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 radiationThe 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 (νT) dν as the energy density (per unit volume) of the radiation with frequencies between ν and ν+dν uS(νT) is the spectral energy density The internal energy of the photon gas

( ) ( ) νν dTuTu Sintinfin

=0

In equilibrium uS (νT) 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 densitywhich is unchanged the number of photons is not conserved but proportional to volume in an isothermal changeA real surface absorbs only a fraction of the radiation falling on it The absorptivity αis a function of ν and T a surface for which α(ν ) =1 for all frequencies is called a black body

PhotonsThe 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 hν (similar to a harmonic oscillators)

( ) νε 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 masslessbosons of spin 1 (in units ħ) They move with the speed of light

ch

cE

p

cpE

hE

phph

phph

ph

ν

ν

==

=

=

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 emissionof 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

0=phμThus in equilibrium the chemical potential for a photon gas is zero

0

=⎟⎟⎠

⎞⎜⎜⎝

⎛partpart

VTNF

On the other hand phVTN

F μ=⎟⎟⎠

⎞⎜⎜⎝

⎛partpart

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

VTNF

⎟⎟⎠

⎞⎜⎜⎝

⎛partpart

- does not exist for the photon gas

Instead we can use μNG = PVFG += ( )V

VTFVFP

T

minus=⎟

⎠⎞

⎜⎝⎛partpart

minus=

- by increasing the volume at T=const we proportionally scale F

0=minus= VVFFGThus - the Gibbs free energy of an

equilibrium photon gas is 0 0==NG

phμ

For μ = 0 the BE distribution reduces to the Planckrsquos distribution

( )1exp

1

1exp

1minus⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==

Tkh

Tk

Tfn

BB

phph νεε Planckrsquos distribution provides the average

number of photons in a single mode of frequency ν = εh

Density of States for Photons

ky

kx

kz

( ) ( ) ( ) ( ) 2

3

2

33

6634

81

ππππππ kkGvolumek

LLL

kkN

zyx

==timestimes

=

( )

extra factor of 2 due to two polarizations

( ) ( )( )

( )( )32

23

32

3

26 hhh

cg

cGkccp

ddGg D

ph πεε

πεεε

εεε =====

( ) ( ) ( )( ) 3

2

32

233 8

cchh

ddgg D

phD

phπν

πν

νεεν ===

h( ) 3

23 8

cg D

phπνν =

1exp minus⎟⎟⎠

⎞⎜⎜⎝

⎛==

Tkhhhn

B

νννε

In the classical (hν ltlt kBT) limit TkB=ε

The average energy in the mode

In order to calculate the average number of photons per small energy interval dε the average energy of photons per small energy interval dε 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

83

3

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=

Tkhc

Tu

B

εεπε

The average energy of photons with frequency between ν and ν+dν (per unit volume)

u(εT) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T

( ) ( ) ( ) νννννν dTudngh S =

- the spectral density of the black-body radiation

(the Plankrsquos radiation law)( ) ( ) ( ) ( ) 1exp

83

3 minus==

νβνπνννν

hchfghTus

( ) ( ) ( ) ( ) ( ) hThuddTuTudTudTu times=== ννεενννεεu as a function of the energy

3νprop

hνg(ν

)

ν

times =n(ν )

average number of photons

photon energy

( ) 3

23 8

cg D

phπνν =

ν νhνg(ν

) n(ν

)

Classical Limit (small f large λ) 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 ( ) νβνβνβ hhh congminusltlt 1exp1

( ) ( ) Tkchc

hTu Bs 3

23

3

81exp

8 νπνβ

νπν congminus

=- purely classical result (no h) can be obtained directly from equipartition

Rayleigh-Jeans Law (cont)

4

1λ

In the classical limit of large λ

( ) 4 large8λπλ λ

TkTu Basymp

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

u as a function of the wavelength

High ν limit Wienrsquos Displacement Law

The maximum of u(ν) shifts toward higher frequencies with increasing temperature The position of maximum

( ) 011

3

1exp2

32

3

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

minusminus

minustimes=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛times=

x

x

x

B

B

B

eex

exconst

TkhTk

h

Tkhd

dconstddu

ν

ν

νν

( ) 8233 asymprarr=minus 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(ν

T)

ν

82max asympTk

h

B

ν

hTkB82max asympν

Nobel 1911

At high frequencieslow temperatures ( ) ( )νβνβνβ hhh exp1exp1 congminusgtgt

( ) ( )νβνπν hc

hTus minus= exp8 33

νmax hArr λmax

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

82max asympTk

h

B

ν- does this mean that

82max

asympλTk

hc

B

maxν maxλ

No

( ) ( ) ( ) ( )

( ) 0

11exp1exp

11exp5

11exp1

25

2

65 =⎥⎦

⎤⎢⎣

⎡

minusminus

minusminus

minustimes=⎥⎦

⎤⎢⎣

⎡minus

times=minus

xxxx

xxconst

xxdxdconst

dfdu

( ) ( )xxx 1exp11exp5 =minus rarr Tkhc

B5max asympλ

ldquonight visionrdquo devices

T = 300 K rarr λmax asymp 10 μm

( )Tu ν ( )Tu λ

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max μλ asymp=

Tkhc

B

eVTkhu B 4182maxmax asymp== ν

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 RadiationThe total number of photons per unit volume

The total energy of photons per unit volume (the energy density of a photon gas) ( ) ( )

( )( )( )3

45

0 158

1exp hcTkdg

VUTu Bπε

βεεε

=minus

times=equiv int

infin

the Stefan-Boltzmann Law23

45

152

chkBπσ =

the Stefan-Boltzmann constant ( ) 44 T

cTu σ

=

( ) ( ) 4281

8

1exp

8 33

0

23

30

2

30

times⎟⎠⎞

⎜⎝⎛=

minus⎟⎠⎞

⎜⎝⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==equiv intintint

infininfininfin

Thck

edxx

hTk

cd

Tkhc

dgnVNn B

xB

B

ππνν

νπεεε

- increases as T 3

The average energy per photon ( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant ( )248 10765 mKWminustimes=σ

Consider a black body at 310K The power emitted by the body 24 500 mWT asympσ

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 times 1s

energy density u1m2

T

424 TR σπ=

uctimes=

Integration over all angles provides a factor of frac14 uctimes=

41areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions ( ) 444

4c

4cpower TT

cTu σσ

=times==

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 λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

