Blog #16: The Great Debate

The Great Debate Between Shapley and Curtis

In 1920, Heber Curtis and Harlow Shapley, two notable astronomers at the time, debated the size and the constitution of the universe. Curtis argued that the universe contained many galaxies, whereas Shapley believed that the universe only contained one galaxy. They each based parts of their arguments on what were then called "spiral nebulae."

Consistent with his views of the universe's size, Curtis thought the spiral nebulae were located outside of our own Galaxy. He thought the Galaxy was small (compared to Shapley's estimate) and that the nebulae, which he believed to be about the same size as the Galaxy, were thus located closer to us than Shapely thought. He also believed that the nebulae were similar in composition to the Milky Way, based on their appearance and their optical spectra, which were the same as that of our Galaxy.

Shapley, on the other hand, argued that they were simply large collections of dust within our Galaxy. Shapley based this idea on a rough calculation of the distance to the Magellanic Clouds, which yielded a distance smaller than the diameter of the Galaxy, and on measured rotational speeds of other nearby objects that would have to be rotating faster than the speed of light if they were outside of the Galaxy (this measured rotation speed turned out to be incorrect).

Neither of the two can really be considered as the "winner" of the debate, since each was right on some points and wrong on other points. Curtis was correct in that the nebulae are external to the Milky Way, but Shapley was right in determining that our Galaxy is larger than previously assumed and that our Sun is not at its center.

Soon after the debate, Cepheids were identified in the Andromeda Galaxy, revealing that it was a galaxy similar to our own, and far enough away to be well outside of the Milky Way. The next decade, the understanding of interstellar absorption of light allowed for better understanding of the sheer scale of the Milky Way and the position of the Sun. We now have better technology, and our understanding of galactic and extragalactic distances, as well as our knowledge of other galaxies far from our own, has improved vastly.

Sources: http://www.nasonline.org/about-nas/history/archives/milestones-in-NAS-history/the-great-debate-of-1920.html

https://en.wikipedia.org/wiki/Great_Debate_(astronomy)

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

http://apod.nasa.gov/diamond_jubilee/1920/cs_why.html

http://encyclopedia2.thefreedictionary.com/Interstellar+Absorption

Blog #15: WS5.1, #3

3. A class of stars named Cepheid Variables reside in a special location in the Hertzsprung-Russell Diagram known as the instability strip. They have properties that make them crucial to the determination of large interstellar distances. 
a) Light curves (i.e. brightness time series) are often useful in studying time-variable phenomena in astrophysics. Below are a set of light curves of Cepheid Variables. Using plotting and fitting programs if necessary, find a rough relation obeyed by all the Cepheids. This relationship should relate the brightness (magnitude) of the Cepheid with another property. In other words, your relationship should have the form \(M_V = A*feature+M_0\), where A and \(M_0\) are variables you should find. 

We can use these light curves to determine the period and the \(M_V\) (mean magnitude) and of each Cepheid. 

Ashley: 

• period: 50 days
• \(M_V\): -5.7

Moiya: 

• period: 15 days
• \(M_V\): -4.4

J Arbanus: 

• period: 20 days
• \(M_V\): -4.8

Yutong: 

• period: 100 days
• \(M_V\): -6.2

Graphing \(M_V\) vs. period appears to give us a logarithmic relationship: 

So we graph \(M_V\) vs. log(period) and get a best-fit line: 

A = -2.17 and \(M_0\) = 1.925, where the "feature" is log(period): 

\(M_V = -2.17log(period)-1.925\)

b) Describe how you would measure the distance to a Cepheid.

Based on the light curve, we should know the apparent magnitude of the Cepheid. We can then plug that into the equation from question 2: 

\(d = D*10^{0.2(m-M)}\) where D = 10pc

c) Below is an image of a Cepheid Variable in the Large Magellanic Cloud. Determine its period, mean absolute magnitude, and distance. 

The period is the distance between the two peaks, which is approximately 5 weeks or 35 days. 

We can calculate the mean absolute magnitude using the equation from part a.

\(M_V = -2.17log(35)-1.925 = -5.28\) 

Using this absolute magnitude, we can calculate the distance using the equation from part b, where m is the mean apparent magnitude of 15.6 and M is the mean absolute magnitude of -5.28. 

\(d = 10*10^{0.2(15.6+5.28)} = 1.5 * 10^5pc \rightarrow 150kpc\)

Blog #14: WS5.1, #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? 

