SOLUTION. The equations of motion are x+ wo2x = ay, j + wo2y = ax. The substitution (23.6) gives Ax(wo2-w2) = aAy, (1) The characteristic equation is (wo2-w2)2= a2, whence w12 = wo2-a, w22 = wo2-+x. For w = W1, the equations (1) give Ax = Ay, and for w = w2, Ax = -Ay. Hence x = (Q1+Q2)/V2, y = (Q1-Q2)/V2, the coefficients 1/V2 resulting from the normalisation of the normal co-ordinates as in equation (23.13). For a < wo2 (weak coupling) we have W1 all wo-1x, W2 ill wotla. The variation of x and y is in this case a superposition of two oscillations with almost equal frequencies, i.e. beats of frequency W2-W1 = a (see $22). The amplitude of y is a minimum when that of x is a maximum, and vice versa. PROBLEM 2. Determine the small oscillations of a coplanar double pendulum (Fig. 1, $5). SOLUTION. For small oscillations (01 < 1, 02 < 1), the Lagrangian derived in §5, Problem 1, becomes L = The equations of motion are = 0, lio +1202+802 = 0. Substitution of (23.6) gives 41(m1+m2)(g-h1w2)-A2w2m2l2 0, = 0. The roots of the characteristic equation are ((ma3 As m1 8 the frequencies tend to the values (g/l1) and /(g/l2), corresponding to indepen- dent oscillations of the two pendulums. PROBLEM 3. Find the path of a particle in a central field U = 1kr2 (called a space oscillator). SOLUTION. As in any central field, the path lies in a plane, which we take as the xy-plane. The variation of each co-ordinate x,y is a simple oscillation with the same frequency = v(k/m): x = a cos(wt+a), y=b cos(wt+), or x = a cos , y = b cos(+8) = b cos 8 cos -b sin 8 sin , where = wt +a, 8 = B-a. Solving for cos o and sin o and equating the sum of their squares to unity, we find the equation of the path: This is an ellipse with its centre at the origin.t When 8 = 0 or IT, the path degenerates to a segment of a straight line. $24. Vibrations of molecules If we have a system of interacting particles not in an external field, not all of its degrees of freedom relate to oscillations. A typical example is that of molecules. Besides motions in which the atoms oscillate about their positions of equilibrium in the molecule, the whole molecule can execute translational and rotational motions. Three degrees of freedom correspond to translational motion, and in general the same number to rotation, so that, of the 3n degrees of freedom of a mole- cule containing n atoms, 3n-6 - correspond to vibration. An exception is formed t The fact that the path in a field with potential energy U = 1kr2 is a closed curve has already been mentioned in $14. §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