tag:blogger.com,1999:blog-32890508949741698642024-02-19T07:21:06.927-08:00Looking UpJohnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.comBlogger24125tag:blogger.com,1999:blog-3289050894974169864.post-23542485824691453292011-11-29T00:50:00.000-08:002011-11-29T00:50:18.368-08:00Interview with an AstronomerToday, David and I conducted an interview with <a href="http://www.astro.caltech.edu/people/faculty/Richard_Ellis.html">Richard Ellis</a>, an astrophysics professor at the California Institute of Technology and my adviser. Professor Ellis didn't require much in the way of guiding questions; he covered a lot on his own, so instead of writing this up in a question-answer format, I'm going to try to write up his stories based on my notes. [I may interject every now and then, but rest assured my statements will be safely walled off behind brackets. Other than that, all that follows should be a paraphrase of Prof. Ellis, as transcribed from my notes. I may write my reflection on the interview in another post]<br />
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To start things off, Richard Ellis wanted to be an astronomer since he was about 6 years old. Upon reading a book about astronomy, he was hooked, developing an interest he would continue to have for the rest of his life. This interest continued throughout high school, and consequently, for college he decided to look into universities with undergraduate astrophysics programs.<br />
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At the time, England had a grand total of 4 universities with such programs. Most people who would end up in astronomy studied physics or math for their undergraduate degree, only moving into astronomy for grad school.<br />
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Not-yet-Dr. Ellis decided to study at London University, which featured six people in astronomy in his class, in a program taught by professional astronomers. As his studies progressed, his professors encouraged him to continue his studies in a PhD program in graduate school. A career in research was considered the highest goal among soon-to-be-graduates in science at the time; this has been somewhat supplanted by finance and other high-paying fields in recent times. Furthermore, Dr. Ellis' senior research project inspired him to continue his studies (especially realizing that people could get paid to study astronomy, something he enjoyed immensely).<br />
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So, soon-to-be-Dr. Ellis began graduate studies at Oxford. During this time, he was somewhat disillusioned by the difficulty of the work, and by his lack of interest in his research project, which involved determining the composition of the Sun through the study of the spectral lines of heavy elements. [When he was describing this, I actually thought the topic was kind of interesting, though maybe it wouldn't be fun to actually run the experiments, which involved testing lots of spectral lines for things like titanium gas.]<br />
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At one point, he actually decided to quit astronomy, and applied for positions at IBM, <i>Nature</i>, and an advertising firm. But then, at the last minute, he changed his mind; he stayed on to complete his PhD, and then moved into a postdoctoral position at Durham University, though he notes that this actually offered a little over half the pay he would have made, had he taken the advertising job.<br />
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He stayed at Durham for 19 years, transitioning from postdoc to professor. Then in 1993, he was invited to be a professor at Cambridge, where he stayed until moving to Caltech in 1999.<br />
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[Having concluded his biography, Prof. Ellis moved on to talk about some aspects of academia and research, like funding.] The search for funding for research projects can be constant, and it has gotten markedly worse in the past 5 years. Postdocs and research equipment can be very expensive, so some professors can be applying for grants and funding virtually all the time. Sometimes, postdoctoral or professorial positions will come with a "start-up package," an initial sum of money that can cover funding for a few years.<br />
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Good researchers aren't just good at collecting data: they have to be able to conceive an important, feasible idea. In astronomy, there is a danger of collecting a lot of data to no overall point, lacking a sort of "big picture" goal. Data collection for a good goal will have the potential to push back the frontier of understanding, rather than simply amass more data for its own sake.<br />
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Also, there are other choices researchers must make. If they have PhD students, are they generous enough to share ideas with them? That is to say, if a professor has an idea for a project, is he willing to "donate" that idea for someone else's paper or thesis? Professors may also have to choose between focusing intently on one specific project or area of research, or multitasking between many projects and areas.<br />
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[On the differences between academia in the US and the UK] In the UK, there tends to be more camaraderie between universities, since they're all part of the same system, whereas in the US, there tends to be more competition and rivalry [Can you think of <a href="http://www.caltech.edu/">any</a> <a href="http://mit.edu/">examples</a>?]. Also, access to facilities may be very different, with many US institutions, such as Caltech, having very good access to observatories like Palomar. The situation in Europe is, however, dramatically improving, especially in the UK, Holland, and Germany.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com3tag:blogger.com,1999:blog-3289050894974169864.post-54646573982744288392011-11-20T23:08:00.000-08:002011-11-20T23:08:55.204-08:00Water, Water EverywhereWhile I was looking things up for my previous post about water on Europa, I ran across <a href="http://en.wikipedia.org/wiki/Extraterrestrial_water">this article</a>. It basically summarizes the search for liquid water, both in the solar system and outside it. I haven't looked through the article thoroughly, so I can't really vouch for the accuracy or the up-to-date-ness of its data.<br />
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It did, however, remind me of <a href="http://www.astronomynow.com/news/n1107/27quasar/">another article</a> that I'd read this past summer:<br />
<blockquote class="tr_bq">A water cloud containing the equivalent of 140 trillion times the water held in Earth's oceans has been detected around a quasar powered by a giant black hole 12 billion light years away.</blockquote>Note that this is water vapor, not liquid water or ice. Also note that this cloud is huge: it spans <i>hundreds</i> of lightyears around a supermassive black hole. Pretty cool.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com0tag:blogger.com,1999:blog-3289050894974169864.post-38007930502743826742011-11-20T21:43:00.000-08:002011-11-20T21:43:19.898-08:00EuropaJupiter has 65 satellites, according to the Carnegie Institution for Science. They vary greatly (four of them make up almost all the mass, and Ganymede makes up a third of "almost all" by itself). It is these four, known as the Main Group or the Galilean Moons, that have captured the most interest, perhaps mostly because, as the largest, the are the easiest to see. They are known as the Galilean Moons because they were discovered in 1609 and 1610 by Galileo Galilei, who also famously recorded their orbits around Jupiter, bringing more support to the heliocentric view of the solar system.<br />
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As an aside, Galileo originally wanted to name the moons after the four brothers at the head of the powerful Medici family: Cosimo, Francesco, Carlo, and Lorenzo. This idea, though, proved to be less popular outside Florence, leading to the names we know today: Io, Ganymede, Europa, and Callisto, four lovers of the god Zeus in Greek mythology.<br />
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Europa, a Galilean moon and the sixth from Jupiter, is a bit smaller than Earth's moon, has a surface made up mostly of silicate rock, and probably has an iron core. It even has a thin atmosphere of oxygen. These and other geological features of the moon have led some to speculate that Europa holds large reserves of subterranean water, possibly entire oceans underground. Because of this, Europa is one of the <a href="http://en.wikipedia.org/wiki/Europa_(moon)#Potential_for_extraterrestrial_life">leading candidates</a> in the solar system for potential habitability and the presence of extraterrestrial life (well, at least since Mars let us down).<br />
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Which leads me into this <a href="http://www.jsg.utexas.edu/news/2011/11/scientists-find-evidence-for-great-lake-on-europa/">new development</a>. Geophysicists at UT Austin have possibly discovered a large (about the volume of the Great Lakes), surface lake of liquid water on Europa (and by surface, I of course mean several kilometers below the surface). The lake, like much of Europa's surface is mostly obscured by a thick ice shell, and it's not really known what is below it. A large subsurface ocean is one possibility, but another model claims that the top layer, composed of cold and brittle ice, covers a much thicker layer of warmer, convecting ice.<br />
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As far as I can tell, this new find doesn't necessarily support either model. <a href="http://www.utexas.edu/know/2011/11/16/europa_great_lake/">This simulation</a> of the formation of the lake seems to ignore that which lies far beneath the surface, instead showing how the lake can form just beneath the top ice sheet. It's possible, though, that I'm not interpreting this correctly.<br />
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This certainly isn't enough to get into the idea of actual life on Europa (though apparently some people have even been talking about <a href="http://en.wikipedia.org/wiki/Colonization_of_Europa">colonizing it</a>). Liquid water is, however, a step in that direction; it is also just interesting to learn more about the features of our solar system.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com0tag:blogger.com,1999:blog-3289050894974169864.post-11588769930945965332011-11-15T01:07:00.000-08:002011-11-16T23:25:04.881-08:00Stars and the Virial TheoremThe Virial Theorem is a theorem in mechanics that relates the kinetic energy of particles in a system to the potential of the system. This can be useful when talking about stars, which balance both an immense gravitational collapsing force from their huge mass with the outward force from the internal particles heated from fusion at the star's core.<br />
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The Virial Theorem gives us this:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20E_G=\left%20|%20E_T_h%20\right%20|" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white E_G=\left | E_T_h \right |" title="\bg_white E_G=\left | E_T_h \right |" /></a><br />
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So the magnitude of the gravitational potential energy of the star is equal to the average kinetic (ie, thermal) energy of the electrons in the star's core. Why electrons? This has to do with degeneracy pressure, which is the tendency for particles to try to not occupy the same space at the same time. Think about the core of a star: it's very very dense, with a lots of electrons flying around. Furthermore, the inter-particle spacing is given by the De Broglie wavelength:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\lambda%20=\frac{h}{p}=\frac{h}{mv}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \lambda =\frac{h}{p}=\frac{h}{mv}" title="\bg_white \lambda =\frac{h}{p}=\frac{h}{mv}" /></a><br />
