Maandag 30 juni 2014, 14.00 - faculteit Technische Natuurkunde

Faculteit Technische Natuurkunde - TU Eindhoven
Tentamen Theoretische Klassieke Mechanica (3EMX0)
Maandag 30 juni 2014, 14.00–17.00.
Het tentamen levert max. 100 punten op, waarvan de verdeling hieronder is aangegeven.
(1) Korte vraagstukken
Geef bij jouw antwoord of uitwerking een korte argumentatie. Bij een ernstige fout
in de argumentatie worden geen punten toegekend.
(3 ptn)
a) When do we have H = T + V = Etot ?
(3 ptn)
b) What are cyclic variables and why are they advantagous in the context of the
Hamilton formalism?
(3 ptn)
c) Explain how you can obtain the Feigenbaum constant δ for the chaotic oscillator.
What is the difference between the values for the chaotic oscillator and the “logistic
map”?
(3 ptn)
d) What is the definition of the Lyapunov exponent?
(3 ptn)
e) What are the two postulates of Einstein?
(3 ptn)
f) Wat zijn de definities van holonome, skleronome en rheonome beperkingen?
(3 ptn)
g) Wat is de essentie van het equivalentieprincipe?
(3 ptn)
h) Wat is de dimensie van de gegeneraliseerde impuls? Geef twee voorbeelden.
(3 ptn)
i) Wat is de essentie van de ijk invariantie van de Lagrangiaan?
(3 ptn)
j) Wat is bedoeld met de “action-at-a-distance” hypothese van Newton?
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(2) Hamilton’s equations for a pointmass on a cone
z
ρ=cz
o
g
y
x
Consider a mass m which is constrained to move on the frictionless surface of a
vertical cone ρ = cz (in cylindrical polar coordinates ρ, φ, z, with z > 0) in a uniform
gravitational field g pointing vertically down.
(5 pnt)
a) Determine Hamilton’s equations using z and φ as generalized coodinates.
(3 pnt)
b) Argue that for any given solution there are maximum and minimum heights zmax
and zmin between which the motion is confined.
(3 pnt)
c) Use this result to describe the motion of the mass on the cone.
(4 pnt)
d) Show that for any given value of z > 0 there is a solution in which the mass moves
in a circular path at fixed height z.
(3) Special relativity
(5 pnt)
a) A train and a tunnel both have proper lengths L. The train drives with velocity v
towards the tunnel. At the front of the train a bomb is placed which is designed to
explode when the front of the train passes the far end of the tunnel. A deactivation
sensor is located at the back of the train. When the back of the train passes the
near end of the tunnel, this sensor tells the bomb to disarm itself. Does the bomb
explode?
Sketch the problem in the train and tunnel frames and argue that the result has
to be the same in both frames by also taking the speed of the signal into account.
b) The train with proper length L now moves at speed c/2 with respect to the ground.
A ball is thrown from the back to the front, at speed c/3 with respect to the train.
How much time does this take, and what distance does the ball cover, in:
(2 pnt)
(4 pnt)
i. The train frame?
ii. The ground frame? Solve this by using a velocity-addition argument and by
using the Lorentz transformations to go from the train frame to the ground
frame.
2
(3 pnt)
(3 pnt)
(3 pnt)
iii. The ball frame?
iv. Verify that the invariant interval is indeed the same in all three frames.
v. Show that the times in the ball frame and ground frame are related by the
relevant γ factor and that the times in the train frame and ground frame are
not related by the relevant γ factor. Why not?
(4) Coupled oscillators
x=0
k1
k2
k1
m1
x=a
m2
x = 2a
wall
wall
Consider the system of two pointmasses m1 and m2 and three linear Hookian springs
with springconstants k1 and k2 sketched in Fig. 1. We restrict ourselves to longitudinal
displacements in the x-direction.
x
x = 3a
Figuur 1: Equilibrium conformation of two pointmasses m1 and m2 located between two
rigid walls at x = 0 and x = 3a and coupled by means of three linear Hookian springs
with spring constants k1 and k2 .
(2 pnt)
a) How many degrees of freedom does the system have? Hint: We restrict ourselves
to longitudinal displacements.
(5 pnt)
b) Determine the Lagrangian of the system in terms of the displacements relative to
the equilibrium positions u1 (t) ≡ x1 (t) − x1,eq and u2 (t) ≡ x2 (t) − x2,eq .
(6 pnt)
c) Determine the matrices B and T corresponding to the approximate expressions
for the kinetic and potential energy: T (app) = ~u˙ t · T · ~u˙ and V (app) = ~u t · B · ~u.
