1. No category

January 14, 2015
Advanced Analytical Mechanics-1
Analytical Mechanics
( AM )
lecture notes part 10, Summary
Damped & Driven Oscillators
Driving force F (t)
= m A0 cos(ωt), eom is real part of
x
¨ + 2β x˙ + ωr2 x = A0 eiωt
Olaf Scholten
KVI, kamer v3.008
tel. nr. 363-3552
email: [email protected]
Web page: http://www.kvi.nl/˜scholten
trial solution:
A=
x(t) = Aeiωt gives special solution:
A0
−iδ
=
|A|e
(ωr2 − ω 2 ) + 2iβω
Full sol. = Special sol. + sol. homogeneous eq.
Non-linear effects
Book
Classical Dynamics of Particles and Systems,
Stephen T. Thornton & Jerry B. Marion
5th Edition
ISBN-10: 0534408966 — ISBN-13: 9780534408961
Frequency doubling (tripling) & Hysteresis & Chaos
Oscillator with memory effect
x
¨ + 2β x˙ + ωr2 x + ωr2 C x3 = A0 cos(ωd t)
0
2
3
4
5
6
. . . . . . . . ...
. . . . . . . . . . . . . . .... ..........
3
2
1
0
-1
-2
...
.
.
.
.
.
.
.
. . .... .... ...... ...... .. . .. .
... ....
Tune up
Tune down
4
Max Ampl
Introduce an abstract reference-frame independent formulation of Mechanics via the
Hamiltonian and Lagrangian formalisms. This lends itself to generalizations to problems
in Statistical Mechanics, Quantum Mechanics, Relativistic Mechanics, Field Theory, .....
Phase
Intro
1
3
....
.....
2
...... .
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.....
... ............
.
1
0
0
1
2
3
....
3
. . . . . . . . .....
. x. . .......
.oscil
. . . . ....
.
4
5
6
Frequency
Initial conditions at slowly tune up or down frequency
pendulum
Tue Mar 31 2009
23:19:23
- Summary.1 -
January 14, 2015
Advanced Analytical Mechanics-2
- Summary.2 -
January 14, 2015
Advanced Analytical Mechanics-2
Fourier transform
Chaos
Computer simulations-6
Driven and damped non-linear Oscillator inverted linear part
EOM of the type:
x
¨ + 2β x˙ + ωr2 x = F (t)/m
|
{z
}
L[x]
x
¨ + 2β x˙ + ωr2 x + ωr2 C x3 = A0 cos(ωd t)
(dx/dt)/
parameters ωr
= 2, β = 0.1, A0 = 4., ωd = 4.3, C = −1.1
2
2
... ..........................
...........................................................................................
...............................................................................................................................................
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0
0
-2
Phase
-3 -2 -1 0 1 2 3
-3 -2 -1 0 1 2 3
x
amplitude
0
3
2
1
0
-1
-2
1000
Poincare´
-4
2000
3000
4000
..
..
.
..
...... ..........
......
.......... ............. ............ ............ .........
...
.......
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. . . . . . . . . ... ..
...
. . .
.......
.... .. . ... .. ..... ..... .... ..... ..... ... ........... .... ..... ............
... .... ...... .... ...... ....
... . .
.... ..
... ...
... ...
... ..
... .
... ...
... ..
... ...
... ..
... ..
...........
........
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....
....
....
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....
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3
2
.
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.
.
..
....
......
......
...... ...... .......
...... .......
1 ......
... .... ....
... .... .... .... ... .... .... ..... .... ....
.. ....
.
.
. ... ..
. .. .. .. . .. . .. .. ..
.
. ..
0 ..... ............. ... ..... ..... ...... ..... ....... .... ..... .... ...... .... ......... .... ...... ........... ... ..... .....
.. .. .. .. .. . .. . . . .. . .. ..... .. . .. . .. .. .. .
... .
... ..
-1 .......... ......... ........ ......... .......... ........ ........ ....... ........ ......... ........
