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Astro 300B: Jan. 26, 2011 Thermal radiation and Thermal Equilibrium Thermal Radiation, and Thermodynamic Equilibrium Thermal radiation is radiation emitted by matter in thermodynamic equilibrium. When radiation is in thermal equilibrium, Iν is a universal function of frequency ν and temperature T – the Planck function, Bν. I B Blackbody Radiation: In a very optically thick media, recall the SOURCE FUNCTION S So thermal radiation has j I S B and j B And the equation of radiative transfer becomes dI dl I B or dI d I B (T ) THERMODYNAMIC EQUILIBRIUM When astronomers speak of thermodynamic equilibrium, they mean a lot more than dT/dt = 0, i.e. temperature is a constant. DETAILED BALANCE: rate of every reaction = rate of inverse reaction on a microprocess level If DETAILED BALANCE holds, then one can describe (1) The radiation field by the Planck function (2) The ionization of atoms by the SAHA equation (3) The excitation of electroms in atoms by the Boltzman distribution (4) The velocity distribution of particles by the Maxwell-Boltzman distribution ALL WITH THE SAME TEMPERATURE, T When (1)-(4) are described with a single temperature, T, then the system is said to be in THERMODYNAMIC EQUILIBRIUM. In thermodynamic equilibrium, the radiation and matter have the same temperature, i.e. there is a very high level of coupling between matter and radiation Very high optical depth By contrast, a system can be in statistical equilibrium, or in a steady state, but not be in thermodynamic equilibrium. So it could be that measurable quantities are constant with time, but there are 4 different temperatures: T(ionization) T(excitation) T(radiation) T(kinetic) given by the Saha equation given by the Boltzman equation given by the Planck Function given by the Maxwell-Boltzmann distribution Where T(ionization) ≠ T(excitation) ≠ T(radiation) ≠ T(kinetic) LOCAL THERMODYNAMIC EQUILIBRIUM (LTE) If locally, T(ion) = T(exc) = T(rad) = T(kinetic) Then the system is in LOCAL THERMODYNAMIC EQUILIBRIUM, or LTE This can be a good approximation if the mean free path for particle-photon interactions << scale upon which T changes Example: H II Region (e.g. Orion Nebula, Eagle Nebula, etc) Ionized region of interstellar gas around a very hot star Radiation field is essentially a black-body at the temperature of the central Star, T~50,000 – 100,000 K However, the gas cools to Te ~ 10,000 K (Te = kinetic temperature of electrons) O star H II HI Q.: Is this room in thermodynamic equilibrium? FYI, we write down the following functions, without deriving them: (1) The Boltzman Equation Boltzman showed that the probability of finding an atom with an electron, e-, in an excited state with energy χn above the ground state decreases exponentially with χn increases exponentially with temperature T Nn N1 and n exp g1 kT gn Where Nn = # atoms in excited state n / volume N1 = # atoms in ground state /volume gn = 2n2 the statistical weight of level n = number of different angular momentum quantum numbers in energy level n (2) The Planck Function I B 2h c 2 3 1 e h / kT 1 (3) The Maxwell-Boltzman distribution of speeds of electrons me f ( v ) 4 2 kT e 3/2 2 v e mev 2 2 kT e = fraction of electrons with velocity between v, v+dv where me = mass of the electron Te = temperature of the electrons (4) The Saha Equation ne N m 1 Nm Z m 1 2 m e kT 2 3 Zm h 3/2 e m kT Where ne = number density of free electrons Nm = number density of atoms in the mth ionization state Zm = partition function of the mth ionization state Zm i 1 g ie i kT Thermodynamics of Blackbody Radiation: The Stefan-Boltzman Law Consider a piston containing black-body radiation: Inside the piston: T, v, p u Move blue wall extract or perform work First Law of Thermodynamics: dQ = dU + p dV where dQ = change in heat dU = total change in energy p = pressure dV = change in volume Second Law of Thermodynamics: dS = dQ/T S = entropy Recall, U = uV p = 1/3 u u 4 c So… u = energy density energy/volume p = radiation pressure in piston J d dS J B and dQ T dU p T d ( uV ) T dV (substitute dQ=dU+pdV) T 1 3 u dV T (substitute U=uV, p=1/3 u) Vdu dS dV T T V dT du 1 u dV 3 4 u T dT 3T V du dT 4 u T dT So... u T dV dV 3T V du dS T dT dT V and Differentiate these…. 