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Chapter 1 Physics of Thermodynamic Equilibrium 1.1 Thermodynamic Equilibrium (T.E.) Any system, which is left alone for sufficiently long time, will reach an equilibrium state known as thermal equilibrium. Its properties then are characterized by a single parameter called temperature. For example, when a system of gas atoms and photons is said to be in T.E., then the particles satisfy the following properties: (1) The velocity distribution is Maxwellian velocity distribution 4m 3 mv 2 f M (v ) 2kT , v 2 exp 2mkT 2 3 f M v dv 1 , or 0 2 E 1 f M (E) E 2 exp , kT 2 kT 3 f M E dE 1 0 where v is the speed (magnitude of velocity) of particle, in non-relativistic limit mv 2 E and f (v)dv f ( E )dE . 2 1 (2) The distribution of atoms in each energy level obeys the Boltzmann law: nj gj E j Ei exp ni gi kT gj h j i exp gi kT Boltzman law E ni g i exp i only valid when kT the temperature is very high where n j ni , g j g i and h j i are the population, statistical weight factor (internal degree of freedom), and energy of the j (i)th energy level respectively, 13.6eV e.g. Hydrogen atom g l 2l 1 , h n n2 (3) The distribution of ionized and neutral atoms is given by the Saha equation ne n g 0 22me kT 2 3 exp I A0 A e n0 g0 h 3 kT n0 n ne where I is the ionization potential , g 0 and g 0 are the statistical weights for the ion and the atom in ground states respectively. 2 We can prove the Saha equation as follows: (a) According to the Boltzmann law , 1 I me v 2 dN 0 (v) g exp 2 (note: dN 0 and N 0 are per unit volume) N0 g0 kT where dN 0 (v ) is the differential number of ions in the ground state per unit volume, with the free electron in velocity regime between v, v dv , and N 0 is the number of atoms in the ground state per unit volume. g 0 is the statistical weight factor of the ion in its ground state and the differential electron statistical factor is g e .We have g g0 ge By definition 1 p e me v , dV (Ne: electron number density) Ne We have chosen the volume element dV s.t. dVNe =1. where N e is the electron density and dV is the volume consisting of one electron. We also have p = mv dVd 3 p 2 4me v 2 dv 3 d3p = dp2dp ge 2 h3 Ne h3 = 4(mv)2d(mv) Hence we have 3 1 I me v 2 dN 0 (v) 8me g 0 3 3 exp 2 v 2 dv N0 h Ne g0 kT N0 Ne dN N 0 e 8me g 0 3 m v2 exp I exp e v 2 dv Not v is the absolute 2kT N0 N0 h3 g0 kT 0 magnitude, 0 3 1 8m g 2 2kT 3 m v2 2 exp I e x2 e 0 2 x dx x e 3 h g0 kT me 2kT 0 3 2me kT 2 2 g 0 exp I Q.E.D. h g0 kT 2 where e x x 2 dx 2 has been used. In fact this is derived in the class. 0 4 Example. Some neutral hydrogen is first injected into a container at Ti 0 . If the container is heated up to T f , what is the ratio of ionized hydrogen to neutral hydrogen Np N ? N.B. g p 1 , g H 2 . (In ground state there are two states and ). H Np Let . Using the Saha equation NH 3 N p Ne 2me kT 2 2 g p N e exp I h gH kT 2 NH Recall 4 N H mH N e me N p m p N H N p mH mH N p 1 1 mH N e 1 1 (N.B. ρ is constant if the size of the container does not change, N e N p , m p m H 1 2me kT 2 , m p m e ) is called the thermal de Broglie wavelength. h 2 Thus 2 N e 3 exp I 1 mH kT 2 m H3 exp I , T 1 kT As T 0 , we can see that 0 , and as T , we find too. For 1, we need to solve a quadratic equation to obtain the exact result. (4) The radiation field emitted by a thermal object with temperature T is isotropic, homogeneous and stationary. The frequency distribution is called a blackbody form (Planck function). 1 2h 3 h I B 2 exp kT 1 c where I = specific intensity (power/area/freq/solid angle=erg/sec Hz cm2 ster) and B is the source function (a specific emission mechanism). 5 (5) The principle of detailed balancing applies N ANB gAgB E A E B E C A B C exp NC gC kT (6) Kirchoff’s law – which relates the emission and absorption coefficients to the Planck function. B or 2 h 3 1 S B 2 c h e kT 1 where is the emission coefficient, is the absorption coefficient ( cm 1 ) and S is the source function (same as B ). N.B. In most astronomical situations, the (global) T.E. cannot be reached. Instead, T.E. could be achieved in a small region. In this case, this small region is said to have local thermodynamic equilibrium (LTE). Optical depth ds Line of sight ds 6 1.2 Connection between Maxwellian Distribution and the Fundamental Distributions Spin 1/2 particles (e.g. electrons, protons, neutrons, etc.) satisfy the Fermi-Dirac distribution 1 nj / kT 1 j e where n j is the occupation number in the energy level with energy j , is the chemical potential and T is the temperature. Integer spin particles (e.g. 1 photons, helium, etc.) satisfy the n j j / kT 1 Bose-Einstein distribution e j / kT If e 1 (classical limit), both distributions reduce to j / kT nj e j . This is called Maxwell-Boltzman (Maxwellian) distribution. We can defined the probability distribution for the Maxwellian distribution as j / kT j / kT e e fj . e e j / kT j / kT j j In the continuous limit, 7 g E e E / kT f j f E g E e E / kT dE 0 where g E dN is called the density of state. For non-relativistic particle dE ( E p 2 / 2m ), d 3 xd 3 p V 4p 2 dp V 4 2m E 1 / 2 dE 3/ 2 N.B. dN h3 h3 h3 dN 4V 2m E 1 / 2 3/ 2 . dE h3 Therefore, the Maxwellian distribution becomes E 1 / 2 e E / kT f M E 0 E 1 / 2 e E / kT dE E 1 / 2 e E / kT kT 3 / 2 0 x 1 / 2 e x dx 2E 1 / 2 e E / kT . kT 3 / 2 Classical limit j / kT Since e 1 must hold for all j , it must be true for e 0 / kT 1 too. Let 0 0 , we obtain e / kT 1 . From statistical mechanics, we can obtain V e / N3 is the thermal de-Broglie wavelength. Since mkT h where 1/ 2 ~ p of 2mkT 1/ 2 V the particle, so 3 represents a minimum volume occupied by a particle by Q.M. and N represents the average volume for each particle. So, the classical limit will be satisfied when the gas is extremely dilute. Alternatively, as T , the classical limit is obviously satisfied. These two conditions ensure us when the Maxwellian distribution can be used. 8 1.3 Blackbody Radiation A blackbody is an object that absorbs all radiation incident on it. Its radiation satisfies the Planck function: 1 2h 3 h I B 2 exp kT 1 c The followings are some useful properties of the Planck function. (A) Stefan-Boltzmann Law The intensity of the total radiation emitted by a blackbody is: 4 2h 3 d 2h kT x 3 dx 2 4 k 4 4 I I d 2 x T c2 h c h 0 e 1 15h 3 c 2 0 exp 1 kT 4 15 h where x , k 1.38 10 16 erg s K-1, h 6.6 10 27 erg s, c 3 1010 cm s-1 and kT x 3 dx 4 e x 1 15 0 outgoing energy The total flux area A 2 Fk 2 5 k 4 4 1 Fn I cos I cosd cos d I d T BT 4 ˆ ˆ k n d 0 0 15h 3 c 2 with B 5.67 10 5 erg cm 2 s 1 K 4 (the Stefan-Boltzmann constant). 9 The total radiation energy density 1 4 8 5 k 4 4 R Id 15h 3 c 3 T aT I 4 c c (R velocity = energy flux) 4 B with a 7.56 10 15 ergcm 3 K 4 . c Some books also call it Stefan-Boltzmann constant, but B is more popular to be called as Stefan-Boltzmann constant. (B) Limiting cases: h (i) Wien’s distribution ( 1 )-high frequency limit (low temperature case) kT 2 h 3 h B ( Wien ) 2 exp c kT h (ii) Rayleigh-Jeans distribution ( 1 )- low frequency limit ( e x 1 x ) kT 1 2h 3 h B ( RJ ) 2 c kT (C) Monotonicity with temperature B T1 B T2 if T1 T2 noticing that for a given frequency, B 0 can prove this result. T 10 h B T1 exp B 2h 2 4 kT N.B. 2 2 0, T c kT h 2 exp kT 1 B T2 (D) Wiens’ law: T1 m axT 0.29 cmK first come out as an experimental law d c 2hc 2 1 T2 Proof. B B B 2 5 d hc B d B d exp 1 kT wavelength frequency distribution distribution B The maximum occurs at 0 hc 1 hc exp B 2 5 kT 2 kT 0 , x hc 2hc 2 6 hc 5 hc kT exp 1 exp 1 kT kT 5(e x 1) xe x x 1 e x 5 iteration method x 4.9 4.92 4.94 4.96 4.98 5 x 1 ex 0.99255 0.9927 0.9928 0.993 0.9932 1 1-e-x x 5 0.98 0.984 0.988 0.992 0.996 x x m 4.96 xm 5 11 hc maxT 0.2902cmK 4.96k H.W. Calculate m ax , note that m ax m ax c . Don’t worry because B (T ) B (T ) . peak position of the two graphs are not at the same point (E) Mean photon energy and number density x4 hB (T )d exp(x) 1 dx (5) (5) h 0 kT 0 kT 3.8kT x 3 (4) (4) B (T )d 0 exp(x) 1 dx 0 1 B 8k 3 n d d d 3 3 (3) (3)T 3 20.42T 3 h c 0 h c 0 h is the energy density per unit frequency and (n) is called the Riemann zeta function. (c.f. Mathematical Results in appendix) 1.4 Applications to Stellar Spectra The continuous spectrum of a star is conventionally compared with a blackbody of intensity B (T ) in three different ways. (a) Brightness temperature: The brightness temperature Tb of an object at a given wavelength (frequency) is the temperature of a blackbody having the same brightness (specific intensity) at that wavelength (frequency): B Tb I (the specific intensity observed at a specific frequency ) 12 h In radio wave band, usually 1 , then kT 2 2 k