Lecture 26 Blackbody Radiation (Ch 7)

Two types of bosons (a) Composite particles which contain an even

number of fermions These number of these particles is conserved if the energy does not exceed the dissociation energy (~ MeV in the case of the nucleus)

(b) particles associated with a field of which the most important example is the photon These particles are not conserved if the total energy of the field changes particles appear and disappear Wersquoll see that the chemical potential of such particles is zero in equilibrium regardless of density

Radiation in Equilibrium with MatterTypically radiation emitted by a hot body or from a laser is not in equilibrium energy is flowing outwards and must be replenished from some source The first step towards understanding of radiation being in equilibrium with matter was made by Kirchhoff who considered a cavity filled with radiation the walls can be regarded as a heat bath for radiation

The walls emit and absorb e-m waves In equilibrium the walls and radiation must have the same temperature T The energy of radiation is spread over a range of frequencies and we define uS (T) d as the energy density (per unit volume) of the radiation with frequencies between and +d uS(T) is the spectral energy density The internal energy of the photon gas

dTuTu S

0

In equilibrium uS (T) is the same everywhere in the cavity and is a function of frequency and temperature only If the cavity volume increases at T=const the internal energy U = u (T) V also increases The essential difference between the photon gas and the ideal gas of molecules for an ideal gas an isothermal expansion would conserve the gas energy whereas for the photon gas it is the energy density which is unchanged the number of photons is not conserved but proportional to volume in an isothermal change

A real surface absorbs only a fraction of the radiation falling on it The absorptivity is a function of and T a surface for which ( ) =1 for all frequencies is called a black body

Photons

The electromagnetic field has an infinite number of modes (standing waves) in the cavity The black-body radiation field is a superposition of plane waves of different frequencies The characteristic feature of the radiation is that a mode may be excited only in units of the quantum of energy h (similar to a harmonic oscillators)

hnii21

This fact leads to the concept of photons as quanta of the electromagnetic field The state of the el-mag field is specified by the number n for each of the modes or in other words by enumerating the number of photons with each frequency

According to the quantum theory of radiation photons are massless bosons of spin 1 (in units ħ) They move with the speed of light

ch

c

Ep

cpE

hE

phph

phph

ph

The linearity of Maxwell equations implies that the photons do not interact with each other (Non-linear optical phenomena are observed when a large-intensity radiation interacts with matter)

T

The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter Presence of a small amount of matter is essential for establishing equilibrium in the photon gas Wersquoll treat a system of photons as an ideal photon gas and in particular wersquoll apply the BE statistics to this system

Chemical Potential of Photons = 0The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter The textbook suggests that N can be found from the equilibrium condition

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

0

VTN

F

On the other hand ph

VTN

F

However we cannot use the usual expression for the chemical potential because one cannot increase N (ie add photons to the system) at constant volume and at the same time keep the temperature constant

VTN

F

- does not exist for the photon gas

Instead we can use NG PVFG V

VTF

V

FP

T

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

0 VV

FFGThus - the Gibbs free energy of an

equilibrium photon gas is 0 0

N

Gph

For = 0 the BE distribution reduces to the Planckrsquos distribution

1exp

1

1exp

1

Tkh

Tk

Tfn

BB

phph Planckrsquos distribution provides the average

number of photons in a single mode of frequency = h

Density of States for Photons

ky

kx

kz

2

3

2

33

66

34

8

1

k

kGvolumek

LLL

kkN

zyx

extra factor of 2 due to two polarizations

32

23

32

3

26

cg

cGkccp

d

dGg D

ph

3

2

32

233 8

cc

hh

d

dgg Dph

Dph

3

23 8

cg Dph

1exp

Tkh

hhn

B

In the classical (h ltlt kBT) limit TkB

The average energy in the mode

In order to calculate the average number of photons per small energy interval d the average energy of photons per small energy interval d etc as well as the total average number of photons in a photon gas and its total energy we need to know the density of states for photons as a function of photon energy

Spectrum of Blackbody Radiation

1exp

8

3

3

Tk

hcTu

B

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

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

dTudngh S

- the spectral density of the black-body radiation

(the Plankrsquos radiation law) 1exp

8

3

3

hc

hfghTus

hThud

dTuTudTudTu

u as a function of the energy

3

h

g(

)

=n( )

average number of photons

photon energy

3

23 8

cg Dph

h

g(

) n

( )

Classical Limit (small f large ) Rayleigh-Jeans Law

This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths

Rayleigh-Jeans Law

At low frequencies or high temperatures hhh 1exp1

Tkchc

hTu Bs 3

23

3

8

1exp

8

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

Rayleigh-Jeans Law (cont)

