HW
Problem set 1, Due Jan 15. Total 80 points. The problem marked by * will not
be graded.
1. Find an equation describing the shortest path between the ( x,y,z ) two points on the
conical surface z = 1
x2 + y 2 , and its general solution
−
Solution:

Cylindrical coordinates: x = r sin φ; y = r cos φ which gives us z = 1
L=
 
ds =
−r
dr 2 + r 2 dφ2 + dz 2 =

dL/dφ = 0.
2 + r 2 φr2dr
(0.1)
Euler equation:
d
dr
r2 φ

2 + r 2 φ2
r 2 φ = C
φ=
√
2C

=0

2 + r 2 φ2
2C
φ =
2
r r
C2
dr
+ C1 =
2
r r
C2
C
21/2 arccos
+ C1
r
√
√ −
√ −
| |
(0.2)
2
. Find the curve y(x) that passes through the endpoints (0 , 0) and (1 , 1) and minimizes
the functional
I [y] =
1 dy(x) 2
[( dx )
0

2
− y (x)]dx
Solution:
f = (y  )2
−y
2
(0.3)
Euler equation:
2y(x) + 2y  = 0
y = C 1 sin t + C2 cos t
C2 = 0, C1 = 1/ sin 1
(0.4)
3. Consider a spherical pendulum of a mass
m and a length l. Write expressions for
a. the Lagrangian in terms of generalized coordinates, and the Lagrange equations
L=
1 2 dθ 2 1 2 2 dφ 2
ml ( ) + ml sin θ( ) + mgl cos θ
2
dt
2
dt
1 2 d2 θ
ml 2
2
dt
dφ 2
) + mgl sin=0θ
dt
1 2d
dφ
ml (sin2 θ =
0)
2
dt
dt
− ml
2
cos θ(
(0.5)
(0.6)
b. the generalized momenta,
dθ
dt
2
2 dφ
pφ = ml sin θ
dt
pθ = ml 2
(0.7)
c. generalized forces
Fφ = 0
Fθ =
2 2
−mgl sin φ + ml φ̇
sin θ cos θ
(0.8)
4. A double pendulum cons ists of two simpl e pendule, with one pendulum suspen ded
from the the blob of the other . The two p endule have equal lengt hs and b obs have equal
masses. They are confined to move in the same plane. Find Lagrange’s equations of motion
for the system. Do not assume small angles.
Solution:
The Lagrangian has a form:
L = ml 2 φ̇21 +
m 2 2
l φ̇2 + ml 2 φ̇1 φ̇2 cos(φ1 φ2 ) + 2mgl cos φ1 + mgl cos φ2
2
∂L
= 2mgl sin φ1 ml2 φ̇1 φ̇2 sin(φ1 φ2 ),
∂φ 1
∂L
= mgl sin φ2 + ml 2 φ̇1 φ̇2 sin(φ1 φ2 ),
∂φ 2
∂L
= 2ml2 φ̇1 + ml 2 φ̇2 cos(φ1 φ2 )
˙
1
∂ φ∂L
= ml 2 φ̇2 + ml 2 φ̇1 cos(φ1 φ2 )
∂ φ̇2
−
−
−
−
−
−
−
−
2
(0.9)
Lagrange equations are:
d
[2ml2 φ̇1 + ml 2 φ̇2 cos(φ1 φ2 )] + 2mgl sin φ1 + ml 2 φ̇1 φ̇2 sin(φ1
dt
d
[ml2 φ̇2 + ml 2 φ̇1 cos(φ1 φ2 )] + mgl sin φ2 ml2 φ̇1 φ̇2 sin(φ1
dt
−
−
−
−φ )=0
−φ )=0
2
2
(0.10)
.
5. A pendu lum consists of a mass m suspended by massless spring with unextended
length b and spring constant k.
.
a. Find Lagrange’s equations of motion.
Solution:
x = l sin φ; y = l cos φ
ẋ = l˙ sin φ + l φ̇ cos φ; ẏ = l˙ cos φ l φ̇ sin φ;
m
k
L = ( l˙2 + l 2 φ̇2 ) + mgl cos φ
(l b)2
2
2
−
− −
(0.11)
Lagrange equations:
m¨l
2
− mlφ̇ − mg cos φ + k(l − b) = 0
ml2 φ̈ + 2mlφ̇l˙ + mgl sin
=0φ
(0.12)
b. Suppose that the point of support of the pendulum is movi ng in vertical direction
according the low y = a sin ct Fig.1). Find the Lagrange equations.
Solution:
y=
−l cos φ + a sin ct = ỹ + a sin ct,
L=
−l˙ cos φ − lφ̇ sin φ + ac cos ct = ỹ˙ + ac cos ct
˙ cos ct + mgl cos φ − k(l − b)(0.13)
+ l φ̇ ) + mỹac
ẏ =
m ˙2
(l
2
2 2
2
The term which depends only on time was omitted from the Lagrangian. It is convenient
to use the identity:
d
dỹ
(ỹ cos ct) =
cos ct
dt
dt
3
− cỹ sin ct
(0.14)
FIG. 1:
Omitting the total derivative from the Lagrangian we have
m 2
2 2
2
2
L = 2 ( l˙ + l φ̇ ) + mac ỹ sin ct + mgl cos φ k(l b) =
m ˙2
( l + l2 φ̇2 ) + mac2 l cos φ sin ct + mgl cos φ k(l b)2
2
− −
− −
(0.15)
(You do not have to use this trick. Lagrange Eqs. will be the same)
Lagrange Eqs:
m¨l
2
2
− mlφ̇ − mg cos φ + k(l − b) + mac cos φ sin ct = 0
ml φ̈ + 2mlφ̇l˙ + mgl sin φ − mac l sin φ sin ct
=0
2
2
6. A spherical pendulum is rotating about the vertical axis with frequency
Write a Lagrange equation.
(0.16)
ω (See Fig.2).
Solution:
L=
1 2 dθ 2 1 2 2 2
ml ( ) + ml ω sin θ + mgl cos θ
2
dt
2
dφ
=ω
dt
d2 θ
g
ω 2 sin θ cos θ + sin=0θ
2
dt
l
−
(0.17)
7. Two identical pendule of mass m and length l are attached to a bar of a mass M , which
can slide without friction in horizontal direction. (see Fig. 3). Write Lagrange equations.
4
FIG. 2:
FIG. 3:
Solution:
x1,2 = x + l sin θ1,2 ,
2
L=
y = l cos θ
2
M + 2m 2 ml 2 ml 2
ẋ +
θ̇ +
θ̇ + ml ẋ cos θ1 θ̇1 + ml ẋ cos θ2 θ̇2 + mgl cos θ1 + mgl cos θ2
2
2 1
2 2
d
(M + 2m)ẍ + ml (cos θ1 θ̇1 + cos θ2 θ̇2 ) = 0
dt
d
mlθ̈1 + (ẋ cos θ1 ) + g sin θ1 = 0
dt
d
θ̈2 + (ẋ cos θ2 ) + g sin θ2 =
(0.18)
0
dt
8. A point particle with a mass m moves along a circle of radius
l in a vertical plane
under the influen ce of the gravity field (mathematical pendulum). Estimate the period of
5
FIG. 4:
the pendulum if
a. E
 2mgl. Here E is the energy of the pendulum.
Solution:
T =
b. 0 < 2mgl

