Post on 21-Dec-2015
Lecture 26 Blackbody Radiation (Ch 7)
Two types of bosons (a) Composite particles which contain an even
number of fermions These number of these particles is conserved if the energy does not exceed the dissociation energy (~ MeV in the case of the nucleus)
(b) particles associated with a field of which the most important example is the photon These particles are not conserved if the total energy of the field changes particles appear and disappear Wersquoll see that the chemical potential of such particles is zero in equilibrium regardless of density
Radiation in Equilibrium with MatterTypically radiation emitted by a hot body or from a laser is not in equilibrium energy is flowing outwards and must be replenished from some source The first step towards understanding of radiation being in equilibrium with matter was made by Kirchhoff who considered a cavity filled with radiation the walls can be regarded as a heat bath for radiation
The walls emit and absorb e-m waves In equilibrium the walls and radiation must have the same temperature T The energy of radiation is spread over a range of frequencies and we define uS (T) d as the energy density (per unit volume) of the radiation with frequencies between and +d uS(T) is the spectral energy density The internal energy of the photon gas
dTuTu S
0
In equilibrium uS (T) is the same everywhere in the cavity and is a function of frequency and temperature only If the cavity volume increases at T=const the internal energy U = u (T) V also increases The essential difference between the photon gas and the ideal gas of molecules for an ideal gas an isothermal expansion would conserve the gas energy whereas for the photon gas it is the energy density which is unchanged the number of photons is not conserved but proportional to volume in an isothermal change
A real surface absorbs only a fraction of the radiation falling on it The absorptivity is a function of and T a surface for which ( ) =1 for all frequencies is called a black body
Photons
The electromagnetic field has an infinite number of modes (standing waves) in the cavity The black-body radiation field is a superposition of plane waves of different frequencies The characteristic feature of the radiation is that a mode may be excited only in units of the quantum of energy h (similar to a harmonic oscillators)
hnii21
This fact leads to the concept of photons as quanta of the electromagnetic field The state of the el-mag field is specified by the number n for each of the modes or in other words by enumerating the number of photons with each frequency
According to the quantum theory of radiation photons are massless bosons of spin 1 (in units ħ) They move with the speed of light
ch
c
Ep
cpE
hE
phph
phph
ph
The linearity of Maxwell equations implies that the photons do not interact with each other (Non-linear optical phenomena are observed when a large-intensity radiation interacts with matter)
T
The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter Presence of a small amount of matter is essential for establishing equilibrium in the photon gas Wersquoll treat a system of photons as an ideal photon gas and in particular wersquoll apply the BE statistics to this system
Chemical Potential of Photons = 0The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter The textbook suggests that N can be found from the equilibrium condition
0phThus in equilibrium the chemical potential for a photon gas is zero
0
VTN
F
On the other hand ph
VTN
F
However we cannot use the usual expression for the chemical potential because one cannot increase N (ie add photons to the system) at constant volume and at the same time keep the temperature constant
VTN
F
- does not exist for the photon gas
Instead we can use NG PVFG V
VTF
V
FP
T
- by increasing the volume at T=const we proportionally scale F
0 VV
FFGThus - the Gibbs free energy of an
equilibrium photon gas is 0 0
N
Gph
For = 0 the BE distribution reduces to the Planckrsquos distribution
1exp
1
1exp
1
Tkh
Tk
Tfn
BB
phph Planckrsquos distribution provides the average
number of photons in a single mode of frequency = h
Density of States for Photons
ky
kx
kz
2
3
2
33
66
34
8
1
k
kGvolumek
LLL
kkN
zyx
extra factor of 2 due to two polarizations
32
23
32
3
26
cg
cGkccp
d
dGg D
ph
3
2
32
233 8
cc
hh
d
dgg Dph
Dph
3
23 8
cg Dph
1exp
Tkh
hhn
B
In the classical (h ltlt kBT) limit TkB
The average energy in the mode
In order to calculate the average number of photons per small energy interval d the average energy of photons per small energy interval d etc as well as the total average number of photons in a photon gas and its total energy we need to know the density of states for photons as a function of photon energy
Spectrum of Blackbody Radiation
1exp
8
3
3
Tk
hcTu
B
The average energy of photons with frequency between and +d (per unit volume)
u(T) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T
dTudngh S
- the spectral density of the black-body radiation
(the Plankrsquos radiation law) 1exp
8
3
3
hc
hfghTus
hThud
dTuTudTudTu
u as a function of the energy
3
h
g(
)
=n( )
average number of photons
photon energy
3
23 8
cg Dph
h
g(
) n
( )
Classical Limit (small f large ) Rayleigh-Jeans Law
This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths
Rayleigh-Jeans Law
At low frequencies or high temperatures hhh 1exp1
Tkchc
hTu Bs 3
23
3
8
1exp
8
- purely classical result (no h) can be obtained directly from equipartition
Rayleigh-Jeans Law (cont)
4
1
In the classical limit of large
4 large
8
TkTu B
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
u as a function of the wavelength
High limit Wienrsquos Displacement Law
The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum
011
3
1exp2
32
3
x
x
x
B
B
B
e
ex
e
xconst
Tkh
Tkh
Tkh
d
dconst
d
du
8233 xex x
Wienrsquos displacement law- discovered experimentally
by Wilhelm Wien
Numerous applications (eg non-contact radiation thermometry)
- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T
u(
