as-500/hw4/text

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(1)
The approximate distances D[feature] and paths from the Solar System of the following features are relevant.
1 cm = 1.057001 × 10⁻¹⁸ ly.
D[Proxima Centauri] = D[PC] = 4.25 ly = 4.02 × 10¹⁸ cm.
D[Orion Nebula] = D[ON] = 1,344 ± 2 ly ≈ 1.272 × 10²¹ cm.
D[Center of Galaxy] = D[CG] = 2428.4 kly ≈ 2.48 × 10²² cm.
I'll assume the following interstellar number densities n[species], and assume isotropic densities across the paths.
n[H⁺] = 1 cm⁻³.
n[e-] = 1 cm⁻³.
n[H⁰] = 0.1 cm⁻³.
n[H⁻] = 10⁻¹¹ cm⁻³.
n[H₂] = 10⁻³ cm⁻³.
n[H⁰ + H⁺ +H⁻] = 1 + 0.1 + 10⁻¹¹ cm⁻³ ≈ 1.100 cm⁻³.
(2)
To find column densities N[species,feature] from Sol to these features, I integrate the (constant) number densities across the path lengths D[feature].
N[species,feature] = n[species] × D[feature].
N[H⁰,feature] = n[H⁰] × D[feature]
= 0.1 cm⁻³ × D[feature].
N[H⁺,feature] = n[H⁺] × D[feature]
= 1 cm⁻³ × D[feature].
N[H⁻,feature] = n[H⁻] × D[feature]
= 10⁻¹¹ cm⁻³ × D[feature].
N[H⁰ + H⁺ +H⁻,feature] = n[H⁰ + H⁺ +H⁻] × D[feature]
≈ 1.100 cm⁻³ × D[feature].
N[e⁻,feature] = n[e⁻] × D[feature]
= 1 cm⁻³ × D[feature].
N[H⁰,PC] = 0.1 cm⁻³ × 4.02 × 10¹⁸ cm
= 4.02 × 10¹⁷ cm⁻².
N[H⁺,PC] = 1 cm⁻³ × 4.02 × 10¹⁸ cm
= 4.02 × 10¹⁸ cm⁻².
N[H⁻,PC] = 10⁻¹¹ cm⁻³ × 4.02 × 10¹⁸ cm
= 4.02 × 10⁷ cm⁻².
N[H⁰ + H⁺ +H⁻,PC] = 1.100 cm⁻³ × 4.02 × 10¹⁸ cm
= 4.42 × 10¹⁸ cm⁻².
N[e⁻,PC] = 1 cm⁻³ × 4.02 × 10¹⁸ cm
= 4.02 × 10¹⁸ cm⁻².
N[H⁰,ON] = 0.1 cm⁻³ × 1.272 × 10²¹ cm
= 1.272 × 10²⁰ cm⁻².
N[H⁺,ON] = 1 cm⁻³ × 1.272 × 10²¹ cm
= 1.272 × 10²¹ cm⁻².
N[H⁻,ON] = 10⁻¹¹ cm⁻³ × 1.272 × 10²¹ cm
= 1.272 × 10¹⁰ cm⁻².
N[H⁰ + H⁺ +H⁻,ON] = 1.100 cm⁻³ × 1.272 × 10²¹ cm
= 1.399 × 10²¹ cm⁻².
N[e⁻,ON] = 1 cm⁻³ × 1.272 × 10²¹ cm
= 1.272 × 10²¹ cm⁻².
N[H⁰,CG] = 0.1 cm⁻³ × 2.48 × 10²² cm
= 2.48 × 10²¹ cm⁻².
N[H⁺,CG] = 1 cm⁻³ × 2.48 × 10²² cm
= 2.48 × 10²² cm⁻².
N[H⁻,CG] = 10⁻¹¹ cm⁻³ × 2.48 × 10²² cm
= 2.48 × 10¹¹ cm⁻².
N[H⁰ + H⁺ +H⁻,CG] = 1.100 cm⁻³ × 2.48 × 10²² cm
= 2.79 × 10²² cm⁻².
N[e⁻,CG] = 1 cm⁻³ × 2.48 × 10²² cm
= 2.48 × 10²² cm⁻².
(3)
Assuming an appropriate choice of wavelength and process, the column density N is related to the optical depth τ by the opacity κ for a process, with κ defined in terms of the cross section of interaction σ and the number density n of the target species of that interaction. I compute κ[target] and then find the optical depth between Sol and the features of interest for each process.
κ[process] = n[target] σ[process].
κ[Thomson Scattering] = n[e⁻] × σ[e⁻]
= 1 cm⁻³ × 6 × 10⁻²⁵ cm²
= 6 × 10⁻²⁵ cm⁻¹.
