Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Bremsstrahlung wikipedia , lookup
Van Allen radiation belt wikipedia , lookup
X-ray astronomy detector wikipedia , lookup
Background radiation wikipedia , lookup
Microplasma wikipedia , lookup
Cosmic microwave background wikipedia , lookup
Photon polarization wikipedia , lookup
Lecture 6. Blackbody Radiation Gustav Robert Kirchhoff (12 March 1824 – 17 October 1887) Review Particle physics Fermion mass charge spin name Boson 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 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: u T uS , T d 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) : T ni 1 / 2 h 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 bosons of spin 1 (in units ħ). They move with the speed of light : 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). E ph h E ph cp ph p ph E ph c h 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. c Chemical Potential of Photons = 0 The 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: On the other hand, F ph N T ,V F 0 N T ,V Thus, in equilibrium, the chemical potential for a photon gas is zero: ph 0 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 T ,V G N Instead, we can use G F PV F T ,V F P V V T - by increasing the volume at T=const, we proportionally scale F Thus, GF F V 0 V - the Gibbs free energy of an equilibrium photon gas is 0 ! ph G 0 N For = 0, the BE distribution reduces to the Planck’s distribution: n ph f ph , T 1 1 exp k T B 1 Planck’s distribution provides the average h 1 number of photons in a single mode of exp frequency = /h. k BT The average energy in the mode: n h h h 1 exp k T B k BT In the classical (h << kBT) limit: 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. Density of States for Photons kz 1 4 / 3 k 3 k 3 volume N k 8 6 2 Lx Ly Lz kx dG g d ky extra factor of 2 due to two polarizations: cp ck G 6 2 c d h 8 2 g g h 2 3 d c c 3 2 3D ph 3D ph 3 3 k3 G k 6 2 g 3D ph 2 2 2 c 8 2 g 3 c 3D ph 3 h g n d uS , T d 8 2 3D g ph 3 c h g( ) The average energy of photons with frequency between and +d (per unit volume): 3 photon energy n( ) = 8 h 3 us , T h g f 3 c exp h 1 u as a function of the energy: average number of photons h g( ) n( ) Spectrum of Blackbody Radiation - the spectral density of the black-body radiation (the Planck’s radiation law) u , T d u , T d u , T u , T 8 u , T hc 3 d u h , T h d 3 1 exp k T B u(,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 ), Rayleigh-Jeans Law At low frequencies: h 1 exp h 1 h 8 h 3 8 2 us , T 3 3 k BT c exp h 1 c - purely classical result (no h), can be obtained directly from equipartition Rayleigh-Jeans Law This equation predicts the so-called ultraviolet catastrophe – an infinite amount of energy being radiated at high frequencies or short wavelengths. Planck curve Rayleigh-Jeans Law (cont.) u as a function of the wavelength: 3 u , T d u , T d c h hc 8 1 d hc 8 hc u , T d 2 hc 3 exp hc 1 2 5 exp hc 1 k T k T B B In the classical limit of large : u , T large 8 k BT 4 1 4 High limit, Wien’s Displacement Law At high frequencies/low temperatures: h 1 exp h 1 exp h u s , T 8 h 3 exp h c3 Nobel 1911 The maximum of u() shifts toward higher frequencies with increasing temperature. The 3 position of maximum: h 3x 2 k T du d x 3e x B const const x 0 2 x d e 1 h h e 1 exp 1 d k BT k BT max 2.8 k BT h 3 x e x 3 x 2.8 u(,T) h max 2.8 k BT Wien’s displacement law - discovered experimentally by Wilhelm Wien - the “most likely” frequency of a photon in a blackbody radiation with temperature T Numerous applications (e.g., non-contact radiation thermometry) max max u , T u , T max max h max 2.8 - does this mean that k BT hc No! 2.8 ? k BT max 3 u , T d u , T d c h hc 8 1 d hc 8 hc u , T d 2 hc 3 exp hc 1 2 5 exp hc 1 k T k T B B du d 1 5 x 2 exp 1 / x const 5 const 6 5 0 2 df dx x exp 1 / x 1 x exp 1 / x 1 x exp 1 / x 1 5xexp 1/ x 1 exp 1/ x T = 300 K “night vision” devices max hc 5 k BT max 10 m Solar Radiation The surface temperature of the Sun - 5,800K. max As a function of energy, the spectrum of sunlight peaks at a photon energy of hc 0.5 m 5 k BT umax h max 2.8k BT 1.4 eV - 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: N 8 n n g d 3 V 0 c 2 h 8 d 0 c3 1 exp k BT The total energy of photons per unit volume : (the energy density of a photon gas) 2 5 k B 15h 3c 2 4 the Stefan-Boltzmann constant k BT h 3 - increases as T 3 U g 8 5 k BT uT d 3 V 0 exp 1 15hc u T 4 4 4 T c the StefanBoltzmann Law uT 8 5 k BT hc 4 k BT 2.7 k BT 3 3 N 15hc 8 k BT 2.4 15 2.4 4 The average energy per photon: 3 x 2 dx kB 3 8 0 e x 1 hc T 2.4 3 (just slightly less than the “most” probable energy) Some numbers: 8 4 2 The value of the Stefan-Boltzmann constant: 5.76 10 W / K m Consider a black body at 310K. The power emitted by the body: T 4 500 W / m2 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). Temperature of the Sun • Stefan also determined the temperature of the Sun's surface. He learned from the data of Charles Soret (1854–1904) that the energy flux density from the Sun is 29 times greater than the energy flux density of a warmed metal lamella. A round lamella was placed at such a distance from the measuring device that it would be seen at the same angle as the Sun. Soret estimated the temperature of the lamella to be approximately 1900 °C to 2000 °C. Stefan surmised that ⅓ of the energy flux from the Sun is absorbed by the Earth's atmosphere, so he took for the correct Sun's energy flux a value 3/2 times greater, namely 29 × 3/2 = 43.5. • Precise measurements of atmospheric absorption were not made until 1888 and 1904. The temperature Stefan obtained was a median value of previous ones, 1950 °C and the absolute thermodynamic one 2200 K. As 2.574 = 43.5, it follows from the law that the temperature of the Sun is 2.57 times greater than the temperature of a lamella, so Stefan got a value of 5430 °C or 5700 K (modern value is 5780 K). This was the first sensible value for the temperature of the Sun. Before this, values ranging from as low as 1800 °C to as high as 13,000,000 °C were claimed. The lower value of 1800 °C was determined by Claude Servais Mathias Pouillet (1790-1868) in 1838 using the Dulong-Petit law. Pouilett also took just half the value of the Sun's correct energy flux. Perhaps this result reminded Stefan that the Dulong-Petit law could break down at large temperatures. Power Emitted by a Black Body For the “uni-directional” motion, the flux of energy per unit area c u T energy density u 1m2 c 1s Integration over all angles provides a factor of ¼: power emitted by unit area 1 cu 4 (the hole size must be >> the wavelength) Thus, the power emitted by a unit-area surface at temperature T in all directions: power c c 4 4 u T T T4 4 4 c The total power emitted by a black-body sphere of radius R: 4R 2 T 4 Sun’s Mass Loss The 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 max hc 5 k BT T hc 5 k B max 6.6 1034 3 108 K 5 , 740 K 23 6 5 1.38 10 0.5 10 2 5 k B W 8 5 . 7 10 15h 3c 2 m2K 4 4 P power emitted by a sphere 4R 2 T 4 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). hc 2 P 4 RSun 2.8 k B max 4 2 W 4 7 108 m 5.7 10 8 2 4 5,740K 4 3.8 10 26 W mK the mass loss per one second 1% of Sun’s mass will be lost in dm P 3.8 1026 W 9 2 4 . 