Blog #27, Worksheet #8.1, Problem #3, Hubble Time

3. It is not strictly correct to associate this ubiquitous distance-dependent redshift we observe with the velocity of the galaxies (at very large separations, Hubble’s Law gives ‘velocities’ that exceeds the speed of light and becomes poorly defined). What we have measured is the cosmological redshift, which is actually due to the overall expansion of the universe itself. This phenomenon is dubbed the Hubble Flow, and it is due to space itself being stretched in an expanding universe. 
Since everything seems to be getting away from us, you might be tempted to imagine we are located at the centre of this expansion. But, as you explored in the opening thought experiment, in actuality, everything is rushing away from everything else, everywhere in the universe, in the same way. So, an alien astronomer observing the motion of galaxies in its locality would arrive at the same conclusions we do. 
In cosmology, the scale factor, \(a(t)\), is a dimensionless parameter that characterizes the size of the universe and the amount of space in between grid points in the universe at time t. In the current epoch, t = t0 and \(a(t_0) = 1\). \(a(t)\) is a function of time. It changes over time, and it was smaller in the past (since the universe is expanding). This means that two galaxies in the Hubble Flow separated by distance \(d_0 = d(t_0)\) in the present were \(d(t) = a(t)d_0\) apart at time t. The Hubble Constant is also a function of time, and is defined so as to characterize the fractional rate of change of the scale factor: \[H(t) = \frac{1}{a(t)} \frac{da}{dt}\]and the Hubble Law is locally valid for any t: \[v = H(t)d\]where v is the relative recessional velocity between two points and d the distance that separates them.
a) Assume the rate of expansion, \(\dot{a} = \frac{da}{dt}\), has been constant for all time. How long ago was the Big Bang (i.e. when a(t = 0)  0)? How does this compare with the age of the oldest globular clusters (~12 Gyr)? What you will calculate is known as the Hubble Time.
We want to solve for the time  We are given that \(a_0\) is the base at which we characterize the size of the universe from the current epoch and assumes the number 1. We also know that \(H_0\) is a constant for the the current epoch found in question #2 of the worksheet. Using this information we can solve show that \[H_0 = \frac{\dot{a}}{a_0}\]Because we said that a = 1 at the current epoch we can say that \[H_0 = \dot{a} = \frac{da}{dt}\]We then get \[\int_{0}^{t_0} da = \int_{0}^{t_0} H_0 dt\]Which gives us \[a(t_0) = H_0 t_0\]We are looking to solve for \(t_0\) and once we rearrange we get \[t_0 = \frac{1}{H_0}\]From question 2 we know that \[H_0 = 67 \frac{km}{MPc \cdot s}\]We need to convert his by getting rid of the distance component and turning seconds into years. \[67  \frac{km}{MPc \cdot s} \thickapprox 7 x 10^{-11} \frac{1}{yr}\]And we know that \[t_0 = \frac{1}{H_0} = \frac{1}{7x10^{-11}} = 1.4x10^{12} \, years = 14 \, billions \, years\]
b) What is the size of the observable universe? What you will calculate is known as the Hubble Length.
The Hubble Length is very easy to find once we have the relative time since the big bang. We know that we may only observe as far as light has traveled. Thus the Hubble Length is simply \[d_{hubble} = 14x10^{11} \cdot c = 14 \, billion \, lightyears = 1.3x10^26 \, meters = 4.3x10^9 \, parsecs\]

Blog #26, Worksheet #8.1, Problem #1, Hubble Flow Thought Experiment

1. Before we dive into the Hubble Flow, let’s do a thought experiment. Pretend that there is an infinitely long series of balls sitting in a row. Imagine that during a time interval \(\Delta t\) the space between each ball increases by \(\Delta x\) 

