The Black Body Radiation

The Black Body Radiation = Chapter 4 of Kittel and Kroemer The Planck distribution Derivation Black Body Radiation Cosmic Microwave Background The genius of Max Planck Other derivations Stefan Boltzmann law Flux => Stefan- Boltzmann Example of application: star diameter Detailed Balance: Kirchhoff laws Another example: Phonons in a solid Examples of applications Study of Cosmic Microwave Background Phys 2 (F06) 5 Black Body Radiation

The Planck Distribution*** Photons in a cavity Mode characterized by Number of photons s in a mode => energy s is an integer! Quantification Similar to harmonic oscillator Photons on same "orbital" cannot be distinguished. This is a quantum state, not s systems in interactions! Occupation number ***Planck Distribution Frequency ν Angular frequency ω = 2πν! System = mode, in contact with reservoir of temperature Partition function Z = e s ω = s =0 exp ω Mean number of photons s = sp( s) = se s ω = log Z exp ω s Z s ω = exp ω s = exp ω ε ω = ε = shν = s ω ω ( exp ω ) Phys 2 (F06) 5 Black Body Radiation 2

2 quantifications st Quantification Wave like 2nd Quantification Discrete particles Massive Particles Discrete states Obvious Photons Obvious Cavity modes Planck Phys 2 (F06) 5 Black Body Radiation 3

Maxwell equations in vacuum Photon properties E = ρ = 0 E = B ε o t B = 0 B E = µ o j + εo t = µ E oε o t = E ( j = 0!) c 2 t Using identity E = ( E ) 2 E Solutions E = p = k = ω c = ε c => Photon has zero mass and 2 polarizations Quantifications st quantification: Photons in a cubic perfectly conductive box Electric field should be normal to the surfaces L 2nd quantification: E o exp i k. x ωt 0 πx E x = E x sin ωt cos n x sin n πy y sin n πz z L L L E = E 0 πx sin ωt sin n x y y cos n πy y sin n πz z L L L 0 πx E z = E z sin ωt sin n x sin n πy y cos n πz z L L 2 E = ( ( )) with k = ω c and 2 E c 2 2 (wave equation) t k. E o = 0 n x,n y,n z should be integers that should satisfy the wave equation π 2 n 2 L 2 x + n 2 2 ( y + n z ) = ω 2 c or h 2 n 2 2 4L 2 x + n 2 2 y + n z This the superposition of momenta p x = ± n xh 2L, p = ± n yh y 2L, p = ± n xh y such that 2L p L ( ) = 2 ω 2 c 2 y 2 x + p 2 y + p 2 z = ω c U = E D + B ε ( H ) d 3 x =U = o ( space 2 time 2 E 2 + c 2 B 2 ) d 3 x = ε oe 02 L 3 = s ω sum of 4 cos2 space 3 time 6n x n y n z π Moreover, E = 0 n E = 0 n x E 0 x + n y E 0 y + n z E 0 z = 0 => for each n x,n y,n z,s, 2 independent ways to choose E 0 (2 polarizations) L x Phys 2 (F06) 5 Black Body Radiation 4

Black Body Radiation Radiation energy density between ω and ω+dω? Note that the first quantification give the same prescription on how momentum states are distributed in phase space as for massive particles. Taking into account that we have 2 different polarizations, the state (=mode)density in phase space is The density of energy flowing in the direction (θ,ϕ ) is therefore u ω ( θ,ϕ )d 3 xdωdω = ω energy /photon exp ω 2 d 3 xd 3 p ω = h 3 exp ω density of states or u ω ( θ,ϕ )dωdω = 2 d 3 xd 3 p = d 3 x p 2 dpdω = d 3 x ω 2 dωdω h 3 4π 3 3 4π 3 c 3 mean number of photons ω 3 dωdω 4π 3 c 3 exp ω d 3 x ω 2 dωdω 4π 3 c 3 If we are not interested in the direction of the photons, we integrate on solid angle and we get ω 3 dω 8πhv 3 dv u ω dω = π 2 c 3 exp ω = u ν dν = c 3 exp hν Phys 2 (F06) 5 Black Body Radiation 5

Cosmic Microwave Radiation Big Bang=> very high temperatures! When T 3000K, p+e recombine =>H and universe becomes transparent Redshifted by expansion of the universe: 2.73K Conclusions: Very efficient thermalization In particular no late release of energy No significant ionization of universe since! e.g., photon-electron interactions= Sunyaev-Zel'dovich effect Fluctuations of T => density fluctuations Phys 2 (F06) 5 Black Body Radiation 6

