

and a successive application of Equation 2.122 gives

r r r r r r r

r

r

a a v rd

dtrs b b= × + × × + ×+ 2( ) ( )ω ω ω ω

. (2.124)

If the angular velocity of the body is constant Newton's law is

r

r rr

rr r

r

F ma ma m v m rs b b= + ( ) + ( )= 2 ω ω ω× × × . (2.125)

To an observer in the rotating system it appears as though the particle ismoving under an effective force

r

r

rr r r r r

F ma F m v m reff b b= = ( ) ( )− × − × ×2 ω ω ω . (2.126)

The second term on the right hand side is called the Coriolis force and the lastterm on the right hand side is called the centrifugal force.

Example 2.3

Show that if a particle is thrown up vertically with initial speed vo and reacheda height h, that it will experiment a Coriolis deflection that is opposite indirection, and four times greater in magnitude, than the deflection it wouldexperience if it were dropped at rest from the same maximum height.

The rate of rotation of the Earth,ω, is identical for each scenario.The difference is the initialvelocity and position of theparticle. From equation 2.126, theCoriolis force is 2

r rω × vbb g.

Defining the coordinate axes asshown, we have ω ωx y= = 0 and

ω ωz = , while vx = 0,v vy = cosθ , and v vz = sinθ .

xs

ys

zs zb

yb

xb

θ

ω

As a result,

− × = + + +2 2 2 0 2 0r rω ω θv v i j kbb g b g b g b gcos $ $

$, (2.127)

As a result, the acceleration due to the Coriolis force is simply

&& cosx v= 2ω θ. (2.128)

A body falling under the influence of gravity satisfies the condition

v v gt= −0b g , (2.129)

where v(0) is defined as the velocity at time t = 0. Substituting this intoEquation 2.128 and integrating gives

& cosx v t gt= −L

NMO

QP2 0

1

22ω θ b g . (2.130)

Integrating a second time gives

x v t gt= −L

NMO

QP2

1

20

1

62 3ω θcos b g , (2.131)

To solve the problem at hand, we let t be the time it takes for the particle tocomplete the trip. The position of the particle at any time is given by

h t h v t gtb g b g b g= + −0 01

22, (2.132)

In one case, the particle starts at rest, v(0) = 0, at height h. From the definitionof the problem, it can be seen that the time required to complete the drop to theground is

th

g

v

go

12= = , (2.133)

Consequently, the deflection when dropped from rest at height h is

x gv

g

v

ggo o

1

3 3

21

6

1

3= −

FHG

IKJ

L

N

MM

O

Q

PP

= −FHG

IKJFHG

IKJ

ω θ ω θcos cos . (2.134)

If instead, we let t2 be the time it takes to fall from height h, we have vi = voand t2 = 2t1. The deflection is

x vv

gg

v

g

v

go

o o o2

2 3 3

22

1

2

2 1

6

2 4

3=

FHG

IKJ

−FHG

IKJ

L

N

MM

O

Q

PP

= +FHG

IKJ

ω θ ω θcos cos . (2.135)

Thus, the deflection when thrown from the ground up is opposite in direction,and four times greater in magnitude, than then deflection when dropped fromrest.

2.2.3 The Rigid Body Equations of Motion

We have previously seen that the total kinetic energy of a rigid body may beexpressed as the sum of the kinetic energy of the entire body as if concentratedat the center of mass, plus the kinetic energy of the motion about the center ofmass. If a rigid body moves with one point stationary, the angular momentumabout the center of mass is

r

r r

L m r vi i i= ( )× . (2.136)

Since ri is a fixed vector in the body

r r r

v ri i= ω × . (2.137)

Thus

r

r r r r r r r

L m r r m r r ri i i i i i i= ( ( )) = ( ( ))× × − ⋅ω ω ω2 . (2.138)

The x-component is

L m r x m x y m x zx x i i i y i i i z i i i= ( )ω ω ω2 2− − − , (2.139)

or

L I I Ix xx x xy y xz z= + +ω ω ω . (2.140)