(6 pnt)
d) Determine the eigenfrequencies of the system.
(4 pnt)
e) Determine the eigenfrequencies for the special case m1 = m2 = m.
(6 pnt)
f) Determine the longitudinal eigenmodes (normal modes), i.e. the displacement
vectors corresponding to the eigenoscillations of the system for the special case
m1 = m2 = m. How many longitudinal eigenmodes are there? Make sketches of
the eigenmodes.
(6 pnt)
g) Determine the normal coordinates for the special case m1 = m2 = m.
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Faculteit Technische Natuurkunde - TU Eindhoven
Uitwerkingen Tentamen Theoretische Klassieke Mechanica (3EMX0)
Maandag 14 april 2014.
1) Korte vraagstukken
a) This only holds if the equations defining general coordinates do not explicitly depend on time and if the forces are derivable from a conservative potential V .
b) Met welk soort van transformaties kun jij variabelen cyclisch maken?
Canonical transformations. Cyclic coordinates:
∂L
=0
∂qi
∂H
=0
∂qi
Cyclic coordinates are not part of the Hamilton function: “ignorable”
⇒ p˙i = 0 ⇒
c) Leg uit hoe jij de Feigenbaum constante δ van de chaotische slinger kunt bepalen.
Wat is het verschil voor de waarde van δ voor de chaotische slinger en de “logistic
map”?
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γn+1 − γn ≈ (γn − γn−1 )
δ
γn : “drijfsterkte” waar n-te verdubbeling van de periode gebeurt. δ = 4.6692016 is
universeel als n → ∞.
d) What is the definition of the Lyapunov exponent?
|∆φ(t)| ≈ K exp(λt). λ < 0: not chaotic, λ > 0: chaotic.
e) Wat zijn de twee postulaten van Einstein?
• Alle inertiaal-waarnemers zijn gelijkwaardig. Of: natuurwetten zien er in alle
inertiaalstelses hetzelfde uit; of de fysica moet voor elke inertiaal-waarnemer
dezelfde zijn.
• De lichtsnelheid is voor elke inertial-waarnemer gelijk, ongeacht de relatieve
snelheid van de lichtbron.
f) Beperkingen die in de vorm F (q1 , q2 , . . . , q3N , t) = 0 kunnen worden geschreven,
worden holonoom genoemd. Rheonome beperkingen hangen expliciet van de tijd
af. Skleronome beperkingen hangen niet expliciet van de tijd af.
g) Volgens het equivalentieprincipe van Einstein is de inertiale massa precies gelijk
aan de zware massa. Een waarnemer heeft dus principi¨eel geen mogelijkheid te
bepalen/onderscheiden of hij 1) in een zwaartekrachtveld ~g zit of 2) een versnelling
~a = ~g ondergaat.
h) De gegeneraliseerde impuls pi ≡ ∂∂L
kan b.v. de dimensie van een lineaire impuls
q˙i
of van een impulsmoment hebben. Bijhorende voorbeelden: een vrij deeltje zonder
externe krachten L = 12 m~v2 en pi = mvi ; een deeltje zonder externe krachten dat
wrijvingsvrij langs een circel met straal R kan bewegen L = 12 mR2 ϕ˙ 2 en pi = mR2 ϕ.
˙
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i) Ijk invariantie van de Lagrangiaan: de Lagrangiaan L is niet uniek. B.v. een additieve constante (L0 = L + C1 ) of vermenigvuldigen met een constante factor
(L0 = C2 L) laten de vgln. van Lagrange invariant. Algemener, als we twee Lagrangiaanse functies L en L0 beschouwen, die zich van elkaar door een zogenoemde ijk
waarbij F = F (q1 , q2 , . . . , q3N , t) onafhantransformatie onderscheiden L0 = L + dF
dt
kelijk van de gegeneraliseerde snelheden is, dan leiden beide functies L en L0 tot
dezelfde bewegingsvgln.
j) Newton veronderstelde dat de interactie krachten tussen twee deeltjes die op een
bepaalde onderlinge afstand met elkaar wisselwerken, instantaan op beide deeltjes werken, d.w.z. interacties tussen voorwerpen worden oneindig snel door de
lege ruimte overgebracht. Deze veronderstelling staat bekend onder “action-at-adistance”.
2) Hamilton’s equations for a pointmass on a cone
a) Determine Hamilton’s equations using z and φ as generalized coodinates.
