BEZOUT IDENTITIES WITH INEQUALITY CONSTRAINTS

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Transcript BEZOUT IDENTITIES WITH INEQUALITY CONSTRAINTS

MA4248 Weeks 1-3.
Topics Coordinate Systems, Kinematics, Newton’s
Laws, Inertial Mass, Force, Momentum, Energy,
Harmonic Oscillations (Springs and Pendulums)
Mechanics a drama authored by physical law
whose stage is four dimensional space-time
Space-time has an affine structure, and additional
structure for either classical or relativistic mechanics
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VECTOR SPACE
Definition a set, whose elements are called vectors,
together with two operations, called vector addition
and scalar multiplication (by elements of R = reals)
- each operation must satisfy certain properties
- the two operations must be related
Examples d-tuples of real numbers, real valued
functions on a specified set S, set of functions on R
having the form x  a cos (x+b)
2
AFFINE SPACE
Definition a set A, whose elements are called points,
together with a vector space V and an operation
AV  A
(called translation), that associates to every p in A
and u in V an element in A (denoted by p+v), and
that satisfies the following two properties:
- for every pair (p, q) of points in A there exists
a unique element u in V such that p  u  q
- for all p  A, u, w  V the following holds
(p  u)  w  p  (u  w)
Examples lines, planes, space without the Euclidean
structure (dot product and derived angle and length)
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EUCLIDEAN VECTOR SPACE
Definition: A vector space together with an operation
V x V  V, that associates to a pair (a,b) of vectors
a, b in V a real number (called their dot product and
denoted by a  b ), that satisfies the following
a  a  0 if and onlyif a  0
ab  ba
a  (1b1   2 b 2 )  1a  b1   2a  b 2
Definition: Length and Angle between | a | aa
a b
nonzero vectors are then defined by
cos  
|a| |b|
Euclidean Affine Space or Euclidean Space is an
Affine Space whose Associated Vector Space
has a Euclidean Vector Space structure
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VECTOR ALGEBRA
Vectors can be represented by their coordinates


with respect to a choice of basis
u  u1b1  u 2 b 2

u ~ (u1 , u 2 )
b1

and then so can
b2
vector operations
 
u  v ~ (u1  v1 , u 2  v 2 )

c u ~ (cu1 , cu2 )
If the basis orthonormal,
 
bi  b j  ij then
 
u  v  u1v1  u 2 v 2
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KINEMATICS
A trajectory in an affine space is a function
f : [c, d ]  A,
Smooth trajectories in affine space define trajectories
in the associated vector space, called velocities
f : [c, d ]  V ,
f
(
t


t
)

f
(
t
)
f ( t )  lim
t
t  0
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KINEMATICS
Example 1 Choose p, q, r be points in A and construct
f ( t )  r  t (q  p),
f ( s  s )  f ( s )
f ( s ) 
lim
s
s  0
( r  ( t  t )( p  q )  ( r  t ( p  q ))

lim
t
t  0

t  [c, d]
t ( p  q )
lim
t  0
t

lim
t  0
(p  q)  p  q
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KINEMATICS
Smooth trajectories in affine space define trajectories
in the associated vector space, called accelerations
 ( t  t )  f ( t )
f
f( t )  lim
t
t  0
Remark Here, in contrast to the situation for velocity,
the numerator is the difference between two vectors
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KINEMATICS
Example 2 Harmonic Oscillation of a small
body, “particle”, along a line is described by
f ( t )  r  a cos(t  b)( q  p)
f ( t )  a sin(t  b)(q  p)
f( t )  a2 cos(t  b)( q  p)
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KINEMATICS
Example 3 Circular Motion of a small body is
described with by orthogonal unit vectors u, v
f (t )  r  a cos(t   )u  a sin(t   )v
f (t )  a sin(t   )u  a cos(t   )v
  ( f (t   / 2 )  r )
f(t )  a 2 (cos(t   )u  sin(t   )v)
  f (t )
2
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NEWTON’s FIRST LAW
The velocity of an isolated body is constant
Criticize the following versions of this law given
by Halliday, Resnick and Walker:
page 73 If no force acts on a body, then the body’s
velocity cannot change; that is, the body cannot
accelerate
page 74 If no net force acts on a body, then the
body’s velocity cannot change; that is, the body
cannot accelerate
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NEWTON’s FIRST LAW
Inertial Frames are preferred coordinate systems for
space-time for which Newtons laws hold
Given an inertial frame, we can obtain others by
translating the original in space and time, by rotating
the original through some angle about an axis, and by
f new ( t )  forig ( t )  u 0  ( t ) v 0
fnew ( t )  forig ( t )  v0
f
f
(
t
)