..
.
.
-2
-3
4000
5000
6000
7000
8000
time [0.01 sec]
NL-pendulum
Wed Apr 01 2009
23:00:02
An ei nω0 t
n=−∞
= 2π/τ , An = A∗−n and
Z
1 τ /2
F (t) e−i nω0 t dt
An =
τ −τ /2
with ω0
..........................
.. .
-2
-4
F (t) =
∞
X
gives the special solution
∞
1 X
An
i nω0 t
x(t) =
e
m n=−∞ (ωr2 − n2 ω02 ) + 2i nβω0
Greens function
The Greens function is the response to F (t)
0
G(t; t ) =
1
−β(t−t0 )
ω1 m e
sin ω1 (t − t0 )
0
dt0 δ(t − t0 )F (t0 ) then
Z
x(t) = dt0 G(t; t0 )F (t0 )
If F (t)
=
R
= δ(t − t0 )
> t0
for t < t0
for t
- Summary.3 -
January 14, 2015
Advanced Analytical Mechanics-3
- Summary.4 -
January 14, 2015
Advanced Analytical Mechanics-3
Hamilton’s principle of least action
Variational Calculus
Fully equivalent to Newtonian Dynamics
Brachystochrone & Soapfilms & Geodesic
Z
R2
A = 1 dA where dA = 2πxds
p
√
√
2
2
02
ds = dx + dy = dx 1 + Y = dy 1 + X 02
Lagrange multiplier λ(x) with G(Y1 , · · · , YN ; x)
d ∂F
∂G
∂F
−
+ λ(x)
=0
∂Yi
dx ∂Yi0
∂Yi
Alternative form
i ∂F
d h
0 ∂F
F −Y
=
0
dx
∂Y
∂x
=0
i = 1, · · · , N
t2
S1,2 =
L(x, x;
˙ t) dt
t1
Lagrangian
L(x, x;
˙ t) = T (x)
˙ − V (x)
Constraint g(xi , t)
= 0 gives
∂L
d ∂L
∂g
−
+ λ(t)
= 0 with g(qi ; t) = 0
∂qi
dt ∂ q˙i
∂qi
where force exerted by the constraint is
Qi = λ(t)
∂g
∂qi
Hamiltonian eq. of motion, momentum pi
=
∂L
∂ q˙i
def
H(q, p, t) = pi q˙i − L(q, q,
˙ t)
∂H
∂H
q˙i =
; p˙i = −
;
∂pi
∂qi
∂L
∂H
dH
=−
=−
∂t
∂t
dt
Poisson Brackets
∂F ∂G
∂F ∂G
{F, G} =
−
∂qn ∂pn
∂pn ∂qn
- Summary.5 -
January 14, 2015
Advanced Analytical Mechanics-4
- Summary.6 -
January 14, 2015
Advanced Analytical Mechanics-4
Central potential
Conservation Laws
If ∂L
∂t
= 0 then
dE
dt
= 0 with
E = q˙i pi − L = T + V
total energy is conserved
L = 21 µ(r˙ 2 + r2 θ˙2 ) − V (r)
L cyclic in θ, conjugate momentum is constant of motion
pθ = ∂L/∂ θ˙ = µr2 θ˙ = Jz
L invariant under translation ~r → ~r + δ~r then
N
X
d
p~a = 0
dt a=1
Z
d X ~N
~ = ~r × p~
Ma=1 = 0 ; M
dt a
r
t − t0 =
2
1 l
2 µr 2
dr0
q
r0
total linear momentum is constant of motion
~ × ~r
L invariant under rotation δ~r = δ φ
2
A) Conservation of energy: E = µr˙ +
1
2
2
µ (E
− V (r0 )) −
l2
µ2 r 02
B) find θ(r): dθ = θr˙˙ dr gives
Z
θ(r) = ±
r
l/µr02 dr0
q
2
