1. No category

PHY 3003 SPRING 2015: SOLUTIONS, PROBLEM SET 8
As previously announced, this problem set does not count double. There was a misprint
on the problem set.
(1) A particle of mass m moves in the helical potential
2πz
U (r, φ, z) = U0 r cos φ −
b
The parameter b is the pitch of the helix.
(a) Write down the Lagrangian using as coordinates r, φ and z.
The relation between cylindrical and Cartesian coordinates is
x = rcosφ
x˙ = rcosφ
˙
− rsinφ φ˙
y = rsinφ
z = z
(1)
y˙ = rsinφ
˙
+ rcosφ φ˙
z˙ = z˙
implying
m 2
r˙ + r2 φ˙ 2 + z˙ 2
2
so
m 2
2πz
2 ˙2
2
L=
r˙ + r φ + z˙ − U0 r cos φ −
2
b
(b) Write down the equations of motion.
KE =
2πz
2
˙
m¨
r = mrφ − U0 cos φ −
b
2πz
d
r2 φ˙
= U0 r sin φ −
m
dt
b
2πU0 r
2πz
m¨
z = −
sin φ −
b
b
(c) Show that there is a continuous one-parameter family of transformations that
leaves U invariant and find the associated conserved quantity. Are there any other
conserved quantities?
The transformation
2πζ
z → z + ζ;
φ→φ+
b
leaves the potential invariant (the change in z and the change in φ cancel each
other).
1
2
PHY 3003 SPRING 2015: SOLUTIONS, PROBLEM SET 8
The corresponding conserved quantity is
δL δz δL δφ
2πr2 ˙
+
= m z˙ +
φ
δ z˙ δζ
b
δ φ˙ δζ
Inspection of the equations of motion shows that the time derivative of this combination is indeed zero.
There is one other conserved quantity, the energy (since the Lagrangian is timeindependent). This is
2πz
m 2
r˙ + r2 φ˙ 2 + z˙ 2 + U0 r cos φ −
E=
2
b
(2) Two particles of equal mass move in two dimensions and are connected by a spring
of spring constant k and natural length L so the potential energy is
U (~r1 , ~r2 ) =
k
(|~r1 − ~r2 | − L)2
2
(a) Please write the Lagrangian using as coordinates the center of mass position
~
~ , the magnitude of the relative distance, r and the angle φ made
R and velocity V
by ~r1 − ~r2 relative to some arbitrarily chosen x axis.
Using
1
(~r1 + ~r2 )
2
~r = ~r1 − ~r2
~ =
R
we have
~ + 1 ~r
r~1 = R
2
~ − 1 ~r
r~2 = R
2
so the kinetic energy K = m
r˙12 + ~r˙22 is
2 ~
~˙ 2 + m ~r˙ 2
K = mR
4
Now using x = rcosφ and y = rsinφ we finally obtain
2
2
~˙ 2 + m r˙ 2 + r2 φ˙ 2 − kr
L = mR
4
2
(b) Please give the Euler-Lagrange equations for the center of mass and relative
coordinates
PHY 3003 SPRING 2015: SOLUTIONS, PROBLEM SET 8
3
~¨ = 0
2mR
m
m ˙2
r¨ =
rφ − kr
2
2
m d 2 ˙
r φ
= 0
2 dt
(c) Show that the relative angular momentum is conserved and from this obtain
an equation for the time dependence of the magnitude of the interparticle distance.
The conservation of relative angular momentum follows from the lack of dependence of U on φ and, similarly, from the last of the three equations above.
Defining
m 2 ˙
Lrel =
r φ
2
we have
L2
m
r¨ = 2 rel3 − kr
2
mr
(3) 7.34
(a) If the spring stretches uniformly then a point at distance x0 from the attachment point moves with velocity proportional to x0 . If the end point (distance x0 = x)
moves with velocity x˙ then the point at distance x0 moves with velocity x0 x/x.
˙
The
mass density per unit length of an infinitesimal interval at distance x0 is ρ = M/x
so the kinetic energy is
0 2
Z x
x x˙
M x˙ 2
01 M
dx
K=
=
2 x
x
6
0
so that
M x˙ 2 mx˙ 2 k
+
− (x − L)2
6
2
2
note: here I use a different convention than the text: x is the total length
of the spring, not the length measured relative to the equilibrium length
L.
