Symmetry and normal modes

NORMAL MODES
The final and most complicated of the three normal modes has angular frequency
ω = {[(µ + 2)/µ](k/m)}1/2 , and involves a motion of the central particle which is in
antiphase with that of the two outer ones and which has an amplitude 2/µ times as great.
In this motion (see figure 9.3(c)) the two springs are compressed and extended in turn. We
also note that in the second and third normal modes the centre of mass of the molecule
remains stationary. 9.2 Symmetry and normal modes
It will have been noticed that the system in the above example has an obvious
symmetry under the interchange of coordinates 1 and 3: the matrices A and B,
the equations of motion and the normal modes illustrated in figure 9.3 are all
unaltered by the interchange of x1 and −x3 . This reflects the more general result
that for each physical symmetry possessed by a system, there is at least one
normal mode with the same symmetry.
The general question of the relationship between the symmetries possessed by
a physical system and those of its normal modes will be taken up more formally
in chapter 29 where the representation theory of groups is considered. However,
we can show here how an appreciation of a system’s symmetry properties will
sometimes allow its normal modes to be guessed (and then verified), something
that is particularly helpful if the number of coordinates involved is greater than
two and the corresponding eigenvalue equation (9.10) is a cubic or higher-degree
polynomial equation.
Consider the problem of determining the normal modes of a system consisting of four equal masses M at the corners of a square of side 2L, each pair
of masses being connected by a light spring of modulus k that is unstretched
in the equilibrium situation. As shown in figure 9.4, we introduce Cartesian
coordinates xn , yn , with n = 1, 2, 3, 4, for the positions of the masses and denote their displacements from their equilibrium positions Rn by qn = xn i + yn j.
Thus
rn = Rn + qn
with
Rn = ±Li ± Lj.
The coordinates for the system are thus x1 , y1 , x2 , . . . , y4 and the kinetic energy matrix A is given trivially by MI8 , where I8 is the 8 × 8 identity matrix.
The potential energy matrix B is much more difficult to calculate and involves,
for each pair of values m, n, evaluating the quadratic approximation to the
expression
2
bmn = 12 k |rm − rn | − |Rm − Rn | .
Expressing each ri in terms of qi and Ri and making the normal assumption that
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9.2 SYMMETRY AND NORMAL MODES
y1
y2
k
M
M
x2
x1
k
k
k
k
y3
y4
x3
M
M
k
x4
Figure 9.4 The arrangement of four equal masses and six equal springs
discussed in the text. The coordinate systems xn , yn for n = 1, 2, 3, 4 measure
the displacements of the masses from their equilibrium positions.
|Rm − Rn | |qm − qn |, we obtain bmn (= bnm ):
2
bmn = 12 k |(Rm − Rn ) + (qm − qn )| − |Rm − Rn |
!
"2
1/2
= 12 k |Rm − Rn |2 + 2(qm − qn ) · (RM − Rn ) + |qm − qn )|2
− |Rm − Rn |
.2
#
1/2
2(qm − qn ) · (RM − Rn )
2
1
= 2 k|Rm − Rn |
+ ···
−1
1+
|Rm − Rn |2
(qm − qn ) · (RM − Rn ) 2
≈ 12 k
.
|Rm − Rn |
This final expression is readily interpretable as the potential energy stored in the
spring when it is extended by an amount equal to the component, along the
equilibrium direction of the spring, of the relative displacement of its two ends.
Applying this result to each spring in turn gives the following expressions for
the elements of the potential matrix.
m
1
1
1
2
2
3
n
2
3
4
3
4
4
2bmn /k
(x1 − x2 )2
(y1 − y3 )2
1
2
(−x
1 + x4 + y1 − y4 )
2
1
2
(x
−
x
+
y
−
y
)
3
2
3
2 2
(y2 − y4 )2
(x3 − x4 )2 .
323
NORMAL MODES
The potential matrix is thus constructed as

3 −1 −2
0
0
0 −1
1
 −1
3
0
0
0
−2
1
−1

 −2
0
3
1 −1 −1
0
0

k
0
1
3 −1 −1
0 −2
 0
B= 
0 −1 −1
3
1 −2
0
4 0

 0 −2 −1 −1
1
3
0
0

 −1
1
0
0 −2
0
3 −1
1 −1
0 −2
0
0 −1
3







.