l
g
(0.19)
− E  2mgl.
Solution:
T =

l
2mgl
ln
g 2mgl E
−
(0.20)
9*. A particle is oscillating on a curve y(x) with a frequency ω, which is independent of
the amplitude. Find the curve.
Solution:
Let s be a length of the curve. (See fig.4)
ds
=
dt
−mg sin θ = −mg dy
ds
ds
= −ω s
dt
2
dy
ds
2
s = (ω /2g)s2
g
s = ω 2 sin ω
Thus the curve has maximum hight Y = g/2ω 2 .
ω2 s = g
6
(0.21)
Problem set 2 , 3 0 poi nts. Due. Jan. 22. The problem, mar ked by *, wil l not be
graded.
1. Find the deflection angle of fast particles ( E
 V ) moving in a potential
r
U (r) = V exp(− )
R
2
2
as a function of the impact parameter ρ.
Solution:
∆p =
−
where x = ρ/R
√
∞
∂
2V π −x2
U ( ρ + vt )dt =
xe
∂ρ −∞
v

θ=
| |
√πvρ
2 /R2
e− ρ
(0.22)
RE
2. For what values of the angular momentum M it is possible to have finite orbits in the
potential ? The particle have a mass m.
U=
−α exp(−κr)
(0.23)
.
Solution:
M2
(0.24)
2m2 r 2
To have finite orbits the effective potential should have minimum. Eq. Uef f = 0 can
Uef f = U (r) +

be reduced to the form f (x) = κM 3 /αm, where f = x(x + 1)e−x . This equation has rea l
roots only if κM 2 /αm is less than the max. of f (x). (x > 0). This maximum has a value
(2 + 5 1/2 )exp[ 1/2(1 + 5 1/2 )].
−
l. The masses are coupled
3. Two masses hang from a support by string of equal length
by a spring of spring constant k, and upstretched length l. (See Fig.5.)
a. Write Lagrange equations.
Solution:
L=
1 2 2 1 2 2
ml θ̇1 + ml θ̇2
2
2
− 12 kl (θ − θ ) − mglθ − mglθ
ml θ̈ + kl (θ − θ ) + mglθ = 0
ml θ̈ + kl (θ − θ ) + mglθ=0
2
2
1
2 1
2
2
2
2
7
2
2
1
2
2
1
2
1
2
1
2
(0.25)
FIG. 5:
b. What are frequencies of small oscillations of the system?
Solution:
θ1,2 = a 1,2 eiωt .
We a looking for a solution in the form
2
2
2
2
−ml ω a
−ml ω a
1
2
+ kl 2 (a1
− a ) + mgla = 0
+ kl (a − a ) + mgla = 0
(ml ω − kl − mgl) = k
2
2
2
2
1
1
2
2
2
2
2
ω12 = (2kl 2 + mgl)/ml2 , ω22 = g/l
(0.26)
4. Find the freque ncy of the small oscil lations for particles moving in the following 1D
potential:
U (x) = V cos αx
.
− Fx
Solution:
U =
−V α sin αx − F = 0
sin αx = −F/Vα
U  = −V α cos αx
0
0
2
2
V α2
ω = m
0
F
−
1
8
2
(V α)
(0.27)
5. Find frequencies of small oscillations of the double pendulu m considered in HW1,
problem 4.
Solution:
Linearized Lagrange Eqs:
−2mglφ − dtd [2ml φ̇
−mglφ − dtd [ml φ̇
2
1
2
2
1
+ ml 2 φ̇2 ] = 0
2
+ ml 2 φ̇1 ] = 0
(0.28)
2
ω1,2
√2 g
=√
2±1l
(0.29)
6. Consider the system shown in Fig.3. What are frequencies of small oscillations?
Lianearizing Lagrange Eqs. we have
(M + 2m)ẍ + mθ̈1 + θ̇2 ) = 0
ml2 θ̈1 + mẍ + mgθ 1 = 0
=
0 2
ml2 θ̈2 + mẍ + mgθ
g
ω12 = ,
l
ω22 =
(0.30)
g M + 2m
l
M
7.* Find the inaccessible region of space for a beam of particles with a mass
(0.31)
m flying
parallel to the z-axis with a velocity v and being scattered by a potential U (r) = α/r.
.
Solution:
The solution of the Kepler problem for the orbit is
P
= e cos(φ
r
where p = M 2 /mα; e =
that φ = 0 at r =
∞.

−φ )−1
0
1 + 2EM 2 /mα2 , while φ0 = 1/e is determined by the condition
The inaccesable region is bounded be the envelope of the family of
orbits. To find it we differentiate the equation for the orbit
M2
+1
mαr
− cos φ −
9
M
α

2E
sin φ = 0
m
with respect to M
2M
mr
−

2E
sin φ = 0
m
Eliminating M from the two Eqs. for r, φ, M we get the equation of caustics:
2α
= 1 + cos φ
Er
and the inaccessible region is
r<
2α
E (1 + cos φ)
is bounded by a paraboloid of revolution.
10
Problem set 3 Due Jan.29. 10 points each. Problems marked by * will not be
graded.
1. A two-dimensional oscillator has kinetic and potential energies
1
T = m(ẋ2 + ẏ 2 )
2
U = 12 k(x2 + y 2 ) + αxy
(0.32)
a. Show by a coordinate transformatio n that this oscillator is equivalent to an anisotropic
oscillator with lagrangian
L=
1
m(η̇ 2 + ζ̇ 2 )
2
− A2 η
2
+
B 2
ζ
2
(0.33)
Solution:
x = η 1 cos φ
−η
2
sin φ
y = η sin φ + η cos φ
1
k
L = m(η̇ 2 + ζ̇ 2 ) + (η12 + η22 ) + α[(η12
2
2
1
−
η12 )sin
2
2
φ cos φ + η1 η2 (cos φ
− sin
2
φ)] (0.34)
If cos2 φ = 0 we have
L=L=
1
k+α
k α
m(η̇ 2 + ζ̇ 2 ) + η12
+ η22
2
2
2
−
(0.35)
b. Find eigenfrequencies of the oscillator
Solution:
k
2
ω1,2 =
2. The Lagrangian of a system has a form
L=
α
±m
1
(m1 ẋ2 + m2 ẏ 2 ) + β ẋẏ
2
(0.36)
− 12 (x
2
+ y2)
a Find the eigenfrequencies of the system.
Solution:
Lagrange equations:
m1 ẍ + β ÿ + x = 0
m2 ÿ + β ẍ + y = 0
(0.37)
11
(m1 ω 2
2
2
− 1)(m ω − 1) − β ω
2
4
=0

