Mechanics
Physics 151
Lecture 2
Elementary Principles
(Goldstein Chapter 1)
Administravia
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First Problem Set
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If you haven’t filled the Survey, please do it
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3 problems for the section
! Work on them before coming to your section!
3 for the report (due next week)
We need it for sectioning and study-grouping
Section time: Tue. 6 PM, 7 PM and Wed. 5 PM
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If none of these slots works for you, let me know
Sectioning will be announced on Monday by email
! We will also assign you into study groups (~6 each)
What We Did Last Time
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Reviewed basic principles of Newtonian Mechanics
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Define standard notations and usages
Momenta, conservation laws, kinetic & potential energies
Concentrated on the motion of a single particle
Goals for Today
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Single " multi-particle system
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Newton’s 3rd Law
Introduce constraints
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Force between particles
! Laws of action and reaction
Holonomic and nonholonomic constraints
Introduce Lagrange’s Equation
System of Particles
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More than one particles? " Just add indices!
Fi = p! i
!
N i = L! i
Subtlety: F may be working between particles
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Distinguish between internal and external forces
Fi = ∑ F ji + Fi( e )
j
Force acting
on particle i
Force from outside
Force from particle j
!
Now add up over i to see the overall picture
Sum of Particles
(e)
(e)
F
=
F
+
F
=
F
+
F
+
F
∑ i ∑ ji ∑ i ∑ ( ji ij ) ∑ i
i
!
i, j
i≠ j
i
i< j
This term vanishes if Fij = −F ji
!
Weak law of action and reaction
i
(e)
F
=
F
∑ i ∑ i
i
i
Forces two particle exert on each other are equal and opposite
C.f. the strong law of action and reaction
Forces two particle exert on each other are equal, opposite,
and along the line joining the particles
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Sum of Particles
!
Now consider the equations of motion
∑F = ∑F
(e)
i
i
i
!
i
d2
= ∑ p! i = 2
dt
i
Define center of mass
∑m r
i i
i
mr ∑mr
∑
R≡
=
M
∑m
i i
i i
i
!! = ∑ F ( e ) ≡ F ( e )
MR
i
i
Center of mass moves like a particle of mass M under total
external force F(e)
Total Linear Momentum
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The sum of the linear momenta is
!
P = ∑ p i = ∑ mi r!i = MR
i
!
!
i
Taking the time derivative
!! = F ( e )
P! = MR
Newton’s equation
of motion for the
center of mass
Conservation of total linear momentum
If the total external force F(e) is zero, the total linear momentum
P is conserved
Assumed weak law of action & reaction
Total Angular Momentum
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The sum of the angular momenta is
L = ∑ L i = ∑ ri × p i
i
!
i
Take time derivative and use p! i = Fi = ∑ F ji + Fi(e )
j
L! = ∑ ri × F ji + ∑ ri × Fi( e )
i, j
i≠ j
!
i
Total external
torque
This term vanishes only if Fji satisfies the strong law of
action and reaction
Total Angular Momentum
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Assuming strong law of action and reaction
L! = ∑ ri × Fi( e ) = ∑ N i( e ) = N ( e )
i
i
" Conservation of total angular momentum
If the total external torque N(e) is zero, the total angular
momentum L is conserved
A multi-particle system (= extended object) can be treated
as if it were a single particle if the internal forces obey the
strong law of action and reaction
Laws of Action and Reaction
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F12
E.g. Lorenz force felt by moving charges
Conservation of linear & angular momenta fails
Take into account the EM field
Particles exchange forces with the field
! The field itself has linear & angular momenta
" Conservation laws restored
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F21 = 0
Gravity, electrostatic force
There are rare exceptions
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!
+Q
Most forces we know obey strong law of
action and reaction
v2
+Q
v1
Conservation Laws
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Weak law of action-reaction
Conservation of P
Strong law of action-reaction
Conservation of L
We will see (in 2 lectures) that P and L must be
conserved if the laws of physics are isotropic in space
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No special origin
No special orientation
If we accept these symmetries as fundamental
principles, all forces must satisfy the action-reaction
laws " “Proof” of Newton’s 3rd Law
Total Angular Momentum
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Define particle i’s position from the center of mass
ri′ = ri − R
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Also define the velocities v′i = r!i′ v = R
Calculate the total angular momentum
L = ∑ ri × p i = ∑ (ri′ + R )× mi (v′i + v )
i
i
L = R × Mv + ∑ ri′ × mi v′i
i
Angular momentum of
motion concentrated at
the center of mass
Angular momentum
of motion around the
center of mass
Kinetic Energy
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The work done by force W12 = ∑ ∫1 Fi ⋅ ds i
2
i
!
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Positions 1 and 2 are now configurations (sets of positions)
Use equations of motion to derive
1
W12 = T2 − T1 where T = ∑ mi vi2
i 2
!
One can split T into two pieces
1
1
1
T = ∑ mi ( v + v′i ) ⋅ ( v + v′i ) → Mv 2 + ∑ mi vi′2
2
i 2
i 2
Motion concentrated at
the center of mass
Motion around the
center of mass
Potential Energy
!
∂
∂yi
∂ 