4

1

In the classical limit of large

4 large

8

TkTu B

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

u as a function of the wavelength

High limit Wienrsquos Displacement Law

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

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures hhh exp1exp1

hc

hTus exp

8 3

3

max max

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

82max Tk

h

B

- does this mean that

82max

Tkhc

B

max max

No

0

11exp

1exp

11exp

5

11exp

125

2

65

xx

xx

xxconst

xxdx

dconst

df

du

xxx 1exp11exp5 Tk

hc

B5max

ldquonight visionrdquo devices

T = 300 K max 10 m

Tu Tu

Solar Radiation

The surface temperature of the Sun - 5800K

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

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The total number of photons per unit volume

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

3

45

0 15

8

1exp hc

Tkd

g

V

UTu B

the Stefan-Boltzmann Law23

45

15

2

ch

kB the Stefan-Boltzmann constant 44

Tc

Tu

4281

8

1exp

8 33

0

23

30

2

30

Thc

k

e

dxx

h

Tk

cd

Tk

hcdgn

V

Nn B

xB

B

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant 248 10765 mKW

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

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Power Emitted by a Black Body

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

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

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

4

c

4

cpower TT

cTu

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Radiation in Equilibrium with MatterTypically radiation emitted by a hot body or from a laser is not in equilibrium energy is flowing outwards and must be replenished from some source The first step towards understanding of radiation being in equilibrium with matter was made by Kirchhoff who considered a cavity filled with radiation the walls can be regarded as a heat bath for radiation

The walls emit and absorb e-m waves In equilibrium the walls and radiation must have the same temperature T The energy of radiation is spread over a range of frequencies and we define uS (T) d as the energy density (per unit volume) of the radiation with frequencies between and +d uS(T) is the spectral energy density The internal energy of the photon gas

dTuTu S

0

In equilibrium uS (T) is the same everywhere in the cavity and is a function of frequency and temperature only If the cavity volume increases at T=const the internal energy U = u (T) V also increases The essential difference between the photon gas and the ideal gas of molecules for an ideal gas an isothermal expansion would conserve the gas energy whereas for the photon gas it is the energy density which is unchanged the number of photons is not conserved but proportional to volume in an isothermal change

A real surface absorbs only a fraction of the radiation falling on it The absorptivity is a function of and T a surface for which ( ) =1 for all frequencies is called a black body

Photons

The electromagnetic field has an infinite number of modes (standing waves) in the cavity The black-body radiation field is a superposition of plane waves of different frequencies The characteristic feature of the radiation is that a mode may be excited only in units of the quantum of energy h (similar to a harmonic oscillators)

hnii21

This fact leads to the concept of photons as quanta of the electromagnetic field The state of the el-mag field is specified by the number n for each of the modes or in other words by enumerating the number of photons with each frequency

According to the quantum theory of radiation photons are massless bosons of spin 1 (in units ħ) They move with the speed of light

ch

c

Ep

cpE

hE

phph

phph

ph

The linearity of Maxwell equations implies that the photons do not interact with each other (Non-linear optical phenomena are observed when a large-intensity radiation interacts with matter)

T

The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter Presence of a small amount of matter is essential for establishing equilibrium in the photon gas Wersquoll treat a system of photons as an ideal photon gas and in particular wersquoll apply the BE statistics to this system

Chemical Potential of Photons = 0The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter The textbook suggests that N can be found from the equilibrium condition

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

0

VTN

F

On the other hand ph

VTN

F

However we cannot use the usual expression for the chemical potential because one cannot increase N (ie add photons to the system) at constant volume and at the same time keep the temperature constant

VTN

F

- does not exist for the photon gas

Instead we can use NG PVFG V

VTF

V

FP

T

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

0 VV

FFGThus - the Gibbs free energy of an

equilibrium photon gas is 0 0

N

Gph

For = 0 the BE distribution reduces to the Planckrsquos distribution

1exp

1

1exp

1

Tkh

Tk

Tfn

BB

phph Planckrsquos distribution provides the average

number of photons in a single mode of frequency = h

Density of States for Photons

ky

kx

kz

2

3

2

33

66

34

8

1

k

kGvolumek

LLL

kkN

zyx

extra factor of 2 due to two polarizations

32

23

32

3

26

cg

cGkccp

d

dGg D

ph

3

2

32

233 8

cc

hh

d

dgg Dph

Dph

3

23 8

cg Dph

1exp

Tkh

hhn

B

In the classical (h ltlt kBT) limit TkB

The average energy in the mode

In order to calculate the average number of photons per small energy interval d the average energy of photons per small energy interval d etc as well as the total average number of photons in a photon gas and its total energy we need to know the density of states for photons as a function of photon energy