new
orig ( t )
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NEWTON’s THIRD LAW
Criticize the following version of this law given
by Halliday, Resnick and Walker:
page 84 When two bodies interact, the forces on
the bodies from each other are always equal in
magnitude and opposite in direction
Logically formulate this law by using Newton’s
second law on page 84
The net force on a body is equal to the product of
the bodies mass and the acceleration of the body
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NEWTON’s SECOND and THIRD LAWS
Deal with the effects of interactions between bodies
on their motion that cause them to accelerate
When bodies i and j interact (only) with each
other, their acceleration magnitudes satisfy
{i , j} {i , j}
is independent of the interaction
ai
/aj
and for bodies i, j, k (interacting pairwise), these ratios
satisfy the equation (this is not an algebraic identity)



{i , j} {i , j} { j,k} { j,k} {k ,i} {k ,i}
ai
/aj
aj
/ ak
ak
/ ai
1
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DEFINITION OF INERTIAL MASS
Choose a standard body and assign it a mass, for
example the SI standard of 1 kilogram mass is that of
the paltinum-iridium cylinder kept at the International
Bureau of Weights and Measures near Paris
Define the mass of any body i to be
{i ,cyl} {i ,cyl}
mi  (a cyl / a i
) kilograms
{i , j}
{i , j}
then the three body equation implies a
/aj 
i
 { j,cyl} { j,cyl}  {i ,cyl} {i ,cyl} 
/aj
/ a cyl  
 a cyl
 a i



m j / mi
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DEFINITION OF FORCE
Define the force on a body to have magnitude ma
and direction given by direction of its acceleration
Then Newton’s second law can be expressed as
F1on2   F2on1
Remark: this is a consequence of Newton’s laws,
as discussed in Calkin’s book, together with the
definitions of mass and force
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NEWTON’s SECOND LAW
Deals with the interaction of three or more bodies
Law is an empirical observation that says the net
acceleration of any body is the sum of the acclerations
that it experience from its interaction with each of the
other bodies individually

a
or, equivalently


aj
j

ma  Fnet 
 Fj
j
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BOUND COMBINATIONS
Suppose that particles 1, 2 and 3 interact. Then

m1a1  F2on1  F3on1

m 2a 2  F3on2  F1on2

m3a 3  F1on3  F2on3
If particles 2 and 3 interact so that they are bound
together as a single particle, then

m 23a 23  F1on2  F1on3  F1on23



where m 23  m 2  m3 , a 23  a 2  a 3
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MOMENTUM
Suppose that particles 1 and 2 interact over time



m1v1 ( t  t )  m1v1 ( t )  m1a1 ( t ) t

 F2on1t   F1on2 t   m 2 a 2 ( t ) t


 ( m 2 v 2 ( t  t )  m 2 v 2 ( t ))
Therefore the momentum of the system, defined by


p( t )  m1v1 ( t )  m2 v 2 ( t )
is constant or invariant. This is the case for any
system of particles.
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ENERGY
Suppose that a particle accelerates under a force


mv( t )  ma ( t )  F( x ( t ), t )
Further assume that the force is conservative, which
by definition means that there is a real valued
potential energy function U(x) on space such that
F  grad U
1
Then energy E ( t )  U( x ( t ))  mv ( t )  v( t )
2
is constant since
E( t )  grad U( x( t ))  v( t )  mv( t )  v ( t )  0
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HARMONIC OSCILLATIONS
Consider an object attached to a
spring that moves horizontally
near equilibrium x 0 Then
2
1
F  kx  grad kx
2
2
2
1
1
 E  kx  mx
2
2
x0  x
 x ( t )  a cos ( t  )
a  2E k ,
  k m,
R
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HARMONIC OSCILLATIONS
Consider a pendulum - an object on
a swinging lever. Then for small θ

2
2
Lm

E
gθ  Lθ
2
θ

L
 θ( t )  a cos (t  )
a
2E ,
Lmg

g
,
L
R
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IN LINE COLLISIONS
Consider the collision of two objects
m1
m2
v1
v2
v1
v2
Since momentum is conserved
m1v1  m2 v 2  m1v1  m2 v2
Since kinetic energy is conserved
2
2
2
2
m1v1  m 2 v 2  m1v1  m 2 v2
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INLINE COLLISIONS
From the two equations we derive
 1  1   v1   v 2  v1 
m m  v   m v  m v 
2  2 
2 2
 1
 1 1
m1  m2
2m2
v1 
v1 
v2
m1  m2
m1  m2
2m1
m2  m1
v2 
v1 
v2
m1  m2
m1  m2
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STATICS
Compute the force that each string exerts on the body


mg
Hint: The direction of each strings force is
along the string and away from the body
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