4u dS 3T dV T 2 d V du 1 du dTdV dV T dT T dT d S (Eqn.1) 2 d 4u 4 u 4 1 du 2 dVdT dT 3T 3T 3 T dT d S Combining (1) and (2) Multiply by T 1 du 4 u T dT 3T du 4 u dT du dT (Eqn. 2) 2 3T 4 u T 4 1 du 3 T dT 4 du 3 dT du 4 dT u T log u 4 log T log a u (T ) aT a=constant of integration Energy density ~T4 4 u can be related to the Planck Function u So… u 4 c J u d For isotropic radiation, 4 c B (T ) d 4 c B (T ) I J B Where B(T) = the integrated Planck function B d ac 4 T 4 For a uniform, isotropically emitting surface, we showed that the flux F F d ac 4 ac T T B d B (T ) 4 4 4 OR…. F T Where 4 ac Stefan-Boltzmann Law = 5.67x10-5 ergs cm-2 deg-4 sec-1 4 [flux] = ergs cm-2 sec-1 also a 4 c flux integrated over frequency, per area per sec = 7.56x10-15 ergs cm-3 deg-4 Blackbody Radiation; The Planck Spectrum • The spectrum of thermal radiation, i.e. radiation in equilibrium with material at temperature T, was known experimentally before Planck • Rayleigh & Jeans derived their relation for the blackbody spectrum for long wavelengths, • Wien derived the spectrum at short wavelengths • But, classical physics failed to explain the shape of the spectrum. • Planck’s derivation involved the consideration of quantized electromagnetic oscillators, which are in equilibrium with the radiation field inside a cavity the derivation launched Quantum Mechanics See Feynmann Lectures, Vol. III, Chapt.4; R&L pp. 20-21 Result: B 2h c 3 3 1 e h / kT 1 ergs s-1 cm-2 Hz-1 ster-1 Or in terms of Bλ recall I d I d d c so B 2 hc 5 d 2 c 2 1 e hc / kT 1 ergs s-1 cm-2 A-1 ster-1 The Cosmic Microwave Background The most famous (and perfect) blackbody spectrum is the “Cosmic Microwave Background.” Until a few hundred thousand years after the Big Bang, the Universe was extremely hot, all hydrogen was ionized, and because of Thomson scattering by free electrons, the Universe was OPAQUE. Then hydrogen recombined and the Universe became transparent. The relict radiation, which was last in thermodynamic equilibrium with matter at the “surface of last scattering” is the CMB. Currently the CMB radiation has the spectrum of a blackbody with T=2.73 K. It is cooling as the Universe expands. The first accurate measurement of the spectrum of the CMB was obtained with the FIRAS instrument aboard the Cosmic Background Explorer (COBE), from space: See Mather + 1990 ApJLetters 354, L37 The smooth curve is the theoretical Planck Law. This plot was made using the first year of data; in subsequent plots the error bars are smaller than the width of the lines! Properties of the Planck Law Two limits simplify the Planck Law (and make it simpler to integrate): Rayleigh-Jeans: hν << kT Wien hν >> kT (Radio Astronomy) Rayleigh-Jeans Law h kT so e h /kT 1 h kT so I (T ) 2h c 3 2 1 e h / kT 1 becomes I (T ) RJ 2 c 2 2 kT The Ultraviolet Catastrophe If the Rayleigh-Jean’s form for the spectrum of a blackbody held for all frequencies, then I d as And the total energy in the radiation field Wien’s Law h kT 1 so e I (T ) W 2h c 2 h / kT 1 1 e h / kT 3 e h / kT Very steep decrease in brightness for peak Monotonicity with Temperature If T1 > T2, then Bν(T1) > Bν(T2) for all frequencies Of 2 blackbody curves, the one with higher temperature lies entirely above the other. dB (T ) dT d 2h 2 dT c 2h 2 2 c kT 4 2 3 e e e 1 h / kT 1 h / kT h / kT 1 2 >0 always Wien Displacement Law At what frequency does the Planck Law Bν(T) peak? dB Bν(T) peaks at νmax, given by d d 2h 2 d c e h / kT 3 1 e h / kT 0 max 0 1 d 2h 1 2 d c 3 2h c2 3 d h / kT 1 0 d e e h / kT 6 h 2 2 h 3 h h / kT 1 2 2 e c c kT Divide by exp(hν/kT), cancel some terms 3 1 e Let x h / kT h kT h max kT Need to solve 3 1 e Solution is x=2.82. x x Need to solve graphically or iteratively. 