If the fluxes of Star A and Star B are \(F_A\) and \(F_B\), their magnitudes (\(m_A\) and (\m_B\)) should obey the relationship \(\frac{F_A}{F_B} = 2.5^{m_B-m_A}\). Since Star B is 3 magnitudes brighter, \(\frac{F_A}{F_B} = 2.5^3 = 15.6\). We'd need to observe Star A for about 16 times longer to get the same amount of energy.

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 = 10pc. How does the apparent magnitude, m, of a star with absolute magnitude M, depend on its distance, d, from you? 

The flux of the star at distance d would be \(F_d = \frac{L}{4\pi d^2}\), so the flux at the standard distance of 10 pc would be \(F_D = \frac{L}{4\pi (10pc)^2}\). We can then take the ratio of the two and use that to solve for m.
\(10^{0.4(m-M)} = \frac{F_D}{F_d}\) which simplifies to \(\frac{d^2}{D^2}\)
\(0.4(m-M) = log(\frac{d^2}{D^2})\)
\(m-M = 2.5log(\frac{d^2}{D^2})\)
\(m = M+5log(\frac{d}{D})\) where D = 10pc.

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

We can rewrite the equation above to solve for d instead of m.
\(d = D*10^{0.2(m-M)}\)
We can then plug this into the parallax equation, \(\theta = \frac{1AU}{d}\).
\(\theta = \frac{1AU}{D*10^{0.2(m-M)}}\) where D = 10pc.

Blog #13: We Come in Peace

The big news for this week is that NASA has discovered running water on Mars. As the liquid water flows across Mars, it absorbs chemicals from Mars's atmosphere, which in turn lower the water's freezing temperature so that it can stay liquid. Understanding the absorbance spectrum of these dissolved chemicals is how NASA was able to determine that the water was there and that it was liquid. As cool as that is, the question that's on everyone's mind is whether this means there could be life on Mars.

This past summer, I read a book called Alone in the Universe, by John Gribbin (not to be confused with the song of the same name). Gribbin asks the same question, but on a larger scale -- are we the only intelligent life in the Galaxy? Before I read the book, I probably would have said no. The Galaxy is a big place, and since we're here, it means that intelligent life isn't impossible. Statistically speaking, then, it should follow that with sheer quantity of stars, some other intelligence should have evolved.

The story that Gribbin presents, however, is much more complicated. He talks about the Drake equation, which I had heard of before, but he investigates factors that I hadn't ever considered before. (Side note: I read the book a while ago, so unfortunately a lot of this is from memory.) For example, he talked about a habitable region in the Galaxy, not only determined by the density of stars, but also by their metallicities. As we've learned so far in Astro 17, the Milky Way has a greater number density of stars towards the center of the Galaxy than it does closer to the outskirts, so this factor alone cuts down pretty heavily on the sheer numbers I had been counting on initially. Among other factors, he also goes on to talk about the set-up of our solar system; I believe he brought up that our planets are spaced out just right and that having a gas giant like Jupiter positioned properly may be vital to protecting rockier planets from constant bombardment from meteors.

He also goes on to talk about Earth's geology (less interesting to me) and the processes that life itself on Earth had to go through to get to the point where I'd be sitting on a futon in the wee hours of the morning writing an increasingly-existential blog post. Gribbin argues (although I don't think it was initially his idea) that there are certain "checkpoints" that life reaches in its evolution, but that at least one of them acts as a bottleneck, restricting the development of intelligent life. The problem is we don't know if we've already gotten lucky and that that bottleneck is behind us, or whether it'll be basically impossible for us to make it to the next stage of our evolution. (Cue suspenseful music.)

Anyway, since this is starting to move away from astronomy and towards biology (I'm a biologist at heart, sorry), I'll wrap up. Basically, Alone in the Universe presented a lot of interesting ideas about the search for extraterrestrial intelligence that I hadn't considered before. Although Gribbin didn't extrapolate for other galaxies, I'm no longer quite so sure what to think about our own Galaxy. I guess being alone in the Galaxy is pretty incredible in that we've made it this far, but definitely sobering to think about just how alone we are. Either way, I think it's safe to bet that the next update on the Mars situation probably won't look like this.