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 radiationThe 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 (νT) dν as the energy density (per unit volume) of the radiation with frequencies between ν and ν+dν uS(νT) is the spectral energy density The internal energy of the photon gas

( ) ( ) νν dTuTu Sintinfin

=0

In equilibrium uS (νT) 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 densitywhich is unchanged the number of photons is not conserved but proportional to volume in an isothermal changeA real surface absorbs only a fraction of the radiation falling on it The absorptivity αis a function of ν and T a surface for which α(ν ) =1 for all frequencies is called a black body

PhotonsThe 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 hν (similar to a harmonic oscillators)

( ) νε 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 masslessbosons of spin 1 (in units ħ) They move with the speed of light

ch

cE

p

cpE

hE

phph

phph

ph

ν

ν

==

=

=

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 emissionof 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

0=phμThus in equilibrium the chemical potential for a photon gas is zero

0

=⎟⎟⎠

⎞⎜⎜⎝

⎛partpart

VTNF

On the other hand phVTN

F μ=⎟⎟⎠

⎞⎜⎜⎝

⎛partpart

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

VTNF

⎟⎟⎠

⎞⎜⎜⎝

⎛partpart

- does not exist for the photon gas

Instead we can use μNG = PVFG += ( )V

VTFVFP

T

minus=⎟

⎠⎞

⎜⎝⎛partpart

minus=

- by increasing the volume at T=const we proportionally scale F

0=minus= VVFFGThus - the Gibbs free energy of an

equilibrium photon gas is 0 0==NG

phμ

For μ = 0 the BE distribution reduces to the Planckrsquos distribution

( )1exp

1

1exp

1minus⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==

Tkh

Tk

Tfn

BB

phph νεε Planckrsquos distribution provides the average

number of photons in a single mode of frequency ν = εh

Density of States for Photons

ky

kx

kz

( ) ( ) ( ) ( ) 2

3

2

33

6634

81

ππππππ kkGvolumek

LLL

kkN

zyx

==timestimes

=

( )

extra factor of 2 due to two polarizations

( ) ( )( )

( )( )32

23

32

3

26 hhh

cg

cGkccp

ddGg D

ph πεε

πεεε

εεε =====

( ) ( ) ( )( ) 3

2

32

233 8

cchh

ddgg D

phD

phπν

πν

νεεν ===

h( ) 3

23 8

cg D

phπνν =

1exp minus⎟⎟⎠

⎞⎜⎜⎝

⎛==

Tkhhhn

B

νννε

In the classical (hν ltlt kBT) limit TkB=ε

The average energy in the mode

In order to calculate the average number of photons per small energy interval dε the average energy of photons per small energy interval dε 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

83

3

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=

Tkhc

Tu

B

εεπε

The average energy of photons with frequency between ν and ν+dν (per unit volume)

u(εT) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T

( ) ( ) ( ) νννννν dTudngh S =

- the spectral density of the black-body radiation

(the Plankrsquos radiation law)( ) ( ) ( ) ( ) 1exp

83

3 minus==

νβνπνννν

hchfghTus

( ) ( ) ( ) ( ) ( ) hThuddTuTudTudTu times=== ννεενννεεu as a function of the energy

3νprop

hνg(ν

)

ν

times =n(ν )

average number of photons

photon energy

( ) 3

23 8

cg D

phπνν =

ν νhνg(ν

) n(ν

)

Classical Limit (small f large λ) 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 ( ) νβνβνβ hhh congminusltlt 1exp1

( ) ( ) Tkchc

hTu Bs 3

23

3

81exp

8 νπνβ

νπν congminus

=- purely classical result (no h) can be obtained directly from equipartition

Rayleigh-Jeans Law (cont)

4

1λ

In the classical limit of large λ

( ) 4 large8λπλ λ

TkTu Basymp

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

u as a function of the wavelength

High ν limit Wienrsquos Displacement Law

The maximum of u(ν) shifts toward higher frequencies with increasing temperature The position of maximum

( ) 011

3

1exp2

32

3

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

minusminus

minustimes=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛times=

x

x

x

B

B

B

eex

exconst

TkhTk

h

Tkhd

dconstddu

ν

ν

νν

( ) 8233 asymprarr=minus 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(ν

T)

ν

82max asympTk

h

B

ν

hTkB82max asympν

Nobel 1911

At high frequencieslow temperatures ( ) ( )νβνβνβ hhh exp1exp1 congminusgtgt

( ) ( )νβνπν hc

hTus minus= exp8 33

νmax hArr λmax

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

82max asympTk

h

B

ν- does this mean that

82max

asympλTk

hc

B

maxν maxλ

No

( ) ( ) ( ) ( )

( ) 0

11exp1exp

11exp5

11exp1

25

2

65 =⎥⎦

⎤⎢⎣

⎡

minusminus

minusminus

minustimes=⎥⎦

⎤⎢⎣

⎡minus

times=minus

xxxx

xxconst

xxdxdconst

dfdu

( ) ( )xxx 1exp11exp5 =minus rarr Tkhc

B5max asympλ

ldquonight visionrdquo devices

T = 300 K rarr λmax asymp 10 μm

( )Tu ν ( )Tu λ

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max μλ asymp=

Tkhc

B

eVTkhu B 4182maxmax asymp== ν

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 RadiationThe total number of photons per unit volume

The total energy of photons per unit volume (the energy density of a photon gas) ( ) ( )

( )( )( )3

45

0 158

1exp hcTkdg

VUTu Bπε

βεεε

=minus

times=equiv int

infin

the Stefan-Boltzmann Law23

45

152

chkBπσ =

the Stefan-Boltzmann constant ( ) 44 T

cTu σ

=

( ) ( ) 4281

8

1exp

8 33

0

23

30

2

30

times⎟⎠⎞

⎜⎝⎛=

minus⎟⎠⎞

⎜⎝⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==equiv intintint

infininfininfin

Thck

edxx

hTk

cd

Tkhc

dgnVNn B

xB

B

ππνν

νπεεε

- increases as T 3

The average energy per photon ( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant ( )248 10765 mKWminustimes=σ

Consider a black body at 310K The power emitted by the body 24 500 mWT asympσ

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 times 1s

energy density u1m2

T

424 TR σπ=

uctimes=

Integration over all angles provides a factor of frac14 uctimes=

41areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions ( ) 444

4c

4cpower TT

cTu σσ

=times==

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 λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

PhotonsThe 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 hν (similar to a harmonic oscillators)

