Table Of ContentChapter 16
Solutions to Exercises
16.1 Introduction
No exercises
16.2 Symmetry and Physics
Exercise 1. Calculate the length D of roadway in Fig. 2.3 as a function
of l and a, and verify that it is minimized by the value of l given in Eq.(2.1)
above.
Solution:
Each diagonal segment on the right hand side of Fig. 2.3 has length:
(cid:115)
(cid:16)a(cid:17)2 (cid:18)a−l(cid:19)2
d(l) = + (16.1)
2 2
The total length of pavement D(l) as a function of l is:
(cid:115)
(cid:16)a(cid:17)2 (cid:18)a−l(cid:19)2
D(l) = l+4d = l+4 + (16.2)
2 2
496
CHAPTER 16. SOLUTIONS TO EXERCISES 497
To find the value of l that minimizes the function D(l) we have to solve:
min
(cid:12)
dD(l)(cid:12) 2(a−l)(−1)
(cid:12) = 1+ (cid:112)
dl (cid:12) a2 +(l−a)2
lmin
= 0 (16.3)
A bit of algebra yields:
a2
(a−l )2 = (16.4)
min
3
Since l must be less than a the relevant solution is:
(cid:18) (cid:19)
1
l = a 1− √ (16.5)
min
3
Plugging this into the expression for D(l) we find:
(cid:18) (cid:19)
1
D(l ) = a 1+ √ (16.6)
min
3
which is less than the length of pavement required by building two diagonal
roads directly through the center of the square, namely:
√ √
D = 2 a2 +a2 = 2 2a (16.7)
diagonals
A straightforward calculation verifies that
(cid:12)
d2D(l)(cid:12)
(cid:12) < 0 (16.8)
dl2 (cid:12)
lmin
so that l is indeed a minimum as required.
min
Exercise 2. Consider three towns, called N (for North), SW and SE,
respectively, located a distance a apart at the vertices of an equilateral tri-
angle, as shown in Fig.[2.4]. We wish to build a network of roads connecting
all three towns such that the roads (shown in blue) meet at an arbitrary
point P along the line joining the center of the triangle to the northern town
at some distance l from N. Show that the minimum total length for such a
CHAPTER 16. SOLUTIONS TO EXERCISES 498
√
configuration of roads occurs when l = a/ 3, so that the least expensive way
to join the towns is to have the three segments of road meet at the center,
C. Is there symmetry breaking in this case? Explain.
Hint: Use the law of cosines to figure out the distance from P to the other
two towns SW and SE. This should give you an expression for the total
length of the three roads that join P to N, SW and SE. Finally, minimize
the total length of pavement with respect to the parameter l and show that
√
the minimum occurs at l = a/ 3.
Figure 16.1: Three towns to be joined by shortest road.
Solution:
Consider the triangle with vertices N-P-SE. The angle at each vertex in the
equilateral triangle is 60◦, so that the angle between the lines N-P and N-SE
is 30◦. Applying the cosine law:
d2(l) = l2 +a2 −2alcos(30)
(cid:113) √
→ d(l) = l2 +a2 − 3al (16.9)
The total length of road D(l) is:
D(l) = 2d(l)+l
(cid:113) √
= 2 l2 +a2 − 3al+l (16.10)
CHAPTER 16. SOLUTIONS TO EXERCISES 499
We want to find the value l of l that minimizes D(l):
min
√
(cid:12)
dD(l)(cid:12) 2l− 3a
(cid:12) = 2 (cid:112) √ +1
dl (cid:12) 2 l2 +a2 − 3al
lmin
= 0
√ (cid:113) √
→ 2l − 3a = l2 +a2 − 3al (16.11)
min min
Squaring both sides of the above gives a quadratic equation for l in terms
min
of a whose solution is:
√
(3±1) 3
l = a (16.12)
min
2
Since we need l to be less than a the physical solution is the minus sign,
min
so that:
√
l = 3a (16.13)
min
A straightforward calculation verifies that
(cid:12)
d2D(l)(cid:12)
(cid:12) > 0 (16.14)
dl2 (cid:12)
lmin
so that l is indeed a minimum as required.
min
Exercise 3. Derive Equation (2.2) using the diagram in Fig. 2.5. The
distances a, b and c are given. You can use the fact that the total time T
tot
taken along Path 2 is:
(cid:112) (cid:112)
c2 +b2 c2 +a2
T = 1 + 2 (16.15)
tot
V V
G S
and find the equation for c that minimizes T . Hint: in order to solve the
1 tot
resulting equation, you may need:
c
1
= tan(θ )
G
b
c
2
= tan(θ ) (16.16)
S
a
CHAPTER 16. SOLUTIONS TO EXERCISES 500
Solution:
We wish to minimize the time take to get to swimmer with respect to c , so
1
require:
dT 1 2c 1 2(−1)(c−c )
tot 1 1
= +
(cid:112) (cid:112)
dc 2V c2 +b2 2 (c −a)2 +a2
1 G 1 1
(cid:112)
c V c2 +b2
→ 1 = G 1 (16.17)
(cid:112)
c−c V (c−c )2 +a2
1 S 1
Now square both sides of the above, and divide both sides by a2b2 to get:
c2 (cid:18)c2 (cid:19) V c2 (cid:18)c2 (cid:19)
1 2 +1 = G 2 1 +1 (16.18)
b2 a2 V a2 b2
S
where we have replaced (c−c ) by c . Next use Eq. (16.16) to replace c /b
1 2 1
and c /a by tan(V ) and tan(V ), respectively to obtain Snell’s law.