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The smaller the momentum of a particle, the greater the inter-particle distance as given by the De Broglie relation. And electrons, being very tiny, tend to have a very small momentum. When such electrons are packed too tightly, this gives rise to degeneracy pressure, which wants to push them outward, counteracting the inward pull of gravitation. For more on this, check <a href="http://tomesngnomes.blogspot.com/">David's blog</a>. Also, incidentally, I think degeneracy pressure has a hand in the formation of <a href="http://en.wikipedia.org/wiki/Neutron_star">neutron stars</a>: if the degeneracy pressure is unable to hold the collapsing force of gravitation at bay, it will be unable to force electrons to their proper inter-particle spacings. So, instead, it drives the nuclear reaction of <a href="http://en.wikipedia.org/wiki/Electron_capture">electron capture</a>, wherein a proton and electron form a neutron and an electron neutrino. Overall, this results in an extremely dense object made up mostly of neutrons.<br />
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But anyway, I digress. Based on the Virial Theorem and Degeneracy pressure, what can we find out about stars? Think of <a href="http://en.wikipedia.org/wiki/Heisenberg_uncertainty_principle">Heisenberg's Uncertainty Principle</a>:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\Delta%20x\Delta%20p\geq%20\frac{h}{4\pi%20}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \Delta x\Delta p\geq \frac{h}{4\pi }" title="\bg_white \Delta x\Delta p\geq \frac{h}{4\pi }" /></a><br />
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This says that the product of the uncertainty in a particle's position and the uncertainty in it's momentum must be greater than or equal to Planck's constant divided by four pi. How does this relate to the Virial Theorem? Well, we know that the position of a particle should be within the De Broglie wavelength. If that's so, then we can write the number density N as<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20N%20=%20\frac{1}{\lambda%20^3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white N = \frac{1}{\lambda ^3}" title="\bg_white N = \frac{1}{\lambda ^3}" /></a><br />
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since the cube of lambda should correspond to the volume taken up by one electron. Now we can use the uncertainty equation to solve for the De Broglie wavelength, and thereby find N in terms of properties of the star:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\Delta%20x\Delta%20p\geq%20\frac{h}{4\pi}\rightarrow%20\lambda%20m_e\Delta%20v\geq%20\frac{h}{4\pi}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \Delta x\Delta p\geq \frac{h}{4\pi}\rightarrow \lambda m_e\Delta v\geq \frac{h}{4\pi}" title="\bg_white \Delta x\Delta p\geq \frac{h}{4\pi}\rightarrow \lambda m_e\Delta v\geq \frac{h}{4\pi}" /></a><br />
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In order to use the Virial Theorem, it would be nice to get lambda in terms of the kinetic energy of the electrons. We can do this by squaring both sides of the equation:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\lambda^2%20m_e^2\Delta%20v^2\geq%20\frac{h^2}{16\pi^2}\rightarrow%20\lambda%20^2m_e(m_ev^2)\geq%20\frac{h^2}{16\pi^2}\rightarrow%20\lambda^2m_e(KE)\geq%20\frac{h^2}{16\pi^2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \lambda^2 m_e^2\Delta v^2\geq \frac{h^2}{16\pi^2}\rightarrow \lambda ^2m_e(m_ev^2)\geq \frac{h^2}{16\pi^2}\rightarrow \lambda^2m_e(KE)\geq \frac{h^2}{16\pi^2}" title="\bg_white \lambda^2 m_e^2\Delta v^2\geq \frac{h^2}{16\pi^2}\rightarrow \lambda ^2m_e(m_ev^2)\geq \frac{h^2}{16\pi^2}\rightarrow \lambda^2m_e(KE)\geq \frac{h^2}{16\pi^2}" /></a><br />
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Now solve for lambda squared:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\lambda^2%20\geq%20\frac{h^2}{16\pi^2m_e(KE)}%20\rightarrow%20\lambda^3%20\geq%20\frac{h^3}{64\pi^3(m_eKE)^{3/2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \lambda^2 \geq \frac{h^2}{16\pi^2m_e(KE)} \rightarrow \lambda^3 \geq \frac{h^3}{64\pi^3(m_eKE)^{3/2}}" title="\bg_white \lambda^2 \geq \frac{h^2}{16\pi^2m_e(KE)} \rightarrow \lambda^3 \geq \frac{h^3}{64\pi^3(m_eKE)^{3/2}}" /></a><br />
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Now we can find N:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20N%20=%20\frac{64\pi^3(m_eKE)^{3/2}}{h^3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white N = \frac{64\pi^3(m_eKE)^{3/2}}{h^3}" title="\bg_white N = \frac{64\pi^3(m_eKE)^{3/2}}{h^3}" /></a><br />
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Awesome. So now we know the number density of electrons in the star. The next question is, how is this related to the star's mass and radius? Well, the mass and radius are related to each other like this:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20M%20=%20\frac{4}{3}\pi%20R^2\rho" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white M = \frac{4}{3}\pi R^3\rho" title="\bg_white M = \frac{4}{3}\pi R^3\rho" /></a><br />
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where rho is the mass density. What is the mass density, you say? Well, I'm glad you asked:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\rho%20=%20Nm_en" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \rho = Nm_en" title="\bg_white \rho = Nm_en" /></a><br />
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where N is the number density as calculated above, m is the mass of an electron, and n is the number of electrons per atom in the star. for the sake of simplicity, let's assume that the star is made up purely of hydrogen, so n = 1:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20M%20=%20\frac{4}{3}\pi%20R^3\rho%20=%20\frac{4}{3}\pi%20R^3Nm_e%20\rightarrow%20N%20=%20\frac{3M}{4\pi%20m_eR^3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white M = \frac{4}{3}\pi R^3\rho = \frac{4}{3}\pi R^3Nm_e \rightarrow N = \frac{3M}{4\pi m_eR^3}" title="\bg_white M = \frac{4}{3}\pi R^3\rho = \frac{4}{3}\pi R^3Nm_e \rightarrow N = \frac{3M}{4\pi m_eR^3}" /></a><br />
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OK, so where does this get us? Well, N is in terms of the kinetic energy, which from the Virial Theorem we relate to gravitational potential energy, which is in terms of R and M. So, by substituting that in, we can then solve for the relationship between R and M. Now, normally this relationship would be fraught with messy constants, making everything more complicated. But we don't really care about the constants (what are we, mathematicians?). So let's just ignore them, and find out generally how R scales with M:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20KE%20=%20\frac{GM^2}{R}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white KE = \frac{GM^2}{R}" title="\bg_white KE = \frac{GM^2}{R}" /></a><br />
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So, without the constants like G getting in the way, we have<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20KE^{3/2}=N=\frac{M}{R^3}%20\rightarrow%20\frac{M^{7/2}}{R^{3/2}}=\frac{M}{R^3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white KE^{3/2}=N=\frac{M}{R^3} \rightarrow \frac{M^{7/2}}{R^{3/2}}=\frac{M}{R^3}" title="\bg_white KE^{3/2}=N=\frac{M}{R^3} \rightarrow \frac{M^{7/2}}{R^{3/2}}=\frac{M}{R^3}" /></a><br />
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Which gives<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20M^{5/2}%20=%20R^{-3/2}%20\rightarrow%20M^5%20\sim%20\frac{1}{R^2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white M^{5/2} = R^{-3/2} \rightarrow M^5 \sim \frac{1}{R^2}" title="\bg_white M^{5/2} = R^{-3/2} \rightarrow M^5 \sim \frac{1}{R^2}" /></a>Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com4tag:blogger.com,1999:blog-3289050894974169864.post-71934537238831025682011-11-13T16:35:00.000-08:002011-11-13T16:35:59.236-08:00Professional Astronomy IIHow does one become a professional astronomer?<br />
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The process begins, I assume, by getting a Bachelors Degree, probably in physics from some accredited college or university. From what I can tell from perusing a few graduate astronomy program websites (like <a href="http://www.astro.caltech.edu/academics/graduate_admissions_faq.html#prereqs">this</a> and <a href="http://astro.uchicago.edu/gradprogram/index.shtml">this</a>), strong backgrounds in physics are usually expected and preferred, but experience in astronomy or studies in astrophysics are not (necessarily). This is interesting, but I think it makes sense: most colleges do not provide degrees in astrophysics, instead offering a more general physics program, maybe with the possibility of specializing or concentrating in astronomy (this is based on my experiences in applying for undergrad; don't worry too much about sample size, I applied to a lot of schools).<br />
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By asking for a background in physics, universities are probably ensuring that their new grad students have a firm grounding in the basic principles on which astronomy rests; if they don't have to worry about that, then they can jump right into teaching and researching the fun stuff. Though, it should be noted, lack of a really strong physics background does not seem to preclude one from becoming an astronomy grad student; indeed, Caltech's site noted that they sometimes liked to accept some students with varying backgrounds, like math or chemistry.<br />
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I guess I don't really have any data to prove this, but my assumption is that working as a professional astronomer or astrophysicist probably requires a PhD, so I'm basing my statements here on the presupposition that if one wants to be a professional astronomer, one will go to grad school (which we all know is <a href="http://www.phdcomics.com/comics.php">all sorts of fun</a>). My limited exposure to the subject informs me that grad school is composed of research, some classes, research, possibly some teaching, and research; this culminates (hopefully) in the reception of a doctoral degree. (As an aside, I found <a href="http://www.cs.indiana.edu/docproject/grad.stuff.html">this</a> list of useful info for grad students at Indiana University's website, though I'm not really qualified to appraise its value.)<br />
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What happens then? Here are Caltech's statistics:<br />
<blockquote class="tr_bq"><span class="Apple-style-span" style="font-family: Frutiger, Arial, sans-serif; font-size: 14px; line-height: 21px;">Since 1980, of the students departing the Caltech graduate astronomy program, 1% of students withdrew with no degree, 10% left with an MS degree, and 88% graduated with PhDs, typically within 5 years. As of December 1996: of the 42 students who graduated from Caltech with PhDs in Astronomy in the years 1986-1996, 41 are still alive, and 36 (88%) are working in astronomy. Of these, 33% are professors, 31% are permanent scientific staff (mainly at national observatories and laboratories), and 36% (the most recent graduates) are in postdoctoral positions at major astronomical research institutes. Only 5 (12%) are employed outside astronomy. This is a better record than holds at most other major schools.</span></blockquote>I have no idea how typical this is, though Caltech claims to have a "better" record (presumably judged by the percentage of graduates still working in astronomy?) than other major universities. Though I suppose this is also a relatively small and select sample. But let's suppose it's roughly representative, and that it suggests that PhD graduates working in astronomy are likely going to become professors, scientific staff, or postdocs.<br />
