Assignments for Physics 5103
-
Reading in Classical Mechanics with a BANG! and Lectures
Assignment 2 Read Unit 1 Chapters 1 thru 6.
Ex. 1.5.5 and 1.6.5 are due Thursday Sept. 1, 2016
Reflections on reflections ( A lesson in group representation theory)
Exercise 1.5.5 This exercise is intended to introduce matrix reflection and rotation operators and the groups they
form. It involves the circular (V1, V2) plots (“l’Etrangian space”) introduced in Fig. 5.2b by eqs. (5.7)-(5.13) and the
mirror diagrams in Fig. 5.3. All elastic (m1 - m2) collisions map an initial VIN=(V1, V2) vector into a VFIN of the same
length so all collisions map to unit vectors. All elastic m1:m2 =3:1 collisions are described by a D6~C6v group of
products of three reflection matrices F, C, and M described around eq. (5.3) of Unit 1. Note also inversion
operation F·C=I that commutes with all 12 operators in this D6~C6v group given in Lect. 3 p. 61 and described by
operations as matrices and as reflections in (V1, V2) space.
(a) As far as the 2-D hex-plane is concerned the inversion I is a rotation. By how much?____° Explain.
(b) Note how each reflection like σ30 is labeled by giving angle 30° of a mirror-plane-slope and each rotation like
r60 by the angle 60° it turns vectors. Show effect of σ30 and r60 on unit initial vectors (V1, V2)=(1, 0)=ex and (0, 1)=ey.
(This derives their matrix representations. See examples in Lect. 3 p. 56 thru p. 59.)
(c) Finish the D6~C6v group product table on page 61 of Lect. 3. First do sub-group D3~C3v whose table is upper-left
6-by-6 block of the D6~C6v table. Note that most C3v elements do not commute with each other. ( ab ≠ ba )
Then note each of the remaining 6-by-6 blocks follow a predictable pattern based on defining reflections σ as
product σ=I·r=r·I of inversion I and C3v rotation r. (This shows that C6v reduces to outer product C6v = C3v×Ci of
subgroup C3v with inversion group Ci={1, I} since Ci commutes with C3v.)
KE becomes PE
Exercise 1.6.5 In Fig. 6.3 (See also Lect. 4 p.45 to 48.†) a mass m1=50kg ball is trapped between two smaller mass
m2=0.1kg balls of relatively high speed (v2(0)=20m/s at t=0) that provide m1 with an effective force law F(x) based
on isothermal approximation (6.11). We assume m1 moves only moderately far or fast from equilibrium at x=0.
(We idealize “balls” as point masses here and in many other CM problems.)
(a) A further approximation is the one-Dimensional Harmonic Oscillator (1D-HO) force and PE in (6.12). If each
mass m2 starts in an interval Y0=3.5m, derive isothermal approximate 1D-HO frequency and period for mass m1.
(b) What if the adiabatic approximation is used instead? Does the frequency decrease, increase, or just become
anharmonic? Compare isothermal and adiabatic quantitative results for m1=50kg ball being hit by two m2=0.1kg
balls each having speed of v2(0)=20m/s as each starts bouncing in its space of about Y0=3.5m on either side of the
equilibrium point x=0 for the 1kg ball. Which seems to give best approximation to Lect. 4 p.45-48† results?
† Lect. 4 p.45-48 was upgraded after class with additional details.
Assignments for Physics 5103
-
Reading in Classical Mechanics with a BANG! and Lectures
Unfinished group product table for Exercise 1.5.5 from Lect. 3
D6
1
1
r120
1
r120
σ 60
σ 60
σz
r120
r120
σ 60 σ 60
σz
r60
r60
σ 30 σ 30
σz
1
1
1
1
Note: r60 = Ir120 = r120 I = r−60 and: I = r±180
1
I
r60
r120 = Ir60 = r60 I = r−120 and: I 2 = 1
σ 60 = Iσ 30 = σ 30 I
r60
1
σ 60 = Iσ 30 = σ 30 I
σ z = Iσ z = σ z I
1
r60
1
σ 30
σ 30
I
1
r60
σz
For Exercise 1.6.5 from Lect. 4 (revised)
1
1
Assignments for Physics 5103
-
Reading in Classical Mechanics with a BANG! and Lectures
Assignment 2- Solutions to Ex. 1.5.5.
(a) Rotation by I is ±180°
(b) D6 or C6v group product table with D3 or C3v subgroup table in upper lefthand 6-by-6 quadrant.
D6
1
r120
r120
! 60 ! 60
!z
I
r60
r60
! 30 ! 30
!z
1
1
r120
r120
! 60 ! 60
!z
I
r60
r60
! 30 ! 30
!z
r120
r120
1
r120
!z
! 60 ! 60
r60
I
r60
!z
! 30 ! 30
r120
r120
r120
1
! 60
!z
! 60
r60
r60
I
! 30
!z
! 30
! 60 ! 60
!z
! 60
1
r120
r120
! 30
!z
! 30
I
r60
r60
! 60 ! 60 ! 60
!z
r120
1
r120
! 30 ! 30
!z
r60
I
r60
! 60 ! 60
r120
r120
1
! 30 ! 30
r60
r60
I
r60
! 30
r120
r120
! 60 ! 60
!z
!z
!z
I
I
r60
r60
I
r120
1
r120
!z
! 60 ! 60
! 30
r120
r120
1
! 60
!z
! 60
r60
! 60
!z
! 60
1
r120
r120
I
r60
! 60 ! 60
!z
r120
1
r120
r60
I
!z
! 60 ! 60
r120
r120
1
I
! 30 ! 30
! 30
1
! 30
r60
r60
I
! 30
!z
r60
! 30
!z
Note: r60 = Ir120 = r120 I = r!60 and: I = r±180
r120 = Ir60 = r60 I = r!120 and: I 2 = 1
" 60 = I" 30 = " 30 I
" 60 = I" 30 = " 30 I
" z = I" z = " z I
-!z = !90°
!-60°
reflection
plane
reflection
plane
r120° =
! 60° =
reflection
plane
! 30° =
rotated state
!-30°
reflected state
! 30° 1
r120° 1
r60° =
!+30°
rotated state
reflected state
! 60° 1
reflection
plane
!+60°
r60° 1
reflection
plane
Easy to make hexagonal (D6) symmetry group table:
Operation: !30°|!-60°! =|I! Product:
!30°•!-60° =I.