(a) Look at the shaded ball, Ball C, in the figure above. Imagine that Ball C is sitting still (so we are in the reference frame of Ball C). What is the distance to Ball D after time ∆t? What about Ball B?
Let's call the original distance between Ball C and Ball D, x. Then we know that after a time \(\Delta t\) the ball travels a distance \(\Delta X\). Thus our total distance after a time \(\Delta t\) is \[d = x + \Delta x\]Because every ball is evenly spaced, we can assume that Ball B travels the same distance as Ball D.
(b) What are the distances from Ball C to Ball A and Ball E? 
Both Ball A and Ball E begin at a distance \(2x\) away from Ball C and over the course of time \(\Delta t\), the each ball has traveled a distance \(\Delta x\) away from the ball next to them which has also traveled a distance \(\Delta x\) from Ball C. Giving the total distance of Ball A/E away from Ball C as \[d = 2x + 2 \Delta x\]
(c) Write a general expression for the distance to a ball N balls away from Ball C after time ∆t. Interpret your finding. 
Seeing the pattern from part a) and b) we can easily generalize our equation for a ball N balls away as \[d = Nx + N \delta x\]
(d) Write the velocity of a ball N balls away from Ball C during ∆t. Interpret your finding.
Velocity is simply distance over time and we know that the distance a Ball N balls away from travels is \[d = N \Delta x \]Thus our velocity comes out to be \[v = \frac{N \Delta x}{\Delta t} = Nv\]

Blog #25, Worksheet #7.2, Problem #5, Red Shifts of Different Spectrums

5. You may also have noticed some weak “dips” (or absorption features) in the spectrum:

(a) Suggest some plausible origins for these features. By way of inspiration, you may want to consider what might occur if the bright light from this quasar’s accretion disk encounters some gaseous material on its way to Earth. That gaseous material will definitely contain hydrogen, and those hydrogen atoms will probably have electrons occupying the lowest allowed energy state. 

As somewhat explained in the question, on it's way to Earth, the emitted light from the quasar may hit very thick patches of gasses, notably hydrogen. As we explored earlier in the worksheet, these hydrogen molecules absorb some of the emitted energy and use it to move through energy levels. They then output the LyA wavelength as output and we use this wavelength change to measure the speed at which the the quasar, or other object, is moving away from us at. The dips in the light spectrum curve symbolize the patches of hydrogen gas that the light encounters at certain wavelengths. Instead of the steady dispersion of different wavelengths on the spectrum, certain ones are absorbed by the clouds of hydrogen and then account for a much smaller flux being received on Earth.

(b) A spectrum of a different quasar is shown below. Assuming the strongest emission line you see here is due to Lyα, what is the approximate redshift of this object? 

We use the equation for red shift that we saw earlier \[z = \frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}}\]We assume that the peak in the light spectrum offers a good estimate for the observed wavelength. We also know that the emitted wavelength in a redshift is the LyA shift where \(\lambda = 1215.67 \, angstroms\). We find that the peak is around \(\lambda = 5650 \, angstroms\). Plugging into our equation and solving for z, we get a redshift of \[z = 3.65\]This shift is much more significant than the first spectrum we looked at. This larger redshift means that the quasar is moving much faster and must be significantly further away from Earth.

(c) What is the most noticeable difference between this spectrum and the spectrum of 3C 273? What conclusion might we draw regarding the incidence of gas in the early Universe as compared to the nearby Universe?

It is clear that on this spectrum, compared to the first, that there are far more peaks and valleys. Far more emission jumps and absorption lines. As we discovered in part a) these are created by light hitting patches of hydrogen along the journey from the quasar to Earth. Given that we found the second quasar to be much further than the original quasar, and the fact that there are far more absorption lines in the second spectrum, we can assume there is much more gaseous material in the earlier universe than there is in the nearby, developed universe. The larger the redshift and older the quasar is, the more absorption lines it should have.

Blog #24, Worksheet #7.2, Question #2, Eddington Luminosity

2. There is actually a hard upper limit to the luminosity of this system – and to the luminosity of any accreting compact object. Consider that the photons being emitted in this scenario will interact with the surrounding material (which has yet to accrete onto the black hole). These photons will undergo Thomson scattering off of electrons in this material. In detail, the electric field of the incident lightwave (i.e., the photon) will accelerate an electron, causing it to then re-emit radiation. The photons are therefore able to transfer some of their momentum to the infalling gas. The energy flux of these photons at a distance r from the black hole is \[F = \frac{L}{4 \pi r^2}\]

Then, recall that the momentum of a photon of energy E is simply p = E/c. Therefore, the momentum flux at r from the black hole is \(\frac{L}{4πcr^2}\).