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At this point, we've gotten pretty much beyond my knowledge on the subject. I don't fully understand the differences in these positions, or why some people end up in one and others in another. I have heard that professorial positions are increasingly rare relative to the numbers of people with PhD's, and that a lot of these people end up in short-term postdoctoral positions (I think I read an article on this, but unfortunately I can't find it; I'll provide a link if I do).<br />
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My research this past summer involved working with a team comprised of the latter groups: research scientists and postdocs. As far as I can tell, the research portion of their jobs was similar to that for professors: they wrote grant proposals, coordinated with research groups at other universities, and published papers. So the most striking distinction was that teaching was not a part of their job. They were full-time researchers. <br />
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A big factor for anyone doing research in astronomy is telescope time, ie, access to some observatory technology so one can collect new data. This could be difficult, if there are lots of research groups who all need access to an observatory at specific times; astronomical resources are often pricey, and as such, can be rare. This research I helped with required access to the Spitzer Space Telescope, on one of its remaining cryogenic cameras. This is where writing proposals comes in; in order to get access to the telescope to get the data they need, they must submit a proposal outlining what they plan to do and why it's important to do it.<br />
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This brings up one final point; at first thought, many (myself included) would assume that most of what an astrophysicist or astronomer does would involve doing math, and that therefore math is crucial to for professional astronomers and astrophysicists. This is probably true (perhaps especially for theorists), but it ignores another crucial aspect: writing. Writing proposals, writing papers... so much of what professional astronomers do depends on their ability to write at least relatively clearly about what they're doing and why. The better a communicator one is, the more likely (it seems) they will be able to do successful research.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com1tag:blogger.com,1999:blog-3289050894974169864.post-84576429086982479642011-11-09T21:59:00.000-08:002011-11-09T21:59:51.853-08:00Happy Birthday, Carl Sagan<div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/MnFMrNdj1yY?feature=player_embedded' frameborder='0'></iframe></div><br />
Today would be Carl Sagan's 77th birthday. He was an American astronomer and popular science writer, known for such works as <i>Cosmos</i>, <i>The Demon Haunted World</i>, and the novel <i>Contact</i>, which was later adapted into a film. But my personal favorite book of his is <i>Pale Blue Dot, </i>from which the above speech was taken. He is referencing this picture:<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh9MSXUNG-q8GvxyJFbj6DcuMYr4g8YAgpBdO29W5ismP5nC66E6x-efuUb8WeSzMOvEo7IQeS1oiXceCVUYgMxqsgYa4Ube_xIFHwFUm5TMFP-N2oF6DnuGxEU2WBlyWugi7SuzQbtQKY/s1600/pale-blue-dot1.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="313" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh9MSXUNG-q8GvxyJFbj6DcuMYr4g8YAgpBdO29W5ismP5nC66E6x-efuUb8WeSzMOvEo7IQeS1oiXceCVUYgMxqsgYa4Ube_xIFHwFUm5TMFP-N2oF6DnuGxEU2WBlyWugi7SuzQbtQKY/s400/pale-blue-dot1.jpg" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Picture from www.realscience.us</td></tr>
</tbody></table>I don't know how well you'll be able to make it out, but in the line of light on the right-hand side of the picture, there is a little blue dot. That is the Earth, as seen from the <a href="http://en.wikipedia.org/wiki/Voyager_1">Voyager 1</a> spacecraft 6 billion kilometers away, at the edge of the solar system. This joins another poignant picture of the Earth from space, <a href="http://en.wikipedia.org/wiki/Earthrise">Earthrise</a>:<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjpj2sFaPlGov8kkYmt2rgVg73b5zEtZXSvj7PHSLjFk6lobkg7Ppko6amEiNaNi9QGKvLcaHI4BRdxdtqwgyxSNapjrgTiDIiKA-wgiKVU1IL8ZHAMnVM4RlZYRNLs0bOu5TzrDDX-pus/s1600/NASA-Apollo8-Dec24-Earthrise.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjpj2sFaPlGov8kkYmt2rgVg73b5zEtZXSvj7PHSLjFk6lobkg7Ppko6amEiNaNi9QGKvLcaHI4BRdxdtqwgyxSNapjrgTiDIiKA-wgiKVU1IL8ZHAMnVM4RlZYRNLs0bOu5TzrDDX-pus/s400/NASA-Apollo8-Dec24-Earthrise.jpg" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Picture from Wikipedia</td></tr>
</tbody></table><br />
This one was taken by Apollo 8 crewmember Bill Anders, on December 24, 1968.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com2tag:blogger.com,1999:blog-3289050894974169864.post-49250780682721291032011-11-09T21:27:00.000-08:002011-11-09T21:27:21.624-08:00The Initial Mass Function II: The Luminosity Strikes BackContinuing the problem from the previous post (and ignoring the lame Star Wars reference), we now have 3 mass categories of stars in the cluster, and the total masses contained in each of these categories in the cluster.<br />
<br />
Now suppose we want to find the total luminosity of the cluster. If the luminosity scaled with mass the same way for all masses of stars, then this would be easy. We could just use the total mass and convert it to luminosity using that relation.<br />
<br />
It's not quite that simple, though, which is why in the previous post it was necessary to break the stars into categories: different masses of stars have different mass-luminosity relations. Here they are for low, intermediate, and high mass stars, respectively:<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20L=M^5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white L=M^5" title="\bg_white L=M^5" /></a><br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20L=M^3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white L=M^3" title="\bg_white L=M^3" /></a><br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20L=64M" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white L=64M" title="\bg_white L=64M" /></a><br />
<br />
So, with each of the masses calculated in the previous post, we can use each of these relations to find the total luminosity for each category. The sum of these will be the total luminosity of the cluster.<br />
<br />
I left those masses in terms of the solar mass, so finding the luminosity in erg/s will take some calculation. Help me, WolframAlpha, you are my only hope...<br />
<br />
Once the luminosities are found, one can use them and <a href="http://en.wikipedia.org/wiki/Wien_displacement">Wien's Displacement Law</a> to find the maximum-intensity wavelengths of the stars. Because younger stars are generally more luminous than older ones (they have more fuel which they use more wuickly), the average output radiation will be toward the blue end of the spectrum; older, cooler stars tend to be red.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com1tag:blogger.com,1999:blog-3289050894974169864.post-22699836912482609892011-11-08T00:21:00.000-08:002011-11-08T00:21:50.073-08:00The Initial Mass FunctionThe <a href="http://en.wikipedia.org/wiki/Initial_mass_function">Initial Mass Function</a> describes the mass distribution of a population of stars based on their theoretical initial masses (the masses they had at "birth"). The IMF is an empirically derived function; ie, it has been found to be true in numerous actual sets of stars, and was derived from that data, rather than being theoretically predicted. The IMF has this form:<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\frac{dN}{dM}=AM^-^2^.^3^5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \frac{dN}{dM}=AM^-^2^.^3^5" title="\bg_white \frac{dN}{dM}=AM^-^2^.^3^5" /></a><br />
<br />
where N is the number of stars, M is the mass of a star, and A is a proportionality constant.<br />
<br />
Suppose we have discovered a newly formed <a href="http://en.wikipedia.org/wiki/Globular_cluster">globular cluster</a> with a total mass of a million solar masses, and the masses of the stars it contains range from .1 solar masses to 20 solar masses. What can we find out about the cluster given this information?<br />
<br />
First, we ought to find A. The first step is to integrate, so we have N as a function of M. (And we'll be playing a little fast and loose with the rules of differential equations, so don't hate me, mathematicians.)<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\int_{0}^{N}dN=\int_{.1M_S}^{20M_S}AM^-^2^.^3^5dM=\frac{-A}{1.35}M^-^1^.^3^5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \int_{0}^{N}dN=\int_{.1M_S}^{20M_S}AM^-^2^.^3^5dM=\frac{-A}{1.35}M^-^1^.^3^5" title="\bg_white \int_{0}^{N}dN=\int_{.1M_S}^{20M_S}AM^-^2^.^3^5dM=\frac{-A}{1.35}M^-^1^.^3^5" /></a><br />
<br />
from one tenth of a solar mass to 20 solar masses. Don't get too hung up on the limits of integration, because we'll be integrating again right away. This will give us the total mass of the system, which, fortunately, we know:<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20M_T=10^6M_S=\int%20NdM=\frac{-A}{1.35}\int_{.1M_S}^{20M_S}M^-^1^.^3^5dM=\frac{-A}{(1.35)(-.35)}[(\frac{1}{20M_S})^.^3^5-(\frac{1}{.1M_S})^.^3^5]" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white M_T=10^6M_S=\int NdM=\frac{-A}{1.35}\int_{.1M_S}^{20M_S}M^-^1^.^3^5dM=\frac{-A}{(1.35)(-.35)}[(\frac{1}{20M_S})^.^3^5-(\frac{1}{.1M_S})^.^3^5]" title="\bg_white M_T=10^6M_S=\int NdM=\frac{-A}{1.35}\int_{.1M_S}^{20M_S}M^-^1^.^3^5dM=\frac{-A}{(1.35)(-.35)}[(\frac{1}{20M_S})^.^3^5-(\frac{1}{.1M_S})^.^3^5]" /></a><br />
<br />
Substituting in the values for the variables (and rounding a bit, and approximating), we get<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20A%20=%20-6.66%20*10^4^5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white A = -6.66 *10^4^5" title="\bg_white A = -6.66 *10^4^5" /></a><br />
<br />
Cool. Now, suppose we divide the stars into categories: massive (between 8 and 20 solar masses), intermediate (between 1 and 8 solar masses) and low-mass (between .1 and 1 solar masses).What is the fraction of total stars in the cluster from each of these categories?<br />
<br />
First, we can find the total number in each category. Now that we have the value of A, we can use the first integral above (where we found N as a function of M), except changing the limits of integration to be one of the sets of boundaries for a mass category (eg, 8 masses to 20 masses for massive stars). Solving these integrals gives around 3540 massive stars, 77,600 intermediate stars, and 1,767,000 low-mass stars. This makes massive stars about .2 % of the cluster by number, intermediate stars about 4.2 %, and low-mass stars 95.6 %.<br />
<br />
One can also find the total mass contained in each of these categories in this particular cluster by using similar limits on the second integral. This gives about 75,000 solar masses; 280,000 solar masses; and 650,000 solar masses, respectively.<br />
<br />
I will continue discussing the Initial Mass Function in a second post, covering more of this problem.<br />
<br />
