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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 2
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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 3
2 quantifications
Massive Particles
Photons
1st Quantification
Wave like
Discretestates
Obvious Cavity modes
2nd Quantification
Discrete particles
Obvious Planck
B.SadouletPhys 112 (S12) 6 Black Body Radiation 4
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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 5
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
⎛ ⎝ ⎜
⎞ ⎠ ⎟
B.SadouletPhys 112 (S12) 6 Black Body Radiation 6
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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 7
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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 8
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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 10
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 !
B.SadouletPhys 112 (S12) 6 Black Body Radiation 11
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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 12
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
⎛ ⎝ ⎜
⎞ ⎠ ⎟
B.SadouletPhys 112 (S12) 6 Black Body Radiation 13
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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 14
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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 15
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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 16
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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 18
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 !
B.SadouletPhys 112 (S12) 6 Black Body Radiation 19
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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 20
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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 21
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
=
B.SadouletPhys 112 (S12) 6 Black Body Radiation 22
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
B.SadouletPhys 112 (S12) 6 Black Body Radiation 23
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