Blog Post 23, Astronomy and Religion

The intersection of astronomy and religion is one that goes back millennia – for as long as humans have been around, they have looked to the stars for the answers to our species’ most profound questions – Why are we here? How are we here? What are we looking at? This fall, I am taking a class called Humans in Space: Past, Present, and Future (History of Science 181), taught by Professor Matthew Hersch. In this course, we have covered many topics relating to the history of spaceflight and our views of the universe. A reoccurring theme that has appeared throughout this class, and through history, is the constant push-and-pull between astronomers and the religious views of the times. While this topic could spur an endless blog post covering many cultures as they have developed through time, I just want to focus on a few points in history that I found really interesting.

For many centuries, the universe was thought to be centered on the Earth, with all objects including the Sun, Moon, planets, and stars orbiting around the Earth. Known as the geocentric model or the Ptolemaic system, this model was accepted in many ancient civilizations, including the ancient Greeks. Prior to the Renaissance, the European Church had adopted this model, as developed by Aristotle and Ptolemy, as part of their own doctrine.

Source: https://upload.wikimedia.org/wikipedia/en/2/24/Ptolemaic-geocentric-model.jpg

Other theories were brewing, however. Although heliocentric theory had been first developed much earlier by Aristarchus (ca. 270 B.C.), Nicolaus Copernicus truly took the theory under his wing, publishing his book De revolutionibus orbium coelestium (On the Revolutions of the Celestial Spheres) shortly before his death in 1543, and forcing the world to question their views of the universe. Geocentric theory had agreed very nicely with Church doctrine, placing the Earth at the center of the universe as suggested in the Bible. Heliocentric theory called the Bible and those in power because of it into question – as you can imagine, the Church wasn’t all too happy with this new world view.

This unhappiness continued to worsen, and by the time of Giordano Bruno, they fully rejected the Copernican model. Bruno upset the Church even further by suggesting a plurality of worlds – not only a heliocentric universe, but a universe in which all of the stars we look up at are in fact their own universe. While this idea may seem harmless at first, Bruno framed it in a way that severely threatened the Church’s power – if the heavens to us on Earth are in fact individual universes, then the Earth must be the heavens to those other universes. And if the Earth is heaven to other universes, how can the Church have the power to determine who goes to heaven, if our world is already heaven to another world? Bruno went so far as to suggest that the Pope can’t in fact have any power, and was literally burned at the stake in 1600 for his ideas. Think about that – burned at the stake, for attempting to advance our scientific world view.

Eventually the Church did mold its accepted model closer to what we now know to be correct, but it took some steps to get there. Tycho Brahe (1546-1601) developed a model (the Tychonic system) in which the Sun continues to orbit the Earth, but all planets orbit around the Sun. This model agreed more closely with observation, yet also satisfied the Church's view that the Earth is at the center of the universe. For a while, the Church adopted the Tychonic system as their official accepted view of the universe. 

Source: https://upload.wikimedia.org/wikipedia/commons/a/a6/Tychonian_system.svg

Eventually the world came to accept heliocentrism as the correct model for our solar system, and to believe that there is no actual center of the universe, according to Einstein's principle of relativity. Yet looking at this history makes one really marvel at the path our world view took to develop - it makes me glad to be alive at a time and in a place where science is in a much more peaceful and separate relationship with religion, yet also raises a question: what do we believe now based on our observations, that is in actuality completely and utterly wrong? What will our next big shift in world view occur?

Blog Post 22, Worksheet 7.1, Problem 3

3.  A white dwarf that exceeds the Chandrasekhar mass will start to fuse carbon in its interior, which releases a great deal of heat, which increases the internal pressure of the white dwarf. However, because the white dwarf is "trying" to support itself using degeneracy pressure, and increasing this pressure doesn't change the star's radius, the increasing temperature leads to more fusion, more energy, and a run-away fusion process is initiated. 