(m1 + m2 )
(m1 + m2 )2 + 4β 2
ω =
2(β 2 m1 m2 )
±
2
−
b Find the normal coordinates of the system.
.
x = Q 1 cos φ
−Q
2
sin φ
y = Q 1 sin φ + Q2 cos φ
(0.38)
This represent a rotation of the syste m of coordinates. The potent ial energ y does not
change its form under the rotation, while the coefficient of Q̇1 Q̇2 is β cos 2φ+(m2 m1 )sin2 φ.
−
Thus to diagonalise the problem we determine the parameter φ from the condition cot 2φ = 0.
3.
A particle moves in a central potential of the form U (r) =
−k/(n − 1)r
(n 1)
− , where k and
n are constants.
a. Find a criterium of existence of a circular orbit.
Solution:
Uef f =
− (n
dUef f
dr
k
M2
+
1)r (n−1) 2mr2
k
M2
= n
=0
R
mR3
mk
R(n−3) = 2
M
| −
r=R
−
(0.39)
b.Find a criterium of stability of the circular orbit.
Solution:
d2 Uef f
dr 2
|
r=R
=
2
nk
R(n 1)
−
(3
12
M
− mR
3
>0
2
− n) Mm0
>
(0.40)
4. Consider a mechanical system described by a Lagrangian
L(q, q̇ ) (for simplicity we
assume that L does not have explicit time dependence). Consider another Lagrangian,
L̃(q, q̇ ) = L(q, q̇ ) +
d
f (q ) = L(q, q̇ ) + f  (q )q̇
dt
where f (q ) is some function of the generalized coordinate q .
a. Write the Euler-Lagrange equation for the new Lagrangian L̃, and show that it coincides with the Euler-Lagrange equation for the old Lagrangian L.
d d(f  (q )q̇ )
dt
dq̇

(q )q̇
= f  q̇ − f=0q̇
− f dq
(0.41)
b. Explain this coincidence using Hami lton’s action princ iple. (What is the difference
between the action for L̃ and L?).
See a comment in LL, p4 after Eq. 2.8.
(b/r2
5. Let the central force be F(r) =
−
unit vector in the radial direction.
c/r4 )n , where b > 0, c > 0, and n is the
r
r
−
a. Find a radius r of a circular orbit.
Solution:
dUef f
d
M2
= [U +
]=
dr
dr
2mr2
M2
(b/r 2 c/r4 )r
=0
mr3
2
M r
br2 + c = 0
m
M2
1 M2 2
1/2
−
−
−
r = R = 2bm + 2b [( m ) + 4bc]
(0.42)
b. Check stability of the orbit.
Solution:
d2 Uef f
dr2
|
r=R
=
3M
mR4
− R2b + R4c = K/2 =
3
5
2
bR = M 2 R/m + c
K/2 = 3M 3 R/m
− 2bR
2
+ 4c = M 2 R/m + 2c0 >
c. What is the angular frequency of the motion as a function of
the angular momentum?
13
(0.43)
M, b and c, where M is
Solution:
M = R2Ω
(0.44)
d. Find eigenfrequency of small oscillations of r(t).
Solution:
ω 2 = k/m
(0.45)
6.
a. Find the integrals of motion (other than the total energy E ) for the potential U =
1
κr 2
2
= 12 κ(x2 + y 2 ).
Solution:
1
Ex = m ẋ2 + κx2 ,
2
1
Ey = m ẏ 2 + κy 2
2
(0.46)
b. Is Mz is an integral which is independent of E, Ex , Ey ?
Solution: No
7. Find the frequency of the small oscillat ions for particles mov ing in the followin g 1D
potential:
U (x) = V cos αx
− Fx
.
Solution:
U =
−V α sin αx − F = 0
sin αx = −F/Vα
U  = −V α cos αx
Vα
F
ω =
1−(
)
m
Vα
0
0
2
2
2

0
2
(0.47)
8. Find frequencies of small oscillations of the double pendulum considered in HW1,
problem 4.
Solution:
Linearized Lagrange Eqs:
14
−2mglφ − dtd [2ml φ̇
−mglφ − dtd [ml φ̇
2
1
2
2
1
+ ml 2 φ̇2 ] = 0
2
+ ml 2 φ̇1 ] = 0
(0.48)
2
ω1,2
√2 g
=√
2±1l
(0.49)
9. Consider the system shown in Fig.3. What are frequencies of small oscillations?
Solution:
Linearizing Lagrange Eqs. we have
(M + 2m)ẍ + mθ̈1 + θ̇2 ) = 0
ml2 θ̈1 + mẍ + mgθ 1 = 0
ml2 θ̈2 + mẍ + mgθ
=
0 2
g
ω12 = ,
l
ω22 =
(0.50)
g M + 2m
l
M
(0.51)
10*.
A particle of mass m is moving in the presence of a force
F =
−
K
e r/a .
r2
−
Determine
conditions for a such that the circular motion is stable.
Solution:
F =
−dU/dr,
dU ef f /dr = 0, dUef f /dr > 0
K
M2
=0
dUef f /dr = F + d/dr(M 2 /2mr2 ) = 2 e−r/a
r
mr3
kRe −R/a = M 2 /m
2
1 2M
KR2
d2 Uef f /dr2 r=R = 4 (
2KRe−R/a
>0
R
m
a
M2
R
(1
)0>
4
mR
a
−
−
|
−
−
−
(0.52)
11*.
A comet is moving alon g a parabolic trajectory. (See fig. 6). U =
−GmM/r, where
M is the mass of sun, and m is the mass of the parti cle. Find the time the comet spends
15
within the orbit of the Earth E. Assume that the Earth’s orbit is circular. (The value of
M
is given).
Solution:
In the case of parabolic trajectory
E=
If r = r min
mṙ2
M2
+
=
0U
2
2mr2
= p, then ˙r = 0, U + M 2 /2mr2 = 0, and M = (2mαp)1/2
+

  −
 −
−
dr
=
dt
rmax
t= 2
a

12*.
p
rdr
23/2
3
2
√r − p = 3

M2
)
2mr2
2 α
(
m r
dr
M2
)
2mr2
2 α
(
m r
rmin
=
(0.53)
=
ma3
2p
p
(1 + ) 1
α
a
a
rmin = p, rmax = a
ma3
2p
(1
+
α
a) 1
−
p/a
Calculate the rate of precession of the planetary perihelion, to lowest order in
−
GM m
r
+

r2
(0.54)
, if U =
= U 0 + δU .
Solution:
At  = 0 the trajectory of the planet is an ellipse and ∆
∆φ = 2
rmin
rmax
∂
−2 ∂M
2m
Since ṙ = φ̇(dr/dφ) =
rmax


rmax
M
(dr/dφ),
mr2
2
2
− − M /r
2m[E − u] − M /r =

 



rmin
∂
∂M
M dr
2m[E u]
r2
dr
φ = 0.
2
2
2mδUdr
2m[E U0 ] M 2 /r 2
rmin
−
−
(0.55)
and
M 2 dr
1
=
mr2 dφ
m
we have
∆φ =
2m[E
∂ 2m
[
∂M M
∂ [ 2m
∂M M
π
φ
2
2
− U ] − M /r ,
0
(0.56)
r 2 δUdφ] =
0
dφ] =
0
16
− 2πm
M
2
(0.57)
FIG. 6:
Problem set 4 Due Feb. 5. 10 point each. Problems marked by * will not be graded.
1.
Consider a pendulum of a length l and mass m. Find a correction ω (2) to the frequency
of small oscillations ω 0 =
Solution:

g/l, which is quadratic in the amplitude of oscillations φ0 .
φ̈ =
2
0 sin φ
−ω
≈ ω (φ + 16 φ
2
0
3
+ ...)
(0.58)
Since we are interested in a correction to ω 0 proportional to φ 20 we do not need higher powers
on Φ is the expansion of sin. We can rewrite this equation as
ω02
φ̈ =
ω2
and look for a solution in a form
2
0
2
−ω (φ + 16 φ ) + ( ωω − 1)φ̈
2
0
3
(0.59)
φ = φ(1) + φ(2) + .... , and ω = ω0 + ω (1) + ω (2) +
....Here superscripts (1), (2), ... indicate that corresponding terms are proportional to
φ0 , φ20 , .... respectively.
In the first order in φ0 we have
φ̈(1) =
2 (1)
0
−ω φ
φ(1) = φ 0 cos ω0 t
(0.60)
In the second order in φ0 we have
φ̈(2) + ω02 φ(2) =
ω (1)
φ̈1 =
ω0
17
−ω
(1)
ω0 φ0 cos ω0 t
(0.61)
FIG. 7:
The requirement of the absence of resonance terms in rhs of the equation gives as
ω (1) = 0.
This is a natural result: the freq uency should be an even function of the ampl itude of
oscillations. Then φ(1) = 0.
In the third order we have
φ̈(3) + ω02 φ(3) =
cos ω0 t[
(cos3 α =
3
4
φ30 ω02
8
−
2
0
ω (1) (1)
φ̈ =
ω0
ω2
ω0 ω (2)φ0 ] + φ20 0 cos 3ω0 t
24
− ω6 (φ
(1) 3
) +
(0.62)
cos α + 14 cos 3α). The requirement of the absence of resona nce terms in rhs of
the equation gives as
ω (2) =
φ20
ω0
8
(0.63)
.
2.
Consider a system shown in Fig.7. The point of support of the oscil lator oscillates according to the low y = a sin ct. Assume that c

.

g/l.
a. Write down equations of motion averaged over the period of the oscillations 2 π/c.
.
Solution:
18
y = a cos ct
− l cos φ,
X = x + l sin φ
ẏ =
L=
1 2 2 m+M 2
ml φ̇ +
ẋ
2
2
Using the fact that (cos φ sin ct) =
−ac sin ct + l sin φφ̇, Ẋ = ẋ + l cos φφ̇
+ ml ẋφ̇ cos φ − acl sin ct sin φφ̇ + mgl cos φ
(0.64)
− sin φφ̇ sin ct + c cos φ cos ct we get, and neglecting a
total derivative we get
L=
1 2 2 M +m 2
ml φ̇ +
ẋ + ml ẋφ̇ cos φ
2
2
− mlac
2
cos ct cos φ + mgl cos φ
(0.65)
Thus the expression for the oscillating in time generalized force is
f = mlac2 sin φ cos ct
(0.66)
(mgl 2 plays the role of the effective mass.) Averaging over the period 2 π/c (See LL. Eq.
Uef f = mgl[ cos φ +
−
30.8) we get an expression for an effective potential
for a Lagrangian
Lef f =
ml2 2 M + m 2
φ̇ +
ẋ + ml ẋφ̇ cos φ
2
2
−U
(ac)2
4gl
sin2 φ] and
(0.67)
ef f
and Lagrange equations
ml2 φ̈ + ml
d
(ẋ cos φ) =
dt
−dU
ef f /dφ
(m + M )ẍ + ml
+ ml ẋφ̇ sin φ
d
(φ̇ cos φ)=0
dt
(0.68)
b. How many stationary solutions do these equations have?
Solution:
In a stationary state all time derivatives are zero. So we have mgl[sin φ +
mgl sin φ(1 +
(ac)2
4gl
2
sin2 φ] =
cos φ) = 0, which gives us sin φ = 0. Thus there are two solutions φ = 0, π
a. φ = 0 is always stable solution. (( d2 Uef f /d2 φ) φ=0 > 0)
2
(ac)2
4gl
|
b. (d Uef f /d φ) φ=π > 0, and φ = π is stable provided ( ac)2 > 2gl.
|
c. What are frequencies of oscillations near the stationary solutions?
4.
19
FIG. 8:
A ladder of mass m and length 2 l stands against a frictionless wall with its feet on a
frictionless floor. (See Fig.8). The initial angle is α0 . It is let go
.
a. Write Lagrange equations.
Solution:
The kinetic energy is
m 2
1
2
( ẋc + ẏc2 ) + I α̇2 = ml2 α̇2
2
2
3
(0.69)
Here xc = l cos α and yc = l sin α are coordinates of the center of mass, and
I = ml2 /3 is
the moment of inertia relative to the center of mass.
L=
2 2 2
ml α̇
3
− mgl sin α
43 lα̈ = −g cos α
(0.70)
.
b. What will be the angle when the ladder loses contact with the wall?
Solution:
The energy conservation low
2 2 2
ml α̇ = mgl sin α0 mgl sin α
3
ẍc = l α̇2 l α̈ sin α = N/m
−
− −
(0.71)
Here N is the the force acting from the vertical wall to the lader. It the pint when the ladder
loses contact with the wall N = 0, which gives us ¨ α =
20
−α̇
2
cot α, and using the lagrange
FIG. 9:
Eq. we get 4 lα̇2 /3 = g sin α. than from the energy conservation low we get
sin α =
2
sin α0
3
(0.72)
5.
A homogeneous cylinder of radius r and mass m is rolling inside a cylindrical surface of
radius R. (There is no sliding.) (See Fig.9)
a. Write Lagrange equation.
Solution:
Vc = φ̇(R
m
L = (R
2
−
1
r)2 φ̇2 + I (R
2
2 2
− r) φ̇ /r
2
+ mg(R
−
3m
r)cos φ =
(R
4
2 2
− a) φ̇
− r),
ω = V /r
+ mg(R
− r)cos(0.73)
φ
Here I = mR 2 /2 is the moment of inertia of the cylinder.
.
b. Find a frequency of small oscillations about the stable equilibrium position.
. 6.
A uniform thin cylinder rod of length
l and mass m is supported at its ends by two
massless springs with identical spring constants k. (See Fig.10). Find eigenfrequencies of
small oscillations. (Springs can move only vertically.)
21
FIG. 10:
Solution:
Equations for the displacement of the center of mass and for small amplitude rotation
about the center of mass
1
(ÿ1 + ÿ2 ) = k(y1 + y2 )
2
1
I θ̈ =
lk((y1 + y2 )
2
−
−
(0.74)
with I = ml 2 /12.
Symmetric mode: ys = y 1 + y2 , ω 2 = 2k/m.
Asymmetric mode: ya = y 1
−y , ω
2
2
= 6k/m.
.
7* .
A particle of mass m is attached to a rigid support by a spring with force constant
k.
(See Fig.1 1) To the mass of this oscillator an identical oscillator is attached. The whole
system is in the gravitation field of the Earth. Find the normal frequencies and the normal
modes of the system. Consider vertical and horizontal motions separately.
Solution:
Denote the unstretched length of the spring as l. The equilibrium length of the sprin gs
will then be l1 = l + 2mg/κ for the upper spring, and l2 = l + mg/κ for the lower springs.
There are many wa ys to introduce the genera lized coordinates. The result shoul d not
depend on the cho ice of the coord inates. We shall use the devi ations of the mass point
22
FIG. 11:
from the equilibrium position. Let the x axis be directed vertically, and y horizontally. The
coordinates of the first (upper) mass point is ( x1 , y1 ), and of the second (lower) mass point
is (x2 , y2 ), respectively.
The kinetic energy of the system is
T =
m 2
( ẋ + ẏ12 + ẋ22 + ẏ22 )
2 1
The potential energy is
V =
κ
2