Spectrum of Blackbody Radiation

1exp

8

3

3

Tk

hcTu

B

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

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

dTudngh S

- the spectral density of the black-body radiation

(the Plankrsquos radiation law) 1exp

8

3

3

hc

hfghTus

hThud

dTuTudTudTu

u as a function of the energy

3

h

g(

)

=n( )

average number of photons

photon energy

3

23 8

cg Dph

h

g(

) n

( )

Classical Limit (small f large ) Rayleigh-Jeans Law

This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths

Rayleigh-Jeans Law

At low frequencies or high temperatures hhh 1exp1

Tkchc

hTu Bs 3

23

3

8

1exp

8

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

Rayleigh-Jeans Law (cont)

4

1

In the classical limit of large

4 large

8

TkTu B

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

u as a function of the wavelength

High limit Wienrsquos Displacement Law

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

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures hhh exp1exp1

hc

hTus exp

8 3

3

max max

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

82max Tk

h

B

- does this mean that

82max

Tkhc

B

max max

No

0

11exp

1exp

11exp

5

11exp

125

2

65

xx

xx

xxconst

xxdx

dconst

df

du

xxx 1exp11exp5 Tk

hc

B5max

ldquonight visionrdquo devices

T = 300 K max 10 m

Tu Tu

Solar Radiation

The surface temperature of the Sun - 5800K

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

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The total number of photons per unit volume

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

3

45

0 15

8

1exp hc

Tkd

g

V

UTu B

the Stefan-Boltzmann Law23

45

15

2

ch

kB the Stefan-Boltzmann constant 44

Tc

Tu

4281

8

1exp

8 33

0

23

30

2

30

Thc

k

e

dxx

h

Tk

cd

Tk

hcdgn

V

Nn B

xB

B

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant 248 10765 mKW

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

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Power Emitted by a Black Body

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

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

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

4

c

4

cpower TT

cTu

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Photons

The electromagnetic field has an infinite number of modes (standing waves) in the cavity The black-body radiation field is a superposition of plane waves of different frequencies The characteristic feature of the radiation is that a mode may be excited only in units of the quantum of energy h (similar to a harmonic oscillators)

hnii21

This fact leads to the concept of photons as quanta of the electromagnetic field The state of the el-mag field is specified by the number n for each of the modes or in other words by enumerating the number of photons with each frequency

According to the quantum theory of radiation photons are massless bosons of spin 1 (in units ħ) They move with the speed of light

ch

c

Ep

cpE

hE

phph

phph

ph

The linearity of Maxwell equations implies that the photons do not interact with each other (Non-linear optical phenomena are observed when a large-intensity radiation interacts with matter)

T

The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter Presence of a small amount of matter is essential for establishing equilibrium in the photon gas Wersquoll treat a system of photons as an ideal photon gas and in particular wersquoll apply the BE statistics to this system

Chemical Potential of Photons = 0The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter The textbook suggests that N can be found from the equilibrium condition

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

0

VTN

F

On the other hand ph

VTN

F

However we cannot use the usual expression for the chemical potential because one cannot increase N (ie add photons to the system) at constant volume and at the same time keep the temperature constant

VTN

F

- does not exist for the photon gas

Instead we can use NG PVFG V

VTF

V

FP

T

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

0 VV

FFGThus - the Gibbs free energy of an

equilibrium photon gas is 0 0

N

Gph

For = 0 the BE distribution reduces to the Planckrsquos distribution

1exp

1

1exp

1

Tkh

Tk

Tfn

BB

phph Planckrsquos distribution provides the average

number of photons in a single mode of frequency = h

Density of States for Photons

ky

kx

kz

2

3

2

33

66

34

8

1

k

kGvolumek

LLL

kkN

zyx

extra factor of 2 due to two polarizations

32

23

32

3

26

cg

cGkccp

d

dGg D

ph

3

2

32

233 8

cc

hh

d

dgg Dph

Dph

3

23 8

cg Dph

1exp

Tkh

hhn

B

In the classical (h ltlt kBT) limit TkB

The average energy in the mode

In order to calculate the average number of photons per small energy interval d the average energy of photons per small energy interval d etc as well as the total average number of photons in a photon gas and its total energy we need to know the density of states for photons as a function of photon energy

