tag:blogger.com,1999:blog-58492104314022340732024-03-12T23:11:45.529-07:00StarstruckJuliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.comBlogger26125tag:blogger.com,1999:blog-5849210431402234073.post-65900339676373866152012-08-22T14:28:00.004-07:002012-08-22T14:28:58.394-07:00Fit those spectra!!!Ignoring the continuum normalization (that part of my code still has issues... more on that later) and giving my "data" spectra (yellow) a doppler shift of 15 km/s, SNR (signal-to-noise ration) of 3000, and a line broadening of 0.1:<br />
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Before any fitting:<br />
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So, what I do in the fitting is I transform the model as if it had been broadened and shifted by some given amount, then evaluate the "likelihood" (this process is the Bayesian methodology that allows minimizing chi squared as a fitting mechanism...) at those values. Then, the Levenburg-Marquardt algorithm finds the highest likelihood value, which should be the correct values of doppler shift and line broadening.<br />
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After cross-correlation (to find an initial guess) and Levenburg-Marquardt:<br />
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<span style="text-align: start;">Let's zoom in on that fit:</span></div>
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What??? I can't see the data, the values for the transformed model are practically the same!! The fit is too good! Where is it? Zoom in further to see the slight discrepancy due to noise:<br />
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<br />
<br />
Fitting algorithm estimates a doppler shift of 14.9999878 (I expected 15) and a line broadening of 0.0999971716 (with no prior information!! I started the line broadening at 0, and expected to get 0.1).<br />
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Not bad!<br />
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<br />Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com0tag:blogger.com,1999:blog-5849210431402234073.post-87244325831736430302012-08-08T14:57:00.001-07:002012-08-08T14:57:10.285-07:00Analogous Inspiration<br />
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When I was a young kid (4-10), I wanted to be an astronaut.
Then, suddenly, my ‘dream’ changed – I started telling people that I wanted to
be a lawyer. I always thought that the reason for this change was that my dad
always said that being a lawyer was a good career, but going through old papers
while I was at home made me realize maybe this wasn’t the whole story.</div>
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<o:p></o:p></div>
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I found a “newspaper” that I had written in fourth grade.
Our class had a variety of student-run “businesses”: some people made crafts
out of paper and traded them for rocks others collected on the playground. To
be a part of this economy, I wrote a sporadic “newspaper,” proudly typed up in
Microsoft Publisher and printed on long paper (to be more authentic), which I
traded to get my share of rocks and paper bracelets. The teacher tried to stop
us, as we used a lot of classroom supplies, but, I digress…<o:p></o:p></div>
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One of these “newspaper” “issues” contained a story about
the space shuttle Challenger and its 1986 explosion. Reading it now, I am
brought back memories of how shocked I was when I discovered that this had
happened, that seven people had died. Even though I was reading about it
fourteen years after the fact, I felt as if those people had just died. With my
father away in Afghanistan, I was profoundly struck by the mortality of these
people I saw as heroes (no doubt there was a fearful analogy in my fourth-grade
mind to my father, who I also saw as my hero). <o:p></o:p></div>
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I am reminded of this shred of my past by the resoundingly
successful landing of Curiosity on Mars, an emotional event I watched while
observing at Palomar Observatory. Early in the evening on the day of the
scheduled landing I excitedly texted my uncle, whose ten-year-old son has
repeatedly told me that he wants to go to Caltech someday, to let him know
about the landing.<o:p></o:p></div>
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<br /></div>
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After the message was sent, however, I worried, perhaps
irrationally, about what would happen if Curiosity were to crash to the surface
in a fiery mess instead of land as it was supposed to. Would my cousin lose
faith in Caltech? NASA? Science? I remembered all too clearly my own first
impression of science’s imperfections, and I was worried that were Curiosity to
fail, it would have some momentous effect on my cousin’s life plans. <o:p></o:p></div>
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My fears were unfounded. Curiosity landed on the surface of
Mars safely, and as I watched the jubilant celebration of the involved
scientists and engineers on a laptop, in the data room of the Hale 200-inch, I
imagined my cousin watching this celebration and how much this success would
inspire him. I imagined him tracking the progress of the MSL over the next
weeks, months, or even years as I voraciously consumed everything I could about
Voyager when I learned about it as a child. I was fascinated with the ideal
that somewhere, something we humans had made was doing science we couldn’t do
ourselves. As I traced my fingers along the a map of our solar system with
Voyager’s trajectory overlaid, so I hope will my cousin will someday have a map
of Mars along which to trace a similar path. <o:p></o:p></div>
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Curiosity is inspirational. Now, I don’t need an inspiration
to spur me towards science, having found my own way, but the awe Curiosity
inspires might lead many children to follow it into a field of science. <o:p></o:p></div>
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There is also something else that I have realized in the ten
years since I wrote that sad article about Challenger. There can be no progress
without risk. Those imperfections that manifest as disaster are not failure,
but merely poorly aimed steps forward. <o:p></o:p></div>
<div class="MsoNormal">
<br /></div>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com1tag:blogger.com,1999:blog-5849210431402234073.post-78184425329280306502012-08-01T11:49:00.003-07:002012-08-01T12:17:00.450-07:00Coding (more) Efficiently in Python! Part 1: prunSince it’s been a while, I think I will write about how to make your python code more efficient! (Also, writing this up means I can stop being stuck on my project for a few minutes… hence, thus I posit my ulterior motive…)
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<br />
Having never had formal training in coding (barring a high school course, in which we spent several months on ‘Hello World’…. So, no formal training), I am what I term a ‘functional coder.’ I can write code. But that’s about all. It might run and get the job done, but when you look at the actual raw code you will see recursive for loops and a stark lack of commenting, variable names like ‘bunny’ and ‘widget’ (hey, it seemed cute at the time! Who knew I would need to come back to this code later?) and almost every other mark of a rookie you can think of. This summer, this has begun to haunt me; I need to run little chunks of code thousands of times as I approach my correct parameters in a Markov Chain Monte Carlo (this process I am also trying to make more efficient, but, more on that later).