1 ˙ 2
˙ 2 + z˙ 2 = 1 m (c2 + 1)z˙ 2 + (cz φ)
T = m ρ˙ 2 + (ρφ)
2
2
U = mgz
∂T
∂T
= m(c2 + 1)z,
˙ pφ =
= mc2 z 2 φ˙
˙
∂ z˙
∂φ
2
p2φ
1
pz
+
H=T +U =
+ mgz
2m c2 + 1 c2 z 2
pz =
Hamilton’s equations:
z˙ =
pz
∂H
=
,
∂pz
m(c2 + 1)
∂H
pφ
φ˙ =
=
,
∂pφ
mc2 z 2
p˙z = −
p˙φ = −
p2φ
∂H
=
− mg
∂z
mc2 z 3
∂H
=0
∂φ
b) Argue that for any given solution there are maximum and minimum heights zmax
and zmin between which the motion is confined.
Remember that H is the total energy Etot and is conserved. Thus for any possible
solution H is equal to a fixe constant energy. H as given above is a function of 3
terms. As z → ∞ the last term also → ∞. Thus, there must be a zmax which z
cannot exceed. Similarly, the second term → ∞ as z → 0, so that there has to be
a zmin > 0 below which z cannot go. In particular this means that the mass can
never fall all the into the bottom of the cone.
c) Use this result to describe the motion of the mass on the cone.
˙
The mass moves around the z axis with constant angular momentum pφ = mc2 z 2 φ.
Since pφ is constant, the angular velocity φ˙ varies – increasing as z gets smaller and
decreasing as z gets larger. At the same time the mass’s height z oscillates between
zmin and zmax .
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d) Show that for any given value of z > 0 there is a solution in which the mass moves
in a circular path at fixed height z.
This requires z˙ = 0 to be always true, which means pz = 0 and hence p˙z = 0. From
the 2nd Hamilton equation for z we see that p˙z = 0 if and only if
p
pφ = ± m2 c2 gz 3 .
If, for any chosen initial height z, we start with pz = 0 and pφ equal to one of these
two values, then since p˙z = 0, pz and z˙ both remain 0, and the mass continues to
move at its initial height around a horizontal circle.
3) Special relativity
a) Yes, the bomb explodes. This is clear in the frame of the train. In this frame, the
train has length L, and the tunnel speeds past it. The tunnel is length- contracted
down to L/γ . Therefore, the far end of the tunnel passes the front of the train
before the near end passes the back, so the bomb explodes.
We can, however, also look at things in the frame of the tunnel. Here the tunnel
has length L, and the train is length-contracted down to L/γ. Therefore, the
deactivation device gets triggered before the front of the train passes the far end
of the tunnel, so you might think that the bomb does not explode. We appear to
have a paradox.
The resolution to this paradox is that the deactivation device cannot instantaneously tell the bomb to deactivate itself. It takes a finite time for the signal to
travel the length of the train from the sensor to the bomb. And it turns out that
this transmission time makes it impossible for the deactivation signal to get to the
bomb before the bomb gets to the far end of the tunnel, no matter how fast the
train is moving. Lets show this.
The signal has the best chance of winning this “race” if it has speed c, so lets
assume this is the case. Now, the signal gets to the bomb before the bomb gets to
the far end of the tunnel if and only if a light pulse emitted from the near end of
the tunnel (at the instant the back of the train goes by) reaches the far end of the
tunnel before the front of the train does. The former takes a time L/c. The latter
takes a time L(1 − 1/γ)/v , because the front of the train is already a distance L/γ
through the tunnel. So if the bomb is not to explode, we must have
L/c < L(1 − 1/γ)/v
p
⇒ β < 1 − 1 − β2
p
⇒ 1 − β2 < 1 − β
p
p
⇒ 1+β < 1−β
This is never true. Therefore, the signal always arrives too late, and the bomb
always explodes.
b) Throwing on a train
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i. In the train frame, the distance is simply d = L. And the time is t = L/(c/3) =
3L/c.
ii. Velocity addition argument: The velocity of the ball with respect to the ground
is (with u = c/3 and v = c/2):
Vg =
c/3 + c/2
u+v
5c
=
uv =
1 + c2
1 + 1/3 · 1/2
7
√
The length of the train in the ground frame is√L/γ1/2 = 3L/2. Therefore, at
time t the position of the front of the train is 3L/2 + vt. And the position of
the ball is Vg t. These two positions are equal when
√
√
3L
3L/2
7L
(Vg − v)t =
⇒t=
=√ .
2
5c/7 − c/2
3c
Equivalently,
√ this time is obtained by noting that the ball closes the initial head
relative speed
start of 3L/2 that the front of the train had, at a √
√ of Vg − v.
