The Age and Size of the Universe: Blog 27: Worksheet 8.1, Problem 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 = \(t_0\) 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: \[H(t) = \frac{1}{a(t)} \frac{ da}{ dt}|_t \] 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.

Part A: Assume the rate of expansion, a = 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.

Since we know \[v = H(t)d\] \[H(t) = \frac{v}{d} = \frac{1}{t} \] So \[ t_0 = \frac{1}{H(t)} = \frac{1}{68.816 \frac{km/s}{Mpc}} = 0.01453 \frac{Mpc \cdot s}{km} \] This is not a very useful number for us, so we can convert it to years: \[ 0.01453 \frac{Mpc \cdot s}{km} \cdot \frac{ 3.086 \times 10^{19} \: km}{Mpc} \cdot \frac{ 1 year }{3.15 \times 10^7}  = 1.385 \times 10^{10} \: years \] This is pretty close! The actual age of the universe is 13.82 billion years. This means the oldest globular clusters formed less than 2 billion years after the big bang.

Part B:  What is the size of the observable universe? What you will calculate is known as the Hubble Length.

Once again, we know \[ v = H(t)d\] So, if the universe is expanding at the speed of light and the speed of light in km/s is \( 3 \times 10^{5} \). \[ d = \frac{c}{H(t)} \] \[ d = \frac{3 \times 10^{5}}{68.816} = 4,356 Mpc \approx 4 \times 10^{3} Mpc \] The actual size of the universe is about \( 3 \times 10^{3} \) Mpc, so we aren't too far off.

Our old friends, Lyman-alphas: Blog 24 & 25, Worksheet 7.2, Problem 4 & 5

This week, we learned about active galactic nuclei and all of their fun applications. One of those applications in particular is especially dear to me, because the summer before tenth grade, I did a research project (and wrote a blog post about it) in a lab where I calculated the distance to ancient galaxies using redshift from Lyman-alpha emitting active galactic nuclei. And wouldn't you know, that's what we're doing in class this week:

Problem 4

One feature you surely noticed in a spectrum 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 λ = 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 Lyα emission from material orbiting around the central black hole.

Part A: Recall the Doppler equation: \[ \frac{ \lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} = z \approx \frac{v}{c} \]Using the data provided, calculate the redshift of this quasar. \[ z =  \frac{ 1410 - 1215.67}{1215.67} \approx \frac{1}{6}  \] So that \[ v \approx \frac{c}{6}\]

Part 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. 

Ideally, our emission lines would be infinitely thin and tall, but they have width. This happens because there is some dispersion due to the rotation of the galaxy. This means some of the galaxy will be redshifted and some will be blueshifted. 

So we can use this to find rotational velocity.

\[ z = \frac{ \lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} = \frac{1415 - 1406}{1406} \approx 0.00640 \]

\[ z = \frac{1415 - 1406}{1406} = 0.0064 \] \[ v \approx 0.0064 c \]

Now we can use the virial theorem: \[ K = - \frac{1}{2} U \] \[ Mv^2 = \frac{G M^2}{R}\] \[M_{BH} = \frac{ v^2 R}{G} = \frac{( 0.0064 \cdot 3 \times 10^{10})^2 (10^{17})}{6.67 \times 10^{-8}} = 5.5 \times 10^{40} \: g \approx 3 \times 10^7 M_{\odot} \] 

Problem 5

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

Part 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 the question implies, we are probably looking at instances where light is being absorbed or emitted by hydrogen gas on its way to us. These dips themselves, occurring before the peak, imply that this gas is between the galaxy and us, therefore they are not as redshifted as the distant galaxy. 

Part 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?

\[ \frac{ \lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} = z \approx \frac{v}{c} \]

\[ z = \frac{5650 - 1215.67}{1215.67} = 3.347 \approx 3\]

Part 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?