Worked out with <a href="http://ay20-mee.blogspot.com/">Mee</a> and <a href="http://tomesngnomes.blogspot.com/">David</a>.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com2tag:blogger.com,1999:blog-3289050894974169864.post-15294705937100940662011-11-07T00:22:00.000-08:002011-11-07T00:22:29.310-08:00Professional Astronomy<blockquote class="tr_bq">"<span class="Apple-style-span" style="font-family: Frutiger, Arial, sans-serif; font-size: 14px; line-height: 21px;">The primary mission of our department is twofold: the performance of cutting-edge research in astronomy and astrophysics, including theory, observation, and experiment, as well as the education of undergraduate and graduate students and training of postdoctoral research associates who will comprise the scientists and leaders of tomorrow."</span></blockquote><span class="Apple-style-span" style="font-family: Frutiger, Arial, sans-serif; font-size: 14px; line-height: 21px;">This is how the Caltech Astronomy Department defines its purpose, as given on its <a href="http://www.astro.caltech.edu/">website</a>. This is, of course, a pretty broad definition, so it can include a lot of different people doing a lot of different things. Furthermore, the teaching faculty page breaks down the astrophysics faculty into theoretical, observational, and experimental astrophysicists; each may encompass a different aspect of what astronomers do (though the more literal-minded among us may consider only observational astrophysics to "really" be astronomy).</span><br />
<span class="Apple-style-span" style="font-family: Frutiger, Arial, sans-serif; font-size: 14px; line-height: 21px;"><br />
</span><br />
<span class="Apple-style-span" style="font-family: Frutiger, Arial, sans-serif; font-size: 14px; line-height: 21px;">So, if one defines a professional astronomer as a person who has a faculty position in astrophysics, the above definition may be pretty sound. But this is not the only avenue for astronomy. For example, I did research this summer with scientists at the Spitzer Science Center. Though the Center is affiliated with and on the campus of Caltech, the people who do research there are not professors; they do not teach courses. So one can be an astronomer outside of (albeit still close to) academia.</span><br />
<span class="Apple-style-span" style="font-family: Frutiger, Arial, sans-serif; font-size: 14px; line-height: 21px;"><br />
</span><br />
<span class="Apple-style-span" style="font-family: Frutiger, Arial, sans-serif; font-size: 14px; line-height: 21px;">The only astronomy research I am personally familiar with is observational. Last summer, as I've mentioned before, I used data from bright, infrared galaxies to draw conclusions about infrared power sources. And a year before that, I used photometry to calculate the rotational periods of asteroids. As a result, I don't really know much about the more theoretical side of astronomy. Learning more about that is one of my main goals for my time here at Caltech.</span><br />
<span class="Apple-style-span" style="font-family: Frutiger, Arial, sans-serif; font-size: 14px; line-height: 21px;"><br />
</span>Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com4tag:blogger.com,1999:blog-3289050894974169864.post-90759116667481191172011-10-31T17:28:00.000-07:002011-11-03T21:02:25.533-07:00Happy HalloweenIt's a ghost!<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhsBPwlhI99FsCJ7EwYIh66K5H81i-rZpb5gNCz_yDDsXU8CgI7Z8Cj6LSoT1v7eKbSe_bp0Q8G1TYlwdGgpawL0vvJvg0xteDgkpbvWfA4ozA86YSPV_jPRxDlvB1hDKMKUY4UB77jRc8/s1600/Ghosts+of+the+Cepheus+Flare.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="326" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhsBPwlhI99FsCJ7EwYIh66K5H81i-rZpb5gNCz_yDDsXU8CgI7Z8Cj6LSoT1v7eKbSe_bp0Q8G1TYlwdGgpawL0vvJvg0xteDgkpbvWfA4ozA86YSPV_jPRxDlvB1hDKMKUY4UB77jRc8/s400/Ghosts+of+the+Cepheus+Flare.jpg" width="400" /></a></div><br />
This is actually known as the <a href="http://apod.nasa.gov/apod/ap111031.html">Ghost of the Cepheus Flare</a>. The Cepheus Flare is a molecular cloud about 1200 lightyears away. This "ghost" is actually a 2-lightyear-wide dust cloud at the very edge of the flare, seen by reflected starlight. The bulges in the cloud's shape are nebula regions; the section of dust on the far right of the image is collapsing in on itself, and is likely in the first stages of forming a binary star.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com2tag:blogger.com,1999:blog-3289050894974169864.post-86862763208938371012011-10-31T00:31:00.000-07:002011-10-31T00:31:24.745-07:00Can You See the Seahorse?<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJ8mAmXYcsCObe_w863Fdbzn0ByvgFYlplliW2jf3wejefGH6LY5IT6-YyRPpAo8wjuFjIupX4Z2FN_8qLpSUeON6qzez3a-McQkgCyoOeFOP_TnRu_fMh6dvKXaSuwLzUxoetGv63ivg/s1600/magellanic.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="266" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJ8mAmXYcsCObe_w863Fdbzn0ByvgFYlplliW2jf3wejefGH6LY5IT6-YyRPpAo8wjuFjIupX4Z2FN_8qLpSUeON6qzez3a-McQkgCyoOeFOP_TnRu_fMh6dvKXaSuwLzUxoetGv63ivg/s400/magellanic.jpg" width="400" /></a></div><div class="separator" style="clear: both; text-align: center;"><br />
</div><div class="separator" style="clear: both; text-align: left;">Above is a picture of a <a href="http://apod.nasa.gov/apod/ap090323.html">seahorse</a> in the <a href="http://en.wikipedia.org/wiki/Large_Magellanic_Cloud">Large Magellanic Cloud</a>, a nearby satellite galaxy of the Milky Way. What looks to us like a seahorse (an example of <a href="http://en.wikipedia.org/wiki/Pareidolia">pareidolia</a>, the same phenomenon that lets us see constellations in groups of stars and the man in the moon) is actually a pillar of dust <i>20 lightyears long.</i></div><div class="separator" style="clear: both; text-align: left;"><i><br />
</i></div><div class="separator" style="clear: both; text-align: left;">The seahorse is very close to another well-known object in the LMC, the Tarantula Nebula:</div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg1qemxw5ZaYwqdq9ioHDoaGooNkqapy1ybrHMsLhUzJN3jGlbuvVPy0LRy3WA2MQNkQFawmOk3gk6-CVpDeEUxKNqfd80X4ikM0a0HjX_CFmbuSSWL-F7tPKfXBoGVYPjlcr8lNwVx_e0/s1600/tarantula2_hst.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="300" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg1qemxw5ZaYwqdq9ioHDoaGooNkqapy1ybrHMsLhUzJN3jGlbuvVPy0LRy3WA2MQNkQFawmOk3gk6-CVpDeEUxKNqfd80X4ikM0a0HjX_CFmbuSSWL-F7tPKfXBoGVYPjlcr8lNwVx_e0/s400/tarantula2_hst.jpg" width="400" /></a></div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: left;">The Tarantula Nebula is very energetic, and is in the process of forming a star cluster, the center of which is just above the cutoff of the Seahorse image. As stars form in the cluster, the resulting solar wind will slowly erode the Seahorse dust pillars over the next million years or so. </div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: left;">I think it is often pictures like this that first get people interested in astronomy. There are several reasons for this. The first is rather obvious: they look really cool. Pictures of astronomical objects can be really artistic, and sometimes breathtaking in their patterns and complexity. This can be enhanced when one is able to see something familiar in the picture, like the Magellanic Cloud's Seahorse. </div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: left;">But then there's another dimension. It's fun to imagine a seahorse in the picture, but it's even more interesting to learn what the image <i>really is. </i>A towering pillar of smoky dust, billowing and reforming as the nearby nebula grows and grasps it ever more tightly. It's a monument taller than the highest skyscraper we will ever build, and it's over 150,000 lightyears away, a distance greater than we could possibly imagine. Yet through the technology of the Hubble Telescope, we are able to gaze into the deep, dark night, and find a seahorse sculpted by gravity and eons. </div>Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com1tag:blogger.com,1999:blog-3289050894974169864.post-68325537970597346812011-10-25T00:25:00.000-07:002011-10-25T00:25:35.754-07:00Nuclear Winter<div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/K-NIwIvdcBM?feature=player_embedded' frameborder='0'></iframe></div><br />
So, in Battlestar Galactica, the Cylons have just launched a nuclear attack on the Twelve Colonies, including the Earth-like world of Caprica. This is, of course, terribly sad, but we will assuage our grief with science.<br />
<br />
Each bomb the Cylons use kicks up a lot of dust into the planet's atmosphere. And increasing the amount of optically thick dust in the atmosphere reduces the sunlight that reaches the planet's surface, which results in lower surface temperatures. Can we figure out the relationship between the amount of dust in the atmosphere and the cooling of the surface?<br />
<br />
OK, so suppose the Cylons drop N bombs, and each bomb moves a mass M of dust into the atmosphere. Each dust particle has a a mass density rho and a radius r (assume a spherical dust particle).<br />
<br />
First, we should find the total number of particles moved to the atmosphere. Note that<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\rho%20=\frac{m_p}{V_p}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \rho =\frac{m_p}{V_p}" title="\bg_white \rho =\frac{m_p}{V_p}" /></a><br />
<br />
where mp is the mass per particle and Vp is the volume per particle. So we can write<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20m_p%20=%20\rho%20V_p=\rho%20(\frac{4}{3})\pi%20r^3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white m_p = \rho V_p=\rho (\frac{4}{3})\pi r^3" title="\bg_white m_p = \rho V_p=\rho (\frac{4}{3})\pi r^3" /></a><br />
<br />
where the units of mp are kg/particle. NM is the total mass of particles. So if we take<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\frac{NM}{m_p}=\frac{NM}{\rho%20V_p}=\frac{3NM}{4\pi%20\rho%20r^3%20}=N_p" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \frac{NM}{m_p}=\frac{NM}{\rho V_p}=\frac{3NM}{4\pi \rho r^3 }=N_p" title="\bg_white \frac{NM}{m_p}=\frac{NM}{\rho V_p}=\frac{3NM}{4\pi \rho r^3 }=N_p" /></a><br />
<br />
we have units of particles, so Np is the total number of particles.<br />
<br />
Next we need to find the fraction of flux that gets through the dust and to the planet's surface. We'll assume that<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\sigma%20=\pi%20r^2" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \sigma =\pi r^2" title="\bg_white \sigma =\pi r^2" /></a><br />
<br />
is the surface area of a particle of dust, and<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%204\pi%20R^2" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white 4\pi R^2" title="\bg_white 4\pi R^2" /></a><br />
<br />
is the surface area of the planet, with R being comparable to the Earth's radius. So then we have<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\frac{F}{F_0}=\frac{4\pi%20R^2-\frac{3NM}{4r\rho%20}}{4\pi%20R^2}=1-\frac{3NM}{16\pi%20r\rho%20R^2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \frac{F}{F_0}=\frac{4\pi R^2-\frac{3NM}{4r\rho }}{4\pi R^2}=1-\frac{3NM}{16\pi r\rho R^2}" title="\bg_white \frac{F}{F_0}=\frac{4\pi R^2-\frac{3NM}{4r\rho }}{4\pi R^2}=1-\frac{3NM}{16\pi r\rho R^2}" /></a><br />
<br />
for the ratio of the flux that reaches the planet's surface to the flux that reaches the planet. Since temperature is related to flux by<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20F=\sigma%20_*T^4" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white F=\sigma _*T^4" title="\bg_white F=\sigma _*T^4" /></a><br />
<br />
we can find the factor relating the original temperature to the new with<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20T=T_0(\frac{F}{F_0})^\frac{1}{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white T=T_0(\frac{F}{F_0})^\frac{1}{4}" title="\bg_white T=T_0(\frac{F}{F_0})^\frac{1}{4}" /></a><br />
<br />
The only problem with this formulation is with the flux ratio. When the second term is greater than one, which occurs for large numbers of bombs or masses of dust, the result is a negative ratio, which doesn't make sense.<br />
<br />
Worked out with<a href="http://tomesngnomes.blogspot.com/"> David</a> and <a href="http://ay20-mee.blogspot.com/">Mee</a>.<br />