κ[Grain Absorption] = n[H⁰ + H⁺ +H⁻] × σ[Grain Absorption]
= 1.100 cm⁻³ × 10⁻²¹ cm²
= 1.100 × 10⁻²¹ cm⁻¹.
κ[H⁰ Photoionization] = n[H⁰] × σ[H⁰ Photoionization]
= 0.1 cm⁻³ × 6 × 10⁻¹⁸ cm²
= 6 × 10⁻¹⁹ cm⁻¹.
κ[Lyα Scattering] = n[H⁰] × σ[Lyα Scattering]
= 0.1 cm⁻³ × 10⁻¹² cm²
= 10⁻¹³ cm⁻¹.
κ[H⁻ Absorption] = n[H⁻] × σ[H⁻ Absorption]
= 10⁻¹¹ cm⁻³ × 3 × 10⁻¹⁷ cm²
= 3 × 10⁻²⁸ cm⁻¹.
τ[process,feature] = κ[process] × N[target,feature].
τ[Thomson Scattering,PC] = 6 × 10⁻²⁵ cm⁻¹ × 4.02 × 10¹⁸ cm⁻²
= 2.41 × 10⁻⁶ cm⁻³.
τ[Thomson Scattering,ON] = 6 × 10⁻²⁵ cm⁻¹ × 1.272 × 10²¹ cm⁻²
= 7.632 × 10⁻⁴ cm⁻³.
τ[Thomson Scattering,GC] = 6 × 10⁻²⁵ cm⁻¹ × 2.48 × 10²² cm⁻²
= 1.49 × 10⁻² cm⁻³.
(4)
τ[Grain Absorption,PC] = 1.100 × 10⁻²¹ cm⁻¹ × 4.42 × 10¹⁸ cm⁻²
= 4.86 × 10⁻³ cm⁻³.
τ[Grain Absorption,ON] = 1.100 × 10⁻²¹ cm⁻¹ × 1.399 × 10²¹ cm⁻²¹
= 0.1539 × cm⁻³.
τ[Grain Absorption,GC] = 1.100 × 10⁻²¹ cm⁻¹ × 2.79 × 10²² cm⁻²
= 3.07 cm⁻³.
(5)
τ[H⁰ Photoionization,PC] = 6 × 10⁻¹⁹ cm⁻¹ × 4.02 × 10¹⁷ cm⁻²
= 0.241 cm⁻³.
τ[H⁰ Photoionization,ON] = 6 × 10⁻¹⁹ cm⁻¹ × 1.272 × 10²⁰ cm⁻²
= 76.32 × 10² cm⁻³.
τ[H⁰ Photoionization,GC] = 6 × 10⁻¹⁹ cm⁻¹ × 2.48 × 10²¹ cm⁻²
= 1.49 × 10³ cm⁻³.
(6)
τ[Lyα Scattering,PC] = 10⁻¹³ cm⁻¹ × 4.02 × 10¹⁷ cm⁻²
= 4.02 × 10⁴ cm⁻³.
τ[Lyα Scattering,ON] = 10⁻¹³ cm⁻¹ × 1.272 × 10²⁰ cm⁻²
= 1.272 × 10⁷ cm⁻³.
τ[Lyα Scattering,GC] = 10⁻¹³ cm⁻¹ × 2.48 × 10²¹ cm⁻²
= 2.48 × 10⁸ cm⁻³.
(7)
τ[H⁻ Absorption,PC] = 3 × 10⁻²⁸ cm⁻¹ × 4.02 × 10⁷ cm⁻²
= 1.21 × 10⁻²⁰ cm⁻³.
τ[H⁻ Absorption,ON] = 3 × 10⁻²⁸ cm⁻¹ × 1.272 × 10¹⁰ cm⁻²
= 3.816 × 10⁻¹⁸ cm⁻³.
τ[H⁻ Absorption,GC] = 3 × 10⁻²⁸ cm⁻¹ × 2.48 × 10¹¹ cm⁻²
= 7.44 × 10⁻¹⁷ cm⁻³.
(8)
The density of Earth's atmosphere is obviously not isotropic, but to a first approximation I will assume it has a particle number density ρ and thickness T. The column density is then computed.
ρ = 5 × 10¹⁹ cm⁻³.
T = 10 km = 10⁶ cm.
N[Earth's Atmosphere] = ρ × T
= 5 × 10¹⁹ cm⁻³ × 10⁶ cm⁻³
= 5 × 10²⁵ cm⁻².
The greatest number density I see in the ISM is that of the sum of hydrogen species. Even given that sum of species, the column density to the center of the galaxy is on the order of 10²², whereas the column density seen across 10km of the Earth's atmosphere is on the order of 10²⁵. So, we see ~1000 times the optical thickness across the Earth's atmosphere than across a quarter of the galaxy!