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Home , Black body, Emissivity, Photon gas

Lecture 26. Blackbody (Ch. 7)

Two types of :

(a) Composite particles which contain an even number of fermions. These number of these particles is conserved if the 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 . These particles are not conserved: if the total energy of the field changes, particles appear and disappear. We’ll see that the of such particles is zero in equilibrium, regardless of density. Radiation in Equilibrium with Matter Typically, 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 T. The energy of radiation is spread over a range of , and we define uS (ν,T) dν as the energy density (per unit volume) of the radiation with frequencies between ν and ν+dν. uS(ν,T) is the spectral energy density. The of the photon : ∞ ()= , (dTuTu νν ) ∫ S 0

In equilibrium, uS (ν,T) is the same everywhere in the cavity, and is a function of 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 and the 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 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 α is a function of ν and T; a surface for which α(ν ) =1 for all frequencies is called a body. Photons

The electromagnetic field has an infinite number of modes (standing waves) in the cavity. The black-body radiation field is a superposition T of plane waves of different frequencies. The characteristic feature of the radiation is that a mode may be excited only in units of the quantum of energy hν (similar to a harmonic oscillators) : ε = ( + 2/1 )hn ν i i 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 ph = hE ν bosons of 1 (in units ħ). They move with the speed of : ph = cpE ph The linearity of Maxwell equations implies that the photons do not E ν p ph == h interact with each other. (Non-linear optical phenomena are ph c c observed when a large-intensity radiation interacts with matter).

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. We’ll treat a system of photons as an ideal photon gas, and, in particular, we’ll apply the BE statistics to this system. Chemical Potential of Photons = 0 The mechanism of establishing equilibrium in a photon gas is absorption ⎛ ∂ F ⎞ and emission of photons by matter. The textbook suggests that N can be ⎜ ⎟ = 0 ⎝ ∂ N ⎠ found from the equilibrium condition: ,VT

⎛ ∂ F ⎞ Thus, in equilibrium, the chemical On the other hand, ⎜ ⎟ = μ μ = 0 ⎜ ⎟ ph potential for a photon gas is zero: ph ⎝ ∂ N ⎠ ,VT However, we cannot use the usual expression for the chemical potential, because one cannot increase N (i.e., add photons to the system) at constant volume and at the same time keep the temperature constant: ⎛ ∂ F ⎞ ⎜ ⎟ - does not exist for the photon gas ⎝ ∂ N ⎠ ,VT ⎛ ∂F ⎞ ( ,VTF ) Instead, we can use = NG μ = + PVFG P −= ⎜ ⎟ −= ⎝ ∂V ⎠T V - by increasing the volume at T=const, we proportionally scale F

F - the of an G Thus, FG V =−= 0 μ == 0 V equilibrium photon gas is 0 ! ph N For μ = 0, the BE distribution reduces to the Planck’s distribution:

1 1 Planck’s distribution provides the average = phph ()ε,Tfn = = ⎛ ε ⎞ ⎛ hν ⎞ number of photons in a single mode of exp⎜ ⎟ −1 exp⎜ ⎟ −1 ⎜ ⎟ ⎜ ⎟ frequency ν = ε/h. ⎝ BTk ⎠ ⎝ BTk ⎠ hν hn νε== The average energy in the mode: ⎛ hν ⎞ exp⎜ ⎟ −1 ⎝ BTk ⎠

In the classical (hν << kBT) limit: ε = BTk In order to calculate the average number of photons per small energy interval dε, the average energy of photons per small energy interval dε, etc., as well as the total average number of photons in a photon gas and its total energy, we need to know the density of states for photons as a function of photon energy.

kz Density of States for Photons

1 ( 3/4 )π k 3 3 (volumek ) k 3 ()kN = = ()kG = πππ 2 2 8 ×× 6π 6π LLL zyx kx ky dG(ε ) ε 3 ε 2 g()ε = ε == Gkccp ()ε = g 3D ()ε = h 2 3 ph 2 3 dε 6π ()ch 2π ()ch

extra factor of 2 due dε (hν )2 8πν 2 8πν 2 3D ()= gg 3D εν ()= h = g 3D ()ν = to two polarizations: ph ph 2 3 3 ph 3 dν π ()ch c c ν (ν ) (ν ) ν = (ν , )dTudngh ν Spectrum of Blackbody Radiation S The average energy of photons with frequency photon average number between ν and ν+dν (per unit volume): energy of photons ) ν ) 2 n(ν ) ( ν n 3D 8πν ( 3 )

g ∝ ν g ph ()ν = 3 ν ν c × = ( h g ν ν ν h ν

8 h νπ3 - the spectral density of the s (), = fghTu ()()νννν= 3 black-body radiation exp()hc νβ−1 (the Plank’s radiation law) dε u as a function of the energy: (), = , ()= ()()ενννεε,, TuTudTudTu ()ν , ×= hThu dν