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<br />
I started out knowing very little about making code efficient. Well, I did realize in Ay117 (the astrostats course I took this past spring) that my five-deep for loops could cause laughter when shown to others. I don’t want to be a comedian, though: I want to be a competent coder!
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<br />
Tim (a graduate student in Professor Johnson’s group) showed me a couple cool tricks earlier this summer in making code more efficient, all of which I have been using ever since! I will show here my personal favorite diagnostic tool, prun.
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prun is a function that is part of the ‘core.magic’ module (so, no need to import anything! It’s super easy!). You call is by simply writing, either on your command line on in a developing environment:
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<br />
%prun functionname
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When functionname is the function you want to test. There are also some options that you can insert as follows, when 'o' is the identifier of your option:
<a api="api" generated="generated" highlight="prun#IPython.core.magics.execution.ExecutionMagics.prun" href="http://www.blogger.com/=" http:="http:" ipython-doc="ipython-doc" ipython.org="ipython.org" stable="stable" ython.core.magics.execution.html="ython.core.magics.execution.html">
documentation of prun.</a>
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This is what you might see when running prun from the command line.
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For me, since I will be calling this function many, many times, I want to try and eliminate whatever is taking the most time. I see that the total time for this function is 1.584 seconds, which seems pretty good, right? Only two seconds? You can wait for two seconds for your function to run, right?
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<br />
That is what I thought at first, too. 1.5 seconds seems fine at first glance, but the nature of my project is such that I will need to run this particular method hundreds or even thousands of times. Suddenly, a couple seconds isn’t sounding so good anymore… luckily, prun separates out for me which functions are taking the most time to run, and how many times they are being called. By looking at this chart generated by prun, I can see that the functions taking the most time are the interpolation and shift_lhood, I function I wrote. There are two easy ways to fix that! Firstly, you might notice that the interpolation it being called 11 times. By examining the code, I notice that the same function is being interpolated 11 times. Silly me, I hadn’t realized that I did that… well, that’s an easy fix!
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<br />
The second thing I can do to improve the efficiency of this function is to improve my shift_likelihood function. The old version took, as according to the profile shown above, 0.121 seconds PER CALL to run (see the fourth row, fifth column). Here is the improved version of that function, shift_lhood, called on its own. To improve it, I tried using matrices instead of for loops to do element by element manipulations. Now it only takes 0.093 seconds! What an improvement!
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prun is a useful function, and it’s my first step on the long journey to becoming a better coder (and not spending days and unneeded days running code)!Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com2tag:blogger.com,1999:blog-5849210431402234073.post-71355537416059786632012-06-21T13:21:00.002-07:002012-06-21T13:32:16.024-07:00I miss my running buddies!!Without Clara and Stephanie (off at Berkeley and in Florida, respectively) I've been running on my own a lot recently... and when I say "recently" I mean "in the last three days." The loneliness is getting to me already!!<br />
<br />
Fun run today, I thought it would be 6 miles, but apparently it was not. It was pretty safe, didn't run through too many sketchy areas (I think Orange Grove is a pretty road).<br />
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<a href="http://www.mapmyrun.com/routes/view/105245019/"> See the route I ran! East Orange Grove on MapMyRun. </a>
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I also started my SURF this week, with a (luckily) very direct application of the methods I learned last semester in Ay 117! So far, I have been writing code that generates a mock noisy spectra of a B-Star (which is basically a very big star!), then playing with this spectra and attempting to extract from it the shift I put it. This part I've done.<br />
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Now, the hard part - I need to write code to extract a doppler shift. This is tricky because a doppler shift or a given magnitude has different effects on different parts of the spectrum. If you don't know what a doppler shift is, imagine a firetruck, sirens blazing, passing you as you stand on the sidewalk. As the truck moves toward you, you will hear a shrill siren, which then becomes deeper after the truck passes you. The reason that this modulation occurs is that the sound waves are experiencing a doppler shift. When the truck is moving towards you, each successive sound wave is being emitted at a decreasing distance from you, meaning that this later wave has a "head start" on the earlier one. The frequency of sound waves reaching your ear is thus higher than the frequency at which they are emitted. The opposite is true if the truck is moving away from you - each successive wave has to travel even farther than the one before it.<br />
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The same thing happens with light. If a source emitting photons is moving away from us, then the distance between each photon is greater as time goes on. The effect of a doppler shift on the observed wavelength is given in this relationship:<br />
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<img src="http://latex.codecogs.com/gif.latex?\bg_black Z(\lambda)=\frac{\Delta\lambda}{\lambda}" title="\bg_black Z(\lambda)=\frac{\Delta\lambda}{\lambda}" /><br />
Which is dependent on the wavelength, represented by lambda. At higher wavelengths, there will be more space between successive wave fronts. This is why we cannot say that with a doppler shift, a 10 nm wave becomes a 12 nm wave and a 16 nm wave becomes a 18 nm wave. the 16 nm wave will actually be larger than 18 nm, since as the wave itself is longer, the shift must also be bigger.<br />