Finally, the rate of momentum transfer to the surrounding electrons (or the force due to photons \((f_{rad})\) is modulated by the Thomson cross section, \(\sigma_t = 6.6524 \, x \, 10^{-25} \, cm^2\) (i.e., the effective area of an electron interacting with a photon): \[f_{rad} = \sigma_t \frac{L}{4 \pi c r^2}\]

(a) When this force from radiation pressure exceeds the force of gravity, accretion is halted and all the gas is blown away. For a black hole mass \(M_{BH}\), derive the maximum possible luminosity due to accretion. This is called the Eddington Luminosity.

We begin with the fact that we know the momentum flux and the force of the radiation \[Flux_{\rho} = \frac{L}{4 \pi cr^2} \qquad and \qquad f_{rad} = \sigma_t \frac{L}{4 \pi c r^2}\]Then to solve for the maximum luminosity, we set the force of the radiation equal to the force of gravity from the black hole on the protons. When the force of radiation is larger, all accretion is halted and the gas leaves the disk. \[f_{rad} = \sigma_t \frac{L}{4 \pi c r^2} = \frac{G M_{BH} M_{proton}}{r^2}\]And finally we solve for our Eddington Luminosity \[L_{max} = \frac{4 \pi c G M_{BH} M_{proton}}{\sigma_t}\]

Bonus: Express the Eddington Luminosity as a number of Solar luminosities, L , and the black hole mass in solar masses \((\frac{M_{BH}}{M_{\odot}})\), such that \[L_{Edd} = X(\frac{M_{BH}}{M_{\odot}}) \cdot L_{\odot} \]

We need our Eddison Luminosity from part a) to equal the equation on the right side. Thus, we have \[\frac{4 \pi c G M_{BH} M_{proton}}{\sigma_t} = X(\frac{M_{BH}}{M_{\odot}}) \cdot L_{\odot} \]Then we simply simplify and solve for X to give us \[X = \frac{4 \pi c G M_{proton}}{\sigma_t} \cdot \frac {M_{\odot}}{L_{\odot}} \]

(b) If the SMBH in Andromeda were accreting at 20% of its Eddington luminosity, how bright would it be? 

We know the Eddington Luminosity can be found with the equation \[L_{Edd} = \frac {4 \pi c G M_{BH} M{proton}}{\sigma_t}\]From Wikipedia we estimate that the mass of the SMBH at the center of Andromeda is about \(10^8 \, M_{\odot} \). We also know that the mass of a proton is \(1.67 x 10^{-27} \, kg\). 

Plugging in and solving for the Eddison Luminosity, we get: \[L_{Edd} = 1.257 \, x \, 10^{39} \frac{J}{s}\]However, we are given that the SMBH is operating at 20% of its Eddison Luminosity, giving us \[L = 2.515 \, x \, 10^{38} \frac{J}{s}\]

Sources Used

https://en.wikipedia.org/wiki/Supermassive_black_hole

Blog #23, Free Form Post, Asteroid Mining

There is quite a bit of intersection when it comes to astronomy and economics. Unfortunately, usually that pertains to how much money should be cut from national budgets funding space programs. Luckily all of the research and discovery over the past few decades has led to a new a branch of space exploration that may actually bring in huge profits to space companies. Asteroid mining is a hot new branch of space exploration focused on extracting rare metals and gasses from asteroids or other near-earth space objects and bringing using the goods to turn a substantial profit.

The two main companies focused on asteroid mining right now are Planetary Resources and Deep Space Industries. Each is a venture backed startup that looks to put its own mining devices onto asteroids and mine for precious materials. The first, and most immediately exciting, prospect of asteroid mining is the money. Researchers believe that in a single asteroid, 1km in diameter, could contain enough nickel ore to supply the Earth for the next 70 million years.