Once this run-away fusion inside the white dwarf is "ignited," it propagates as a wave traveling outward at the speed of sound, \(c_s\). How much time does it take the flame to sweep outward across the radius of the white dwarf? This is also known as the "sound-crossing timescale." 

How does this time scale relate to the density of the white dwarf?

From problem 2, we know that the relationship between the speed of sound (\(c_s\)), pressure (P), and density (\(\rho\)) is given by: \[c_s \sim (\frac{P}{\rho})^{1/2}\] The speed of sound is a speed, or \(\frac{\text{distance}}{\text{time}}\). To solve for time, we thus have: \[\text{time} = \frac{\text{distance}}{c_s} = \frac{R}{(\frac{P}{\rho})^{1/2}}\] From problem 1, we know that the pressure of the white dwarf is given by: \[P = \frac{GM^2}{4πR^4},\] and from the definition of density as \(\frac{\text{mass}}{\text{volume}}\), we know that \[\rho = \frac{\text{mass}}{\text{volume}} = \frac{M}{\frac{4}{3}πR^3}\] Plugging in these expression for P and \(\rho\), we have: \[t = R (\frac{4πR^4}{GM^2})^{1/2} (\frac{M}{\frac{4}{3}πR^3})^{1/2}\] \[t = R(\frac{3R}{GM})^{1/2}\] \[t = (\frac{3R^3}{GM})^{1/2}\] We know that density, \(\rho\), is given by mass/volume, or \(\frac{M}{\frac{4}{3}πR^3}\). If we ignore the constants, we can thus replace \(\frac{R^3}{M}\) in the expression for time with \(\frac{1}{\rho}\): \[t \sim (\frac{1}{G\rho})^{1/2}\] From this, we can see that the relationship between time scale and density is given by: \[t \sim \rho^{-1/2}\]

Blog Post 21, Worksheet 7.1, Problem 1

1. White dwarfs are supported internally against the force of gravity by "electron degeneracy" pressure. The maximum mass that can be supported by this exotic form of pressure is the 1.4 \(M_\odot\) (also known as the Chandrasekhar Mass). The radius of our white dwarf is approximately twice the radius of the Earth or ~ \(12 * 10^8\) cm. 

Given this mass, M, and radius, R, derive an algebraic expression 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 = \(1.4*10^{-16}\) erg \(\text{K}^{-1}\) is the Boltzmann constant. You can also assume the interior of the white dwarf is an ideal gass, and its mass is uniformly distributed. 

Starting with the Virial theorem as suggested, we know that the total kinetic energy of the white dwarf will be equal to the opposite of half of the potential energy: \[K = -\frac{1}{2}U\] We know that the potential energy of the system comes from gravity, so the total (gravitational) potential energy of the system is given by \[U = -\frac{GM^2}{R}\] We are also given that the kinetic energy of each particle in the system is \(\frac{3}{2}kT\). If we say there are N total particles in the system, the total kinetic energy in the system is given by: \[K = N(\frac{3}{2}kT)\] Plugging these values for K and U into the Virial theorem above, we have: \[N(\frac{3}{2}kT) = \frac{1}{2}\frac{GM^2}{R}\] If we rearrange the constants, we get the following relation: \[NkT = \frac{GM^2}{3R}\] Now, the left hand side of the equation should look a little familiar, especially because we are told that we can assume the interior of the white dwarf is an ideal gas. One form of the ideal gas law gives us the following equation: \[PV = NkT,\] where P is the absolute pressure of the gas, N is the number of molecules in the given volume V, k is the Boltzmann constant, and T is the absolute temperature. Because the right hand side of this equation is equal to the left hand side of our previous equation, we can set the remaining two sides equal to one another to get: \[PV = \frac{GM^2}{3R}\] If we solve for V, we get: \[P = \frac{GM^2}{3VR}\] We know that V is the volume of a sphere with a radius of R, or \(V = \frac{4}{3}πR^3\), so we have \[P = \frac{GM^2}{3(\frac{4}{3}πR^3)R}\] \[P = \frac{GM^2}{4πR^4}\]

Blog Post 20, Hubble Tuning Fork

My Hubble Tuning Fork:

(Click on image for larger version)

The "Hubble Tuning Fork" is the classification system for galaxies developed by Edwin Hubble that groups galaxies into categories based on their appearance - how round or flat they are, whether or not they have a central bar structure, if they have spiral arms, and how tightly wound those arms are.

