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Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Microscopic PictureSlide 14Slide 15Slide 16Solutions to the Equation of Radiative TransferSlide 18Slide 19General SolutionSlide 21Slide 22Astro 300B: Jan. 21, 2011Equation of Radiative TransferSign Attendance SheetPick up HW #2, due FridayTurn in HW #1Chromey,Gaches,Patel: Do Doodle pollFirst Talks: Donnerstein, Burleigh, Sukhbold Fri., Feb 4Radiation Energy Densityspecific energy density uν = energy per volume, per frequency rangeConsider a cylinder with length ds = c dt c = speed of lightdAds uν(Ω) = specific energy density per solid angleThen dE = uν(Ω) dV dΩ dνuνenergyHz steradian volenergyvolumesteradianHzBut dV = dA c dt for the cylinder, so dE = uν(Ω) dA c dt dΩ dνRecall that dE = Iν dA dΩ dt dν so….€ uν(Ω) = IνcIntegrate uν(Ω) over all solid angle, to get the energy density uν   duu  dI c1 Recallsoergs cm-3 Hz-1 € Jν = 14π Iν dΩ04π∫€ uν = 4πc JνRadiation Pressure of an isotropic radiation field inside an enclosureWhat is the pressure exerted by each photon when it reflects off the wall?Each photon transfers 2x its normal component of momentumphotoninout+ p┴- p┴dIcp cos 2 2dIcp cos 2 2Since the radiation field is isotropic, Jν = IνddJcp sin cos 2 2 sin cos 2 020ddJc3cos 30 2 JcIntegrate over2π steradiansonlyNot 2π31 4 JcBut, recall that the energy densitySo….Radiation pressure of an isotropic radiation field = 1/3 of its energy density€ pν = 13 uν€ uν = 4πc JνExample: Flux from a uniformly bright sphere (e.g. HII region)At point P, Iν from the sphere is a constant (= B) if the ray intersects the sphere, and Iν = 0 otherwise. RPrθθc sin cos B 020ddcrR 1-sin wherecAnd…looking towards the sphere from point P, in the plane of the paperSo we integrate dφ from 2π to 0dφ€ ν F ≡ Iν∫ cosθ dΩFν = π B ( 1 – cos2 θc ) = π B sin2 θcOr…. 2rR  BFNote: at r = R€ Fν = π BEquation of Radiative TransferWhen photons pass through material, Iν changes due to (a) absorption (b) emission (c) scatteringdsIνIν + dIνdIν = dIν+ - dIν- - dIνscIν added by emissionIν subtracted by absorptionIν subtracted by scatteringEMISSION: dIν+DEFINE jν = volume emission coefficientdsjdI jν = energy emitted in direction Ĩ per volume dV per time dt per frequency interval dν per solid angle dΩUnits: ergs cm-3 sec-1 Hz-1 steradians-1Sometimes people write emissivity εν = energy emitted per mass per frequency per time integrated over all solid angleSo you can write:ddtddVjdE or4 dddtdVdEMass densityFraction of energy radiated into solid angle dΩABSORPTIONdIExperimental fact: dsIdI Define ABSORPTION COEFFICIENT =such that dsIdI has units of cm-1Microscopic PictureN absorbers / cm3Each absorber has cross-section for absorption has units cm2; is a function of frequencyIdII ASSUME:(1) Randomly distributed, independent absorbers(2) No shadowing: 3/12/1 distance cleinterpartimean  NThendsdAN  dsdANTotal area presented by the absorbers = So, the energy absorbed when light passes through the volume isTotal # absorbers in the volume = ddtddsdANIdE dsINdI dsIdI In other words,€ αν = N σν is often derivable from first principlesCan also define the mass absorption coefficient Where ρ = mass density, ( g cm-3 )has units cm2 g-1Sometimes is denoted So… the Equation of Radiative Transfer isdIdIdIdsIdsj ORIjdsdI emissionabsorptionAmount of Iν removed by absorption is proportional to IνAmount of Iν added by emission is independent of IνTASK: find αν and jν for appropriate physical processesSolutions to the Equation of Radiative TransferIjd sdI (1) Pure Emission(2) Pure absorption(3) Emission + AbsorptionIjdsdI (1)Pure Emission Only0Absorption coefficient = 0So,jdsdI sssdsjsIsIoo)( )()(Increase in brightness = The emission coefficientintegrated along the line ofsight.Incident specific intensity(2) Pure Absorption Only0jEmission coefficient = 0IdsdI ssoosdssIsI )(exp)()(Factor by which Iν decreases = exp of the absorption coefficient integrated along the line of sightincidentGeneral Solution)(1LI)(2LIIjdldI L1L2Eqn. 1Multiply Eqn. 1 by lLdl2expAnd rearrange€ dIν+ ανIνdl[ ]exp ανdlL2l∫ ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟= jνexp ανdlL2l∫ ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟dlEqn. 2l€ d Iνexp ανdlL2l∫ ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟= jνexp ανdlL2l∫ ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟dlNow integrate Eqn. 2 from L1 to L221221 2expexpLLlLLLlLdlIdlId LHS=12exp)()(12LLdlLILI21exp)()(12LLdlLILISo…dldljdlLILILlLLLL22121expexp)()(12dlldjdlLILILlLLLL22121expexp)()(12at L2 is equal to the incident specific intensity, decreased by a factor of plus the integral of jν along the line of sight, decreased by a factor ofI)(1LI21expLLdl2expLlld= the integral of αν from l to


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UA ASTR 300B - Lecture Notes

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