We define

I r r x x dVjk jk j k

v

= −zρ δ( )( )r 2 , (2.141)

to be the inertia tensor so that

r tr

L I= ω . (2.142)

The kinetic energy of motion about a point is

T m v m v ri i i i i=1

2

1

22 = ×⋅

r r r

( )ω , (2.143)

or

TL

I=r

r

rt

rω ω ω⋅ = ⋅ ⋅2

1

2. (2.144)

The moment of inertia, t

I , aboutsome given axis is related to themoment about a parallel axisthrough the center of mass. Letthe vector from the origin to thecenter of mass be

r

R and let theradii vectors from the origin andthe center of mass to the ith

particle be r

ri and r

ri ' respectively.That is

CoM

riri’

R

a b

r

r

r

r R ri i= + '. (2.145)

The moment of inertial about axis a is

I m r n m r na i i i i= ( ) = [(R + ' ) ]r

r

r× ×$ $

2 2, (2.146)

or

I M R n m r n m R n r na i i i i= ( ) + ( ' + ) + ( )( ' )r

r

r

r× × ×$ $ $ $

2 2 2 . (2.147)

The last term is 2( ) ( ' )r

r

R n n m ri i× ⋅ ×$ $ , but m ri ii

'∑ = 0 and t

r

I m r nb i i= ( ' )× $

2 so

t t r

I I M R na b= + ( )× $

2. (2.148)

2.2.3.1 The Euler Equations and Torque-Free Motion

By definition

dL

dt

dL

dtL N

s b

r r

rr rF

HGI

KJ=

F

HGI

KJ+ × =ω . (2.149)

In the body system, using Li = I ii ωi, we find

Id

dtI Ni

iijk j k k i

ωε ω ω+ = . (2.150)

The Euler equations are

I I I N1 1 2 3 2 3 1&ω ω ω− − ( ) = , (2.151)

I I I N2 2 3 1 3 1 2&ω ω ω− − ( ) = , (2.152)

I I I N3 3 1 2 1 2 3&ω ω ω− − ( ) = . (2.153)

The principle axes of a system are those where the tensor t

I is diagonal. If thisis the case,

TL

I

L

I

L

I= + +1

2

1

2

1

212

1

22

2

32

3. (2.154)

Because T is constant, the relation defines an ellipsoid fixed in the body axes.Because angular momentum is conserved

r

L must be on a fixed sphere definedby

L L L L212

22

32= + + . (2.155)

For the given initial conditions, kinetic energy and angular momentum, the

path of r

L is constrained to be the intersection of the sphere, L2 , and theellipsoid, T. If we have a body symmetrical about the z axis so that I1 = I2 theEuler equations are, in the absence of torque's,

I I I1 1 1 3 3 1&ω ω ω= ( ) − , (2.156)

I I I2 2 1 3 3 1&ω ω ω= ( ) − − , (2.157)

I 3 3 0&ω = . (2.158)

We have ω3 = constant, and

&ω ω1 = 2−Ω , &ω ω2 = 1Ω , (2.159)

where

Ω =−I I

I3 1

13ω . (2.160)

Thus,

&&ω ω1 1= −Ω , (2.161)

which has solutions ω1 = A tcosΩ and ω2 = A tsinΩ . Hence, the total angular

velocity is constant in magnitude but precesses with frequency Ω about the ζ-axis. We may solve for A and ω3 by noting that

T I A I= +1

2

1

212

3 32ω , (2.162)

and

L I A I212 2

32

32= + ω . (2.163)

2.2.3.2 The Heavy Symmetrical Top

Consider a heavy top with a symmetry axis taken to be the z-axis of thecoordinate system fixed in the body as shown. The 3 Euler angles are: θ =inclination of the z-axis from the vertical, φ = azimuth of the top about thevertical, and ψ = rotation angle of the top about its own z-axis. The kineticenergy is

T I I= + +1

2

1

21 12

22

3 32( )ω ω ω , (2.164)

or

T I I= + + +1

2

1

212 2 2

32( & & sin ) ( &

& cos )θ φ θ ψ φ θ . (2.165)

