123B_1_EE 123B W11 lecture 6, chapter 4 part 2

# 123B_1_EE 123B W11 lecture 6, chapter 4 part 2 - Chapter 4...

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Unformatted text preview: Chapter 4. Phonons Lecture 6 1/20/11 Last Lecture: Chapter 4 – Phonons • Hooke’s Law – force, displacement • Newton’s Law – force, mass, acceleration • Oscillation modes – longitudinal, transverse • Dispersion relation • Group velocity • Brillouin zone boundaries and limit • Diatomic chains This Lecture – Chapter 4 continued Reminder: Homework #3 due Feb. 2 Tuesday • 4.1: For a monatomic chain, detemine total energy in an elastic wave. • 4.3 For a diatomic basis, calculate wave amplitude ratio , u/v at K=π/a . • Extra – what is the dispersion relation for a diatomic basis if m1=m2? Sketch. • 4.5 For diatomic basis, find dispersion relationship at K=0 and K=π/a. Phonon • What are they? – Propagation lattice vibrations that carry energy through a crystal Crystal can be considered an array of mass centers connected by springs Harmonic Oscillator – Hooke’s law • M oscillates with kinetic and potential energy. n=0 n=1 n=2 ħw k E M E n μ ϖ E n = (ν + 1 2 29 ηϖ mode number n ≡ ϖ = 2πγ = 2π χ λ ϖ ↑, Ε ↑,λ ↓ ϖ = κ / Μ , k = Μϖ 2 • We use Newton’s law to derive ϖ(κ29 Energy in a Crystal Definition: • Kinetic energy is the sum of the individual kinetic energies each of the form E μ 1 2 Μυ σ 2 σ ∑ What is u? How do we relate F to E? How do we relate u to ω (amplitude to frequency)? •Force between atoms s and s+1 is –C(u s – u s +1) • Potential energy associated with the stretching of this bond is C(u s −u s+1 ) 2 Newton’s Law F Ma = 2 2 ( ) d r t M dt = v ( , ), r t φ = ∇ v X=0 ∆X ˆ r x = v ( , ) ( , ) d F r t r t dx φ = v v ( ) is position of atom r t v In one dimension: potential energy φ ≡ ( 0) x φ = = ( ) x x k x φ = ∆ ≠ = ∆ Ф arises from interaction between atom and rest of crystal. displacement φ μ 2 3 ( ) 1 ... r r r r φ = + + + + Liner nearest neighbor interaction Monatomic chain - Phonons • Derive equation of motion S S+1 S-1 S+2 u S u S+1 u S-1 u S-u S- 1 u S- u S+1 2 1 1 2 ( 2 ) S S S S d u C u u u M dt +- +- = 2.) Equation of motion – Hooke meets Newton 1 ( ) S S S S nn F C u u ± =- ∑ = Χ Σ , Σ + 1 (υ Σ- υ Σ + 1 29 + Χ Σ , Σ-1 (υ Σ- υ Σ-1 29 “spring” constant C S,S+1 =C S,S-1 =C 1.) Force- similar to Hookes law We assumed • Nearest neighbors • Uniform force constant u S ( t ) μ εξπ(-ιϖτ 29 d 2 u S dt 2 = -ϖ 2 υ Σ 3.) Write wave solution – Find Dispersion Relation – monatomic chain. 2 1 1 2 ( 2 ) S S S S d u C u u u M dt +- +- = d 2 u S dt 2 = -ϖ 2 υ Σ C ( u S + 1 + υ Σ-1- 2υ Σ 29 = -Μϖ 2 υ Σ 1 exp( )exp( ) S u u isKa iKa ± = ± 5.) Use wave solution: a: spacing between atomic planes K : the wavevector 4.) Combine equation of motion and wave solution: Acoustic, Optical Phonons - contd....
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## This note was uploaded on 02/20/2012 for the course EE 123B taught by Professor Dianahuffaker during the Winter '11 term at UCLA.

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123B_1_EE 123B W11 lecture 6, chapter 4 part 2 - Chapter 4...

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