Beyond nickel, iron, and gold, asteroid miners are most excited about the vast abundance of platinum in space. Platinum is relatively rare on the Earth but makes up a significant portion of many asteroids. In a recent article from Space.com, it was predicted that asteroid 2011 UW158 missed the Earth by about 1.5 million miles, roughly the distance from the Earth to the Moon. Although not a very close miss, the asteroid got potential miners very excited as it was predicted to hold up to $20 trillion dollars worth of platinum at the current market price. Trillion. With a t. These kinds of prices are what gets Planetary Resources' engineers to work every morning and what keeps thousands of investors interested in the somewhat crazy sounding business plan.

Still there are skeptics that criticize the idea of mining as a significant waste of money. They argue that the cost of a trip in addition the experimentation and all of the materials required to actually land on an asteroid and extract anything useful far outweigh the current market for the materials they look to bring back to Earth. Economists also argue that such an influx in precious metals may saturate the market, deeming the metals brought back essentially useless, and certainly not worth $20 trillion.

Regardless of who is correct in the economic debate, the prospect of consistently bringing material back to Earth for commercial use is certainly an entertaining prospect. It is exciting to live in a time where the technology is available and ever progressing, allowing humans to do things that were previously considered a dream or a goal far out of reach.

Sources:
http://www.planetaryresources.com/company/overview/
http://www.space.com/30074-trillion-dollar-asteroid-2011-uw158-earth-flyby.html
https://www.youtube.com/watch?v=VLouRKHknOU
http://www.wired.com/images_blogs/wiredscience/2012/04/SEP_Airlock_Concept.jpg
https://62e528761d0685343e1c-f3d1b99a743ffa4142d9d7f1978d9686.ssl.cf2.rackcdn.com/files/10153/area14mp/35vqmr6c-1335830904.jpg

Blog #22, Worksheet #7.1, Problem #2, Sound Speed of White Dwarf

2. Use the previous result to derive the sound speed, \(c_s\), within the white dwarf, where \(c_s\) can be related to pressure, \(P\), and mass density, \(\rho\), using dimensional analysis (make the units work).

We begin with our expression from question 1 for Pressure \[P = \frac{GM^2}{4 \pi R^4}\]We then look at the units for this expression \[P = \frac{[F]}{m^2} = \frac{kg \cdot m}{m^2 \cdot s^2} = \frac{kg}{m \cdot s^2}\]We want to solve for \(c_s\) which has units \(\frac{m}{s}\), so we must find some exponent for both density and pressure that gives us the \(\frac{m}{s}\) relationship we need. Density, \(\rho\) has the units \[\rho = \frac{kg}{m^3}\]So we end up with the relationship \[(\frac{kg}{m \cdot s^2})^{\alpha} \cdot (\frac{kg}{m^3})^{\beta} = \frac{m}{s}\]We then need to figure out what exponents for \(\alpha\) and \(\beta\) lead us the the correct units for velocity. After a quick inspection we can clearly see the the density should be in the denominator of our equation and after canceling, we need to take the square root of both equations to get the proper units for velocity. Thus \[\alpha = \frac{1}{2} \, and \, \beta = -\frac{1}{2}\]Thus our final answer gives us \[c_s \propto (\frac{P}{\rho})^{\frac{1}{2}}\]

Blog #21, Worksheet #7.1, #1, Electron Degeneracy Pressure

1. Given this mass, M and radius, R, derive an algebraic expression for for the internal pressure of a white dwarf with these properties. Start with the Virial theorem, recall that the internal kinetic energy per particle is \(\frac{3}{2}kT\), where \(k = 1.4 x 10^{-16} \, erg \, K^{-1} \) is the Boltzmann constant. You can also assume the interior of the white dwarf is an ideal gas, and its mass is uniformly distributed. 

We begin by using the fact that the internal structure of the white dwarf acts as an ideal gas. We can use the standard formula for pressure \[P = \frac{NkT}{V}\] Where N is the number of molecules, k is the Boltzmann constant \((k = 1.4 x 10^{-16} \, erg \, K^{-1})\), T is temperature, and V is the volume. In our case, \[V = \frac{4}{3} \pi R^3\]Combining these equations we get an equation for P. \[P = \frac{3NkT}{4 \pi R^3} \]Next we use the Virial theorem to try and get rid of the N in our Pressure equation. We know that \[K = - \frac{1}{2} U = \frac {3}{2} NkT\]Where \[U = \frac{GM^2}{R}\]Setting the equations equal to one another and simplifying to solve for N, we get \[N = \frac{GM^2}{3RkT}\]We then substitute this variable into our equation for P and get a final answer of \[P = \frac{3kT}{4 \pi R^3} \cdot \frac{GM^2}{3RkT} = \frac{GM^2}{4 \pi R^4}\]