In this galaxy, there are many more variation, especially a larger number of deeper "dips." Based on the large redshift of the galaxy, we know it is quite old and far away. This means the light had to have traveled through a lot more gas to get here, and some of that gas had hydrogen atoms occupying the lowest energy state. Additionally, this could tell us that there was a lot more gas in the early universe. This fits in well with our current ideas of the early universe. 

Deriving Hubble's Law: Blog 26: Worksheet 8.1: Problem 2

In 1929, astronomer Edwin Hubble discovered that almost all distant galaxies exhibit a positive redshift. Furthermore, it appeared that the farther the galaxy, the larger its redshift. Here we will rediscover Hubble’s Law using modern spectroscopic data and supernovae Ia as our distance indicator to these galaxies. The data we will use come from the Sloan Digital Sky Survey (SDSS), a project that aims to comprehensively map the universe. You can access the relevant data products for this exercise at http://goo.gl/fmIvqc

Part A: Below is a list of supernovae observed between 2004 and 2007 and their positions in RA and Dec. You can find the images and spectra of their host galaxies by entering their coordinates in the respective fields. Explore the functions available, including magnifying the image, reading off the photometric measurements (magnitudes in wavebands u, g, r, i, z) of your selected object, and using the ‘Explore’ button to access more quantitative measurements for these objects. In particular, familiarize yourself with the ‘interactive spectrum’ feature.

For each galaxy, we have a pretty extensive spectrum.

Part B: One of the features for determining distances to Type Ia supernovae is its peak absolute magnitude. You explored the peak bolometric luminosities of SN Ia’s in Worksheet 7.1. The peak V-band magnitude for SN Ia’s is about -19.3. Use the apparent peak magnitudes given in the table above to calculate the distance of these supernovae in unit of Mpc.

To find distance, we can use our trusty distance formula :\[d = 10^{ \frac{m - M + 5}{5}} \] for each magnitude. The fourth galaxy did not have a spectrum associated with it, so we did not analyze it.

Part C: We can use the absorption or emission lines of the host galaxy to find their redshifts which, as you found in Question 1), roughly equals the recessional velocity as a fraction of the speed of light. To measure the redshift to each host galaxies, click on ‘Explore’ and then on the link ‘Interactive Spectrum’. Uncheck the boxes Best Fit and Mark Emission Lines. Zoom in on the absorption line labeled Hα, and move your mouse over to the center of the line to read the observed wavelength in Angstroms. The Hα has a rest (i.e. emitted) wavelength of 6563.0 Angstroms. Calculate the redshift, and then derive the radial velocity in kilometers per second, using the relation z = v/c. How close does your redshift measurements compare to the one SDSS reports in the table under the Interactive Spectrum link? Repeat for all the galaxies.

We found redshift and velocity using \[ \frac{ \lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} = z \approx \frac{v}{c} \] to find: 

Part D: Make a plot of your findings, with distance on the x-axis and velocity on the y-axis. Report the slope of your line in appropriate units. This is the Hubble Constant, H0.

Result: \(H_0 = 68.816 \frac{km/s}{Mpc} \)

Part E: Write an equation for this line in the form of v = ___ D, where v is an object’s recessional velocity and D is the distance to that object. Express your Hubble Constant in terms of units km/s/Mpc. Congratulations, you have arrived at Hubble’s Law!
\[ v = 68.816 \frac{km/s}{Mpc} D \] Now we have a basic law of the universe (kinda) down!

Blog 23: Astronomy and the Harvard Art Museums

This week, we are going to explore the intersection of Astronomy and another field because while Astronomy is the study of the universe, it is not the only subject to study in the universe. Last semester, I talked about Astronomy and Mystery-Solving. This semester I started a job as a tour guide at the Harvard Art Museums (HAM). At HAM, tour guides have to design their own tours based on a central theme connecting 3-5 objects on display in the museums. Themes vary from political to religious to technical. For this blog post I am going to try to propose an astronomy-themed tour based on our collection. All images and information about the objects can be found at the museum's website.