Spectrum of Blackbody Radiation

1exp

8

3

3

Tk

hcTu

B

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

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

dTudngh S

- the spectral density of the black-body radiation

(the Plankrsquos radiation law) 1exp

8

3

3

hc

hfghTus

hThud

dTuTudTudTu

u as a function of the energy

3

h

g(

)

=n( )

average number of photons

photon energy

3

23 8

cg Dph

h

g(

) n

( )

Classical Limit (small f large ) Rayleigh-Jeans Law

This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths

Rayleigh-Jeans Law

At low frequencies or high temperatures hhh 1exp1

Tkchc

hTu Bs 3

23

3

8

1exp

8

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

Rayleigh-Jeans Law (cont)

4

1

In the classical limit of large

4 large

8

TkTu B

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

u as a function of the wavelength

High limit Wienrsquos Displacement Law

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

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures hhh exp1exp1

hc

hTus exp

8 3

3

max max

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

82max Tk

h

B

- does this mean that

82max

Tkhc

B

max max

No

0

11exp

1exp

11exp

5

11exp

125

2

65

xx

xx

xxconst

xxdx

dconst

df

du

xxx 1exp11exp5 Tk

hc

B5max

ldquonight visionrdquo devices

T = 300 K max 10 m

Tu Tu

Solar Radiation

The surface temperature of the Sun - 5800K

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

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The total number of photons per unit volume

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

3

45

0 15

8

1exp hc

Tkd

g

V

UTu B

the Stefan-Boltzmann Law23

45

15

2

ch

kB the Stefan-Boltzmann constant 44

Tc

Tu

4281

8

1exp

8 33

0

23

30

2

30

Thc

k

e

dxx

h

Tk

cd

Tk

hcdgn

V

Nn B

xB

B

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant 248 10765 mKW

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

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Power Emitted by a Black Body

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

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

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

4

c

4

cpower TT

cTu

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Chemical Potential of Photons = 0The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter The textbook suggests that N can be found from the equilibrium condition

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

0

VTN

F

On the other hand ph

VTN

F

However we cannot use the usual expression for the chemical potential because one cannot increase N (ie add photons to the system) at constant volume and at the same time keep the temperature constant

VTN

F

- does not exist for the photon gas

Instead we can use NG PVFG V

VTF

V

FP

T

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

0 VV

FFGThus - the Gibbs free energy of an

equilibrium photon gas is 0 0

N

Gph

For = 0 the BE distribution reduces to the Planckrsquos distribution

1exp

1

1exp

1

Tkh

Tk

Tfn

BB

phph Planckrsquos distribution provides the average

number of photons in a single mode of frequency = h

Density of States for Photons

ky

kx

kz

2

3

2

33

66

34

8

1

k

kGvolumek

LLL

kkN

zyx

extra factor of 2 due to two polarizations

32

23

32

3

26

cg

cGkccp

d

dGg D

ph

3

2

32

233 8

cc

hh

d

dgg Dph

Dph

3

23 8

cg Dph

1exp

Tkh

hhn

B

In the classical (h ltlt kBT) limit TkB

The average energy in the mode

In order to calculate the average number of photons per small energy interval d the average energy of photons per small energy interval d etc as well as the total average number of photons in a photon gas and its total energy we need to know the density of states for photons as a function of photon energy

Spectrum of Blackbody Radiation

1exp

8

3

3

Tk

hcTu

B

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

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

dTudngh S

- the spectral density of the black-body radiation

(the Plankrsquos radiation law) 1exp

8

3

3

hc

hfghTus

hThud

dTuTudTudTu

u as a function of the energy

3

h

g(

)

=n( )

average number of photons

photon energy

3

23 8

cg Dph

h

g(

) n

( )

Classical Limit (small f large ) Rayleigh-Jeans Law

This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths

Rayleigh-Jeans Law

At low frequencies or high temperatures hhh 1exp1

Tkchc

hTu Bs 3

23

3

8

1exp

8

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

Rayleigh-Jeans Law (cont)

4

1

In the classical limit of large

4 large

8

TkTu B

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

u as a function of the wavelength

High limit Wienrsquos Displacement Law

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

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures hhh exp1exp1

hc

hTus exp

8 3

3

max max

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

82max Tk

h

B

- does this mean that

82max

Tkhc

B

max max

No

0

11exp

1exp

11exp

5

11exp

125

2

65

xx

xx

xxconst

xxdx

dconst

df

du

xxx 1exp11exp5 Tk

hc

B5max

ldquonight visionrdquo devices

T = 300 K max 10 m

Tu Tu

Solar Radiation

The surface temperature of the Sun - 5800K

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

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The total number of photons per unit volume