Elliptical galaxies (those that begin with an "E" in the diagram) are ellipsoidal with no central structure, and are classified by their eccentricity, or how long/thin they are. E0 galaxies (far left) are essentially spherical, whereas E7 galaxies are extremely long and thin. Elliptical galaxies generally have many older stars, and tend to have many globular clusters surrounding them.

Spiral galaxies have a central bulge surrounded by a flattened disk structure with arms spiraling outwards from the center. A faint halo surrounds the disk, in which there are some globular clusters. Spiral galaxies are split into two groups - barred spirals and non-barred spirals - and then into a further three groups, types a, b, and c. Barred spiral galaxies (SBa, SBb, SBc) have a central bar structure from which the spiral arms begin, whereas non-barred spirals (Sa, Sb, Sc) do not. Type a spiral galaxies have loosely-wound arms, whereas type c has very tightly wound arms.

Lenticular galaxies (S0) is an intermediate between an elliptical galaxy and a spiral galaxy. They have no spiral structure, but have a rotating disk and central bulge. They do sometimes have a central bar structure.

Irregular galaxies are those that do not fall under any of the above categories - they have a unique shape, and thus do not appear on the tuning fork. They often form from the collision of two galaxies. Here is an example of an irregular galaxy:

Image sources:
http://burro.case.edu/Academics/Astr222/Galaxies/Intro/M87.jpg
http://freestarcharts.com/images/Articles/Messier/M86_NASA_AURA_STScI.jpg
https://www.le.ac.uk/ph/faulkes/web/images/m59.jpg
http://cdn.iopscience.com/images/0067-0049/190/1/147/Full/apjs351534f1_lr.jpg
http://chandra.as.utexas.edu/bh/NGC3115.jpg
http://www.noao.edu/image_gallery/images/d4/m81y.jpg
http://cs.astronomy.com/cfs-file.ashx/__key/telligent-evolution-components-attachments/13-58-00-00-00-43-91-97/NGC1365_5F00_master_5F00_LRGB_5F00_n_5F00_blue_5F00_sb.jpg
http://www2.lowell.edu/rsch/LMI/gallery/n6946_2.jpg
https://upload.wikimedia.org/wikipedia/commons/1/15/NGC3486-hst-R814GB450.jpg
http://cse.ssl.berkeley.edu/SegwayEd/lessons/classifying_galaxies/SBa.htm
http://photojournal.jpl.nasa.gov/catalog/PIA07901
http://www.starcorral.com/html/observing_at_oil_capital.html
http://hubblesite.org/newscenter/archive/releases/2005/09/image/a/

Blog Post 19, Free Form Post, A Cepheid Anomaly

For this blog post, I had to read an article from either Astrobites or astro-ph that I found interesting, and summarize and discuss the result. I chose this article, titled A Cepheid Anomaly, that summarizes the paper The Strange Evolution of Large Magellanic Cloud Cepheid OGLE-LMC-CEP1812 by Hilding R. Neilson, Robert G. Izzard, Norbert Langer, and Richard Ignace.

On Worksheet 5.2, we looked at Cepheid Variable stars, a class of stars that pulsate radially in a predictable way. As there is a relationship between the Cepheid's luminosity and period of pulsation, Cepheid variable stars can be used as standard candles to determine the distances to objects. This article discusses an apparent discrepancy between predicted masses of Cepheids: the mass predicted by stellar pulsation models of a Cepheid with a set luminosity and temperature varies from that predicted by stellar evolution models. This discrepancy has been difficult to resolve, as it is not so easy to measure the mass of a star. However, one rather interesting method of measuring star masses has come in handy, known as an eclipsing binary.

