
§24
Vibrations of molecules
71
by molecules in which the atoms are collinear, for which there are only two
rotational degrees of freedom (since rotation about the line of atoms is of no
significance), and therefore 3n-5 vibrational degrees of freedom.
In solving a mechanical problem of molecular oscillations, it is convenient
to eliminate immediately the translational and rotational degrees of freedom.
The former can be removed by equating to zero the total momentum of the
molecule. Since this condition implies that the centre of mass of the molecule
is at rest, it can be expressed by saying that the three co-ordinates of the
centre of mass are constant. Putting ra = rao+Ua, where ra0 is the radius
vector of the equilibrium position of the ath atom, and Ua its deviation from
this position, we have the condition = constant = or
= 0.
(24.1)
To eliminate the rotation of the molecule, its total angular momentum
must be equated to zero. Since the angular momentum is not the total time
derivative of a function of the co-ordinates, the condition that it is zero can-
not in general be expressed by saying that some such function is zero. For
small oscillations, however, this can in fact be done. Putting again
ra = rao+ua and neglecting small quantities of the second order in the
displacements Ua, we can write the angular momentum of the molecule as
= .
The condition for this to be zero is therefore, in the same approximation,
0,
(24.2)
in which the origin may be chosen arbitrarily.
The normal vibrations of the molecule may be classified according to the
corresponding motion of the atoms on the basis of a consideration of the sym-
metry of the equilibrium positions of the atoms in the molecule. There is
a general method of doing so, based on the use of group theory, which we
discuss elsewhere. Here we shall consider only some elementary examples.
If all n atoms in a molecule lie in one plane, we can distinguish normal
vibrations in which the atoms remain in that plane from those where they
do not. The number of each kind is readily determined. Since, for motion
in a plane, there are 2n degrees of freedom, of which two are translational
and one rotational, the number of normal vibrations which leave the atoms
in the plane is 2n-3. The remaining (3n-6)-(2n-3) = n-3 vibrational
degrees of freedom correspond to vibrations in which the atoms move out
of the plane.
For a linear molecule we can distinguish longitudinal vibrations, which
maintain the linear form, from vibrations which bring the atoms out of line.
Since a motion of n particles in a line corresponds to n degrees of freedom,
of which one is translational, the number of vibrations which leave the atoms
t See Quantum Mechanics, $100, Pergamon Press, Oxford 1965.
72
Small Oscillations
§24
in line is n - 1. Since the total number of vibrational degrees of freedom of a
linear molecule is 3n - 5, there are 2n-4 which bring the atoms out of line.
These 2n-4 vibrations, however, correspond to only n-2 different fre-
quencies, since each such vibration can occur in two mutually perpendicular
planes through the axis of the molecule. It is evident from symmetry that
each such pair of normal vibrations have equal frequencies.
PROBLEMS
PROBLEM 1. Determine the frequencies of vibrations of a symmetrical linear triatomic
molecule ABA (Fig. 28). It is assumed that the potential energy of the molecule depends
only on the distances AB and BA and the angle ABA.
3
B
A
(o)
(b)
(c)
FIG. 28
SOLUTION. The longitudinal displacements X1, X2, X3 of the atoms are related, according
to (24.1), by MA(X1+x3) +mBX2 = 0. Using this, we eliminate X2 from the Lagrangian of the
longitudinal motion
L =
and use new co-ordinatesQa=x1tx,Qx1-x3.The result
where u = 2mA+mB is the mass of the molecule. Hence we see that Qa and Qs are normal
co-ordinates (not yet normalised). The co-ordinate Qa corresponds to a vibration anti-
symmetrical about the centre of the molecule (x1 = x3; Fig. 28a), with frequency
wa = (k1u/mAmB). The co-ordinate Q8 corresponds to a symmetrical vibration (x1 = -x3;
Fig. 28b), with frequency
The transverse displacements y1, y2, y3 of the atoms are, according to (24.1) and (24.2),
related by mA(y1+y2) +mBy2 = 0, y1 = y3 (a symmetrical bending of the molecule; Fig. 28c).
The potential energy of this vibration can be written as 1/22/282, where 8 is the deviation of the
angle ABA from the value IT, given in terms of the displacements by 8 = [(y1-yy)+(ys-y2)]/L.