Operation: r60° |!-60°!=|!-30°! Product: r60°•!-60° =!-30°
11th
row
1
r120
! 30 ! 30 ! 30
r120
r120 ! 60 ! 60 ! z
!z
r60
I
r60
I
r60
r60
!z
!z
!z=!0°
r120
reflection plane
reflection
plane
# cos" 120° " sin" 120° & # "1 / 2 + 3 / 2
r120 = %
=%
$ sin" 120° cos" 120° (' %$ " 3 / 2 "1 / 2
! 60
# cos" 60° sin" 60° & # +1 / 2 " 3 / 2
=%
=%
$ sin" 60° " cos" 60° (' %$ " 3 / 2 "1 / 2
!z =!z 1
I =I1
r120
1
# cos120° " sin120° & # "1 / 2 " 3 / 2 &
=%
=%
( = Ir60
$ sin120° cos120° (' $% + 3 / 2 "1 / 2 ('
! 60
reflected state
! 30 ! 30
! 60 ! 60 ! z
# cos 60° sin 60° & # +1 / 2 + 3 / 2
=%
=%
$ sin 60° " cos 60° (' %$ + 3 / 2 "1 / 2
180°
flipped state
&
( = Ir60
('
&
( = I! 30
('
&
( = I! 30
('
# cos 0° sin 0° & # +1 0 &
# "1 0 & # "1 0 &
!z = %
=
= I! z = %
$ sin 0° " cos 0° (' %$ 0 "1 ('
$ 0 "1 (' %$ 0 +1 ('
# "1 0 &
# +1 0 &
# "1 0 &
I =%
, !z = %
, !z = %
$ 0 "1 ('
$ 0 "1 ('
$ 0 +1 ('
1
!z =!z 1
Initial state
reflected state
!+30°
reflection
plane
!-30°
r120° =
! 30° =
rotated state
r!120° 1
reflected state
! 60° =
reflected state
! "60° 1
!+60°
reflection
plane
r60° =
! "30° 1
rotated state
r!60° 1
-!z
!-60°
reflection
plane
reflection
plane
reflection
plane
Assignments for Physics 5103
-
Reading in Classical Mechanics with a BANG! and Lectures
Assignment 2- Solutions to Ex. 1.6.5.
(a) Isothermal approximations
m v2
m v 2 (0)
const.
f
For isothermal we set: v2 = const. = v2IN = v2 (0) to give: F isoth (1wall Y )=± 2 2 = ± 2 2
=±
=±
Y
Y
Y
Y
For Y0=3.5m there is an isothermal spring constant k ≡ 2 f = 2·
F isoth =
⎤ f
f
f
f ⎡
x
−
= ⎢1 − + ...⎥ −
Y0 + x Y0 − x Y0 ⎣⎢ Y0
⎥⎦ Y0
m2 v22 (0)
due to left (Y0+x) AND right (Y0-x) walls.
Y0 2
⎡
⎤
x
x
x
2
⎢1 + + ...⎥ ≈ −2 f · 2 = −2m2 v2 (0)· 2 = −k·x …
Y
⎢⎣
⎥⎦
Y0
Y0
0
Given Y0=3.5m, m2=0.1kg=10-4kg has speed v2=20m/s so k= 2m2 v22 (0) / Y02 =2(0.1)(20)2/3.52=6.53.
Frequency of m1=50kg is ω=√(k/m1)=√(6.53/50)=0.361 rad/sec, or υ=0.361/2π=0/0575Hz or period τ=17.38sec.
(a) Adiabatic approximations
For adiabatic v2 is not constant but inversely dependent on Y so we set:
(
IN
v2IN Y0
m2 v22
const. v2 Y0
adibad
v2 =
=
, F
(1 wall)=±
± m2
Y
Y
Y
Y3
adibad
F2−wall
)
2
=±
g
Y
3
(
where: g=m2 v2IN Y0
) ~f
2
(for:Y~1).
⎡
⎤
⎢
⎥
⎡
⎤
⎡
⎤
g ⎢
g
g
⎥ = g ⎢1− 3 x + ...⎥ − g ⎢1+ 3 x − ...⎥ = − 6g ·x + ...
=
−
=
−
3
3
3⎢
3
3
Y0
Y0
(Y0 + x) (Y0 − x)
Y 0 (1+ x )3 (1− x )3 ⎥ Y 0 ⎣
Y 04
⎦ Y0 ⎣
⎦
⎢
⎥
Y0
Y0 ⎦
⎣
g
=−
g
6·m2v22 (0)Y 02
Y 04
·x + ... = −
6·m2v22 (0)
Y 02
·x + ...
(Effective spring constant increases by factor of 3)
Given Y0=3.5m, m2=0.1kg=10-4kg has speed v2=20m/s so k= 6m2 v22 (0) / Y02 =6(0.1)(20)2/3.52=19.59.
Frequency of m1=50kg is ω=√(k/m1)=√(19.59/50)=0.626 rad/sec, or υ=0.361/2π=0/0996Hz or period τ=10.03sec.
So adiabatic frequency increases by factor of √3 over the isothermal values while period decreases by 1/√3 .