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Chapter 7

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November 30, 2014
Advanced Analytical Mechanics-1
Analytical Mechanics
( AM )
Lecture notes part 4, Action
Olaf Scholten
KVI, kamer v2.008
tel. nr. 363-3552
email: scholten@kvi.nl
Web page: http://www.kvi.nl/˜scholten
- LeastAction.1 -
Hamilton’s principle of least action
Simpler, more general, fully equivalent to Newtonian Dynamics
Action
t2
S1,2 =
L(x, x;
˙ t) dt
t1
defined in terms of Lagrangian
L(x, x)
˙ = T (x)
˙ − V (x)
with x(t) and velocity v
=
d
dt x
= x˙
Principle of least action gives, through the Euler-Lagrange equations, the equations of
motion
∂L
d ∂L
−
=0
dt ∂ x˙
∂x
When T
= 12 mx˙ x˙ and V (x) then E-L equations give
Newton’s second law:
− ∂V
∂x = F =
d
Note: dt
f (p, q, t)
∂f
∂p p˙
=
+
∂f
∂q q˙
+
d
˙
dt [mx]
∂f
∂t
= m¨
x
November 30, 2014
Advanced Analytical Mechanics-2
- LeastAction.2 -
November 30, 2014
Advanced Analytical Mechanics-2
Generalized Coordinates
Lagrangian
Example
[example 7.11]
L(q1 , q2 , · · · , qn , q˙1 , q˙2 , · · · , q˙n , t) = T − V
Particle, mass m, constrained to move on a cylinder,
x = R cos θ, y = R sin θ, z
t2
L(q1 , · · · , qn , q˙1 , · · · , q˙n , t) dt
S1,2 =
t1
with central force F = −krˆ
r
—————————————————————————–
U = 21 kr2 = 21 k(x2 + y 2 + z 2 ) = 12 k(R2 + z 2 )
Euler-Lagrange equations
∂L
d ∂L
−
= 0 , i = 1, 2, · · · , n
dt ∂ q˙i
∂qi
[ problem 7.2–17]
Since x˙
= −Rθ˙ sin θ,
T = 21 mv 2 = 12 m(x˙ 2 + y˙ 2 + z˙ 2 ) = 21 m(R2 θ˙2 + z˙ 2 )
L = T − U = 12 m(R2 θ˙2 + z˙ 2 ) − 21 k(R2 + z 2 )
d
E-L eqn. for z gives dt
mz˙ + kz = 0
Solution: z(t) = A cos (ωt + φ), ω =
d
E-L eqn. for θ gives dt
mR2 θ˙
Solution: θ(t) = at + b
=0
k/m
- LeastAction.3 -
November 30, 2014
Advanced Analytical Mechanics-3
- LeastAction.4 -
November 30, 2014
Advanced Analytical Mechanics-3
- LeastAction.5 -
y
Example
ω
Example
[example 7.5]
[ Example 7.4]
r
Particle, mass m, moving in vertical
cone, opening angle α and thus r =
z tan α
V = mgz and T = 21 mv 2
z
r
θ
gives
=0
mz 2 θ˙ tan2 α = C
EL in z :
mz θ˙2 tan2 α − mg − m¨
z (1 + tan2 α) = 0
b
m
T = 21 m a2 ω 2 + b2 θ˙2 + 2abω θ˙ (sin θ cos ωt − cos θ sin ωt)
sin (θ−ωt)
U = mg(a sin ωt − b cos θ)
L = T − U and E.L.
∂θ
x
θ
= z˙ tan α and thus
L = 12 m(z˙ 2 tan2 α + z 2 θ˙2 tan2 α + z˙ 2 ) − mgz
d ∂L
EL equation in θ : dt
˙
ωt
Support of pendulum (length b, mass
m) fixed on rim rotating disk (radius a
angular velocity ω ).
Coordinates mass m:
x = a cos ωt + b sin θ
y = a sin ωt − b cos θ
L = T − V = 21 m(r˙ 2 + r2 θ˙2 + z˙ 2 ) − mgz
relation z and r gives r˙
α
a
∂L
∂θ
−
d ∂L
dt ∂ θ˙
= 0 give
bθ¨ = ω 2 a cos (θ − ωt) − g sin θ
Check solution numerically!
Go through all examples in the book !
November 30, 2014
Advanced Analytical Mechanics-4
- LeastAction.6 -
November 30, 2014
Advanced Analytical Mechanics-4
Example
Constraints & Lagrange multipliers
If motion is constraint by g(xi , t)
= 0 we get
- LeastAction.7 -
Disk rolling down inclined plane not slipping, angle α
T = 21 M y˙ 2 + 21 ( 12 M R2 )θ˙2
∂L
d ∂L
∂g
−
+ λ(t)
= 0 with g(qi ; t) = 0
∂qi
dt ∂ q˙i
∂qi
U = M g(l − y) sin α
L = T − U , constraint
Since Fi
= m¨
qi force exerted by the constraint is
Qi = λ(t)
∂g
∂qi
g(y, θ; t) = y − Rθ = 0
d ∂L
∂g
∂L
−
+ λ(t)
=0
∂qi
dt ∂ q˙i
∂qi
gives
Ai q˙i + B = 0
∂g
∂g
with Ai = ∂q ; B = ∂t then rewrite constraint
i
If constraint
i
d
or dt
g
i
∂g d
∂g
qi +
=0
∂qi dt
∂t
= 0 giving
g(qi ; t) − C = 0
Note: repeated indices imply summation i.e.