Blog #18, Worksheet #6.1, Problem #4, Tully-fisher

4) Over time, from measurements of the photometric and kinematic properties of normal galaxies, it became apparent that there exist correlations between the amount of motion of objects in the galaxy and the galaxy’s luminosity. In this problem, we’ll explore one of these relationships. Spiral galaxies obey the Tully-Fisher Relation: \[L \propto v^4\]where L is total luminosity, and vmax is the maximum observed rotational velocity. This relation was initially discovered observationally, but it is not hard to derive in a crude way: 

(a) Assume that \(v_{max} \thickapprox σ\) (is this a good assumption?). Given what you know about the Virial Theorem, how should \(v_{max} \)relate to the mass and radius of the Galaxy?

We can assume that \(v_{max} \thickapprox σ\) because the mean velocity scatter will of course be lower than the maximum observed velocity. However, because the grouping of stars are roughly similar, the maximum observed velocity will not be much higher than the mean unless there is an anomaly. Thus it will not be too far off to approximate the scatter velocity as the vmax of the system.

We can use knowledge of the Virial Theorem to then relate mass and radius to the vmax using what we found from problem #3. \[M = \frac{\sigma ^2 R}{G}\] Thus \[\sigma = (\frac{GM}{R})^{\frac{1}{2}} \thickapprox v_{max}\]

(b) To proceed from here, you need some handy observational facts. First, all spiral galaxies have a similar disk surface brightnesses \(I = \frac{L}{R^2}\) Freeman’s Law. Second, they also have similar total mass-to-light ratios \(\frac{M}{L}\).

(c) Use some squiggle math (drop the constants and use \(\thickapprox \) instead of =) to find the Tully-Fisher relationship.

We will use general proportionality to find the Tully-Fisher relation. We know from part one that \[v_{max}^2 \propto \frac{M}{R} \] We know from Freeman's law that \[R \propto L^{\frac{1}{2}}\]We also know that \(M \propto L\) from the mass-to-light ratio. Plugging in this info, we get \[v_{max}^2 \propto \frac{L}{L^{\frac{1}{2}}}\]If we square this, we get to the Tully-Fisher relation \[L \propto v^4\]

(d) It turns out the Tully-Fisher Relation is so well-obeyed that it can be used as a standard candle, just like the Cepheids and Supernova Ia you saw in the last worksheet. In the B-band \(λ_{cen} \thickapprox 445 nm \) blue light, this relation is approximately: \[M_B = -10 \, log(\frac{v_{max}}{km/s}) + 3 \]

Suppose you observe a spiral galaxy with apparent, extinction-corrected magnitude B = 13 mag. You perform longslit optical spectroscopy (ask a TF what that is), obtaining a maximum rotational velocity of 400 km/s for this galaxy. How distant do you infer this spiral galaxy to be?

We first solve for \(M_B \) and obtain \[M_B = -10 \, log(\frac{400}{km/s}) + 3 = -23.02\]We then use the distance modulus equation we obtained ob a previous worksheet which we gives us the relationship of distance to apparent and absolute magnitude. \[d = 10^{\frac{m - M_B + 5}{5}} \]giving us \[d = 10^{\frac{13 + 23 + 5}{5}} \]and a final answer of \[d = 10^{8.2} \, parsecs = 1.58 x 10^8 \, parsecs \]

Blog #19, Astrobite sumamry, Earth 2.0

In July, Nasa's Kepler Mission discovered the first exoplanet to exhibit a similar orbit to Earth, rotating its star in a roughly annual pattern. The planet, known as Kepler 452b is considered a small planet, meaning it is less than twice the size of the Earth. The year long orbit pattern led researchers to explore more about Kepler 452b, one of 5000 exoplanets discovered over the past 4 years by the Kepler program. The Nasa crew described Kepler 452b as an "older, bigger cousin to Earth."