8 επ3 ()ε ,Tu = ()hc 3 ⎛ ε ⎞ exp⎜ ⎟ −1 ⎝ BTk ⎠ u(ε,T) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T. Classical Limit (small f, large λ), Rayleigh-Jeans Law

At low frequencies or high : β hν << exp1 (β hν )−1 ≅ β hν

8 h νπ3 8 νπ2 - purely classical result (no h), can be ()ν ,Tu = ≅ Tk s 3 exp()νβ−1 chc 3 B obtained directly from equipartition

Rayleigh-Jeans Law

This equation predicts the so-called catastrophe –an infinite amount of energy being radiated at high frequencies or short wavelengths. Rayleigh-Jeans Law (cont.)

u as a function of the wavelength:

3 ⎛ c ⎞ ⎜h ⎟ ⎡ dε hc ⎤ 8π ⎝ λ ⎠ ⎛ ⎞ 8π hchc 1 (), −= , ()dTudTu εελλ−= ()λ,Tu = ⎜ ⎟ = ⎣⎢d λλ2 ⎦⎥ ()hc 3 ⎛ hc ⎞ ⎝ λ2 ⎠ λ5 ⎛ hc ⎞ exp⎜ ⎟ −1 exp⎜ ⎟ −1 ⎝ λ BTk ⎠ ⎝ λ BTk ⎠

In the classical limit of large λ:

8π Tk ()λ,Tu ≈ B 1 large λ λ4 λ4 High ν limit, Wien’s Displacement Law

At high frequencies/low temperatures: β hν >> exp1 ()β hν − ≅ exp1 (β hν )

8π h 3 s ()ν ,Tu = exp()− hνβν Nobel 1911 c3

The maximum of u(ν) shifts toward higher frequencies with increasing temperature. The position of maximum: 3 ⎡ ⎛ hν ⎞ ⎤ ⎢ ⎜ ⎟ ⎥ ⎜ ⎟ ⎡ 2 3 x ⎤ du d ⎢ ⎝ BTk ⎠ ⎥ 3x ex const ×= const ×= ⎢ − ⎥ = 0 ⎢ ⎥ x x 2 dν ⎛ hν ⎞ ⎛ hν ⎞ ⎣⎢e −1 ()e −1 ⎦⎥ d⎜ ⎟ ⎢exp⎜ ⎟ −1⎥ Tk ⎜ Tk ⎢ Tk ⎥ ν ≈ 8.2 B ⎝ B ⎠ ⎣ ⎝ B ⎠ ⎦ max h ( )ex x 33 x ≈→=− 8.2

hν Wien’s displacement law max ≈ 8.2 Tk - discovered experimentally B by ,T) ν (

u - the “most likely” frequency of a photon in a blackbody radiation with temperature T Numerous applications (e.g., non-contact radiation thermometry) ν νmax ⇔λmax hν ()ν ,Tu (λ,Tu ) max ≈ 8.2 - does this mean that BTk hc ≈ 8.2 ? No! ν max λmax BTk λmax

3 ⎛ c ⎞ ⎜h ⎟ ⎡ dε hc ⎤ 8π ⎝ λ ⎠ ⎛ ⎞ 8π hchc 1 (), −= , ()dTudTu εελλ−= ()λ,Tu = ⎜ ⎟ = ⎣⎢d λλ2 ⎦⎥ ()hc 3 ⎛ hc ⎞ ⎝ λ2 ⎠ λ5 ⎛ hc ⎞ exp⎜ ⎟ −1 exp⎜ ⎟ −1 ⎝ λ BTk ⎠ ⎝ λ BTk ⎠ du d ⎡ 1 ⎤ ⎡ 5 (− −2 ) ( /1exp xx ) ⎤ const ×= const −×= − = 0 ⎢ 5 ⎥ ⎢ 6 5 2 ⎥ df ⎣ {}()xxdx −1/1exp ⎦ ⎣ {}()xx −1/1exp {}()xx −1/1exp ⎦

hc { ()xx − }= ( /1exp1/1exp5 x) → λmax ≈ 5 BTk

T = 300 K → λmax ≈ 10 μm

“night vision” devices Solar Radiation

The surface temperature of the - 5,800K.

hc λmax ≈= 5.0 μm 5 BTk

As a function of energy, the spectrum of max = hu ν max = B ≈ 4.18.2 eVTk peaks at a photon energy of - close to the energy gap in Si, ~1.1 eV, which has been so far the best material for solar cells

Spectral sensitivity of human eye: Stefan-Boltzmann Law of Radiation

The total number of photons per unit volume:

3 3 N ∞ 8 ∞ νπ2 8π ⎛ Tk ⎞ ∞ 2dxx ⎛ k ⎞ n =≡ ()()dgn εεε= dν = ⎜ B ⎟ = 8π ⎜ B ⎟ T 3 × 4.2 V ∫ c3 ∫ ⎛ hν ⎞ c3 ⎝ h ⎠ ∫ ex −1 ⎝ hc ⎠ 0 0 exp⎜ ⎟ −1 0 ⎜ ⎟ 3 ⎝ BTk ⎠ - increases as T

The total energy of photons per unit volume : U ∞ × g(εε) 8π 5 ( Tk )4 ()Tu =≡ dε = B (the energy density of a photon gas) ∫ 3 V 0 ()βε −1exp 15()hc

2π 5k 4 the Stefan-Boltzmann 4σ the Stefan- σ = B ()Tu = T 4 15 ch 23 constant c Boltzmann Law

(Tu ) 8π 5 ()(hcTk )34 π 4 The average energy per photon: B ε == 3 3 = B ≈ 7.2 BTkTk N ()815 π B (Tkhc × )4.2 × 4.215 (just slightly less than the “most” probable energy)

Some numbers: The value of the Stefan-: σ ×= −8 /1076.5 ( mKW 24 ) Consider a at 310K. The power emitted by the body: σ 4 ≈ /500 mWT 2 While the of skin is considerably less than 1, it still emits a considerable power in the range. For example, this radiation is easily detectable by modern techniques (night vision). Power Emitted by a Black Body

For the “uni-directional” motion, the of energy per unit area = ×uc T

energy density u 1m2 c × 1s

1 Integration over all angles power emitted by areaunit ×= uc provides a factor of ¼: 4 (the hole size must be >> the wavelength)

Thus, the power emitted by a unit-area c c 4σ power ()Tu 4 =×== σ TT 4 surface at temperature T in all directions: 4 4 c

42 The total power emitted by a black-body sphere of radius R: = 4 σπTR Sun’s Mass Loss The spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength λ = 0.5 μm. Find the mass loss for the Sun in one second. How long it takes for the Sun to loose 1% of its mass due to radiation? Radius of the Sun: 7·108 m, mass - 2 ·1030 kg. hc hc ⎛ −34 ⋅×⋅ 103106.6 8 ⎞ λ = T =→ ⎜= = 740,5 KK ⎟ λmax = 0.5 μm → max ⎜ −23 −6 ⎟ 5 BTk 5kBλmax ⎝ ⋅×⋅× 105.01038.15 ⎠

2π 5k 4 W P ()power emitted aby = 4sphere σπTR 42 σ B ⋅≈= 107.5 −8 15 ch 23 Km 42

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). 4 ⎛ hc ⎞ 2 W 4 = 4 ()RP 2σπ⎜ ⎟ π ()1074 8 ⋅×⋅= 107.5m −8 ×()4 ⋅= 26 W108.3,740K5 Sun ⎜ ⎟ 42 ⎝ 8.2 kBλmax ⎠ Km dm P ⋅ 26 W108.3 the mass loss per one second 9 2 == 2 ⋅= 102.4 kg/s dt c ()⋅103 8 m 01.0 M ⋅102 28 kg 1% of Sun’s mass will be lost in t ==Δ 107.4 18 ⋅=⋅= 101.5s 11 yr / dtdm ⋅102.4 9 kg/s Dewar Radiative Energy Transfer

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 is small, energy can still be transferred by radiation. Both surfaces, and warm, radiate at a rate:

4 2 i=a for the outer (hot) wall, i=b for the inner (cold) wall, ()1−= σ / mWTrJ rad i r – the coefficient of reflection, (1-r) – the coefficient of emission

Let the total ingoing flux be J, (1 )σ 4 +−= ′′ (1 )σ 4 +−= rJTrJJrTrJ and the total outgoing flux be J’: a b

1− r The net ingoing flux: − JJ ′ = σ ()−TT 44 1+ r ba

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). Superinsulation

Two 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 T3 T2

0 4 4 Without the third plane, the energy flux per unit area is: σ (1 −= TTJ 2 )

The equilibrium temperature of the third plane can be found from the energy balance: 4/1 ⎛ 4 +TT 4 ⎞ σ ()()4 4 σ 4 −=− 4 4 4 =+ 2 4 TTTTTTTT = ⎜ 1 2 ⎟ 1 3 3 2 1 2 3 3 ⎜ ⎟ ⎝ 2 ⎠ The energy flux between the 1st and 2nd planes in the presence of the third plane: ⎛ 4 +TT 4 ⎞ 1 1 σ ()4 4 σ ⎜TTTJ 4 −=−= 1 2 ⎟ σ ()4 4 =−= JTT 0 - cut in half 1 3 ⎜ 1 ⎟ 1 2 ⎝ 2 ⎠ 2 2 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: J 0 J = N N +1 The