I c 2 B (T ) T Tb c2 2k 2 If the deduced brightness temperature is unreasonably high, then Tb will imply that some non-thermal emission processes are taking place in the source, e.g. Tb 10 30 K for pulsar in radio wave. (b) Effective temperature ( Teff ): L L d 4R 2 BTeff 4 There are two main differences between Tb and Teff . i) Tb depends on but Teff does not. ii) In order to determine Teff , the size of the source must be known (If the spectrum also has a blackbody shape, Teff = Tbb R) (c) Color temperature ( Tc ) (usually, only a finite range of spectrum can be obtained) The observed spectrum of a source may often have a shape which is similar to that of a blackbody, but not necessarily with the same absolute value, then the color temperature is defined as 13 hc exp 1 f i Bi , Tc j 5 kT j c f j B j , Tc i hc kT 1 exp observed spectra i c Note that while both Tb and Teff depend on the magnitude of the source intensity, Tc depends only on the shape of the observed spectrum. In general Tc varies from one spectral regime to another. 1.5 The Magnitude System The magnitude system has its origin in the classification of stars according to their apparent visual brightness used by the Greek astronomers. In the 19th century, this classification was quantified as a logarithmic law . Nowadays, magnitudes are relative measures of monochromatic flux of a source. If f ( ) is the monochromatic flux due to a source and f 0 is a reference monochromatic flux, then the magnitude is f ( ) m 2.5 log 2.5 log f ( ) q0 f0 with q0 2.5 log f 0 . q 0 defines the zero point of the magnitude scale. Sometimes, another magnitude system is used, it is the Bolometric Magnitude which gives the flux integrated over all wavelengths: 14 f ( )d mb 2.5 log 0 Fb and Fb 2.52 10 8 Wm 2 2.52 10 5 erg sec1 cm 2 . Since the Sun is the most visible stellar object in the sky, the sun’s luminosity is often used as a reference mb 0.23 5 log D 2.5 log( L / L ) where D is the distance in parsec from the source to the Earth, L is the luminosity of the source and L 3.8 10 33 ergs 1 is the solar luminosity. Remark : for a very distance object the correction due to interstellar absorption is necessary to include in the above expression. So far, we have just calculated the apparent magnitude that is a measure of the flux received on Earth. To compare the intrinsic luminosity, we introduce the absolute magnitude, which is the apparent magnitude the source would have at a distance of 10 pc (N.B. 1 parsec 3 10 18 cm ), corrected for any effects of interstellar absorption A : M m 5 5 log D A N.B.: A is a complicated correction function which depends on interstellar medium (ISM) Our sun has an absolute magnitude 4.6 and visual or apparent magnitude –26.73. (A for Sun is very small because there is very little ISM.) 15 Example Estimate the apparent magnitude of the moon. The moon light comes from L reflecting the sunlight. power received at the surface of the moon 4D 2 L the Moon like a flat mirror area of the moonthe power received by the whole moon RMoon 2 4D 2 LMoon L 1.2 10 23 ergs 1 Moon 4D 2 Total surface area at d distance D from the Earth Sun Sun d 1 pc 2.5 log( LMoon / L ) mb 0.23 5 log D 13.5 where D 1AU 1.5 1013 cm , d 3.8 1010 cm and RMoon 1.7 10 8 cm have been used. The real observed value is –12.7. The discrepancy is due to the fact that the Moon absorbs some sunlight and only reflects 60% of the incident light. The amount of sunlight being reflected depends on the surface properties of the planet. reflected LR luminosity percentage of sunlight reflected: Albedo LI incident luminosity 16 Appendix: Mathematical Results A useful result for evaluating the frequency moment of the Planck function is 2k 4 B( , T )d c 2 h 3 4 4T 4 0 where (n 1) n! is the gamma function and 1 u n1 du m n (n) e u 1 ( n) 0 m 1 is the Riemann zeta function. N.B. (I) u n1 e u u n1 eu 1 du u du u n1e mu du m n x n1e x dx (n) m n 0 1 e m 1 0 m 1 m 1 0 0 n where the change of variable mu x has been used. (II) 4 4 2h 3 d 2h kT x 3 2h kT B( , T )d h c 2 c 2 h e x 1 dx c 2 h ( 4) 4 0 0 exp 1 0 kT h x zero moment =0 kT l moment = l Some useful numerical values of zeta function 17 n (n) 2 2 /6 3 1.2 4 4 / 90 5 1.04 6 1.02 (n) 1 for n>4 18

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posted: | 3/27/2011 |

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