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sol1-f5 - s along the line-of-sight Then the optical depth...

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Astro 405/505, fall semester 2005 Homework, 1st set, solutions Problem 1: equation of state First of all the integral ~ V = Z d 3 p ~v f = 0 because the integrand is antisymmetric. Then the pressure integral π ij = Z d 3 p u i u j f = Z d 3 p v i v j f = 0 i 6 = j R d 3 p v 2 i f i = j because in the first case we are again dealing with an antisymmetric integrand. The diagonal elements ( i = j ) are all the same because the distribution function doesn’t depend on direction. Setting then P = m Z d 3 p v 2 i f = m 3 Z d 3 p v 2 f = m 3 Z d 3 p u 2 f = 2 3 One equation of state is the ideal gas law P = ρ k m T which applies in high-density envi- ronments in thermodynamical equilibrium. Problem 2: Cygnus OB2 No.5 a) Use spherical coordinates. The mass conservation equation then reads ∂ρ ∂t + ~ ∇ · ( ~ V ρ ) = 1 r 2 ∂r ( r 2 V r ρ ) = 0 We assume V r = const. and therefore ρ r - 2 . b) First of all, the source function is independent of location. S ν = j ν α ν ν 2 An arbitrary line-of-sight passes the star with an impact parameter r 0 , so we can relate the radial coordinate r in our spherical coordinate system to a pathlength coordinate
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Unformatted text preview: s along the line-of-sight. Then the optical depth for the line-of-sight in question is r 2 = s 2 + r 2 ⇒ τ = Z ∞-∞ ds α ν = α ρ 2 ν 2 Z ∞-∞ ds 1 ( s 2 + r 2 ) 2 As recommended use the transformation s = r tan φ to obtain τ = 2 α ρ 2 r 3 ν 2 Z π/ 2 dφ cos 2 φ = π α ρ 2 2 r 3 ν 2 The intensity therefore is I ν = S ν [1-exp (-τ )] ∝ ± ν 2 τ À 1 ν τ ¿ 1 c) I replace the solid angle integral by one over the surface area, d Ω = (1 /D 2 ) dA . Then the Fux is F ν = 1 D 2 Z ∞ dr 2 π r I ν = 2 π j ν 2 α D 2 Z ∞ dr r " 1-exp-π α ρ 2 2 r 3 ν 2 !# We may use τ as new integration variable. Then F ν = 2 π j ν 2 3 α D 2 π α ρ 2 2 ! 2 / 3 ν 2 / 3 Z ∞ dτ τ-7 / 3 [1-exp (-τ )] ∝ ν 2 / 3 So we can reproduce the quiet phase spectrum....
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