The potential energy is

V mgl= cosθ , (2.166)

so that the Lagrangian is given by

L I I mgl= + + + −1

2

1

212 2 2

32( & & sin ) ( & & cos ) cosθ φ θ ψ φ θ θ . (2.167)

We have

pL

I I I aψ ω∂∂ψ

ψ φ θ= = + = =&

( &

& cos )3 3 3 1 , (2.168)

pL

I I I I bφ ψ∂∂φ

θ θ φ θ= = + + =&

( sin cos )& cos&12

32

3 1 , (2.169)

and

E T V I I Mgl= + = + + +1

2

1

212 2 2

32

32( & & sin ) cosθ φ θ ω θ . (2.170)

Combining these first two expressions gives

&

cos

sinφ θ

θ= −b a

2, (2.171)

and

& coscos

sinψ θ θ

θ= − −I a

I

b a1

32

. (2.172)

2.3 OSCILLATORY MOTION

2.3.1 Oscillations

Consider a system that is subjected to a potential that is only a function ofcoordinates. If we expand the coordinates qi about their equilibrium positionaccording to

q qi io i= + η , (2.173)

and then perform a Taylor series expansion on the potential we obtain

V q V qV

q

V

q qi io i oi

i jo

i j( ) = ( ) +∂∂

η∂

η η∂

∂

F

HGI

KJ+

F

HG

I

KJ +

1

2

2K (2.174)

The first term on the right hand side may be redefined to be zero by shifting thezero potential to be the equilibrium value. Similarly, if the generalized forcesare zero the second term is also zero. As a result, the potential can beapproximated by the matrix relation

V qV

q qii j

o

i j( ) ≈F

HG

I

KJ

1

2

2∂

∂∂η η . (2.175)

Similarly, we can define

T m q q m Tij i j ij i j ij i j= = =1

2

1

2

1

2& &

& & & &η η η η , (2.176)

so that the Lagrangian is

L T Vij i j ij i j=1

2& &η η η η−d i . (2.177)

The equations of motion are given by

T Vij j ij j + =&&η η 0. (2.178)

If we try a solution of the form η ωi

i tC= ai exp− we find that

V a T aij j ij j =− ω 2 0, (2.179)

which is equivalent to

V T− ω2 = 0. (2.180)

Example 2.4

Two particles move in onedimension at the junction of threesprings as shown. The springs allhave unstretched length a andforce constants as shown. Find thefrequencies of the normal modes ofoscillation.

x1x2

m1 m2

We define the coordinates x1 and x2 to describe the displacement of the twoblocks, relative to the left attachment point. In these coordinates

T mx mx= +1

2

1

212

22

& & . (2.181)

Expanding about equilibrium, we define x1 = a + η1, x2 = 2a + η2. Equation2.181 is equivalent to

T m m Tij i j= + =1

2

1

2

1

212

22

& & & &η η η η , (2.182)

where

Tm

mij =FHG

IKJ

0

0. (2.183)

Similarly,

V k x a k x x a k a x= − + − − + −1

2

1

23

1

221

22 1

22

2b g b g , (2.184)

or

V Vij i j= 1

2η η , (2.185)

with

VV

x xiji j

a

=F

HG

I

KJ

∂∂ ∂

2. (2.186)

By examination

∂∂V

xk x a k x x a

11 2 13= − − − −b g b g , (2.187)

∂∂

2

12

3 4V

xk k k= + = , (2.188)

and

∂∂

V

xk x x a k a x

22 1 23 2= − − − −b g b g , (2.189)

∂∂

2

22

3 4V

xk k k= + = . (2.190)

Likewise, the cross terms are

∂∂ ∂

∂∂ ∂

2

1 2

2

2 13

V

x x

V

x xk= = − , (2.191)

so that

Vk k

k kij =−

−FHG

IKJ

4 3

3 4. (2.192)

The secular equation is

V Tk m k

k k m− = − −

− −=ω ω

ω2

2

24 3

3 40, (2.193)

which reduces to

4 9 8 7 02 2 2 4 2 2k m k m km k− − = − + =ω ω ωe j . (2.194)