WMAP 3 years 3/6/06 Power Spectrum of temperature fluctuations TT TT TE EE Fitted Cosmological Parameters Phys 2 (F06) 5 Black Body Radiation 7 BB Correlation Power Spectra of temperature fluctuations and E/B polarization

The genius of Max Planck Before Physicists only considered continuous set of states (no 2nd quantization!) Partition function of mode ω dε Z = ω 0 exp ε ε 0 = ε 0 Mean energy in mode ω ε ω = 2 log Z = independent of ω Sum on modes => infinity = ultraviolet catastrophe Max Planck Quantized! ω << exp ω Z = ε ω exp ω ω ε ω = exp ω ω + ω ε ω >> exp ω >> ε ω exp ω ω ω ω ω Phys 2 (F06) 5 Black Body Radiation 8 0

Have we solve everything? No: Zero point energy Phys 2 (F06) 5 Black Body Radiation 9

A technical remark Important to sum with series instead of integral If we had used a continuous approximation Z ω = dx exp ω x = 0 ω instead of e ω ω >> we would have add the same problems as the classical physicists We would have underestimated zero occupancy for large ω! exp x ω For large ω big drop in first bin But discrete sum still gives! x ω Phys 2 (F06) 5 Black Body Radiation 0

Other derivations () Grand canonical method Consider N photons ( Z = exp µ ω )N = N exp µ ω => s = N = ( logz ) = µ exp ω µ What is µ? There is no exchange of photons with reservoir (only exchange of energy) => entropy of reservoir does not change with the number of photons in the black body (at constant energy) Zero chemical potential! => s = exp ω Microscopic picture Photons are emitted and absorbed by electrons on walls of cavity We have the equilibrium γ + e e A + B C µ γ + µ e µ e = 0 µ γ = 0 Special case of equilibrium (cf Notes chapter 3 Kittel & Kroemer Chap. 9 p. 247) this is due to the fact that the number of photons is not conserved σ R = µ γ N γbb U = 0 Phys 2 (F06) 5 Black Body Radiation

Other derivations (2) Microcanonical method To compute mean number photons in one mode, consider ensemble of N oscillators at same temperature and compute total energy U U => ε = lim s = ε N N ω = lim U N N ω Microcanonical= compute entropy σ ( U ) with U = n ω (n photons in combined system) and define temperature at equilibrium as = σ U ( U ) U ( ) s ( ) We have to compute the multiplicity g(n,n) of number of states with energy U= number of combinations of N positive integers such that their sum is n Same problem as coefficient of t n in expansion (cf Kittel chap. ) N t m ( ) N = = g( n, N)t n => n m = 0 t n ( )! g( n, N) = n! ( ) N t= 0 t n t Using Stirling approximation σ ( n) = log( g( n, N) ) N log + n N N n i = n = U i= ω n i 0 => or = σ U = N ω log + = ω U ω log + s Phys 2 (F06) 5 Black Body Radiation 2 = n! N ( N + ) ( N + 2) N + n ( ) = N + n n! ( N )! ( ) + n log ( + N ) n = N log ( + U ) ωn + U ω s = exp ω ( ) log + N ω U

Counting Number of States From first principles (Kittel-Kroemer) In cubic box of side L (perfectly conductive=>e perpendicular to surface) 0 πx E x = E x sin ωt cos nx sin n πy y sin n πz z L L L E = E 0 πx sin ωt sin n x y y cos n πy y sin n πz z L L L 0 πx E z = E z sin ωt sin nx sin n πy y cos n πz z L L L with the wave equation constraint π 2 L 2 n x 2 + n 2 2 ( y + n z ) = ω2 c 2 n2 = ω2 L 2 π 2 c 2 => ω has specific values! ( st quantification ) 2 degrees of freedom (massless spin particle) ("2nd quantification") E = 0 n E = 0 n x E 0 x + n y E 0 y + n z E 0 z = 0 U = E D + B ε ( H ) d 3 x =U = o ( space 2 time 2 E 2 + c 2 B 2 ) d 3 x = ε oe 02 L 3 = s ω sum of 4 cos2 space 3 time 6n x n y n z π => number of states in mode ω is 2xnumber of integers satisfying wave equation constraints n 2 x + n 2 y + n 2 z = ω 2 L 2 Sum on states can be replaced by integral on quadrant of sphere of radius squared π 2 c 2 Energy in ω ω+dω ω ω U ω dω = 2 ( exp ω ) = 2 ( n U ω dω = 2 4π ω n 2 dn = πω 2 L 3 ωdω x,n y,n z exp ω ) d 3 n quadrant ω 3 dω 8 π 3 c 3 ( exp ω ) ( exp ω ) Phys 2 (F06) 5 Black Body Radiation 3 0 πx B x = B x cos ωt sin nx cos n πy y cos n πz z L L L B = B 0 πx c cos ωt cos n x y y sin n πy y cos n πz z L L L 0 πx B z = B z cos ωt cos nx cos n πy y sin n πz z L L L B 0 = π n E 0 ωl B 0 2 = n 2 π 2 B 0 2 ω 2 L 2 u ω dω = U ω L 3 dω = π 2 c 3 exp ω