( ) νε 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 masslessbosons of spin 1 (in units ħ) They move with the speed of light

ch

cE

p

cpE

hE

phph

phph

ph

ν

ν

==

=

=

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 emissionof 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

0=phμThus in equilibrium the chemical potential for a photon gas is zero

0

=⎟⎟⎠

⎞⎜⎜⎝

⎛partpart

VTNF

On the other hand phVTN

F μ=⎟⎟⎠

⎞⎜⎜⎝

⎛partpart

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

VTNF

⎟⎟⎠

⎞⎜⎜⎝

⎛partpart

- does not exist for the photon gas

Instead we can use μNG = PVFG += ( )V

VTFVFP

T

minus=⎟

⎠⎞

⎜⎝⎛partpart

minus=

- by increasing the volume at T=const we proportionally scale F

0=minus= VVFFGThus - the Gibbs free energy of an

equilibrium photon gas is 0 0==NG

phμ

For μ = 0 the BE distribution reduces to the Planckrsquos distribution

( )1exp

1

1exp

1minus⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==

Tkh

Tk

Tfn

BB

phph νεε Planckrsquos distribution provides the average

number of photons in a single mode of frequency ν = εh

Density of States for Photons

ky

kx

kz

( ) ( ) ( ) ( ) 2

3

2

33

6634

81

ππππππ kkGvolumek

LLL

kkN

zyx

==timestimes

=

( )

extra factor of 2 due to two polarizations

( ) ( )( )

( )( )32

23

32

3

26 hhh

cg

cGkccp

ddGg D

ph πεε

πεεε

εεε =====

( ) ( ) ( )( ) 3

2

32

233 8

cchh

ddgg D

phD

phπν

πν

νεεν ===

h( ) 3

23 8

cg D

phπνν =

1exp minus⎟⎟⎠

⎞⎜⎜⎝

⎛==

Tkhhhn

B

νννε

In the classical (hν ltlt kBT) limit TkB=ε

The average energy in the mode

In order to calculate the average number of photons per small energy interval dε the average energy of photons per small energy interval dε 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

83

3

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=

Tkhc

Tu

B

εεπε

The average energy of photons with frequency between ν and ν+dν (per unit volume)

u(εT) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T

( ) ( ) ( ) νννννν dTudngh S =

- the spectral density of the black-body radiation

(the Plankrsquos radiation law)( ) ( ) ( ) ( ) 1exp

83

3 minus==

νβνπνννν

hchfghTus

( ) ( ) ( ) ( ) ( ) hThuddTuTudTudTu times=== ννεενννεεu as a function of the energy

3νprop

hνg(ν

)

ν

times =n(ν )

average number of photons

photon energy

( ) 3

23 8

cg D

phπνν =

ν νhνg(ν

) n(ν

)

Classical Limit (small f large λ) 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 ( ) νβνβνβ hhh congminusltlt 1exp1

( ) ( ) Tkchc

hTu Bs 3

23

3

81exp

8 νπνβ

νπν congminus

=- purely classical result (no h) can be obtained directly from equipartition

Rayleigh-Jeans Law (cont)

4

1λ

In the classical limit of large λ

( ) 4 large8λπλ λ

TkTu Basymp

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

u as a function of the wavelength

High ν limit Wienrsquos Displacement Law

The maximum of u(ν) shifts toward higher frequencies with increasing temperature The position of maximum

( ) 011

3

1exp2

32

3

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

minusminus

minustimes=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛times=

x

x

x

B

B

B

eex

exconst

TkhTk

h

Tkhd

dconstddu

ν

ν

νν

( ) 8233 asymprarr=minus 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(ν

T)

ν

82max asympTk

h

B

ν

hTkB82max asympν

Nobel 1911

At high frequencieslow temperatures ( ) ( )νβνβνβ hhh exp1exp1 congminusgtgt

( ) ( )νβνπν hc

hTus minus= exp8 33

νmax hArr λmax

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

82max asympTk

h

B

ν- does this mean that

82max

asympλTk

hc

B

maxν maxλ

No

( ) ( ) ( ) ( )

( ) 0

11exp1exp

11exp5

11exp1

25

2

65 =⎥⎦

⎤⎢⎣

⎡

minusminus

minusminus

minustimes=⎥⎦

⎤⎢⎣

⎡minus

times=minus

xxxx

xxconst

xxdxdconst

dfdu

( ) ( )xxx 1exp11exp5 =minus rarr Tkhc

B5max asympλ

ldquonight visionrdquo devices

T = 300 K rarr λmax asymp 10 μm

( )Tu ν ( )Tu λ

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max μλ asymp=

Tkhc

B

eVTkhu B 4182maxmax asymp== ν

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 RadiationThe total number of photons per unit volume

The total energy of photons per unit volume (the energy density of a photon gas) ( ) ( )

( )( )( )3

45

0 158

1exp hcTkdg

VUTu Bπε

βεεε

=minus

times=equiv int

infin

the Stefan-Boltzmann Law23

45

152

chkBπσ =

the Stefan-Boltzmann constant ( ) 44 T

cTu σ

=

( ) ( ) 4281

8

1exp

8 33

0

23

30

2

30

times⎟⎠⎞

⎜⎝⎛=

minus⎟⎠⎞

⎜⎝⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==equiv intintint

infininfininfin

Thck

edxx

hTk

cd

Tkhc

dgnVNn B

xB

B

ππνν

νπεεε

- increases as T 3

The average energy per photon ( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant ( )248 10765 mKWminustimes=σ

Consider a black body at 310K The power emitted by the body 24 500 mWT asympσ

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 times 1s

energy density u1m2

T

424 TR σπ=

uctimes=

Integration over all angles provides a factor of frac14 uctimes=

41areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions ( ) 444

4c

4cpower TT

cTu σσ

=times==

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 λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