From the quadratic equation, this has solutions

ω1 = k

m, (2.195)

and

ω27= k

m. (2.196)

2.4 HAMILTON'S EQUATIONS

2.4.1 Legendre Transformations and Hamilton's Equations of Motion

Consider a function f of two variables such that

df udx vdy= + , (2.197)

where uf

x= ∂

∂ and v

f

y= ∂

∂. We wish to change from the variables x, y to the

variables u, y. We define a function

g f ux= − . (2.198)

We have

dg df dux udx vdy xdu= − − = − , (2.199)

so that xg

u= − ∂

∂ and v

g

y= ∂

∂. This is an example of a Legendre transformation.

It is frequently used in thermodynamics.

Recall that Lagrange's equations are

d

dt

L

q

L

qi i

∂∂

∂∂&

F

HGI

KJ− = 0, (2.200)

where L L q q ti i= ( , & , ) , and the conjugate momenta are

pL q q t

qii i

i=

∂∂

( , & , )&

. (2.201)

We can transform from the variables ( ,& , )q q ti i to the variables ( , , )q p ti i by theuse of

H q p t q p L q q ti i i i i i( , , ) & ( , & , )= − , (2.202)

where H is called the Hamiltonian. We have

dHH

qdq

H

pdp

H

tdt

ii

ii= + +∂

∂∂∂

∂∂

. (2.203)

However, from the definition

dH p dq q dpL

qdq

L

qdq

L

tdti i i i

ii

ii= + − − −& &

&

&

∂∂

∂∂

∂∂

, (2.204)

or simply

dH q dpL

qdq

L

tdti i

ii= − −&

∂∂

∂∂

. (2.205)

Comparing the two expressions for dH we obtain Hamilton's equations

∂∂

∂∂

H

q

L

qp

i ii= − = − & , (2.206)

∂∂H

pq

ii= & , (2.207)

∂∂

∂∂

H

t

L

t= − . (2.208)

If the equations defining the generalized coordinates don't depend on timeexplicitly, and if the forces are derivable from a conservative potential V, then

H T V E= + = . (2.209)

If we define a column matrix r

η with 2n elements such that ηi iq= , ηi n ip+ = ,for i n≤ , then

∂∂η

∂∂

H H

qi i

F

HGI

KJ= , (2.210)

and

∂∂η

∂∂

H H

pi n ir

+

F

HGI

KJ= . (2.211)

If we define t

J =FHG

IKJ

0 1

1 0, then Hamilton's equations may be written in the form

r t

r&η ∂

∂η= J

H. (2.212)

This is called symplectic notation.

Example 2.5

The Lagrangian for a simple spring is given by

L mx kx= −1

2

1

22 2

& .

Find the Hamiltonian and the equations of motion using the Hamiltonianformulation. Identify any conserved quantities.

From the definition of pi we have

pL

xmxx = =∂

∂ &

& . (2.213)

From the definition of the Hamiltonian, equation 2.227, we see that

H x mx mx kx= − +& & &b g1

2

1

22 2, (2.214)

or simply

Hm

p kxx= +1

2

1

22 2. (2.215)

From Hamilton's equations we have

∂∂H

qkx p

ix= = − & , (2.216)

∂∂H

p

p

mx

i

x= = & , (2.217)

∂∂H

t= 0. (2.218)

The first equation is the equation of motion in one dimension,

mx kx&&+ = 0, (2.219)

the second equation is the definition of momentum, and the last equation is thestatement of conservation of energy.