Admittedly, astronomy is not one of the most centralizing themes in art history. However, it has been a constant part of our lives before, throughout, and long after history. The night sky has always been round us, both inciting our imaginations and giving us the tools to understand the world we live in. The oldest astronomy-themed object on this tour is the Long-Case Musical Clock, created by the Dutch craftsman Otto Van Meurs in 1750. The grand, imposing clock sits in a seconds floor gallery of the museum, still keeping accurate time - with a little adjusting from conservationists every set number of years. That is not all the clock does; it also tracks tides, the current position of the sun and moon, gives the date, and plays one of dozens of charming tunes every half hour. Mechanical clocks like these are one of the first iterations of modern computers. This "computer" in particular, heavily showcases society's uses for the skies above. Our entire conception of time, of calendars and schedules, of past, present, and future is based on the motion of stars and planets. We can see that in this piece, a representation of years of evolution of the clock and calendar.

Yet this piece isn't purely functional. It is highly symbolic, covered in classical allusions such as Atlas carrying Earth at the very top of the clock. This piece, used to show the high societal status of its owner, is inherently intertwined with modern cultural norms. This culminates in a symbolic, functional, and aesthetic piece that dominates any room it is in.

More than a century later, Paul Manship (considered the most prominent American Art Deco artist) created about a dozen celestial spheres as a preparation for a grand memorial to Woodrow Wilson. This historically influential tool was also highly functional, allowing for navigation, charting, and seasonal predictions. Even today, it is our foremost model of the night sky.

Celestial Sphere - Paul Manship

However, this piece itself is also not purely functional, it is not even very useful. Its chief function is to memorialize a political figure and does so with many cultural symbols. The constellations and base figures are a mix of Chinese, Babylonian, Assyrian, Latin, and Greek zodiacs to emphasize the international political harmony the artist wanted a viewer to associate with Woodrow Wilson. Once again, Astronomy is used here as a seamless part of history and modern society, and not just for one culture, but for many. 

There are definitely cultural levels of astronomy that move beyond the scientific into the even more cultural and aesthetic. For example, the second floor has a beautiful painting with a puzzling composition that signifies the coming of the dawn, the daily transition from dark to light and a pagan goddess. The astronomical phenomena has become a personified deity, with her own gossip, lovers, and enemies.

The Dawn - John LaFarge

Even Christian religions have used Astronomy as cultural markers. In this stained glass window, angels are represented by the morning stars in a reference to a biblical passage. The stars symbolized heavenly light, purity, and truth.

When the Morning Stars Sang Together and All the Sons of God Shouted for Joy

Even in contemporary art today, these themes persist. De Kooning's piece, Untitled (The Cow Jumps Over the Moon), plays upon a common cultural anchor, a childhood nursery rhyme. From there, De Kooning's questions a viewer's perspective of traditional art and accepted ways of portraying cows and emotions alike. 

Willem de Kooning - Untitled (The Cow Jumps Over the Moon)

Even in art's evolution, Astronomy is used as a grounding point. The night sky has become irrevocably intertwined with our perception of the world, and what better way is there to portray our world than through art?

How far away are those supernovae? Blog 22, Worksheet 7.1, Problem 6

If your telescope can detect optical magnitudes \(m_V < 21\), how far away, in parsecs, can you detect a Type Ia supernova with your telescope? (HINT: The Sun’s absolute magnitude is \(M_V = 4.83 \) .)

