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Year 2 Condensed Matter > Phonons > Flashcards

Flashcards in Phonons Deck (23):
1

Simple 1D Model of Atoms
Description

-atoms with mass M separated by a distance a and connected by springs with spring constant K

2

Simple 1D Model of Atoms
Equation of Motion

-an equation of motion can be written down in terms of the displacement of the nth atom, un:
M d²un/dt² = -2K*un + K(un+! + un-1)
-we therefore expect a time dependence proportional to e^(iωt)
-so:
un'' = -ω²*un

3

Simple 1D Model of Atoms
Difference Equation for Displacements

-sub un'' = -ω²*un into the equation of motion:
-M*ω²*un = K(un+1 + un-1 - 2un)
-therefore we can expect travelling wave solutions of the form:
u±1 = u*e^(inqa)*e^(±iqa)
-where a is the atomic separation distance and q is the phonon wave vector

4

Simple 1D Model of Atoms
Derivation of the Dispersion Relation for Phonons

-substitute the travelling wave solutions for un, un+1 and un-1 into the equation of motion:
-M*ω²*u*e^(inqa) = Ku[e^(i(n+1)qa) + e^(i(n-1)qa) - 2*e^(inqa)]
-cancel u*e^(inqa) and use trigonometric double angle formula for cos to find the dispersion relation:
ω = √[4K/M] * sin(qa/2)

5

Phonon Dispersion Relation

ω = √[4K/M] * sin(qa/2)

-where M is the mass of an atom, K is the spring constant, q is the phonon wave vector and a is the interatomic distance

6

What are phonons?

-the normal modes of a collection of harmonic oscillators characterised by wavevectors k and angular frequencies ω, and have a dispersion relation given by;
ω = √[2K/M] * sin(qa/2)

7

Energies of Phonons

-the same as the energies of the quantum harmonic oscillator:
En = (n + 1/2)*ℏω

8

Distribution Function for Phonons

-phonons are bosons so are governed by the Bose-Einstein distribution function;
n(ω) = 1 / [exp(ℏω/kb*T) - 1]
-note that the chemical potential is missing from the numerator of the Boltzmann factor since bosons are not conserved

9

Relationship Between Phonon Wavevector and Wavelength

q = 2π/λ

10

Minimum and Maximum Phonon Wavelengths

-there is no maximum phonon wavelength since it will be as big as the size of the sample of the metal
-the minimum allowed wavelength is the distance between atoms ~3Å, this corresponds to maximum energy

11

Phonon Heat Capacity
Einstein

-the first person to expain heat capacity due to lattic vibrations was Einstein
-he modelled each atom as a harmonic oscillator independent from the others and calculated the occupation of each energy level using a Boltzmann factor and then summed over all levels
-although this works at high temperatures predicting 3R, it predicts a low temperature heat capacity which does not agree to well with experiment
-the problem with this method is that the atoms don't vibrate independently but collectively

12

Number of States as a Function of Wavevector

-consider a linear chain with spacing L/2π = Na/2π
-and a sphere of radius q for the phonon wave vector
-multiply by three for the three polarisations of the phonons:
Ν(q) = q³V/2π²
-where Ν is greek letter Nu

13

Debye Wavevector Equation

-starting with the number of states in terms of q:
Ν(q) = q³V/2π²
-but Ν=3N, where 3 is the number of degrees of translational freedom and N is the number of atoms
-taking qD to be the Debye wavevector, analogous to kf, we can write:
6π²n = qD³

14

Alternative name for 3N

-3N, the number of translational degrees of freedom multiplied by the number of atoms, is also called the number of normal modes of the system

15

Phonon Density of States Derivation

-start with the number of states in terms of q:
Ν(q) = q³V/2π²
-the velocity of sound is given by vs=ω/q, so:
Ν(q) = V/2π² * ω³/vs³
-to find the density of states, differentiate with respect to ω:
D(ω) = V*3ω²/2π² * 1/vs³
-as for electrons the density of states is defined to be the number of states in the range ω -> ω+dω and is written:
D(ω)dω = V*3ω²/2π²vs³ dω

16

Phonon Derivation of States
Formula

D(ω)dω = V 3ω²/2π²vs³ dω

-the integral of this is obviously equal to 3N

17

Phonon Total Internal Energy

U(T) = ∫ ℏω D(ω) fb(ω,T) dω
-where the integral is taken between 0 and ωD
-after substitution:
U(T) = 3Vℏ/2π²vs³ * ∫ω³/[e^(ℏω/kbT)-1] dω
-where the integral is taken between 0 and ωD

18

Phonon Heat Capacity
Derivation

C = d/dT U(T)
-sub in for U(T):
C = 3Vℏ/2π²vs³kbT² * ∫[e^(ℏω/kbT)*ω^4]/[e^(ℏω/kbT)-1]² dω
-changing variable x=ℏω/kbT we can write:
C = 9Nkb(T/θD)³ *
∫[x^4*e^x]/[e^(x)-1]² dx

19

Debye Temperature Formula

θD = ℏvs/kb * (6π²n)^(1/3)

20

Phonon Heat Capacity at Low Temperatures

-at low temperatures (T<

21

Phonon Heat Capacity at High Temperatures

-at high temperatures (T>>θD), i.e. x<1 sp the integrand becomes ~x²:
C = 3R
-which is the Dulong-Petit Law

22

Heat Capacity From Electrons and Phonons

C = γT + αT³
-the linear term is the electron contribution and the cubic term is the phonon contribution
-data is normally plotted as C/T against T² giving γ as the y intercept and α as the gradient

23

What is the relationship between the velocity of sound and the Debye temperature

-velocity of sound is proportional to the Debye temperature