2 10 kg/s 2 8 dt c 3 10 m 0.01M 2 1028 kg t 4.7 1018 s 1.5 1011 yr 9 dm / dt 4.2 10 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 pressure is small, energy can still be transferred by radiation. Both surfaces, cold and warm, radiate at a rate: J rad 1 r Ti 4 W / m2 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 Let the total ingoing flux be J, and the total outgoing flux be J’: The net ingoing flux: J 1 r Ta4 rJ J 1 r Tb4 rJ J J 1 r Ta4 Tb4 1 r 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 J 0 T1 T2 Without the third plane, the energy flux per unit area is: 4 4 T3 T2 The equilibrium temperature of the third plane can be found from the energy balance: T14 T3 4 T3 4 T2 4 1/ 4 T1 T2 2T3 4 4 T14 T2 4 T3 2 4 The energy flux between the 1st and 2nd planes in the presence of the third plane: J T1 T3 4 4 4 T14 T2 4 1 1 T14 T2 4 J 0 T1 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: J0 JN N 1 - cut in half The Greenhouse Effect Absorption: 2 4 R Power in RE TSun Sun Rorbit 2 the flux of the solar radiation energy received by the Earth ~ 1370 W/m2 Emission: Power out 4 RE TE 2 4 1/ 4 R 2 TE Sun TSun 4 Rorbit Rorbit = 1.5·1011 m Transmittance of the Earth atmosphere = 1 – TEarth = 280K In reality = 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: RSun = 7·108 m Thermodynamic Functions of Blackbody Radiation The heat capacity of a photon gas at constant volume: 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: 16 u T cV VT 3 c T V c T 16V 16 2 3 S T V dT T d T VT T c 0 3c 0 T T Now we can derive all thermodynamic functions of blackbody radiation: the Helmholtz free energy: F U TS 4 16 4 VT 4 VT 4 VT 4 c 3c 3c G U TS PV F PV 0 the Gibbs free energy: the pressure of a photon gas (radiation pressure) 4 4 1 F P T u 3 V T , N 3c N 1 PV U 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. In terms of the average density of phonons: 1 1 1 PV U V n V n 2.7k BT 0.9 N k BT 3 3 3 Radiation in the Universe Approximately 98% of all the photons emitted since the Big Bang are observed now in the submillimeter/THz range. In the spectrum of the Milky Way galaxy, at least one-half of the luminous power is emitted at sub-mm wavelengths 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 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: S T 16 VT 3 const 3c Since V R3, the isentropic expansion leads to : T R 1 CMBR (cont.) “… for their 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 stars 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. Results: WMAP 3 years 3/16/06 Power Spectrum of temperature fluctuations Correlation Power Spectra of temperature fluctuations and E/B polarization Fitted Cosmological Parameters WMAP results • Big Bang=> very high temperatures! – When T≈3000K, p+e recombine =>H and universe becomes transparent – Redshifted by expansion of the universe: 2.73K • Very efficient thermalization – In particular no late release of energy • No significant ionization of universe since! – e.g., photon-electron interactions= Sunyaev-Zel'dovich effect • Fluctuations of T => density fluctuations Problem The frequency peak in the spectral density of radiation for a certain distant star is at 1.7 x 1014 Hz. The star is at a distance of 1.9 x 1017 m away from the Earth and the energy flux of its radiation as measured on Earth is 3.5x10-5 W/m2. a) b) c) d) (5) What is the surface temperature of the star? (5) What is the total power emitted by 1 m2 of the surface of the star? (5) What is the total electromagnetic power emitted by the star? (5) What is the radius of the star? h 6.6 10 34 1.7 1014 T 3000 K 23 2.7 k B 1.38 10 2.7 (a) (b) J T 4 5.7 10 8 W / K 4 m 2 3000 K 4 4.6 106 W / m 2 (c) power 4 r 2 J r 4 1.9 1017 m2 3.5 105W / m2 1.6 1031W (d) 4 R J RS 4 r J r 2 S 2 4 J r 3.5 10 5 17 11 RS r 1.9 10 m 5 . 2 10 m 6 J RS 4.6 10 Problem 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? (d) (15) What is approximately the number of CMBR photons hitting the earth per second per square meter [i.e. photons/(s·m2)]? (a) (b) (c) max hc 6.6 1034 3 108 3 1 . 1 10 m 1.1mm 23 5 k BT 5 1.38 10 2.7 max k BT 2.8 1.38 10 23 2.7 11 2.8 1 . 