as-500/hw5/text

(1)

Light is visible to approximately an optical depth τ = 1.

τ = ∫ κ dL, with κ the opacity and ∫ dL the integral over some path with length L.

κ is wavelength dependent and known for the spectrum, so to find where on the spectrum I expect to be able to see across 100 PC of the ISM, I substitute the following and compute. κ is considered constant, per wavelength, across a region where only scattering from the ISM is considered.

1 = ∫ κ dL = κ ∫ dL = κ (100 PC).

κ = 10⁻² PC⁻¹ = 3.1 × 10⁻²⁰ cm ≈ 10⁻²⁰.

I've made this very approximate, since we're just looking for understanding.

This is the limiting value of κ, so any wavelengths with greater κ will not be visible to the observer.

κ < 3.1 × 10⁻²⁰ cm.

Consider both the cross-section of absorption and scattering. The opacity of absorption is significantly greater through most of the spectrum than that of scattering. The scattering opacity does not reach strength sufficient to "tip" the opacity above the threshold when summed with that of absorption. Therefore, I only consider opacity of absorption.

The regions of interest are:

κ[λ → 0 μm] → 0
κ[10⁻² μm < λ < 10⁻¹ μm] > 10⁻²⁰ cm⁻¹
κ[10⁻¹ μm < λ < 10⁷ μm] < 10⁻²⁰ cm⁻¹
κ[λ → ∞] → ∞

The opacity is sufficiently weak through the regions 

	λ = {0 .. 10⁻² μm } and λ = { 10⁻¹ .. 10⁷ μm }.


(2)

(a)
n[e⁻] = n[H⁺] = 10⁹ cm⁻³.

(b)
R[sol] = 6.957 × 10¹⁰ cm.

L = R[sol]/4 = 1.740 × 10¹⁰ cm.

N = n[e⁻] × L 
  = 10⁹ cm⁻³ × 1.740 × 10¹⁰ cm
  = 1.740 × 10¹⁹ cm⁻².

(c)
τ = ∫ κ dL = κ × L.

κ = n[e⁻] × σ[e⁻] 
  = 10⁹ cm⁻³ × 6 × 10⁻²⁵ cm²
  = 6 × 10⁻¹⁶ cm⁻¹.

τ = 6 × 10⁻¹⁶ cm⁻¹ × 1.740 × 10¹⁰ cm
  = 1.05 × 10⁻⁵.

(d)
L[sol] = 3.848 × 10³³ erg s⁻¹.

τ < 1, so this cloud is optically thin.

L[scattered] ≈ L[sol] × τ
       		 = 3.848 × 10³³ erg s⁻¹ × 10⁻⁵
       		 = 3.848 × 10²⁸ erg s⁻¹.



(3)

All good here. I can login, but I haven't compiled cloudy or anything.