2
(l1 + x1 )2 + y12
−l
 
+
κ
2
(l2 + x2
2
−x )
1

2
2
− y ) − l − mg(x
+ (y2
1
1
+ x2 )
Expanding the potential energy in Taylor series over x 1,2 and y 1,2 , keeping only terms up to
the second order inclusively, we find
2
1
V =κ
2 [(l
2
− l) + (l − l) ]
+ [κ(l − l) − mg]x + [κ(l − l ) − mg]x
κ
κ l −l
l −l
+ [x + (x − x ) ] +
y +
(y − y )
2
2
l
l
2
1
1
2
1
2
1
2

2
1
1
1
2
2
2
1
2
2
1
2

The const ant term (firs t line) can be ignored. The terms line ar in x1 and x2 (second line)
vanish due to the equilibrium condition. The last line is all that has to be taken into account.
Clearly, the vertical and the horizontal motions separate.
For the vertical motion ( x1 , x2 ) the characteristic equation is
det

2κ
− mω
κ
2
κ
23
κ
mω 2
−

=0
The two normal frequencies are
2
ω±
=
and the two normal modes are
κ
(3
2m
±
√
5)
 
− ± 
1
√5
1
2
For the horizontal motion, let us introduce the notations
α1 =
l1
−l =
l1
2mg
κl + 2mg
α2 =
l2
−l =
l2
mg
κl + mg
The characteristic equation is then
det
which gives two solutions


(α1 + α2 )κ
2
α2 κ
κ
ω± =
2m
2
− mω

α2 κ
α2 κ

±
− mω
α12
+
α12 + 4α22


α1 + 2α2
The two corresponding normal modes are

− ± 
α2
α1
2
1
2
24
2


4α22

=0
FIG. 12:
Problem set 5 Due Feb. 12. 10 point each. Problems marked by * will not be graded.
1.
A torsion pendulum consists of a vertical wire attached to a mass which may rotate about
the vertical. Consider three torsion pendulums whic h consist of identical wires from which
identical homogeneous solid cubes are hung . Cube A is hung from a corner, cube B from
midway along an edge, and cube C from the midd le of a face (See Fig.12). What are the
ratios of periods of the three pendulums? Briefly explain or derive your answer.
Solution:
I1 = I 2 = I 3 . The ratios equal to one.
2.
A solid ball of radius r rolling with velocity v, collides inelastically with a step of height
h < r. (See Fig.13). Assume no slipping. What is the minimum velocity for which the ball
will ”trip” up over the step?
Solution:
M = mv(r
− h) + Iω = 75 mvr − mvh
M  = I  ω  = (I + mr 2 )ω  =
7 2 
mr ω
5
(0.75)
Here M and M  are angular momenta about the point of impact before and after the
collision. As the center of mass of the ball is momenterely at rest after the colli sion, the
25
FIG. 13:
angular momentum conservation low gives us
ω  = v/r
− 5hv/7r
2
(0.76)
The kinetic energy must be sufficient to provide for the increase in potential energy. I  ω 2 /2 =
mgh.
v=
√
−
r 70gh
7r 5h
3.
(0.77)
Consider a nonlinear pendulum described by an equation
ẍ + ω02 x + αx2 + βx3 = f sin
ω0
t
2
(0.78)
a. What is the lowest order in f in which a resonance takes place?
Solution:
In the absence of nonlinear terms
x(0) = a sin(ω0 t + γ ) +
ω02
f
ω0
sin t
2
ω0 /4
2
(0.79)
−
Nonlinear terms generate harmonics with frequency ω0 . So if a = 0, in lowest in f order
ẍ(2) + ω02 x(2) =
α(
−
(1) 2
−α(x
4f
3ω02
)2
) =
1
4f
−α( 3ω
)
cos ω0 t
−
2
2
0
2
sin2
= F (1
ω 02
t=
2
cos ω t)
−
There is a component of the force at the resonance frequency ω0 .
26
0
(0.80)
b. Estimate the amplitude of the oscillati ons at which the resonance satura tes. Neglect
numerical factors.
As the amplitude a of the harmonics with the frequency ω0 grows, the frequency of the
oscillations changes. The a- dependence of the frequency is quadratic in a (See Eq.28.13 in
LL)
2
∆ω = [ 3β
8ω0
5α
− 12ω
3
0
]a2
(0.81)
As the frequency changes, the system is moving out of the resonance and the amplitude of
oscillations with the frequency ω0 saturates at a value (See Eq.22.4 in LL)
a=
(ω0
 ∆ω) Finally
a
F
ω0 ∆ω
∼ [ −F
3β
8ω0
5α2
12ω03
(0.82)
]1/3
(0.83)
4. Consider the prob lem 1 in $ 35 of LL (Fig . 48) and ass ume that the lowest poin t
of the symmetrical top oscillates vertically according to the low
Z = a sin γt. Write the
Lagrangian of the system.
27
MechFig1MT.pdf
FIG. 14:
MT
1. 50 points.
A particle of mass m1 is constrained to move on a horizontal plane . A second partic le
of mass m, is constrained to a vertical line. The two particles are connected by a massless
string which passes through a hole in the plane. (See Fig. 1.)
a. Write a Lagrange equation.
Solution:
1
L = m (ṙ2 + r 2 θ̇2 ) + 1 mṙ2 mgr
2
2
d
(m1 r2 θ̇ 2 ) = 0
dt
(m1 + m)r̈ m1 r θ̇ 2 + =
mg
0
−
−
(0.84)
b. Find the dependence of the radius r = R(M ) as a function of the angular momentum
of the system M in a stationary state ( ṙ = 0).
Solution:
28
d
d
M2
Uef f = (mgr +
m1 r 2 ) = 0
2
dr
dr
M 2 1/3
r=R= (
)
gm 1 m
(0.85)
c. Find a frequency of oscillations.
Solution:
ω2 =
1
d2 Uef f
m1 + m dr2
|
r=R
=
1
(3g/R)
1 + m1 /m
(0.86)
2. 50 points
A particle of mass m = 1 moves in a potential
1
r2
V =
−
where r is the distance to the center.
a. Write down the Lagrangian of the system, using the polar coordinates
r and φ.
Solution:
L=
1 2
1
(ṙ + r 2 φ̇2 ) + 2
2
r
b. The particle moves from r =
∞ along an orbit with angular momentum
√
that if M < 2 then the particle falls into the center of force (capture), and if
M . Show
M >
√2 it
will escape to infinity.
Solution:
The radial motion is determined by the effective potential, which is equal to the sum of
the potential energy and the “centrifugal potential,”
Veff (r) =
− r1 + M2r
2
2
2
=
When M 2 > 2, the effective potential increases to +
M2 2
2r2
−
∞ as r → 0. A particle moving from
r = ∞ will bounce back. When M < 2 the effective potential is negative and goes to −∞
as r → 0. A particle moving from r = ∞ will fall to the center r = 0.
c. Suppose the velocity of the particle at t = −∞ is v ∞ . Write the condition for capture
2
in term of the impact parameter b.
29
Solution:
The angular momentum is conserved, and is equal to the angular momentum at
which is bv∞ . Therefore capture occurs when bv∞
√
< 2
t=
−∞,
d. Express ˙r and φ̇ in terms of E , M , and r.
Solution:
The angular momentum and energy are
l = r 2 φ̇,
therefore
−
ṙ 2 M 2 2
+
2
2r2
−
2
M
r2
in ṙ to correspond to a particle moving toward the center).
ṙ =
(I choose the sign

