The Black Body Radiation - Berkeley Cosmology Groupcosmology.berkeley.edu/Classes/S2012/Physics_112/lectures/chapter... · Phys 112 (S12) 6 Black Body Radiation 1 B.Sadoulet The Black

B.SadouletPhys 112 (S12) 6 Black Body Radiation 1

The Black Body Radiation= Chapter 4 of Kittel and Kroemer

The Planck distributionDerivation

Black Body RadiationCosmic Microwave BackgroundThe genius of Max Planck

Other derivationsStefan Boltzmann law

Flux => Stefan- BoltzmannExample of application: star diameter

Detailed Balance: Kirchhoff lawsAnother example: Phonons in a solid

Examples of applicationsStudy of Cosmic Microwave Background


The Planck Distribution***Photons in a cavity

Mode characterized by

Number of photons s in a mode => energys is an integer ! Quantification

Similar to harmonic oscillatorPhotons on same "orbital" cannot be distinguished. This is a quantum state, not s systems in

interactions!

Occupation number

! System =1 mode , in contact with reservoir of temperaturePartition function

Mean number of photons

***Planck Distribution

Frequency νAngular frequency ω = 2πν

ε = shν = sω

Z = e− sω

τ

s =0

∞

∑ =1

1 − exp − ωτ

⎛ ⎝

⎞ ⎠

s = sp s( ) =1Zs

∑ ss∑ e

− sωτ = −

∂ log Z

∂ ωτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟

=exp − ω

τ⎛ ⎝ ⎜ ⎞

⎠ ⎟

1− exp − ωτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟

s =1

exp ωτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1

εω =ω

expω

τ( ) − 1


2 quantifications

Massive Particles

Photons

1st Quantification

Wave like

Discretestates

Obvious Cavity modes

2nd Quantification

Discrete particles

Obvious Planck


Photon propertiesMaxwell equations in vacuum

=> Photon has zero mass and 2 polarizations

Quantifications1st quantification: Photons in a cubic perfectly conductive box Electric field should be normal to the surfaces +

2nd quantification:

∇ ⋅ E =

ρεo

= 0 ∇ × E = −

∂ B ∂t

∇ ⋅B = 0

∇ ×B = µo

j + εo

∂E∂t

⎛⎝⎜

⎞⎠⎟= µoεo

∂E∂t

=1c2

∂E∂t

(j = 0!)

�

Using identity ∇ × ∇ × E = ∇ ⋅ ∇ ⋅ E ( )−∇2

E

⇒∇2 E = 1

c2∂2 E ∂t2 (wave equation)

Solutions

E = E o exp i

k . x −ωt( )( ) with k =

ωc

and k . E o = 0

p = k = ω

c=εc

�

Ex = Ex0

sin ωt cos nxπx

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ sin ny

πy

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ sin nz

πz

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

Ey

= Ey0 sin ωt sin nx

πx

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ cos ny

πy

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ sin nz

πz

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

Ez = Ez0

sin ωt sin nxπx

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ sin ny

πy

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ cos nz

πz

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

�

nx ,ny ,nz should be integers that should satisfy the wave equationπ2

L2 nx2 + ny

2 + nz2( ) = ω 2

c2 or h2

4L2 nx2 + ny

2 + nz2( ) =

2ω 2

c2

This the superposition of momenta

px = ± nxh2L

, py = ±nyh2L

, py = ± nxh2L

such that px2 + py

2 + pz2 = ω

c

L

L

y

x

∇iE = 0

U =

12E ⋅D +B ⋅H( )

timespace∫ d 3x =U =

εo2

E2 + c2B2( )time

space∫ d 3x = εoE02V16

= sω ⇐ sum of 4 cos2 or sin2

∇iE = 0⇔ E

x

0nx + Ey

0ny + Ez0nz = 0 (This is why we need the cosines): 2 polarizations


Black Body RadiationRadiation 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

or

If we are not interested in the direction of the photons, we integrate on solid angle and we get