∂zi 
Assume conservative external force Fi( e ) = −∇ iVi
∑∫
i
!
 ∂
∇ i ≡ 
 ∂xi
2
1
F ⋅ dsi = −∑ ∫ ∇iVi ⋅ dsi = −∑ Vi 1
2
(e)
i
i
1
2
i
Assume also conservative internal forces F ji = −∇ iVij
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To satisfy strong law of action/reaction
Potential depends only
Vij = Vij (| ri − r j |)
on the distance
∑∫
i, j
i≠ j
2
1
F ji ⋅ dsi = −∑ ∫ ∇iVij ⋅ dsi
2
i, j
i≠ j
1
Bit of work
2
1
− ∑ Vij
1
2 i, j
i≠ j
Energy Conservation
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If all forces are conservative, one can define total
potential energy: V = V + 1 V
∑
i
!
!
i
∑
2 i, j
ij
i≠ j
Then the total energy T + V is conserved
The second term is internal potential energy
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It depends on the distances between all pairs of particles
Constant if particles’ relative configuration is fixed
" Rigid bodies
Constraints
!
(e)
!
!
m
r
=
F
=
F
+ ∑ F ji assumes
Equation of motion i i
i
i
j
that particles can
move anywhere in space
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!
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Not generally true
! In fact never true – Free space is an idealization
Amusement-park ride constrained (hopefully) on a rail
Billiard balls on a pool table
How can we accommodate constraints in the equation
of motion?
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Depends on the type of the constraint
Holonomic Constraints
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Constraints may be expressed by
f (r1 , r2 , r3 , … , t ) = 0
!
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Particle on the x-y plane z = 0
!
Rigid body (ri − r j ) 2 − cij2 = 0
All other cases are called nonholonomic
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A holonomic
constraint
It means “we don’t really want to mess with it”
May be inequalities such as z ≥ 0
May depend on derivatives such as r!i
We will deal only with holonomic constraints
Independent Variables
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A holonomic constraint reduces the number of
independent variables by 1
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If z = 0, you are left with only x and y
You may be able to solve the constraint for one variable
f (r1 , r2 , r3 , … , t ) = 0
x1 = g ( y1 , z1 , r2 , r3 , … , t )
Then you can drop this variable
You may have to switch to a different set of variables
2
2
2
2
! For a particle on a sphere x + y + z = c
a good choice is (θ, φ)
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New set of variables " Generalized coordinates
Generalized Coordinates
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N particles have 3N degrees of freedom
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Introducing k holonomic constraints reduces it to 3N – k
Using generalized coordinates q1, q2,…, q3N – k
ri = ri (q1 , q2 , q3 , …, t )
Transformation equations from (rl) to (ql)
!
 x = c sin θ cos φ

Example:  y = c sin θ sin φ
 z = c cos θ

Transformation
from (x, y, z)
to (θ, φ)
Now What?
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We know the equations of motion for (ri)
mi!r!i = Fi = Fi( e ) + ∑ F ji
j
!
We know how to include constraints by switching to
generalized coordinates
ri = ri (q1 , q2 , q3 , …, t )
!
How can we transform the equation of motion to the
generalized coordinates?
Lagrange’s Equations
Why Constraints?
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Constraint is an idealized classical concept
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Nothing is perfectly constrained in QM Uncertainty
How useful is it to switch between coordinates?
More than it seems
Constraints and Force
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A holonomic constraint is an infinitely strong force
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Or an infinitely high potential wall
Reality is always smoother
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V(r)
E.g. electron of a hydrogen atom
free
V(r)
r
It’s still true that the
electron feels strong
radial (binding) force,
while it can move
freely around the
nucleus
constrained
Binding force
makes the r
direction special
Force and Symmetry
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Without forces, all coordinate systems are equal
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Forces break the symmetry
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x-y-z system is the simplest
Some coordinate system works better than others
Generalized coordinates offer natural way of handing
systems with such forces
Constraints are extreme cases
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We develop our technique with them
OK, back to the business…
Lagrange’s Equations
d  ∂L

dt  ∂q! j
 ∂L
=0
 −
 ∂q j
Recipe
Kinetic energy
L(q, q! , t ) ≡ T − V
Potential energy
Lagrangian
!
Express L = T – V in terms of generalized coordinates
{q j } , their time-derivatives {q! j }, and time t
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The potential V = V(q, t) must exist
i.e. all forces must be conservative
Let’s do a quick example to see how it works
d  ∂L

dt  ∂q! j
Ex: Particle on a Line
!
 ∂L
=0
 −
 ∂q j
A particle moving on the x-axis x = x(t ), y = 0, z = 0
!
Kinetic and potential energies:
m
T = x! 2 V = V ( x)
2
Lagrange’s Eqn
!
L=
m 2
x! − V ( x)
2
∂V
=0
mx!! +
∂x
∂V
Equivalent to Newton’s Eqn given that Fx = −
∂x
OK, it works
Summary
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Discussed multi-particle systems
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Introduced constraints
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Internal and external forces
! Laws of action and reaction
Momenta, conservation laws, kinetic & potential energies
Holonomic and nonholonomic constraints
Generalized coordinates
Introduced Lagrange’s Equations
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Next: Prove that Lagrange’s and Newton’s Equations are
equivalent