µ (E
− V (r0 )) −
l2
µ2 r 02
total angular momentum is Constant Of Motion
C) Euler-Lagrange: µ(¨r − rθ˙2 ) = − ∂V
∂r = F (r)
−µr2
d2 1 1
+ = 2 F (r)
2
dθ r
r
l
Stability
Small amplitude expansion
+ V (r)
- Summary.7 -
January 14, 2015
Advanced Analytical Mechanics-5
- Summary.8 -
January 14, 2015
Advanced Analytical Mechanics-5
Kepler’s problem; Planetary motion
F~ = − rk2 rˆ or V = −k/r with k > 0
Inertial frame e
ˆi , rotating frame, eˆ0i , same origine
Vectors ~
rI = (r1 , r2 , r3 ) and ~rB = (r10 , r20 , r30 ) point to point
α
= 1 + cos θ
r
with α
2
p = (r10 eˆ01 , r20 eˆ02 , r30 eˆ03 ) = (r1 eˆ1 , r2 eˆ2 , r3 eˆ3 )
= l /µk and eccentricity =
p
1 + 2Eα/k
E
E
E
E
k
= − 2α
<0
=0
>0
~vI = ~vB + ω
~ × ~rB
~aI = ~aB + 2~
ω × ~vB + ω
~˙ × ~rB + ω
~ × (~
ω × ~rB )
Kepler’s first law:
=0
<1
=1
>1
Non inertial frame
~B
Thus F
circle
ellipse
parabola
hyperbola
= F~inert − 2m~
ω × ~vB − mω
~˙ × ~rB − m~
ω × (~
ω × ~rB )
Deviation of falling mass from plumb line
F~B⊥ = F~B − m~g = −2m~
ω × ~vB
Interplanetary travel & stability circular orbit
Effective potential
Absorb parts of kinetic energy in an Effective Potential
Foucault pendulum:
g
Solution x = Ax (t)eiω0 t , y = Ay (t)eiω0 t with ω02 = l
Ax = A sin ωr t & Ay = A cos ωr t with ωr = ωE sin λ
- Summary.9 -
January 14, 2015
Advanced Analytical Mechanics-6
Inertial Tensor
P
1
1
2
~
T = 2 M V + Tr with Tr = 2 i,j Iij ωi ωj and
h
i
X
X
2
Iij =
mα δij (
rα,k
) − rα,i rα,j
α
k
R
Integral from: Iij = ρ(~
r) δij r2 − ri rj d3 r
h
i
O
cm
Steiner parallel axis theorem: Iij
= Iij
+ M δij~a2 − ai aj
Li =
X
Iij ωj
j
- Summary.10 -
January 14, 2015
Advanced Analytical Mechanics-6
Rotating Body
Equation of motion
~˙ = N
~
L
in Body-fixed (rotating) system:
~ =N
~
{I} · ω
~˙ + ω
~ ×L
Use principal axes
I1 ω˙ 1 + (I3 − I2 )ω2 ω3 = N1
I2 ω˙ 2 + (I1 − I3 )ω3 ω1 = N2
I3 ω˙ 3 + (I2 − I1 )ω1 ω2 = N3
Euler’s equations for Rigid Body
Coordinate Transformation:
P
P
0
ri = j λij rj and ri = j
˜ ij r0 with
λ
j
˜ ij = eˆi · eˆ0 = λji thus
λ21 = eˆ02 · eˆ1 = |r0 ihr| and λ
j
˜ = {λ}−1 = {λ}T = [|r0 ihr|]T
|rihr0 | = {λ}
˜
~ 0 = {λ}L
~ and {I 0 } = {λ}{I}{λ}
L
Euler angles
~r = λψ λθ λφ~r0 = λ~r0
Symmetric top, N=0