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

3

45

0 15

8

1exp hc

Tkd

g

V

UTu B

the Stefan-Boltzmann Law23

45

15

2

ch

kB the Stefan-Boltzmann constant 44

Tc

Tu

4281

8

1exp

8 33

0

23

30

2

30

Thc

k

e

dxx

h

Tk

cd

Tk

hcdgn

V

Nn B

xB

B

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant 248 10765 mKW

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

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Power Emitted by a Black Body

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

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

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

4

c

4

cpower TT

cTu

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Density of States for Photons

ky

kx

kz

2

3

2

33

66

34

8

1

k

kGvolumek

LLL

kkN

zyx

extra factor of 2 due to two polarizations

32

23

32

3

26

cg

cGkccp

d

dGg D

ph

3

2

32

233 8

cc

hh

d

dgg Dph

Dph

3

23 8

cg Dph

1exp

Tkh

hhn

B

In the classical (h ltlt kBT) limit TkB

The average energy in the mode

In order to calculate the average number of photons per small energy interval d the average energy of photons per small energy interval d etc as well as the total average number of photons in a photon gas and its total energy we need to know the density of states for photons as a function of photon energy

Spectrum of Blackbody Radiation

1exp

8

3

3

Tk

hcTu

B

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

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

dTudngh S

- the spectral density of the black-body radiation

(the Plankrsquos radiation law) 1exp

8

3

3

hc

hfghTus

hThud

dTuTudTudTu

u as a function of the energy

3

h

g(

)

=n( )

average number of photons

photon energy

3

23 8

cg Dph

h

g(

) n

( )

Classical Limit (small f large ) Rayleigh-Jeans Law

This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths

Rayleigh-Jeans Law

At low frequencies or high temperatures hhh 1exp1

Tkchc

hTu Bs 3

23

3

8

1exp

8

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

Rayleigh-Jeans Law (cont)

4

1

In the classical limit of large

4 large

8

TkTu B

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

u as a function of the wavelength

High limit Wienrsquos Displacement Law

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

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures hhh exp1exp1

hc

hTus exp

8 3

3

max max

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

82max Tk

h

B

- does this mean that

82max

Tkhc

B

max max

No

0

11exp

1exp

11exp

5

11exp

125

2

65

xx

xx

xxconst

xxdx

dconst

df

du

xxx 1exp11exp5 Tk

hc

B5max

ldquonight visionrdquo devices

T = 300 K max 10 m

Tu Tu

Solar Radiation

The surface temperature of the Sun - 5800K

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

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The total number of photons per unit volume

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

3

45

0 15

8

1exp hc

Tkd

g

V

UTu B

the Stefan-Boltzmann Law23

45

15

2

ch

kB the Stefan-Boltzmann constant 44

Tc

Tu

4281

8

1exp

8 33

0

23

30

2

30

Thc

k

e

dxx

h

Tk

cd

Tk

hcdgn

V

Nn B

xB

B

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant 248 10765 mKW

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

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Power Emitted by a Black Body

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

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

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

4

c

4

cpower TT

cTu

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Spectrum of Blackbody Radiation

1exp

8

3

3

Tk

hcTu

B

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

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

dTudngh S

- the spectral density of the black-body radiation

(the Plankrsquos radiation law) 1exp

8

3

3

hc

hfghTus

hThud

dTuTudTudTu

u as a function of the energy

3

h

g(

)

=n( )

average number of photons

photon energy

3

23 8

cg Dph

h

g(

) n

( )

Classical Limit (small f large ) Rayleigh-Jeans Law

This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths

Rayleigh-Jeans Law

At low frequencies or high temperatures hhh 1exp1

Tkchc

hTu Bs 3

23

3

8

1exp

8

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

Rayleigh-Jeans Law (cont)

4

1

In the classical limit of large

4 large

8

TkTu B

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

u as a function of the wavelength

High limit Wienrsquos Displacement Law

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

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures hhh exp1exp1

hc

hTus exp

8 3

3

max max

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

82max Tk

h

B

- does this mean that

82max

Tkhc

B

max max

No

0

11exp

1exp

11exp

5

11exp

125

2

65

xx

xx

xxconst

xxdx

dconst

df

du

xxx 1exp11exp5 Tk

hc

B5max

ldquonight visionrdquo devices

T = 300 K max 10 m

Tu Tu

Solar Radiation

The surface temperature of the Sun - 5800K

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

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The total number of photons per unit volume