T)
82max Tk
h
B
h
TkB82max
Nobel 1911
At high frequencieslow temperatures hhh exp1exp1
hc
hTus exp
8 3
3
max max
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
82max Tk
h
B
- does this mean that
82max
Tkhc
B
max max
No
0
11exp
1exp
11exp
5
11exp
125
2
65
xx
xx
xxconst
xxdx
dconst
df
du
xxx 1exp11exp5 Tk
hc
B5max
ldquonight visionrdquo devices
T = 300 K max 10 m
Tu Tu
Solar Radiation
The surface temperature of the Sun - 5800K
As a function of energy the spectrum of sunlight peaks at a photon energy of
m 505max
Tk
hc
B
eVTkhu B 4182maxmax
Spectral sensitivity of human eye
- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells
Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume
The total energy of photons per unit volume (the energy density of a photon gas)
3
45
0 15
8
1exp hc
Tkd
g
V
UTu B
the Stefan-Boltzmann Law23
45
15
2
ch
kB the Stefan-Boltzmann constant 44
Tc
Tu
4281
8
1exp
8 33
0
23
30
2
30
Thc
k
e
dxx
h
Tk
cd
Tk
hcdgn
V
Nn B
xB
B
- increases as T 3
The average energy per photon
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(just slightly less than the ldquomostrdquo probable energy)
The value of the Stefan-Boltzmann constant 248 10765 mKW
Consider a black body at 310K The power emitted by the body 24 500 mWT
While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)
Some numbers
Power Emitted by a Black Body
For the ldquouni-directionalrdquo motion the flux of energy per unit area
c 1s
energy density u
1m2
T
424 TR
uc
Integration over all angles provides a factor of frac14
uc4
1areaunit by emittedpower
Thus the power emitted by a unit-area surface at temperature T in all directions 444
4
c
4
cpower TT
cTu
The total power emitted by a black-body sphere of radius R
(the hole size must be gtgt the wavelength)
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Radiation in Equilibrium with MatterTypically radiation emitted by a hot body or from a laser is not in equilibrium energy is flowing outwards and must be replenished from some source The first step towards understanding of radiation being in equilibrium with matter was made by Kirchhoff who considered a cavity filled with radiation the walls can be regarded as a heat bath for radiation
The walls emit and absorb e-m waves In equilibrium the walls and radiation must have the same temperature T The energy of radiation is spread over a range of frequencies and we define uS (T) d as the energy density (per unit volume) of the radiation with frequencies between and +d uS(T) is the spectral energy density The internal energy of the photon gas
dTuTu S
0
In equilibrium uS (T) is the same everywhere in the cavity and is a function of frequency and temperature only If the cavity volume increases at T=const the internal energy U = u (T) V also increases The essential difference between the photon gas and the ideal gas of molecules for an ideal gas an isothermal expansion would conserve the gas energy whereas for the photon gas it is the energy density which is unchanged the number of photons is not conserved but proportional to volume in an isothermal change
A real surface absorbs only a fraction of the radiation falling on it The absorptivity is a function of and T a surface for which ( ) =1 for all frequencies is called a black body
Photons
The electromagnetic field has an infinite number of modes (standing waves) in the cavity The black-body radiation field is a superposition of plane waves of different frequencies The characteristic feature of the radiation is that a mode may be excited only in units of the quantum of energy h (similar to a harmonic oscillators)
hnii21
This fact leads to the concept of photons as quanta of the electromagnetic field The state of the el-mag field is specified by the number n for each of the modes or in other words by enumerating the number of photons with each frequency
According to the quantum theory of radiation photons are massless bosons of spin 1 (in units ħ) They move with the speed of light
ch
c
Ep
cpE
hE
phph
phph
ph
The linearity of Maxwell equations implies that the photons do not interact with each other (Non-linear optical phenomena are observed when a large-intensity radiation interacts with matter)
T
The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter Presence of a small amount of matter is essential for establishing equilibrium in the photon gas Wersquoll treat a system of photons as an ideal photon gas and in particular wersquoll apply the BE statistics to this system
Chemical Potential of Photons = 0The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter The textbook suggests that N can be found from the equilibrium condition
0phThus in equilibrium the chemical potential for a photon gas is zero
0
VTN
F
On the other hand ph
VTN
F
However we cannot use the usual expression for the chemical potential because one cannot increase N (ie add photons to the system) at constant volume and at the same time keep the temperature constant
VTN
F
- does not exist for the photon gas
Instead we can use NG PVFG V
VTF
V
FP
T
- by increasing the volume at T=const we proportionally scale F
0 VV
FFGThus - the Gibbs free energy of an
equilibrium photon gas is 0 0
N
Gph
For = 0 the BE distribution reduces to the Planckrsquos distribution
1exp
1
1exp
1
Tkh
Tk
Tfn
BB
phph Planckrsquos distribution provides the average
number of photons in a single mode of frequency = h
Density of States for Photons
ky
kx
kz
2
3
2
33
66
34
8
1
k
kGvolumek
LLL
kkN
zyx
extra factor of 2 due to two polarizations
32
23
32
3
26
cg
cGkccp
d
dGg D
ph
3
2
32
233 8
cc
hh
d
dgg Dph
Dph
3
23 8
cg Dph
1exp
Tkh
hhn
B
In the classical (h ltlt kBT) limit TkB
The average energy in the mode
In order to calculate the average number of photons per small energy interval d the average energy of photons per small energy interval d etc as well as the total average number of photons in a photon gas and its total energy we need to know the density of states for photons as a function of photon energy
Spectrum of Blackbody Radiation
