Lecture 24. 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. We’ll see that the chemical potential of such particles is zero in equilibrium, regardless of density.

Ultraviolet Catastrophe (classical E&M)

Normal modes (standing waves) of EM wave in a box: , 1, 2,3,i i iL

Each mode has degeneracy 2 degrees of freedom (polarization). From equipartition theorem:

12

2i B Bk T k T

Total energy of the EM standing waves: 1

!i Bi

U k T

Catastrophe!

Solution: Light consist of quantized energy units, called photons with energy and momentum p:

2,

hc hpc p k k

And equipartition theorem is only applicable when i Bk T

The partition function of photon gas in a box: 0

1exp

1 expi ij i

Z j

explnln 1 exp

1 exp exp 1i ii i

i ii i

ZE

Average photon energy on ith level:

1

Since mean occupancy of photon: exp 1i i i i

i

E n n

Chemical Potential of Photons = 0

1

exp 1n

1

exp 1BEn

Recall Bose-Einstein distribution: For photon:

Photons are bosons with =0. Why ph=0?

Photons are not conserved because they can be emitted or absorbed by matters. Therefore, free energy cannot depends on total number (Nph) of photons, i.e.

ph matter ph matter phph ,

0 or + = =0 T P

G

N

The Planck spectrum (photon gas)

The total energy of photon gas at temperature T:

3

0 using continuum approx.

exp 1D

i i phi

U n g d

3 23 3 2

322 4

2D Dph ph

dk L dk Vg g k k

d d c

Spin (polarization) ck

2 3

3 320 0

3 34

3 34 0 0

8

exp 1 exp 1

8 1 8

1 1Bx x

V VU d d

c hc

V x V xdx k T dx

e ehc hc

x The energy density per unit photon energy :

(Planck spectrum)

3

3

8

exp 1

Vu

hc

The Planck spectrum (photon gas)

3

3

8,

exp 1

Vu T

hc

0 2 4 6 8 100.0

0.5

1.0

1.5

x3 /(ex -1

)

x =

peak 2.82 Bk T Wien’s law

Planck spectrum

Numerous applications (e.g., non-contact radiation thermometry)

3 4

0 Pr. B19

1 15x

xdx

e

Total energy of photo gas (T, V)

54

3

8

15B

VU k T

hc

Heat capacity

53

3

32

15V B B

V

U VC k T k

T hc

Entropy

53

30

' 32'

' 45

T VB B

C T VS dT k T k

T hc

54

3

8

15B

Uk T

V hc

energy density (unit volume)

max max

1exp

18

1exp

8,,,

52

3

32

Tkhc

hchc

Tkhc

ch

hcTu

hc

d

ddTudTu

BB

8.2max Tk

h

B

- does this mean that

8.2max

Tkhc

B

? max max

No!

0

1/1exp

/1exp

1/1exp

5

1/1exp

125

2

65

xx

xx

xxconst

xxdx

dconst

df

du

xxx /1exp1/1exp5 Tk

hc

B5max

“night vision” devices

T = 300 K max 10 m

Tu , Tu ,

Solar Radiation

The surface temperature of the Sun - 5,800K.

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

m 5.05max

Tk

hc

B

eVTkhu B 4.18.2maxmax

Spectral sensitivity of human eye:

- close to the energy gap in Si, ~1.1 eV, which has been so far the best material for solar cells

Blackbody radiation (photons escape from a hole)

Blackbody: absorb all EM radiation falls on it. Blackbody also emits photons with Planck spectrum. (Blackbody could be very bright, e.g. Sun)

The photons that come out of the hole (i.e. emitted by a blackbody with area A) come from the photo gas inside the box that is in thermal equilibrium of the wall of box at temperature T.

AA physical mode of a blackbody with surface area A a hole of area A of a very large box (so that incoming photo get absorbed before escape).

Blackbody radiation = photon flux (EM energy) of a photon gas at temperature T through surface area A (the hole).

Blackbody radiation (photons escape from a hole)

A

powerenergy flux

unit area

c 1s

energy density u

1m2

22

cos, ,

4

U AdE d dt A cdt R d

V R

1 sincos

4 4

dU U d d cUc

A dt V V

d

Blackbody radiation (Stefan’s law )

5

4 43 2

1 2

15 B

dEk T T

A dt h c

Stefan-Boltzmann constant5 4

3 2

2

15Bk

h c

8 2 45.67 10 W/ m K

Real materials4power e AT

Emissivity: e

Reflectivity: 1-e

Sun’s Mass LossThe spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength = 0.5 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: 7·108 m, mass - 2 ·1030 kg.

max = 0.5 m

KKk

hcT

Tk

hc

BB

740,5105.01038.15

103106.6

55 623

834

maxmax

This result is consistent with the flux of the solar radiation energy received by the Earth (1370 W/m2) being multiplied by the area of a sphere with radius 1.5·1011 m (Sun-Earth distance).

428

23

45

Km

W 107.5

15

2 ch

kB 424sphere aby emittedpower TRP

4

22 48 8 262 4

max

W 4 4 7 10 m 5.7 10 5,740K 3.8 10 W

5 m KSunB

hcP R

k

kg/s 102.4m 103

W108.3 928

26

2

c

P

dt

dmthe mass loss per one second

1% of Sun’s mass will be lost in yr 101.5s 107.4kg/s 102.4

kg 102

/

01.0 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 J’:

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

44

1

1ba TT

r

rJJ

If r=0.98 (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:

4/14

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:

Emission:424outPower EE TR

the flux of the solar radiation energy received by the Earth ~ 1370 W/m2

2

42ower in SunE Sun

orbit

RP e R T

R

1/ 42

4Sun

E Sunorbit

ReT T

R

e = 1 – TEarth = 280K

Rorbit = 1.5·1011 m RSun = 7·108 m

In reality e = 0.7 – 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 – PV = (2/3)U. The difference – because the energy-momentum relationship for photons is ultra-relativistic, and the number of photon depends on T.

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

approximately 2.7 K.(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density

u(λ,T) 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(λ,T) and u(,T) correspond to the same photon energy? If

not, why?

mmmTk

hc

B

1.1101.17.21038.15

103106.6

53

23

834

max

(a)

(b)

(c)

Hzh

TkB 1134

23

max 1058.1106.6

7.21038.18.28.2

meVh 65.0max

meVhc

1.1max

the maxima u(λ,T) 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