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

3

45

0 15

8

1exp hc

Tkd

g

V

UTu B

the Stefan-Boltzmann Law23

45

15

2

ch

kB the Stefan-Boltzmann constant 44

Tc

Tu

4281

8

1exp

8 33

0

23

30

2

30

Thc

k

e

dxx

h

Tk

cd

Tk

hcdgn

V

Nn B

xB

B

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant 248 10765 mKW

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

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Power Emitted by a Black Body

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

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

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

4

c

4

cpower TT

cTu

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Classical Limit (small f large ) Rayleigh-Jeans Law

This equation predicts the so-called ultraviolet catastrophe ndash an infinite amount of energy being radiated at high frequencies or short wavelengths

Rayleigh-Jeans Law

At low frequencies or high temperatures hhh 1exp1

Tkchc

hTu Bs 3

23

3

8

1exp

8

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

Rayleigh-Jeans Law (cont)

4

1

In the classical limit of large

4 large

8

TkTu B

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

u as a function of the wavelength

High limit Wienrsquos Displacement Law

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

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures hhh exp1exp1

hc

hTus exp

8 3

3

max max

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

82max Tk

h

B

- does this mean that

82max

Tkhc

B

max max

No

0

11exp

1exp

11exp

5

11exp

125

2

65

xx

xx

xxconst

xxdx

dconst

df

du

xxx 1exp11exp5 Tk

hc

B5max

ldquonight visionrdquo devices

T = 300 K max 10 m

Tu Tu

Solar Radiation

The surface temperature of the Sun - 5800K

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

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The total number of photons per unit volume

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

3

45

0 15

8

1exp hc

Tkd

g

V

UTu B

the Stefan-Boltzmann Law23

45

15

2

ch

kB the Stefan-Boltzmann constant 44

Tc

Tu

4281

8

1exp

8 33

0

23

30

2

30

Thc

k

e

dxx

h

Tk

cd

Tk

hcdgn

V

Nn B

xB

B

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant 248 10765 mKW

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

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Power Emitted by a Black Body

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

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

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

4

c

4

cpower TT

cTu

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Rayleigh-Jeans Law (cont)

4

1

In the classical limit of large

4 large

8

TkTu B

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

u as a function of the wavelength

High limit Wienrsquos Displacement Law

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

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures hhh exp1exp1

hc

hTus exp

8 3

3

max max

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

82max Tk

h

B

- does this mean that

82max

Tkhc

B

max max

No

0

11exp

1exp

11exp

5

11exp

125

2

65

xx

xx

xxconst

xxdx

dconst

df

du

xxx 1exp11exp5 Tk

hc

B5max

ldquonight visionrdquo devices

T = 300 K max 10 m

Tu Tu

Solar Radiation

The surface temperature of the Sun - 5800K

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

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The total number of photons per unit volume

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

3

45

0 15

8

1exp hc

Tkd

g

V

UTu B

the Stefan-Boltzmann Law23

45

15

2

ch

kB the Stefan-Boltzmann constant 44

Tc

Tu

4281

8

1exp

8 33

0

23

30

2

30

Thc

k

e

dxx

h

Tk

cd

Tk

hcdgn

V

Nn B

xB

B

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant 248 10765 mKW

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

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Power Emitted by a Black Body

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

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

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

4

c

4

cpower TT

cTu

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

High limit Wienrsquos Displacement Law

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

011

3

1exp2

32

3

x

x

x

B

B

B

e

ex

e

xconst

Tkh

Tkh

Tkh

d

dconst

d

du

8233 xex x

Wienrsquos displacement law- discovered experimentally

by Wilhelm Wien

Numerous applications (eg non-contact radiation thermometry)

- the ldquomost likelyrdquo frequency of a photon in a blackbody radiation with temperature T

u(

T)

82max Tk

h

B

h

TkB82max

Nobel 1911

At high frequencieslow temperatures hhh exp1exp1

hc

hTus exp

8 3

3

max max

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

82max Tk

h

B

- does this mean that

82max

Tkhc

B

max max

No

0

11exp

1exp

11exp

5

11exp

125

2

65

xx

xx

xxconst

xxdx

dconst

df

du

xxx 1exp11exp5 Tk

hc

B5max

ldquonight visionrdquo devices

T = 300 K max 10 m

Tu Tu

Solar Radiation

The surface temperature of the Sun - 5800K

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

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The total number of photons per unit volume

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

3

45

0 15

8

1exp hc

Tkd

g

V

UTu B

the Stefan-Boltzmann Law23

45

15

2

ch

kB the Stefan-Boltzmann constant 44

Tc

Tu

4281

8

1exp

8 33

0

23

30

2

30

Thc

k

e

dxx

h

Tk

cd

Tk

hcdgn

V

Nn B

xB

B

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant 248 10765 mKW

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

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Power Emitted by a Black Body