<br />
Also, on a semi-related note, if you are a fan of nuclear winter and/or post-apocalyptic fiction, I recommend <i>The Road</i> by Cormac McCarthy.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com0tag:blogger.com,1999:blog-3289050894974169864.post-25051496254398934602011-10-24T01:20:00.000-07:002011-10-24T01:20:10.193-07:00The Big Bang<div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/ItIdeKg5kt4?feature=player_embedded' frameborder='0'></iframe></div><div class="separator" style="clear: both; text-align: center;"><br />
</div><div class="separator" style="clear: both; text-align: left;">If you watch the TV show "The Big Bang Theory," you may be somewhat familiar with this song. </div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: left;">One quibble, though. The members of the band Barenaked Ladies appear to be proponents of the <a href="http://en.wikipedia.org/wiki/Big_Crunch">Big Crunch</a> hypothesis for the <a href="http://en.wikipedia.org/wiki/Ultimate_fate_of_the_universe">Ultimate Fate of the Universe</a>, but all current evidence shows that the universe is not only expanding, but accelerating in its expansion. So, if things keep going the way they are, the universe won't reconstitute itself into a singularity, but will keep expanding outward. Most of the universe will move so far away so fast the it will pass beyond our local event horizon (its light won't be able to reach us because it is expanding away too quickly).</div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: left;">The most commonly held hypothesis for the fate of the universe is the Big Freeze, or <a href="http://en.wikipedia.org/wiki/Heat_death_of_the_universe">heat death</a>. This occurs when the universe achieves maximum entropy, leaving the universe bereft of free energy capable of doing work. Sadly, this would put an end to most of the interesting parts of the universe.</div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: left;">That could be a bit depressing, so let's end with something more optimistic:</div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: left;"><br />
<iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/IOXMjCnKwb4?feature=player_embedded' frameborder='0'></iframe></div>Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com1tag:blogger.com,1999:blog-3289050894974169864.post-16824748490834429212011-10-19T00:48:00.000-07:002011-10-19T02:07:25.361-07:00The Distance to the Sun<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzGmKUW6IVyudBd1wLDxesxqkJWDMX1dn3-qFZ7KBXgLc_klNPaDynjXhTXj4XzoTnT3jEeGi6GpUMC3q30RQ1s78xNSfSr1wU9LPn2e_eTRqzI6aX-JnTHsZbw651xtzrSvNkVzU2wFU/s1600/Mercury2003_combo.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="102" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzGmKUW6IVyudBd1wLDxesxqkJWDMX1dn3-qFZ7KBXgLc_klNPaDynjXhTXj4XzoTnT3jEeGi6GpUMC3q30RQ1s78xNSfSr1wU9LPn2e_eTRqzI6aX-JnTHsZbw651xtzrSvNkVzU2wFU/s400/Mercury2003_combo.gif" width="400" /></a></div><div class="separator" style="clear: both; text-align: center;"><br />
</div><div class="separator" style="clear: both; text-align: left;">The above image is a series of pictures of Mercury passing between the Earth and the Sun. It was taken from a satellite in polar orbit around the Earth. Is there any way we can use information from this picture, and some other basic astronomy knowledge, to find the distance from the Earth to the Sun?</div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGOhBFfy64VIskF5njwNwuyI2kGx-Faa9DUuPI-P3H_EOSyVTeJ343Tah8fcXnjX53M-oPvQY_kBP0QjMtnNlmvzxPHzvvLlUgIYTUNwjx60K_T9IswB8eoNgERkYl1dZ7EX_m1OMzNeA/s1600/AU.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="268" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGOhBFfy64VIskF5njwNwuyI2kGx-Faa9DUuPI-P3H_EOSyVTeJ343Tah8fcXnjX53M-oPvQY_kBP0QjMtnNlmvzxPHzvvLlUgIYTUNwjx60K_T9IswB8eoNgERkYl1dZ7EX_m1OMzNeA/s640/AU.png" width="640" /></a></div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: left;">The above diagram can help us find everything we need to know. We want to find a, a variable that stands for the astronomical unit, or the distance from the Earth to the Sun. Supposing we have only the orbital periods of the Earth and Mercury, and the two pictures above, can we find it? Kepler's Laws and trigonometry say yes.</div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: left;">First, we can use the picture of Mercury's motion to find the angle alpha in the upper left of the diagram. If you look at the picture, you may notice that Mercury's "height" relative to the Sun oscillates up and down. This is due to the movement of the satellite around the Earth: when it's near the North Pole, Mercury's position relative to the Sun appears different compared to when viewed from the South Pole. The angle between Mercury's high and low points is alpha.</div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: left;">We know that the height of the Sun (the diameter) is 0.5 degrees relative to the total sky. By comparing Mercury to the Sun in the above picture, one can estimate Mercury's oscillation in degrees. It's not very big: about 0.0005 degrees. This is alpha. In radians, alpha is 0.00000876. </div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: left;">OK, what next? Take a look at the top triangle, whose angles are alpha, beta, and pi minus theta. Since the sum of these angles is equal to pi (using radians), we have</div><br />
<a href="http://www.codecogs.com/eqnedit.php?latex=0%20=%20\alpha%20@plus;\beta%20-\Theta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white 0 = \alpha +\beta -\Theta" title="0 = \alpha +\beta -\Theta" /></a><br />
<br />
Using trigonometry and small-angle approximations, we have<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=0%20=%20\alpha%20@plus;\frac{R_E}{a}-\frac{R_E}{\Delta%20a}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white 0 = \alpha +\frac{R_E}{a}-\frac{R_E}{\Delta a}" title="0 = \alpha +\frac{R_E}{a}-\frac{R_E}{\Delta a}" /></a><br />
<br />
Solving for delta-a, we get<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\Delta%20a%20=%20\frac{R_E}{\alpha%20@plus;\frac{R_E}{a}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \Delta a = \frac{R_E}{\alpha +\frac{R_E}{a}}" title="\Delta a = \frac{R_E}{\alpha +\frac{R_E}{a}}" /></a><br />
<br />
Since the total distance a is the sum of delta-a and the distance from Mercury to the Sun, we can write<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=a_M=\frac{R_E}{\alpha%20@plus;\frac{R_E}{a}}-a" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white a_M=\frac{R_E}{\alpha +\frac{R_E}{a}}-a" title="a_M=\frac{R_E}{\alpha +\frac{R_E}{a}}-a" /></a><br />
<br />
So, how does this help us? Fortunately, Kepler's Third Law of Planetary Motion says that the square of a planet's orbital period is proportional to the cube of the planet's semimajor axis, which we call a. So we have<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=(\frac{P_E}{P_M})^2=(\frac{a}{a_M})^3=(\frac{a}{\frac{R_E}{\alpha%20@plus;\frac{R_E}{a}}-a})^3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white (\frac{P_E}{P_M})^2=(\frac{a}{a_M})^3=(\frac{a}{\frac{R_E}{\alpha +\frac{R_E}{a}}-a})^3" title="(\frac{P_E}{P_M})^2=(\frac{a}{a_M})^3=(\frac{a}{\frac{R_E}{\alpha +\frac{R_E}{a}}-a})^3" /></a><br />
<br />
We know the radius of the Earth (using 6300 km), and that the Earth's orbital period is 365 days, so if we also happen to know that Mercury's orbital period is 87 days, we can solve for a. Though I must warn you, it involves some pretty hideous algebra. Do not attempt at home.<br />
<br />
But, if you do, you get around 200 million kilometers. How much is it really? About <a href="http://maia.usno.navy.mil/NSFA/IAU2009_consts.html">150 million kilometers</a>. Cool.<br />
<br />
Given this, we can use a refined version of Kepler's Third law to find the mass of the Sun:<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=P^2%20=%20\frac{4\pi%20^3a^2}{G(M_S@plus;M_P)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white P^2 = \frac{4\pi ^3a^2}{G(M_S+M_P)}" title="P^2 = \frac{4\pi ^3a^2}{G(M_S+M_P)}" /></a><br />
<br />
Assuming we're using Earth, we know everything in this equation now except for the mass of the Sun (We can neglect the Earth's mass, as it is ridiculously tiny compared to that of the Sun). Using the values we've calculated, I get<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=5.58*10^2^9kg" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white 5.58*10^2^9kg" title="5.58*10^2^9kg" /></a><br />
<br />
for the mass of the Sun. What is it <a href="http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html">really</a>?<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%201.99*10^3^0kg" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white 1.99*10^3^0kg" title="\bg_white 1.99*10^3^0kg" /></a><br />
<br />
So we end up a bit under on our estimate, coming up with only about 30% of the mass of the Sun.<br />
<br />
Worked on with <a href="http://ay20-mee.blogspot.com/">Mee</a> and <a href="http://tomesngnomes.blogspot.com/">David</a>.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com1tag:blogger.com,1999:blog-3289050894974169864.post-65203496955066662582011-10-18T02:34:00.000-07:002011-10-18T02:34:12.140-07:00The Sun<div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/3JdWlSF195Y?feature=player_embedded' frameborder='0'></iframe></div><div class="separator" style="clear: both; text-align: center;"><br />
</div>For your edification and enjoyment. And now, just for your enjoyment:<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/Bj1AesMfIf8?feature=player_embedded' frameborder='0'></iframe></div>Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com2tag:blogger.com,1999:blog-3289050894974169864.post-83244730345409234592011-10-14T22:32:00.000-07:002011-10-15T12:21:03.986-07:00More on BlackbodiesPreviously on Looking Up, we explored the Blackbody properties of a Y Dwarf orbiting a Sun-like star, and we found the peak wavelength emitted by the Y Dwarf given its temperature. From this, we were able to calculate the photons per second per area from the Y Dwarf that reached an observer 30 lightyears away. But many questions were left unanswered: how many photons came from the Sun-like star? What is the flux ratio of the Y Dwarf to this star? Tonight, the thrilling conclusion:<br />
<br />
<a name='more'></a><br />
<br />
To determine the photons arriving from the Sun-like star (henceforth, for the sake of convenience, referred to as <a href="http://en.wikipedia.org/wiki/51_Pegasi">Pegasi</a>, for a somewhat similar system), one need only repeat the process used for the dwarf, but for a higher temperature. Being Sun-like, Pegasi has a temperature of about 5800 K. So, using the Wein Displacement Law, we get<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%20\lambda%20_m_a_x%20=%204.98%20*%2010^-^7m" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn \lambda _m_a_x = 4.98 * 10^-^7m" title="\bg_white \fn_jvn \lambda _m_a_x = 4.98 * 10^-^7m" /></a><br />
<br />
This is about 500 nm, which corresponds to green light. As a side note, this does not make the star appear green to the human eye (which is good, since it's supposed to be Sun-like, and the Sun is not green). Though the intensity peaks at 500 nm, the high-output portion of Pegasi's emission spectrum covers a wide range of wavelengths, and the visible spectrum is very narrow. This means that there are also plenty of blue and yellow and red photons produced, and when all of the wavelengths of visible light are produced together, you get white light. So, when viewed in the visible spectrum, Pegasi will appear white, not green.<br />
<br />