The distance the ball travels is d = Vg t = (5c/7)(7L/ 3c) = 5L/ 3.
Lorentz-Transformation: In the train frame, the space and time intervals are
0
x0 = L and
√ t = 3L/c, from part (a). The γ factor between the frames is
γ1/2 = 2/ 3, so the Lorentz transformations give the coordinates in the ground
frame as
c 3L
5L
2
0
0
=√
x = γ(x + vt ) = √ L +
2 c
3
3
2
3L c/2 L
7L
t = γ(t0 + vx0 /c2 ) = √
+ 2
=√
c
3 c
3c
√
iii. In the ball frame, the train has length, L/γ1/3 = 8L/3.
√ Therefore, the√time it
takes the train to fly past the ball at speed c/3 is t = ( 8L/3)/(c/3) = 2 2L/c.
And the distance is d = 0, of course, because the ball does not move in the ball
frame.
iv. The values of c2 t2 − x2 in the three frames are:
• Train frame: c2 (3L/c)2 − L2 = 8L2
√
√
• Ground frame: c2 (7L/ 3c)2 − (5L/ 3)2 = 8L2
√
• Ball frame: c2 (2 2L/c)2 − (0)2 = 8L2
v. The relative
√ speed of the ball frame and the ground frame is 5c/7. Therefore,
γ5/7 = 7/2 6, and the times are indeed related by
√ !
7
2 2L
7L
tg = γtb ⇒ √ = √
.
c
3c
2 6
The relative
√ speed of the train frame and the ground frame is c/2. Therefore,
γ1/2 = 2/ 3, and the times are not related by a simple time-dilation factor,
because
7L
2
3L
tg 6= γtt ⇒ √ 6= √
3c
3 c
7
We do not obtain an equality because time dilation is legal to use only if the
two events happen at the same place in one of the frames. Mathematically,
the Lorentz transformation ∆t = γ(∆t0 + (v/c2 )∆x0 ) leads to ∆t = γ∆t0 only
if ∆x0 = 0. In this problem, the “ball leaving the back” and “ball hitting the
front” events happen at the same place in the ball frame, but in neither the
train frame nor the ground frame. Equivalently, neither the train frame nor
the ground frame is any more special than the other, as far as these two events
go. So if someone insisted on trying to use time dilation, he would have a hard
time deciding which side of the equation the γ should go on.
4) Coupled oscillators
a) 2.
b) Kinetic energy T = 12 m1 x˙ 21 + 12 m2 x˙ 22 = 12 m1 u˙ 21 + 21 m2 u˙ 22 .
Potential energy V = V1 + V2 + V3 = 12 k1 (x1 − a)2 + 12 k2 (x2 − x1 − a)2 + 21 k1 (x2 − 2a)2
= 21 k1 u21 + 21 k2 (u2 − u1 )2 + 21 k1 u22 = 21 (k1 + k2 )u21 − k2 u1 u2 + 21 (k1 + k2 )u22
L = T − V = 12 m1 u˙ 21 + 21 m2 u˙ 22 − 12 (k1 + k2 )u21 + k2 u1 u2 − 21 (k1 + k2 )u22
c) The T and B matrices are given by
T =
1
2
m1 0
0 m2
,
B=
and
1
2
k1 + k2
−k2
−k2
k1 + k2
d) The eigenfrequencies are determined by the condition det B − ω 2 T
which we find
√
(m +m )(k +k )− (m1 +m2 )2 (k1 +k2 )2 −4m1 m2 k1 (k1 +2k2 )
ω12 = 1 2 1 2
2m1 m2
√
(m +m )(k +k )+ (m1 +m2 )2 (k1 +k2 )2 −4m1 m2 k1 (k1 +2k2 )
ω22 = 1 2 1 2
2m1 m2
= 0, from
e) In the special case m1 = m2 = m, the expressions for the eigenfrequencies reduce
to ω12 = k1 /m and ω22 = (k1 + 2k2 )/m.
f) In the special case m1 = m2 = m, the normalized amplitude vectors corresponding
~ (2) = √12 (1, −1)t .
~ (1) = √12 (1, 1)t and w
to the 2 longitudinal eigenoscillations are w
m
m
m
m
m
m
m
m
Figuur 2: Left: first eigenmode. Right: second eigenmode.
g) The normalcoordinates can be determined using the relation ~η = P −1 · ~u, where
~ (1) |~
P≡ w
w(2) . We find η1 = u1 + u2 and η2 = u1 − u2 .
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