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

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

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

4

c

4

cpower TT

cTu

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

max max

1exp

18

1exp

8

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

82max Tk

h

B

- does this mean that

82max

Tkhc

B

max max

No

0

11exp

1exp

11exp

5

11exp

125

2

65

xx

xx

xxconst

xxdx

dconst

df

du

xxx 1exp11exp5 Tk

hc

B5max

ldquonight visionrdquo devices

T = 300 K max 10 m

Tu Tu

Solar Radiation

The surface temperature of the Sun - 5800K

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

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The total number of photons per unit volume

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

3

45

0 15

8

1exp hc

Tkd

g

V

UTu B

the Stefan-Boltzmann Law23

45

15

2

ch

kB the Stefan-Boltzmann constant 44

Tc

Tu

4281

8

1exp

8 33

0

23

30

2

30

Thc

k

e

dxx

h

Tk

cd

Tk

hcdgn

V

Nn B

xB

B

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant 248 10765 mKW

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

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Power Emitted by a Black Body

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

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

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

4

c

4

cpower TT

cTu

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Solar Radiation

The surface temperature of the Sun - 5800K

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

m 505max

Tk

hc

B

eVTkhu B 4182maxmax

Spectral sensitivity of human eye

- close to the energy gap in Si ~11 eV which has been so far the best material for solar cells

Stefan-Boltzmann Law of Radiation

The total number of photons per unit volume

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

3

45

0 15

8

1exp hc

Tkd

g

V

UTu B

the Stefan-Boltzmann Law23

45

15

2

ch

kB the Stefan-Boltzmann constant 44

Tc

Tu

4281

8

1exp

8 33

0

23

30

2

30

Thc

k

e

dxx

h

Tk

cd

Tk

hcdgn

V

Nn B

xB

B

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant 248 10765 mKW

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

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Power Emitted by a Black Body

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

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

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

4

c

4

cpower TT

cTu

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Stefan-Boltzmann Law of Radiation

The total number of photons per unit volume

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

3

45

0 15

8

1exp hc

Tkd

g

V

UTu B

the Stefan-Boltzmann Law23

45

15

2

ch

kB the Stefan-Boltzmann constant 44

Tc

Tu

4281

8

1exp

8 33

0

23

30

2

30

Thc

k

e

dxx

h

Tk

cd

Tk

hcdgn

V

Nn B

xB

B

- increases as T 3

The average energy per photon

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(just slightly less than the ldquomostrdquo probable energy)

The value of the Stefan-Boltzmann constant 248 10765 mKW

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

While the emissivity of skin is considerably less than 1 it still emits a considerable power in the infrared range For example this radiation is easily detectable by modern techniques (night vision)

Some numbers

Power Emitted by a Black Body

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

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

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

4

c

4

cpower TT

cTu

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Power Emitted by a Black Body

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

c 1s

energy density u

1m2

T

424 TR

uc

Integration over all angles provides a factor of frac14

uc4

1areaunit by emittedpower

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

4

c

4

cpower TT

cTu

The total power emitted by a black-body sphere of radius R

(the hole size must be gtgt the wavelength)

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Sunrsquos Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 05 m Find the mass loss for the Sun in one second How long it takes for the Sun to loose 1 of its mass due to radiation Radius of the Sun 7middot108 m mass - 2 middot1030 kg

max = 05 m

KKk

hcT

Tk

hc

BB

74051050103815

1031066

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 Wm2) being multiplied by the area of a sphere with radius 15middot1011 m (Sun-Earth distance)

428

23

45

Km

W 1075

15

2 ch

kB 424sphere aby emittedpower TRP

W1083740K5Km

W 1075m 1074

824 264

42828

4

max

2

BSun k

hcRP

kgs 1024m 103

W1083 928

26

2

c

P

dt

dmthe mass loss per one second

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

kg 102

010 11189

28

dtdm

Mt

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Radiative Energy Transfer Dewar

Liquid nitrogen and helium are stored in a vacuum or Dewar flask a container surrounded by a thin evacuated jacket While the thermal conductivity of gas at very low pressure is small energy can still be transferred by radiation Both surfaces cold and warm radiate at a rate

24 1 mWTrJ irad

The net ingoing flux

rJTrJJrTrJ ba 44 11 Let the total ingoing flux be J and the total outgoing flux be Jrsquo

i=a for the outer (hot) wall i=b for the inner (cold) wallr ndash the coefficient of reflection (1-r) ndash the coefficient of emission