�

2 d3xd 3 ph 3 = d

3x p 2dpdΩ4π 3 3 = d 3x ω 2dωdΩ

4π 3c3

uωdω = ω 3dω

π2c3 exp ωτ

⎛⎝⎜

⎞⎠⎟ −1⎛

⎝⎜⎞⎠⎟

= uνdν = 8πhv3dv

c3 exp hντ

⎛⎝⎜

⎞⎠⎟ −1⎛

⎝⎜⎞⎠⎟

�

uω θ ,ϕ( )d 3xdωdΩ = ωenergy/photon

1

exp ωτ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

⎛ ⎝ ⎜

⎞ ⎠ ⎟

mean numberof photons

2 d3xd 3 ph 3

density ofstates

= ω

exp ωτ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

⎛ ⎝ ⎜

⎞ ⎠ ⎟

d 3x ω 2dωdΩ4π 3c3

�

uω θ ,ϕ( )dωdΩ = ω 3dωdΩ

4π3c3 exp ωτ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

⎛ ⎝ ⎜

⎞ ⎠ ⎟


Cosmic Microwave Radiation

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

Big Bang=> very high temperatures!When T≈3000K, p+e recombine =>H and universe becomes transparentRedshifted by expansion of the universe: 2.73K


WMAP 5 years 3/5/08

EE

TT

Power Spectrum of temperature fluctuations

Fitted Cosmological Parameters

TE

EE

Correlation Power Spectra of temperature fluctuations and E polarization


BeforePhysicists only considered continuous set of states (no 2nd quantization!)

Partition function of mode ω

Mean energy in mode ω independent of ω

Sum on modes => infinity =“ultraviolet catastrophe”

Max PlanckQuantized!

Cut off at The energy of 1 photon cannot be smaller than ! => finite integral

The genius of Max Planck

Zω =dεε00

∞∫ exp −ετ

⎛⎝⎜

⎞⎠⎟=

τε0

εω =τ 2∂log Z

∂τ= τ

Z =1

1− exp − ωτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟

εω =ω

exp ωτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1

�

εω

ω

�

τ

�

τ

�

ω << τ exp ωτ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ≈ 1 + ω

τ ⇒ εω ≈ τ ω >> τ exp ω

τ⎛ ⎝ ⎜

⎞ ⎠ ⎟ >> 1⇒ εω ≈ ω exp − ω

τ⎛ ⎝ ⎜

⎞ ⎠ ⎟ → 0

B.SadouletPhys 112 (S12) 6 Black Body Radiation

Have we solved everything?No: Zero point energy

We have again an infinite sum!

This is related to the problem of the vacuum energy, dark energy and the cosmological constant!

9

Each mode will contribute 1

2ω


A technical remark

Important to sum with series instead of integral

xω

For large ω big drop in first binBut discrete sum still gives 1!

exp −xω

τ⎛⎝⎜

⎞⎠⎟

1

If we had used a continuous approximation

Zω = dx exp −ωxτ

⎛⎝⎜

⎞⎠⎟0

∞

∫ =τω

instead of 1

1− e−ωτ

ω >>τ⎯ →⎯⎯ 1

we would have add the same problems as the classical physicistsWe would have underestimated zero occupancy for large ω !


Other derivations (1)Grand canonical method

Consider N photons

=>

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! =>

Microscopic picturePhotons are emitted and absorbed by electrons on walls of cavityWe have the equilibriumSpecial 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

�

Z = expµ− ω( )N

τ⎛

⎝ ⎜

⎞

⎠ ⎟

N∑ = 1

1− exp µ− ωτ

⎛ ⎝ ⎜

⎞ ⎠ ⎟

�

s = N =∂ τ logZ( )

∂µ= 1

exp ω − µτ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

�

∂σ R

∂NγBB U

=−µγ

τ= 0

s =1

exp ωτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1

γ + e↔ e

�

A +B↔Cµγ + µe − µe = 0⇒ µγ = 0


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

=>Microcanonical= compute entropy

and define temperature at equilibrium as

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 tn in expansion (cf Kittel chap. 1)

=>

Using Stirling approximation

=> or

ε = lim

N→∞

UN

s =εω

= limN→∞

UNω

σ U( ) with U = nω (n photons in combined system)

g n, N( ) =1

n!