1exp
8
3
3
Tk
hcTu
B
The average energy of photons with frequency between and +d (per unit volume)
u(T) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T
dTudngh S
- the spectral density of the black-body radiation
(the Plankrsquos radiation law) 1exp
8
3
3
hc
hfghTus
hThud
dTuTudTudTu
u as a function of the energy
3
h
g(
)
=n( )
average number of photons
photon energy
3
23 8
cg Dph
h
g(
) n
( )
Classical Limit (small f large ) Rayleigh-Jeans Law
This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths
Rayleigh-Jeans Law
At low frequencies or high temperatures hhh 1exp1
Tkchc
hTu Bs 3
23
3
8
1exp
8
- purely classical result (no h) can be obtained directly from equipartition
Rayleigh-Jeans Law (cont)
4
1
In the classical limit of large
4 large
8
TkTu B
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
u as a function of the wavelength
High limit Wienrsquos Displacement Law
The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum
011
3
1exp2
32
3
x
x
x
B
B
B
e
ex
e
xconst
Tkh
Tkh
Tkh
d
dconst
d
du
8233 xex x
Wienrsquos displacement law- discovered experimentally
by Wilhelm Wien
Numerous applications (eg non-contact radiation thermometry)
- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T
u(
T)
82max Tk
h
B
h
TkB82max
Nobel 1911
At high frequencieslow temperatures hhh exp1exp1
hc
hTus exp
8 3
3
max max
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
82max Tk
h
B
- does this mean that
82max
Tkhc
B
max max
No
0
11exp
1exp
11exp
5
11exp
125
2
65
xx
xx
xxconst
xxdx
dconst
df
du
xxx 1exp11exp5 Tk
hc
B5max
ldquonight visionrdquo devices
T = 300 K max 10 m
Tu Tu
Solar Radiation
The surface temperature of the Sun - 5800K
As a function of energy the spectrum of sunlight peaks at a photon energy of
m 505max
Tk
hc
B
eVTkhu B 4182maxmax
Spectral sensitivity of human eye
- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells
Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume
The total energy of photons per unit volume (the energy density of a photon gas)
3
45
0 15
8
1exp hc
Tkd
g
V
UTu B
the Stefan-Boltzmann Law23
45
15
2
ch
kB the Stefan-Boltzmann constant 44
Tc
Tu
4281
8
1exp
8 33
0
23
30
2
30
Thc
k
e
dxx
h
Tk
cd
Tk
hcdgn
V
Nn B
xB
B
- increases as T 3
The average energy per photon
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(just slightly less than the ldquomostrdquo probable energy)
The value of the Stefan-Boltzmann constant 248 10765 mKW
Consider a black body at 310K The power emitted by the body 24 500 mWT
While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)
Some numbers
Power Emitted by a Black Body
For the ldquouni-directionalrdquo motion the flux of energy per unit area
c 1s
energy density u
1m2
T
424 TR
uc
Integration over all angles provides a factor of frac14
uc4
1areaunit by emittedpower
Thus the power emitted by a unit-area surface at temperature T in all directions 444
4
c
4
cpower TT
cTu
The total power emitted by a black-body sphere of radius R
(the hole size must be gtgt the wavelength)
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Photons
The electromagnetic field has an infinite number of modes (standing waves) in the cavity The black-body radiation field is a superposition of plane waves of different frequencies The characteristic feature of the radiation is that a mode may be excited only in units of the quantum of energy h (similar to a harmonic oscillators)
hnii21
This fact leads to the concept of photons as quanta of the electromagnetic field The state of the el-mag field is specified by the number n for each of the modes or in other words by enumerating the number of photons with each frequency
According to the quantum theory of radiation photons are massless bosons of spin 1 (in units ħ) They move with the speed of light
ch
c
Ep
cpE
hE
phph
phph
ph
The linearity of Maxwell equations implies that the photons do not interact with each other (Non-linear optical phenomena are observed when a large-intensity radiation interacts with matter)
T
The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter Presence of a small amount of matter is essential for establishing equilibrium in the photon gas Wersquoll treat a system of photons as an ideal photon gas and in particular wersquoll apply the BE statistics to this system
Chemical Potential of Photons = 0The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter The textbook suggests that N can be found from the equilibrium condition
0phThus in equilibrium the chemical potential for a photon gas is zero
0
VTN
F
On the other hand ph
VTN
F
However we cannot use the usual expression for the chemical potential because one cannot increase N (ie add photons to the system) at constant volume and at the same time keep the temperature constant
VTN
F
- does not exist for the photon gas
Instead we can use NG PVFG V
VTF
V
FP
T
- by increasing the volume at T=const we proportionally scale F
0 VV
FFGThus - the Gibbs free energy of an
equilibrium photon gas is 0 0
N
Gph
For = 0 the BE distribution reduces to the Planckrsquos distribution
1exp
1
1exp
1
Tkh
Tk
Tfn
BB
phph Planckrsquos distribution provides the average
number of photons in a single mode of frequency = h
Density of States for Photons
ky
kx
kz
2
3
2
33
66
34
8
1
k
kGvolumek
LLL
kkN
zyx
extra factor of 2 due to two polarizations
32
23
32
3
26
cg
cGkccp
d
dGg D
ph
3
2
32
233 8
cc
hh
d
dgg Dph
Dph
3
23 8
cg Dph
1exp
Tkh
hhn
B
In the classical (h ltlt kBT) limit TkB
The average energy in the mode
In order to calculate the average number of photons per small energy interval d the average energy of photons per small energy interval d etc as well as the total average number of photons in a photon gas and its total energy we need to know the density of states for photons as a function of photon energy
Spectrum of Blackbody Radiation
1exp
8
3
3
Tk
hcTu
B
The average energy of photons with frequency between and +d (per unit volume)
u(T) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T
dTudngh S