Blog #12: William Herschel

William Herschel was born November 15, 1738 in Hanover, although he eventually moved to England and did most of his work there. Over his long career, he catalogued over 800 binary star systems and over 2400 nebulae, discovered a planet and four moons in the solar system, and discovered infrared radiation, as well as making his own telescopes (including one that was the longest in the world for around 50 years) and composing over 20 symphonies.

Herschel's 40-foot telescope

His work with binary star systems is particularly interesting. Many at the time believed that the perceived motion was due to the Earth's rotation around the Sun, rather than the actual motion of the stars that were being observed. Herschel, however, hypothesized that the stars in binary systems did move, orbiting each other because of their mutual gravitational attraction. He later confirmed his hypothesis and continued to discover binary and multiple star systems.

During Herschel's time, probably the only binary systems that could be detected were visual binaries -- systems in which the stars are far enough apart to be visually distinguished. However, pairs of stars that are closer together can sometimes be detected by measuring the Doppler effect on the system as each star moves closer or farther from our point of observation. These are called spectroscopic binaries. A third type of binary star system is the eclipsing binary. If the plane of the stars' orbit is properly aligned with the observer such that the stars eclipse each other as they orbit, this will be observable and will indicate that the stars form a system.

Binary star system

He worked closely with his sister Caroline, who acted as his secretary, recording his descriptions of the objects he saw in the night sky. They must have made a pretty good team, because they tracked the relative positions of pairs of stars for 25 years carefully enough that they were able to determine that the stars were orbiting each other. In fact, the first recorded orbit of a pair of stars actually wasn't completed until 5 years after Herschel died.

For these and his other discoveries, Herschel became quite well known as an astronomer. After discovering a planet and trying to name it after King George III, Herschel was knighted (although I'm curious as to whether he got that honor before or after the name "Uranus" was picked instead). He died at 83, but his son John continued his work and eventually became a prominent astronomer in his own right.

"That's Sir William Herschel to you!" 

Sources: 

http://listverse.com/2011/09/08/top-10-most-important-astronomers/

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

https://en.wikipedia.org/wiki/40-foot_telescope

http://www.space.com/17432-william-herschel.html

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

https://upload.wikimedia.org/wikipedia/commons/3/36/William_Herschel01.jpg

https://upload.wikimedia.org/wikipedia/commons/6/62/Herschel_40_foot.jpg

http://www.dailygalaxy.com/.a/6a00d8341bf7f753ef01bb08126160970d-pi

Blog #11: WS4.1, #5

5. Let's think about the time dependence of magnification of a lensing event by considering the lens and the source stars as they move through the Galaxy. The stars have a non-zero relative, projected speed of \(v_{rel}\), which appears as a relative angular motion \(\mu_{rel}\) on the sky. A typical \(\mu_{rel}\) might be 10.5 mas/yr. As a result of this relative proper motion, the projected angular distance between the objects, u, will change with time. We'll call the projected distance of closest approach \(u_0\). It is analogous to an impact parameter. 
a) Give an equation for u(t), the instantaneous projected separation as a function of time, in terms of \(u_0, t_0\), and anything else relevant. \(t_0\) is the time at which the closest projected approach occurs, so \(u(t_0) = u_0\). Recall that u(t) is measured in units of the Einstein radius. 

The source star is moving along the arrow labeled \(\mu_{rel}\), and how far it has moved on that line is \(\mu_{rel}\) multiplied by a ∆t. We know that at \(t_0\), the source star is at its minimum distance \(u_0\), so we can use that time point to calibrate the time point we use. By using the right triangle, we can find the value of u(t). \(\theta_E\) is used to normalize \(\mu_{rel}\). 

\([u(t)]^2 = [u_0]^2+[\frac{\mu_{rel}(t-t_0)}{\theta_E}]^2\)

\(u(t) = [u_0^2+(\frac{\mu_{rel}(t-t_0)}{\theta_E})^2]^{1/2}\)

b) What might be a good timescale (\(t_E\) in this problem? Rewrite part a in terms of this timescale. Express this timescale algebraically, involving 'typical' values for \(M_L, \pi_{rel}\), and \(\mu_{rel}\). Find a typical \(t_E\) we might encounter for microlensing events in the Milky Way. 