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

0=phμThus in equilibrium the chemical potential for a photon gas is zero

0

=⎟⎟⎠

⎞⎜⎜⎝

⎛partpart

VTNF

On the other hand phVTN

F μ=⎟⎟⎠

⎞⎜⎜⎝

⎛partpart

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

VTNF

⎟⎟⎠

⎞⎜⎜⎝

⎛partpart

- does not exist for the photon gas

Instead we can use μNG = PVFG += ( )V

VTFVFP

T

minus=⎟

⎠⎞

⎜⎝⎛partpart

minus=

- by increasing the volume at T=const we proportionally scale F

0=minus= VVFFGThus - the Gibbs free energy of an

equilibrium photon gas is 0 0==NG

phμ

For μ = 0 the BE distribution reduces to the Planckrsquos distribution

( )1exp

1

1exp

1minus⎟⎟

⎠

⎞⎜⎜⎝

⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==

Tkh

Tk

Tfn

BB

phph νεε Planckrsquos distribution provides the average

number of photons in a single mode of frequency ν = εh

Density of States for Photons

ky

kx

kz

( ) ( ) ( ) ( ) 2

3

2

33

6634

81

ππππππ kkGvolumek

LLL

kkN

zyx

==timestimes

=

( )

extra factor of 2 due to two polarizations

( ) ( )( )

( )( )32

23

32

3

26 hhh

cg

cGkccp

ddGg D

ph πεε

πεεε

εεε =====

( ) ( ) ( )( ) 3

2

32

233 8

cchh

ddgg D

phD

phπν

πν

νεεν ===

h( ) 3

23 8

cg D

phπνν =

1exp minus⎟⎟⎠

⎞⎜⎜⎝

⎛==

Tkhhhn

B

νννε

In the classical (hν ltlt kBT) limit TkB=ε

The average energy in the mode

In order to calculate the average number of photons per small energy interval dε the average energy of photons per small energy interval dε 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

83

3

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=

Tkhc

Tu

B

εεπε

The average energy of photons with frequency between ν and ν+dν (per unit volume)

u(εT) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T

( ) ( ) ( ) νννννν dTudngh S =

- the spectral density of the black-body radiation

(the Plankrsquos radiation law)( ) ( ) ( ) ( ) 1exp

83

3 minus==

νβνπνννν

hchfghTus

( ) ( ) ( ) ( ) ( ) hThuddTuTudTudTu times=== ννεενννεεu as a function of the energy

3νprop

hνg(ν

)

ν

times =n(ν )

average number of photons

photon energy

( ) 3

23 8

cg D

phπνν =

ν νhνg(ν

) n(ν

)

Classical Limit (small f large λ) 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 ( ) νβνβνβ hhh congminusltlt 1exp1

( ) ( ) Tkchc

hTu Bs 3

23

3

81exp

8 νπνβ

νπν congminus

=- purely classical result (no h) can be obtained directly from equipartition

Rayleigh-Jeans Law (cont)

4

1λ

In the classical limit of large λ

( ) 4 large8λπλ λ

TkTu Basymp

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

u as a function of the wavelength

High ν limit Wienrsquos Displacement Law

The maximum of u(ν) shifts toward higher frequencies with increasing temperature The position of maximum

( ) 011

3

1exp2

32

3

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

minusminus

minustimes=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛times=

x

x

x

B

B

B

eex

exconst

TkhTk

h

Tkhd

dconstddu

ν

ν

νν

( ) 8233 asymprarr=minus 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(ν

T)

ν

82max asympTk

h

B

ν

hTkB82max asympν

Nobel 1911

At high frequencieslow temperatures ( ) ( )νβνβνβ hhh exp1exp1 congminusgtgt

( ) ( )νβνπν hc

hTus minus= exp8 33

νmax hArr λmax

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

82max asympTk

h

B

ν- does this mean that

82max

asympλTk

hc

B

maxν maxλ

No

( ) ( ) ( ) ( )

( ) 0

11exp1exp

11exp5

11exp1

25

2

65 =⎥⎦

⎤⎢⎣

⎡

minusminus

minusminus

minustimes=⎥⎦

⎤⎢⎣

⎡minus

times=minus

xxxx

xxconst

xxdxdconst

dfdu

( ) ( )xxx 1exp11exp5 =minus rarr Tkhc

B5max asympλ

ldquonight visionrdquo devices

T = 300 K rarr λmax asymp 10 μm

( )Tu ν ( )Tu λ

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max μλ asymp=

Tkhc

B

eVTkhu B 4182maxmax asymp== ν

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 RadiationThe total number of photons per unit volume

The total energy of photons per unit volume (the energy density of a photon gas) ( ) ( )

( )( )( )3

45

0 158

1exp hcTkdg

VUTu Bπε

βεεε

=minus

times=equiv int

infin

the Stefan-Boltzmann Law23

45

152

chkBπσ =

the Stefan-Boltzmann constant ( ) 44 T

cTu σ

=

( ) ( ) 4281

8

1exp

8 33

0

23

30

2

30

times⎟⎠⎞

⎜⎝⎛=

minus⎟⎠⎞

⎜⎝⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==equiv intintint

infininfininfin

Thck

edxx

hTk

cd

Tkhc

dgnVNn B

xB

B

ππνν

νπεεε

- increases as T 3

The average energy per photon ( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant ( )248 10765 mKWminustimes=σ

Consider a black body at 310K The power emitted by the body 24 500 mWT asympσ

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 times 1s

energy density u1m2

T

424 TR σπ=

uctimes=

Integration over all angles provides a factor of frac14 uctimes=

41areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions ( ) 444

4c

4cpower TT

cTu σσ

=times==

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 λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

Density of States for Photons

ky

kx

kz

( ) ( ) ( ) ( ) 2

3

2

33

6634

81

ππππππ kkGvolumek

LLL

kkN

zyx

==timestimes

=

( )

extra factor of 2 due to two polarizations

( ) ( )( )

( )( )32

23

32

3

26 hhh

cg

cGkccp

ddGg D

ph πεε

πεεε

εεε =====

( ) ( ) ( )( ) 3

2

32

233 8

cchh

ddgg D

phD

phπν

πν

νεεν ===

h( ) 3

23 8

cg D

phπνν =

1exp minus⎟⎟⎠

⎞⎜⎜⎝

⎛==

Tkhhhn

B

νννε

In the classical (hν ltlt kBT) limit TkB=ε

The average energy in the mode

In order to calculate the average number of photons per small energy interval dε the average energy of photons per small energy interval dε 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

83

3

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=

Tkhc

Tu

B

εεπε

The average energy of photons with frequency between ν and ν+dν (per unit volume)

u(εT) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T

( ) ( ) ( ) νννννν dTudngh S =

- the spectral density of the black-body radiation

(the Plankrsquos radiation law)( ) ( ) ( ) ( ) 1exp

83

3 minus==

νβνπνννν

hchfghTus

( ) ( ) ( ) ( ) ( ) hThuddTuTudTudTu times=== ννεενννεεu as a function of the energy

3νprop

hνg(ν

)