Absorption: 2 ⎛ R ⎞ 2 4 ⎜ Sun ⎟ Power in = ()E σπα()TR Sun ⎜ ⎟ ⎝ Rorbit ⎠ the flux of the solar radiation energy received by the Earth ~ 1370 W/m2 42 Emission: = 4outPower σπTR EE

2 4/1 ⎡α ⎛ R ⎞ ⎤ T = ⎢ ⎜ Sun ⎟ ⎥ T E 4 ⎜ R ⎟ Sun ⎣⎢ ⎝ orbit ⎠ ⎦⎥

11 8 Rorbit = 1.5·10 m RSun = 7·10 m of the Earth atmosphere

α = 1 – TEarth = 280K

In reality α = 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

∂ (Tu ) 16σ The of a photon gas at constant volume: ⎡ ⎤ 3 cV ≡ ⎢ ⎥ = VT ⎣ ∂ ⎦V cT

This equation holds for all T (it agrees with T T (Tc ′) 16σV 2 16σ 3 the Nernst theorem), and we can integrate it ()TS = V Td ′ = TdT ′′ = VT ∫ T′ c ∫ 3c to get the of a photon gas: 0 0

Now we can derive all thermodynamic functions of blackbody radiation: σ 164 σ 4σ the : TSUF VT 4 −=−= VT 4 −= VT 4 c 3c 3c the Gibbs free energy: = − + = + PVFPVTSUG = 0 (= μN )

⎛ ∂F ⎞ 4σ 4 1 1 the pressure of a photon gas P −= ⎜ ⎟ == uT = UPV () ⎝ ∂ ⎠ ,NT 3cV 3 3

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. 1 1 1 In terms of the average ε =××=××== 9.07.2 TkNTknVnVUPV density of : 3 3 3 B B Radiation in the

Approximately 98% of all the photons emitted since the are observed now in the submillimeter/THz range.

The dependence of the radiated energy In the spectrum of the Milky Way galaxy, versus wavelength illustrates the main at least one-half of the luminous power is sources of the THz radiation: the emitted at sub-mm wavelengths interstellar dust, emission from light and heavy molecules, and the 2.7-K cosmic background radiation. Cosmic Microwave Background

A. Penzias R. Wilson Nobel 1978

In the standard Big Bang model, the radiation is decoupled from the matter in the Universe about 300,000 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 “primordial” photons. The further expansion of the Universe can be considered as quasistatic adiabatic (isentropic) for the radiation: 16σ ()TS = 3 = constVT 3c Since V ∝ R3, the isentropic expansion leads to : T ∝ R−1 “… for their CMBR (cont.) discovery of the blackbody form and anisotropy of the CMBR”. Mather, Smoot, Nobel 2006

At present, the temperature of the Planck’s distribution for the CMBR photons is 2.735 K. The radiation is coming from all directions and is quite distinct from the radiation from and galaxies.

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. 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 νmax (in Hz) corresponds to the maximum spectral density u(ν,T) of the cosmic background radiation? (c) (5) Do the maxima u(λ,T) and u(ν,T) correspond to the same photon energy? If not, why?

hc −34 ⋅×⋅ 103106.6 8 hc (a) λ =≈ −3 =⋅= 1.1101.1 mmm = 1.1 meV max −23 λ 5 BTk ×⋅× 7.21038.15 max

−23 BTk ×⋅× 7.21038.18.2 11 (b) ν 8.2 =≈ ⋅= 1058.1 Hz hν max = 65.0 meV max h ⋅106.6 −34

(c) the maxima u(λ,T) and u(ν,T) do not correspond to the same photon energy. The reason of that is

⎡dν c ⎤ 8π hc 1 (), −= , ()dTudTu ννλλ−= ()λ,Tu = ⎢ 2 ⎥ 5 ⎣ d λλ⎦ λ ⎛ hc ⎞ exp⎜ ⎟ −1 ⎝ λ BTk ⎠ Problem 2006 (blackbody radiation)

(d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [i.e. photons/(s·m2)]?

4 −8 24 4 4 −6 2 (d) σ TJ CMBR ⋅== ( /107.5 mKW )×⋅ ()K ⋅= /1037.2 mW

5 34 4 ()Tu 8π ( B ) (hcTk ) π ε == 3 3 = B ≈ 7.2 BTkTk N ()815 π B (Tkhc × )4.2 × 4.215

⎛ W ⎞ J⎜ 2 ⎟ −6 ⎛ photons ⎞ ⎝ m ⎠ ⋅103 16 photons N⎜ ⎟ = ≈ ⋅≈ 103 ⎝ ⋅ ms 2 ⎠ ε ()J −23 ×⋅× 7.21038.17.2 ⋅ ms 2