∂n

∂tn1

1− t( )Nt= 0

=1

n!N ⋅ N + 1( ) ⋅ N + 2( ) ⋅ ⋅ ⋅ N + n − 1( ) =

N + n − 1( )!

n! N − 1( )!

t m

m= 0

∞

∑⎛ ⎝ ⎜

⎞ ⎠ ⎟ N

=1

1 − t( )N = g n, N( )tn

n∑

σ n( ) = log g n, N( )( ) ≈ N log 1+

n

N( ) + n log 1 + N

n( ) = N log 1 +U

ωN( ) + U

ωlog 1 +

Nω

U( )

s =1

exp ωτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1

nii=1

N

∑ = n = Uω

ni ≥ 0

1τ=∂σ∂U

⇒ τ U( ) ⇒U τ( )⇒ s τ( )

1τ

=∂σ∂U

=1ω

log 1+NωU

⎛ ⎝ ⎜ ⎞

⎠ ⎟ =

1ω

log 1+1s

⎛ ⎝ ⎜

⎞ ⎠ ⎟


Counting Number of States From first principles (Kittel-Kroemer)

In cubic box of side L (perfectly conductive=>E perpendicular to surface)

with the wave equation constraint

=> ω has specific values! (“1st quantification”) 2 degrees of freedom (massless spin 1 particle)("2nd quantification")

=> number of states in mode ω is 2xnumber of integers satisfying wave equation constraints

Sum on states can be replaced by integral on quadrant of sphere of radius squared Energy in ω ω+dω

π2

L2nx2 + ny2 + nz2( ) = ω2

c2⇔ n2 = ω2L2

π2c2

�

∇ ⋅ E = 0⇔ n ⋅

E = 0⇔ nxEx

0 + nyEy0 + nz Ez

0 = 0

�

B 0 = π

ωL n × E 0

⇒ B 0

2= n 2π2

ω 2L2 B 0

2

Uωdω = 2

4π

8n

2

dnω

expω

τ( ) − 1

=πω 2L3

π3

c3

ωdω

expω

τ( ) − 1

Uω dω = 2ω

expω

τ( ) − 1nx ,ny ,n z

∑ = 2ω

expω

τ( )− 1quadrant∫ d 3n

uωdω =UωL3

dω =1

π2c3ω3dω

exp ωτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1

nx2 + ny2 + nz2 =ω 2L2

π2c2

�

Ex = Ex0

sin ωt cos nxπx

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ sin ny

πy

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ sin nz

πz

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ Bx = Bx

0cos ωt sin nx

πx

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ cos ny

πy

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ cos nz

πz

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

Ey

= Ey0 sin ωt sin nx

πx

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ cos ny

πy

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ sin nz

πz

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ By

= By0c cos ωt cos nx

πx

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ sin ny

πy

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ cos nz

πz

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

Ez = Ez0

sin ωt sin nxπx

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ sin ny

πy

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ cos nz

πz

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ Bz = Bz

0cos ωt cos nx

πx

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ cos ny

πy

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ sin nz

πz

L

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

�

U = 12 E ⋅ D + B ⋅

H ( )time

space∫ d 3 x = U = ε o

2E 2 + c2B 2( )

timespace∫ d 3 x = ε o E 02L3

16nxnynzπ3 = sω ⇐ sum of 4 cos2


FluxesEnergy 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

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

Flux density through a fixed opening (Energy /unit time, area,frequency)

�

uω θ ,ϕ( )dωdΩ = ω 3dωdΩ

4π3c3 exp ωτ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

⎛ ⎝ ⎜

⎞ ⎠ ⎟

�

uν θ ,ϕ( )dνdΩ = 2hν 3dνdΩ

c3 exp hντ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

⎛ ⎝ ⎜

⎞ ⎠ ⎟

�

Iν dνdAdΩ = 1dtdν uν cdtdA

energy incylinder

dΩ = 2hν 3dνdAdΩ

c2 exp hντ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

⎛ ⎝ ⎜

⎞ ⎠ ⎟

�

Jν dνdA = 1dtdν dϕ uν cdt cosθdA

energy in cylinder d cosθ

0

1∫0

2π∫ = 2πhν 3dνdA

c2 exp hντ

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

⎛ ⎝ ⎜

⎞ ⎠ ⎟

= c4uν dvdA

!