The Kepler program images a particular area of the night sky and tracks the flickering of the light from the stars to track the planets orbiting the respective stars. Kepler 452b's star is roughly 1400 lightyears away from Earth making tracking the exact orbit a very difficult challenge. Although the light curves are very accurate, there are many other objects that can cause the same .02% drop in light from the star that the exoplanet would. To try and confirm the existence of a planet, researchers compared typical light curves of other potential objects such as larger planets or background stars, to the expected light curve of the exoplanet's passing. They placed 424:1 odds, pretty good for astronomical events.

After further experimentation and data analysis, researchers tried to narrow down similarities between Kepler 452b and Earth. They found the Keplerian planet to have a radius 1.6 the size of Earth's. More excitingly, they claimed that there was a 96% chance that the planet was in the habitable zone of its solar system, meaning it sits at a distance from its star similar to the distance that Earth sits at in our solar system. Researchers also believe that the surface of the planet could be rocky, similar to Earth's.

While we should by no means pack our suitcases and get ready to travel the 1400 lightyears to Kepler 452b, the article does provide a very optimistic and interesting opportunity to look at habitable exoplanets.

Work used:
http://astrobites.org/2015/08/03/have-we-really-found-earth-2-0/
http://i.space.com/images/i/000/049/219/i02/452b_ArtistConcept.jpg?1438295759

Blog #20, Hubble Tuning Fork

The Hubble tuning fork is a very generic tool used by astronomers to try and identify a category for galaxies. Although often considered too simple to properly categorize most galaxies, the tuning fork does allow astronomers to get off to a start when discovering new galaxies.

Ellipticals

Ellipticals are some of the oldest and brightest galaxies. They are categorized by how much shape they have. On a scale from 0 to 7, the higher the number, the more elliptical the galaxy is. An E0 galaxy is essentially circular.

Lenticulars 
Lenticulars are a less specific than spirals and ellipticals and act as a transition phase. Most lenticulars look like a hazy central bulge surrounded by an elliptical cloud.

Spirals 
Spiral galaxies are the most common galaxies in the universe. There are two types of Spiral galaxies, normal spirals and barred spirals. Each category is defined by level a, b, and c where the later the letter, the less tightly wrapped the spiral is. A level spirals are also brighter than b and c spirals. Barred spirals, as expected, have a bar structure that goes through the center and the galaxies are usually a bit more elongated than standard spirals.


Work used:

http://www.astro.cornell.edu/academics/courses/astro201/galaxies/images/m87_ell.gif
http://freestarcharts.com/images/Articles/Messier/M86_NASA_AURA_STScI.jpg
http://cas.sdss.org/dr6/en/proj/basic/galaxies/images/ellipt3.jpg
http://bama.ua.edu/~rbuta/nearirs0/plate014c.jpg
http://www.noao.edu/image_gallery/images/d4/m81y.jpg
http://www2.lowell.edu/rsch/LMI/gallery/n6946_2.jpg
https://upload.wikimedia.org/wikipedia/commons/thumb/5/52/Hubble2005-01-barred-spiral-galaxy-NGC1300.jpg/350px-Hubble2005-01-barred-spiral-galaxy-NGC1300.jpg
http://wind.caspercollege.edu/~marquard/astronomy/images/ngc1365_vlt.jpg
https://www.noao.edu/image_gallery/images/d4/m109a.jpg
http://scienceblogs.com/startswithabang/files/2010/09/File-Ngc253-2mass-barred-spiral.jpeg
http://jtgnew.sjrdesign.net/images/galaxies_elliptical_ngc1132_hst.jpg

Blog #17, Worksheet #6.1, Question 3, Virial Theorem

3. One of the most useful equations in astronomy is an extremely simple relationship known as the Virial Theorem. It can be used to derive Kepler’s Third Law, measure the mass of a cluster of stars, or the temperature and brightness of a newly-formed planet. The Virial Theorem applies to a system of particles in equilibrium that are bound by a force that is defined by an inverse central-force law. It relates the kinetic (or thermal) energy of a system, K, to the potential energy, U, giving \[K = - \frac{1}{2} U\] (a) Consider a spherical distribution of N particles, each with a mass m. The distribution has total mass M and total radius R. Convince yourself that the total potential energy, U, is approximately \[U \thickapprox - \frac{GM^2}{R}\] You can derive or look up the actual numerical constant out front. But in general in astronomy, you don’t need this prefactor, which is of order unity.