ν

times =n(ν )

average number of photons

photon energy

( ) 3

23 8

cg D

phπνν =

ν νhνg(ν

) n(ν

)

Classical Limit (small f large λ) 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 ( ) νβνβνβ hhh congminusltlt 1exp1

( ) ( ) Tkchc

hTu Bs 3

23

3

81exp

8 νπνβ

νπν congminus

=- purely classical result (no h) can be obtained directly from equipartition

Rayleigh-Jeans Law (cont)

4

1λ

In the classical limit of large λ

( ) 4 large8λπλ λ

TkTu Basymp

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

u as a function of the wavelength

High ν limit Wienrsquos Displacement Law

The maximum of u(ν) shifts toward higher frequencies with increasing temperature The position of maximum

( ) 011

3

1exp2

32

3

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

minusminus

minustimes=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛times=

x

x

x

B

B

B

eex

exconst

TkhTk

h

Tkhd

dconstddu

ν

ν

νν

( ) 8233 asymprarr=minus 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(ν

T)

ν

82max asympTk

h

B

ν

hTkB82max asympν

Nobel 1911

At high frequencieslow temperatures ( ) ( )νβνβνβ hhh exp1exp1 congminusgtgt

( ) ( )νβνπν hc

hTus minus= exp8 33

νmax hArr λmax

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

82max asympTk

h

B

ν- does this mean that

82max

asympλTk

hc

B

maxν maxλ

No

( ) ( ) ( ) ( )

( ) 0

11exp1exp

11exp5

11exp1

25

2

65 =⎥⎦

⎤⎢⎣

⎡

minusminus

minusminus

minustimes=⎥⎦

⎤⎢⎣

⎡minus

times=minus

xxxx

xxconst

xxdxdconst

dfdu

( ) ( )xxx 1exp11exp5 =minus rarr Tkhc

B5max asympλ

ldquonight visionrdquo devices

T = 300 K rarr λmax asymp 10 μm

( )Tu ν ( )Tu λ

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max μλ asymp=

Tkhc

B

eVTkhu B 4182maxmax asymp== ν

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 RadiationThe total number of photons per unit volume

The total energy of photons per unit volume (the energy density of a photon gas) ( ) ( )

( )( )( )3

45

0 158

1exp hcTkdg

VUTu Bπε

βεεε

=minus

times=equiv int

infin

the Stefan-Boltzmann Law23

45

152

chkBπσ =

the Stefan-Boltzmann constant ( ) 44 T

cTu σ

=

( ) ( ) 4281

8

1exp

8 33

0

23

30

2

30

times⎟⎠⎞

⎜⎝⎛=

minus⎟⎠⎞

⎜⎝⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==equiv intintint

infininfininfin

Thck

edxx

hTk

cd

Tkhc

dgnVNn B

xB

B

ππνν

νπεεε

- increases as T 3

The average energy per photon ( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant ( )248 10765 mKWminustimes=σ

Consider a black body at 310K The power emitted by the body 24 500 mWT asympσ

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 times 1s

energy density u1m2

T

424 TR σπ=

uctimes=

Integration over all angles provides a factor of frac14 uctimes=

41areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions ( ) 444

4c

4cpower TT

cTu σσ

=times==

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 λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

Spectrum of Blackbody Radiation

( )( ) 1exp

83

3

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=

Tkhc

Tu

B

εεπε

The average energy of photons with frequency between ν and ν+dν (per unit volume)

u(εT) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T

( ) ( ) ( ) νννννν dTudngh S =

- the spectral density of the black-body radiation

(the Plankrsquos radiation law)( ) ( ) ( ) ( ) 1exp

83

3 minus==

νβνπνννν

hchfghTus

( ) ( ) ( ) ( ) ( ) hThuddTuTudTudTu times=== ννεενννεεu as a function of the energy

3νprop

hνg(ν

)

ν

times =n(ν )

average number of photons

photon energy

( ) 3

23 8

cg D

phπνν =

ν νhνg(ν

) n(ν

)

Classical Limit (small f large λ) 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 ( ) νβνβνβ hhh congminusltlt 1exp1

( ) ( ) Tkchc

hTu Bs 3

23

3

81exp

8 νπνβ

νπν congminus

=- purely classical result (no h) can be obtained directly from equipartition

Rayleigh-Jeans Law (cont)

4

1λ

In the classical limit of large λ

( ) 4 large8λπλ λ

TkTu Basymp

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

u as a function of the wavelength

High ν limit Wienrsquos Displacement Law

The maximum of u(ν) shifts toward higher frequencies with increasing temperature The position of maximum

( ) 011

3

1exp2

32

3

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

minusminus

minustimes=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛times=

x

x

x

B

B

B

eex

exconst

TkhTk

h

Tkhd

dconstddu

ν

ν

νν

( ) 8233 asymprarr=minus 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(ν

T)

ν

82max asympTk

h

B

ν

hTkB82max asympν

Nobel 1911

At high frequencieslow temperatures ( ) ( )νβνβνβ hhh exp1exp1 congminusgtgt

( ) ( )νβνπν hc

hTus minus= exp8 33

νmax hArr λmax

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

82max asympTk

h

B

ν- does this mean that

82max

asympλTk

hc

B

maxν maxλ

No

( ) ( ) ( ) ( )

( ) 0

11exp1exp

11exp5

11exp1

25

2

65 =⎥⎦

⎤⎢⎣

⎡

minusminus

minusminus

minustimes=⎥⎦

⎤⎢⎣

⎡minus

times=minus

xxxx

xxconst

xxdxdconst

dfdu

( ) ( )xxx 1exp11exp5 =minus rarr Tkhc

B5max asympλ

ldquonight visionrdquo devices

T = 300 K rarr λmax asymp 10 μm

( )Tu ν ( )Tu λ

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max μλ asymp=

Tkhc

B

eVTkhu B 4182maxmax asymp== ν

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 RadiationThe total number of photons per unit volume

The total energy of photons per unit volume (the energy density of a photon gas) ( ) ( )

( )( )( )3

45

0 158

1exp hcTkdg

VUTu Bπε

βεεε

=minus

times=equiv int

infin

the Stefan-Boltzmann Law23

45

152

chkBπσ =

the Stefan-Boltzmann constant ( ) 44 T

cTu σ

=

( ) ( ) 4281

8

1exp

8 33

0

23

30

2

30

times⎟⎠⎞

⎜⎝⎛=

minus⎟⎠⎞

⎜⎝⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==equiv intintint

infininfininfin

Thck

edxx

hTk

cd

Tkhc

dgnVNn B

xB

B

ππνν

νπεεε

- increases as T 3

The average energy per photon ( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant ( )248 10765 mKWminustimes=σ