- the spectral density of the black-body radiation
(the Plankrsquos radiation law) 1exp
8
3
3
hc
hfghTus
hThud
dTuTudTudTu
u as a function of the energy
3
h
g(
)
=n( )
average number of photons
photon energy
3
23 8
cg Dph
h
g(
) n
( )
Classical Limit (small f large ) Rayleigh-Jeans Law
This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths
Rayleigh-Jeans Law
At low frequencies or high temperatures hhh 1exp1
Tkchc
hTu Bs 3
23
3
8
1exp
8
- purely classical result (no h) can be obtained directly from equipartition
Rayleigh-Jeans Law (cont)
4
1
In the classical limit of large
4 large
8
TkTu B
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
u as a function of the wavelength
High limit Wienrsquos Displacement Law
The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum
011
3
1exp2
32
3
x
x
x
B
B
B
e
ex
e
xconst
Tkh
Tkh
Tkh
d
dconst
d
du
8233 xex x
Wienrsquos displacement law- discovered experimentally
by Wilhelm Wien
Numerous applications (eg non-contact radiation thermometry)
- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T
u(
T)
82max Tk
h
B
h
TkB82max
Nobel 1911
At high frequencieslow temperatures hhh exp1exp1
hc
hTus exp
8 3
3
max max
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
82max Tk
h
B
- does this mean that
82max
Tkhc
B
max max
No
0
11exp
1exp
11exp
5
11exp
125
2
65
xx
xx
xxconst
xxdx
dconst
df
du
xxx 1exp11exp5 Tk
hc
B5max
ldquonight visionrdquo devices
T = 300 K max 10 m
Tu Tu
Solar Radiation
The surface temperature of the Sun - 5800K
As a function of energy the spectrum of sunlight peaks at a photon energy of
m 505max
Tk
hc
B
eVTkhu B 4182maxmax
Spectral sensitivity of human eye
- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells
Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume
The total energy of photons per unit volume (the energy density of a photon gas)
3
45
0 15
8
1exp hc
Tkd
g
V
UTu B
the Stefan-Boltzmann Law23
45
15
2
ch
kB the Stefan-Boltzmann constant 44
Tc
Tu
4281
8
1exp
8 33
0
23
30
2
30
Thc
k
e
dxx
h
Tk
cd
Tk
hcdgn
V
Nn B
xB
B
- increases as T 3
The average energy per photon
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(just slightly less than the ldquomostrdquo probable energy)
The value of the Stefan-Boltzmann constant 248 10765 mKW
Consider a black body at 310K The power emitted by the body 24 500 mWT
While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)
Some numbers
Power Emitted by a Black Body
For the ldquouni-directionalrdquo motion the flux of energy per unit area
c 1s
energy density u
1m2
T
424 TR
uc
Integration over all angles provides a factor of frac14
uc4
1areaunit by emittedpower
Thus the power emitted by a unit-area surface at temperature T in all directions 444
4
c
4
cpower TT
cTu
The total power emitted by a black-body sphere of radius R
(the hole size must be gtgt the wavelength)
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Chemical Potential of Photons = 0The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter The textbook suggests that N can be found from the equilibrium condition
0phThus in equilibrium the chemical potential for a photon gas is zero
0
VTN
F
On the other hand ph
VTN
F
However we cannot use the usual expression for the chemical potential because one cannot increase N (ie add photons to the system) at constant volume and at the same time keep the temperature constant
VTN
F
- does not exist for the photon gas
Instead we can use NG PVFG V
VTF
V
FP
T
- by increasing the volume at T=const we proportionally scale F
0 VV
FFGThus - the Gibbs free energy of an
equilibrium photon gas is 0 0
N
Gph
For = 0 the BE distribution reduces to the Planckrsquos distribution
1exp
1
1exp
1
Tkh
Tk
Tfn
BB
phph Planckrsquos distribution provides the average
number of photons in a single mode of frequency = h
Density of States for Photons
ky
kx
kz
2
3
2
33
66
34
8
1
k
kGvolumek
LLL
kkN
zyx
extra factor of 2 due to two polarizations
32
23
32
3
26
cg
cGkccp
d
dGg D
ph
3
2
32
233 8
cc
hh
d
dgg Dph
Dph
3
23 8
cg Dph
1exp
Tkh
hhn
B
In the classical (h ltlt kBT) limit TkB
The average energy in the mode
In order to calculate the average number of photons per small energy interval d the average energy of photons per small energy interval d etc as well as the total average number of photons in a photon gas and its total energy we need to know the density of states for photons as a function of photon energy
Spectrum of Blackbody Radiation
1exp
8
3
3
Tk
hcTu
B
The average energy of photons with frequency between and +d (per unit volume)
u(T) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T
dTudngh S
- the spectral density of the black-body radiation
(the Plankrsquos radiation law) 1exp
8
3
3
hc
hfghTus
hThud
dTuTudTudTu
u as a function of the energy
3
h
g(
)
=n( )
average number of photons
photon energy
3
23 8
cg Dph
h
g(
) n
( )
Classical Limit (small f large ) Rayleigh-Jeans Law
This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths
Rayleigh-Jeans Law
At low frequencies or high temperatures hhh 1exp1
Tkchc
hTu Bs 3
23
3
8
1exp
8
- purely classical result (no h) can be obtained directly from equipartition
Rayleigh-Jeans Law (cont)
4
1
In the classical limit of large
4 large
8
TkTu B
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
u as a function of the wavelength
High limit Wienrsquos Displacement Law
The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum
011
3
1exp2
32
3
x
x
x
B
B
B
e
ex
e
xconst
Tkh
Tkh
Tkh
d
dconst
d
du
8233 xex x
Wienrsquos displacement law- discovered experimentally
by Wilhelm Wien
Numerous applications (eg non-contact radiation thermometry)
- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T
u(
T)
82max Tk
h
B
h
TkB82max
Nobel 1911
At high frequencieslow temperatures hhh exp1exp1
hc
hTus exp
8 3
3
max max
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
82max Tk
h
B
- does this mean that
82max
Tkhc
B
max max
No
0
11exp
1exp
11exp
5
11exp
125
2
65
xx
xx
xxconst
xxdx
dconst
df
du
xxx 1exp11exp5 Tk
hc
B5max
ldquonight visionrdquo devices
T = 300 K max 10 m
Tu Tu
Solar Radiation