2.5.2 Canonical Transformations

If a generalized coordinate qi has constant conjugate momenta it is said to becyclic. If this is the case, then pi = 0, which tells us that the Hamiltonian isindependent of that pi. If all coordinates qi are cyclic the conjugate momentacan be defined by pi = αi. Consequently,

&qH

ii

i= =∂∂α

ω , (2.220)

or

q ti i i= +ω β . (2.221)

A problem is often easier to solve if we can find a system where the number ofcyclic coordinates is maximum. How do we transform to this set of

coordinates? We need a new set of coordinates Qi, Pi, where Q Q q p ti i i i= , ,b g,

P P q p ti i i i= , ,b g . We require Qi, Pi to be canonical coordinates. Therefore,

some function K K Q P ti i= , ,b g exists such that

&QK

Pii

= ∂∂

, (2.222)

and

&PK

Qii

= ∂∂

. (2.223)

If Qi, Pi are canonical coordinates they must satisfy a modified Hamilton'sprinciple that can be put in the form

δ PQ K Q P t dti i i i

t

t

& ( , , )− =z d i1

2

0 , (2.224)

because the old coordinates satisfy

δ p q H q p t dti i i i

t

t

& ( , , )− =z b g

1

2

0. (2.225)

Both requirements can be satisfied if we require a relation

λ p q H PQ KdF

dti i i i&

&− = − +b g , (2.226)

where λ = constant and F is any function of the phase space coordinatescontinuous through the second derivative. λ is related to a scaletransformation. If λ = 1 the relation defines a canonical transformation. Thefunction F is termed the generating function. It may be a function of qi, pi, Qi,Pi, t and defines the transformation.

2.5.3 Symplectic Transformations and Poisson Brackets

Recall that Hamilton's equations can be written in the from

rt

r&η ∂

∂η= J

H. (2.227)

If we have a canonical transformation from η ξ ξ η→ = ( ) , then

&

&ξ∂ξ∂η

ηii

jj= . (2.228)

In matrix form,

r t r& &ξ η= M , (2.229)

where Miji

j=

∂ξ∂η

. We have

r t t

r&ξ ∂

∂η= MJ

H, (2.230)

also

∂∂η

∂∂ξ

∂ξ∂η

H H

i j

j

i= , (2.231)

and

∂∂η

∂∂ξ

HM

Hr

t

r= % . (2.232)

Consequently,

r t t tr

tr

& %ξ ∂∂ξ

∂∂ξ

= =MJMH

JH

. (2.233)

Therefore, a transformation is canonical if

t t t t

MJM J% = . (2.234)

The Poisson bracket of a function is defined by

u vu

q

v

p

u

p

v

qPBi i i i

, = −∂∂

∂∂

∂∂

∂∂

. (2.235)

We have

r rt

η η, PB J= , (2.236)

and

r r

r

r

t

r

r

t t t t

ξ ξ ∂ξ∂η

∂ξ∂η

,%

%

PBJ MJM J= = = . (2.237)

In other words, the fundamental Poisson brackets are invariant under canonicaltransformations. We also have

du

dt

u

qq

u

pp

u

tii

ii= + +∂

∂∂∂

∂∂

& & , (2.238)

or

du

dtu H

u

tPB= +,

∂∂

. (2.239)

Similarly,

& ,q q Hi i PB= , & ,p p Hi i PB

= . (2.240)

2.6 CONTINUOUS SYSTEMS

2.6.1 The Transition from a Discrete to a Continuous System

Consider an infinitely long elastic rod that can undergo small longitudinalvibrations. We approximate this by an infinite chain of equal mass points adistance a apart and connected by uniform massless springs having forceconstraints k. The kinetic energy is

T m ii

= ∑1

22

&η , (2.241)

where m is the mass of each particle and ηi is the location of the ith particle.The potential energy is

V k i ii

= −+∑1

2 12η ηb g . (2.242)

We have

L m ki i ii

= − −LNM

OQP+∑1

22

12

&η η ηb g , (2.243)

or

L am

aka

aii i

i

= − −FHG

IKJ

L

NMM

O

QPP

+∑1

22 1

2&η η η

. (2.244)

The equations of motion are

m

aka

aka

ai

i i i i&&η

η η η η−

−FHG

IKJ

+−F

HGIKJ

=+ −12

12

0 . (2.245)

We have m/a = µ = mass/unit length. Hooke's law states that the extension of arod/unit length is proportional to the force, i.e.