So, with the maximum wavelength in hand, we can now use the specific intensity formula, the solid angle, and the equation for the energy of a photon to find the number of photons arriving from Pegasi per second per area. Using 2*pi for the solid angle (again assuming the observer receives half the photons), we have<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%20\frac{2c}{\lambda%20^4(e^(\frac{hc}{kT\lambda%20})-1)}=\frac{2kT}{\lambda%20^3h}=1.23%20*10^3^4photons/s(cm^2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn \frac{2c}{\lambda ^4(e^(\frac{hc}{kT\lambda })-1)}=\frac{2kT}{\lambda ^3h}=1.23 *10^3^4photons/s(cm^2)" title="\bg_white \fn_jvn \frac{2c}{\lambda ^4(e^(\frac{hc}{kT\lambda })-1)}=\frac{2kT}{\lambda ^3h}=1.23 *10^3^4photons/s(cm^2)" /></a><br />
<br />
Okay, cool. So, what's the flux ratio of the Y Dwarf to Pegasi? Flux is given in units of energy per time per frequency per area, so unlike specific intensity, it does not depend on the solid angle. To get to this from the values we have now, we need to convert photons back to energy. Which we can do with this equation:<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%20E_\gamma%20=h(\frac{c}{\lambda%20})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn E_\gamma =h(\frac{c}{\lambda })" title="\bg_white \fn_jvn E_\gamma =h(\frac{c}{\lambda })" /></a><br />
<br />
So, after converting our values back into terms of energy per time per area and then dividing to get the ratio, we get<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%206.98*10^-^1^2" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn 6.98*10^-^1^2" title="\bg_white \fn_jvn 6.98*10^-^1^2" /></a><br />
<br />
for the ratio of the Y Dwarf's flux to Pegasi's.<br />
<br />
Once again, this was worked out in collaboration with <a href="http://ay20-mee.blogspot.com/">Mee</a> and <a href="http://tomesngnomes.blogspot.com/">David</a>.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com3tag:blogger.com,1999:blog-3289050894974169864.post-33115266401606829882011-10-13T22:46:00.000-07:002011-10-13T22:46:40.341-07:00Blackbody RadiationAnything with a temperature above absolute 0 (which, considering that the background temperature of space is 2.7 K, is pretty much everything) emits lights with varying degrees of efficiency. Stars, for instance, are some of the most efficient emitters. Ideal emitters (an object that absorbs all light energy that hits it and reradiates that energy in a certain spectrum) reflect no light, and are thus known as blackbodies, and the reradiated energy is called blackbody radiation. Stars are not actually ideal emitters, but they are close enough to roughly approximate one. Some example blackbody spectra are shown on this graph:<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgBfdPK1_zkLfaiwLjTnbPTvYymFBVwSgjOWywKg8Rk4Edw7jyumEga2TZH7URIrXgpNoAPpnXy4HPs9uB9zh3FzL9WxN08NKa9BGTj-gshyphenhyphenVkOMtlUDqkA_wE0ds5WyXCWWc219HfrXQA/s1600/Black-Body-Spectrum.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="270" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgBfdPK1_zkLfaiwLjTnbPTvYymFBVwSgjOWywKg8Rk4Edw7jyumEga2TZH7URIrXgpNoAPpnXy4HPs9uB9zh3FzL9WxN08NKa9BGTj-gshyphenhyphenVkOMtlUDqkA_wE0ds5WyXCWWc219HfrXQA/s400/Black-Body-Spectrum.png" width="400" /></a></div><br />
As you can see from this graph, the blackbody spectrum varies with temperature. If you want to see an example of some of the things you can do with this fact, keep reading.<br />
<br />
<br />
<a name='more'></a><br />
<br />
Suppose you have a system with a Sun-like star and a <a href="http://en.wikipedia.org/wiki/Y_dwarf#Spectral_class_Y">Y Dwarf</a>. This is a kind of Brown Dwarf, which is like a star, only too small to sustain Hydrogen fusion at its core. Consequently, they are much cooler than stars, and Y Dwarfs are the coolest of them all, with temperatures generally less than 500 K (We'll say this one has a temperature of 350 K). To understand how cool this is, consider that the room you're in now is probably 293-298 K.<br />
<br />
Once you've wrapped your mind around this, consider this question: what is the best wavelength of light in which to observe this particular Y Dwarf? Well, if we treat this object as a blackbody emitter, then we can relate it's temperature to its wavelength output. If you look at the graph above, which plots intensity vs wavelength, you can probably guess that the best wavelength for seeing the object is the wavelength at which the intensity peaks.<br />
<br />
The relation between temperature and this maximum wavelength is given by Wien's Displacement Law:<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%20\lambda%20T%20=%200.002897755" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn \lambda T = 0.002897755" title="\bg_white \fn_jvn \lambda T = 0.002897755" /></a><br />
<br />
Well, that's handy. For T = 350 K, we have a wavelength of about 8.3 microns, which falls in the near-infrared electromagnetic spectrum.<br />
<br />
Okay, so now assume that the Y Dwarf has a radius equal to that of Jupiter, and that the nearby star is once again about the same as the Sun. Can we figure out how many photons emitted from the Y Dwarf at the maximum wavelength per second, per unit area, will reach an observer 30 lightyears away?<br />
<br />
To do this, we'll need this equation:<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%20B_\lambda%20(T)%20=%20\frac{2hc^2}{\lambda%20^5(e^(\frac{hc%20}{kT\lambda%20})-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn B_\lambda (T) = \frac{2hc^2}{\lambda ^5(e^(\frac{hc }{kT\lambda })-1)}" title="\bg_white \fn_jvn B_\lambda (T) = \frac{2hc^2}{\lambda ^5(e^(\frac{hc }{kT\lambda })-1)}" /></a><br />
<br />
This is the formula for the specific intensity of the object, which is the energy emitted per time, per area, per frequency, per solid angle. Solid angle is an element of a sphere's surface.<br />
<br />
Now, suppose the observer can see half of the sphere. Thus, we'll assume that that half of the sphere's photons make it to the observer. This corresponds to a solid angle of 2 times pi.<br />
<br />
With that out of the way, we need to change units from energy to photons. These are related by this equation:<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%20E_\gamma%20=%20h(\frac{c}{\lambda%20})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn E_\gamma = h(\frac{c}{\lambda })" title="\bg_white \fn_jvn E_\gamma = h(\frac{c}{\lambda })" /></a><br />
<br />
We can use this to reduce the specific intensity equation to this:<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%20B_\lambda%20(T)%20=%20\frac{2c}{\lambda%20^4(e^(\frac{hc%20}{kT\lambda%20})-1)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn B_\lambda (T) = \frac{2c}{\lambda ^4(e^(\frac{hc }{kT\lambda })-1)}" title="\bg_white \fn_jvn B_\lambda (T) = \frac{2c}{\lambda ^4(e^(\frac{hc }{kT\lambda })-1)}" /></a>\<br />
<br />
This is in units of photons per time per area per wavelength per solid angle. Multiplying by 2 times pi will remove the solid angle component, and we have the wavelength necessary, so we essentially have the relation we want. But we still have that complicated exponential in the denominator. Is there any way to eliminate or simplify that?<br />
<br />
Well, the Taylor expansion of e^x to the first order is 1 + x, where x is the expression e is raised to. If we do this, we get the much friendlier<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%20B_\lambda%20(T)%20=%20\frac{2c}{\lambda%20^4(\frac{hc%20}{kT\lambda%20})}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn B_\lambda (T) = \frac{2c}{\lambda ^4(\frac{hc }{kT\lambda })}" title="\bg_white \fn_jvn B_\lambda (T) = \frac{2c}{\lambda ^4(\frac{hc }{kT\lambda })}" /></a><br />
<br />
Which reduces to<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%20B_\lambda%20(T)%20=%20\frac{2kT}{\lambda%20^3h}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn B_\lambda (T) = \frac{2kT}{\lambda ^3h}" title="\bg_white \fn_jvn B_\lambda (T) = \frac{2kT}{\lambda ^3h}" /></a><br />
<br />
Which is entirely in terms of things we know (k is Boltzmann's constant). Multiply this by 2*pi, and we get<br />
<br />
<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%201.43%20*%2010^(^2^4^)%20photons/s(cm^2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn 1.43 * 10^(^2^4^) photons/s(cm^2)" title="\bg_white \fn_jvn 1.43 * 10^(^2^4^) photons/s(cm^2)" /></a><br />
<br />
Worked out with <a href="http://tomesngnomes.blogspot.com/">David</a> and <a href="http://ay20-mee.blogspot.com/">Mee</a>.<br />
<br />
This was the first two parts of 2.d on the Blackbody worksheet. I will add the next two parts soon.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com0tag:blogger.com,1999:blog-3289050894974169864.post-38764818943981492842011-10-10T01:03:00.000-07:002011-10-10T01:03:51.609-07:00Universe SandboxI recently purchased a program called <a href="http://universesandbox.com/">Universe Sandbox</a> for $5. It is billed as an interactive space gravity simulator, featuring much of the known universe. You can focus on the solar system, nearby stars, the Milky Way, even the Local Cluster. You can click on each major object, and it will provide you with a list of pertinent information: mass, density, radius, orbit eccentricity, etc. That's cool, but it's not the best part. Having selected the pertinent information, you are then free to <i>change it and see what happens!</i><br />
<i><br />
</i><br />
Pretty much nothing is off-limits, not even the gravitational constant. You can add moons, or take them away. You can build your own solar system from scratch, or mess with the one we have. It's pretty awesome.<br />
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The program also can save screenshots, so I'll post images here if I discover anything interesting. So far, the main thing I've learned is that if you increase the radius of the Sun by a factor of 100, the Earth will die a fiery death. But that's not too surprising, so in the meantime, here's a simulation of the collision of the Milky Way and the Andromeda Galaxy.<br />
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<div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/IxxJ2z5ImPU?feature=player_embedded' frameborder='0'></iframe></div><br />
It's due to happen in 4.5 billion years. Mark your calendars.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com1tag:blogger.com,1999:blog-3289050894974169864.post-27425453440575485552011-10-10T00:42:00.000-07:002011-10-10T00:42:52.145-07:00More on ULIRGs<span class="Apple-style-span" style="font-size: large;">My presentation on my research is coming up, and there were some questions after my <a href="http://ay20-john.blogspot.com/2011/09/looking-up.html">first post</a>, so I thought I'd elaborate.</span><br />
