Ay121 Problem Set 1. TA: Kathryn Plant. October Some parts are based on solutions by previous TAs Yuguang Chen and Rachel Theios.

Transcription

1 Ay2 Problem Set TA: Kathryn Plant October 28 Some parts are based on solutions by previous TAs Yuguang Chen and Rachel Theios. Temperatures One point for each part. (a) Since we are observing at MHz and talking about brightness temperature, we are in the Rayleigh-Jeans limit. In more detail: The Rayleigh-Jeans limit is the limit in which hν << k b T. Thus, at MHz we are in the Rayleigh-Jeans limit if the temperature is well above hν/k b =.5K. The supernova is definitely much much hotter than this. T b = c2 I ν 2k b ν 2 We need I ν. Since the angle subtended by the source is small, the solid angle is approximately π(θ/2) 2. I ν = F π(θ/2) =.6 9 erg cm 2 s Hz 2 π(4. /2) =. erg cm 2 s Hz steradian 2 Thus, T b = 4 7 K. (b) If the source is more compact, the solid angle is smaller and thus the specific intensity is larger and so the brightness temperature is larger than the result from (a). (c) Using the Wien-Displacement Law, ν max = 58.8GHzK T b = 2. 8 Hz. (d) For thermal radiation, S ν = B ν and so the radiative transfer equation in the Rayleigh-Jeans tail becomes dt b dτ = T T b Solving this, T b (τ) = T ( e τ ) and thus T b < T. The brightness temperature underestimates the physical temperature, and approaches the physical temperature if the source becomes completely opaque (we have assumed that there are no background sources of radiation). An extremely large brightness temperature can imply that the radiation has a non-thermal source.

2 2 Blackbody Luminosity One point for setting up the calculation, one point for the flux at point p, one point for the luminosity. Viewed from point p, the sphere subtends an angle θ c R/r (this is shown as angle θ c in Rybicki and Lightman Figure.6). The flux density is F ν = 2π θc I ν cos(θ)dω For black body radiation, I ν = B ν and is isotropic, so F ν = B ν 2π The total flux is θc cos(θ) sin(θ)dθdφ = πb ν sin 2 (θ) ] θ c πb ν F ν dν = π ( ) 2 R B ν dν = r, Where the integral over B ν was done in class. Thus, Limb Darkening L = r 2 F = R 2 σt 4 ( ) 2 R σt 4 r ( ) 2 R r The gist: This problem was a way to get familiar with the radiative transfer equation, the moments of the specific intensity, and their physical interpretations. Paying attention to the bounds of the integrals (i.e. when you are considering half a sphere versus a whole sphere) was key. (a) points: for mean intensity, for flux, for energy density and radiation pressure. To get a point, the answer needs the result as well as the correct set up with bounds on the integrals. Note that dω = sin θdθdφ = dµdφ and for the unit sphere, φ, 2π] and θ, pi] so µ, ]. Mean intensity J IdΩ = (I o + µi )dω = 2π (I o + µi )dµ = 2 µ + I ] 4 µ2 ) = I o Flux Density The problem specifies to integrate over, because often we are interested in the flux from a particular direction (e.g. all the flux going outward) but here we are interested in the net flux from all directions. The isotropic component will cancel from each hemisphere but the non-isotropic component will 2

3 not. F µidω = (µi o + µ 2 I )dω = 2π (µi o + µ 2 I )dµ Energy Density = 2π 2 µ2 + I ] µ ) = I u = c IdΩ = c J = c I o p = c cos 2 (θ)idω = c (µ 2 I o +µ I )dω = 2π c (µ 2 I o +µ I )dµ = c µ + I ] 4 µ4 ) = c I o = u This is the same result as for an isotropic intensity field, because the symmetry of this problem made the contribution from the non-isotropic component go to zero. (b) Integrate the radiative transfer equation over all angles. One point each for correct treatment of each of the terms in the radiative transfer equation. Left hand side: µ I τ dω = IdΩ µ I F dω = τ τ SdΩ, which is zero because we are making the approximation that F is independent of τ. In a real star, the temperature increases as you go deeper. Right hand side: IdΩ = J by the definition of mean intensity, and SdΩ = S because S is independent of angle. Thus, J S = J = S (c) We want to solve for I o and I in terms of the temperature and optical depth. One point for I and two points for I. Using results from (a) and F = σt 4 eff, I = F = σt 4 eff

4 To go after I o, we take the first moment of the radiative transfer equation. µ 2 I τ dω = µidω µsdω The first term on the right hand side is the definition of F. S is independent of angle (and from (b) is equal to the mean intensity) and so the second term on the right hand side becomes S µdω = Using the definition of radiation pressure, the left hand side is Iµ 2 dω = τ τ pc Thus we have τ pc = F Substituting results from part (a) and using F = σteff 4, this becomes Finally, I o τ = σt 4 eff I o = σt 4 τ + constant I = I o + µi = σt 4 (τ + µ + C) (d) Use a boundary condition to determine the integration constant. One point for setting up the integral (over a hemisphere) and determining the constant, one point for a plot with labelled axes and a numerical scale. Assume the flux is zero at the surface. In other words, I(τ =, µ)µdω = Substituting the result from (d) and setting τ = : 2π σt 4 (µ + C)µdΩ = (µ 2 + Cµ)dΩ = (µ 2 + Cµ)dµ = 4

5 (θ goes from π to zero for the half. θ = π is down.) 2π + πc = C = 2/ At τ =, the normalized intensity simplifies to I(τ =, µ) I(τ =, µ = ) = 5 µ