{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

26_580ln_fa08

# 26_580ln_fa08 - MIT OpenCourseWare http/ocw.mit.edu 5.80...

This preview shows pages 1–4. Sign up to view the full content.

MIT OpenCourseWare http://ocw.mit.edu 5.80 Small-Molecule Spectroscopy and Dynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
5.80 Lecture #26 Fall, 2008 Page 1 of 7 pages Lecture #26: Polyatomic Vibrations II: s-Vectors, G-matrix, and Eckart Condition Last time: ξ , q, S, Q Q = L 1 DM 1/2 ξ  q  S G –1 ( D –1 ) D –1 (because q = = D 1 S and D q S ) 2T = S G 1 S 2 V 2V = S F S F ij S i S j 0 L ({ S }, { S }) = T({ S }) – V({ S }) d L L secular equation dT S j S j = 0 F − λ G 1 = 0 assuming that S j (t) = A j cos ( λ 1/2 t + ε ) all j = 1, 2, … 3N–6 i.e. that all internal coordinates oscillate at same frequency and relative phase (but with λ 1/2 different amplitudes) ν = 2 π 3N–6 possible different values of λ obtained from (3N – 6) × (3N – 6) secular equation. TODAY: Finish discussion of * secular equation * forms of various transformations * descriptions of each normal coordinate s -Vectors * definition and properties * imposition of translational and rotational constraints * derivation of G from s -Vectors NEXT TIME - SOME EXAMPLES OF s -VECTOR CALCULATIONS Last time, we derived 0 = | F λ G –1 | left multiply by | G | 0 = | GF λ 1 | must diagonalize GF to get 3N – 6 eigenvalues { λ k } Similarity (not unitary) transformation
5.80 Lecture #26 Fall, 2008 Page 2 of 7 pages λ 1 0 0 0 0 0 0 λ 3N 6 L 1 GFL = = (Although G and F are both real and symmetric, GF is not symmetric, so the diagonalizing transformation is not unitary: L –1 L ) For example, the product of two real and symmetric matrices a b ⎞ ⎛ c d ⎞ ⎛ ac + bd ad bc b a d c bc ad bd + ac is a matrix that is not symmetric. What do we already know about L from prior requirements that T and V must be put into separable forms? Want 2 T = = Q k 2 where Q = L 1 S Q Q k 2 T = q q = S ( D 1 ) D 1 S = S G 1 S [ G 1 ( D 1 ) D 1 ] We had, previously = Q L G 1 L Q So L G –1 L = 1 is required thus L –1 = L G –1 is needed to keep

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

26_580ln_fa08 - MIT OpenCourseWare http/ocw.mit.edu 5.80...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online