Expressing y1,y2, y3 in terms of 8, we obtain the Lagrangian of the transverse motion:
L =
whence the frequency is = V(2k2u/mAmB).
t Calculations of the vibrations of more complex molecules are given by M. V. VOL'KENSH-
TEIN, M. A. EL'YASHEVICH and B. I. STEPANOV, Molecular Vibrations (Kolebaniya molekul),
Moscow 1949; G. HERZBERG, Molecular Spectra and Molecular Structure: Infra-red and
Raman Spectra of Polyatomic Molecules, Van Nostrand, New York 1945,
§24
Vibrations of molecules
73
PROBLEM 2. The same as Problem 1, but for a triangular molecule ABA (Fig. 29).
y
A
A
3
2a
I
2
B
(a)
(b)
(c)
FIG. 29
SOLUTION. By (24.1) and (24.2) the x and y components of the displacements u of the
atoms are related by
0,
0,
(y1-y3) sin x-(x1+x3) cos a = 0.
The changes 8l1 and Sl2 in the distances AB and BA are obtained by taking the components
along these lines of the vectors U1-U2 and U3-U2:
8l1 = (x1-x2) sin cos a,
8l2 = -(x3-x2) sin at(y3-y2) cos a.
The change in the angle ABA is obtained by taking the components of those vectors per-
pendicular to AB and BA:
sin sin
a].
The Lagrangian of the molecule is
L
We use the new co-ordinates Qa = 1+x3, q81 = X1-x3, q82 = y1+y3. The components
of the vectors u are given in terms of these co-ordinates by X1 = 1(Qa+q x3 = 1(Qa-981),
X2
= -MAQa/MB, = -MAQ82/MB. The
Lagrangian becomes
L
=
qksici +
sin
a
cos
a.
74
Small Oscillations
§25
Hence we see that the co-ordinate Qa corresponds to a normal vibration antisymmetrical
about the y-axis (x1 = x3, y1 = -y3; Fig. 29a) with frequency
The co-ordinates qs1, qs2 together correspond to two vibrations symmetrical about the
y-axis (x1 = -X3, y1 y3; Fig. 29b, c), whose frequencies Ws1, W82 are given by the roots
of the quadratic (in w2) characteristic equation
1
When 2 x = 75, all three frequencies become equal to those derived in Problem 1.
PROBLEM 3. The same as Problem 1, but for an unsymmetrical linear molecule ABC
(Fig. 30).
A
FIG. 30
SOLUTION. The longitudinal (x) and transverse (y) displacements of the atoms are related
by
mAX1+mBX2+mcx3 = 0, mAy1tmBy2+mcy3= 0,
MAhy1 = mcl2y3.
The potential energy of stretching and bending can be written
where 2l = li+l2. Calculations similar to those in Problem 1 give
for the transverse vibrations and the quadratic (in w2) equation
=
for the frequencies wil, W12 of the longitudinal vibrations.
$25. Damped oscillations
So far we have implied that all motion takes place in a vacuum, or else that
the effect of the surrounding medium on the motion may be neglected. In
reality, when a body moves in a medium, the latter exerts a resistance which
tends to retard the motion. The energy of the moving body is finally dissipated
by being converted into heat.
Motion under these conditions is no longer a purely mechanical process,
and allowance must be made for the motion of the medium itself and for the
internal thermal state of both the medium and the body. In particular, we
cannot in general assert that the acceleration of a moving body is a function
only of its co-ordinates and velocity at the instant considered; that is, there
are no equations of motion in the mechanical sense. Thus the problem of the
motion of a body in a medium is not one of mechanics.
There exists, however, a class of cases where motion in a medium can be
approximately described by including certain additional terms in the
§25
Damped oscillations
75
mechanical equations of motion. Such cases include oscillations with fre-
quencies small compared with those of the dissipative processes in the
medium. When this condition is fulfilled we may regard the body as being
acted on by a force of friction which depends (for a given homogeneous
medium) only on its velocity.
If, in addition, this velocity is sufficiently small, then the frictional force
can be expanded in powers of the velocity. The zero-order term in the expan-
sion is zero, since no friction acts on a body at rest, and so the first non-
vanishing term is proportional to the velocity. Thus the generalised frictional
force fir acting on a system executing small oscillations in one dimension
(co-ordinate x) may be written fir = - ax, where a is a positive coefficient
and the minus sign indicates that the force acts in the direction opposite to
that of the velocity. Adding this force on the right-hand side of the equation
of motion, we obtain (see (21.4))
mx = -kx-ax.
(25.1)
We divide this by m and put
k/m= wo2, a/m=2x; =
(25.2)
wo is the frequency of free oscillations of the system in the absence of friction,
and A is called the damping coefficient or damping decrement.
Thus the equation is
(25.3)
We again seek a solution x = exp(rt) and obtain r for the characteristic
equation r2+2xr + wo2 = 0, whence ¥1,2 = The general
solution of equation (25.3) is
c1exp(rit)+c2 exp(r2t).
