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Planck Blackbody Radiation

Stefan-Boltzmann Law

2 4 Z3 2 Radiation energy flux density JU = σ BT σ B ≡ π kB 60 c

Entropy of thermal photons:

Peak frequency in spectrum tells temperature Zω τ ≈ 2.82

Big Bang: background radiation from expansion of universe

Today Chapter 4 (part 2) Johnson noise, Debye model of phonons

Johnson noise (thermal electrical noise) Phonons in solid: Debye model Heat capacity of a solid Thermal Electrical Noise

V

Discovered by J.B. Johnson – noise ~ R (resistance)

Explained by H. Nyquist (1928) Planck law in one dimension

Thermal Electrical Noise

Nyquist theorem: V 2 = 4Rτ∆f

Frequency bandwidth

V For R=R’ the noise power is I = maximum in respect to R’ R + R ' 2 2 2 V R V V R ' P = = P = I 2 R ' = 2 2 (2R) 4R (R + R ') The origin of “4” Thermal Electrical Noise

Assume lossless transmission along line L – electromagnetic system in one dimension.

One mode in frequency range δ f = c '/ 2L For τ > > Z ω energy ε ≈τ Transmission line photon modes: And energy within ∆f is: (τ /δ f )∆f = 2τ L∆f / c '

This energy is absorbed by resistors f1 = c '/ 2L in time ∆t = L/c’, and power within ∆f

f2 = 2c '/ 2L 2τ∆f

fn = nc '/ 2L

Thermal Electrical Noise

This energy is absorbed by resistors in time ∆t = L/c’, and power density is 2τ∆f This power is absorbed by two resistors, i.e. one resistor absorbs: τ∆f Transmission line photon modes: In thermal equilibrium the absorbed energy should equal to the emitted: V 2 P = =τ∆f 4R

2 Nyquist theorem: V = 4Rτ∆f Thermal Electrical Noise

Measured Johnson noise Gaussian distribution

Time dependence

Flat frequency response

http://www.physics.ucdavis.edu/Classes/Physics116/P116C_Notes/Noise.pdf

Phonons in Solids

I’m moving! Phonons in Solids

Two transverse modes

+ one longitudinal mode

Phonon: Quantum of energy of an elastic wave (sound particles) Thermal average of number if phonons at frequency ω is given by Planck: 1 s(ω) = exp(Zω τ ) −1

Phonons in Solids

Analogy with photons:

Waves below represent local displacement of wiggling atoms in the solid.

one phonon two phonons

one phonon two phonons Lattice Vibrations

Harmonic oscillations a Force from the left “spring”

FL = K (un−1 − un )

Force from the right “sping”:

FR = K (un+1 − un ) un-2 un-1 un un+1 un+2 Total force on nth atom: x xn = na F = K (un−1 − 2un + un−1 )

Newton’s law of motion:

2 d un "" ma ≡ m ≡ mun = K (un+1 − 2un + un−1 ) dt 2

travelling wave solutions:

Lattice Vibrations "" mun = K (un+1 − 2un + un−1 )

Every atom is equivalent: travelling wave solutions Plug this into the differential equation:

Cancel the common factor

Dispersion of a linear chain of atoms Lattice Vibrations

Dispersion of a linear chain of atoms

Wavelength λ = 2π/k or k = 2π/λ

Debye Approximation

Velocity does not depend on frequency Debye: (1) Pretend ω = v|k| all the way 1884-1966 (2) Set cutoff ωD to get the right number of modes.

Not all frequencies are available This ensures the high and low T results are correct! If ω = v|k| | all the way up, then the energy levels of phonons are just like photons, with v in place of c....

There is limiting ω Do Statistical Mechanics On The Photons

Start with one photon mode

Recognize this as:

Partition function of one photon mode 6 e at energy ur ct Le

Stat Mech on one Photon Mode

Planck Distribution

Function 6 e ur ct Le Phonons in Solids

two polarizations:

∞ Zω Counting U 2 = ƒ Zω τ Photons nx ,ny ,nz e −1

nmax Zω Counting U = 3 ƒ Zω τ Phonons nx ,ny ,nz e −1

Hthroewe pmolarnizya tmionosd: e tws?o transverse; one longitudinal

(Countably) infinite number of photon modes. They can have as high frequency as they like.

Finite number of Phonon modes. Cut off at small wavelength (high frequency) by the lattice spacing!

Convert the Sum to an Integral

3 N 3 N 3 N

− 3 N − 3 N − 3 N 3 N

n Surface area of D a sphere

PROBLEM: it is OK for infinite limits, but for finite limits cube does not look like a sphere Convert the Sum to an Integral

3 N 3 N 3 N

− 3 N − 3 N − 3 N

nD

3 N

Debye: use a sphere of radius nD that contains all 8N particles:

6N 4 3 3 π n = 8N nD = Debye Constant 3 D π

How good is this approximation?

Counting phonon modes

Lets count al modes:

All modes

Weight of each mode is 1

1 6N = π = 3N 2 π

Number of modes: 3 N = 3 x (atoms) – as expected! Total Energy

Sum the average energy in each phonon mode

x = Zω τ

Spot the Debye approximation! v = speed of sound in the solid It’s the low frequency velocity -- consistent with a low temperature approximation

Total Energy

Debye Temperature Total Energy: Low Temperature Limit

Low temperature approximation:

Phonon Heat Capacity

At low temperatures:

Phonon Heat Capacity Low Temperature Limit

Debye T3 Law Phonon Heat Capacity

At high temperatures: τ >>τ D

ex ≈1+ x xD x3 — x2dx = D 3 3 2 ≈Z v 6π N ’ 0 3 ∆ ÷ « 3Vτ 3 ◊ U = 3Nτ

C 3N Phonon Heat Capacity V = High Temperature Limit

Phonon Heat Capacity

High Temperature: Equipartition!

Degrees of freedom: 3 N atoms x 2 = 6N (They're harmonic oscillators)

Phonons High Temperature Equipartition

Why did phonons go to the equipartition limit, when our box of photons did not? Phonon Heat Capacity

3N Debye TD: C 2230 K

Si 645 K 2N Fe 470 K Al 428 K Mg 400 K 1N Cu 343 K

τ τ D Low temperature High temperature

Phonons in Solids

Average thermal occupation of a phonon mode

Phonon heat capacity (low temperature), Debye law

Phonon heat capacity (high temperature is equipartition)

We’ll show both of these heat capacity limits. To calculate the total energy, count phonons… What We Did Today

Johnson’s noise in resistors: Planck law in one dimension

V 2 = 4Rτ∆f Debye model of solids: phonons

Debye temperature:

Low temperature limit: τ <<τ D

(Debye law)

High temperature limit: τ >>τ D