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To get this working, I need to tackle interpolating functions. AH! Wish me luck!<br />
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</a>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com0tag:blogger.com,1999:blog-5849210431402234073.post-48102009859661917802012-01-16T19:02:00.001-08:002012-01-16T19:03:41.669-08:00Fun With Kepler!<!--[if gte mso 9]><xml> <o:documentproperties> <o:revision>0</o:Revision> <o:totaltime>0</o:TotalTime> <o:pages>1</o:Pages> <o:words>499</o:Words> <o:characters>2847</o:Characters> <o:company>Caltech</o:Company> <o:lines>23</o:Lines> <o:paragraphs>6</o:Paragraphs> <o:characterswithspaces>3340</o:CharactersWithSpaces> <o:version>14.0</o:Version> </o:DocumentProperties> <o:officedocumentsettings> <o:allowpng/> </o:OfficeDocumentSettings> </xml><![endif]--> <!--[if gte mso 9]><xml> <w:worddocument> <w:view>Normal</w:View> <w:zoom>0</w:Zoom> <w:trackmoves/> <w:trackformatting/> <w:punctuationkerning/> <w:validateagainstschemas/> <w:saveifxmlinvalid>false</w:SaveIfXMLInvalid> <w:ignoremixedcontent>false</w:IgnoreMixedContent> 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priority="39" qformat="true" name="TOC Heading"> </w:LatentStyles> </xml><![endif]--> <!--[if gte mso 10]> <style> /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:Cambria; mso-ascii-font-family:Cambria; mso-ascii-theme-font:minor-latin; mso-hansi-font-family:Cambria; mso-hansi-theme-font:minor-latin;} </style> <![endif]--> <!--StartFragment--> <p class="MsoNormal"><b><o:p> </o:p></b><b>What is Kepler?</b></p> <p class="MsoNormal">The easiest way to imagine what Kepler is doing is to think of it as staring at the sky, taking measurements at how much light is being received from different spots of the sky at different times. Kepler is focused on one part of the sky, observing continuously, measuring for changes in the flux of light reaching the telescope.<o:p></o:p></p> <p class="MsoNormal"><b><o:p> </o:p></b></p> <p class="MsoNormal"><b>What is Kepler looking for?<o:p></o:p></b></p> <p class="MsoNormal">Kepler’s primary goal is to find planets. Specifically, Kepler’s aim is to identify planets and planet candidates that will help scientists understand the types of and distribution of planets in the universe. In order to find these planets, Kepler uses the transit method, with measures dips in light as a planet crosses a distant star. <o:p></o:p></p> <p class="MsoNormal"><b><o:p> </o:p></b></p> <p class="MsoNormal"><b>What do these fluctuations in light mean?<o:p></o:p></b></p> <p class="MsoNormal">Kepler is looking for stars that have a dip in luminosities, because this could indicate a planet transit. For a planet to ‘transit’ a star means that the planet is crossing in between us (the observers) and the star. One example of a transit closer to home is Venus. When Venus transited the sun several years ago, you might remember pictures of the transit showed a dark circle on the sun. That was the planet blocked part of the sun’s light from reaching earth. <o:p></o:p></p> <p class="MsoNormal">When a planet transits a distant star, it decreases the amount of light reaching observers here. So, when Kepler measures a temporary decrease in light from a star, it could mean that a planet is transiting across the star. <o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><b>Problems with the transit method<o:p></o:p></b></p> <p class="MsoNormal">Although the transit method is conceptually simple, there are many problems that accompany it. For example, not every dim in light indicates a planet. Other possible causes for the dimming include a transiting binary star, three-star system, or two stars blending their light together. So, if Kepler detects a dip in light from a source, it COULD be, but is not necessarily, a planet. This is why the “planets” Kepler identifies are called “planet candidates.” Scientists have estimates the “false positive” rate – that is, how many sources that are not actually planets are accidentally identified as such by the Kepler mission<a href="http://www.blogger.com/post-edit.g?blogID=5849210431402234073&postID=4810200985966191780&from=pencil#_edn1" name="_ednref1" title=""><span class="MsoEndnoteReference"><!--[if !supportFootnotes]--><span class="MsoEndnoteReference"><span style="font-size:12.0pt;font-family:Cambria; mso-ascii-theme-font:minor-latin;mso-fareast-font-family:"MS 明朝";mso-fareast-theme-font: minor-fareast;mso-hansi-theme-font:minor-latin;mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;mso-ansi-language:EN-US;mso-fareast-language: EN-US;mso-bidi-language:AR-SA">[i]</span></span><!--[endif]--></span></a>. This rate, although small, is not insignificant, thus the differentiation between confirmed planets and planet candidates. <o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><b>In an ideal world...<o:p></o:p></b></p> <p class="MsoNormal">As the name “direct imaging” implies, the best way to identify an exoplanet is to actually see it. There are two main reasons that this is hard to do, however:<o:p></o:p></p> <p class="MsoNormal">High contrast ratio between luminous sun and planet (which reflects only a very small amount of light)<o:p></o:p></p> <p class="MsoNormal">The atmospheric distortion blends light from distinct sources into what appears to be one source<o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><b>Adaptive Optics can really help!<o:p></o:p></b></p> <p class="MsoNormal">This is where adaptive optics comes in! Adaptive optics counteracts the atmospheric distortion. As a result, photons are received in a pattern closer to that by which they were emitted. If there were a dark spot on the star (say… caused by a transiting planet!) then ideally, that spot would also be received by the mirror. Without adaptive optics, it is very difficult if not impossible to see this spot.<o:p></o:p></p> <p class="MsoNormal">Adaptive optics also allow the resolving of very bright stars and the planets next to them. Adaptive optics can measure sources within one arcsecond of each other, even if one of the soruces (a planet!) is 10 million times less luminous than the other (a star!)<a href="http://www.blogger.com/post-edit.g?blogID=5849210431402234073&postID=4810200985966191780&from=pencil#_edn2" name="_ednref2" title=""><span class="MsoEndnoteReference"><!