What we know is \(M_{ V \odot}  = 4.83 \) and \(L_{WD} = 10^{10} L_{\odot} \).
To solve, we can revisit week 5 to solve this problem. We know that absolute magnitude is related to luminosity: \[ M_{WD} - M_{\odot} = -2.5log \left( \frac{L_{WD}}{L_{\odot}} \right) \] So \[ M_{WD} =  -2.5log \left( \frac{10^{10} L_{\odot}}{L_{\odot}} \right) + M_{\odot}  \]  \[ M_{WD} =  -2.5log \left( 10^{10} \right) + 8.83 = -25 + 4.83 = -20.17  \] Now that we have a (rather large) absolute magnitude and a limit on our ability to detect apparent magnitudes, we can find distance. \[ m - M = 5log(d) - 5 \] \[ d = 10^{\frac{m-M +5}{5}} \] \[ d = 10^{\frac{21 + 20.17+5}{5}} = 1.7 \times 10^9 \: pc \] By comparison, the center of the galaxy is about 8 kilo-parsecs away so we would get a decent range within our galaxy.

Source

Supernovae can be amazingly bright.

Boom: Blog 21, Worksheet 7.1, Problem 4

This week we are going to talk about one of the brightest, yet most transient, objects in our sky: supernovae.

Source

Astronomers believe that most supernovae are the result of 1) the gravitational collapse of a massive star at the end of its main-sequence life or 2) the explosion of one or more white dwarfs, likely caused by the collision between two white dwarfs. In this problem, we’ll focus on the second type of supernovae, or Type Ia supernovae which we are familiar with.

Calculate the total energy output, in ergs, of the explosion, assuming that the white dwarf’s mass is converted to output energy via fusion of carbon into nickel. Note that the process of carbon fusion is not entirely efficient, and only about 0.1% of this mass will be radiated away as electromagnetic radiation (light!). How does this compare to the total binding energy, in ergs, of the original white dwarf? Does the white dwarf completely explode, or is some mass left over in the form of a highly concentrated remnant?

To calculate energy output we can call upon possibly the most famous equation in science: \[ E=mc^2 \] Well, 0.1% of that energy anyway. Because we know the speed of light \( c = 3 \times 10^8)^2 \) and the mass of a white dwarf is \( M_{WD} = 1.4 M_{\odot}\). Plugging everything in gives us: \[E = \frac{1}{1000} (1.4 \times 2 \times 10^{32} ) ( 3 \times 10^{10})^2  \frac{1}{1000} (1.4 M_{\odot} ) ( 3 \times 10^8)^2 = 2.7 \times 10^{51} \: ergs \]

We can find binding energy from the potential energy of the star: \[U = E = \frac{GM^2}{R} \] We also know the radius of the White Dwarf is about twice the radius of the Earth. \[ E_{binding} = \frac{ c_s  M_{WD}^2 }{R_{WD}} = \frac{ (6.67 \times 10^{-8}) (1.4 \times 2 \times 10^{33})^2 }{2 \times 6.4 \times 10^8} \approx 4.1 \times 10^{50} \: ergs \]

There is something off here, the energy output is not equal to the energy input. So how can that be? Thanks to Einstein (e.g. \(E = mc^2 \) ) we know that mass can be converted to energy. In the explosion, some of the mass of the white dwarf gets turned into energy, and the rest is flung out throughout the galaxy. Explosions like these can even help explain why we have heavier elements dispersed through the Milky Way.

Blog 20: The Hubble Tuning Fork

The Hubble Tuning Fork

The Hubble Tuning Fork is a classification system developed by Edwin Hubble in 1926 as a theory of galaxy evolution. Today we view it as outdated, as there is evidence that spiral galaxies do not evolve from ellipticals, but it is a useful visual for galaxy types.

Ellipticals (E0-E7): 

Elliptical galaxies are spheroid or elongated spheres where stars have no uniform rotation around the center of the galaxy. Elliptical galaxies tend to have older, and therefore redder, stars. Elliptical galaxies are classified due to their shape. E0 galaxies appear almost perfectly spherical, at least from our perspective on Earth. E3 galaxies appear as slightly elongated ellipsoids, E5 slightly more, and E7 galaxies are extremely elongated.