E=
−
2E
− M r− 2,
φ̇ =
2
M 2 = 2.
e. Find a trajectory of the particle in the case
dφ
dr =
or
M
φ = φ0 +
or
1
− √2E r
2
√M2E 1r
√2E
1
M φ φ0
The trajectory is a spiral that makes an infinite number of rotations before approaching
r=
r = 0 (when φ
−
→ ∞, r → 0).
2.
A particle of mass m is moving in a central potential U = Ar n /n with n non-zero integer
and A > 0.
a. Find a radius of a circular orbit as a function of
A, m, n, M .
Solution:
A circular motion corresponds to a minimum of the effective potential
d2 Uef f
dr2
|
r=R
=
d2
M2
[Arn /n +
] r=R
2
2mr2
dr
1
R = (M 2 /mA) ( +2)
|
n
(0.87)
b. What is the angular frequency of this motion?
mR2 Ω = M
30
(0.88)
c. What is the frequency of small oscillations of r(t) near the circular orbit?
d2 Uef f
r=R
dr 2
2
2
M M −4/(n+2)
K = (n + 2)
[
]
m mA
ω 2 = k/m
K/2 =
|
(0.89)
d. At what value of n = non-circular orbits are closed?
Solution: n =
−1, 2.
3.
A point of support of a pendulum of a length l and a mass m is moving horizontally with
an acceleration a.
Write a lagrange equation.
Solution:
x = l sin φ + at2 /2,
ẋ = l cos φφ̇ + at
y = l cos φ. ẏ = l sin φ
ml2 2
L=
φ̇ + mlat cos φφ̇ mgl(1 cos φ)
2
mlat sin φφ̇ + mlat cos φ mlat sin φφ̇ + sin φ
−
ml2 φ̈
−
−
31
−
−
(0.90)
Problem set 5. Due Feb.26. (70 points). Problems marked by * will not be graded.
1.
Obtain the Hamilton equations by calculating
dqi
= [H ; qi ];
dt
dpi = [H ; pi ]
dt
(0.91)
where [..,..] are the Poiss on bracket. The form of the Hamilton functi on H should not b e
specified.
Solution:
Since q and p are independent variables ∂q i /∂p j = 0 and ∂p i /∂q j = 0 and ∂q i /∂q j = δ ij
one gets
[H ; qi ] =
[H ; pi ] =
∂H
∂p i
(0.92)
− ∂H
∂q
(0.93)
i
2.
Find the canonical transformation generating by the generating function
F = (r P) + ( a [r
· × P])
·
Solution:
p=
∂F
= P + [a r];
r
R = r + [ a r]
×
×
The transformation is a rotation of the system of coordinates over the angle δ a
32
(0.94)
3.
Is the following transformation is canonical one?
Q = ln(1 + q 1/2 cos p)
P = 2(1 + q 1/2 cos p)q 1/2 sin p
(0.95)
Solution:
[Q, Q] = 0, [P, P ] = 0, [Q, P ] =
∂Q ∂P
∂P ∂Q
=
=
∂q ∂p
∂q ∂p
−
q −1/2 cos p
1+
q 1/2
[ q sin2 p + (1 + q 1/2 cos p)q 1/2 cos p] +
cos p
q 1/2 sin2 p
+
[cos p + (1 + q 1/2 cos p)q −1/2 ]=1
1 + q 1/2 cos p
−
(0.96)
So the transformation is canonical.
4.
Consider a top with principle moments of inertia I3 > I2 > I1 . Initially it rotate s with
τ 2 = τ 0 co s Ωt is applied about axis 2.
frequency ω1 bout the axis 1. A a small torque
Use Euler’s equations and calculate ω1 , ω2 , ω3 in the first order in τ 0 .
Solution:
dω1
+ (I3
dt
dω2
I2
+ (I1
dt
dω3
I
+ (I2
dt
I1
− I )ω ω
− I )ω ω
− I )ω ω
2
2 3
=0
3
1 3
=τ
1
1 2
=0
33
Since we expect that ω2 , ω3
ω
1
we can leanearise the eqs.
I1
dω2
+ (I3
dt
dω3
I3
+ (I2
dt
I2
dω1
=0
dt
− I )ω ω
2
1 3
=τ
I1 )ω1 ω2 = 0
−
(0)
According to the first eq. ω1 = const. Solving the eqs. in a form ω1,2 (t) = ω 1,2 eiΩt we get
iΩI3
(0)
ω2 = τ 0
ω1 (I3
2
(0.97)
− I )(I − I ) − Ω I I
1
2
3
2 3
(To solution of the uniform eq. should be added to this eq.)
5.
The surface of a sphere is vibrating SLOWLY in such a way that the principal moments
of inertia are harmonic functions of time.
2mr2
(1 +  cos ωt)
5
2
2mr

