SuSe 2018
Classical Theoretical Physics II
Lecture: Prof. Dr. K. Melnikov – Exercises: Dr. H. Frellesvig, Dr. R. Rietkerk
Exam 1
1. August 2018, 14:00-16:00
Name(in capital letters)
Student number
Exercise 1 / 20 pt
Exercise 2 / 28 pt
Exercise 3 / 30 pt
Exercise 4 / 22 pt
Total / 100 pt
This table of points serves the correction. Please do not write on it.
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Exercise 1: Independent Questions 20 points This exercise consists of questions that can be answered independently.
(a) 1 pt If the Lagrangian describing a physical problem is independent of time, what quantity is conserved?
(b) 1 pt A particle is moving in two dimensions described by Cartesian coordina- tes xandy. If thex-component of the momentum of the particle is conserved, what is the associated property of the Lagrangian?
(c) 1 pt A harmonic oscillator with eigenfrequencyω is subject to a force of the form F(t) =F0cos(Ωt). What will happen qualitatively with the amplitude of the oscillations if Ω is getting closer to ω?
(d) 1 pt Give one example of a physical problem that may be described by the Lagrangian L= 12m~r˙2+kr.
(e) 2 pt State in words the content of Liouville’s theorem.
(f) 2 pt If δRt2
t1 f(q,q, t)dt˙ = 0 and q at times t1 and t2 is held fixed, what differential equation does the function f have to fulfill?
(g) 2 pt A canonical transformation from coordinates (q, p) to coordinates (Q, P) is given by the generating function F(q, Q) = q Q. What transformation is generated?
(h) 3 pt A massless square plate has a particle of mass mattached to each of its four corners, see Figure 1. The plate is placed in thex, y-plane with its center at the origin. What are the moments of inertia of the plate with respect to the z-axis, and with respect to a diagonal axis going through the origin and (a, a,0)?
(i) 3 pt The LagrangianL= 12mx˙2−U(x) describes a particle with mass m in one dimension. At time t= 0 the particle is at position x1. Derive an integral representation for the time t at which the particle is at position x2.
(j) 4 pt A harmonic oscillator with mass m and eigenfrequency ω, is subject to a constant force that is turned on at t= 0, so F(t) =F0ϑ(t). (The function ϑ(t) is the Heaviside function, that is 0 for t < 0 and 1 for t > 0). If the oscillator is at rest when t <0, what is x(t) at t >0?
(a,a,0) (-a,a,0) (a,-a,0)
(-a,-a,0) z
y x
Figure 1: Massless square plate with a particle of massm attached to each corner.
Exercise 2: Particle in central potential 28 points Consider a particle of massm, which is located at the position~rin a three-dimensional space. The particle is subject to a central potential, described by the Lagrangian
L= 12m~r˙2−U(r) , r =|~r|. (1) (a) 4 pt What are the symmetries of this Lagrangian? What are the correspon-
ding conserved quantities?
(b) 3 pt Demonstrate briefly (using one of the conserved quantities) the reduction of this three-dimensional problem to a two-dimensional problem.
(c) 3 pt Give the expression for the energy of the particle and write it as E = 12mr˙2+Ueff(r). (2)
Consider the situation where the potential vanishes at infinity and the particle comes from infinity with energy E and impact parameter ρ.
(d) 3 pt Show that the effective potential may be written as Ueff(r) =Eρ2
r2 +U(r) . (3)
Consider now an explicit potential
U(r) = b r2 − c
r4 , (4)
with positive constants b and c.
(e) 4 pt Sketch the corresponding effective potential Ueff(r). Describe briefly and qualitatively the different types of motion that this potential allows for.
(f) 4 pt Calculate the maximum valueUmax of the effective potential.
(g) 7 pt Compute the total capture cross section σtot for the process whereby the particle comes from infinity and reaches the center of the potential.
Exercise 3: Spring-Pendulum System 30 points
M
θ m x
k
l
Figure 2: Spring-Pendulum system.