The surface temperature of the Sun - 5800K
As a function of energy the spectrum of sunlight peaks at a photon energy of
m 505max
Tk
hc
B
eVTkhu B 4182maxmax
Spectral sensitivity of human eye
- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells
Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume
The total energy of photons per unit volume (the energy density of a photon gas)
3
45
0 15
8
1exp hc
Tkd
g
V
UTu B
the Stefan-Boltzmann Law23
45
15
2
ch
kB the Stefan-Boltzmann constant 44
Tc
Tu
4281
8
1exp
8 33
0
23
30
2
30
Thc
k
e
dxx
h
Tk
cd
Tk
hcdgn
V
Nn B
xB
B
- increases as T 3
The average energy per photon
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(just slightly less than the ldquomostrdquo probable energy)
The value of the Stefan-Boltzmann constant 248 10765 mKW
Consider a black body at 310K The power emitted by the body 24 500 mWT
While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)
Some numbers
Power Emitted by a Black Body
For the ldquouni-directionalrdquo motion the flux of energy per unit area
c 1s
energy density u
1m2
T
424 TR
uc
Integration over all angles provides a factor of frac14
uc4
1areaunit by emittedpower
Thus the power emitted by a unit-area surface at temperature T in all directions 444
4
c
4
cpower TT
cTu
The total power emitted by a black-body sphere of radius R
(the hole size must be gtgt the wavelength)
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Density of States for Photons
ky
kx
kz
2
3
2
33
66
34
8
1
k
kGvolumek
LLL
kkN
zyx
extra factor of 2 due to two polarizations
32
23
32
3
26
cg
cGkccp
d
dGg D
ph
3
2
32
233 8
cc
hh
d
dgg Dph
Dph
3
23 8
cg Dph
1exp
Tkh
hhn
B
In the classical (h ltlt kBT) limit TkB
The average energy in the mode
In order to calculate the average number of photons per small energy interval d the average energy of photons per small energy interval d etc as well as the total average number of photons in a photon gas and its total energy we need to know the density of states for photons as a function of photon energy
Spectrum of Blackbody Radiation
1exp
8
3
3
Tk
hcTu
B
The average energy of photons with frequency between and +d (per unit volume)
u(T) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T
dTudngh S
- the spectral density of the black-body radiation
(the Plankrsquos radiation law) 1exp
8
3
3
hc
hfghTus
hThud
dTuTudTudTu
u as a function of the energy
3
h
g(
)
=n( )
average number of photons
photon energy
3
23 8
cg Dph
h
g(
) n
( )
Classical Limit (small f large ) Rayleigh-Jeans Law
This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths
Rayleigh-Jeans Law
At low frequencies or high temperatures hhh 1exp1
Tkchc
hTu Bs 3
23
3
8
1exp
8
- purely classical result (no h) can be obtained directly from equipartition
Rayleigh-Jeans Law (cont)
4
1
In the classical limit of large
4 large
8
TkTu B
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
u as a function of the wavelength
High limit Wienrsquos Displacement Law
The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum
011
3
1exp2
32
3
x
x
x
B
B
B
e
ex
e
xconst
Tkh
Tkh
Tkh
d
dconst
d
du
8233 xex x
Wienrsquos displacement law- discovered experimentally
by Wilhelm Wien
Numerous applications (eg non-contact radiation thermometry)
- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T
u(
T)
82max Tk
h
B
h
TkB82max
Nobel 1911
At high frequencieslow temperatures hhh exp1exp1
hc
hTus exp
8 3
3
max max
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
82max Tk
h
B
- does this mean that
82max
Tkhc
B
max max
No
0
11exp
1exp
11exp
5
11exp
125
2
65
xx
xx
xxconst
xxdx
dconst
df
du
xxx 1exp11exp5 Tk
hc
B5max
ldquonight visionrdquo devices
T = 300 K max 10 m
Tu Tu
Solar Radiation
The surface temperature of the Sun - 5800K
As a function of energy the spectrum of sunlight peaks at a photon energy of
m 505max
Tk
hc
B
eVTkhu B 4182maxmax
Spectral sensitivity of human eye
- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells
Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume
The total energy of photons per unit volume (the energy density of a photon gas)
3
45
0 15
8
1exp hc
Tkd
g
V
UTu B
the Stefan-Boltzmann Law23
45
15
2
ch
kB the Stefan-Boltzmann constant 44
Tc
Tu
4281
8
1exp
8 33
0
23
30
2
30
Thc
k
e
dxx
h
Tk
cd
Tk
hcdgn
V
Nn B
xB
B
- increases as T 3
The average energy per photon
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(just slightly less than the ldquomostrdquo probable energy)
The value of the Stefan-Boltzmann constant 248 10765 mKW
Consider a black body at 310K The power emitted by the body 24 500 mWT
While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)
Some numbers
Power Emitted by a Black Body
For the ldquouni-directionalrdquo motion the flux of energy per unit area
c 1s
energy density u
1m2
T
424 TR
uc
Integration over all angles provides a factor of frac14
uc4
1areaunit by emittedpower
Thus the power emitted by a unit-area surface at temperature T in all directions 444
4
c
4
cpower TT
cTu
The total power emitted by a black-body sphere of radius R
(the hole size must be gtgt the wavelength)
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Spectrum of Blackbody Radiation
1exp
8
3
3
Tk
hcTu
B
The average energy of photons with frequency between and +d (per unit volume)
u(T) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T
dTudngh S
- the spectral density of the black-body radiation
(the Plankrsquos radiation law) 1exp
8
3
3
hc
hfghTus
hThud
dTuTudTudTu
u as a function of the energy
3
h
g(
)
=n( )
average number of photons
photon energy
3
23 8
cg Dph
h
g(
) n
( )
Classical Limit (small f large ) Rayleigh-Jeans Law
This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths
Rayleigh-Jeans Law
At low frequencies or high temperatures hhh 1exp1
Tkchc
hTu Bs 3
23
3
8
1exp
8
- purely classical result (no h) can be obtained directly from equipartition
Rayleigh-Jeans Law (cont)
4
1
In the classical limit of large
4 large
8
TkTu B
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
u as a function of the wavelength