Fluxes Energy density traveling in a certain direction So far energy density integrated over solid angle. If we are interested in energy density traveling traveling in a certain direction, isotropy implies u ω ( θ,ϕ )dωdω = Note if we use ν instead of ω Flux density in a certain direction=brightness (Energy /unit time, area, solid angle,frequency) opening is perpendicular to direction I ν dνdadω = dt dν u cdtda ν dω = 2hν 3 dνdadω energy in c 2 exp hν cdt cylinder Flux density through a fixed opening (Energy /unit time, area,frequency) Phys 2 (F06) 5 Black Body Radiation 4 u ν ( θ,ϕ )dνdω = J ν dνda = dt dν 2π dϕ u ν cdt cosθda d cosθ = 0 0 energy in cylinder! ω 3 dωdω 4π 3 c 3 exp ω c 3 2hν 3 dνdω exp hν c 2 2πhν 3 dνda exp hν = c 4 u dvda ν cdt da da θ n

Total Energy Density Integrate on ω Stefan-Boltzmann Law*** u 0 ω dω = 0 ω 3 π 2 c 3 exp ω dω Change of variable 4 u 0 ω dω = 3 π 2 c 3 0 x 3 dx e x u = ω x = ω π 2 5 3 c 3 4 = a B T 4 0 x 3 dx e x = π4 5 with a B = π2 k 4 B 5 3 c 3 Total flux through an aperture Integrate Jn = multiply above result by c/4 π 2 J = 60 3 c 2 4 = σ B T 4 with σ B = π2 k 4 B 60 3 c 2 = 5.67 0 8 W/m 2 /K 4 Stefan-Boltzmann constant! Phys 2 (F06) 5 Black Body Radiation 5

Detailed Balance Definition: A body is black if it absorbs all electromagnetic radiation incident on it Usually true only in a range of frequency e.g. A cavity with a small hole appears black to the outside Detailed balance: in thermal equilibrium, power emitted by a system=power received by this system! Otherwise temperature would change! Consequence: The spectrum of radiation emitted by a black body is the black body spectrum calculated before Cavity Black Body Proof : c 4 Au BB = c 4 Au cavity u BB = u cavity We could have inserted a filter u BB ( ω)dω = u cavity ( ω)dω Phys 2 (F06) 5 Black Body Radiation

Absorptivity, Emissivity: Absorptivity =fraction of radiation absorbed by body Emissivity = ratio of emitted spectral density to black body spectral density. Kirchhoff law: Emissivity=Absorptivity Black Body Grey Body received power = a( ω)u BB ( ω)dω = emitted power = e( ω )u BB ( ω)dω a( ω) = e( ω) Applications: Johnson noise of a resistor at temperature T We will come back to this! Kirchhoff law Rayleigh Jeans =>Spectral density = k B T d V 2 noise dν Superinsulation (shinny) = 4k B TR Phys 2 (F06) 5 Black Body Radiation 7

Entropy Entropy, Number of photons Method (Kittel) use thermodynamic identity+ 3rd law du du = dσ at constant volume dσ = d V Third law : σ = 0 d Method 2 : sum of entropy of each mode d du d V = dσ = 4π2 V 45 3 c 3 3 = 4 3 = 4π 2 V 5 3 c 3 2 a B k B VT 3 σ = σ ω = p s log p s = log Z ω + s ω = log exp ω + Remembering density of states σ ω Finally Number of photons Starting from s e Classical integral x dx = ( s )!ζ( s) with ζ ( 3).22 0 = N γ = V2ζ ( 3) π 2 c 3 3 3 = 30ζ ( 3) a B π 4 VT 3 0.37 a B VT 3 k B k B Note that entropy is proportional to number of photons! Phys 2 (F06) 5 Black Body Radiation 8 ( ( )) d 3 x ω 2 dω π 2 c 3 = V u 3 + u (integrating first term by part) s = exp ω σ = 4 u 3 V = 4 3 x s N γ = a B k B VT 3 V π 2 c 3 0 ω 2 dω exp ω ω ( ( exp ω ) ) σ 3.6N γ