An eclipsing binary is a system of two stars in which one star crosses in front of our line of sight to the other. As it turns out, if the orbits of the two stars are very close to edge-on to our point of view, we can use the variations in the light observed during this eclipse to determine the period, speed of stars in the orbit, and the size of the orbit, which can be used to find the mass using Kepler's laws! (Read more about it here) Unfortunately, eclipsing binary systems in the right orientation are very rare, and the chances of finding one with a Cepheid are even lower - there have only been four found in the LMC.

This paper focuses on a specific Cepheid, CEP1812, in a system with a red giant. While both stars are expected to be the same age, the Cepheid instead appears to be approximately 100 million years younger than the red giant. This paper suggests that this age discrepancy could be because of the Cepheid's history - they suggest that the Cepheid formed from a merging event of two smaller, main sequence stars. This would make the Cepheid appear younger, as the star formed as a result of the merger would evolve like a star that was born with the combined mass of the two smaller stars, but look younger.

The authors believe that if this is indeed the case - the Cepheid formed from a merger of two stars - then it should have a relatively short period with a mass of approximately 1 \(M_\odot\). However, using the above method of determining the Cepheid's mass, it has been found that CEP1812 has a mass of around 3.8 \(M_\odot\)! This discovery will force astronomers to reevaluate their Cepheid classifications, and proves that while we may know a lot about Cepheid variable stars and their characteristics, there is still much to learn.

See the article here: http://astrobites.org/2015/08/28/a-cepheid-anomaly/

Blog Post 18, Worksheet 6.1, Problem 3

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 (\(F\propto1/r^\alpha\)). 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.

In this spherical distribution, there are N particles, each with a mass m, contributing to a total mass M. Thus, Nm = M. If we think of the gravitational force between a single particle, of mass m, and the rest of the particles in the distribution, of essentially mass M (as m is significantly smaller than M, M-m = M), we can see where this expression for potential energy comes from. If we treat the mass M as a point mass at an approximate distance R away, we have: \[F = -\frac{GMm}{R^2}\] Then, we can use the relationship between potential energy and force to find the gravitational potential energy for this particle: \[F = -\frac{dU}{dx}\] \[-\int F\:dx = U(x)\] \[-\int\frac{-GMm}{R^2}dR = U(R)\] \[U = \frac{-GMm}{R}\] If we add the potential energies for each of the N particles, we get a total potential energy of: \[N(\frac{-GMm}{R}) = -\frac{GM^2}{R}\]
(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 \(v_i\) 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 of the system along our line of sight, the velocity scatter \(\sigma^2\). Show that the kinetic energy of the system is: \[K = N\frac{3}{2}m\sigma^2\]
The kinetic energy of an individual particle with mass m and velocity v is given by \(K = \frac{1}{2}mv^2\). As we can't measure the individual velocity vector of each particle, we can instead look at the kinetic energy of the entire system, given by \[K = \frac{1}{2}Mv_{\text{mean}}^2\] As \(v_{\text{mean}}\) is given by \(\sigma\), and M = Nm, we can write this as: \[K = N\frac{1}{2}m\sigma^2\] Lastly, as velocity is a vector, this expression actually gives the kinetic energy of the particle for its motion in one dimension/direction. As this is a three-dimensional system, we must add the kinetic energy for motion in each direction to get a total kinetic energy of: \[3(N\frac{1}{2}m\sigma^2) = N\frac{3}{2}m\sigma^2\]
(c) Use the Virial Theorem to showt hat the total mass of, say, a globular cluster of radius R and stellar velocity dispersion \(\sigma\) is (to some prefactor of order unity): \[M \approx \frac{\sigma^2R}{G}\]
The Virial Theorem gives a relationship between the kinetic and potential energies of a system. As we know the total kinetic energy of the system (\(K = N\frac{3}{2}m\sigma^2 = \frac{3}{2}M\sigma^2\)) and the potential energy of the system (\(U \approx -\frac{GM^2}{R}\)), we can plug these values into the Virial Theorem: \[K = -\frac{1}{2}U\] \[\frac{3}{2}M\sigma^2 = -\frac{1}{2}(-\frac{GM^2}{R})\] \[\frac{3}{2}M\sigma^2 = \frac{GM^2}{2R}\] \[GM^2 = 3M\sigma^2\] \[M = \frac{3R\sigma^2}{G}\] \[M \sim \frac{R\sigma^2}{G}\]

