PHY4222/Spring 08: CLASSICAL MECHANICS II
HOMEWORK ASSIGNMENT #5: SOLUTIONS
due by 12:50 p.m. Wed 02/13
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly. Every (algebraic) final result
must be supplemented by a check of units. Without such a check, no more
than 75% of the credit will be given even for an otherwise correct solution.
1. A two-dimensional (“flat”) beam of particles is scattered by a hard disk of radius R in a planeR(see Fig.) Find the
dσ
(1 − cos θ) ,
differential, total, and transport cross-sections (the transport cross-section is defined as σtr = dΩ dΩ
where θ is the scattering angle.) Comment on the differences between scattering by a disk and by a sphere.
Solution
The flux of the incoming beam, I, is the number of particles crossing the unit length per unit time. The number
of particles scattered into an angular interval dθ per unit time is dN. Then the differential cross-section (which
has units of length) is
dσ = dN/I.
For a given relation between the scattering angle and impact parameter, b (θ) , particles with impact parameters
in the interval (b, b + db) will be scattered into an angular interval (θ, θ + dθ) . Hence,
db dσ = 2db = 2 dθ
dθ
and
db dσ
= 2 .
dθ
dθ
(Factor of two is from the symmetry with respect to reflection about the line parallel to the beam and going
through the scattering center. We define b to be positive). The relation between b and θ is the same as for a
sphere
θ
2
dσ
θ
= R sin
dθ
2
b = R cos
Total cross-section
Z
σ=
π
dθ
0
dσ
= 2R
dθ
Transport cross-section
Z
σtr =
π
dθ
0
dσ
(1 − cos θ) = R
dθ
Z
π
dθ sin
0
θ
8
(1 − cos θ) = R
2
3
The differential cross-section for a hard sphere does not depend on the angle. In contract, the differential
cross-section increases with θ and reaches a maximum value at θ = π. The total cross-sections are equal to
the geometric one: the equatorial circle for a sphere and diameter for a disk. The transport cross-section for a
sphere is equal to the total one. The transport cross-section for a disk is larger than the total one.
2. Determine the effective cross-section for a particle to fall on the center of a central force F (r) = −(k/3)r−4
(k > 0).
2
Solution
The potential energy corresponding to this force is U (r) = −k/9r3 . effective potential energy is
Uef f (r) = −k/9r3 + M 2 /2mr2 = −k/9r3 + Eb2 /r2 .
For r → 0, Uef f is large and negative. For r → ∞, Uef f is small and positive. In between, Uef f goes through a
maximum determined from
b2
k
k
dUef f
.
= 0 → 4 − 2E 3 = 0 → rm =
dr
3rm
rm
6Eb2
Indermediate unit check
rm = b
k
.
6Eb3
The units of k/b3 is energy, energy over energy is dimensionless. The units of rm is the same as of b (m) .
OK.
The maximum value of Uef f
3
2
6Eb2
63 E 3 b6
E 3 b6
k 6Eb2
+ Eb2
=−
+ 62 2
2
9
k
k
9 k
k
3 6
3 6
E b
2 E b
= 12 2
= 36 1 −
3
k2
k
max
Uef
f = −
Intermediate unit check
max
Uef
f = 12E
Since k/Eb3
2
E 2 b6
k2
max
is dimensionless, the units of Uef
f are the same as of E (energy) . OK.
max
A particle coming from infinity can reach the center (r = 0) only if it is energy is larger than Uef
f
k2
E 2 b6
1/3
E ≥ 12E 2 → b6 ≤ 12 2 → b2 ≤ b2max = (12)
k
E
k
E
2/3
The cross-section is
σc =
πb2max
= π (12)
2/3
Final units check: [k] /[E] = m3 , [σc ] = ([k] /[E])
1/3
k
E
2/3
.
= m2 . OK.
3. T&M 9.52
Solution
The number of events scattering incident particles by angle θ in the C.O.M. system
dN ∝ σ (θ) sin θdθ = σ (θ) d (− cos θ)
From Eq.(9.87a) in the book,
T2
2m1 m2
=
2 (1 − cos θ) .
T1
(m1 + m2 )
Solving for cos θ,
cos θ =
Tm − 2T2
,
Tm
.
3
where
Tm =
4m1 m2
2 T1
(m1 + m2 )
is the maximum kinetic energy of the recoil particle.
d (− cos θ) =
2dT2
Tm
Hence the distribution of recoil particles is given by
dT2
−1 Tm − 2T2
dN2 ∝ σ (θ) d (− cos θ) = 2σ cos
Tm
Tm
4. Bonus: Find the differential scattering cross-section in a potential U = α/r2 (α > 0).
Solution
For any central field, the energy conservation gives
r
m
dr
q
dt =
2 E − M2 − U
2mr 2
The angular momentum is M = r2 φ̇ or dt = (r2 /M )dφ or
r
M/r2
m
q
dφ =
dr
2 E − M2 − U
2mr 2
Integrating this relation from the distance of maximum approach to infinity, we obtain
r Z ∞
M/r2
m
q
φ0 =
dr,
2 rmin E − M 2 − U
2
2mr
where φ0 is the angle to the asymptote of the trajectory. Using M = Eb2 /r2 , we find
Z ∞
b/r2 dr
q
φ0 =
.
2
U
rmin
1 − rb2 − E
rmin is the root of the determinant. For U = α/r2 , we introduce a new variable x = b/r and the integral reduces
to
Z xmin
dx
p
φ0 =
,
2
1 − x [1 + α/b2 E]
0
p
where xmin is found from 1 = x2min 1 + α/b2 E → xmin = 1/ 1 + α/b2 E. Re-writing the integral as
Z xmin
dx
p
φ0 =
1
−
x2 /x2min
0
and introducing yet another variable y = x/xmin , we obtain
Z 1
π
1
dy
π
p
φ0 = xmin
= xmin = p
2
2
2
1
+
α/b2 E
1−y
0
The scattering angle
θ = π − 2φ0 = π 1 − p
1
1 + k/b2 E
!
.
4
Solving this for b2 = b2 (θ) gives
b2 =
α/E
1 − (1 − θ/π)
−2
Differentiating the last result gives
db 2α
1 − θ/π
2b =
dθ
πE (θ/π)2 (2 − θ/π)2
The differential cross-section is
σ (θ) =
b
sin θ
db α
1 − θ/π
=
dθ πE sin θ (θ/π)2 (2 − θ/π)2