F Y= ξ , (2.246)

where ξη η

=−F

HGIKJ

+i i

a1 is the extension/unit length. The force necessary to

stretch the string by an amount x is

F k kaai i

i i= − =−F

HGIKJ+

+η ηη η

11b g . (2.247)

Consequently, ka = Y = Young's modulus. Note that

η η η η ηi i

a

x a x

a

d

dx+ −

= + − →1 ( ) ( ), (2.248)

as a → 0. Also as a → 0 the summation over the particles becomes an integraland Equation 2.244 becomes

L Yd

dxdx= − F

HGIKJ

F

HG

I

KJz

1

22

2

µη η& . (2.249)

The equation of motion is

µ η ηd

dtY

d

dx

2

2

2

20− = . (2.250)

This is a wave equation with wave propagation velocity vY=µ

. Equation

2.249 is said to define a Lagrangian density

%Ld

dtY

d

dx= F

HGIKJ

− FHG

IKJ

µ η η2 2

. (2.251)

2.6.2 The Lagrangian Formulation

Consider a Lagrangian density

% % , , , ,L Ld

dx

d

dtx t= F

HGIKJ

η η η. (2.252)

Hamilton's principle is

δ δI Ldxdt= =zz % 0

1

2

. (2.253)

We choose value of η such that

η α η αξ( , ; ) ( , ; ) ( , )x t x t x t= +0 , (2.254)

where η(x,t;0) is the function that satisfies Hamilton's principle. We have

δα α

IdI

d= F

HGIKJ

==0

0 , (2.255)

where

dI

ddxdt

L Lddx

d

dx

Lddt

d

dtx

x

t

t

α∂∂η

∂η∂α

∂

∂ η∂

∂αη ∂

∂ η∂

∂αη= +

FHG

IKJ

FHG

IKJ

+FHG

IKJ

FHG

IKJ

R

S||

T||

U

V||

W||

zz% % %

1

2

1

2

. (2.256)

This expression may be simplified as follows. First,

dtLd

dt

d

dt

d

dt

Ld

dt

dt

t

t

t

t∂

∂ η∂

∂αη ∂

∂ η∂η∂α

% %

FHG

IKJ

FHG

IKJ

R

S||

T||

U

V||

W||

= −FHG

IKJ

L

N

MMMM

O

Q

PPPP

z z1

2

1

2

, (2.257)

plus a boundary term that goes to zero. Also,

dxLd

dx

d

dx

d

dx

Ld

dx

dx

x

x

x

x∂

∂ η∂

∂αη ∂

∂ η∂η∂α

% %

FHG

IKJ

FHG

IKJ

R

S||

T||

U

V||

W||

= −FHG

IKJ

L

N

MMMM

O

Q

PPPP

z z1

2

1

2

, (2.258)

plus another boundary term that goes to zero. Therefore,

δ ∂∂η

∂

∂ η∂

∂ η∂η∂α α

I dxdtL d

dt

L

d

dx

d

dx

L

d

dtx

x

t

t

= −FHG

IKJ

F

H

GGGG

I

K

JJJJ

−FHG

IKJ

F

H

GGGG

I

K

JJJJ

R

S||

T||

U

V||

W||

FHG

IKJ

==

zz% % %

01

2

1

2

0, (2.259)

and

d

dt

Ld

dx

d

dx

Ld

dt

L∂

∂ η∂

∂ η∂∂η

% % %

FHG

IKJ

F

H

GGGG

I

K

JJJJ

+FHG

IKJ

F

H

GGGG

I

K

JJJJ

− = 0. (2.260)

Similarly for other Lagrangians.

2.6.3 Noether's Theorem

A formal description of the connection between invariance or symmetryproperties and conserved quantities is contained in Noether's theorem. Weconsider transformations where

x x x xµ µ µ µδ→ = +' , (2.261)

η η η η δρ µ ρ µ ρ µ ρ µx x x xd i e j d i d i→ = +' ' , (2.262)

% , , % ' , ,,' '

,' ' 'L x x x L x x xη η η ηρ µ ρ ν µ µ ρ µ ρ ν µ µd i d ie j e j e j→ F

HIK . (2.263)

We make three assumptions:

1.) 4-space is Euclidean,

2.) The Lagrangian density has the same functional form aftertransformation,

3.) The magnitude of the action integral is invariant under thetransformation.