∂L
∂qi
δqi ≡
n
∂L
i=1 ∂qi
δqi
M g sin α − M y¨ + λ = 0
− 1 M R2 θ¨ − λR = 0
2
constraint y
= Rθ gives
2
Mg
λ=−
sin α and y¨ = g sin α
3
3
Without constraint result would be y
¨ = g sin α
Force due to constraint Fc
∂g
= λ ∂q
i
Direct substitution of constraint would yield motion faster, however without force of constraint
check example 7.10
End lecture Tuesday, Nov 25 2014
November 30, 2014
Advanced Analytical Mechanics-5
- LeastAction.8 -
November 30, 2014
Advanced Analytical Mechanics-5
Conservation Laws
Momentum:
Physics is invariant if r → r + δr for all particles thus
d
∂L
δra since δ x˙ = dt
0 = δL = ∂r
δx = 0 ∀ δra =
[ch. 7.9]
Definition of conjugated momentum
def ∂L
pi =
∂ q˙i
For L(x, x)
˙
a
E-L:
= 12 m(x)
˙ 2 − V (x) consistent with usual definition momentum.
Conservation of total energy
If ∂L
∂t
∂L
∂t
d
∂L
∂L E−L d
∂L
L=
q˙i +
q¨i =
q˙i
dt
∂qi
∂ q˙i
dt
∂ q˙i
= 0 with
∂L
E = q˙i
− L = q˙i pi − L = T + V
∂ q˙i
If L does not depend on time,
total energy is Constant Of Motion
Note: repeated indices imply summation i.e.
∂L
∂qi
qi ≡
n
∂L
i=1 ∂qi
qi
d ∂L
a dt ∂ r˙ a
δr
= 0 thus
N
d
pa = 0
dt a=1
=0
= 0 then
giving dE
dt
Conservation Laws, cont’d
If L does not depend on translation,
total momentum is Constant Of Motion
Angular momentum
Rotation δ φ gives
δr = δ φ × r & δ r˙ =
0 = δL =
Thus
d
dt
∂L
∂ra
δ φ × r˙
δra + ∂∂L
δ r˙ a = p˙a · (δ φ × r) + pa · (δ φ × r)
˙
r˙
a
N
Ma=1
=0 ; M =r×p
a
If L does not depend on rotation,
total angular momentum is Constant Of Motion
Noether theorem:
Symmetry in problem ⇒ conserved quantity
= constant of motion = integral of motion
- LeastAction.9 -
November 30, 2014
Advanced Analytical Mechanics-6
- LeastAction.10 -
November 30, 2014
Advanced Analytical Mechanics-6
Canonical equation of motion
Definition of conjugate momentum
[chapter 7.10]
Poisson Brackets
Poisson brackets
def
{F, G} =
∂L
pi =
∂ q˙i
- LeastAction.11 -
[M: problem 7.30]
∂F ∂G
∂F ∂G
−
∂qn ∂pn
∂pn ∂qn
The time-derivative of the arbitrary function, F , can be written
∂L
− L = q˙i pi − L = T + V
E = q˙i
∂ q˙i
dF
dt
=
=
Hamiltonian
def
H(q, p, t) = pi q˙i − L(q, q,
˙ t)
Hamilton Equations
∂H
q˙i =
∂pi
=
∂F
∂F
∂F
q˙n +
p˙n +
∂qn
∂pn
∂t
∂F ∂H
∂F
∂F ∂H
−
+
∂qn ∂pn
∂pn ∂qn
∂t
∂F
{F, H} +
∂t
The Hamiltonian generates time-translations. In particular,
q˙n = {qn , H}
p˙n = {pn , H}
The first two of the Hamilton equations.
∂H
p˙i = −
∂qi
∂L
∂H
dH
=−
=−
∂t
∂t
dt
The Hamilton equations, an alternative to Euler-Lagrange
{qk , pl } = δkl
To quantum mechanics:
represent dynamical variables by linear operators on a Hilbert space, and Poisson brackets by commutators (multiplied by 2π/(ih)).
November 30, 2014
Advanced Analytical Mechanics-7
Liouville theorem
System of N particles, (qi , pi ), i = 1 · · · N
Consider a statistical ensemble,
phase-space density ρ dq1 dp1 · · · dqN dpN
What is density at time t + ∆t ??