=
(1
cos ωt)
5
2
Izz =
Ixx = I yy
where 
 1. Find Ω
z,y,z
−
(0.98)
(0.99)
in the first order in .
Solution:
The orbital moment is conserved. However, the principe moments of inertia are defined
in the body frame. Using Euler’s equations we get:
dIzz Ωz
=0
dt
dIxx Ωx 3
+ I0 Ωy Ωz cos ωt = 0
dt
2
dIyy Ωy 3
I0 Ωx Ωz cos=0ωt
dt
2
−
(0.100)
where I0 = 2mr2 /5. Thus
Ωz0
Ωz = 1 + cosωt
34
(0.101)
FIG. 15:
Since 
 1, in our approximation Ω
z
= const. Then, in the first in  order, one can neglect
dIxx,yy /dt, and the last two eqs. are
x
y z
I0 dΩ
dt + 32 I Ω Ω cos ωt = 0
dΩy 3
I0
I Ωx Ωz cos=0ωt
dt
2
−
If Ωz
(0.102)
 ω one can regard ω in these eqs as a constant. Then a solution of these equations
is Ωx = Ω0 sin ω0 t, Ωx = Ω0 cos ω0 t, where ω0 = 32 Ωz .
6.
Consider a pendulum shown in Fig.15 Assume that the length of the pendulum
l0 + l1 cos ωt changes in time, and l 1
l0 .
l =

a. Find the interval of frequencies where the resonance takes place.
Solution:
The Lagrangian of the system is
L=
m 2
l (t)(θ̇)2 + mgl(t)cos θ
2
(0.103)
(The term proportional to l˙2 can be dropped from the Lagrangian. A Lagrang e equation
has a form:
l˙
θ̈ + 2 θ̇ =
l
2
−Ω (t)sin θ
35
(0.104)
The second term in the Eq. 0.89 makes this problem slightly different from that in LL
Ch.30. (If g(t) oscillates in time there is no second term in the Eq. and the problem would
be exactly the same as in LL. In both cases the parametric resonances exists.)
Expanding the frequency of oscillations in powers of l1 (keeping only lowest in h terms),
linearizing the Eq. and introducing ω = 2Ω0 +  we get
θ̈
Ω20 =
− 8Ω h sin(2Ω
0
0
g
; Ω2 (t) = Ω20 [1
l0
+ )tθ̇ =
2
−Ω (t)θ
− h cos(2Ω
0
(0.105)
+ )t],
h = l 1 /2l0 ,
Let us look for a solution in the form
(0.106)
θ = a sin(Ω0 + 12 )t + b cos(Ω0 + 12 )t, At the ends of
the interval of  where the resonance takes place a and b are time independent, and and one
can drop their time derivatives. (In other words, one can differe ntiate over time only sin
2
and cos. The terms proportional to Ω 0 cancel. Using the fact that
1
1
1
1
cos(Ω0 + )t cos(2Ω0 + )t = [cos 3(Ω0 + )t + cos(Ω0 + )t]
2
2
2
2
1
1
1
1
1
sin(Ω0 + )t cos(2Ω0 + )t = [sin 3(Ω0 + )t + sin[ (Ω0 + )]t]
2
2
2
2
2
1
1
1
1
1
sin(Ω0 + )t sin(2Ω0 + )t = [cos(Ω0 + )t cos 3(Ω0 + )t]
2
2
2
2
2
−
−
and dropping the harmonics oscillating with the frequency close to 3Ω
0
(0.107)
we get
9
1
b( + hω0 )sin( ω0 + )t +
2
2
9
1
a(  + 2 hω0 )(cos(ω0 + 2)t
=0
−
This Eq. , should valid at arbitrary t, and we get for the endpoints  =
(0.108)
±
9
hΩ0 ),
2
and for
the interval:
− 92 hΩ
b. Assume that ω
0
9
<  < hΩ0
2
(0.109)
g/l and write a Lagrange equation averaged over the period 2 π/ω


36
Solution:
The Lagran ge equati on is different from that consi dered in Ch. 30 LL. by the second
term. However, the form of the p otential energy is the same. (After the averaging the second
term does not contribute to the Lagrange Eq. ) Thus, in notations of LL
Uef f =
−
f = mgl1 sin θ2 cos ωt
(mgl 1 )
mgl 0 cos θ +
sin2 θ
4
dUef f
ml02 θ̈ =
dθ
−
c. Now consider the case ω


(0.110)
g/l Write an expression for the adiabatic invariant. How
the energy of the pendulum depends on time?
7.
Calculate the following Poisson brackets:
a.[Mi , Mj ]; b.[Mi ; pj ]; c.[Mi , xj ], d.[(a M), (b M)].
·
·
Solution:
[Mi , Mj ] =
−ijkM
k;
[Mi ; pj ] =
−ijkp
k,
[Mi ; pj ] =
−ijkp
k;
[(a M)(b M)] =
·
·
−([a × b]·M)
(0.111)
8.
Consider an infinite system of of pendulums in a gravitational field, shown in Fig.16
(Only a spring between the first and the second pendulums is shown.)
a.
Find a Lagrangian of the system.
Solution:
L=

n
b.
[
ml2 2 k
φ̇ + (φn+1
2 n 2
−φ )
n
2
+ mgl cos φn ]
Write the Lagrangian in a continuous limit where φn+1
|
(0.112)
− φ |  1. For simplicity use the
n
system of units where l,m , g, a, k = 1. Here a and k are spacing between springs and the
spring constant.
37
FIG. 16:
Solution:
L=
1
2
c.
dx[(φ̇(x, t))2 + (∂x φ)2 + 2 cos φ]
(0.113)

Write Lagrange equations both in both discrete and continues limits.
solution:
ml2 φ̈n + k(2φn
− φ − φ − ) + mgl sin φ = 0
φ̈ − ∇ φ − sin=φ0
n+1
n 1
n
2
(0.114)
d.
Write the expressions for generalized momenta and for the Hamiltonians in the continues
limit.
Solution:
∂L
= φ̇
∂ φ̇
p2 + ( φ)2 + 2(1 cos φ)
dx
2
p=
H (p, φ) =

dxpφ̇
−L=

∇
−
(0.115)
8*.
Find a soliton (solitary wave) solutions of the Lagrang e eq. in the previous problem in
the continues limit. Use the boundary condi tions φ( ) = 0, φ(
) = 2π This solution
∞
38
−∞
should have a form Φ( y) = φ(x
− ut) which does not change its shape in time and move with
a velocity u. Substitute this form int o the lagrange eq., and reduce it to the probl em of a
single particle moving in an effective potential U (φ). So φ plays a role of coordinate, and
y plays a role of time. Estimate (up to a numerical factor) a relation between the soliton’s
width and it’s velocity u.
Solution:
(1
2
− u )( dd Φy )
2
2
2
+ 2 cos φ = 2E
(0.116)
the boundary conditions gives E = 1. One can introduce a variable z = y/(1
eliminates u from the eq. So the width of the soliton scales as (1
to an eq. describing a particle with a mass (1
the role of time. If (1
2
2
2 1/2
−u )
2 1/2
−u )
which
. The eq. is identical
− u ) moving in a well potential . Here
y plays
− u ) > 0 the ”particle’s” motion is finite, and it oscillates.
y = y 0 (Φ(y0 ) = 0 is
Solution of this eq. with a center at
y
1
−
y0
u2 =
Φ/2

π/2
dθ
sin θ = ln tan(Φ/4)
(0.117)
BTW, Sin-Gordon eq. has infini te number of conservation lows. It is totaly integrable
system, and all exact solutions can be obtained. The famou s method of obtain ing these
solutions is a ”inverse scattering” method. please read about it.
39
Problem set 6 Due March. 5 Problems marked by * will not be graded.
1 A fast particle ( v