Consider a black body at 310K The power emitted by the body 24 500 mWT asympσ

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 times 1s

energy density u1m2

T

424 TR σπ=

uctimes=

Integration over all angles provides a factor of frac14 uctimes=

41areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions ( ) 444

4c

4cpower TT

cTu σσ

=times==

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 λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

Classical Limit (small f large λ) 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 ( ) νβνβνβ hhh congminusltlt 1exp1

( ) ( ) Tkchc

hTu Bs 3

23

3

81exp

8 νπνβ

νπν congminus

=- purely classical result (no h) can be obtained directly from equipartition

Rayleigh-Jeans Law (cont)

4

1λ

In the classical limit of large λ

( ) 4 large8λπλ λ

TkTu Basymp

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

u as a function of the wavelength

High ν limit Wienrsquos Displacement Law

The maximum of u(ν) shifts toward higher frequencies with increasing temperature The position of maximum

( ) 011

3

1exp2

32

3

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

minusminus

minustimes=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛times=

x

x

x

B

B

B

eex

exconst

TkhTk

h

Tkhd

dconstddu

ν

ν

νν

( ) 8233 asymprarr=minus 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(ν

T)

ν

82max asympTk

h

B

ν

hTkB82max asympν

Nobel 1911

At high frequencieslow temperatures ( ) ( )νβνβνβ hhh exp1exp1 congminusgtgt

( ) ( )νβνπν hc

hTus minus= exp8 33

νmax hArr λmax

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

82max asympTk

h

B

ν- does this mean that

82max

asympλTk

hc

B

maxν maxλ

No

( ) ( ) ( ) ( )

( ) 0

11exp1exp

11exp5

11exp1

25

2

65 =⎥⎦

⎤⎢⎣

⎡

minusminus

minusminus

minustimes=⎥⎦

⎤⎢⎣

⎡minus

times=minus

xxxx

xxconst

xxdxdconst

dfdu

( ) ( )xxx 1exp11exp5 =minus rarr Tkhc

B5max asympλ

ldquonight visionrdquo devices

T = 300 K rarr λmax asymp 10 μm

( )Tu ν ( )Tu λ

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max μλ asymp=

Tkhc

B

eVTkhu B 4182maxmax asymp== ν

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 RadiationThe total number of photons per unit volume

The total energy of photons per unit volume (the energy density of a photon gas) ( ) ( )

( )( )( )3

45

0 158

1exp hcTkdg

VUTu Bπε

βεεε

=minus

times=equiv int

infin

the Stefan-Boltzmann Law23

45

152

chkBπσ =

the Stefan-Boltzmann constant ( ) 44 T

cTu σ

=

( ) ( ) 4281

8

1exp

8 33

0

23

30

2

30

times⎟⎠⎞

⎜⎝⎛=

minus⎟⎠⎞

⎜⎝⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==equiv intintint

infininfininfin

Thck

edxx

hTk

cd

Tkhc

dgnVNn B

xB

B

ππνν

νπεεε

- increases as T 3

The average energy per photon ( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant ( )248 10765 mKWminustimes=σ

Consider a black body at 310K The power emitted by the body 24 500 mWT asympσ

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 times 1s

energy density u1m2

T

424 TR σπ=

uctimes=

Integration over all angles provides a factor of frac14 uctimes=

41areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions ( ) 444

4c

4cpower TT

cTu σσ

=times==

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 λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

Rayleigh-Jeans Law (cont)

4

1λ

In the classical limit of large λ

( ) 4 large8λπλ λ

TkTu Basymp

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

u as a function of the wavelength

High ν limit Wienrsquos Displacement Law

The maximum of u(ν) shifts toward higher frequencies with increasing temperature The position of maximum

( ) 011

3

1exp2

32

3

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

minusminus

minustimes=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛times=

x

x

x

B

B

B

eex

exconst

TkhTk

h

Tkhd

dconstddu

ν

ν

νν

( ) 8233 asymprarr=minus 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(ν

T)

ν

82max asympTk

h

B

ν

hTkB82max asympν

Nobel 1911

At high frequencieslow temperatures ( ) ( )νβνβνβ hhh exp1exp1 congminusgtgt

( ) ( )νβνπν hc

hTus minus= exp8 33

νmax hArr λmax

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

82max asympTk

h

B

ν- does this mean that

82max

asympλTk

hc

B

maxν maxλ

No

( ) ( ) ( ) ( )

( ) 0

11exp1exp

11exp5

11exp1

25

2

65 =⎥⎦

⎤⎢⎣

⎡

minusminus

minusminus

minustimes=⎥⎦

⎤⎢⎣

⎡minus

times=minus

xxxx

xxconst

xxdxdconst

dfdu

( ) ( )xxx 1exp11exp5 =minus rarr Tkhc

B5max asympλ

ldquonight visionrdquo devices

T = 300 K rarr λmax asymp 10 μm

( )Tu ν ( )Tu λ

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max μλ asymp=

Tkhc

B

eVTkhu B 4182maxmax asymp== ν

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 RadiationThe total number of photons per unit volume

The total energy of photons per unit volume (the energy density of a photon gas) ( ) ( )

( )( )( )3

45

0 158

1exp hcTkdg

VUTu Bπε

βεεε

=minus

times=equiv int

infin

the Stefan-Boltzmann Law23

45

152

chkBπσ =

the Stefan-Boltzmann constant ( ) 44 T

cTu σ

=

( ) ( ) 4281

8

1exp

8 33

0

23

30

2

30

times⎟⎠⎞

⎜⎝⎛=

minus⎟⎠⎞

⎜⎝⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==equiv intintint

infininfininfin

Thck

edxx

hTk

cd

Tkhc

dgnVNn B

xB

B

ππνν

νπεεε

- increases as T 3

The average energy per photon ( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant ( )248 10765 mKWminustimes=σ

Consider a black body at 310K The power emitted by the body 24 500 mWT asympσ

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 times 1s

energy density u1m2

T

424 TR σπ=

uctimes=

Integration over all angles provides a factor of frac14 uctimes=

41areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions ( ) 444

4c

4cpower TT

cTu σσ

=times==

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 λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