--[if !supportFootnotes]--><span class="MsoEndnoteReference"><span style="font-size:12.0pt;font-family:Cambria; mso-ascii-theme-font:minor-latin;mso-fareast-font-family:"MS 明朝";mso-fareast-theme-font: minor-fareast;mso-hansi-theme-font:minor-latin;mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;mso-ansi-language:EN-US;mso-fareast-language: EN-US;mso-bidi-language:AR-SA">[ii]</span></span><!--[endif]--></span></a>.<o:p></o:p></p><p class="MsoNormal">Adaptive optics is really the most reliable way to confirm a planet candidate is actually a planet (or to show that it is NOT a planet). </p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <div><!--[if !supportEndnotes]--><br clear="all"> <hr align="left" size="1" width="33%"> <!--[endif]--> <div id="edn1"> <p class="MsoNormal"><a href="http://www.blogger.com/post-edit.g?blogID=5849210431402234073&postID=4810200985966191780&from=pencil#_ednref1" name="_edn1" title=""><span class="MsoEndnoteReference"><!--[if !supportFootnotes]--><span class="MsoEndnoteReference"><span style="font-size:12.0pt;font-family:Cambria;mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"MS 明朝";mso-fareast-theme-font:minor-fareast; mso-hansi-theme-font:minor-latin;mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;mso-ansi-language:EN-US;mso-fareast-language: EN-US;mso-bidi-language:AR-SA">[i]</span></span><!--[endif]--></span></a> http://iopscience.iop.org/0004-637X/644/2/1237/pdf/64043.web.pdf<o:p></o:p></p> <p class="MsoEndnoteText"><o:p> </o:p></p> </div> <div id="edn2"> <p class="MsoNormal"><a href="http://www.blogger.com/post-edit.g?blogID=5849210431402234073&postID=4810200985966191780&from=pencil#_ednref2" name="_edn2" title=""><span class="MsoEndnoteReference"><!--[if !supportFootnotes]--><span class="MsoEndnoteReference"><span style="font-size:12.0pt;font-family:Cambria;mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"MS 明朝";mso-fareast-theme-font:minor-fareast; mso-hansi-theme-font:minor-latin;mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;mso-ansi-language:EN-US;mso-fareast-language: EN-US;mso-bidi-language:AR-SA">[ii]</span></span><!--[endif]--></span></a> http://spie.org/documents/Newsroom/Imported/003471/003471_10.pdf<o:p></o:p></p> <p class="MsoEndnoteText"><o:p> </o:p></p> </div> </div> <!--EndFragment-->Juliette 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priority="39" qformat="true" name="TOC Heading"> </w:LatentStyles> </xml><![endif]--> <!--[if gte mso 10]> <style> /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:Cambria; mso-ascii-font-family:Cambria; mso-ascii-theme-font:minor-latin; mso-hansi-font-family:Cambria; mso-hansi-theme-font:minor-latin;} </style> <![endif]--> <!--StartFragment--> <p class="MsoNormal">Since I am waiting for code to run anyway, I thought I would look up some topics from the first Ay 21 lecture that were mentioned but that I wasn’t extremely familiar with/ didn’t know what they were, and briefly summarize what I glean from a quick, superficial review. <o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><b>String theory<o:p></o:p></b></p> <p class="MsoNormal">I’ve always thought of string theory as being some crazy theory, but never actually looked up what it was. String theory posits that the universe is made up of one-dimensional vibrating strings, which serve as the most basic particle. The theory also requires extra dimensions – I am very interested in how a theory could require extra dimensions, and how these dimensions are defined. String theory is an attempt to reconcile general relativity and quantum mechanics, which is why it was mentioned with respect to the Age of Quantum Gravity (that bit of time at the beginging of the universe when GR and QM were both important, and you couldn’t describe the physics of a discrete event with only one of the two theories. <o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><b>M-theory<o:p></o:p></b></p> <p class="MsoNormal">M-theory is an extension of string theory. There are different ‘types’ of string theory, and M-theory is a unifying theory meant to connect them all. <o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><b>Curvature of the Universe<o:p></o:p></b></p> <p class="MsoNormal">It is hard to imagine space itself being curved. Best I can understand it, the exact curvature of the universe is as of yet an unsolved problem. The concepts of curvature, density, and expansion of the universe are all realted: there is a critical density at which expansion would stop, and similarly the density also deteremines whether the universe is curved like a sphere, like a hyperbola, or not at all. In a curved universe, light would bend as it travels. <o:p></o:p></p> <p class="MsoNormal">Carrol and Ostlie has a section on curved spacetime (it beings on page 1185). This sections says that the curvature must be constant throughout the universe for a given point in time, although the curvature may vary as the time elapsed since the big bang changes. Curvature is a time-dependant function.<o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><b>Dark Energy<o:p></o:p></b></p> <p class="MsoNormal">In class, dark energy was mentioned as being (possibly) one of the “extra” components of mass in the universe, holding 73% of the mass. Dark energy is something that Einstein predicted when he developed his cosmological costant. Dark energy, different than dark matter, exists throughout the universe, although we are unsure what form it may take. It could be a constant energy density (the cosmological constant) or it could exist in fields or some other form. Dark energy also contributes to the expansion of the universe. <o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><b>Weak/Strong Forces<o:p></o:p></b></p> <p class="MsoNormal">There are four fundamental kinds of forces in the universe: weak, strong, gravity, and electromagnetic. Unlike gravity and EM forces, the effects of which dwindle but do not disappear as distance increases, weak and strong forces only act on a atomically small scale. <o:p></o:p></p> <p class="MsoNormal">The strong force is what holds a nucleas together, even as the positively charged protons attempt to repell each other. I can’t believe I never heard about this before! When the protons and neutrons are considered to be composed of quarks, the strong force can be attributed to the “color force” between the quarks. The weak force allows one “flavor” of quark to change to another, allowing nuclear fusion to take place, and is thus necessary for every star that burns. <o:p></o:p></p> <p class="MsoNormal"><b><o:p> </o:p></b></p> <p class="MsoNormal"><b>Quarks<o:p></o:p></b></p> <p class="MsoNormal">This is another subject I have read about briefly but never really considered until now. I had thought that quarks were as “imaginary” as string theory; however, this is not true. Quarks are the elementary particles of the universe in that the 6 flaovrs of quarks make up the protons, neutrons, mesons, and all other particles which then, successively, make up matter itself. Mesons are particles composed of two quarks, and baryons are particles composed of three quarks. Protons and neutrons are thus mesons (constructed from up and down quarks). Quarks exist in the flavors up, down, strange, charm, bottom, and top. <o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">My code is still running… the more I learn about physics, the more I realize I don’t know! I am super-excited about quarks – I need to go check out a book on elementary particle physics from the library because they seem SUPER COOL and I want to learn more about them!<o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">Sources:<o:p></o:p></p> <p class="MsoNormal">Wikipedia, Carrol & Ostlie<o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <!--EndFragment-->Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com2tag:blogger.com,1999:blog-5849210431402234073.post-30459273856541275472012-01-08T18:29:00.000-08:002012-01-08T18:33:32.141-08:00Why is AO imaging important?<p class="MsoNormal">In the last post, I included a cool picture showing the difference in imaging resolution due to AO. AO is very useful in direct imaging, because the higher resolution allows better differentiation between different sources. One complication that has always been as issue in detecting exoplanets accurately is that even if observers measure a light curve indicating that there may be a planet, the light dip might actually be due to something else, such as a transiting binary. This is of particular importance when wide-sky surveys are used, because each individual source has probably not been studied at length; therefore, the superficial appearance of their being a planet might be accepted when it should be rejected. Even when studied at length, the false positives may appear so similar to real planets that they continue to fool observers. This paper shows one specific instance where a source appeared to have a planet (indeed, it passed many of the ‘tests’ of planethood) but was actually a false positive. The fact that there is a false positive rate in all planet-finding surveys is a problem: this leads to the difference between a planet candidate and a confirmed planet. <o:p></o:p></p> <p class="MsoNormal">One good thing about the Kepler false-positive rate (the rate at which things that are not planets are being called planets) is that it is not random. There are very specific reasons as to why a light curve might be altered to appear that there was a planet where one did not exist. Some of these reasons might be an eclipsing binary, “blend” (this is when a less bright star, not cataloged in Kepler, ‘mimics’ a planet) or the existence of a “hierarchal triple” (a multiple – three in this case, as its name implies – star system) (<i>On the Low False Positive Probabilities of Kepler Planet Candidates</i>, Morton and Johnson). Knowing the possible reasons that there might be a false positive in the Kepler data allows a statistical correction for this. Using approximations and creating probability distributions, Morton and Johnson developed a possible model for the false positive rate. Although this will not tell which individuals stars have planets, it makes the Kepler distribution as a whole more helpful, as observers have a way to estimate how many planet candidates, on the whole, are likely planets. <o:p></o:p></p> <p class="MsoNormal">The process for determining, after a planet candidate has been found, whether or not it is a real planet can be lengthy. It also shows the importance of adaptive optics in confirming planets, as AO images can be used in either preliminary or follow-up observations. Many false positives are due to light being misattributed to an incorrect source, and adaptive optics reduces the degree to which this happens. <o:p></o:p></p>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com0tag:blogger.com,1999:blog-5849210431402234073.post-85938209110947668252011-12-25T18:59:00.000-08:002011-12-25T22:39:22.801-08:00Adaptive Optics<div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgdZb6wU5ySFfqrCV2LXU76AVZLHYxghmprgTZY09zET0Nm19WxXTqZLXd2VUuyPR599J7OyC5Omm7Fl1HQ18XIiMsN3MuPh_IkZ-s2Fv2_kKE7zn-5Z4CHzOmO2bRwmPzbXh6Sc74qXv2f/s1600/aooffon.jpg"><span style="color: rgb(0, 0, 0);"></span></a><br /><div><strong>Why do we need adaptive optics?</strong><br /><strong><br /></strong>There are two places for optical telescopes: on the surface of the earth and in space. The effort to put a telescope in space and maintain it is much greater than that required for a ground-based telescope, so why would this avenue of observation have ever been pursued? The answer to this question is, simply, the atmosphere. Earth has an atmosphere, which affects light traveling through it.<br /><br />The variances in the atmosphere consist mainly of differences in temperature from one region to another. The pockets of differing temperatures mean that a path straight from a distant source to a telescope located on the surface of the earth would include regions of higher and lower temperatures? Why does this matter for light?<br /><br />Photons propagate as waves. The speed of a propagating wave is dependent upon the medium through which it is propagating. For photons from distant stars, the majority of the medium they propagate through is the interstellar medium, which can be approximated as a vacuum (this is what I have been taught, but the fact that Caltech has a whole class on the interstellar medium –Ay 101, which I will take next year – makes me wonder if this is merely a simplistic approximation... anyway, for now I consider space to be roughly a vacuum). Once the photon reaches the atmosphere, however, it is propagating through much denser air. Even this air is not uniformly dense, however: regions of hotter and colder air change the index of refraction for different air pockets, creating what are essentially different mediums.