A/k 2 m) of mass m is moving in the potential field
U (r) = A(x2
2
− y )sin kz
(0.118)
at a small angle to the z-axis.
Describe the motion of the particle in xy-plane.
Solution:
In the first approximation z = vt. In the xy-plane there is a fast oscillating force acting
on the particle
fx = 2Ax sin kvt; fy =
−2Ay sin kvt
the corresponding effective potential (See Ch. 30 Landau) has a form
mΩ2 2
(x + y 2 )
2
Uef f (x, y) =
(0.119)
where Ω = A/mkv.
Thus the particle performs harmonic oscillations in the xy plane with the frequency Ω.
Note that Ω
 kv and we can apply results of Ch. 30 to this problem.
2
Consider a configuration of the magnetic field shown in Fig. 19. In the center there is a
source of particles which emits them isotropically with momentum p. Estimate a fraction of
particles which escape the magnetic trap.
Solution:
The adiabatic invariant is
I=
1
2π

Pt dr =
1
2π

p t dr +
e
2πc
Here pt is the component of momentum perpendicular to

Ad r
(0.120)
H. The modulus pt = const,
| |
and r = cp t /eH . Thus
cp2t
= const
(0.121)
2eH
Energy is also conserved ( p2t + p2 )/2m = E . If at x =
p = 0, then the particle do not
I = rpt
− 2ce Hr
2
=
±∞
leave the region. At this point
I=
cE
eH
40
(0.122)
Inside the region
I =I =
where sin α = p t /p . Thus sin α = [(H
cE sin2 α
e(H δH )
(0.123)
−
− δH )/H ]
1/2
.
Thus the fraction of particles which escapes on infinity is
2π(1
−4πcos α)
(0.124)
Solve the Hamiltonian-Jacoby equation for S (q,t,α ) in the case of a single particle moving
under the Hamiltonian H = p 2 /2m.
∂S 1 ∂S 2
+ ( ) =0
2 ∂q
∂t
S = αq
(0.125)
2
− 12 α t
(0.126)
3
Find a solution solution u(x
− vt) of of Koteveg de Vries equation
∂u ∂ 3 u
∂u
+ 3 + u=0
∂t
∂x
∂x
(0.127)
Express u(x, t) in terms of an integral, and estimate the relation between the width and the
velocity of the soliton.
Solution:
u = u(x
Let us look at a solution in the form
y =x
− ct we get
d d2 u
[
dy d2 y
− ct).
2
− cu + u /2=0
∂t u =
−c∂ u, and introducing
x
(0.128)
Integrating on time over y and putting the integration constant to zero we get an equation
describing a non-linear oscillator
2
W =
dd2u
y = dU
dx
cu2 /2 + u3 /6
−
−
41
(0.129)
FIG. 17:
The ”potential” W has a minimum at y = 2c width 3 c. Thus the amplit ude of the soliton
√
is of order c, and its width is of order 1 / c. The exact solution for the soliton is
u = ch2 (x 3cct)/∆)
(0.130)
−
∆2 = 2/c
4.
Consider a symmetrical top I1 = I 2 = I 3 whose lower point is fixed (See Fig. 17) in the

presence of gravity. The mass of the top is m, and the distance from the lower point to the
center of mass is l.
a.
Write an expression for its rotational kinetic energy about the center of mass using
Eulerian co-ordinates.
Solution:
K=

Ii Ω i
i
I1
( θ̇ 2 + φ̇2 sin2 θ) +
I3
2
2
b.
42
( ψ̇ + ψ̇ cos θ)2
(0.131)
Write the expression for a Lagrangian. (Do not forget that the top is rotating about the
fixed point).

L=
(Ii Ωi
i
I1 + ml 2 2
I3
(θ̇ + φ̇2 sin2 θ) + ( ψ̇ + ψ̇ cos θ)2 = mgl cos θ
2
2
5.
Write the adiabatic invariant I for the pendulu m shown in Fig. 15 assuming that l(t)
changes slowly in time. (Assume that the ampli tude of oscill ations is small.) Given the
value of I , how the amplitude of the oscillations
φ0 changes in time ? How the frequency
changes in time?
Solution:
ω2 =
g/l(t).

L=
I=

pdq/2π = Ē/ω =
1 2 2
l φ̇ + mgl cos φ
2m
l2 (t)φ20 ω
= l 3/2 g 1/2 φ20 /2 = const
2
φ0 = [2Il−3/2 (t)g −1/2 ]1/2
(0.132)
6.
The Hamiltonian for the system has the form:
1 1
H = ( 2 + p2 q 4 )
2 q
(0.133)
a. Find equation of motion.
b. Find a canonical transf ormation that reduces the Hamil tonian to the form of harmonic
oscillator.
Solution:
43
P =
2
−pq ,
Q=
1
q
1
H = (P 2 + Q2 )
2
1
(Q, P ) =
( q2 )
p,q
2
q
(0.134)
− −
So the transformation is canonical.
7*
xy) plane. A constant exte rnal magnetic field B is
Consider a particle moving on a (
applied in the dire ction perpendicular to the plane (z direction). Chose Ax =
−By/2;
Ay = Bx/2, show that the Lagrangian is invariant under infinitesimal rotation:
δx = αy,
δy =
−αx
(0.135)
where α is a small number. What is the conserved quantity that can be extracted from this
invariance?
Solution:
The Lagrangian is
L=
m 2
e
m
eB
( ẋ + ẏ 2 ) + A
̇x = ( ẋ2 + ẏ 2 ) +
(xẏ
2
c
2
2c
·
Under an infinitesimal rotation δx = αy, δy =
−αx the change of the Lagrangian is
eB
(δx ẏ + xδ ẏ δy ẋ yδ ẋ)
2c
ẏ ẋ) + eB α(y ẏ xẋ + xẋ y ẏ) = 0
2c
δL = m(ẋδ ẋ + ẏδ ẏ) +
= αm(ẋẏ
−
− yẋ)
−
−
−
−
The corresponding conserved quantity is (the z -component of) the angular momentum
L = xp y
− yp
x
= m(xẏ
− yẋ) + eB
(x
2c
2
+ y 2)
8*
Consider an infinitely thin rectangular of mass m with sides a and b which rotates about
the diagonal. (See Fig.18) . Find a torque applied to the system.
Solution:
44
FIG. 18:
FIG. 19:
Euler’s equations are:
dωx
+ (Iz
dt
dωy
Iy
+ (Ix
dt
dωz
Iz
+ (Iy
dt
Ix
Ix =
1
ma2 ; Iy
12
=
1
mb2 , ωx
12
− I )ω ω
− I )ω ω
− I )ω ω
y
y
z
= τx
z
z
x
= τy
x
y
x
= τz
(0.136)
= ω √a2b+b2 , ωy = ω √a2a+b2 .
All time derivatives of the angular velocities are zero, ωz = 0. Therefore τx,y = 0.
τz = ω x ωy /(Iy
Ix )
−
9. *.
45
(0.137)
Also see LL Ch. 5 Problems 1-3. Ch.9 Problems 1,2, Ch.14. Problem1, Ch. 18 Problam
1, Ch 21; Problems 1,2, Ch.27 Problam 1, Ch 51 Problem 1.
46