High ν limit Wienrsquos Displacement Law

The maximum of u(ν) shifts toward higher frequencies with increasing temperature The position of maximum

( ) 011

3

1exp2

32

3

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

minusminus

minustimes=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛times=

x

x

x

B

B

B

eex

exconst

TkhTk

h

Tkhd

dconstddu

ν

ν

νν

( ) 8233 asymprarr=minus 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(ν

T)

ν

82max asympTk

h

B

ν

hTkB82max asympν

Nobel 1911

At high frequencieslow temperatures ( ) ( )νβνβνβ hhh exp1exp1 congminusgtgt

( ) ( )νβνπν hc

hTus minus= exp8 33

νmax hArr λmax

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

82max asympTk

h

B

ν- does this mean that

82max

asympλTk

hc

B

maxν maxλ

No

( ) ( ) ( ) ( )

( ) 0

11exp1exp

11exp5

11exp1

25

2

65 =⎥⎦

⎤⎢⎣

⎡

minusminus

minusminus

minustimes=⎥⎦

⎤⎢⎣

⎡minus

times=minus

xxxx

xxconst

xxdxdconst

dfdu

( ) ( )xxx 1exp11exp5 =minus rarr Tkhc

B5max asympλ

ldquonight visionrdquo devices

T = 300 K rarr λmax asymp 10 μm

( )Tu ν ( )Tu λ

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max μλ asymp=

Tkhc

B

eVTkhu B 4182maxmax asymp== ν

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 RadiationThe total number of photons per unit volume

The total energy of photons per unit volume (the energy density of a photon gas) ( ) ( )

( )( )( )3

45

0 158

1exp hcTkdg

VUTu Bπε

βεεε

=minus

times=equiv int

infin

the Stefan-Boltzmann Law23

45

152

chkBπσ =

the Stefan-Boltzmann constant ( ) 44 T

cTu σ

=

( ) ( ) 4281

8

1exp

8 33

0

23

30

2

30

times⎟⎠⎞

⎜⎝⎛=

minus⎟⎠⎞

⎜⎝⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==equiv intintint

infininfininfin

Thck

edxx

hTk

cd

Tkhc

dgnVNn B

xB

B

ππνν

νπεεε

- increases as T 3

The average energy per photon ( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant ( )248 10765 mKWminustimes=σ

Consider a black body at 310K The power emitted by the body 24 500 mWT asympσ

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 times 1s

energy density u1m2

T

424 TR σπ=

uctimes=

Integration over all angles provides a factor of frac14 uctimes=

41areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions ( ) 444

4c

4cpower TT

cTu σσ

=times==

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 λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

νmax hArr λmax

( ) ( ) ( )( ) 1exp

18

1exp

8 52

3

32

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

minus⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

=⎥⎦⎤

⎢⎣⎡ minus=minus=

Tkhc

hchc

Tkhc

ch

hcTuhc

dddTudTu

BB λλπ

λλ

λπλλλ

εεελλ

82max asympTk

h

B

ν- does this mean that

82max

asympλTk

hc

B

maxν maxλ

No

( ) ( ) ( ) ( )

( ) 0

11exp1exp

11exp5

11exp1

25

2

65 =⎥⎦

⎤⎢⎣

⎡

minusminus

minusminus

minustimes=⎥⎦

⎤⎢⎣

⎡minus

times=minus

xxxx

xxconst

xxdxdconst

dfdu

( ) ( )xxx 1exp11exp5 =minus rarr Tkhc

B5max asympλ

ldquonight visionrdquo devices

T = 300 K rarr λmax asymp 10 μm

( )Tu ν ( )Tu λ

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max μλ asymp=

Tkhc

B

eVTkhu B 4182maxmax asymp== ν

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 RadiationThe total number of photons per unit volume

The total energy of photons per unit volume (the energy density of a photon gas) ( ) ( )

( )( )( )3

45

0 158

1exp hcTkdg

VUTu Bπε

βεεε

=minus

times=equiv int

infin

the Stefan-Boltzmann Law23

45

152

chkBπσ =

the Stefan-Boltzmann constant ( ) 44 T

cTu σ

=

( ) ( ) 4281

8

1exp

8 33

0

23

30

2

30

times⎟⎠⎞

⎜⎝⎛=

minus⎟⎠⎞

⎜⎝⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==equiv intintint

infininfininfin

Thck

edxx

hTk

cd

Tkhc

dgnVNn B

xB

B

ππνν

νπεεε

- increases as T 3

The average energy per photon ( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant ( )248 10765 mKWminustimes=σ

Consider a black body at 310K The power emitted by the body 24 500 mWT asympσ

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 times 1s

energy density u1m2

T

424 TR σπ=

uctimes=

Integration over all angles provides a factor of frac14 uctimes=

41areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions ( ) 444

4c

4cpower TT

cTu σσ

=times==

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 λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

Solar RadiationThe surface temperature of the Sun - 5800K

As a function of energy the spectrum of sunlight peaks at a photon energy of

m 505max μλ asymp=

Tkhc

B

eVTkhu B 4182maxmax asymp== ν

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 RadiationThe total number of photons per unit volume

The total energy of photons per unit volume (the energy density of a photon gas) ( ) ( )

( )( )( )3

45

0 158

1exp hcTkdg

VUTu Bπε

βεεε

=minus

times=equiv int

infin

the Stefan-Boltzmann Law23

45

152

chkBπσ =

the Stefan-Boltzmann constant ( ) 44 T

cTu σ

=

( ) ( ) 4281

8

1exp

8 33

0

23

30

2

30

times⎟⎠⎞

⎜⎝⎛=

minus⎟⎠⎞

⎜⎝⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==equiv intintint

infininfininfin

Thck

edxx

hTk

cd

Tkhc

dgnVNn B

xB

B

ππνν

νπεεε

- increases as T 3

The average energy per photon ( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant ( )248 10765 mKWminustimes=σ

Consider a black body at 310K The power emitted by the body 24 500 mWT asympσ

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 times 1s

energy density u1m2

T

424 TR σπ=

uctimes=

Integration over all angles provides a factor of frac14 uctimes=

41areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions ( ) 444

4c

4cpower TT

cTu σσ

=times==

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 λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

Stefan-Boltzmann Law of RadiationThe total number of photons per unit volume

The total energy of photons per unit volume (the energy density of a photon gas) ( ) ( )