Two cases must be distinguished. If   wo, we have two complex con-
jugate values of r. The general solution of the equation of motion can then
be written as
where A is an arbitrary complex constant, or as
= aexp(-Xt)cos(wta),
(25.4)
with w = V(w02-2) and a and a real constants. The motion described by
these formulae consists of damped oscillations. It may be regarded as being
harmonic oscillations of exponentially decreasing amplitude. The rate of
decrease of the amplitude is given by the exponent X, and the "frequency"
w is less than that of free oscillations in the absence of friction. For 1 wo,
the difference between w and wo is of the second order of smallness. The
decrease in frequency as a result of friction is to be expected, since friction
retards motion.
t The dimensionless product XT (where T = 2n/w is the period) is called the logarithmic
damping decrement.
76
Small Oscillations
§25
If A < wo, the amplitude of the damped oscillation is almost unchanged
during the period 2n/w. It is then meaningful to consider the mean values
(over the period) of the squared co-ordinates and velocities, neglecting the
change in exp( - At) when taking the mean. These mean squares are evidently
proportional to exp(-2xt). Hence the mean energy of the system decreases
as
(25.5)
where E0 is the initial value of the energy.
Next, let A > wo. Then the values of r are both real and negative. The
general form of the solution is
-
(25.6)
We see that in this case, which occurs when the friction is sufficiently strong,
the motion consists of a decrease in /x/, i.e. an asymptotic approach (as t ->
00)
to the equilibrium position. This type of motion is called aperiodic damping.
Finally, in the special case where A = wo, the characteristic equation has
the double root r = - 1. The general solution of the differential equation is
then
(25.7)
This is a special case of aperiodic damping.
For a system with more than one degree of freedom, the generalised
frictional forces corresponding to the co-ordinates Xi are linear functions of
the velocities, of the form
=
(25.8)
From purely mechanical arguments we can draw no conclusions concerning
the symmetry properties of the coefficients aik as regards the suffixes i and
k, but the methods of statistical physics make it possible to demonstrate
that in all cases
aki.
(25.9)
Hence the expressions (25.8) can be written as the derivatives
=
(25.10)
of the quadratic form
(25.11)
which is called the dissipative function.
The forces (25.10) must be added to the right-hand side of Lagrange's
equations:
(25.12)
t See Statistical Physics, $123, Pergamon Press, Oxford 1969.
§26
Forced oscillations under friction
77
The dissipative function itself has an important physical significance: it
gives the rate of dissipation of energy in the system. This is easily seen by
calculating the time derivative of the mechanical energy of the system. We
have
aL
=
Since F is a quadratic function of the velocities, Euler's theorem on homo-
geneous functions shows that the sum on the right-hand side is equal to 2F.
Thus
dE/dt==2-2F,
(25.13)
i.e. the rate of change of the energy of the system is twice the dissipative
function. Since dissipative processes lead to loss of energy, it follows that
F > 0, i.e. the quadratic form (25.11) is positive definite.
The equations of small oscillations under friction are obtained by adding
the forces (25.8) to the right-hand sides of equations (23.5):
=
(25.14)
Putting in these equations XK = Ak exp(rt), we obtain, on cancelling exp(rt),
a set of linear algebraic equations for the constants Ak:
(25.15)
Equating to zero their determinant, we find the characteristic equation, which
determines the possible values of r:
(25.16)
This is an equation in r of degree 2s. Since all the coefficients are real,
its roots are either real, or complex conjugate pairs. The real roots must be
negative, and the complex roots must have negative real parts, since other-
wise the co-ordinates, velocities and energy of the system would increase
exponentially with time, whereas dissipative forces must lead to a decrease
of the energy.
§26. Forced oscillations under friction
The theory of forced oscillations under friction is entirely analogous to
that given in §22 for oscillations without friction. Here we shall consider
in detail the case of a periodic external force, which is of considerable interest.
78
Small Oscillations
§26
Adding to the right-hand side of equation (25.1) an external force f cos st
and dividing by m, we obtain the equation of motion:
+2*+wox=(fm)cos = yt.
(26.1)
The solution of this equation is more conveniently found in complex form,
and so we replace cos st on the right by exp(iyt):
exp(iyt).
We seek a particular integral in the form x = B exp(iyt), obtaining for B
the value
(26.2)
Writing B = exp(i8), we have
b tan 8 = 2xy/(y2-wo2).
(26.3)
Finally, taking the real part of the expression B exp(iyt) = b exp[i(yt+8)],
we find the particular integral of equation (26.1); adding to this the general
solution of that equation with zero on the right-hand side (and taking for
definiteness the case wo > 1), we have
x = a exp( - At) cos(wtta)+bcos(yt+8)
(26.4)
The first term decreases exponentially with time, so that, after a sufficient
time, only the second term remains:
x = b cos(yt+8).