High limit Wienrsquos Displacement Law
The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum
011
3
1exp2
32
3
x
x
x
B
B
B
e
ex
e
xconst
Tkh
Tkh
Tkh
d
dconst
d
du
8233 xex x
Wienrsquos displacement law- discovered experimentally
by Wilhelm Wien
Numerous applications (eg non-contact radiation thermometry)
- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T
u(
T)
82max Tk
h
B
h
TkB82max
Nobel 1911
At high frequencieslow temperatures hhh exp1exp1
hc
hTus exp
8 3
3
max max
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
82max Tk
h
B
- does this mean that
82max
Tkhc
B
max max
No
0
11exp
1exp
11exp
5
11exp
125
2
65
xx
xx
xxconst
xxdx
dconst
df
du
xxx 1exp11exp5 Tk
hc
B5max
ldquonight visionrdquo devices
T = 300 K max 10 m
Tu Tu
Solar Radiation
The surface temperature of the Sun - 5800K
As a function of energy the spectrum of sunlight peaks at a photon energy of
m 505max
Tk
hc
B
eVTkhu B 4182maxmax
Spectral sensitivity of human eye
- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells
Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume
The total energy of photons per unit volume (the energy density of a photon gas)
3
45
0 15
8
1exp hc
Tkd
g
V
UTu B
the Stefan-Boltzmann Law23
45
15
2
ch
kB the Stefan-Boltzmann constant 44
Tc
Tu
4281
8
1exp
8 33
0
23
30
2
30
Thc
k
e
dxx
h
Tk
cd
Tk
hcdgn
V
Nn B
xB
B
- increases as T 3
The average energy per photon
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(just slightly less than the ldquomostrdquo probable energy)
The value of the Stefan-Boltzmann constant 248 10765 mKW
Consider a black body at 310K The power emitted by the body 24 500 mWT
While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)
Some numbers
Power Emitted by a Black Body
For the ldquouni-directionalrdquo motion the flux of energy per unit area
c 1s
energy density u
1m2
T
424 TR
uc
Integration over all angles provides a factor of frac14
uc4
1areaunit by emittedpower
Thus the power emitted by a unit-area surface at temperature T in all directions 444
4
c
4
cpower TT
cTu
The total power emitted by a black-body sphere of radius R
(the hole size must be gtgt the wavelength)
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Classical Limit (small f large ) Rayleigh-Jeans Law
This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths
Rayleigh-Jeans Law
At low frequencies or high temperatures hhh 1exp1
Tkchc
hTu Bs 3
23
3
8
1exp
8
- purely classical result (no h) can be obtained directly from equipartition
Rayleigh-Jeans Law (cont)
4
1
In the classical limit of large
4 large
8
TkTu B
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
u as a function of the wavelength
High limit Wienrsquos Displacement Law
The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum
011
3
1exp2
32
3
x
x
x
B
B
B
e
ex
e
xconst
Tkh
Tkh
Tkh
d
dconst
d
du
8233 xex x
Wienrsquos displacement law- discovered experimentally
by Wilhelm Wien
Numerous applications (eg non-contact radiation thermometry)
- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T
u(
T)
82max Tk
h
B
h
TkB82max
Nobel 1911
At high frequencieslow temperatures hhh exp1exp1
hc
hTus exp
8 3
3
max max
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
82max Tk
h
B
- does this mean that
82max
Tkhc
B
max max
No
0
11exp
1exp
11exp
5
11exp
125
2
65
xx
xx
xxconst
xxdx
dconst
df
du
xxx 1exp11exp5 Tk
hc
B5max
ldquonight visionrdquo devices
T = 300 K max 10 m
Tu Tu
Solar Radiation
The surface temperature of the Sun - 5800K
As a function of energy the spectrum of sunlight peaks at a photon energy of
m 505max
Tk
hc
B
eVTkhu B 4182maxmax
Spectral sensitivity of human eye
- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells
Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume
The total energy of photons per unit volume (the energy density of a photon gas)
3
45
0 15
8
1exp hc
Tkd
g
V
UTu B
the Stefan-Boltzmann Law23
45
15
2
ch
kB the Stefan-Boltzmann constant 44
Tc
Tu
4281
8
1exp
8 33
0
23
30
2
30
Thc
k
e
dxx
h
Tk
cd
Tk
hcdgn
V
Nn B
xB
B
- increases as T 3
The average energy per photon
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(just slightly less than the ldquomostrdquo probable energy)
The value of the Stefan-Boltzmann constant 248 10765 mKW
Consider a black body at 310K The power emitted by the body 24 500 mWT
While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)
Some numbers
Power Emitted by a Black Body
For the ldquouni-directionalrdquo motion the flux of energy per unit area
c 1s
energy density u
1m2
T
424 TR
uc
Integration over all angles provides a factor of frac14
uc4
1areaunit by emittedpower
Thus the power emitted by a unit-area surface at temperature T in all directions 444
4
c
4
cpower TT
cTu
The total power emitted by a black-body sphere of radius R
(the hole size must be gtgt the wavelength)
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Rayleigh-Jeans Law (cont)
4
1
In the classical limit of large
4 large
8
TkTu B
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
u as a function of the wavelength
High limit Wienrsquos Displacement Law
The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum
011
3
1exp2
32
3
x
x
x
B
B
B
e
ex
e
xconst
Tkh
Tkh
Tkh
d
dconst
d
du
8233 xex x
Wienrsquos displacement law- discovered experimentally
by Wilhelm Wien
Numerous applications (eg non-contact radiation thermometry)
- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T
u(
T)
82max Tk
h
B
h
TkB82max
Nobel 1911
At high frequencieslow temperatures hhh exp1exp1
hc
hTus exp
8 3
3
max max
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
82max Tk
h
B
- does this mean that
82max
Tkhc
B
max max
No
0
11exp
1exp
11exp
5
11exp
125
2
65
xx
xx
xxconst
xxdx
dconst
df
du
xxx 1exp11exp5 Tk
hc
B5max
ldquonight visionrdquo devices
T = 300 K max 10 m
Tu Tu
Solar Radiation
The surface temperature of the Sun - 5800K
As a function of energy the spectrum of sunlight peaks at a photon energy of
m 505max
Tk
hc
B
eVTkhu B 4182maxmax
Spectral sensitivity of human eye
- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells
Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume
The total energy of photons per unit volume (the energy density of a photon gas)
3
45
0 15
8
1exp hc
Tkd
g
V
UTu B
the Stefan-Boltzmann Law23
45
15
2
ch
kB the Stefan-Boltzmann constant 44
Tc
Tu
4281
8
1exp
8 33
0
23
30
2
30
Thc
k
e
dxx
h
Tk
cd
Tk
hcdgn
V
Nn B
xB
B
- increases as T 3
The average energy per photon
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(just slightly less than the ldquomostrdquo probable energy)
The value of the Stefan-Boltzmann constant 248 10765 mKW
Consider a black body at 310K The power emitted by the body 24 500 mWT
While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)
Some numbers
Power Emitted by a Black Body
For the ldquouni-directionalrdquo motion the flux of energy per unit area
c 1s
energy density u
1m2
T
424 TR
uc
Integration over all angles provides a factor of frac14
uc4
1areaunit by emittedpower
Thus the power emitted by a unit-area surface at temperature T in all directions 444
4
c
4
cpower TT
cTu
The total power emitted by a black-body sphere of radius R
(the hole size must be gtgt the wavelength)
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
High limit Wienrsquos Displacement Law
The maximum of u() shifts toward higher frequencies with increasing temperature The position of maximum
011
3
1exp2
32
3
x
x
x
B
B
B
e
ex
e
xconst
Tkh
Tkh
Tkh
d
dconst
d
du
8233 xex x
Wienrsquos displacement law- discovered experimentally
by Wilhelm Wien
Numerous applications (eg non-contact radiation thermometry)
- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T
u(
T)
82max Tk
h
B
h
TkB82max
Nobel 1911
At high frequencieslow temperatures hhh exp1exp1
hc
hTus exp
8 3
3
max max
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
82max Tk
h
B
- does this mean that
82max
Tkhc
B
max max
No
0
11exp
1exp
11exp
5
11exp
125
2
65
xx
xx
xxconst
xxdx
dconst
df
du
xxx 1exp11exp5 Tk
hc
B5max
ldquonight visionrdquo devices
T = 300 K max 10 m
Tu Tu
Solar Radiation
The surface temperature of the Sun - 5800K
As a function of energy the spectrum of sunlight peaks at a photon energy of
m 505max
Tk
hc
B
eVTkhu B 4182maxmax
Spectral sensitivity of human eye
- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells
Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume
The total energy of photons per unit volume (the energy density of a photon gas)
3
45
0 15
8
1exp hc
Tkd
g
V
UTu B
the Stefan-Boltzmann Law23
45
15
2
ch
kB the Stefan-Boltzmann constant 44
Tc
Tu
4281
8
1exp
8 33
0
23
30
2
30
Thc
k
e
dxx
h
Tk
cd
Tk
hcdgn
V
Nn B
xB
B
- increases as T 3
The average energy per photon
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(just slightly less than the ldquomostrdquo probable energy)
The value of the Stefan-Boltzmann constant 248 10765 mKW
Consider a black body at 310K The power emitted by the body 24 500 mWT
While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)
Some numbers
Power Emitted by a Black Body
For the ldquouni-directionalrdquo motion the flux of energy per unit area
c 1s
energy density u
1m2
T
424 TR
uc
Integration over all angles provides a factor of frac14
uc4
1areaunit by emittedpower
Thus the power emitted by a unit-area surface at temperature T in all directions 444
4
c
4
cpower TT
cTu
The total power emitted by a black-body sphere of radius R
(the hole size must be gtgt the wavelength)
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
max max
1exp
18
1exp
8
52
3
32
Tkhc
hchc
Tkhc
ch
hcTu
hc
d
ddTudTu
BB
82max Tk
h
B
- does this mean that
82max
Tkhc
B
max max
No
0
11exp
1exp
11exp
5
11exp
125
2
65
xx
xx
xxconst
xxdx
dconst
df
du
xxx 1exp11exp5 Tk
hc
B5max
ldquonight visionrdquo devices
T = 300 K max 10 m
Tu Tu
Solar Radiation
The surface temperature of the Sun - 5800K
As a function of energy the spectrum of sunlight peaks at a photon energy of
m 505max
Tk
hc
B
eVTkhu B 4182maxmax
Spectral sensitivity of human eye
- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells
Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume
The total energy of photons per unit volume (the energy density of a photon gas)
3
45
0 15
8
1exp hc
Tkd
g
V
UTu B
the Stefan-Boltzmann Law23
45
15
2
ch
kB the Stefan-Boltzmann constant 44
Tc
Tu
4281
8
1exp
8 33
0
23
30
2
30
Thc
k
e
dxx
h
Tk
cd
Tk
hcdgn
V
Nn B
xB
B
- increases as T 3
The average energy per photon
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(just slightly less than the ldquomostrdquo probable energy)
The value of the Stefan-Boltzmann constant 248 10765 mKW
Consider a black body at 310K The power emitted by the body 24 500 mWT
While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)
Some numbers
Power Emitted by a Black Body
For the ldquouni-directionalrdquo motion the flux of energy per unit area
c 1s
energy density u
1m2
T
424 TR
uc
Integration over all angles provides a factor of frac14
uc4
1areaunit by emittedpower
Thus the power emitted by a unit-area surface at temperature T in all directions 444
4
c
4
cpower TT
cTu
The total power emitted by a black-body sphere of radius R
(the hole size must be gtgt the wavelength)
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Solar Radiation
The surface temperature of the Sun - 5800K
As a function of energy the spectrum of sunlight peaks at a photon energy of
m 505max
Tk
hc
B
eVTkhu B 4182maxmax
Spectral sensitivity of human eye
- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells
Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume
The total energy of photons per unit volume (the energy density of a photon gas)
3
45
0 15
8
1exp hc
Tkd
g
V
UTu B
the Stefan-Boltzmann Law23
45
15
2
ch
kB the Stefan-Boltzmann constant 44
Tc
Tu
4281
8
1exp
8 33
0
23
30
2
30
Thc
k
e
dxx
h
Tk
cd
Tk
hcdgn
V
Nn B
xB
B
- increases as T 3
The average energy per photon
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(just slightly less than the ldquomostrdquo probable energy)
The value of the Stefan-Boltzmann constant 248 10765 mKW
Consider a black body at 310K The power emitted by the body 24 500 mWT
