AIT. Blackbody Radiation IAAT

Transcription

1 3 1 Blackbody Radiatio

2 Itroductio 3 2 First radiatio process to look at: radiatio i thermal equilibrium with itself: blackbody radiatio Assumptios: 1. Photos are Bosos, i.e., more tha oe photo per phase space cell possible. 2. Photos are i thermodyamic equilibrium at all frequecies. Outlie of computatio: 1. Compute mea eergy of photos of frequecy ν i phase space cell, E(ν) 2. Compute umber of phase space cells as a fuctio of frequecy, N(ν). 3. Compute photo spectrum as product E(ν) N(ν). Blackbody Radiatio: Derivatio 1

3 Derivatio: Step 1, I 3 3 First step: Mea eergy of photos of frequecy ν i phase space cell. Describe phase space cell as box = Photos: solutio of QM harmoic oscillator = Total eergy of box with photos: E = ( + 1 ) hν (3.1) 2 where 1 hν: groud state eergy (uobservable). 2 Probability that oscillator is i th state from Boltzma: ) hν ) P (ν, T ) = exp ( ( exp ( ( ) ) = + 1 exp( hν/kt ) 2 hν exp( hν/kt ) (3.2) Therefore, average eergy per phase cell: E = = E P (ν, T ) (3.3) {( + 1 ) hν 2 itroducig x = hν/kt } exp( hν/kt ) exp( hν/kt ) (3.4) = kt ( + 1 2) x exp( x) exp( x) { x exp( x) = kt + x } exp( x) 2 (3.5) (3.6) Blackbody Radiatio: Derivatio 2

4 Derivatio: Step 1, II 3 4 To evaluate E, eed to compute the geometric sums exp( x) ad x exp( x). We fid (see hadout) 1 exp( x) = (3.7) 1 exp( x) ad Therefore, x exp( x) = x exp( x) (1 exp( x)) 2 (3.8) ( x e x (1 e x ) 2 E = kt + x (1 e x ) 1 2 hν exp( x) = 1 exp( x) + hν 2 hν = e hν/kt 1 + hν 2 We reiterate: the hν/2 term is uobservable = Reormalize zero-poit of eergy to get rid of it. ) (3.9) (3.10) (3.11) Could have kow this result sice from Bose-Eistei statistics of particles with chemical potetial µ = 0 the occupatio umber is γ (ν, T ) = E hν = 1 exp(hν/kt ) 1 (3.12) Blackbody Radiatio: Derivatio 3

5 3 4 To prove Eqs. (3.7) ad (3.8), look at the Taylor series of f(y) = (1 y) 1. By iductio: ad i geeral f(y) = (1 y) 1 (3.13) df dy = ( 1)( 1) 1 = (1 y) 2 (1 y) 2 (3.14) d 2 f dy = ( 1)( 2) = (1 y) 3 (1 y) 3 (3.15) d f dy =! (1 y) +1 (3.16) Therefore, the Taylor series of f(y) aroud y = 0 is 1 1 y = 1 d f! dy y = y=0 y (3.17) Substitutig y = exp ( x) proves Eq. (3.7). To prove Eq. (3.8), we eed to compute x exp( x) = x exp( x) (3.18) Note that such that by Eq. (3.7) d dx exp( x) = exp( x) = d dx exp( x) (3.19) exp( x) (3.20) = d ( ) 1 dx 1 exp( x) (3.21) exp( x) = (1 exp( x)) 2 (3.22) Multiplyig with x proves Eq. (3.8).

6 Derivatio: Step 2, I 3 5 Secod Step: Computatio of desity of phase space cells i box L x, L y, L z. Wave vector of photo: k = 2π λ = 2πν (3.23) c To get all possible photos: cout distiguishable photos at same frequecy, i.e., photos with differet spi or differet umber of odes (=differet ). Spi is easy: there are 2 polarizatio states Number of odes: i the x, y, or z directio, umber of odes is x = L x λ = k xl x d x = L x 2π λ = L x dk x (3.24) 2π For 1, ca go to cotiuum of states. Eq the implies dn = d x d y d z = L xl y L z d 3 k (2π) 3 = V d3 k (2π) 3 (3.25) Therefore, the umber of states per uit volume per wave umber is Factor 2 from spi. k d 3 k = 2 dn V 1 d 3 k = 2 (3.26) (2π) 3 Blackbody Radiatio: Derivatio 4

7 Derivatio: Step 2, II 3 6 Because of Eq. (3.23), d 3 k = k 2 dk dω = (2π)3 c 3 ν 2 dν dω (3.27) such that the desity of states ρ s = ν dν dω = 2 (2π) (2π)3 3 c 3 ν 2 = 2ν2 c 3 (3.28) (umber of states per solid agle, per volume, per frequecy). Blackbody Radiatio: Derivatio 5