<br /><br />If you consider the wavefront of light emitted from a distant star, you can see why this would be a problem. When the wavefront reaches the atmosphere, each individual part of the wave front has propagated at the same speed (since the entire wave front has been propagating through the same medium). However, this is no longer true once the photons pass through the atmosphere. Photons that pass through hotter packets of air will propagate at a greater speed than those traveling through cooler air.<br /><br />You can imagine what happens to the poor wavefront when each photon travels through different temperature air pockets – the poor thing is in disarray by the time it reaches an earth-based telescope! Some areas of the wavefront have travelled faster while other bits have travelled more slowly – what used to be a smooth wavefront is altered by the atmosphere into a much more “squiggly” form.<br /><br />This is a problem! One solution to this problem is to place your telescope in space, where the wavefronts don’t have to make the perilous journey through the atmosphere. However, in recent years a much more amazing method of avoiding atmospheric distortion has come into being… and this method is adaptive optics!<br /><br /><strong>What does adaptive optics (AO) do?</strong><br /><br />Adaptive optics corrects for distortions due to the atmosphere in the wavefront. The way that is does is REALLY COOL – I hardly believed it was real the first time I heard about it!<br /><br />AO uses a deformable mirror. The mirror itself is deformed!!! Now, this mirror is not the mirror that first receives the image. You aren’t deforming the 5 meter mirror at Palomar, for example. Adaptive optics are not a part of the main mirror, but instead an instrument that you can add onto a telescope. The photons are first received by the ‘big mirror,’ as you might expect, but then refracted to the smaller, deformable mirror, which can correct the image.<br /><br />Basically, AO corrects the image for atmospheric distortion. The goal is that the image received from a ground based telescope using AO would be as good as an image from a space-based telescope.<br /><br /><strong>How do you calibrate AO (find out how air pockets affect the wavefronts)?</strong><br /><br />The distortions by the atmosphere on the wavefronts are not random. They are very directly caused by the varying temperatures in the atmosphere. So, theoretically, if you could map the air pockets of different temperatures, you could write program to subtract the “noise” caused by said air pockets.<br /><br />To measure how much correction is needed, AO systems use mainly two methods: a guide star or a laser. Guide stars are those used as references. We already know what we SHOULD get by observing these stars, so if the observed image differs from the expected image, the difference is the interference due to the atmosphere, and can be corrected in subsequent observations.<br /><br />Lasers are even cooler! The laser used with AO is a sodium laser, which works by emitting a wavelength of light that will excite sodium atoms, which then emit light back to the telescope. This is sort of a fake guide star – it is used in the same way that a real guide star is, but it is much more versatile!<br /><br />The laser/guide star calibration method is used simultaneously with the observation of whatever source the telescope is studying. That way, corrections are as accurate as possible. As the air pockets change and the wavefront’s “squiggles” change, the corrections to the entire image can be altered (based on whatever the light from the guide star says the change should be) so the image is as accurate as possible.<br /><br /><strong>How does a deformable mirror correct the image?</strong><br /><br />Firstly – how is a mirror deformed? The method of deformation is pretty cool! I looked at documentation about the PALM-3000, which is an AO system for Palomar, as a case study of how the method works.<br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiUVrVZRYzDORj5G2fWEHiZhE-e0vYeGNfYA_Tj_kDd6fJ-YXIbuyd4DXpxF4BIfTNwXxsgHjvp8LmNHOgBYRcGgTi1_vIODKqibP02l04YLVQvUAofvSORD7kUNXYum4PCsilsSIrN2u6e/s1600/ao+mirror.jpg"><img style="margin: 0px 10px 10px 0px; width: 314px; height: 320px; float: left; cursor: pointer;" id="BLOGGER_PHOTO_ID_5690267293367337810" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiUVrVZRYzDORj5G2fWEHiZhE-e0vYeGNfYA_Tj_kDd6fJ-YXIbuyd4DXpxF4BIfTNwXxsgHjvp8LmNHOgBYRcGgTi1_vIODKqibP02l04YLVQvUAofvSORD7kUNXYum4PCsilsSIrN2u6e/s320/ao+mirror.jpg" /></a><br />Mirrors are deformed by needle-looking things called actuators. The actuators actually push the surface of the mirror into a pattern that counteracts the effects of the atmosphere. If you don’t think about it too much, the idea that you can just physically counteract the distortions in the wavefront makes sense. Where the wave travelled too slow, push the mirror up so it receives the lagging photons sooner. Where the wave travelled too fast, leave the actuator unpushed so the photons must travel farther and thus all photons arrive at the mirror at the same time. If you start thinking more deeply about this, you will realize how complicated writing the code for this algorithm would be! (Image to the left from 4)<br /><br />Another remarkable fact is the rate of correction – that is, how often AO reads in data from its guide star and adjusts the actuators to counteract the distortions due to the atmosphere. The whole process sounds computationally arduous – can you believe that AO is corrected 2000 times a second (5)?!? This frequency seems CRAZY to me! Imagine the engineering that went into allowing the fast and precise movements of all those actuators – which, on the PALM 3000, is 4356 actuators (4)!</div><div><br /><strong>Why is AO useful?</strong><br /><br />As they say, a picture is worth one thousand words. This Palomar image shows the difference between an observation with adaptive optics (right) and an observation without (left) (image from 2).