2 G S
16.3 Formal Aspects of Symmetry
Exercise 1.
1. Verify that the entries in the multiplication table (3.3) follow from the
actions of the symmetry operations listed in Eq.(16.99).
2. Verify that this multiplication table obeys the four axioms of group
theory (i.e. properties of symmetry operations). Note: this should not
require any calculations once you have the table completed.
3. Is this group of symmetry operations commutative? That is, does the
order of operations matter in some cases? If so, which ones?
Solution:
CHAPTER 16. SOLUTIONS TO EXERCISES 501
We consider the set of permutations S := {I,A,B,C,G,F} of three identical
objects
a
b
c
where the symmetry operations corresponding to the permutations are de-
fined as:
a a a b
I : b → b ; A : b → a (16.19)
c c c c
a a a c
B : b → c ; C : b → b (16.20)
c b c a
a c a b
F : b → a ; G : b → c (16.21)
c b c a
1. Fill in the rest of the multiplication table below for this group of sym-
metry operations:
• Top horizontal row and left vertical row trivial since I ·X = X ·I
for any X.
• A·A = I since A just permutes the top two components a and b
with each other twice so that this is the same as doing nothing at
all.
• Similarly B permutes 2nd and 3rd, and C permutes 1st and 3rd,
so that B ·B = I, and C ·C = I.
• For A·C, remember you act with C and then with A:
a c b
b →C b →A c (16.22)
c a a
Note that the operations move the components based on their
locations, not their letter names. The net effect is the same as G.
Similarly F followed by G is:
CHAPTER 16. SOLUTIONS TO EXERCISES 502
a c a
b →F a →G b (16.23)
c b c
The net effect is the identity.
Continuing as above you get:
I A B C G F
I I A B C G F
A A I F G C B
B B G I F A A
C C F G I B A
G G B C A F I
F F C A B I G
2. Verify that this multiplication table obeys the four axioms of group
theory (i.e. properties of symmetry operations). Note: this should
not require any calculations once you have the table completed. For
associativity, just check on or two cases.
–It is complete because the product of any two symmetry operations
yields another symmetry operation. That is, all the entries in the table
are one of the operations along the top row.
–There is an identity operator, I
–Every element has an inverse. A, B and C are their own inverses.
The inverse of F is G and vice versa.
–Verifying associativity is a bit of a pain, but for example:
(A·C)·G = G·G = F
A·(C ·G) = A·B = F (16.24)
So associativity holds in this case.
3. Is this group of symmetry operations commutative? That is, does the
order of operations matter in some cases? If so, which ones?
No it is not commutative. For example A·G = C whereas G·A = B.
However, F·G = G·F = I. That is about it for commuting operations.
CHAPTER 16. SOLUTIONS TO EXERCISES 503
Exercise 2. Construct the multiplication table for the set of rotations that
leave a hexagon invariant.
Solutions
As can be seen in Fig. 16.2 directly below, rotations by multiples60◦ take
the diagonal lines onto diagonal lines and leave the hexagon invariant.
Figure 16.2: Hexagon
0 60 120 180 240 300
0 0 60 120 180 240 300
60 60 120 180 240 300 0
120 120 180 240 300 0 60
180 180 240 300 0 60 120
240 240 300 0 60 120 180
300 300 0 60 120 180 240
Exercise 3. Show that dE = 0 if x(t) satisfies Eq. (3.13)
dt
CHAPTER 16. SOLUTIONS TO EXERCISES 504
Solution:
Start with the expression for energy:
1
E := K +V = m[x˙(t)]2 +V(x(t)) = constant (16.25)
2
Now calculate its time derivative, noting that all the time dependence is
through x(t):
dE 1 dV(x(t))dx
= m2x˙(t)x¨(t)+
dt 2 dx dt
(cid:18) (cid:19)
dV(x(t))
= v(t) x¨(t)+ (16.26)
dx
whichvanishesifx(t)obeysNewton’sequations. NotethatifV := V(x(t),t)depended
on time explicitly, there would be an extra term on the right hand side, so
that energy would not be conserved..
Exercise 4. Derive the equations of motion for a bead moving on a circular
frictionless wire of radius R (see Fig. 3.3) using the principle of least action.
Figure 16.3: Bead moving along frictionless circular wire of radius R.
CHAPTER 16. SOLUTIONS TO EXERCISES 505
Solutions:
The action is the difference between the kinetic energy and potential energy.
We neglect the transverse forces (i.e. perpendicular to the wire) that keep
the bead moving in a circle since they do not contribute to the motion along
the wire. The Lagrangian is then just equal to the kinetic energy of the bead,
namely:
L = K
1
= mv2
2 t
1
= mR2ω2 (16.27)
2
where v is the transverse speed of the bead along the wire and perpendicular
t
to its position vector and
˙
ω := θ (16.28)
isitsangularvelocity. Theonlyrelevantcoordinateinthiscaseisθ. Applying
Eq. (3.30) with x = θ we find:
p
(cid:18) (cid:19) (cid:18) (cid:19)
∂L d ∂L d ∂L
− = 0−
∂x dt ∂x˙ dt ∂θ˙
p p
= −mR2θ¨= 0 (16.29)
Thus the Euler Lagrange equations tell us, as expected, that the angular
acceleration of the bead is zero.
16.4 Symmetries and Linear Transformations
Exercise 1. Draw the three position vectors A, magnitude 2 m, direction
due East (E), B magnitude 4 m, direction North-North-East (NNE) and
C, magnitude 3 m direction South-West (NW). (SW points between South
and West, whereas NNE points in the direction midway between North and
North-East. North-East points mid-way between North an East)
