Transmittance notes Transmittance notes – cont I What is the difference between transmittance and transmissivity? There really isn’t any: see the AMS glossary entry I Review: prove that the total transmittance for two layers is I Review the relationship between the transmittance, absorption Trtot = Tr1 Tr2: cross section, flux and radiance × I Introduce the optical depth, τλ and the differential F2 transmissivity, dTr = dτλ Trtot = − F0 I Show that Fλ = Fλ0 exp( τλ) for “direct beam” flux − F1 Tr1 = F0 F2 F2 Tr2 = = F1 Tr1F0 F2 Tr1Tr2 = = Trtot F0 Transmittance 1/10 Transmittance 2/10 scattering and absorption cross section flux and itensity I Since we know that we are dealing with flux in a very small I Beer’s law for flux absorption comes from Maxwell’s equations: solid angle, we can divide (1) through by ∆ω, recogize that Iλ Fλ/∆ω in this near-parallel beam case and get Beer’s Fλ = Fλ0exp( βas) (1) ≈ − law for intensity: 2 3 where βa m m− is the volume absorption coefficient. 2 We are ignoring the spherical spreading (1/r ) loss in (1) Iλ = Iλ0exp( βas) (2) under the assumption that the source is very far way, so that − I Take the differential of either (1) or (2) and get something s, the distance traveled, is much much smaller than r, the like this: distance from the source. I This limits the applicability of (1) to direct sunlight and laser dIλ = βadsIλ0exp( βas) = Iλβads (3) light. − − − or I The assumption that the source is far away is the same as assuming that it occupies a very small solid angle ∆ω. dF = β dsF exp( β s) = F β ds (4) λ − a λ0 − a − λ a Transmittance 3/10 Transmittance 4/10 scattering and absorption cross section cross section continued How do we determine βa? 3 3 I If we have N absorbers per m , then the total power Separate the number of absorbers/unit volume, N ( m− ) from 2 2 absorbed in a volume of 1 m ds m long is: their absorption cross section σa ( m ) which is sometimes written × as an efficiency Kλ times the physical cross section σ: Ptotal = F σaNds (8) dI = I K Nσds (5) If we’d rather work with the gas density ρg instead of the λ − λ λ molecular density N, then write the gas density as ρg = NMw , For the flux this is 3 where Mw is the kg/molecule to get kg m− . dF = F σ Nds (6) − a Now (8) is where σa is the absorption cross section (or if we’re working with scattering, σs is the scattering cross section. For a single molecule Mw σa or droplet, the absorbed power is given by: Ptotal absorbed = F σaN ds = F ρg ds = F ρg kλds (9) Mw Mw 2 1 Pabsorbed = F σa (7) where k is the mass absorption coefficient ( m kg ) × λ − Transmittance 5/10 Transmittance 6/10 Absorption coefficient summary differential transmissivity Note that that if Tr = F = exp( β s) then Absorption can be reported as F0 a dF − 2 dTr = = βads exp( βas), or rearranging I K σ: Absorption efficiency a physical cross section ( m ) F0 − − × 2 I σa: Absorption cross section for an individual molecule ( m ) 2 dF I kλ: Aborption cross section per kg absorber ( m kg) dTr = = βads exp( βas) (10) F0 − − I βa: = ρgas kλ = Absorption cross section per unit volume 2 3 1 ( m m− ) or m− But what is the correct value to insert for s in (10)? We are looking at an infinitesimally thin layer of thickness ds, starting at Why both k and β ? k is useful because it separates the λ a λ s = 0, so exp( β s) = 1 and F = F , leaving: information about the absorber (say water vapor concentration) − a 0 from the information about the molecule’s ability to absorb (which can come from lab results, etc.). β is useful because 1/β is on dF a a dTr = = β ds (11) average the distance a photon goes before being absorbed (the F − a mean free path). Can we work backwards from dTr and arrive back at Tr? Transmittance 7/10 Transmittance 8/10 Integrated transmissivity Summary To go the other way, define the optical thickness: I σa and kλ are determined by the atomic bonds (with some dτ = β ds (12) dependence on temperture and pressure) λ − a I β = ρ k combines the absorber amount with its efficiency. Then Beers law becomes: a g λ 1/betaa is the mean free path of a photon – i.e. how far it can travel before it’s absorbed. dI = I dτ (13) λ − λ λ s I The optical depth τλ = 0 ρg kλds is related to the Integrating (13) from Iλ0 (at τ 0 = 0) to Iλ (where τ 0 = τλ) gives: transmittance by λ λ R Iλ(τλ) = Iλ0 exp( τλ) (14) Tr(s) = exp( τ ) (16) − − λ so when τλ=1, Iλ = Iλ0/exp(1) Iλ0/3. and to the direct beam flux (or intensity if we divide through ≈ And similarly: by ∆ω) by: dF dTr = = β ds = dτ (15) F = F exp( τ ) (17) F − a − λ λ λ0 − λ Transmittance 9/10 Transmittance 10/10.