2 . 82 h max kT max 5 . 88 10 10 Hz deg 1 T Similarly, one can find the wavelength λmax at which Bλ(T) peaks dB 0 d max max T 0 . 290 cm deg max max c NOTE: That is to say, Bν and Bλ don’t peak at the same wavelength, or frequency. For the Sun’s spectrum, λmax for Iλ is at about 4500 Å whereas λmax for Iν is at about 8000 Å Why? recall d c 2 d So equal intervals in wavlength correspond to very different intervals of frequency across the spectrum With increasing l, constant dl (the Il case) corresponds to smaller and smaller dn so these smaller dn intervals contain smaller energy, compared to constant dn intervals (the In case) Radiation constants in terms of physical constants Recall the Stefan-Boltzman law for flux of a black body F T Let x h 4 B d 0 x e 1 x So 0 3 dx B d 0 then kT c 2 2 h kT B d 2 c h 4 so... 15 2h 0 2 k 5 4 2 3 15 c h e 0 4 h / kT 0 x 1 3 e 1 x 2 k 4 B d 3 2 15 c h dx 4 3 T 4 d Also, since u 4 c 0 B d aT 8 k 5 a 3 4 15 c h 3 4 As an example of the kind of things you can model with the Planck radiation formulae, consider the following: (see http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/nickel.html) (1) How much radiant energy comes from a nickel at room temperature per second? Measured properties of the nickel are diameter = 2.14 cm, thickness 0.2 cm, mass 5.1 grams. This gives a volume of 0.719 cm3 and a surface area of 8.54 cm 2. The radiation from the nickel's surface can be calculated from the Stefan-Boltzman Law F= σT4 The room temperature will be taken to be 22°C = 295 K. Assuming an ideal radiator for this estimate, the radiated power is P = σAT4 A=surface area of nickel = (5.67 x 10-8 W/m2K4)x(8.54 x 10-4 m2)x(295 K)4 = 0.367 watts. So the radiated power from a nickel at room temperature is about 0.37 watts 2. How many photons per second leave the nickel? Since we know the energy, we can divide it by the average photon energy. We don't know a true average, but the wavelength of the peak of the blackbody radiation curve is a representative value which can be used as an estimate. This may be obtained from the Wien displacement law. lpeak = 0.0029 m K/295 K = 9.83 x 10-6 m = 9830 nm, in the infrared. The energy per photon at this peak can be obtained from the Planck relationship. Ephoton = hν = hc/λ = 1240 eV nm/ 9830 nm = 0.126 eV Then the number of photons per second is very roughly N = (0.367 J)/(0.126 eV x 1.6 x 10-19 J/eV) = 1.82 x 1019 photons Characteristic Temperatures for Blackbodies 1. BRIGHTNESS TEMPERATURE, Tb Instead of stating Iν, one can state Tb, where I B (T B ) i.e. Tb is the temperature of the blackbody having the same specific intensity as the source, at a particular frequency. Notes: 1. TB is often used in radio astronomy, and so you can assume that the Rayleigh-Jeans Law holds, h kT or T B I so c 2 c 2 2 kT B 2 2 k 2 I 2. The source need not be a blackbody, despite being described as a source with brightness temperature TB. 3. Units of TB are easier to remember than units of Iν TB and the equation of Radiative Transfer: dI d I B (T ) I Assume Rayleigh-Jeans, 2 c 2 2 kT I B (T B ) dI d 2 c 2 2 kT B 2 d 2 2 kT B d c So the equation of radiative transfer becomes: dT B d TB T dT B d TB T T B brightness T temperatur e of the material If T is constant w TB TB (0)e If then T B Otherwise, temperatur e describing TB T ith , then T 1 e The brightness temperature = The actual temperature at large optical depth I dT B d TB T T B brightness T temperatur e of the material If T is constant w TB TB (0)e If then T B Otherwise, temperatur e describing TB T ith , then T 1 e The brightness temperature = The actual temperature at large optical depth I (2) Color Temperature, Tc Often one can measure the spectrum of a source, and it is more or less a blackbody of some temperature, Tc. We may not know Iν, but only Fν, if for example the source is unresolved. Tc can be estimated from λ(max), the peak of the spectrum, or the ratio of the spectrum at 2 wavelengths. e.g. B-V colors of stars The solar spectrum vs. blackbody – from Caroll & Ostlie (3) Antenna Temperature, TA A radio telescope mearures the brightness of a source, Often described by T A TB Where S A η = the beam efficiency of the telescope, typically ~0.4-0.8 Ωs= solid angle subtended by the source ΩA= solid angle from which the antenna receives radiation (“beam”) (4) Effective Temperature, Teff If a source has total flux F, integrated over all frequencies we can define Teff such that F T 4 eff The Einstein Coefficients Einstein (1917) related αν and jν to microscopic processes, by considering how a photon interacts with a 2-level atom: E 2 E1 h 0 E2 emission E1 Level 2, statistical weight g2 absorption Level 1, statistical weight g1 Absorption: system goes from Level 1 to Level 2 by absorbing a photon with energy hν0 Emission: system goes from Level 2 to Level 1 and a photon is emitted. Three processes can occur: 1. Spontaneous Emission 2. Absorption 3. Stimulated Emission 1. Spontaneous Emission 2 An atom in Level 2 drops to Level 1, emitting a photon, even in the absence of a radiation field Einstein A coefficient A21 ≡ transition probability per unit time for spontaneous emission [A21]= sec -1 Examples: permitted, dipole transitions A21 ~ 108 sec-1 magnetic dipole, forbidden transitions A21~103 sec-1 electric quadrupole, forbidden transitions A21~1 sec-1 1 2. Absorption An atom in level 1 absorbs a photon and ends up in level 2. 2 Due to the Heisenberg uncertainty principle, ΔE Δt > ħ, the energy levels are not precisely sharp 1 Each level has a “spread” in energy, called the “natural” Line width, a Lorentzian. So let’s parameterize the line profile as φ(ν), Centered on frequency νo. We define φ(ν) so that ( ) d 0 1 φ(ν) Einstein B-coefficients B12 J ≡ transition probability per unit time for absorption Where J J ( ) d 0 Stimulated Emission The presence of a radiation field will stimulate an atom to go from level 2 level 1 B 21 J Transition probability, per unit time for stimulated emission Equation of Statistical Equilibrium If detailed balance holds Number of transitions/sec from Level 1 Level 2 = Number of transitions/sec from Level 2 Level 1 Let n1 = # of atoms / volume in Level 1 n2 = # of atoms / volume in Level 2 Then: n 1 B 12 J Absorption n 2 A 21 n 2 B 21 J Spontaneous emission Stimulated emission hence A J n n 1 2 21 B B B 21 12 21 1 In thermodynamic equilibrium, the Boltzman equation gives n1/n2 n n 1 2 g g 1 exp 2 h o kT A So J 21 B g 1 B 12 exp g 2 B 21 21 h kT 0 1 (1) In thermodynamic equilibrium, J B Since the Lorentzian is narrow, we can approximate J B ( 0 ) 2h 0 c 2 3 1 h 0 exp 1 kT (2) Comparing (1) and (2), we get the EINSTEIN RELATIONS g 1 B 12 A 21 g 2 B 21 2h 3 B 21 2 c Comments: • There’s no “T” in the Einstein Relations, they relate atomic constants only. Hence, they must be true even if T.E. doesn’t hold. • Sometimes people derive the Einstein relations in terms of energy density, uν instead of Jnu, so there’s an extra factor of 4π/c: g 1 B 12 A 21 g 2 B 21 8 B 21 3 c 3 h 0 The Milne Relation Another example of using detailed balance to derive relations which are independent of the LTE assumption Relate photo-ionization cross-section at frequency nu, with cross-section for recombination for electrom with velocity v: g h a g m c v 2 (v ) 2 1 2 2 See derivation in Osterbrock & Ferland 2 2