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

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

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

This note was uploaded on 02/20/2012 for the course EE 123B taught by Professor Dianahuffaker during the Winter '11 term at UCLA.

Page1 / 65

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

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online