To solve the eigenvalue equation |B − λA| = 0 directly would mean solving
an eigth-degree polynomial equation. Fortunately, we can exploit intuition and
the symmetries of the system to obtain the eigenvectors and corresponding
eigenvalues without such labour.
Firstly, we know that bodily translation of the whole system, without any
internal vibration, must be possible and that there will be two independent
solutions of this form, corresponding to translations in the x- and y- directions.
The eigenvector for the first of these (written in row form to save space) is
x(1) = (1 0 1 0 1 0
1 0)T .
Evaluation of Bx(1) gives
Bx(1) = (0
0 0 0 0 0 0 0)T ,
showing that x(1) is a solution of (B − ω 2 A)x = 0 corresponding to the eigenvalue
ω 2 = 0, whatever form Ax may take. Similarly,
x(2) = (0 1 0 1
0 1 0 1)T
is a second eigenvector corresponding to the eigenvalue ω 2 = 0.
The next intuitive solution, again involving no internal vibrations, and, therefore, expected to correspond to ω 2 = 0, is pure rotation of the whole system
about its centre. In this mode each mass moves perpendicularly to the line joining
its position to the centre, and so the relevant eigenvector is
1
x(3) = √ (1 1 1
2
−1
−1 1
−1
− 1)T .
It is easily verified that Bx(3) = 0 thus confirming both the eigenvector and the
corresponding eigenvalue. The three non-oscillatory normal modes are illustrated
in diagrams (a)–(c) of figure 9.5.
We now come to solutions that do involve real internal oscillations, and,
because of the four-fold symmetry of the system, we expect one of them to be a
mode in which all the masses move along radial lines – the so-called ‘breathing
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9.2 SYMMETRY AND NORMAL MODES
(a) ω 2 = 0
(e) ω 2 = k/M
(d) ω 2 = 2k/M
(c) ω 2 = 0
(b) ω 2 = 0
(f) ω 2 = k/M
(h) ω 2 = k/M
(g) ω 2 = k/M
Figure 9.5 The displacements and frequencies of the eight normal modes of
the system shown in figure 9.4. Modes (a), (b) and (c) are not true oscillations:
(a) and (b) are purely translational whilst (c) is a mode of bodily rotation.
Mode (d), the ‘breathing mode’, has the highest frequency and the remaining
four, (e)–(h), of lower frequency, are degenerate.
mode’. Expressing this motion in coordinate form gives as the fourth eigenvector
1
x(4) = √ (−1 1 1 1
2
−1
−1 1
− 1)T .
Evaluation of Bx(4) yields
k
Bx(4) = √ (−8
4 2
8 8 8
−8
−8 8
− 8)T = 2kx(4) ,
i.e. a multiple of x(4) , confirming that it is indeed an eigenvector. Further, since
Ax(4) = Mx(4) , it follows from (B − ω 2 A)x = 0 that ω 2 = 2k/M for this normal
mode. Diagram (d) of the figure illustrates the corresponding motions of the four
masses.
As the next step in exploiting the symmetry properties of the system we note
that, because of its reflection symmetry in the x-axis, the system is invariant under
the double interchange of y1 with −y3 and y2 with −y4 . This leads us to try an
eigenvector of the form
x(5) = (0
α 0 β
0
−α 0
− β)T .
Substituting this trial vector into (B − ω 2 A)x = 0 gives, of course, eight simulta325
NORMAL MODES
neous equations for α and β, but they are all equivalent to just two, namely
α + β = 0,
4Mω 2
α;
5α + β =
k
these have the solution α = −β and ω 2 = k/M. The latter thus gives the frequency
of the mode with eigenvector
x(5) = (0
1 0
−1 0
− 1 0 1)T .
Note that, in this mode, when the spring joining masses 1 and 3 is most stretched,
the one joining masses 2 and 4 is at its most compressed. Similarly, based on
reflection symmetry in the y-axis,
x(6) = (1
0
−1 0
− 1 0 1 0)T
can be shown to be an eigenvector corresponding to the same frequency. These
two modes are shown in diagrams (e) and (f) of figure 9.5.
This accounts for six of the expected eight modes, and the other two could be
found by considering motions that are symmetric about both diagonals of the
square or are invariant under successive reflections in the x- and y- axes. However,
since A is a multiple of the unit matrix, and since we know that (x(j) )T Ax(i) = 0 if
i = j, we can find the two remaining eigenvectors more easily by requiring them
to be orthogonal to each of those found so far.
Let us take the next (seventh) eigenvector, x(7) , to be given by
x(7) = (a b
c d e f
g
h)T .
Then orthogonality with each of the x(n) for n = 1, 2, . . . , 6 yields six equations
satisfied by the unknowns a, b, . . . , h. As the reader may verify, they can be reduced
to the six simple equations
a + g = 0, d + f = 0, a + f = d + g,
b + h = 0, c + e = 0, b + c = e + h.
With six homogeneous equations for eight unknowns, effectively separated into
two groups of four, we may pick one in each group arbitrarily. Taking a = b = 1
gives d = e = 1 and c = f = g = h = −1 as a solution. Substitution of
x(7) = (1 1
−1 1 1
−1
−1
− 1)T .
into the eigenvalue equation checks that it is an eigenvector and shows that the
corresponding eigenfrequency is given by ω 2 = k/M.
We now have the eigenvectors for seven of the eight normal modes and the
eighth can be found by making it simultaneously orthogonal to each of the other
seven. It is left to the reader to show (or verify) that the final solution is
x(8) = (1
−1 1 1
−1
326
−1
− 1 1)T