We can start by relating the potential energy to the work of the system through \[U = - \int_{0}^{R} F dr\] where \[F = \frac{GMm}{r^2}\] and \[M = 4 \pi r^2 \rho dr\ \qquad m = \frac{4}{3} \pi r^3 \rho\]and once we integrate we get \[U = - \frac{16}{15} G \pi^2 \rho^2 R^5\]Finally we replace rho with a general expression for density with \[\rho = \frac{M}{\frac{4}{3} \pi R^3}\]which gives us the desired expression \[U = - \frac{3}{5} \frac{GM^2}{R} \thickapprox - \frac{GM^2}{R}\] 

(b) Now let’s figure out what K is equal to. Consider a bound spherical distribution of N particles (perhaps stars in a globular cluster), each of mass m, and each moving with a velocity of vi with respect to the center of mass. If these stars are far away in space, their individual velocity vectors are very difficult to measure directly. Generally, it is much easier to measure the scatter around the mean velocity if the system along our line of sight, the velocity scatter. Show that the kinetic energy of the system is: \[K = \frac{3}{2} N m \sigma^2\]The proof is quite straightforward for this part of the theorem. We begin with the general form for kinetic energy where \[K = \frac{1}{2} m v^2 \]However, this only accounts for line of sight, one direction of the 3 possible directions that the system can move so we get \[K = \frac{3}{2} m v^2 \] but we have N stars in our system and our velocity profile, \(v = \sigma \). Substituting into our equation we get our desired final answer \[K = \frac{3}{2} N m \sigma^2\](c) Use the Virial Theorem to show that the total mass of, say, a globular cluster of radius R and stellar velocity dispersion σ is (to some prefactor of order unity): 
\[M \thickapprox \frac{\sigma^2 R}{G} \]

We begin with the general Virial Theorem \[K = - \frac{1}{2} U\]We plug in our answers from part a) and b) where \[U = - \frac {GM^2}{R} \qquad K = \frac{3}{2} N m \sigma^2 \]so \[\frac{3}{2} N m \sigma^2 = \frac{1}{2} \frac{GM^2}{R} \]Solving for M we get our final answer of \[M = \frac{3 \sigma^2 R}{G}\] which simplifies to our desired final solution of \[M \thickapprox \frac{\sigma^2 R}{G} \]

Blog #16, The Great Debate, Shapley vs. Curtis

The Great Debate in astronomy was an argument between astronomers Harlow Shapley and Heber Curtis in 1920. The debate encompassed two major issues of the time including the size of our own milky way galaxy and whether there was more than 1 galaxy in the universe. Shapley argued that there was only one massive galaxy in the solar system with millions of stars. He argued that our sun was only part of this huge galaxy and by no means in the center of the galaxy. Curtis on the other hand believed that our Milky Way was one of just many galaxies, or Spiral Nebulae, in the universe but that our sun was at the center of the Milky Way. Shapley believed these other "galaxies" that Curtis observed were simply large clouds of gas within our own very large (Shapley believed the Milky Way to be 300,000 light years in diameter) galaxy. 

Shapley based his claims on the Magellanic Clouds and globular clusters. He believed that by finding the magnitude of a cepheid star, he could approximate the distance of all observable globular clusters and thus prove that they all fell within the supposed boundaries of the Milky Way. With the help of astronomer Adriaan van Maanen, Shapley was able to show that the galaxy M101 had an angular speed and orbital speed that would make it impossible to lie beyond the known size of the Milky Way, thusly "proving" his theories. 

Curtis agreed with Shapley on the idea that the globular clusters surrounded the core of the galaxy, however his estimate showed that the milky way was only 30,000 light years in diameter, only one tenth of Shapley's estimate. Using this claim, Curtis said the luminosity argument that Shapely brought up proved that the observed Galaxies must be "island" galaxies, completely separate and free from our Milky Way. 