8 Blackbody spectrum 3 7 To summarize, we had: Mea eergy of state: State desity: E = hν e hν/kt 1 (3.11) ρ s = 2ν2 c 3 (3.28) The total eergy desity is the u ν (Ω) = E ρ s (3.29) = 2hν3 c 3 1 exp(hν/kt ) 1 (eergy per volume per frequecy per solid agle) (3.30) Because of Eq. (2.30) (u ν = I ν /c), the itesity is give by I ν = 2hν3 c 2 1 exp(hν/kt ) 1 =: B ν (3.31) This is the spectrum of a black body. I λ space, the spectrum is B λ = 2hc 2 /λ 5 exp(hc/λkt ) 1 (3.32) (sice we eed B λ dλ = B ν dν). Blackbody Radiatio: Derivatio 6

9 Spectrum K 10 9 K K 10 6 K I ν (T) [erg s -1 cm -2 Hz -1 sr -1 ] K 10 4 K 10 3 K 10 2 K K Frequecy ν [Hz] Blackbody Radiatio: Properties 1

10 Rayleigh-Jeas Law 3 9 For hν kt (ν T ), ( ) hν exp = 1 + hν kt kt +... (3.33) such that B ν 2ν2 c This is the Rayleigh-Jeas law. 2 kt (3.34) The Rayleigh-Jeas law is used i the radio regime to defie the brightess temperature, T b = I ν c 2 2kν 2 (3.35) where I ν is the measured radio itesity. Blackbody Radiatio: Properties 2

11 Wie Spectrum 3 10 For hν kt, (ν T ), ( ) ( hν hν exp 1 exp kt kt such that B ν 2hν3 c 2 exp ( hν kt the Wie spectrum (or Wie s law). ) ) (3.36) (3.37) Blackbody Radiatio: Properties 3

12 Wie Displacemet Law 3 11 The frequecy of maximum itesity, ν max is obtaied by solvig B ν ν which is equivalet to solvig = 0 (3.38) ν=νmax x = 3(1 exp( x)) (3.39) where x = hν max /kt. Numerically, x = 2.82, therefore hν max = 2.82 kt (3.40) This is the Wie displacemet law. The frequecy of maximum flux is directly proportioal to the black body temperature. Likewise, for B λ, oe fids Note that λ max ν max c! λ max T = cm K (3.41) Do ot cofuse Wie s law ad the Wie displacemet law... Blackbody Radiatio: Properties 4

13 Summary: Rayleigh-Jeas vs. Wie T=10 6 K 10 0 B ν (T) [erg s -1 cm -2 Hz -1 sr -1 ] Rayleigh-Jeas Law Wie La w ν max Frequecy ν [Hz] Rayleigh-Jeas applies for ν ν max Wie applies for ν ν max. Blackbody Radiatio: Properties 5

14 Stefa-Boltzma law 3 13 The total brightess of a black body is obtaied from B(T ) =... substitutig x = hν/kt = 2h ( ) kt 4 c 2 h... the itegral has the value π 4 /15 0 B ν (T ) dν (3.42) 0 x 3 dx exp(x) 1 = 2π4 k 4 15c 2 h T 4 = ac 3 4π T 4 = σ SBT 4 π Covert the brightess to the flux (F = πb, Eq. 2.24), to obtai the Stefa-Boltzma law. (3.43) (3.44) F = σ SB T 4 (3.45) Ad, yes, Boltzma s first ame is Ludwig, while Stefa s first ame is Josef. a is the radiatio desity costat, a := 8π5 k 4 15c 3 h 3 = erg cm 3 K 4 (3.46) also writte as the Stefa-Boltzma costat σ SB := 2π5 k 4 15c 2 h 3 = erg cm 2 K 4 s (3.47) Blackbody Radiatio: Properties 6

15 Effective Temperature G0 V T eff (λ>1000a)=6485k T eff (λ>5000a)=6094k 0.8 I λ /I Wavelegth [A] G0 V spectrum after Pickles (1998), PASP 110, 863 The effective temperature, T eff, of a spectrum I ν is the temperature where F = I ν cos θ dν dω = σt 4 eff (3.48) Sometimes, I ν is oly kow over a certai wavelegth rage, ad depedig o the spectrum the measured T eff will deped o this rage (see figure). Blackbody Radiatio: Properties 7

16 Applicatio: Plaets 3 15 The temperature of a irradiated body is give from eergy equilibrium: L 4πa 2πr2 = σ SB T 4 4πr 2 (3.49) where a: distace to su, r: plaetary radius. Therefore T = ( L 16πσ SB r 2 ) 1/4 = 281 K (a/1 AU) 1/2 (3.50) Last step used L = erg s 1 ad 1 AU = cm. If the plaet reflects part of the radiatio ad if the IR emissivity is oly roughly a BB, the Eq. (3.49) is modified, (1 B) L 4πa 2πr2 = ɛσ SB T 4 4πr 2 = T = 281 K (a/1 AU) 1/2 where B: Bod albedo, ad ɛ: effective emissivity ( ) 1 B 1/4 (3.51) For the Earth, B = 0.39, for Veus, B = Thus, sice T Earth 288 K, ɛ Earth = 0.55 < 1 (greehouse effect). If the plaet is ot a fast rotator, replace 4πr 2 by 2πr 2. Solar System 1 ɛ