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Unformatted text preview: 02Feb2011 Chemistry 21b Spectroscopy Lecture # 13 The Classical Treatment of Molecular Vibrations For a gas phase molecule with N atoms, there will be a total of 3 N degrees of freedom in free space. Three of these will be involved in translation, which may be factored out as demonstrated previously for diatomic systems. Symmetric and asymmetric tops have an additional three degrees of freedom involved in rotation, while diatomic and linear molecules only have two rotational degrees of freedom (because there is no moment of inertia about the internuclear axis). Thus, the vibrational degrees of freedom are 3 N 6 for symmetric or asymmetric tops, and 3 N 5 for linear molecules. We now define the displacement coordinates x = a  a ,e , y = b  b ,e , z = c  c ,e : ( x, y, z ) = displacement coordinates of atom ( a, b, c ) = molecule (or body ) fixed coordinates ( a, b, c ) ,e = molecule fixed equilibrium positions . (13 . 1) Classically, the kinetic energy of vibration is thus T = 1 2 N summationdisplay =1 m bracketleftBigg parenleftbigg dx dt parenrightbigg 2 + parenleftbigg dy dt parenrightbigg 2 + parenleftbigg dz dt parenrightbigg 2 bracketrightBigg (13 . 2) The first step in simplifying this expression is to use mass weighted coordinates , that is q 1 = m 1 / 2 1 x 1 ; q 2 = m 1 / 2 2 y 1 ; ... , q 3 N = m 1 / 2 N z N , which results in T = 1 2 3 N summationdisplay =1 parenleftbigg dq i dt parenrightbigg 2 or T = 1 2 3 N summationdisplay =1 q 2 i 1 2 q 2 = 1 2 q q , (13 . 3) using matrix notation, where the dots over the characters denote time derivatives, and the last expression is the dot product of the row and column vector containing the q i . Next we examine the potential energy of vibration, which well label U ( q 1 , ..., q 3 N ). As we have often done in this class, to make progress it is helpful to expand the potential energy in a Taylor series about the equilibrium bodyfixed positions ( a, b, c ) ,e , or U = U e + 3 N summationdisplay i =1 parenleftbigg U q i parenrightbigg e q i + 1 2 3 N summationdisplay i =1 3 N summationdisplay k =1 parenleftbigg 2 U q i k parenrightbigg e q i q k + 1 6 3 N summationdisplay i =1 3 N summationdisplay j =1 3 N summationdisplay k =1 parenleftbigg 3 U q i q...
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