(b) We see that this is exactly the usual harmonic oscillator Lagrangian, just
with m replaced by m + M/3, so the frequency is shifted accordingly.
L=
(4) 7.36
The usual Cartesian coordinates are given by
z = −(l + r)cosθ;
x = (l + r)sinθ;
z˙ = (l + r)sinθθ˙ − cosθr˙
x˙ = (l + r)cosθθ˙ + sinθr˙
so
L=
m
k
(l + r)2 θ˙2 + r˙ 2 + mg(l + r)cosθ − r2
2
2
4
PHY 3003 SPRING 2015: SOLUTIONS, PROBLEM SET 8
Thus we have
m¨
r = m (l + r) θ˙2 + mgcosθ − k(l + r)
d (l + r)2 θ˙
= −mg(l + r)sinθ
m
dt
The right hand side of the first equation is the component of force along the direction
of the spring including the centripetal force from the angular motion, while the right
hand side of the second is the component of force perpendicular to the direction of
the spring.
Expanding to leading order in r and θ and their derivatives we have
m¨
r = mg − kl − kr
2¨
ml θ = −mglθ
Thus we identify l with the equilibrium length of the spring hanging vertically and
subject to the force of gravity (kl = mg) and then have two independent equations,
one for changes of the spring length r and one for the angle.
(5) 7.40
The usual Cartesian coordinates are given by
z˙ = Rsinθθ˙
z = −Rcosθ;
x˙ = Rcosθcosφθ˙ − Rsinθsinφφ˙
y˙ = Rcosθsinφθ˙ + Rsinθcosφφ˙
x = Rsinθcosφ;
y = Rsinθsinφ;
so
L=
mR2 ˙2
θ + sin2 θφ˙ 2 + mgRcosθ
2
and
mR2 θ¨ = mR2 sinθcosθφ˙ 2 − mgRsinθ
d
mR2
sin2 θφ˙
= 0
dt
The φ equation implies that the component of the angular momentum about the
z axis,
Lz = mR2 sin2 θφ˙
is conserved.
If φ˙ = 0 then the θ equation becomes
mR2 θ¨ = −mgRsinθ
which is just the usual equation for a pendulum moving in a vertical plane.
Using φ˙ = Lz /(mR2 sin2 θ) in the θ equation gives
mR2 θ¨ =
L2z cosθ
− mgRsinθ
mR2 sin3 θ
PHY 3003 SPRING 2015: SOLUTIONS, PROBLEM SET 8
5
Clearly if
L2z cosθ
= mgRsinθ
mR2 sin3 θ
then there is no force on θ so a solution θ¨ = θ˙ = 0 is possible. In this case the
pendulum orbits the z axis at a constant polar angle, so traces out the surface of a
cone of opening angle θ.
Expanding the right hand side about the value θ0 that makes it vanish gives
3L2z cos2 θ0
L2z sinθ0
2
−
− mgRcosθ0 mR ¨ = −
mR2 sin3 θ0 mR2 sin4 θ0
Because each of the terms inside the bracket is negative the motion is stable oscillation with frequency determined by the value of the term in brackets. While the
θ is oscillating, the azimuthal angle φ is precessing at a frequency dictated by Lz .
This leads to orbits in θ, φ space that are not closed, in general.
(6) 7.46
(a) is just a restatement of the conditions of the problem, along with the observation that rotations about the z axis change φ. If L is invariant then dL/d = 0
but since L depends on only via the φa the required result follows immediately by
the chain rule.
(b) The total angular momentum about the z axis is
X
X
~ a × P~a =
Ltot =
R
ma Ra2 sin2 θa φ˙a
a
a
But since (referring to problem 7.40)
X ma 2
L=
R˙ a2 + Ra2 θ˙a + Ra2 sin2 θa φ˙ 2 − U ({Ra , θa , φa })
2
a
we have
δU
d 2
Ra sinθa2 φ˙a =
dt
δφa
Summing up all these equations over a gives the required result.
ma