Blog #11 - Dust and Spectral Lines

Dust and Spectral Lines 

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? Compare your answer in magnitudes.

We follow the the relationships that we previously established between absolute and relative magnitudes and the distance to an observed star. \[M-m = -5*(log(d) - 1)\]With this equation we have every constant given to us besides the observed apparent magnitude. Thus we solve for "m" and get \[2-m = -5*(log(200pc) -1) \rightarrow m = 8.5\] With the expected apparent magnitude as 12, this gives us a magnitude loss of 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:

τ = ln(Iin/Iout)

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 τ in terms of the flux:

τ = ln(Fin/Fout)

Use this latter definition to determine the relationship between τ and apparent magnitude. What is the optical depth along the line of sight to the star in the previous problem?

We are told that \[\tau = ln(\frac{F_{in}}{F_{out}})\]and know that know that \[\frac{F_{in}}{F_{out}} = 2.5^{m-M} \] Thus we can change the equation to \[\tau = ln(2.5^{m_{expected}-m_{observed}}) = (m_{expected}-m_{observed})*ln(2.5)\]With the equation updated, we can plug in with the values found in part A to solve for \(\tau\). \[\tau = (12-8.5)*ln(2.5) \rightarrow \tau \approx 3.5\]

(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 with that cross section. This can be written as:

τ = Nσ

where N is the total number of particles per \(cm^2\) in the line of sight and σ is the cross section of the individual particles.

Assume that each dust grain has the typical size of r = 0.1µm and is spherical. Calculate the geometric cross section of a dust grain in units of \(cm^2\). 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.

First we compute the circular cross section of the spherical particle as \[\sigma = \pi * r^2 = \pi * (1*10^{-5} cm)^2 \approx 3*10^{-10} cm^2\]With this cross sectional area we can find the particle density with the \(\tau\) we solved for earlier \[\tau = N* \sigma \rightarrow 3.5 = N*(3*10^{-10}) \rightarrow N \approx 10^{10} \frac{particles}{cm^2} \]

(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?

To solve for the column number density of the more massive gas, we have to start by solving for the mass column of the dust. To do this we first assume the density of the dust to be somewhat similar to rock on Earth \[\rho_{dust} \approx 1 \frac{g}{cm^3}\] We can then multiply by the volume of a dust particle to get the \(\frac{g}{particle}\) fraction. \[1 \frac{g}{cm^3} * 4*10^{-15} \frac {cm^3}{particle} \approx 4*10^{-15} \frac {g}{particle} \] Now we can multiply by the number density from part C to get the number of grams per cm^2 of the dust. \[4*10^{-15} \frac {g}{particle} * 10^{10} \frac{particles}{cm^2} = 4*10^{-5} \frac{g}{cm^2} \]Which is the mass column of our dust. Now working backwards to get back to to number density of the gas. We first multiply by the mass ratio for the gas to dust. \[4*10^{-5} \frac{g_{dust}}{cm^2} * 100 \frac{g_{gas}}{g_{dust}} = 4*10^{-3} \frac{g_{gas}}{cm^2} \]Finally we assume that the majority of the gas in space is made of hydrogen and thus can approximate the weight of one particle of gas to be the weight of one particle of hydrogen giving us the final number density. \[4*10^{-3} \frac{g_{hydrogen}}{cm^2} * \frac{1 \space particle}{1.7*10^{-24} g_{hydrogen}} \approx 2*10^{21} \frac{N_{gas}}{cm^2}\]

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

To find the average gas density along the line of sight we simply must take our column number density and extend it the length of the column, or the viewing distance from the star, 200pc. \[200pc = 6*10^{20} cm\]Giving us an average density of \[2*10^{21} \frac{N_{gas}}{cm^2} * \frac{1}{6*10^{20} cm} \approx 3 \frac{gas \space particles}{cm^3} \]