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<span class="Apple-style-span" style="font-size: large;">Remember this graph?</span><br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhxTwghbeIy4TQVa2K97gL4U_5m2Bi2uBIrALnvEqcgBDnOCbsTUTUwar_xHZ_531jps88YvXl2OPvrAySvVUnLJGFmqIYKSKNxFFouWFB6UoZ-ygzFTzSOuN1a3xU00UaFz_Eh_wMNVlo/s1600/page0001.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="456" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhxTwghbeIy4TQVa2K97gL4U_5m2Bi2uBIrALnvEqcgBDnOCbsTUTUwar_xHZ_531jps88YvXl2OPvrAySvVUnLJGFmqIYKSKNxFFouWFB6UoZ-ygzFTzSOuN1a3xU00UaFz_Eh_wMNVlo/s640/page0001.jpg" width="640" /></a></div><span class="Apple-style-span" style="font-size: large;">It's called the SED, or Spectral Emission Decomposition. It breaks up the total radiation emitted from a galaxy into its constituent sources. The solid yellow line above all the others is the total emission spectrum, graphed in flux vs wavelength. (Note the wavelength range: near to far infrared)</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">The other lines represent emissions from different sources. The yellow dashed line in the lower left corresponds to light from stars. You can't see it much on this graph, but some have a purple line to represent emissions from Polycyclic Aromatic Hydrocarbons (PAH). These are often a good indicator of widespread star formation.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">The rest represent dust at different temperatures, and the one my project chiefly concerned was the dash-and-dot magenta line in the near-infrared, which is hot dust.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">The greater the fraction of hot dust emissions out of total galactic (observed) radiation, the more an Active Galactic Nucleus (the central supermassive black hole) is probably contributing to the galaxy's luminosity.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">Also notice the data points on the top yellow line. These represent flux measurements taken at specific wavelengths, and they constrain the fit, making the results more accurate (we hope). My main contribution to the project involved adding data points at 3.6 and 4.5 microns. Because it turns out that having constraining data points in that region of the spectrum, where hot dust occurs, can have a dramatic effect.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhRMfr9NudBMXlNvX2ayWpRROEFXp9dX36JffbzGAN4OwjigmiOn1PQEkev3zR26WAbWvk7IYCFsjPGAjdtISOCYTxj9mLmVbCEt1ixdamNEI6z_mcVUpZslr6PqbYSsSx74qTkNsBc9tA/s1600/page0001.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="456" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhRMfr9NudBMXlNvX2ayWpRROEFXp9dX36JffbzGAN4OwjigmiOn1PQEkev3zR26WAbWvk7IYCFsjPGAjdtISOCYTxj9mLmVbCEt1ixdamNEI6z_mcVUpZslr6PqbYSsSx74qTkNsBc9tA/s640/page0001.jpg" width="640" /></a></div><span class="Apple-style-span" style="font-size: large;">This is a fit of the same object, but without the points at 3.6 and 4.5 microns. Notice that the hot dust peak is much lower. This helps demonstrate the significance of collecting those data points.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">Adding the new points doesn't always increase the amount of hot dust, though. It could lower it, or the fraction could stay about the same. But you won't know if your fit at that region is really accurate until you have some actual data backing it up.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">But anyway, I did this process for around 40 objects, including 6 known QSOs (Quasi-stellar objects, likely <a href="http://en.wikipedia.org/wiki/Quasar">quasars</a>). Maybe after my presentation, I'll post something about my results.</span><br />
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</span>Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com2tag:blogger.com,1999:blog-3289050894974169864.post-41414841418058393252011-10-08T22:00:00.000-07:002011-10-08T22:00:32.412-07:00Astronomy at the BeachThis post is a little overdue, as this experiment occurred last week, but whatever. Last Friday, most of the Ay 20 class journeyed to Santa Monica beach, not merely for the cool ocean breeze or the calming sound of waves breaking on the sand, but for science. Read on for the details.<br />
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<b>Abstract</b><br />
The goal of this experiment is to calculate the radius of the Earth to a reasonable amount of accuracy by observing the setting of the Sun. I worked with one of three groups, each taking measurements from different heights. My group observed from the beach, which, for the purposes of our calculations, we define as the surface of the Earth. Though this was essentially a project for the whole class, I specifically worked with <a href="http://ibutsky.blogspot.com/">Iryna</a>, <a href="http://ay20-nathaniel.blogspot.com/">Nathan</a>, <a href="http://ay20-mee.blogspot.com/">Mee</a>, and <a href="http://tomesngnomes.blogspot.com/">David</a>.<br />
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<b>Procedure</b><br />
Data collection for this experiment was relatively simple. Upon arriving at the beach and securing a good vantage point (where the pier wasn't blocking the Sun), we simply waited for the Sun to begin setting. One person (Iryna) observed while lying down. When she first saw the bottom of the Sun touch the horizon, she stood up and we began timing; the timers stopped when she saw the Sun touch the horizon again. We were, however, unsatisfied with this measurement, as the atmosphere made it difficult to tell when exactly the Sun first touched the horizon. So we started our timers again, this time until the Sun disappeared. From this data, we could then calculate the radius of the Earth!<br />
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<b>Solution</b><br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5DU0RHsUfhINYdxc6KBGVQAMzdAdJ86vvaUTITEfvjsqBURFRVH8xcxzjawE3yITidphB3BYSGyua2mHAfzYlzLOdaJS84FR3FRw6H7lC0hY2RFminF1pxn_9E87wNhvEaJP0jrVP2gE/s1600/horizonDiagram.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5DU0RHsUfhINYdxc6KBGVQAMzdAdJ86vvaUTITEfvjsqBURFRVH8xcxzjawE3yITidphB3BYSGyua2mHAfzYlzLOdaJS84FR3FRw6H7lC0hY2RFminF1pxn_9E87wNhvEaJP0jrVP2gE/s1600/horizonDiagram.gif" /></a></div>Starting at the Earth's center, you can draw a line to the point on the Earth's surface to the point of the observer, and another to the point at the observer's horizon. Call this distance R. We can then make a right triangle with the lines R, R + h (where h is the height of the observer when standing up), and a line connecting the point on the horizon with the top of the observer's head. The goal is to calculate R.<br />
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To do this, we'll need the value of the angle created by the two lines from the Earth's center. Call this Theta. We can use the measured time to calculate Theta, if we think of one rotation of the Earth in terms of time. The time we measured while the Sun was setting represents a small fraction of the 24-hour full rotation of the Earth, and it's related to the distance to the horizon, which in on our triangle. What else is on our triangle? The radius of the Earth.<br />
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So, both of our measured times were on the order of ten seconds, which we'll use for simplicity. The ratio of the time we measured to the time in one rotation of the Earth should be equal to the ratio of Theta to one full turn around the Earth in degrees:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%20\frac{\Delta%20t}{24%20*%2060%20*%2060}%20=%20\frac{\Theta%20}{360}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn \frac{\Delta t}{24 * 60 * 60} = \frac{\Theta }{360}" title="\bg_white \fn_jvn \frac{\Delta t}{24 * 60 * 60} = \frac{\Theta }{360}" /></a><br />
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Using 10 s for t, we can solve for Theta in degrees:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%20\Theta%20=%200.042" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn \Theta = 0.042" title="\bg_white \fn_jvn \Theta = 0.042" /></a><br />
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We then have an equation relating Theta to R and h by way of cosine, since cosine of Theta represents the ratio of the side of a triangle adjacent to Theta to the triangle's hypoteneuse:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%20\cos%20\Theta%20=%20\frac{R}{R%20@plus;%20h}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn \cos \Theta = \frac{R}{R + h}" title="\bg_white \fn_jvn \cos \Theta = \frac{R}{R + h}" /></a><br />
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Assuming h = 2 meters (definitely an overestimate, as none of us are very tall), then R is the only unknown variable. So, with some simple algebra, we can solve for the radius:<br />
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<a href="http://www.codecogs.com/eqnedit.php?latex=\bg_white%20\fn_jvn%20R%20=%206.67%20*%2010^{6}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_white \fn_jvn R = 6.67 * 10^{6}" title="\bg_white \fn_jvn R = 6.67 * 10^{6}" /></a><br />
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This value is in meters, and is equal to 6670 kilometers. How close is this to the real value? The <a href="http://asa.usno.navy.mil/SecK/2011/Astronomical_Constants_2011.txt">Astronomical Almanac</a> puts the Earth's radius at 6378.1 km. Not too bad!Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com2tag:blogger.com,1999:blog-3289050894974169864.post-20754404387454812122011-10-05T13:19:00.000-07:002011-10-05T13:19:49.682-07:00Eric Idle and Astronomy<div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/vIy76M-4txo?feature=player_embedded' frameborder='0'></iframe></div><br />
If you haven't seen Monty Python's <i>The Meaning of Life</i>, I highly recommend it.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com2tag:blogger.com,1999:blog-3289050894974169864.post-48397720460665867382011-10-05T02:28:00.000-07:002011-10-05T02:28:45.801-07:00Tau Ceti and Declination<span class="Apple-style-span" style="font-size: large;">Partners:<a href="http://ay20-nathaniel.blogspot.com/"> Nathan</a>, <a href="http://ay20-mee.blogspot.com/">Mee</a>, <a href="http://ay20-eric.blogspot.com/">Eric</a></span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">Imagine you are standing outside on a clear night, far away from any sources of light pollution, yet also in the relative vicinity of Palomar Observatory (this will be important later). When you look up, you see a vast array of stars spread out before you.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">Now, before you get too distracted by the splendor of the night sky, turn to face north, and find Polaris. Polaris is the North Star, and it is fixed in the heavens (for the next several thousand years, anyway), shining above the northern horizon.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">If you watched the area around Polaris for the whole night, you would see the sky appear to rotate around it, with the paths of nearby stars tracing (parts of) circles in the night around the unmoving North Star.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">You see, the Earth’s axis of rotation points toward Polaris; therefore, as the Earth rotates during the night, the angle between Polaris, the point of observation, and the observed star changes. So the star in question appears to be at a different point in the sky.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">If you shift your observation of the sky to a position south of the area around Polaris, you’ll notice that the stars in this area trace larger circles. If you were able to watch their paths for 24 hours (the Sun tends to be a problem here), you would perhaps see that part of the stars’ paths are blocked by the Earth; their movements take place “behind” the horizon.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">Why do the paths get longer as your gaze drifts south? This is a function of declination. This is essentially the same as latitude, and is measured in degrees north or south of the celestial equator (the celestial equator is simply the plane of the Earth’s equator extended out into space). Palomar is at about 30 degrees North; any object directly above it in the sky will have a declination of 30 degrees N (or an object directly above any point on the 30-degree latitude line). </span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">You can also think of the declination as an angle whose vertex is at the Earth’s center, with one line on the celestial equator and one connecting the center of the Earth to the star in question. So Polaris, on the line of Earth’s rotational axis, is perpendicular to the celestial equator. Thus, Polaris’ declination is 90 degrees.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">Take another star, Tau Ceti. It has a declination of negative 15 degrees, so it is in the Southern Hemisphere. But it’s close enough to the declination of the equator (0 degrees) that it should be visible in parts of the Northern Hemisphere. Like, say, Palomar.