Lenticular galaxies (S0):

Lenticular galaxies (charmingly named after the lentil bean) fall between elliptical and spiral galaxies. They have no spiral shape but do contain a bulge and thin disk, similar to a spiral galaxy.

Spiral Galaxies (Sa-Sc) and Barred Spiral Galaxies (SBa-SBc):

Spiral galaxies are galaxies like the Milky Way or Andromeda. They typically contain a disk around which stars revolve, a bulge, and spiral arms. Spiral galaxies contain younger, bluer, stars than ellipticals. Spirals are classified by how tightly their arms are wound with Sa spirals being the tightest wound and Sc being the loosest. Barred spirals have a bar of stars running across the nucleus out of which the arms extrude. Barred spirals are also classified by the distribution of their arms.

Irregular:

Irregular galaxies do not fit into any of these categories and are often the result of two galaxies colliding.

• E0: Image: M87
• E3: Image M86
• E5: Image: M59
• E7: Image
• S0: Image: Messier 102
• SA: Image: NGC 1302
• Sb: Image: M77 
• Sc: Image: NGC 628
• Sba: Image: M83
• Sbb: Image: Ngc 1300
• Sbc: Image: NGC 1365 

Blog 19: Exomoons

You've heard of exoplanets, now get ready for exomoons! If the goal of our search for other planets is to possibly find other life, or just to have more knowledge, the next logical step would be to look at moons. Even within our solar system, we have 182+ moons and of those moons, more have signs of liquid water (e.g. Europa and Enceladus) than planets (e.g. Mars). With all the exoplanets we have been finding lately, exomoons could fructuous line of research, if we have the technology to explore it.

Jupiter and some of its moons. Source

So how do we find exomoons? Some projects are already underway. Here at the Center for Astrophysics, research is already being conducted to look for anomalies in stars' light curves that could point to exomoons in addition to exoplanets. Today, I am going to talk about this astrobite, which discuses future, direct detection of exomoons.

How would one directly detect an exomoon? They key to exomoon detection is the compositional differences between a moon and the planet it orbits. This would lead to distinct spectrographs for a planet and its moon. By analyzing the combined spectrum of the planet-moon system, there could arise evidence for moons. For example, the figure below shows the spectra for an Earth-Moon analog system orbiting Alpha Centuri. The bottom half shows the percent of the flux density due to the moon in this system.

By analyzing a system's flux at different frequencies, different elements of the system can stand out. For example, in the infrared wavelength in the system above, the moon accounts for 99.8% of the total flux in the water band at ~2.7 microns. Analyzing these differences can also give us the spacial separation of a planet and its moon. The offset of the origin of peak flux in different wavelengths can point to the location of a moon (as shown in the figure below).

The one problem with this exciting new research is that we don't quite have the technology to resolve these systems. Therefore, these plans are for future technologies. For example, to find the Earth-Moon analog orbiting Alpha Centauri, one would need a spacial resolution of ~2 milliarcseconds in the infrared. The Hubble Space Telescope, by comparison,  can only resolve ~100 milliarcseconds in similar wavelengths. The image below describes the size of telescope needed to resolve nearby candidates.

If we could achieve this technology, not only could we find these exomoons, but we could check for chemical compositions that could mean the moons are habitable or even containing biosignatures. The cliche saying to shoot for the moon could take on a new meaning in the near future. 

Tully-Fisher: Blog 18, Worksheet 6.1, Problem 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: \[L \sim v_{max}^4 \], 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:

Source

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

The \(v_{max} \sim σ \) assumption is a pretty safe one to make because σ is basically the width of the possible velocities distribution. It has the same units as v_{max} and will always be off just by a constant.

In the previous problem, we derived \[M \approx \frac{σ^2R}{ G}\] a relationship that can be rearranged to show \[ v_{max}^2 \sim σ^2 \approx \frac{GM}{ R} \] So as v increases as mass increases and radius decreases.

(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 (M/L).