44

1

1ba TT

r

rJJ

If r=098 (walls are covered with silver mirror) the net flux is reduced to 1 of the value it would have if the surfaces were black bodies (r=0)

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

SuperinsulationTwo parallel black planes are at the temperatures T1 and T2 respectively The energy flux between these planes in vacuum is due to the blackbody radiation A third black plane is inserted between the other two and is allowed to come to an equilibrium temperature T3 Find T3 and show that the energy flux between planes 1 and 2 is cut in half because of the presence of the third plane T1 T2T3

42

41

0 TTJ Without the third plane the energy flux per unit area is

The equilibrium temperature of the third plane can be found from the energy balance

414

24

13

43

42

41

42

43

43

41 2

2

TTTTTTTTTT

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

042

41

42

414

14

34

1 2

1

2

1

2JTT

TTTTTJ

- cut in half

Superinsulation many layers of aluminized Mylar foil loosely wrapped around the helium bath (in a vacuum space between the walls of a LHe cryostat) The energy flux reduction for N heat shields

1

0

N

JJN

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

The Greenhouse Effect

Transmittance of the Earth atmosphere

Absorption

Emission424outPower EE TR

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

2

42in ower

orbit

SunSunE R

RTRP

Sunorbit

SunE T

R

RT

412

4

= 1 ndash TEarth = 280K

Rorbit = 15middot1011 m RSun = 7middot108 m

In reality = 07 ndash TEarth = 256K

To maintain a comfortable temperature on the Earth we need the Greenhouse Effect

However too much of the greenhouse effect leads to global warming

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Thermodynamic Functions of Blackbody Radiation

444

3

4

3

164VT

cVT

cVT

cTSUF

NPVFPVTSUG 0

Now we can derive all thermodynamic functions of blackbody radiation

the Helmholtz free energy

the Gibbs free energy

The heat capacity of a photon gas at constant volume 316

VTcT

Tuc

VV

This equation holds for all T (it agrees with the Nernst theorem) and we can integrate it to get the entropy of a photon gas

3

0

2

0 3

1616VT

cTdT

c

VTd

T

TcTS

TTV

the pressure of a photon gas(radiation pressure)

uTcV

FP

NT 3

1

3

4 4

UPV3

1

For comparison for a non-relativistic monatomic gas ndash PV = (23)U The difference ndash because the energy-momentum relationship for photons is ultra-relativistic and the number of photon depends on T In terms of the average density of phonons

TkNTknVnVUPV BB 90723

1

3

1

3

1

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Radiation in the Universe

The dependence of the radiated energy versus wavelength illustrates the main sources of the THz radiation the interstellar dust emission from light and heavy molecules and the 27-K cosmic background radiation

In the spectrum of the Milky Way galaxy at least one-half of the luminous power is emitted at sub-mm wavelengths

Approximately 98 of all the photons emitted since the Big Bang are observed now in the submillimeterTHz range

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Cosmic Microwave Background

In the standard Big Bang model the radiation is decoupled from the matter in the Universe about 300000 years after the Big Bang when the temperature dropped to the point where neutral atoms form (T~3000K) At this moment the Universe became transparent for the ldquoprimordialrdquo photons The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation

Nobel 1978A Penzias

R Wilson

constVTc

TS 3

3

16

Since V R3 the isentropic expansion leads to

1 RT

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

CMBR (cont)

Alternatively the later evolution of the radiation temperature may be considered as a result of the red (Doppler) shift (z) Since the CMBR photons were radiated at T~3000K the red shift z~1000

At present the temperature of the Planckrsquos distribution for the CMBR photons is 2735 K The radiation is coming from all directions and is quite distinct from the radiation from stars and galaxies

Mather Smoot Nobel 2006

ldquohellip for their discovery of the blackbody form and anisotropy of the CMBRrdquo

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Problem 2006 (blackbody radiation)The cosmic microwave background radiation (CMBR) has a temperature of

approximately 27 K(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λT) of the cosmic background radiation(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density

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

not why

mmmTk

hc

B

11101172103815

1031066

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 105811066

72103818282

meVh 650max

meVhc

11max

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

1exp

18

52

Tkhc

hcTu

c

d

ddTudTu

B

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23

Problem 2006 (blackbody radiation)

216

23

62

2103

721038172

103

ms

photons

JmW

J

ms

photonsN

TkTkTkhc

hcTk

N

TuBB

B

B 72421542815

8 4

33

345

(d) 26442484 103721075 mWKmKWTJ CMBR

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

• 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)
• Slide 23