—————————————————————————–
q
flow into hypercube side qi : Ii = ρq˙i ∆t∆pi
p
flow into hypercube side pi : Ii = ρp˙ i ∆t∆qi
flow out from hypercube at side qi + ∆qi :
Oiq
∂ρq˙i
= ρq˙i + ∆qi
∆t∆pi
∂qi
Net increase density
∂ρ
∂ρq˙i
∂ρp˙i
I −O =
∆qi ∆pi ∆t = −
+
∆qi ∆pi ∆t
∂t
∂qi
∂pi
∂ q˙i
∂H
∂2H
and
thus
=
∂pi
∂qi
∂qi ∂pi
2
∂ p˙
∂ H
similarly ∂pi = − ∂q
i
i ∂pi
q˙i =
and thus
d
∂ρ
∂ρ
∂ρ
ρ=
+ q˙i
+
p˙i = 0
dt
∂t
∂qi
∂pi
Liouville theorem:
Phase-space density is conserved.
- LeastAction.12 -
November 30, 2014
Advanced Analytical Mechanics-7
- LeastAction.13 -
Liouville theorem application
reference:
paper by Lemaˆıtre and Vallarta, Phys. Rev. 43 (1933) p87
”On Compton’s Latitude Effect of Cosmic Radiation”,
Interesting: Lemaˆıtre is also the inventor of the Big Bang.
Observation by Compton et al. (Phys. Rev. 41, p111):
The intensity of cosmic radiation is nearly constant for latitudes north of 34’ in the American continent and south of 34’ in Australasia, it drops sharply between these latitudes
to a value about 87 percent as great as that for high latitudes, reaching a minimum at or
near the magnetic equator.
Lemaˆıtre:
Liouville’s theorem is applicable. If now we assume that the intensity distribution of the
cosmic radiation at infinity is homogeneous and isotropic, the intensity in all allowed
directions at any point in the earth’s magnetic field is, by Liouville’s theorem, the same.
Thus the question of calculating the intensity at any point on the earth’s surface reduces
to that of finding out in which directions particles coming from infinity can reach that point.
This proves that cosmic rays are high energy charged particles.
November 30, 2014
Advanced Analytical Mechanics-8
Spherical Pendulum
particle of mass
l.
Calculate L
November 30, 2014
U=0
Hamilton eqns. of motion
p2φ
p2θ
H(θ, pθ , φ, pφ ) =
+
− mgl cos θ
2
2
2ml2
2ml sin θ
∂H
˙
= pφ /(ml2 sin2 θ)
φ=
∂pφ
∂H
−p˙φ =
=0
∂φ
˙θ = ∂H = pθ /ml2
∂pθ
p2φ cos θ
∂H
−p˙θ =
= − 2 3 + mgl sin θ
∂θ
ml sin θ
b
g
φ
m
T = 12 m(x˙ 2 + y˙ 2 + z˙ 2 ) = 12 ml2 θ˙2 + φ˙ 2 sin2 θ
V = mgz = −mgl cos θ
L = 12 ml2 θ˙2 + φ˙ 2 sin2 θ + mgl cos θ
∂L
= ml2 θ˙
∂ θ˙
∂L
pφ =
= ml2 φ˙ sin2 θ
∂ φ˙
Stability circular orbit ( θ˙
for pθ = C ,
pθ =
Hamiltonian =?
H = pθ θ˙ + pφ φ˙ − L = 12 ml2 θ˙2 + Vef f (θ, pφ )
0=−
p2φ cos θ0
ml2
3
=0)
End lecture Thursday, November 27, 2014
+ mgl sin θ0
sin θ0
solve for θ0 , thus θ˙ = 0 and pθ = 0 thus circular orbit
p2φ cos θ0 = m2 gl3 sin4 θ0
Small perturbation, θ
= θ0 + α(t), order α terms:
O(α)
with pφ is COM and
Vef f (θ, pφ ) =
- LeastAction.15 -
Spherical Pendulum, cont’d
θ
Constants Of Motion =?
L is independent of φ; pφ is thus COM
L is independent of t; total energy E is thus COM
Canonical momenta =?
Advanced Analytical Mechanics-8
[example 7.12]
m confined to move on sphere radius
x = l sin θ cos φ; y = l sin θ sin φ; z = −l cos θ
(θ = 0 is under)
x˙ = l θ˙ cos θ cos φ − φ˙ sin θ sin φ
- LeastAction.14 -
−ml2 α
¨ ml2 sin3 θ0 = −p2φ cos θ + m2 gl3 sin4 θ =
p2φ
2ml2
2
sin θ
− mgl cos θ
− p2φ (cos θ0 − α sin θ0 ) + m2 gl3 (sin4 θ0 + 4α sin3 θ0 cos θ0 ) = αp2φ sin θ0 +
4 cos2 θ0 / sin θ0
like −¨
x
= ω 2 x with ω 2 > 0
1/--pages
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