This part is really just setting up definitions for part c: \[ <I> =   \frac{L}{R^2} \] can be rearranged to \[ R = \left(\frac{<I>}{L} \right)^{\frac{1}{2}} \] and \[ x = \frac{M}{L}\] which is \[ M = \frac{x}{L}\]

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

\[v_{max}^2 \sim \frac{M}{R} = \frac{ \frac{x}{L}}{\left(\frac{<I>}{L} \right)^{\frac{1}{2}}}= x <I>^{\frac{1}{2}} L^{\frac{1}{2}} \] Dropping the constants gives: \[L \sim v_{max}^4 \]

(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 (λcen ~ 445 nm, blue light), this relation is approximately: \[M_B = -10 log \left( \frac{v_{max}}{ km/s} \right) + 3\] 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?
\[M_B = -10 log \left( \frac{v_{max}}{ km/s} \right) + 3\] \[M_B = -10 log \left( \frac{400}{ km/s} \right) + 3\] \[M_B \approx -23 \] We know the distance formula from last week and can use it to find our answers: \[ d = 10^{\frac{m-M + 5}{5}} \] \[ d = 10^{\frac{13 - (-23) + 5}{5}} = 1.6 \times 10^8 \: pc \]

Good Ol' Virial Theorem: Blog 17, Worksheet 6.1, Problem 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 ( \( F \propto 1/r^α\) ). 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 \[U \approx \frac{GM^2}{R} \] 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 think about this sphere of particles with a uniform density in terms of thin shells, each with a width of dR.

We also know that \[ U = - \frac{GMm}{R} \] We can treat the entire sphere as our mass, M, and the shell as our mass, m. Then, we can rewrite each mass in terms of density: \[ \rho = \frac{M}{V} \] Where \( V = \frac{4}{3} \pi R^3\) So \[ \rho = \frac{M}{\frac{4}{3} \pi R^3} \] And we can rewrite the equation for potential energy of one shell as: \[ dU = - \frac{GMdm}{R} = - \frac{G(\frac{4}{3} \pi R^3 \rho) (4 \pi R^2 \rho) dR}{R}= - G \frac{16}{3} \pi^2 R^4 \rho^2 dR \] But now we have dU in terms of dR, so we can integrate to get U: \[ \int_{0}^{R} - \frac{16}{3} G \pi^2 R^4 \rho^2 dR =  - \frac{16}{3} G \pi^2 \rho^2  \int_{0}^{R} R^4 dR = - \frac{16}{3} \frac{R^5}{5} G \pi^2 \rho^2 - 0 \] But this still is not the answer we want. We need U in terms of M, not density so we can reuse  \[ \rho = \frac{M}{ \frac{4}{3} \pi R^3 } \] to get: \[ U = - \frac{16}{3} \frac{R^5}{5}\left( \frac{M}{\frac{4}{3} \pi R^3 } \right)^2 G \pi^2  \] Amazingly enough, this simplifies to the much nicer expression: \[ U = - \frac{3GM^2}{5R} \] This simplifies to our original expression: \[ U \approx - \frac{GM^2}{R} \]

(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 σ 2 . Show that the kinetic energy of the system is: \[K =  N \frac{3 }{2} m σ^2\]

Our traditional, classical mechanics, definition of kinetic energy is \[ K = \frac{1}{2} mv^2 \] where velocity is the total or average velocity of the system. That is very difficult to measure for a distant object here on Earth, because the entire system (e.g. a globular cluster) is moving away from us with the universe's expansion. To account for this we must change our frame of reference to the system.
 

So our definition for kinetic energy \[ K = \frac{1}{2} m \Sigma (v_o + \sigma_{vi} )^2 \] turns into  \[ K = \frac{1}{2} m \Sigma ( \sigma_{vi} )^2 =   \frac{1}{2} m \sigma_{vi}^2\] in the reference frame. However, this is a 1 dimensional representation of velocity scatter, as that is all we can observe - the cluster in one plane. To account for the 3 dimensions of space.