θdA

n

�

cdt

�

dA

�

cdt


Stefan-Boltzmann Law***Total Energy Density

Integrate on ω

Change of variable

Total flux through an aperture Integrate Jn= multiply above result by c/4

Stefan-Boltzmann constant!

uω0∞∫ dω =

1π2c3

ω 3

exp ωτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −10

∞∫ dω

ω → x = ω

τ

uω0

∞∫ dω =

τ 4

3π2c3x3dxex −10

∞∫

x3dxex −10

∞∫ =

π4

15

u = π2

153c3 τ4 = aBT4 with aB =

π2kB4

153c3

J =

π2

603c2 τ4 =σBT4 with σ B =

π2kB4

603c2 = 5.67 10−8 W/m2/K4


Entropy, Number of photonsEntropy

• Method 1 (Kittel) use thermodynamic identity+ 3rd lawat constant volume

• Method 2 : sum of entropy of each mode

Remembering density of states

Finally Number of photons

Starting from

Classical integral=

Note that entropy is proportional to number of photons!

dU = τdσ

dσ

dτ V=

dU

dττ

=

du

dτV

τ=4π2V

153c3τ 2

Third law : σ =

dσdτ0

τ∫ =

4π2V453c3 τ

3 =43aBkB

VT3

σω = − ps log ps = log Zω +s ω

τ= − log 1 − exp

−ω

τ( )( )s∑ +

ω

τ expω

τ( )− 1( )σ = σω∫ d3xω

2dωπ2c3 =V 1

3uτ

+uτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ (integrating first term by part)

σ =43uτV =

43aBkB

VT3

s =1

exp ωτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1

Nγ =V

π2c3ω 2dω

exp ωτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1⎛

⎝ ⎜ ⎞

⎠ ⎟

0∞∫

xs−1

ex −10∞∫ dx = s −1( )!ζ s( ) with ζ 3( ) ≈1.22

Nγ =

V2ζ 3( )π2c33

τ3 =30ζ 3( )π4

aBkB

VT3 ≈ 0.37aBkBVT3

σ ≈ 3.6Nγ

B.SadouletPhys 112 (S12) 6 Black Body Radiation

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

We could have inserted a frequency filter or a direction selector

Cavity BlackBody

Proof : c4AuBB =

c4Aucavity ⇒ uBB = ucavity

Detailed Balance

uBB ω ,θ,ϕ( )dωdΩ = ucavity ω ,θ,ϕ( )dωdΩ


Absorptivity, Emissivity:Absorptivity =fraction of radiation absorbed by bodyEmissivity = ratio of emitted spectral density to black body spectral density.

Kirchhoff law: Emissivity=Absorptivity

Applications: • Johnson noise of a resistor at temperature T

• Superinsulation (shinny)

received power = a ω( )uBB ω( )dω = emitted power = e ω( )uBB ω( )dω

a ω( ) = e ω( )

Rayleigh − Jeans =>Spectral density = kBT⇒d Vnoise

2

dν= 4kBTR

Kirchhoff law

BlackBody

GreyBody

We will come back to this !


ApplicationsMany!

e.g. Star angular diameterApproximately black body and spherical!

Spectroscopy =>Effective temperatureApparent luminosity l =power received per unit area But power output

=> Baade-Wesserlink distance measurements of varying stars • Oscillating stars (Cepheids, RR Lyrae) • Supernova Luminosity/temperature variation

Doppler

If spherical expansion

l = LΩe4π

1Ar

=L

4πd2

L = 4πr2σ BTeff4

⇒ l = r2

d2σBTeff4

�

angular diameter = r⊥d

= lσ BTeff

4

dθdt

=ddt

lσ BTeff

4 =1ddr⊥dt

velocity v// =Δλλc =

dr//dt

⇒ d =

dr//dtdθdt

Ωe =Ard2

Ωe

Ar

d

dr//dt

=dr⊥dt


Spectrum at low frequency

Rayleigh-Jeans region (= low frequency)