(26.5)
The expression (26.3) for the amplitude b of the forced oscillation increases
as y approaches wo, but does not become infinite as it does in resonance
without friction. For a given amplitude f of the force, the amplitude of the
oscillations is greatest when y = V(w02-2)2); for A < wo, this differs from
wo only by a quantity of the second order of smallness.
Let us consider the range near resonance, putting y = wote with E small,
and suppose also that A < wo. Then we can approximately put, in (26.2),
22 =(y+wo)(y-wo) 22 2woe, 2ixy 22 2ixwo, SO that
B = -f/2m(e-ii))wo
(26.6)
or
b f/2mw01/(22+12),
tan 8 = N/E.
(26.7)
A property of the phase difference 8 between the oscillation and the external
force is that it is always negative, i.e. the oscillation "lags behind" the force.
Far from resonance on the side < wo, 8 0; on the side y > wo, 8
-77.
The change of 8 from zero to - II takes place in a frequency range near wo
which is narrow (of the order of A in width); 8 passes through - 1/2 when
y = wo. In the absence of friction, the phase of the forced oscillation changes
discontinuously by TT at y = wo (the second term in (22.4) changes sign);
when friction is allowed for, this discontinuity is smoothed out.
§26
Forced oscillations under friction
79
In steady motion, when the system executes the forced oscillations given
by (26.5), its energy remains unchanged. Energy is continually absorbed by
the system from the source of the external force and dissipated by friction.
Let I(y) be the mean amount of energy absorbed per unit time, which depends
on the frequency of the external force. By (25.13) we have I(y) = 2F, where
F is the average value (over the period of oscillation) of the dissipative func-
tion. For motion in one dimension, the expression (25.11) for the dissipative
function becomes F = 1ax2 = Amx2. Substituting (26.5), we have
F = mb22 sin2(yt+8).
The time average of the squared sine is 1/2 so that
I(y) = Mmb2y2. =
(26.8)
Near resonance we have, on substituting the amplitude of the oscillation
from (26.7),
I(e) =
(26.9)
This is called a dispersion-type frequency dependence of the absorption.
The half-width of the resonance curve (Fig. 31) is the value of E for which
I(e) is half its maximum value (E = 0). It is evident from (26.9) that in the
present case the half-width is just the damping coefficient A. The height of
the maximum is I(0) = f2/4mx, and is inversely proportional to . Thus,
I/I(O)
/2
€
-1
a
FIG. 31
when the damping coefficient decreases, the resonance curve becomes more
peaked. The area under the curve, however, remains unchanged. This area
is given by the integral
[ ((7) dy = [ I(e) de.
Since I(e) diminishes rapidly with increasing E, the region where |el is
large is of no importance, and the lower limit may be replaced by - 80, and
I(e) taken to have the form given by (26.9). Then we have
"
(26.10)
80
Small Oscillations
§27
PROBLEM
Determine the forced oscillations due to an external force f = fo exp(at) COS st in the
presence of friction.
SOLUTION. We solve the complex equation of motion
2+wo2x = (fo/m) exp(at+iyt)
and then take the real part. The result is a forced oscillation of the form
x=bexp(at)cos(yt+8),
where
b =
tan s =
§27. Parametric resonance
There exist oscillatory systems which are not closed, but in which the
external action amounts only to a time variation of the parameters.t
The parameters of a one-dimensional system are the coefficients m and k
in the Lagrangian (21.3). If these are functions of time, the equation of
motion is
(27.1)
We introduce instead of t a new independent variable T such that
dr = dt/m(t); this reduces the equation to
d2x/d-2+mkx=0.
There is therefore no loss of generality in considering an equation of motion
of the form
(27.2)
obtained from (27.1) if m = constant.
The form of the function w(t) is given by the conditions of the problem.
Let us assume that this function is periodic with some frequency y and period
T = 2n/y. This means that w(t+T) = w(t), and so the equation (27.2) is
invariant under the transformation t t+ T. Hence, if x(t) is a solution of
the equation, so is x(t+T). That is, if x1(t) and x2(t) are two independent
integrals of equation (27.2), they must be transformed into linear combina-
tions of themselves when t is replaced by t + T. It is possible to choose X1
and X2 in such a way that, when t t+T, they are simply multiplied by
t A simple example is that of a pendulum whose point of support executes a given periodic
motion in a vertical direction (see Problem 3).
+ This choice is equivalent to reducing to diagonal form the matrix of the linear trans-
formation of x1(t) and x2(t), which involves the solution of the corresponding quadratic
secular equation. We shall suppose here that the roots of this equation do not coincide.