Consider the system shown in Figure 2, where a block of mass M moves frictionless on a horizontal plane and is attached to a spring with spring constant k. The extension of the spring is calledx. Attached to the block is a pendulum, consisting of a massless rod of length ` with a mass m attached at its endpoint. Gravity acts along the vertical direction.
(a) 4 pt Give a Lagrangian for this system in terms x, θ and their derivatives.
(b) 4 pt Construct the Euler-Lagrange equations.
(c) 3 pt Set the masses equal,M =m, and assume the spring constant is such that g =k`/(2m). Taylor expand the Lagrangian around the equilibrium point and write it as
L= 1 2
2
X
i,j=1
˙
qimˆijq˙j −qiˆkijqj
+O(q3, qq˙2) , (1)
whereq1 =x and q2 =`θ. Verify that the matrices ˆm and ˆk are ˆ
m=m 2 1
1 1
, ˆk = k 2
2 0 0 1
. (2)
(d) 4 pt Derive the eigenfrequencies of small oscillations.
(e) 4 pt Derive the vectors corresponding to the two eigenmodes of the small oscillations.
(f) 3 pt Describe the motion of each eigenmode qualitatively. Make a small sketch.
(g) 4 pt Explain (without calculation) how the eigenvectors can be used to find the general solutions to x(t) and θ(t).
The general solutions to the equations of motion are x(t) = C1sin(ω1t+φ1) +C2sin(ω2t+φ2) , θ(t) =
√2
` h
−C1sin(ω1t+φ1) +C2sin(ω2t+φ2) i
.
(3)
(h) 4 pt Let the system be at rest in the equilibrium position for all timest <0.
At timet= 0, the pendulum is given a small kick, such that it instantaneously acquires an angular velocity ˙θ(t = 0) =v0/`. Find θ(t) andx(t) for allt >0.
Exercise 4: Hamiltonian mechanics 22 points A relativistic free particle of rest-mass m in a one-dimensional space is described by the Lagrangian
L=−mc2 r
1− v2
c2 , (1)
where v = ˙x.
(a) 4 pt Show that the Hamiltonian for the relativistic Lagrangian in Equation (1) is given by H =cp
p2+m2c2.
(b) 5 pt Calculate the Poisson brackets{p, H} and{x, H}. Use these results to find the explicit time dependence of x(t).
The expansion of the Hamiltonian in question (a) around the non-relativistic limit allows for the computation of relativistic corrections. After the addition of a potential, it is similarly possible to study relativistic corrections to the harmonic oscillator.
After an intricate canonical transformation, the resulting Hamiltonian is written as H(x, p) = p2
2m + 1
2mω20x2+λ p2
2m + 1
2mω02x2 2
. (2)
(c) 3 pt Derive the Hamilton equations of motion for the Hamiltonian in Equa- tion (2).
(d) 3 pt Show with the help of Poisson brackets that H0 = p2
2m +1
2mω02x2 (3)
is a conserved quantity.
(e) 3 pt Using the result of the previous question, show thatx satisfies
¨
x+ω2x= 0 , (4)
whereω =ω0(1 + 2λH0).
(f) 4 pt Solve the differential equation in Equation (4). Express ω in terms of λ, m, ω0 and the amplitude of oscillation A.
Solution of exercise 1: Questions
(a) 1 pt Independent of time The energy is conserved (b) 1 pt px conserved
The Lagrangian is independent of x. Or:
The Lagrangian is invariant under translations in thex direction.
(c) 1 pt Force with ω≈Ω
The amplitude will grow, as the system is close to resonance.
(d) 1 pt What is described by L= 12m~x˙ + kr
Kepler problem, Coulomb problem, Planet around sun, classical electron around nucleus, . . .
(e) 2 pt What is Liouville’s theorem
Areas in phase-space are conserved in time.