( )( )( )3

45

0 158

1exp hcTkdg

VUTu Bπε

βεεε

=minus

times=equiv int

infin

the Stefan-Boltzmann Law23

45

152

chkBπσ =

the Stefan-Boltzmann constant ( ) 44 T

cTu σ

=

( ) ( ) 4281

8

1exp

8 33

0

23

30

2

30

times⎟⎠⎞

⎜⎝⎛=

minus⎟⎠⎞

⎜⎝⎛=

minus⎟⎟⎠

⎞⎜⎜⎝

⎛==equiv intintint

infininfininfin

Thck

edxx

hTk

cd

Tkhc

dgnVNn B

xB

B

ππνν

νπεεε

- increases as T 3

The average energy per photon ( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant ( )248 10765 mKWminustimes=σ

Consider a black body at 310K The power emitted by the body 24 500 mWT asympσ

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 times 1s

energy density u1m2

T

424 TR σπ=

uctimes=

Integration over all angles provides a factor of frac14 uctimes=

41areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions ( ) 444

4c

4cpower TT

cTu σσ

=times==

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 λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

Power Emitted by a Black Body

For the ldquouni-directionalrdquo motion the flux of energy per unit area

c times 1s

energy density u1m2

T

424 TR σπ=

uctimes=

Integration over all angles provides a factor of frac14 uctimes=

41areaunit by emittedpower

Thus the power emitted by a unit-area surface at temperature T in all directions ( ) 444

4c

4cpower TT

cTu σσ

=times==

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 λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength λ = 05 μm 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

λmax = 05 μm rarr ⎟⎟⎠

⎞⎜⎜⎝

⎛=

sdottimessdottimessdottimessdot

==rarr= minusminus

minus

KKkhcT

Tkhc

BB

74051050103815

103106655 623

834

maxmax λ

λ

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

KmW 1075

152 minussdotasymp=

chkBπσ( ) 424sphere aby emittedpower TRP σπ=

( ) ( ) ( ) W1083740K5Km

W 1075m 107482

4 26442

8284

max

2 sdot=timessdottimessdot=⎟⎟⎠

⎞⎜⎜⎝

⎛= minusπ

λσπ

BSun k

hcRP

( ) kgs 1024m 103

W1083 928

26

2 sdot=sdot

sdot==

cP

dtdm

the mass loss per one second

1 of Sunrsquos mass will be lost in yr 1015s 1074kgs 1024kg 102

010 1118

9

28

sdot=sdot=sdotsdot

==ΔdtdmMt

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

Radiative Energy TransferDewar

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 σminus=

The net ingoing flux

( ) ( ) rJTrJJrTrJ ba +minus=primeprime+minus= 44 11 σσLet 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

11

ba TTrrJJ minus

+minus

=primeminus σ

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

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 minus=σWithout 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 ⎟⎟⎠

⎞⎜⎜⎝

⎛ +==+minus=minus

TTTTTTTTTT σσ

The energy flux between the 1st and 2nd planes in the presence of the third plane

( ) ( ) 042

41

42

414

14

34

1 21

21

2JTTTTTTTJ =minus=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +minus=minus= σσσ - 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

+=

NJJ N

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission 424outPower EE TR σπ=

the flux of the solar radiation energy received by the Earth ~ 1370 Wm2

( ) ( )2

42in ower ⎟⎟⎠

⎞⎜⎜⎝

⎛=

orbit

SunSunE R

RTRP σπα

Sunorbit

SunE T

RRT

412

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

α

α = 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality α = 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

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

Thermodynamic Functions of Blackbody Radiation

444

34

3164 VT

cVT

cVT

cTSUF σσσ

minus=minus=minus=

( )NPVFPVTSUG μ==+=+minus= 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

TucV

Vσ

=⎥⎦⎤

⎢⎣⎡part

partequiv

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 31616 VT

cTdT

cVTd

TTcTS

TTV σσ

=primeprime=primeprimeprime

= intint

the pressure of a photon gas(radiation pressure)

uTcV

FPNT 3

134 4

==⎟⎠⎞

⎜⎝⎛partpart

minus=σ

UPV31

=

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 907231

31

31

=timestimes=timestimes== ε

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

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

316σ

Since V prop R3 the isentropic expansion leads to

1minusprop 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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

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 νmax (in Hz) corresponds to the maximum spectral density

u(νT) of the cosmic background radiation(c) (5) Do the maxima u(λT) and u(νT) correspond to the same photon energy If

not why

mmmTk

hc

B

11101172103815

10310665

323

834

max =sdot=timessdottimessdottimessdot

=asymp minusminus

minus

λ(a)

(b)

(c)

HzhTkB 11

34

23

max 105811066

72103818282 sdot=sdot

timessdottimes=asymp minus

minus

ν meVh 650max =ν

meVhc 11max

=λ

the maxima u(λT) and u(νT) do not correspond to the same photon energy The reason of that is

( ) ( ) ( )1exp

18 52

minus⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥⎦

⎤⎢⎣⎡ minus=minus=

Tkhc

hcTucdddTudTu

Bλλπλ

λλνννλλ

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)

Problem 2006 (blackbody radiation)

( ) 216

23

62

2 103721038172

103ms

photonsJ

mWJ

msphotonsN

sdotsdotasymp

timessdottimessdot

asymp⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛

sdot minus

minus

ε

( ) ( ) ( )( ) ( )

TkTkTkhchcTk

NTu

BBB

B 72421542815

8 4

33

345

asymptimes

=times

==π

ππε

(d) ( ) ( ) 26442484 103721075 mWKmKWTJ CMBRminusminus sdot=timessdotsdot==σ

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [ie photons(sm2)]

• Lecture 26 Blackbody Radiation (Ch 7)
• Radiation in Equilibrium with Matter
• Photons
• Chemical Potential of Photons = 0
• Density of States for Photons
• Spectrum of Blackbody Radiation
• Classical Limit (small f large ) Rayleigh-Jeans Law
• Rayleigh-Jeans Law (cont)
• High limit Wienrsquos Displacement Law
• max max
• Solar Radiation
• Stefan-Boltzmann Law of Radiation
• Power Emitted by a Black Body
• Sunrsquos Mass Loss
• Radiative Energy Transfer
• Superinsulation
• The Greenhouse Effect
• Thermodynamic Functions of Blackbody Radiation
• Radiation in the Universe
• Cosmic Microwave Background
• CMBR (cont)
• Problem 2006 (blackbody radiation)
• Problem 2006 (blackbody radiation)