58 10 Hz 34 h 6.6 10 hc max 1.8 10 22 J h max 1.04 10 22 J The maxima u(λ,T) and u(,T) do not correspond to the same photon energy. As increases, the frequency range included in unit wavelength interval increases as 2, moving the peak to shorter wavelengths: u , T d u , T d c 8 hc d u , T d 2 5 1 hc 1 exp k T B Problem cont. (d) J TCMBR 5.7 108 W / K 4 m2 2.7 K 4 3 106W / m2 4 4 uT 8 5 k BT hc 4 k BT 2.7 k BT 3 3 N 15hc 8 k BT 2.4 15 2.4 4 3 W J 2 3 106 m photons 16 photons N 3 10 2 J 2.7 1.38 1023 2.7 s m2 sm Problem The planet Jupiter is a distance of 7.78 x 1011 meters from the Sun and is of radius 7.15 x 107 meters. Assume that Jupiter is a blackbody even though this is not entirely correct. Recall that the solar power output of Sun is 4 x 1026 W. a) What is the energy flux of the Sun's radiation at the distance of Jupiter's orbit? PSun 4 1026W J 52.6W / m2 2 2 4Rorbit 4 7.78 1011 m b) Recognizing that Sun's energy falls on the geometric disk presented to the Sun by the planet, what is the total power incident on Jupiter from the Sun? 2 PJupiter J RJupiter 52.6W / m2 7.15 107 m 8.4 1017W 2 c) Jupiter rotates at a rather rapid rate (one revolution per 0.4 earth days) and therefore all portions of the planet absorb energy from the Sun. Hence all portions of the surface of this planet radiate energy outward. On the basis of this information find the surface temperature of Jupiter. 2 4 PJupiter 4RJupiter TJupiter PJupiter TJupiter 4R 2 Jupiter 1/ 4 1/ 4 8.4 1017W 2 7 8 4 2 4 7.15 10 m 5.76 10 W / K m 123K Problem (blackbody radiation) The black body temperature is 3000K. Find the power emitted by this black body within the wavelength interval = 1 nm near the maximum of the spectrum of blackbody radiation. c Jd u , T d 4 hc 8 hc 1 max u , T d 5 d 5 k BT hc 1 exp k BT the power emitted by a unit area in all directions: near the maximum: c c 8 hc5k BT 1 2 c 5k BT 1 J d u , T d d d 5 4 4 4 exp 5 1 exp 5 1 hc hc 5 J d 2 3 108 5 1.38 10 23 3000 6.6 10 34 3 108 4 5 5 1 110 9 3.1103 W/m 2 exp 5 1 Problem (radiation pressure) (a) (b) (c) (d) At what temperature will the pressure of the photon gas be equal to 105 Pa (= one bar )? At what temperature will the pressure of the photon gas be equal to 10-5 Pa? The temperature at the Sun’s center is 107 K. What is the pressure of the radiation? Compare it to the pressure of the gas at the center of the Sun, which is 1011 bar. Calculate the pressure of the Sun’s radiation on the Earth’s surface, given that the power of the radiation reaching earth from the Sun is 1400 W/m2. 4 4 P T 3c (b) (c) 1/ 4 3cP T 4 P=10-5 Pa 3 3 108 1 105 (a) P=105 Pa T 8 4 5.7 10 3 3 108 1 10 5 T 8 4 5.7 10 4 5.7 108 107 P 3 3 108 1/ 4 1.4 105 K 1/ 4 450 K 4 2.5 1012 Pa 2.5 107 bar - at this T, the pressure of radiation is still negligible in the balance for mechanical equilibrium (d) P 1 4 u T u T I 3 c P 4I 4 1400 6 10 6 Pa 8 3c 3 3 10 Problem Tungsten has an emissivity of 0.3 at high temperatures. Tungsten filaments operate at a temperature of 4800 K. a) At what frequency does a tungsten filament radiate the most energy? b) What is the power/unit surface emitted by the tungsten filament? c) If the surface area of a tungsten filament is 0.01 cm2 what is the power output of the bulb? d) What is the energy flux of radiation emitted by the filament 2 meters from the bulb? e) What fraction of radiation incident on a tungsten filament is reflected? [Answers: a) 2.88 x 1014 Hz, b) 9.03 x106 W/m2, c) 9.03 watts, d) 0.180 watts, and e) 70%.] Problem Quiz The spectrum of Sun, plotted as a function of energy, peaks at a photon energy of 1.4 eV. The spectrum for Sirius A, plotted as a function of energy, peaks at a photon energy of 2.4 eV. The luminosity of Sirius A (the total power emitted by its surface) is by a factor of 24 greater than the luminosity of the Sun. How does the radius of Sirius A compare to the Sun’s radius? L 4R T 2 4 R L T 2 The temperature, according to Wien’s law, is proportional to the energy that corresponds to the peak of the photon distribution. LSirius / LSun RSirius 24 1.67 2 2 RSun TSirius / TSun 2.4 / 1.4