(f) 2 pt Variation of action is zero The Euler-Lagrange equation
d dt
∂f
∂q˙ = ∂f
∂q (5)
(g) 2 pt Canonical transformation with F =qQ When F is a function of q and Q, we have
p= ∂F
∂q , P =−∂F
∂Q (6)
so the transformation is
Q=p , P =−q (7)
(h) 3 pt The square plate
I =X
mr2 (8)
so with respect to the two axes we get Iz = 4(m(√
2a)2) = 8ma2 (9)
Idiag = 2(m(√
2a)2) + 2×0 = 4ma2 (i) 3 pt Derive t(x)
E = 1
2mx˙2+U(x)⇔ dx
dt = r2
m
pE−U(x)⇔ dt=
rm 2
1
pE−U(x)dx⇔ (10) t=
rm 2
Z x2
x
dx pE−U(x)
(j) 4 pt Solve the oscillator with constant force The equation of motion is
mx¨+mω2x=F0ϑ(t) (11) For t >0 the general solution is
x=Acos(ωt+θ) + F0
mω2 (12)
which corresponds to
˙
x=−ωAsin(ωt+θ) (13)
The boundary conditions arex(0) = ˙x(0) = 0 implying θ = 0, A=− F0
mω2 (14)
and inserting gives
x= F0
mω2 (1−cos(ωt)) (15)
Solution of exercise 2: ParticleCapture
(a) 4 pt What are the symmetries of the Lagrangian and what are the conserved quantities
The symmetries are invariance under time-translations and under rotations.
The corresponding conserved quantities are the energy and the (three com- ponents of the) angular momentum.
(b) 3 pt How to reduce to a 2d problem
Since M~ = ~r ×~p, ~r will always be perpendicular to M~. We can pick the coordinate system such that the conserved M~ points along the z-axis, and then we get that ~r is confined to the two-dimensionalxy plane.
(c) 3 pt Express E using Ueff
E = 12m~r˙2+U(r)
= 12m( ˙r2+r2θ˙2) +U(r)
= 12mr˙2+ M2
2mr2 +U(r) (16)
= 12mr˙2+Ueff(r) with Ueff(r) = M2
2mr2 +U(r) along the way we used
M~ =mr2θ~˙uz (17) (d) 3 pt Re-express Ueff using ρ and E
Having energy E at infinity, corresponds to the momentum p∞ = √ 2mE. That gives the angular momentumM =p∞ρ =√
2mEρ. Inserting this in the expression for Ueff gives
Ueff(r) = M2
2mr2 +U(r)
= Eρ2
r2 +U(r) (18)
(e) 4 pt Sketch the potential and describe the types of motion We now have
Ueff(r) = Eρ2 +b 1 r2 − c
r4 (19)
The sketch is shown in Figure 3. The types of motion are scattering, circular orbit, capture (from infinity), capture (while bound).
(f) 4 pt Calculate the maximum of Ueff
The maximum is where the derivative is zero.
dUeff
dr = Eρ2+b−2
r3 − −4c
r5 (20)
and that is zero when
−4c=−2 Eρ2+b r2 ⇔ r =±
r 2c
Eρ2+b (21)
Inserting this point gives
Ueff|max= (Eρ2+b)Eρ2+b 2c −c
Eρ2+b 2c
2
= (Eρ2+b)2
4c (22)
(g) 7 pt Calculate the cross section for capture We derive an upper bound on ρ:
E > Ueff|max ⇔E > (Eρ2+b)2
4c ⇔ √
4cE > Eρ2+b ⇔ ρ2 < 2√
cE−b
E ≡ρ2max (23)
Since ρ is by definition positive, we get the refined condition on the minimum energy in terms of the parameters in the potential:
0< 2√ cE−b
E ⇔ E > b2
4c (24)
The lower boundρmin is zero. The capture cross section is then πρ2max if the energy is large enough, and zero otherwise:
σcapture= (π2
√cE−b
E for E > b4c2
0 for E < b4c2 (25)
U_eff
Solution of exercise 3: SpringPendulumSystem
(a) 4 pt Give a Lagrangian
L= 12Mx˙2+12m
˙
x2+`2θ˙2+ 2`cos(θ) ˙xθ˙
− 12kx2−mg`
1−cos(θ) . (26) (b) 4 pt Construct Euler-Lagrange equations
Derive with respect to x and ˙x:
(M +m)¨x+m`cos(θ)¨θ+kx−m`sin(θ) ˙θ2 = 0 . (27) Derive with respect to θ and ˙θ:
m`2θ¨+m`cos(θ)¨x+mg`sin(θ) = 0 . (28) (c) 3 pt Expand Lagrangian
Set M = m and g = k`/(2m). Replace (1−cos(θ)) = θ2/2 and cos(θ) = 1 (because the latter is already multiplied by something small). The Lagrangian
becomes L= 12m
2 ˙x2+`2θ˙2+ 2`x˙θ˙
−12kx2− 12k`2θ2 2
= 12m
2 ˙q12+ ˙q22+ 2 ˙q1q˙2
− 12k
q12+ 12q22
≡ 12
m11q˙12+m22q˙22+ (m12+m21) ˙q1q˙2
− 12
k11q12+k22q22
. (29) From this we read of that
ˆ m =
m11 m12 m21 m22
=m 2 1
1 1
, ˆk=
k11 k12 k21 k22
= k 2
2 0 0 1
. (30) (d) 4 pt Derive eigenfrequencies
Method 1: Set det(ˆk−ω2m) = 0. This gives quadratic equation forˆ ω2, m2(ω2)2−2kmω2+ k2
2 = 0 , (31)
whose solutions are ω12 = k
2m 2 +√ 2
, ω22 = k
2m 2−√ 2
. (32)
Method 2: Calculate eigenvalues of the matrix ˆ
m−1kˆ= k 2m
2 −1
−2 2
(33) since the Euler-Lagrange equations with harmonic ansatz gives (ˆk−ω2m)~aˆ =~0,
ˆ
(e) 4 pt Derive eigenvectors
Method 1: Solve (ˆk−ω1,22 m)ˆ ~a1,2 =~0 for constant vectors~a1,2. For instance, (ˆk−ω12m)ˆ ~a1 =
k−2mω12 −mω21
−mω12 k2 −mω12 a1,1 a1,2
= 0
0
(34) Solve the first equation, (k−2mω21)a1,1−mω21a1,2 = 0,
a1,2 = k−2mω12
mω21 a1,1 = k
mω12 −2
a1,1 = 2 2 +√
2−2 a1,1
= 2(2−√ 2) (2 +√
2)(2−√
2) −2
a1,1 =
(2−√
2)−2
a1,1 =−√
2a1,1 (35) Thus~a1 =a1,1(1,−√
2) is determined up to an overall constant. Similarly,
~a1 =A1
1
−√ 2
, ~a2 =A2
√1 2
, (36)
with undetermined prefactorsA1,2.
Method 2: Compute the eigenvectors for the matrix ˆm−1ˆk.
(f) 3 pt Describe and sketch eigenmodes
The eigenvalueω1 and eigenvector~a1 correspond to the motion where the two masses move in opposite directions.
The eigenvalueω2 and eigenvector~a2 correspond to the motion where the two masses move in the same direction.
(g) 4 pt Explain how to find general solution
Method 1: Decompose ~q in the basis of eigenvectors: ~q = P2
s=1rs~as. The eigenvectors in eq. (36) should be properly normalized:
~as·mˆ ·~as0 =δss0
~as·ˆk·~as0 =ωs2δss0 )
⇒ As = ωs
√2k
. (37)
Then the Lagrangian is, in terms of the normal coordinatesrs, guaranteed to be diagonal and “canonically” normalized:
L=
2
X
s=1
Ls , Ls= 12r˙2s −12ωs2r2s ⇒ rs =Cscos(ωst+φs) . (38)
Find the corresponding Euler-Lagrange equations, m(2−√
2) ¨Q1+kQ1 = 0 ⇒ Q¨1+ω12Q1 = 0 . (41) m(2 +√
2) ¨Q2+kQ2 = 0 ⇒ Q¨2+ω22Q2 = 0 . (42) Solve these equations by
Q1 =C1cos(ω1t+φ1) , Q2 =C2cos(ω2t+φ2) . (43) Insert these solutions into ~q= ˆA·Q~ to find~q. Finally, use q1 =x andq2 =`θ to obtain the general solution for x and θ.