Power / unit area/solid angle/unit frequency =Brightness

Power emitted / unit (fixed) area

Power received from a diffuse source

If telescope is diffraction limited:

Detected Power (1 polarization)

Antenna temperature

uωdω ≈ τω 2dωπ 2c3 uνdν ≈ τ

8πν 2dνc3 if ω << τ εω ≈ τ

IνdνdAdΩ = c 2hν 3dνdAdΩ

c3 exp hντ

⎛⎝⎜

⎞⎠⎟ −1⎛

⎝⎜⎞⎠⎟

≈ τ 2ν 2dνdAdΩc2 = τ 2dνdAdΩ

λ2

Jωdω = Jνdν ≈ τω2dω4π2c2

= τ2πdνλ2

Ωr

Ar

Ωe

AedPdν

= IνΩeAe =2τΩeAeλ2

=2τΩrArλ2

Ωr Ar = λ2

dPdν(1polarization) = τ

TA =1kB

dPdν(1polarization)

Ωe =Ard2

Ae =Ωrd2

d

log v

log Iv

dPdν

= 2τ


Phonons in a solidPhonons:

Quantized vibration of a crystaldescribed in same way as photons

mean occupancySame as for photons but • 3 modes (3 polarizations)

• Maximum energy = Debye EnergyBut half-wave length smaller than lattice spacing is meaningless

<=> solid is finite: If N atoms each with 3 degrees of freedom, at most 3N modes

Debye approximation: • isotropic

•

ε = ω p = k = ω

cs εs = sω

s =1

exp ωτ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1

3d3xd 3ph3 integrating over angles 3d

3x p2dp2π23 = d 3x 3ω 2dω

2π2cs3

D ω( )dω

k

x<

π

a where a is the lattice spacing ⇒ maximum energy ω

D

N

atoms=

L

a( )3 (cubic lattice) ⇒ VD ω( )

0

ωD∫ dω = 3N where D ω( ) is the density of states

k=ωcs

or p= εcs

with velocity cs = constantk

ω

ω D

SoundVelocity

=


Phonons in a Solid

Debye law

VD ω( )0

ωD

∫ dω = V3 ω 2

2π2cs30

ωD

∫ dω = 3N⇒ωD =6NV

π 2cs3

⎛

⎝⎜⎜

⎞

⎠⎟⎟

13= 6π2( )

13 csa

U = d 3x∫ ω

exp ωτ

⎛⎝⎜

⎞⎠⎟−1⎛

⎝⎜⎞⎠⎟

3 k 2dk2π20

kD

∫ =V ω

exp ωτ

⎛⎝⎜

⎞⎠⎟−1⎛

⎝⎜⎞⎠⎟

3 ω 2dω2π2cs

30

ωD

∫

for τ <<ωD U = 3τ 4

23π2cs3V x3dx

ex −10

∞

∫ = π2

103cs3V kB4T 4

Introducing the Debye temperature TD =θ = cs

kB6π2 N

V⎛⎝⎜

⎞⎠⎟

1/3

= ωD

kB

U =35π4kBN

T 4

TD3 CV =

125π4kBN

TTD

⎛

⎝ ⎜

⎞

⎠ ⎟ 3

σ =1215

π4kBNTTD

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 3


ApplicationsCalorimetry: Measure energy deposition by temperature rise

Bolometry: Measure energy flux F by temperature rise

Chopping e.g., between sky and calibration load

Very sensitive!Heat capacity goes to zero at low temperatureFluctuationsWide bandwidth (sense every frequency which couples to bolometer)

Study of cosmic microwave backgroundBolometers

Search for dark matter particles

ΔT =

ΔEC

⇒ need small C

ΔT =

ΔF

G⇒ need small G but time constant=

C

G limited by stability ⇒ small heat capacity C

C ≈ T3

σΔE = kBT2C ∝ T52M

12

ΔF

ΔE

Heat Capacity

Sky

Load

Heat Conductivity

4K → 300mK → 100mK

170g at 20mK

Documents

The Black Body Radiation - Berkeley Cosmology Groupcosmology.berkeley.edu/Classes/S2012/Physics_112/lectures/chapter... · Phys 112 (S12) 6 Black Body Radiation 1 B.Sadoulet The Black