ω1 (t) = A cos(Ωt + φ), ω2 (t) = A sin(Ωt + φ)
with Ω = ω3 (I3 − I1 )/I1
Symmetric top, tip fixed -I
˙ 2 − M gh cos θ
L = 21 I1 (φ˙ 2 sin2 θ + θ˙2 ) + 21 I3 (φ˙ cos θ + ψ)
Euler-Lagrange eq in φ: pφ
=
Euler-Lagrange eq in ψ : pψ
=
∂L
∂ φ˙
∂L
∂ ψ˙
= constant
= constant
- Summary.11 -
January 14, 2015
Advanced Analytical Mechanics-7
- Summary.12 -
January 14, 2015
Advanced Analytical Mechanics-7
Relativistic Kinematics
Continuous string
dτ , time in rest system particle
p
2
2
2
2
2
2
(c dτ ) = (c dt) −(dx) −(dy) −(dz) = (c dt/γ) with γ = 1/ 1 − (~v /c)2 ,
p
˙~r = ~v , and L = −mc2 1 − (~v /c)2
Invariant:
The Euler-Lagrange eqn
∂L
d ∂L
d ∂L
−
−
=0
∂ψ dt ∂ ψ˙
dx ∂ψ 0
p~ =
String: Lagrangian density
ρ ˙ 2 τ 02
0
˙
L(ψ, ψ, ψ ) = ψ − ψ
2
2
ρ
Lorentz invariant xν aν
2
d
d
ψ
−
τ
ψ=0
2
2
dt
dx
with solutions (k/ω)2
Z
ψ(x, t) =
∂L
= γm~v ; H(~r, p~, t) = p~ · ~r˙ − L(~r, ~r˙ , t) = γmc2
∂~r˙
Contravariant four vector xµ = [ct, ~
x]; Covariant four vector xµ
Invariant x2 = xµ g µν xν = xµ gµν xν
Lorentz Transformation, any four vector A0µ (x0 ) = Λµν Aν (x)
resulting finally in the wave equation
2
Four-vector uµ
dk f (k) ei(kx−ωt)
Proca Lagrangian:
2
ρ ˙ 2 τ 02 µ 2
0
˙
L(ψ, ψ, ψ ) = ψ − ψ −
ψ
2
2
2
= ei(kx−ωt) we get dispersion
2
2
−µ + ρω − τ k = 0
∂r µ
∂τ
= x0ν a0ν for any 2 four vectors
µ
r
= γ ddt
= γ[c, ~v ]
~; B
~ =∇
~ ×A
~; E
~ = −∇φ
~ − ∂ A~
E&M 4-vector: Aν (x) = [φ, A]
∂ct
Simplest Invariant action (thus same in any reference system):
S=−
2
=
= gµν xν
pµ = [E/c, p~] = muµ
= ρ/τ
Z
with ψ(x, t)
- EDy.1 -
Lagrangian:
2
µ
mc + eA uµ /c dτ
L(~x, ~v ; t) = −mc
2
p
1
− ~v 2 /c2
~
− e φ(x) − A(x) · ~v /c gives eom
d
e
~
~
γm~v = eE + ~v × B
dt
c
H(~x, P~ , t) =
q
~ x, t)
cP~ − eA(~
2
+ m2 c4 + eφ(~x, t)
January 14, 2015
Advanced Analytical Mechanics-8
- EDy.2 -
Particle in Electromagnetic Field
choice coordinates:
particle at position ~
a(t)
EM field given by Aµ (x) or Fµν (x)
Z
S=
Z
Ldt =
Ld4 x with L = c
Z
Ld3 ~x
L is L-invariant
L = −mc
2δ
3
1
1
(~x − ~a)
− 2 Aµ (x)jµ (x) −
F µν Fµν
γ
c
16πc
Traveling Light.
Euler-Lagrange equation for fields gives Maxwell
∂µ F
µν
4π ν
=
j
c
Aµ − ∂ µ ∂ν Aν =
Lorentz Invariant:
4π µ
j
c
α
= ∂α ∂ =
∂2
∂(ct)2
− O2 Wave equation for EM-field.