A good timescale will simplify the expression and make it easier to tell that the units cancel out, so \(t_E = \frac{\theta_E}{\mu_{rel}}\) 

\(u(t) = [u_0^2+(\frac{t-t_0}{t_E})^2]^{1/2}\)

\(t_E = \frac{2}{c\mu_{rel}}(GM_L\pi_{rel})^{1/2}\) 

\(t_E = \frac{2[GM_{L(typ)}\pi_{rel(typ)}]^{1/2}}{c\mu_{rel(typ)}}(\frac{\mu_{rel(typ)}}{\mu_{rel}})(\frac{M_L}{M_{L(rel)}})^{1/2}(\frac{\pi_L}{\pi_{L(rel)}})^{1/2}\)

\(\theta_E = (\frac{4GM_L}{c^2}\pi_{rel})^{1/2} = (\frac{M}{10^{11.09}M_{\odot}})^{1/2}(\frac{\pi_{rel}^{-1}}{Gpc})^{-1/2} arcsec\) (with help)

If the typical values are: 

\(M_L = 0.3M_{\odot}\)

\(\pi_{rel} = \frac{D_S-D_L}{D_SD_L} = \frac{8kpc-4kpc}{(4kpc)(8kpc)} = 0.125/kpc\)

\(\mu_{rel} = 10.5 mas/yr \rightarrow 10.5*10^{-3}arcsec/yr\)

Then \(\theta_E = (\frac{0.3M_{\odot}}{10^{11.09}M_{\odot}})^{1/2}(\frac{8*10^{-6}Gpc}{Gpc})^{-1/2} arcsec = 5.52*10^{-4}arcsec\)

\(t_E = \frac{\theta_E}{\mu_{rel}} = \frac{5.52*10^{-4}arcsec}{10.5*10^{-3}arcsec/yr} = 0.0525 yr \rightarrow 19 days\)

c) Trace out a microlensing curve. How does changing the the lens mass and \(u_0\) affect the shape of the light curve? 

A is the magnification of the source, and t is the time. The peak occurs when the source is as close to being directly in line with the source as possible. As the mass of the lens increases, the Einstein radius increases (since \(\theta_E = (\frac{4GM_L}{c^2}\pi_{rel})^{1/2}\)), so the time that it takes for the source to pass across the Einstein radius increases, and the peak widens. By increasing the \(u_0\), the source will be farther out of line with the lens, and will therefore not be amplified as effectively; thus the peak will fall. 

Blog #10: WS4.1, #1

1. Mass bends space-time! This is a prediction of general relativity, but fortunately we can heuristically derive the effect (up to a factor of 2) using Newtonian mechanics and some simplifying assumptions. Consider a photon of "mass" \(m_{\gamma}\) passing near an object of mass \(M_L\); we'll call this object a "lens" (the 'L' stands for 'lens', which is the object doing the bending). The closest approach (b) of the photon is known as the impact parameter. We can imagine that the photon feels a gravitational acceleration from this lens which we imagine is vertical. 

a) Give an expression for the gravitational acceleration in the vertical direction in terms of M, b, and G. 

To do this, we can consider two different ways of calculating force and set them equal to each other, then solve for the acceleration. 

\(F = m_{\gamma}a = \frac{GM_Lm_{\gamma}}{b^2}\) 

\(a = \frac{GM_L}{b^2}\)

b) Consider the time of interaction, ∆t. Assume that most of the influence the photon feels occurs in a horizontal distance 2b. Express ∆t in terms of b and the speed of the photon. 

\(v_{photon} = c\) 

\(c = \frac{2b}{∆t}\) (speed = displacement/time)

\(∆t = \frac{2b}{c}\)

c) Solve for the change in velocity, ∆v, in the direction perpendicular to the original photon path over this time of interaction.

We can assume that the photon's gravitational acceleration is constant as it moves past the lens. 

\(∆v = ∆ta = \frac{2b}{c}\frac{GM_L}{b^2} = \frac{2GM_L}{cb}\) 

d) Solve for the deflection angle (\(\alpha\)) in terms of G, \(M_L\), b, and c using the answers from parts a, b, and c. 

\(tan(\alpha) = \frac{∆v}{c}\)

\(tan(\alpha) = \frac{2GM_L}{c^2b}\) 

Because of the small angle approximation, \(tan(\alpha) = \alpha\) 

\(\alpha = \frac{2GM_L}{c^2b}\) 

This result is a factor of 2 smaller than the correct, relativistic result, so \(\alpha = \frac{4GM_L}{c^2b}\)