While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)
Some numbers
Power Emitted by a Black Body
For the ldquouni-directionalrdquo motion the flux of energy per unit area
c 1s
energy density u
1m2
T
424 TR
uc
Integration over all angles provides a factor of frac14
uc4
1areaunit by emittedpower
Thus the power emitted by a unit-area surface at temperature T in all directions 444
4
c
4
cpower TT
cTu
The total power emitted by a black-body sphere of radius R
(the hole size must be gtgt the wavelength)
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Stefan-Boltzmann Law of Radiation
The total number of photons per unit volume
The total energy of photons per unit volume (the energy density of a photon gas)
3
45
0 15
8
1exp hc
Tkd
g
V
UTu B
the Stefan-Boltzmann Law23
45
15
2
ch
kB the Stefan-Boltzmann constant 44
Tc
Tu
4281
8
1exp
8 33
0
23
30
2
30
Thc
k
e
dxx
h
Tk
cd
Tk
hcdgn
V
Nn B
xB
B
- increases as T 3
The average energy per photon
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(just slightly less than the ldquomostrdquo probable energy)
The value of the Stefan-Boltzmann constant 248 10765 mKW
Consider a black body at 310K The power emitted by the body 24 500 mWT
While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)
Some numbers
Power Emitted by a Black Body
For the ldquouni-directionalrdquo motion the flux of energy per unit area
c 1s
energy density u
1m2
T
424 TR
uc
Integration over all angles provides a factor of frac14
uc4
1areaunit by emittedpower
Thus the power emitted by a unit-area surface at temperature T in all directions 444
4
c
4
cpower TT
cTu
The total power emitted by a black-body sphere of radius R
(the hole size must be gtgt the wavelength)
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Power Emitted by a Black Body
For the ldquouni-directionalrdquo motion the flux of energy per unit area
c 1s
energy density u
1m2
T
424 TR
uc
Integration over all angles provides a factor of frac14
uc4
1areaunit by emittedpower
Thus the power emitted by a unit-area surface at temperature T in all directions 444
4
c
4
cpower TT
cTu
The total power emitted by a black-body sphere of radius R
(the hole size must be gtgt the wavelength)
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg
max = 05 m
KKk
hcT
Tk
hc
BB
74051050103815
1031066
55 623
834
maxmax
This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)
428
23
45
Km
W 1075
15
2 ch
kB 424sphere aby emittedpower TRP
W1083740K5Km
W 1075m 1074
824 264
42828
4
max
2
BSun k
hcRP
kgs 1024m 103
W1083 928
26
2
c
P
dt
dmthe mass loss per one second
1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024
kg 102
010 11189
28
dtdm
Mt
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Radiative Energy Transfer Dewar
Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate
24 1 mWTrJ irad
The net ingoing flux
rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo
i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission
44
1
1ba TT
r
rJJ
If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3
42
41
0 TTJ Without the third plane the energy flux per unit area is
The equilibrium temperature of the third plane can be found from the energy balance
414
24
13
43
42
41
42
43
43
41 2
2
TTTTTTTTTT
The energy flux between the 1st and 2nd planes in the presence of the third plane
042
41
42
414
14
34
1 2
1
2
1
2JTT
TTTTTJ
- cut in half
Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields
1
0
N
JJN
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
The Greenhouse Effect
Transmittance of the Earth atmosphere
Absorption
Emission424outPower EE TR
the flux of the solar radiation energy received by the Earth ~ 1370 Wm2
2
42in ower
orbit
SunSunE R
RTRP
Sunorbit
SunE T
R
RT
412
4
= 1 ndash TEarth = 280K
Rorbit = 15middot1011 m RSun = 7middot108 m
In reality = 07 ndash TEarth = 256K
To maintain a comfortable temperature on the Earth we need the Greenhouse Effect
However too much of the greenhouse effect leads to global warming
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Thermodynamic Functions of Blackbody Radiation
444
3
4
3
164VT
cVT
cVT
cTSUF
NPVFPVTSUG 0
Now we can derive all thermodynamic functions of blackbody radiation
the Helmholtz free energy
the Gibbs free energy
The heat capacity of a photon gas at constant volume 316
VTcT
Tuc
VV
This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas
3
0
2
0 3
1616VT
cTdT
c
VTd
T
TcTS
TTV
the pressure of a photon gas(radiation pressure)
uTcV
FP
NT 3
1
3
4 4
UPV3
1
For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons
TkNTknVnVUPV BB 90723
1
3
1
3
1
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Radiation in the Universe
The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation
In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths
Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Cosmic Microwave Background
In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation
Nobel 1978A Penzias
R Wilson
constVTc
TS 3
3
16
Since V R3 the isentropic expansion leads to
1 RT
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
CMBR (cont)
Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000
At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies
Mather Smoot Nobel 2006
ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of
approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(T) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(T) correspond to the same photon energy If
not why
mmmTk
hc
B
11101172103815
1031066
53
23
834
max
(a)
(b)
(c)
Hzh
TkB 1134
23
max 105811066
72103818282
meVh 650max
meVhc
11max
the maxima u(λT) and u(T) do not correspond to the same photon energy The reason of that is
1exp
18
52
Tkhc
hcTu
c
d
ddTudTu
B
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]
Problem 2006 (blackbody radiation)
216
23
62
2103
721038172
103
ms
photons
JmW
J
ms
photonsN
TkTkTkhc
hcTk
N
TuBB
B
B 72421542815
8 4
33
345
(d) 26442484 103721075 mWKmKWTJ CMBR
(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]