Thought the universe was empty? Guess again, it’s actually emptier.

We are bounded in a nutshell of Infinite Space: Worksheet # 8, Problem #1: Thought the universe was empty? Guess again, it’s actually emptier.

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? Compute your answer in magnitudes.

(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 apparent magnitude. What is the optical depth along the line of sight to the star in the previous problem?

(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 \[ \tau = N \times \sigma\] , where N is the total number of particles per \(cm^{-2}\) 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 \(cm^2\). Assume that this geometric cross section is the absorption cross section, which is true for visible light. Determine how many particles per \(cm^2 \) you would need to obtain the calculated optical depth. This value is referred to as a “column number density.”

(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?

(e) What is the average gas density along the same line of sight? Express your answer as the number of particles per \(cm^3\) .

a. Dust is part of the interstellar medium, present in (very) high orders of magnitude, dust affects most astronomical observations, and can lead to drastic changes of the ideal observations. This is the case for the object we are observing which has an apparent magnitude of 12, but according to its abosolute magnitude of 2, we can solve and find that: \[M = m - 5 (\log_{10} (d) - 1)\] \[m = M + 5 (\log_{10} (d) - 1)\]  \[m = 2 + 5 (\log_{10} (200) - 1)\] \[m = 2+ 5 (1.5),\] the apparent magnitude we should be receiving is: \[ m_{calculated} = 8.5,\] and we know that: \[ m_{measured} = 12 ,\] so the \[ m_{discrepancy} = ~Loss~ of ~Flux~ = 3.5 \]

b. Another important component is the relationship of intensity and flux describes as: \[ \tau = \ln(F_{in} / F_{out}) ,\] and we also know that  \[ \frac{F}{F_0} = 10^{-0.4(m - m_0)}, \] as per the basic equations relating magnitudes we have been using thus far. And this is analogous to \[ F_{in} / F_{out} = F~ /~ F_0,\] so \[ \tau = \ln \left(10^{-0.4(m - m_0)} \right) ,\] and in the case of the apparent magnitudes we had calculated and observed, we find that: \[ \tau = \ln \left(10^{-0.4(8.5 - 12)} \right) ,\] \[ \tau = 3.22 ,\] which is the so-called optical depth.

c. Furthermore, from the equation given in the problem, we can see how: \[ \tau = N \times \sigma, \] which describes the optical density as the number of particles in a cross section of the sky. Furthermore, the cross section is simply:  \[ \sigma = \pi r^2 ,\] where the radius is that of a dust particle, this being:  \[ \sigma = \pi (0.1 \mu m )^2 \] \[ \sigma = \pi (10^{-5} cm)^2\] \[ \sigma = \pi 10^{-10} cm^2.\] now we can plug these numbers to find the Column Number Density where the optical depth is what we calculated in the precious problem:  \[ N = \frac{\tau}{\sigma}\] \[ N = \frac{3.22}{\pi 10^{-10} cm^2}\] \[N = 1.02 \times 10^{10} cm^{-2},\] which is the amount of dust particles per optical column (200 pc long in this case) .  

d. Next, we can find the amount of atoms considering their collective mass is 100 times that of the collective of dust particles. To find the dust particles mass, we multiply their volume, density, and number to get the total dust in a column: \[ V_{dust} \cdot N_{dust} \cdot \rho_{dust} = M_{dust/cm^2},\] where the density is the density of most silicates:  \[\rho_{dust} = 3 \frac{g}{cm^3},\]  and the volume is simply: \[ V = \frac{4}{3} \pi r^3 \] \[ V_ dust = \frac{4}{3} \pi (10^{-5} cm)^2 = 4 \times 10^{-15} cm^3,\] and we know the column number density from the last problem: \[N = 1.02 \times 10^{10} cm^{-2}.\] Plugging in, we get: \[M_{dust/cm^2} = V_{dust} \cdot N_{dust} \cdot \rho_{dust} \] \[ M_{dust/cm^2} =  4 \times 10^{-15} cm^3 \cdot 1.02 \times 10^{10} cm^{-2} \cdot 3 \frac{g}{cm^3} \] which tells us the mass of all the dust in the column is: \[ M_{dust/cm^2} =  1.2 \times 10^{-4} \frac{g}{cm^{2}}\]

Now we just multiply this mass by 100 and find the mass of all the gas ( which we can assume is predominantly Hydrogen):  \[ M_{gas/cm^2} = 100 M_{dust/cm^2} \] \[ M_{gas/cm^2}= 100 \cdot 1.2 \times 10^{-4} \frac{g}{cm^{2}}\] \[ M_{gas/cm^2} =  1.2 \times 10^{-2} \frac{g}{cm^{2}}\] and now we can just plug in the mass of a hydrogen atom to find the amount of atoms: \[M_{H ~atoms} = 1.67 \times 10^{-24} \frac{g}{atom},\] so we have: \[ \frac{ M_{gas/cm^2} }{ M_{H ~atom} } = Particles / cm^2\] \[ \frac{ M_{gas/cm^2} }{ M_{H ~atom} } = \frac{ 1.2 \times 10^{-2} \frac{g}{cm^{2}}}{1.67 \times 10^{-24} \frac{g}{atom}},\] which is: \[ Particles / cm^2 = 7.2 \times 10^{21} \frac{atoms}{cm^2},\] in the entire column we are observing.  

e. But, all these calculations have been on the basis that we are talking about the entire column, what about a simple \(cm^3\) ? We know the column is 200 pc long, and this equates to:  \[ 1 pc = 3.1 \times 10^{18} cm\] and there are\[ d = 200 pc,\] therefore: \[ Particles/ cm^3 = \frac{ H ~atoms / cm^2 }{d} \] so it is just a matter of plugging in and we find: \[ Particles / cm^3 = \frac{7.2 \times 10^{21} \frac{atoms}{cm^2}}{200 \times 3.1 \times 10^{18} cm }\]  \[ Particles / cm^3 = 11.6 \frac{atoms}{cm^3},\] which serves to show us how little there actually is in the vastness of space, rather we only see it once we start looking at the objects in incredibly large perspectives.