From assumptions 2, 3 we have

% , , % , ,',

' '

'

',L x dx L x dxη η η ηρ ρ ν µ µ ρ ρ ν µ µe j e j

Ω Ωz z− = 0. (2.264)

xµ' is a dummy variable, so let x xµ µ

' → to obtain

% , , % , ,',

'

'

,L x x x dx L x x x dxη η η ηρ µ ρ ν µ µ µ ρ µ ρ ν µ µ µd i d ie j d i d ie jΩ Ωz z− = 0. (2.265)

Under the transformations the first order difference between the integralsconsists of two parts, one is an integral over Ω and the other is an integral over Ω' – Ω. For example, in one dimension,

f x f x dx f x dx

a a

b b

a

b

( ) ( ) ( )+ −+

+

z zδδ

δ

= + + − +z zz++

δ δ δδδ

f x dx f x f x dx f x f x dx

a

b

a

a a

b

b b

( ) ( ) ( ) ( ) ( ) . (2.266)

To first order, the last two terms are

f x dx f x dx bf b af a

a

a a

b

b b

( ) ( ) ( ) ( )− = −++

zz δ δδδ

. (2.267)

The difference between integrals is

δ δ δ δf x dx f x x f xd

dxxf x dx

ab

a

b

b

b

( ) ( ) ( ) ( )+ = +L

NMO

QPzz b g . (2.268)

Consequently, for the Lagrangians we have

% ' , % ,

'

L x dx L x dxη ηµ µ µ µd i d i− zzΩΩ

= − + =z z% ' , % , %

'

L x L x d x L x dS

S

η η η δµ µ µ µ µd i d i b gΩ

0, (2.269)

or

d x L x L xd

dxL x xµ µ µ

µµ µη η η δ% ' , % , % ,

'

d i d i d ie j− +RS|

T|

UV|

W|=z

Ω

0. (2.270)

To first order,

% ' , % ,% %

,,L x L x

L Lη η ∂∂η

δη ∂∂η

δηµ µρ

ρρ ν

ρ νd i d i− = + , (2.271)

or

% ' , % ,%

,L x L x

d

dx

Lη η ∂∂η

δηµ µν ρ ν

ρd i d i− =F

HG

I

KJ . (2.272)

The invariance condition is

dxd

dx

LL xµ

ν ρ νρ ν

∂∂η

δη δz +RS|

T|

UV|

W|=

%

%

,0 . (2.273)

We define

δ εν νx Xr r= , δη ερ ρ= r rΨ , (2.274)

because

η η δηρ µ ρ µ ρ µ' 'x x xe j d i d i= + , (2.275)

and

δη δη∂η∂

δρ ρρ

σσ= +

xx , (2.276)

we have

δη ε ηρ ρ ρ σ σ= −r r rXΨ ,e j . (2.277)

Equation 2.305 reduces to

ε ∂∂η

η δ ∂∂ην ρ ν

ρ σ νσ σρ ν

ρ µr r rd

dx

LL X

Ldx

%

%

%

,,

,−

F

HG

I

KJ −

RS|

T|

UV|

W|=z Ψ d i 0. (2.278)

The result is Noether's theorem, which states that

d

dx

LL X

Lr r

ν ρ νρ σ νσ σ

ρ νρ

∂∂η

η δ ∂∂η

%

%

%

,,

,−

F

HG

I

KJ −

RS|

T|

UV|

W|=Ψ 0. (2.279)

2.7 Bibliography

Lanczos, C., The Variational Principles of Mechanics, 4th Ed., (Toronto,Canada: University of Toronto Press, 1970).

Goldstein, H., Classical Mechanics, 2nd Ed., (Reading, MA: Addison-WesleyPublishing Co., 1980.)

Symon, K. R., Mechanics, 3rd Ed., (Reading, MA: Addison-WesleyPublishing Co., 1971.)