Blog Post 17, Worksheet 6.1, Problem 2

2. The Hubble Classes have characteristic surface light distribution profiles. They are fairly well-described by parametrized equations in the form I(R) where I = intensity and R = distance from the centre. Often each is scaled to their value at the effective radius, such that \(I_e = I(R_e)\). The most general profile is the Sersic Profile, given by the equation \[I(R) = I_e \:\text{exp}(-b_n[(\frac{R}{R_e})^{1/n}-1])\] The constant \(b_n\) depends on the shape parameter n. n = 4 gives rise to the famous \(r^{1/4}\)-law, or the de Vaucoleur Profile, which approximates elliptical and the bulge of spiral galaxies n = 4, on the other hand, is equivalent to a simple exponential profile, which often corresponds with the outskirts of spiral galaxies. The best fit is often obtained by a combination of the functional forms.  

(a) Another way to write the exponential profile is: \[I(R)=I_0\:\text{exp}(-R/b)\] where \(I_0\) is the central surface brightness and b is a characteristic lengthscale, a constant. 

i) Describe what b is, and suggest how one might measure it for a given spiral galaxy. Be as specific as possible about what kind of data and tools may be required. 

From the equation and given value for the Milky Way, we can infer b is the scale radius of the spiral galaxy, or the radius at which the galaxy's intensity has decreased by a factor of e. To measure the scale radius for a given spiral galaxy, one could use a telescope with a CCD to observe the brightness of the galaxy at different radii. From these data points, one could create the light curve for the galaxy across different radii, and find the radius at which the intensity has decreased by a factor of e.

ii) The Milky Way has an estimated b = 3.5 kpc. Plot the disk surface brightness profile relative to the central brightness and indicate the location of the characteristic lengthscale from the centre. 

The Milky Way's exponential profile is given by: \[I(R) = I_0\text{exp}(-R/3.5 \text{ kpc})\]The following is a plot showing the exponential profile of the Milky Way, given a central surface brightness of \(I_0 = 1.0\). (Not the actual value).

iii) Assuming the entire Milky Way (including the central bulge) can be described by this profile, and that it is circularly symmetric. Approximately what fraction of all its stars are interior to the Sun (at 8 kpc)?

To determine the fraction of the Milky Way's stars that are interior of the Sun, we must find the ratio of the intensity at all radii interior the the Sun to the intensity at all radii, from zero to infinity. As we are assuming that the Milky Way is circularly symmetric, the total intensity between the center of the galaxy and an unspecified radius r is given by the intensity equation above integrated around the circumference of a circle at a given radius: \[\int_{0}^{r}(2πR)I_0e^{-R/b}dR,\] with b = 3.5 kpc, and \(I_0\) = the central surface brightness of the Milky Way. Thus, the ratio we are looking for is: \[\frac{\int_{0}^{8}(2πR)I_0e^{-R/b}dR}{\int_{0}^{\infty}(2πR)I_0e^{-R/b}dR}\] Using u-v integration by parts , with u = R and dv = \(e^{-R/b}dR\), we see that the general integral can be expressed as the following: \[\int_{0}^{r}(2πR)I_0e^{-R/b}dR\ = 2πI_0\int_{0}^{r}Re^{-R/b}dR\] \[=\:2πI_0[-Rbe^{-R/b}\rvert_{0}^{r} + b\int_{0}^{r}e^{-R/b}dR]\] \[= 2πbI_0(b-e^{-r/b}(r+b))\] The limit of this expression as r goes to zero is just \(2πI_0b^2\), so our fraction becomes: \[\frac{2πbI_0(b-e^{-r/b}(r+b))}{2πI_0b^2}\] \[\frac{b-e^{-r/b}(r+b)}{b}\] With b = 3.5 kpc and r = 8 kpc, our fraction is equal to 0.666, or approximately 2/3.