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">Suppose that from Palomar’s 30 degrees N, any object with a declination >30 degrees is visible in the sky throughout the day. Everything between -30 and 30 is visible at some times but not others. Everything below -30 is never seen in the sky above Palomar.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">Tau Ceti is within the sometimes-visible interval. So, when can we see Tau Ceti? Well, obviously we can’t see it during the day, because there’s an annoyingly close star whose light masks the presence of other stars during the daytime. At this latitude, the Sun is visible for just over 14 ours at the summer solstice, and just over 10 at the winter. Give it an extra two hours for the sky to darken, and you have possibly eight hours of time to see Tau Ceti in the summer and 12 in the winter.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">But there is another factor to consider here. As the Earth moves in its orbit around the Sun, the time at which an object reaches the meridian (the highest point it reaches in the sky, and the best point for viewing) changes. Because, when the Earth rotates, it takes 360 degrees to complete one full rotation. But because it is also moving in space, after one full rotation, it must then rotate one extra degree to bring the starting point of the measurement in line with the sun. This extra rotation takes about 4 minutes. Accordingly, a given distant star will rise 4 minutes earlier each subsequent night. </span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">Tau Ceti’s Right Ascension is 1 hour and 44 minutes, so about 2 hours. This is the time after noon (set at 0:00 hours) that a star will reach the meridian in the sky on the Vernal Equinox (known by some as March 20). So in late March, Tau Seti will be in prime viewing position around 2 pm. This is obviously a problem, as the Sun is also in prime viewing position around 2 pm. So when will Tau Ceti peak at a better time?</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">Every 30 days is about 120 minutes of change, or 2 hours. So around April 20, Tau Ceti will peak around noon. Not better. But by August, it will be peaking at around 4 in the morning. Tau Ceti will reach meridian at midnight in October.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">So, if we assume that, because of its declination, Tau Ceti is best viewed around its meridian peak, when it can’t be hidden by the horizon, and that Tau Ceti, like most stars, is best viewed at night, then we can estimate the best times for viewing Tau Ceti as early fall to early winter.</span><br />
<span class="Apple-style-span" style="font-size: large;"><br />
</span><br />
<span class="Apple-style-span" style="font-size: large;">Possibly more to follow, including graphs, if I have time.</span>Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com5tag:blogger.com,1999:blog-3289050894974169864.post-46428967624891483862011-10-02T12:39:00.000-07:002011-10-02T12:39:00.179-07:00Supermassive Black Hole<div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/bBb-J0hcBQA?feature=player_embedded' frameborder='0'></iframe></div><br />
I mentioned supermassive black holes briefly in my last post as one of the potential causes of the power outputs of ultraluminous galaxies. However, I don't actually know a lot about them, so I went to my favorite place for quick information, Wikipedia.<br />
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<a href="http://en.wikipedia.org/wiki/Supermassive_black_hole"><span class="Apple-style-span" style="color: blue;">http://en.wikipedia.org/wiki/Supermassive_black_hole</span></a><br />
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According to this esteemed database (don't worry, the article appears to be properly referenced), while stellar black holes have masses on the order of those of large stars (makes sense), supermassive black holes may have millions of solar masses. More interesting, though, is the claim about the density of supermassive black holes:<br />
<div><br />
</div><blockquote><span class="Apple-style-span" style="background-color: black; font-family: sans-serif; font-size: 13px; line-height: 19px;">The average <a href="http://en.wikipedia.org/wiki/Density" style="background-attachment: initial; background-clip: initial; background-image: none; background-origin: initial; color: #0645ad; text-decoration: none;" title="Density">density</a> of a supermassive black hole (defined as the mass of the black hole divided by the volume within its <a href="http://en.wikipedia.org/wiki/Schwarzschild_radius" style="background-attachment: initial; background-clip: initial; background-image: none; background-origin: initial; color: #0645ad; text-decoration: none;" title="Schwarzschild radius">Schwarzschild radius</a>) can be much less than the density of <a href="http://en.wikipedia.org/wiki/Water" style="background-attachment: initial; background-clip: initial; background-image: none; background-origin: initial; color: #0645ad; text-decoration: none;" title="Water">water</a> (the densities are similar for 10<sup style="line-height: 1em;">8</sup> solar mass black holes<sup class="reference" id="cite_ref-CMS1999_4-0" style="font-style: normal; line-height: 1em;"><a href="http://en.wikipedia.org/wiki/Supermassive_black_hole#cite_note-CMS1999-4" style="background-attachment: initial; background-clip: initial; background-image: none; background-origin: initial; color: #0645ad; text-decoration: none; white-space: nowrap;">[5]</a></sup>). This is because the Schwarzschild radius is <a href="http://en.wikipedia.org/wiki/Proportionality_(mathematics)" style="background-attachment: initial; background-clip: initial; background-image: none; background-origin: initial; color: #0645ad; text-decoration: none;" title="Proportionality (mathematics)">directly proportional</a> to <a href="http://en.wikipedia.org/wiki/Mass" style="background-attachment: initial; background-clip: initial; background-image: none; background-origin: initial; color: #0645ad; text-decoration: none;" title="Mass">mass</a>, while density is inversely proportional to the volume. Since the volume of a spherical object (such as the <a href="http://en.wikipedia.org/wiki/Event_horizon" style="background-attachment: initial; background-clip: initial; background-image: none; background-origin: initial; color: #0645ad; text-decoration: none;" title="Event horizon">event horizon</a> of a non-rotating black hole) is directly proportional to the cube of the radius, average density decreases for larger black holes, being inversely proportional to the square of the mass.</span></blockquote>The Schwarzchild radius is the distance from the center of an object such that, if an object's total mass is contained within the sphere of the Schwarzchild radius, the necessary escape velocity would be the speed of light. Which, unless you are some sort of rule-breaking neutrino, is <a href="http://www.wired.com/geekdad/2011/09/neutrinos-and-the-speed-of-light-a-primer-on-the-cern-study/"><span class="Apple-style-span" style="color: blue;">against the law</span></a>.<br />
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A black hole occurs when the radius of the object is less than that of its Schwarzchild radius. Fortunately for us, most things (like, say, the Milky Way) are not so compact.<br />
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But, the larger the Schwarzchild radius, the less compact (ie, dense) and object needs to be to still fulfill the criteria to be a black hole. And since the Schwarzchild radius is proportional to mass, the more massive an object, the less dense it needs to be. Wow.<br />
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Further <a href="http://hubblesite.org/explore_astronomy/black_holes/home.html"><span class="Apple-style-span" style="color: blue;">links</span></a> in <a href="http://apod.nasa.gov/htmltest/gifcity/bh_pub_faq.html"><span class="Apple-style-span" style="color: blue;">case</span></a> you're <a href="http://www.washingtonpost.com/wp-dyn/content/article/2007/10/30/AR2007103002073.html?nav=most_emailed"><span class="Apple-style-span" style="color: blue;">interested</span></a>.Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com2tag:blogger.com,1999:blog-3289050894974169864.post-8463758720514887502011-09-27T23:38:00.000-07:002011-09-27T23:38:51.432-07:00Looking Up<div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.youtube.com/embed/RQhNZENMG1o?feature=player_embedded' frameborder='0'></iframe></div><br />
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The title of this post, and of my blog, comes from this talk by astronomer Neil DeGrasse Tyson, in which he defends the importance of maintaining the space program. It is important, he says, for people to start "looking up," to realize that there is so much more out there than the things people worry about day to day.<br />
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I don't mean to inflate the importance of astronomy, or to belittle the real and significant problems facing the world today. The point, I think, is that discovery is never a waste of time.<br />
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For example, I did research at Caltech over the summer, and I'll be the first to admit that I don't think my results have any practical value. But I (and hopefully you!) don't see it as a waste of time.<br />
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I did work studying Ultraluminous Infrared Galaxies (or ULIRGs), objects emitting over a trillion times the luminescence of the sun, mostly in the infrared (as you may have guessed). This huge radiation output is generally the result of two things: the accretion of mass by a galaxy's central, supermassive black hole, or the widespread formation of new stars. Both of these phenomena are often caused by merging systems; that is, this is what happens when galaxies run into each other at top speed. Awesome.<br />
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</div>But there's a problem. The galaxies we're studying contain massive amounts of dust floating around. This dust intercepts the radiation from the black holes and new stars, and re-emits in the infrared. So, if you wanted to know whether a particular galaxy was powered mostly by a central black hole or by new stars, how would you find out? By looking at this:<br />
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<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEipwIYIDyDm10CgxAsBkd2s-F_LAgv_GBV-r_-Z5cC8JQwWLRJqdp7ldNs7pD1RqgsOV5aKS0DpqtK7BP6CNqHhm2ksSZWul2qrzItM52z53wVZUHysdR9el7guqLho2DobkP8fd1yFhEA/s1600/page0001.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="228" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEipwIYIDyDm10CgxAsBkd2s-F_LAgv_GBV-r_-Z5cC8JQwWLRJqdp7ldNs7pD1RqgsOV5aKS0DpqtK7BP6CNqHhm2ksSZWul2qrzItM52z53wVZUHysdR9el7guqLho2DobkP8fd1yFhEA/s320/page0001.jpg" width="320" /></a></div><div class="separator" style="clear: both; text-align: left;">The graph is the result of a program that breaks down the emission spectrum into several sources, including stellar light and various temperatures of dust. A galaxy powered by a black hole results in a different decomposed spectrum than one mostly powered by star formation.</div><div class="separator" style="clear: both; text-align: left;"><br />
</div><div class="separator" style="clear: both; text-align: left;">Now, this sort of information has pretty much no impact on everyday life. I don't really see how it could. But we don't seek to learn only to solve the problems standing immediately before us, but also to drive a process of discovery that I think enriches society. Literature won't feed the hungry, but surely there is no doubt of its cultural value. A value that is, I think, similar to that of looking up.</div>Johnhttp://www.blogger.com/profile/05368870183761814217noreply@blogger.com1