F = c where ^ı is a unit vector along the ray. The normal component is. Iν cos 2 θ. d dadt. dp normal (θ,φ) = dpcos θ = df ν

Transcription

1 INTRODUCTION So far, the only information we have been able to get about the universe beyond the solar system is from the eletromagneti radiation that reahes us (and a few osmi rays). So doing Astrophysis is like a big puzzle: look at the radiation and use it to figure out well, everything! Where is the objet, what is it, when and how did it form, how has it evolved sine then, how does it fit in the "astrophysial zoo"?. So our objetive in this ourse is to learn some of the triks for solving the puzzle. We start by looking at the way that radiation is modified as it travels from its soure to the observer (us). Radiative transfer Voabulary Intensity You an think of intensity as measuring the energy traveling along a light ray. Of ourse a single ray arries no energy, so we look at a bundle of rays within a solid angle d of the entral ray. The energy rossing area da normal to the ray in time dt and frequeny range dν is de = I ν dadtdνd (1) The intensity I ν =energy per unit time per unit frequeny per unit area per unit solid angle units (astronomers like gs units) erg se m 2 Hz ster Mean intensity J n (ergs/m 2 s Hz) is obtained by averaging over angles: J ν = 1 Z I ν d (2) 4π An isotropi radiation field has J ν = I ν. Flux F ν measures the energy rossing a surfae, in erg/m 2 s Hz. To get the energy rossing the surfae, we inlude only the omponent of the diretion vetor along the ray that is normal to the surfae. That introdues a fator of osθ. 1

2 Z F ν = I ν osθ d (3) Flux is usually the easiest quantity to measure for a typial astrophysial soure, suh as a star. We an only measure intensity if the soure is resolved, like the sun or the moon. The flux at Earth due to an unresolved soure with luminosity L erg/s a distane D away is F = L 4πD 2 Momentum flux and radiation pressure. The momentum of a photon is its energy/, so we an get momentum flux from energy flux. Pressure is normal fore/unit area=rate of hange of momentum normal to the surfae/area. So we ompute the total momentum rosing area da in time dt due to a bundle of rays at θ,φ as d p(θ,φ) = df ν ^ı dadt = I ν os θ d ^ı dadt where ^ı is a unit vetor along the ray. The normal omponent is dp normal (θ,φ) = dpos θ = df ν dadtosθ = I ν os 2 θ d dadt and integrating over the whole radiation field, we have Z Iν os 2 θ dp normal = d dadt 2

3 and so the pressure on the surfae path is P ν = dp normal = 1 Z I ν os 2 θ d (4) dadt Finally the total radiation energy in a volume dv = dadl ontributed by radiation moving in diretion θ,φ is de = du ν (θ,φ)dadldν All the energy passes out of the volume in a time dt = dl/, so from (1), de = I ν da dl dνdand omparing the two expressions we find du ν (θ,φ) = I ν d To get the total energy density we sum the ontributions from rays in all diretions, to get Z Z Iν u ν = du ν (θ,φ) = d = 4π J ν (5) Let s use these definitions in a few simple examples. Radiation pressure in a perfetly refleting enlosure with an isotropi radiation field. At eah refletion the momentum tranferred is twie the inident momentum, so P ν = 2 Z I ν os 2 θd But an isotropi radiation field means that I ν is independent of diretion, so I ν = J ν and it omes out of the integral. Radiation only travels over half the total solid angle sine no radiation is oming in from outside the enlosure: P ν = 2 J ν Z +1 µ 2 dµ Z 2π Flux from a uniformly bright sphere: dφ = 4πJ ν µ = 4π 3 J ν = 1 3 u ν 3

4 Rays reahing P leave the sphere at angles from θ = to θ = θ max, as shown. The intensity is the same along all rays. The flux at P is Z F = Z 1 I osθd = 2πI µdµ = πi 1 os 2 θ m ax = πi sin 2 θ m ax = πi os θ m ax where in the last step we used the fat that sin θ ' tanθ for small angles. Notie that we annot use the result for r = R (as is done in the book) beause that violates the small angle approximation we made. But at the surfae the integral beomes F = 2πI Z +1 µdµ = πi So the textbook result is orret even though their method of getting it is not. Radiative Transfer. So let s start with the intensity and look at the ray rossing a ylinder of area da and thikness ds. As the ray rosses the ylinder, the intensity will inrease due to emission in the ylinder, and derease due to absorption. (We will omit sattering for the moment, and ome bak to that later.) The emission oe ient is j ν (erg/m 3 s Hz) and the absorption may be written as the density of partiles in the slab n m 3 times the absorption ross setion per partile σ ν (m/hz), or just as α ν = nσ ν m 1 Hz 1. There are ndads partiles in the slab with a total area A part = nσdads. The photons hitting that area will be absorbed, so Energy out = energy in + emission - absorption Dividing a fator of d dtdν from eah term, we have dai ν (s + ds) = dai ν (s) + j ν dads α ν I v dads or di ν ds = j ν α ν I ν (6) This is the equation of radiative transfer. To simplify, we define the optial depth τ ν by dτ ν = α ν ds so the radiative transfer equation may be written as di ν dτ = j ν α ν I ν = S ν I ν µ 2 R r where is the soure funtion. S ν jν α ν (7) 4

5 We may solve the equation formally as follows. e τv and rearrange. Multiply the equation by d dτ ν (e τv I ν ) = e τv I ν + e τv di ν dτ ν = e τν S ν Thus Z τν I ν (τ ν ) = e τν I ν () + e τν e τ νs ν (τ ν)dτ ν (8) 5