Week 4: WS8 #1

1. The matter between stars is known as the interstellar medium and is composed of mostly dust and gas. This dust absorbs stellar light and therefore reduces the apparent brightness of stars as viewed from the Earth. This effect is known as extinction and worsens as we observe more distant stars. 
a) Using trigonometric parallax you measure that the distance to a particular star is 200pc. You know that the absolute magnitude of this star should be 2 and measure an apparent magnitude of 12. How much flux have you lost due to the intervening dust? 

We know the equation relating magnitudes and distances is \(M=m-5[log(d)-1]\), so we can just plug in what we know to calculate what the apparent magnitude should have been.
\(2=m-5[log(200)-1]\)
m = 8.5, so ∆m = 3.5.

b) Optical depth is a measure of the absorption of photons as they travel through a medium. The definition of optical depth is \[\tau=ln(I_{in}/I_{out})\]where I denotes the specific intensity of the source. In this problem, we are looking at a single source at a fixed distance, so we can also express \(\tau\) in terms of the flux: \[\tau=ln(F_{in}/F_{out})\]Use this latter definition to determine the relationship between \(\tau\) and the apparent magnitude. What is the optical depth along the line of sight to the star in the previous problem? 

We can plug the equation relating flux and magnitude (\(\frac{F_1}{F_2}=10^{0.4(m_2-m_1)}\), where \(m_1\) and \(m_2\) correspond to \(m_{in}\) and \(m_{out}\) respectively) into the equation for \(\tau\):
\(\tau=ln[10^{0.4(m_{out}-m_{in})}]\)
\(\tau=ln[10^{0.4(3.5)}]\approx3.2\)

c) We can now compute the amount of dust required along our line of sight to produce the observed extinction. The optical depth along a line of sight can be calculated from the absorption cross section of any intervening material and the number density of the particles within that cross section. This can be written \[\tau=N\times\sigma\]where N is the total number of particles per cm² in the line of sight and \(\sigma\) is the cross section of the individual particles. 
Assume that each dust grain has the typical size of \(r=0.1\mu m\) and is spherical. Calculate the geometric cross section of a dust grain in units of cm². Assume that this geometric cross section is the absorption cross section, which is true for visible light. Determine how many particles per cm² you would need to obtain the calculated optical depth. This value is referred to as a "column number density." 

\(\sigma=\pi r^2=\pi(0.1\times10^{-4})^2=3.14\times10^{-10}cm^2\)
\(N=\frac{\tau}{\sigma}=\frac{3.2}{3.14\times10^{-10}cm^2}\approx1\times10^{10}particles/cm^2\)

d) We know that the mass in gas is about 100 times more than in dust. What is the column number density of gas along the same line of sight?

Since we want to find the number density of gas, we can say \(m_{dust}N_{dust}(100)=m_{gas}N_{gas}\).
We know \(N_{dust}=10^{10}particles/cm^2\).
We can estimate the density \(\rho_{dust}=2.2g/cm^3\) and thus calculate \(m_{dust}=(2.2)(4/3)(\pi)(0.1\times10^{-4})^3=9.2\times10^{-15}g\).
Finally we can estimate the mass of gas by assuming that each gas molecule has the mass of hydrogen gas: \(m_{gas}=3.3\times10^{-24}g\).
\(N_{gas}=\frac{m_{dust}N_{dust}100}{m_{gas}}\approx2.8\times10^{21}/cm^2\)

e) What is the average gas density along the same line of sight? Express your answer as the number of particles per cm³.

We simply have to divide the column number density by the distance to the star (200pc, or \(6.2\times10^{20}cm\)), which gives us approximately 4.5 gas molecules per cm³.