Blog Post 16, The Great Debate

The Shapley-Curtis Debate, also known as the "Great Debate," occurred in 1920 between Harlow Shapley and Heber D. Curtis, both American astronomers attempting to understand our place in the universe.

Shapley (left) and Curtis (right)

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

Curtis's model of the universe argued that the universe is composed of many galaxies, of which the Milky Way is only one of many. Within the Milky Way, the Sun is relatively near the center of the galaxy. His main arguments included:

• The Galaxy is relatively small (compared to Shapley's model), with a diameter of only 30,000 light years. 
• Spiral-shaped areas of light that had been observed by telescopes, known as "spiral nebulae," are galaxies themselves, located outside of the Milky Way. 

Curtis's model were based on observations of the optical spectrum of the "spiral nebulae" - he observed that they had very similar spectra to that of the Milky Way, and thus must actually be galaxies themselves.

Shapley's model argued that the entire universe is composed of only our galaxy, the Milky Way. Within this galaxy, the Sun was very far from the center. His main arguments included:

• The entire Galaxy/Universe is approximately 300,000 light years in diameter. 
• Sun is far from the center of the galaxy. 
• "Spiral nebulae" are actually gas clouds within the Milky Way. 

Shapley's ideas were based on observations made of globular clusters. He used the known distance to a particular globular cluster (M13) to estimate the distance to other globular clusters based on each cluster's apparent size, on the assumption that each cluster is roughly the same size.

Using the discovery made by Henrietta Swan Leavitt that the luminosity and period of Cepheid variable stars - stars in which their luminosity periodically brightens and then dims - positively correlate (as the luminosity of the star brightens, the longer the period), Shapley estimated the distance to the Small and Large Magellanic Clouds. He determined that because they were at a distance much within his estimated diameter of the Milky Way, they must be contained within the Milky  Way.

In reality, neither astronomer was entirely correct, nor entirely wrong. In 1923, Edwin Hubble used the 100-inch Hooker Telescope on Mt. Wilson to locate Cepheid variable stars within M31, the Andromeda Galaxy. Hubble determined that the distance to Andromeda was actually around 1.2 million light years - and thus definitely outside of the Milky Way. Curtis was thus correct in arguing that the universe is composed of many galaxies. However, as we know now, Curtis was much more correct than Shapley in determining the location of the Sun within the Milky Way - the Sun is considerably far from the center of the galaxy, at a distance of approximately 8 kpc. 

Blog Post 15, Worksheet 5.2, Cepheid Relations

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. 

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

Below are the data for the 25 Cepheids:

ID number 1505	Max 14.8	Min 16.1	Period(days) 1.25336
1436	14.8	16.4	1.6637
1446	14.8	16.4	1.762
1506	15.1	16.3	1.87502
1413	14.7	15.6	2.17352
1460	14.4	15.7	2.913
1422	14.7	15.9	3.501
842	14.6	16.1	4.297
1425	14.3	15.3	4.547
1742	14.3	15.5	4.9866
1646	14.4	15.4	5.311
1649	14.3	15.2	5.323
1492	13.8	14.8	6.2926
1400	14.1	14.8	6.65
1355	14	14.8	7.483
1374	13.9	15.2	8.397
818	13.6	14.7	10.336
1610	13.4	14.6	11.645
1365	13.8	14.8	12.417
1351	13.4	14.4	13.08
827	13.4	14.3	13.47
822	13	14.6	16.75
823	12.2	14.1	31.94
824	11.4	12.8	65.8
821	11.2	12.1	127

To calculate the mean apparent magnitude for each Cepheid, we add the maximum and minimum apparent magnitudes for each Cepheid, and then divide by two to get: 

ID number 1505	Mean 15.45
1436	15.6
1446	15.6
1506	15.7
1413	15.15
1460	15.05
1422	15.3
842	15.35
1425	14.8
1742	14.9
1646	14.9
1649	14.75
1492	14.3
1400	14.45
1355	14.4
1374	14.55
818	14.15
1610	14
1365	14.3
1351	13.9
827	13.85
822	13.8
823	13.15
824	12.1
821	11.65

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. 