(h) 4 pt Imposing boundary conditions
The conditions x(0) = 0 andθ(0) = 0 produce
C1sin(φ1) +C2sin(φ2) = 0 , (44)
−C1sin(φ1) +C2sin(φ2) = 0 , (45) which have the solution
φ1 =φ2 = 0 . (46)
The velocity conditions ˙x(0) = 0 and ˙θ(0) =v0/` yield
C1ω1cos(φ1) +C2ω2cos(φ2) = 0 , (47)
√2
` (−C1ω1cos(φ1) +C2ω2cos(φ2)) = v0
` , (48)
Inserting φ1 =φ2 = 0, this becomes
C1ω1+C2ω2 = 0 , (49)
√2
` (−C1ω1+C2ω2) = v0
` , (50)
which have the solution
C1 = −1 2√
2 v0
ω1 , C2 = 1 2√
2 v0
ω2 . (51)
Solution of exercise 4: HamiltonianMechanics
(a) 4 pt Compute Hamiltonian
First calculate the conjugate momentum p≡ ∂L
∂x˙ =−mc2 ∂
∂v r
1− v2
c2 = mv q
1− vc22
.
Note: this clearly gives the correct value mv in the non-relativistic limit.
Solve this for v = ˙x, in order to substitute that later into the Hamiltonian:
v = cp pp2+m2c2 The Hamiltonian is then given by
H ≡pv−L=pv+mc2 r
1−v2
c2 = cp2
pp2 +m2c2 +mc2 s
1− p2 p2+m2c2
=cp
p2+m2c2
(b) 5 pt Calculate the Poisson brackets. Find x(t)
{p, H}= ∂p
∂p
∂H
∂x − ∂p
∂x
∂H
∂p = 0 , this means that p is conserved! (52) {x, H}= ∂x
∂p
∂H
∂x − ∂x
∂x
∂H
∂p =−∂H
∂p = −cp
pp2+m2c2 . (53) Combining this with the Hamilton equation ˙x= ∂H∂p we get
˙
x= cp
pp2+m2c2 = constant. (54) Therefore
x(t) = cpt
. (55)
(d) 3 pt Show H0 is conserved In general we have that
df dt = ∂f
∂t +{H, f} . (57) Inserting f =H0 we find
dH0
dt = ∂H0
∂t +{H, H0} . (58) H0 does not depend on time explicitly, so ∂H0/∂t= 0. The Poisson bracket is also zero:
{H, H0}={H0+λH02, H0}={H0, H0}+λ{H02, H0}= (1 + 2λ){H0, H0}= 0 (59) due to antisymmetry of the Poisson bracket. We conclude that dH0/dt= 0, in other words that H0 is conserved.
(e) 3 pt Show x satisfies harmonic differential equation
Use the Hamilton equations in eq. (56) and the fact that H0 is constant. Take the time derivative of the first equation and insert the second equation.
¨
x= (1 + 2λH0)p˙
m =−(1 + 2λH0)2ω02x≡ −ω2x , (60) whereω ≡ω0(1 + 2λH0).
(f) 4 pt Solve the differential equation. Re-express ω.
The general solution is
x(t) = Acos(ωt+φ) . (61) From this one can calculate p, upon inverting the first Hamilton equation,
p= mx˙ 1 + 2λH0
= −mωAsin(ωt+φ) 1 + 2λH0
=−mω0Asin(ωt+φ) . (62) Now one can calculate H0
H0 = (−mω0Asin(ωt+φ))2
2m + 12mω02(Acos(ωt+φ))2
= 12mω02A2 (63)
Inserting this into the definition of ω yields
ω=ω0(1 +λmω20A2). (64) We see that the frequency receives anharmonic corrections that scale as the amplitude squared, as usual.