The debate was settled in 1923 when Edwin Hubble was able to use a newly constructed 100 in.  telescope on Mt. Wilson to prove that the cluster of stars M31 was over 1.2 million light years away, much further than even Shapley's size of the Milky Way. This proved that Curtis was right about the existence of multiple (one hundred billion) galaxies. However, Shapley was correct about the Sun being on the outskirts of the Milky Way, and his luminosity-period relationship in Cepheid stars was actually correct. It turns out the van Maanen's calculations were simply wrong, leading to the faulty arguments on Shapley's part. 

One interesting take on The Great Debate was how the situation went on to push astronomers to be much more precise and careful with their claims moving forward. Both Shapley and Curtis made incorrect scientific statements with very little hard evidence to backup their claims. After the great debate, most astronomers tried to be much more thorough in their discoveries. 

Sources: 

http://apod.nasa.gov/htmltest/gifcity/cs_why.html

http://astronomy.nmsu.edu/geas/lectures/lecture27/slide01.html

http://img.ezinemark.com/imagemanager2/files/30004252/2011/01/2011-01-20-22-58-37-1-whether-the-debates-result-went-both-scientist.jpeg

https://upload.wikimedia.org/wikipedia/commons/f/ff/Andromeda_galaxy.jpg

Blog #15, Worksheet #5.1, Distance Ladder, Problem #2

2) Okay, let’s use your new-found knowledge of magnitudes. 

(a) Suppose you are observing two stars, Star A and Star B. Star A is 3 magnitudes fainter than Star B. How much longer do you need to observe Star A to collect the same amount of energy in your detector as you do for Star B?

Using our chart from question #1, we can see that for a star with a difference in magnitude \(\delta m\) of 3, we have a difference of flux of about 16. That means we must observe star A 16 times longer to obtain the same amount of energy as star B.

(b) Stars have both an apparent magnitude, m, which is how bright they appear from the Earth. They also have an absolute magnitude, M, which is the apparent magnitude a star would have at d = 10 pc. How does the apparent magnitude, m, of a star with absolute magnitude M, depend on its distance, d away from you?

We know that flux is related to distance through a relationship of F ~ d^2. Thus we take our flux equations from question #1 and come up with the relationship, \[\frac{F_1}{F_2} = (\frac{d}{10})^2 = 10^{.4(M-m)}\]We take the log of both sides and end up with \[2 log (\frac{d}{10}) = .4(M-m)\]Rearrange to solve isolate the M's and get \[5 log(\frac{d}{10}) = (M-m)\] and finally we solve for little m to get \[m = M - 5 log (\frac{d}{10})\]

(c) What is the star’s parallax in terms of its apparent and absolute magnitudes?

From part (b) we know that \[\frac{F_1}{F_2} = (\frac{d}{10})^2 = 10^{.4(M-m)}\]From here we can solve that \[d^2 = 100 pc \cdot 10^{.4(M-m)} \]Then to solve for parallax all we do is solve for d, giving \[d = 10 pc \cdot (10^{.4(M-m)})^{\frac{1}{2}} \]Which when simplified gives us a final answer of \[d = 10 pc \cdot 10^{.2(M-m)} = p\]

Blog #14, Worksheet #5.1, Distance Ladder, Problem #1

1) Many students consider the astronomical magnitude system for measuring stellar fluxes baffling and a bit scary. With this exercise, I’d like to convince you otherwise. Magnitudes are defined such that, for Star One and Star Two with fluxes \(F_1\) and \(F_2\), \[\frac{F_1}{F_2} = 10^{0.4 \cdot (m_1 - m_2)}\]Where \(m_1\) is the magnitude of Star One and \(m_2\) is the magnitude of Star Two. Notice that, to two sig figs, \(10^{0.4}\) is just 2.5. As a result \[\frac{F_1}{F_2} = 2.5^{(m_1 - m_2)}\] Note also that a larger magnitude corresponds to a fainter star. Using this information, fill out the following table, where ∆m = m2 - m1, and identify the simple (approximate) recursion relation: 

The clear recursive relation is that for every 5th term, the pattern increases by a factor of 10^2 which makes sense given that \(10^{0.4(5)} = 10^2\). This process repeats until infinity with the same constants leading the recursive pattern.