Spectrum at low frequency log I v d Rayleigh-Jeans region (= low frequency) A e Ω e Ω r A r if ω << ε u ω dω ω 2 dω u π 2 c 3 ν dν 8πν 2 ω dν c 3 Power / unit area/solid angle/unit frequency =Brightness 2hν 3 dνdadω I ν dνdadω = c c 3 exp hν 2ν 2 dνdadω = 2dνdAdΩ c 2 λ 2 Power emitted / unit (fixed) area J ω dω = J ν dν ω2 dω 4π 2 c 2 = 2πdν λ 2 Power received from a diffuse source A e = Ω r d 2 Ω e = A r d 2 If telescope is diffraction limited: Ω r A r = λ 2 Detected Power ( polarization) Antenna temperature Phys 2 (F06) 5 Black Body Radiation 9 log v dp dν = I νω e A e = 2Ω ea e λ 2 dp dν = 2 = 2Ω ra r λ 2 dp dν (polarization) = T A = dp k B dν (polarization)

Applications Many! e.g. Star angular diameter Approximately black body and spherical! Spectroscopy =>Effective temperature Apparent luminosity l =power received per unit area A r Ω e d Ω e = A r d 2 => Baade-Wesserlink distance measurements of varying stars Oscillating stars (Cepheids, RR Lyrae) Supernova Luminosity/temperature variation Phys 2 (F06) 5 Black Body Radiation 20 l = LΩ e = 4π A But power output r L = 4πr 2 4 σ B T eff l = r 2 d 2 σ 4 BT eff l angular diameter = r d = Doppler velocity v // = Δλ λ c = dr // dt dr // If spherical expansion = dr dt dt dθ dt = d dt 4 σ B T eff l σ B T eff 4 d = L 4πd 2 dr // dt dθ dt = d dr dt

D Phonons in a solid Phonons: Quantized vibration of a crystal described in same way as photons mean occupancy Same as for photons but 3 modes (3 polarizations) ε = ω p = k = ω c s s = exp ω ε s = s ω Sound Velocity 3 d 3 xd 3 p integrating over angles 3 d 3 x p 2 dp h 3 2π 2 3 Maximum energy = Debye Energy But half-wave length smaller than lattice spacing is meaningless = d 3 x 3ω 2 dω 2π 2 3 c s D( ω )dω <=> solid is finite: If N atoms each with 3 degrees of freedom, at most 3N modes Debye ω approximation: ω D k = ( N atoms = L a isotropic Phys 2 (F06) 5 Black Body Radiation 2 )3 k x (cubic lattice) < π a 0 ω where a is the lattice spacing maximum energy ω D ( ) VD ω dω = 3N where D ω ( ) is the density of states k= ω or p= ε with velocity c s = constant c s c s

Debye law Phonons in a Solid ω D VD ( ω ) ω dω = V 3 ω 2 D 0 2π 2 3dω = 3N ω D = 6N 0 c s V π 2 c s 3 ( ) 3 c s a 3 = 6π 2 U = d 3 x 0 k D ω exp ω 3 k 2 dk 2π 2 = V ω D 0 ω exp ω 3 ω 2 dω 2π 2 c s 3 for << ω D U = 3 4 2 3 π 2 c s 3 V x 3 dx = 0 e x π 2 0 3 c V k 4 3 BT 4 s Introducing the Debye temperature T D = θ = c s k B U = 3 5 π4 k B N T 4 3 C V = 2 T D 5 π4 k B N T T D 3 6π 2 N V /3 = ω D k B σ = 2 5 π4 k B N T T D 3 Phys 2 (F06) 5 Black Body Radiation 22

Applications Calorimetry: Measure energy deposition by temperature rise ΔE ΔT = ΔE need small C C Heat Capacity Bolometry: Measure energy flux F by temperature rise ΔT = ΔF G Very sensitive! Sky Load Chopping e.g., between sky and calibration load ΔF need small G but time constant= C G Heat Conductivity limited by stability small heat capacity C Heat capacity goes to zero at low temperature 5 C T 3 Fluctuations σ ΔE = k B T 2 C T 2 M 2 Wide bandwidth (sense every frequency which couples to bolometer) Study of cosmic microwave background Bolometers 4K 300mK 00mK Search for dark matter particles 70g at 20mK Phys 2 (F06) 5 Black Body Radiation 23