So our final expression would be \[ K =   \frac{3}{2} m \sigma_{vi}^2\]

   (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): \[M = \frac{σ^2R}{ G}\]

Using the Virial Theorem, we can combine our two derived equations to find the answer. \[K = - \frac{1}{ 2} U\] \[\frac{3}{2} m \sigma_{v}^2 = -\frac{1}{2} \left( - \frac{GM^2}{R} \right) \] \[3M\sigma_{v}^2 = \frac{GM^2}{R} \]  \[M = \frac{σ^2R}{ 3G}\]  \[M \approx \frac{σ^2R}{ G}\]

The Great Debate: Blog 16

Like many discoveries in science, astronomy discoveries have not come into modern cannon uncontested. Many of our now-accepted theories began out of scientists trying to understand nebulous (get it?) data collected with limited technology. The question of the scale of the universe, which is still debated today, was a hot topic in 1920. 

"The Great Debate" as it has been named, took place between two astronomers on April 26 1920. The "contestants" were Harlow Shapley, a young upcoming astronomer at Mt. Wilson Solar Observatory, and Heber D. Curtis, an older, more established professor at the Lick Observatory. Shapely argued in favor of the Milky Way being the entire scale of the universe while Curtis argues in favor of the universe being composed of multiple, separate galaxies. The debate centered around "nebulae" in the Milky Way, such as Andromeda, and whether on not they were a part of our galaxy or separate entities.

Shapley and Curtis - Source

Both scientists had previously published their findings on the scale and composition of the universe in different journals. Curtis's research focussed on star count analysis and spectral distance estimates. He concluded that the Milky Way is about 10 kpc in diameter and lens shaped. He found the nebulae observed had a similar number of novae to the rest of the galaxy, and therefore must be their own galaxies. On the other hand, Shapley argued for a model where one massive Milky Way (100 kpc across) composed the entire universe. He looked at Cepheid variables in globular clusters to make distance determinations. Shapely used the already known distance to the M31 cluster to estimate the relative distances to other clusters and determined them to be at the edge of the Milky Way. Additionally, the data of Adriaan van Maanen, a friend of Shapley's at Mt. Wilson Observatory, found that the Pinwheel galaxy was rotating so quickly that you could observe one rotation in a matter of years. If this galaxy lay beyond the Milky Way, this would mean the galaxy was rotating so fast that it violated the speed of light! This provided further evidence for Shapley's case.

Source

On April 26, 1920 both scientists gave separate talks proposing their ideas during the day and came together for a discussion at night, where they could provide counter arguments to each other's points. Who "won" is unclear. There is a common misconception that Shapely won, as he became more famous, though many scientists present agreed with Curtis's argument.

Today we know that both scientists were right in some ways and wrong in others. Shapely got the order of magnitude of the universe correct and our relative placement in it. Curtis correctly concluded that the "nebulae" were other galaxies. This was confirmed when Edwin Hubble studied Cepheid variables in Andromeda and concluded it lies outside of the radius of the Milky Way. Alas, Maanen's measurements were proven to be incorrect, the rotation of the Pinwheel galaxy cannot be measured in years and the high recessional velocities observed are actually evidence for the expanding universe, a debate that occurred  a few years later.


Sources:
http://cosmos.phy.tufts.edu/~zirbel/ast21/handouts/Curtis-Shapley.pdf
http://apod.nasa.gov/diamond_jubilee/papers/trimble.htmlhttp://astronomy.nmsu.edu/geas/lectures/lecture27/slide01.html
http://astronomy.nmsu.edu/geas/lectures/lecture27/slide01.html

Graphing Cepheid Magnitudes: Blog 15, Worksheet 5.2

As we previously learned, Cepheid variables are a special class of stars that radially pulsate in a predictable way. In 1908, Henrietta Swan Leavitt discovered that there is a distinct relationship between a Cepheid’s luminosity and pulsation period by examining many stars in the Magellanic Clouds. Henrietta was a member of “Harvard’s computers,” a group of women hired by Edward Pickering to analyze stellar spectra and light curves. In this worksheet, we will use Henrietta’s original data set to find our own Period-Luminosity relation for Cepheid variables. The data files for this activity will be located on Canvas under the name “Cepheid variables.csv.”

