Quantum mechanics and Black Hole Mass, all wrapped up in a nice problem

We are bounded in a nutshell of Infinite space: Blog Post #24, Worksheet # 7.2, Problem #3 & #4: Quantum mechanics and Black Hole Mass, all wrapped up in a nice problem.  

3 & 4. Such bright objects, known as quasars, can be easily observed at great distances, and astronomers started taking spectra of them back in the 1960’s. Here’s a spectrum of the first quasar ever discovered, called 3C 273:

What are the main features you see in this spectrum (ignoring the gap in the data at around 1625 \(\dot{A}\)?

One feature you surely noticed was the strong, broad emission lines. Here is a closer look at the strongest emission line in the spectrum:

This feature arises from hydrogen gas in the accretion disk. The photons radiated during the accretion process are constantly ionizing nearby hydrogen atoms. So there are many free protons and electrons in the disk. When one of these protons comes close enough to an electron, they recombine into a new hydrogen atom, and the electron will lose energy until it reaches the lowest allowed energy state, labeled n = 1 in the model of the hydrogen atom shown below (and called the ground state):

On its way to the ground state, the electron passes through other allowed energy states (called excited states). Technically speaking, atoms have an infinite number of allowed energy states, but electrons spend most of their time occupying those of lowest energies, and so only the n = 2 and n = 3 excited states are shown above for simplicity. Because the difference in energy between, e.g., the n = 2 and n = 1 states are always the same, the electron always loses the same amount of energy when it passes between them. Thus, the photon it emits during this process will always have the same wavelength. For the hydrogen atom, the energy difference between the n = 2 and n = 1 energy levels is 10.19 eV, corresponding to a photon wavelength of \(\lambda\) = 1215.67 Angstroms. This is the most commonly-observed atomic transition in all of astronomy, as hydrogen is by far the most abundant element in the Universe. It is referred to as the Lyman α transition (or Lyα for short). It turns out that that strongest emission feature you observed in the quasar spectrum above arises from \(l \gamma \alpha\)  emission from material orbiting around the central black hole.

(a) Recall the Doppler equation: \[\frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}}  = z ≈ \frac{v}{c}\] Using the data provided, calculate the redshift of this quasar.

(b) Again using the data provided, along with the Virial Theorem, estimate the mass of the black hole in this quasar. It will help to know that the typical accretion disk around a \(10^8 M_\odot\) black hole extends to a radius of \( r = 10^{15} m.\)

(a) As we can see from the graphs, a main characteristic of the emission spectra is the high peak, which we interpret as the main light output from Hydrogen atoms changing energy levels, one of the most common, if not the most common, occurrences in the universe. Knowing this is hydrogen, we can compare its emission spectra to that of what we know it should be from experiments on Earth, and some great conclusions can be drawn from this experiment.

Understanding the nature of how light propagates is essential to this problem. Here, an object, namely another galaxy, is moving away from the Milky Way, understood as evidence of the expanding universe, since all galaxies are continually moving outwards and expanding the space they occupy as a whole. Furthermore, this expansion is incredibly fast, going at speeds comparable to the speed of light, which offers astronomers the opportunity to better understand how galaxies move and how far away they are. This method is easy enough to understand, using the equation for a Doppler Shift, as you might have learned from a physics class (for sound waves and their wavelengths) we can use the same one to compare what we know the wavelengths of light should be and compare it with the observed wavelength. So we take:  \[\frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}}  = z ,\] where z is the Red Shift (the analog of Doppler Shift for light). Knowing Hydrogen’s emission peaks at 1215.67 Angstroms ( \(10^{-10} m \) ) and from the zoomed in graph we know the peak of this emission spectra is around 1406 Angstroms. These can be plugged into the equation and we get: \[\frac{1406 \dot{A} - 1215.67 \dot{A} }{1215.67\dot{A}}  = z ,\] and thus the red shift of this galaxy is: \[z = .1566 ,\] which can be used to approximate the speed of the galaxy as well and, consequently, understand the time its light has taken to reach us.

(b) From previous problems, we know the virial theorem is an equation which permits us to use a simple relationship between the kinetic and potential energies in a galaxy. (See full Virial Theorem explanation in http://ay17-rcordova.blogspot.com/2015/10/energy-in-space-is-anything-but-one.html ). Furthermore, by taking the Virial Theorem and expanding it with the full definition of Flux, Luminosity, and Potential Energy, we get an equation which describes a relationship between Luminosity and the mass of the black hole which creates the accretion disk. The full equation is as follows: \[L = \frac{G M_{SMBH} \cdot 4\pi\cdot c\cdot M_{Proton}}{\sigma_t},\] where G is the gravitational constant, c is the speed of light, and \(\sigma_t\) is the Thomson Cross Section, valued at \(6.6524 \times 10^{-25} cm^2 \), which is the effective area of an electron interacting with a photon. This equation can be rewritten to fit the data given in the problem, Flux. BY dividing both sides by \(4\pi r^2\) we have: \[\frac{L}{4\pi r^2}= \frac{G M_{SMBH}\cdot c\cdot M_{Proton}}{r^2 \sigma_t},\] \[F_{SMBH}= \frac{G M_{SMBH}\cdot c\cdot M_{Proton}}{r^2 \sigma_t}.\] Taking this into account, and knowing the value of the mass of a proton is \(1.6726219 \times 10^{-24} g , \) as well as the problem establishing that the radius of the black hole is  \( r = 10^{15} m\). Plugging in all the values we have accumulated, and solving for the mass of the Black Hole, we have: \[F_{SMBH}= \frac{G M_{SMBH}\cdot c\cdot M_{Proton}}{r^2 \sigma_t},\]  \[\frac{F_{SMBH} \cdot r^2 \cdot \sigma_t }{G \cdot c \cdot M_{Proton} }=  M_{SMBH}, \]  \[  M_{SMBH}= \frac{\left(0.93 \times 10^{12} \frac{erg}{cm^2 s}\right) \cdot (10^{17} cm)^2 \cdot (6.6524 \times 10^{-25} cm^2) }{(6.67 \times 10^{-8} \frac{cm^3}{g~s^2})\cdot (3 \times 10^{10}\frac{cm}{s} )\cdot (1.6726219 \times 10^{-24} g) } , \]   \[  M_{SMBH}= 1.85 \times 10^42 g , \] and this is the mass of the black hole we have been studying.