To determine the mean absolute magnitude, we use the equation from Worksheet 5.1, Problem 2: \[m = 5(\text{log}(d) - 1) + M\]

\[M =  5(\text{log}(d) - 1) - m,\] where d = 60,000 pc. If we plug in each mean apparent magnitude at a distance of 60,000 pc, we get the following graph relating mean absolute magnitudes to period: 

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, we we don't need to adjust that axis. Plot the Cepheid mean absolute magnitudes as a function of \(\text{log}_{10}\)(Period). Verify that the plot now looks linear. 

If we plot the logarithm of period along the x-axis, the plot does indeed look linear:

4. Now that the data looks linear, we can estimate the parameters of a linear relation, \(M_V(P) = A\text{ log}_{10}(\text{Period}) + B\). A and B are "free parameters" that allow the function to match the data. 

Using Python's linear regression function, we can determine that A, the slope of the line, is -2.033, and B, the y-intercept, is -2.726. We get a final function of: \[M_V(P) = -2.033 \text{ log}_{10}(P) - 2.726\]

We can plot this on the same graph as before to see the linear approximation:

Blog Post 14, Worksheet 5.1, Problem 2

2. (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?

We know from Problem 1 that the relationship between Star A and Star B's flux and magnitude is given by the following equation.
\[\frac{F_A}{F_B} = 10^{0.4(m_B-m_B)},\] which can also be written as \[\frac{F_A}{F_B} \approx 2.5^{(m_B-m_A)}\]

We know that Star A is 3 magnitudes fainter than Star B, so the quantity \(m_B-m_A\) is equal to 3. We get: \[ \frac{F_A}{F_B} = 2.5^3\]
\[ F_A = 15.625\: F_B\] In order to collect the same amount of energy for Star A as for Star B, you have to observe Star A for 15.625 times the amount of time you would need to observe 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 the relationship between the fluxes and magnitudes of two stars at a distance 10 kpc and a distance d away is given by the following equation:
\[\frac{F_{10 \text{ pc}}}{F_d} = 10^{0.4(m-M)}\] We also know that we can calculate the flux for each star using the following formula: \[F = \frac{L}{4πd^2}\]. From this, we have: \[F_d = \frac{L}{4πd^2}\] \[F_{10 \text{ pc}} = \frac{L}{4π(10 \text{ pc})^2}\] We can plug these values for each star's flux into the previous equation to relate the two: \[\frac{F_{10 \text{ pc}}}{F_d} = 10^{0.4(m-M)}\] \[\frac{\frac{L}{4π(10 \text{ pc})^2}}{\frac{L}{4πd^2}} = 10^{0.4(m-M)}\] \[\frac{d^2}{100} = 10^{0.4(m-M)}\] \[d^2 = 10^{2+0.4(m-M)}\]If we take the logarithm (base 10) of each side of the equation, we get: \[2 \text{log}(d) = 2 + 0.4(m-M)\] \[\frac{2(\text{log}(d) - 1)}{0.4} = m - M\] \[m = 5(\text{log}(d) - 1) + M\] This equation relates the apparent magnitude, m, of a star with absolute value M to its distance, d, away from the observer.

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

From a previous worksheet, we know that the relationship between parallax and distance of an object is given by the equation: \[\text{parallax [arcseconds]} = \frac{1}{\text{distance[pc]}}\] From part (b), we know: \[d^2 = 10^{2+0.4(m-M)}\] \[d = (10^{2+0.4(m-M)})^{1/2}\] We can plug this value for d into the relationship between parallax and distance to get: \[p = (10^{2+0.4(m-M)})^{-1/2}\]