Question 15.10: Calculate the normal frequencies of a symmetric molecule ABA...
Calculate the normal frequencies of a symmetric molecule ABA of triangula shape:
Conservation of the center of gravity here reads
m_{A}(x_{1} +x_{3})+m_{B}x_{2} = 0, m_{A}(y_{1} + y_{3})+ m_{B}y_{2} = 0.For angular momentum conservation, we go to the rest position of atom B, and because m_{1} = m_{3} = m_{A}, it follows that
r_{10} ×x_{1} +r_{30} ×x_{3} = 0.We have
r_{10} ×x_{1} = |r_{10}|(−x_{1} cos α + y_{1} sin α)e,r_{30} ×x_{3} = |r_{30}|(−x_{3} cos α + y_{3} sin α)e,
Because |r_{10}| = |r_{30}|, the angular momentum conservation law follows:
sin α(y_{1} −y_{3}) = cos α(x_{1} + x_{3}) or y_{1} −y_{3} = cot α(x_{1} +x_{3}).The changes δl_{1} and δl_{2} of the distances AB and BA result by projection of the vectors x_{1} − x_{2} and x_{3} −x_{2} onto the directions of the lines AB and BA:
δl_{1} = (x_{1} − x_{2}) sin α +(y_{1} − y_{2}) cos α,δl_{2} = -(x_{3} − x_{2}) sin α +(y_{3} − y_{2}) cos α,
The change of the angle 2α =∢ (ABA) is found by projection of the vectors x_{1} − x_{2} and x_{3} − x_{2} onto the directions orthogonal to the line segments AB and BA:
δ = \frac{1}{l} \left[ (x_{1} −x_{2}) cos α −(y_{1} −y_{2}) sin α\right] + \frac{1}{l} \left[−(x_{3} − x_{2}) cos α − (y_{3} − y_{2}) sin α\right]We write the Lagrangian of the molecule as
L = \frac{m_{A}}{2} (\dot{x}^{2}_{1}+ \dot{x}^{2}_{3} )+ \frac{m_{B}}{2} \dot{x}^{2}_{2} − \frac{K_{1}}{2}\left[(δl_{1})^{2} +(δl_{2})^{2}\right] − \frac{K_{2}}{2} (lδ)^{2}.Here, (K_{1}/2)[(δl_{1})^{2} +(δl_{2})^{2}] is the potential energy of the rotation, and K_{2}(lδ)^{2}/2 is the potential energy of the bending of the molecule. We adopt as new coordinates
Q_{α} = x_{1} +x_{3}, q_{s 1} = x_{1} − x_{3}, q_{s 2} = y_{1} +y_{3},and then have
x_{1} = \frac{1}{2}(Q_{α} +q_{s 1}), x_{2} =−\frac{m_{A}}{m_{B}}Q_{α}, x_{3} = \frac{1}{2} (Q_{α} −q_{s 1}),y_{1} = \frac{1}{2} (q_{s 2} +Q_{α} cot α), y_{2} =−\frac{m_{A}}{m_{B}}q_{s 2}, y_{3} = \frac{1}{2}(q_{s 2} −Q_{α} cot α).
Because y_{1} −y_{3} =Q_{α} cot α, we find for L
L = \frac{m_{A}}{4} \left(\frac{2m_{A}}{m_{B}}+ \frac{1}{sin^{2} α}\right)\dot{Q}^{2}_{α} + \frac{m_{A}}{4} \dot{q}^{2}_{s 1}+ \frac{m_{A}μ}{4m_{B}} \dot{q}^{2}_{s 2}−Q^{2}_{α} \frac{K_{1}}{4} \left(\frac{2m_{A}}{m_{B}}+ \frac{1}{sin^{2} α}\right) \left(1+ \frac{2m_{A}}{m_{B}} sin^{2} α\right)
− \frac{q^{2}_{s 1}}{4} (K_{1} sin^{2} α + 2K^{2} cos^{2} α) −q^{2}_{s 2} \frac{μ^{2}}{4m^{2}_{B}} (K_{1} cos^{2} α +2K^{2} sin^{2} α)
+ q_{s 1}q_{s 2} \frac{μ}{2m_{B}} (2K_{2} −K_{1}) sin α cos α.
Obviously, Q_{α} is a normal coordinate, with the vibration frequency
ω^{2}_{α} = \frac{K_{1}}{m_{A}} \left(1 + \frac{2m_{A}}{m_{B}} sin^{2} α\right).Pure Q_{α}-vibrations occur for x_{1} = x_{3}, y_{1} = −y_{3}; i.e., Q_{α} describes antisymmetric vibrations with respect to the y-axis in Fig. 15.16.
The eigenfrequencies ω_{s 1} and ω_{s 2} of the normal vibrations for q_{s 1} and q_{s 2} must be determined by the characteristic equation
ω^{4} −ω^{2} \left[ \frac{K_{1}}{m_{A}} \left(1+ \frac{2m_{A}}{m_{B}} cos^{2} α\right) + \frac{2K_{2}}{m_{A}} \left(1+ \frac{2m_{A}}{m_{B}}sin^{2} α\right)\right]+ \frac{2μK_{1}K_{2}}{m_{B}m^{2}_{A}}= 0.
The coordinates q_{s 1} and q_{s 2} correspond to vibrations that are symmetric about the y-axis (Fig. 15.17):
(x_{1} =−x_{3}, Q_{α} = 0 ⇒ y_{1} = y_{3}).