Source

1. The data file, “Cepheid variables.csv,” contains data for 25 Cepheid variables located in the Small Magellanic Cloud (SMC). Each line contains a specific Cepheids: (1) ID number, (2) Maximum apparent magnitude, (3) Minimum apparent magnitude and (4) Period. Calculate the mean apparent magnitude for each Cepheid.

Done! It is pretty easy to do in excel.

2. The distance to the SMC is about 60 kpc, where kpc = 1000 pc. Convert your mean apparent magnitudes into mean absolute magnitudes. Plot the Cepheid mean absolute magnitudes as a function of period. This plot should look exponential.

3. It is often handy to plot exponential (or power-law) functions with one or more logarithmic axes, which “straightens out” the data. Magnitudes are already exponential, so we don’t need to adjust that axis. Plot the Cepheid mean absolute magnitudes as a function of log(Period). Verify that the plot now looks linear.

4. Now that the data look linear, we can estimate the parameters of a linear relation, \(M_V (P) = A log_{10}(Period) + B \). A and B are “free parameters” that allow the function to match the data.

This would make A = -2.033 and B  = -.2782 for a final expression of approximately \[M_V (P) = -2.0 log_{10}(Period) - 2.8 \]

Stellar Magitude: Blog 14, Worksheet 5.1, Problem 2

Okay, let’s use your new-found knowledge of magnitudes. In the first problem of the wroksheet, we derived a relationship that looks something like this:

Source

Now lets put that knowledge to use: 

(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? 

From question 1, we know: \[ \frac{F_1}{F_2} \approx 10^{0.4( m_2 - m_1)} \] For this problem we have  \[ \frac{F_1}{F_2} \approx 10^{0.4 \cdot 3} \approx 15.6 \] So we would have to observe for about 16 times longer.

(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? 

For this problem, we are going to look back to a few weeks ago and use our definition of flux:
\[F_M  = \frac{L}{4 \pi d^2} \] \[F_m = \frac{L}{4 \pi r^2} \]  and  \[ \frac{F_1}{F_2} \approx 10^{0.4( M - m)} \] \[ \frac{d^2}{r^2} = 10^{0.4(M- m)} \] \[ \frac{2}{5}(M - m) = log_{10} \left( \frac{d^2}{r^2} \right) \] We know d=10pc, so  \[ \frac{2}{5}(M - m) = log_{10} \left( \frac{100}{r^2} \right) \]  \[ \frac{2}{5}(M - m) = log_{10}(100) - log_{10}(r^2) \] \[ \frac{2}{5}(M - m) = 2 - 2log_{10}(r) \] \[ m= M + 5log_{10}(r) -5\]

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

We know from blog 2 that, \[ \theta = \frac{1 \: AU}{r}\] And we can find r from our work above \[ r = \left( \frac{100}{10^{0.4(M-m)}} \right)^{\frac{1}{2}}\] So \[ \theta = \frac{1 \: AU}{\left( \frac{100}{10^{0.4(M-m)}} \right)^{\frac{1}{2}}}\] but we need to multiply by 1000 to get our answer in AU/kpc so it is really  \[ \theta = \frac{1 \: AU}{\left( \frac{100}{10^{0.4(M-m)}} \right)^{\frac{1}{2}}}\cdot (1000) \] This simplifies to \[\theta = \frac{1000 \left( 10^{0.4(M-m)} \right)^{\frac{1}{2}}}{10} = 100 \left( 10^{0.4(M-m)} \right)^{\frac{1}{2}} \]