<br /></div><div> <img style="margin: 0px auto 10px; width: 320px; height: 148px; text-align: center; display: block; cursor: pointer;" id="BLOGGER_PHOTO_ID_5690267099183782946" border="0" alt="" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgdZb6wU5ySFfqrCV2LXU76AVZLHYxghmprgTZY09zET0Nm19WxXTqZLXd2VUuyPR599J7OyC5Omm7Fl1HQ18XIiMsN3MuPh_IkZ-s2Fv2_kKE7zn-5Z4CHzOmO2bRwmPzbXh6Sc74qXv2f/s320/aooffon.jpg" /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgdZb6wU5ySFfqrCV2LXU76AVZLHYxghmprgTZY09zET0Nm19WxXTqZLXd2VUuyPR599J7OyC5Omm7Fl1HQ18XIiMsN3MuPh_IkZ-s2Fv2_kKE7zn-5Z4CHzOmO2bRwmPzbXh6Sc74qXv2f/s1600/aooffon.jpg"></a></div><div><br /><strong>Useful links:</strong><br />1. http://www.mtwilson.edu/ao/ - tells about the AO system used at Mt. Wilson; good explanation about AO more generally as well<br />2. http://www.astro.caltech.edu/palomar/AO/ - cool images – AO is important at Palomar!<br />3. http://amazing-space.stsci.edu/resources/explorations/groundup/lesson/basics/g18a/ - good image on what the air pockets do<br />4. http://spie.org/x39226.xml?ArticleID=x39226 – specifically about the PALM-3000, good explanation on how the actuators work and some cool images!<br />5. http://www.oir.caltech.edu/twiki_oir/bin/view/Palomar/Palm3000/WebHome - more on PALM-3000</div></div>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com2tag:blogger.com,1999:blog-5849210431402234073.post-44694253981677869402011-12-24T17:54:00.000-08:002011-12-25T14:07:04.188-08:00Fixing my mistakes… [lab 3]When I said “by Wednesday”… I guess I didn’t specifically say WHICH Wednesday I would fix it by…<br />New information:<br />• the “ marks in the image are arcseconds, not degrees - I should have converted units to account for this!<br />• Keep values in radians<br />• Recheck what K-band really means!<br /><br />SO! Now, I have 0.009 arcseconds as the value for the separation of the binaries, which I can covert to degrees:<br /><a href="http://www.codecogs.com/eqnedit.php?latex=="\bg_black .09(1/3600) = x = 0.000025 \textup{ degrees }" target="_blank"><img src="http://latex.codecogs.com/gif.latex? ="\bg_black .09(1/3600) = x = 0.000025 \textup{ degrees }" title="\bg_black .09(1/3600) = x = 0.000025 \textup{ degrees }" /></a><br />Proceeding with this value, I know that the distance to the first airy ring should be 0.1 arcseconds, not 0.1 degrees. With this knowledge, I can convert again to degrees and then radians (for radians, see summary table way below).<br /><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black .1(1/3600) = x = 0.0000277 \textup{ degrees }" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black .1(1/3600) = x = 0.0000277 \textup{ degrees }" title="\bg_black .1(1/3600) = x = 0.0000277 \textup{ degrees }" /></a> <br />Now, I also need to check the values for the wavelengths of K-band. According to Wikipeida: http://en.wikipedia.org/wiki/Photometric_system <br />The values of K-band in this system of measurement have a center of around 2190 nm. Using these new values, I can use the same methodology as last time, except my value for theta will be different. Theta in the equation below refers to not the distance from the center to the airy ring, but the entire distance across the diameter of the airy ring. So,<br /><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black sin(\Theta ) = 1.22\frac{\lambda }{D}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black sin(\Theta ) = 1.22\frac{\lambda }{D}" title="\bg_black sin(\Theta ) = 1.22\frac{\lambda }{D}" /></a><br /><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black \theta = 2*0.0000277 = 0.0000554 \textup{ degrees} = 9.66912*10^-7 \textup{ radians}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black \theta = 2*0.0000277 = 0.0000554 \textup{ degrees} = 9.66912*10^-7 \textup{ radians}" title="\bg_black \theta = 2*0.0000277 = 0.0000554 \textup{ degrees} = 9.66912*10^-7 \textup{ radians}" /></a><br /><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black sin(9.66912*10^-7 \textup{ radians}) = 1.22\frac{\lambda }{D}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black sin(9.66912*10^-7 \textup{ radians}) = 1.22\frac{\lambda }{D}" title="\bg_black sin(9.66912*10^-7 \textup{ radians}) = 1.22\frac{\lambda }{D}" /></a><br />Using small angle approximation, and the fact that 2190 nm = 2.19*10^-6 meters,<br /><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black 9.66912*10^-7 = 1.22*\frac{2.19*10^{-6}}{D}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black 9.66912*10^-7 = 1.22*\frac{2.19*10^{-6}}{D}" title="\bg_black 9.66912*10^-7 = 1.22*\frac{2.19*10^{-6}}{D}" /></a><br /><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black 0.361895 = \frac{1}{D}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black 0.361895 = \frac{1}{D}" title="\bg_black 0.361895 = \frac{1}{D}" /></a><br /><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black D=2.76323 \textup{ meters}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black D=2.76323 \textup{ meters}" title="\bg_black D=2.76323 \textup{ meters}" /></a><br />This suggests that the aperture is 2.76 meters – not the 10ish meters I previously found.<br />NOW – which telescope could this be? …. I notice that depending on the measurement for the width of the airy ring, I get different results. This is frustrating because depending on from where I measure (the innermost and smallest distance, or from the center of the outer pixel) I get different values for the minimum angular resolution. I can get what look like equally valid measurements varying from 0.09 to 0.23 arcseconds. These correspond to apertures between roughly 2.5 and 6.5 meters. See graph below, which graphs D as a function of the diameter of the smallest airy ring, measured in arcseconds.<br /><br />Choosing what telescope this image is from is difficult because of the uncertainty in my original measurement of the diameter of an airy ring. However, based on considerations of where in the sky this source would be visible (see the uncorrected post) I can say that it would be visible from Hawaii, California, or equivalent areas. Combined with my knowledge that this image came from a Caltech-affiliated source, who likely has most easy access to Palomar and Keck, I can hypothesize that the image was taken using the Hale telescope at Palomar, which has an approximately 5 meter aperture, within my error bounds. <br />Trying to confirm or disprove my hypothesis, I combed through the publications of the two observers listed in the FITS header and found this abstract:<br />http://adsabs.harvard.edu/abs/2010DPS....42.3930L<br />So, K-band observing was going on at Palomar April 25-26…. Looks promising! Unfortunately I can’t seem to find a copy of the entire paper – maybe because I am not on the Caltech network? <br />Also, I found:<br />http://www.palomar.caltech.edu:8000/calendar.tcl?cal_date=2010%2d04%2d25+00%3a00%3a00%2b00<br />Which shows that April 27 was used by a different astronomer… but the nights of the 25th and 26th were used by a consistent observer to the file. I would say that the day had just changed and this observation was from the night of the 27th, but the observation time says 11:28. This does not mean that my hypothesis was wrong; I’m not sure which time zones /sidereal time/PST time were used to mark the file; so although the observation was marked 11:29, April 27th it could possibly have been from the 26th. <br />Also – Knowing that Justin Crepp was the observer, (I feel like such a stalker…) how likely is it that he would, having observed all night on April 26th, then proceed to fly somewhere else and observe AGAIN the next night? I don’t think that would be very likely, and it is always possible that the nights got switched or something else unreported on the observation time website happened. Therefore I will not reject the hypothesis that this image was taken at Palomar on the Hale 200-inch telescope. <br /><br /><br />Fixing my errors: a summary<br /><br />Airy ring distance from center: I used to think this was 0.09 degrees.<br /> Now I know it's 0.000025 degrees, or 4.36332 × 10^-7 radians<br />K-band wavelengths: I used to think it was 7.5*10^-1 to 1.5 cm,<br /> Now I know that it's centered at 2190 nm<br />Aperture: my first value was 10.48 meters,<br /> My new value was 2.9 meters but I also realized that this number comes with some sizeable error bounds!<br />Telescope: I used to think it was the Keck2, NIRC2<br /> Now I think the Hale 200 inch at Palomar was usedJuliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com4tag:blogger.com,1999:blog-5849210431402234073.post-26857990440215745792011-12-04T20:52:00.000-08:002011-12-04T20:53:48.443-08:00Becoming a professional astronomer, the final post<p class="MsoNormal">So, before I explain how my opinion has changed since the start of this project (affected by the interviews my group and I did – see Iryna and Tommy's blogs for our other posts – as well as the various conversations I have had with post-docs throughout the term), I wanted to post some interesting posts I found while perusing resources online, along with just a couple thoughts I had on each one. Most of the most interesting things online are blog posts. I have never read so many blogs before this class, but they are fun and useful!</p><p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal"><a href="http://science-professor.blogspot.com/2011/09/dealing-with-it.html">http://science-professor.blogspot.com/2011/09/dealing-with-it.html</a> - Just what is the relationship between an advisor and a grad student? If you’ve ever seen the PhD comics, then it appears that grad students are like slaves to uncaring, busy advisors. From my SURF, I thought that grad students and professors were more like friends (both of whom too busy for an undergrad like me!). This article implies that the relationship doesn’t have to be chummy, but merely professional.<o:p></o:p></p> <p class="MsoNormal"><a href="http://www.astrobetter.com/valuing-all-kinds-of-astronomy-smarts/">http://www.astrobetter.com/valuing-all-kinds-of-astronomy-smarts/</a> - This is interesting because it discusses the idea of the gap between school skills and practical skills. The people who can do problem sets really well might not be the best scientists…<o:p></o:p></p> <p class="MsoNormal"><a href="http://astrobites.com/2011/11/16/for-your-perusing-pleasure-some-preliminary-results-from-the-social-perceptions-of-astronomy-survey/">http://astrobites.com/2011/11/16/for-your-perusing-pleasure-some-preliminary-results-from-the-social-perceptions-of-astronomy-survey/</a> - on a side note, I tried out Professor Johnson’s idea that telling people on a plane you were an astronomer, as opposed to an astrophysicist, is more conducive to conversation. I had a two-hour conversation with a telephone tech about astronomy on the way home for Thanksgiving… maybe I’ll write one last blog post on this! It was actually pretty cool.<o:p></o:p></p> <p class="MsoNormal"><a href="http://astrobites.com/2011/10/02/applying-for-the-nsf-our-own-experiences/">http://astrobites.com/2011/10/02/applying-for-the-nsf-our-own-experiences/</a> - Oh gosh! I don’t even want to think about this. One of my cross country teammates, a senior, just applied for NSF, and it sounds nerve-wracking.<o:p></o:p></p> <p class="MsoNormal">Now – my own thoughts. I have realized over the course of the semester, after talking to various grad students, that for astronomy there are many different directions you can go. You can be more instrumentation, or theory, but each discipline is different. Also, I have gotten over my hesitation to say the word “astronomy”. I was always worried that “astronomy” was not a real career path. This is what my parents and high school teachers led me to believe; there was always the joke about the student going off to college to pursue something useless like philosophy or astronomy. For that reason, I always answered “astrophysics” when people asked my major. <o:p></o:p></p> <p class="MsoNormal">This term, I have realized that many of the actually parts of astronomy are exciting. Actually observing data is exciting, because of the years of work that lie behind every single observation. It’s not as easy as just looking at the sky; adaptive optics and other modern aspects of telescopes show the field is not static and involves far more than simply looking at stars. <o:p></o:p></p> <p class="MsoNormal">One other contrast that this term has brought to my mind is the difference between astronomy and other fields of science. The research process of biology or chemistry has always seemed a little scary to me; you could spend years working on a project only to see it fail or another group scoop you before you publish. In astronomy, there seem to be far more data than astronomers. I like that idea that even someone like me could comb through public data and find something publishable that simply hasn’t been studied by anyone yet.<o:p></o:p></p> <p class="MsoNormal">The process for becoming a professional astronomer seems less daunting. I have also realized, because of what Jackie said last week, that there is nothing wrong with verbalizing my goal of being a professor, even if I think it might be presumptuous of me to suppose that I “deserve” to have this as a goal. There is no reason to pretend to pursue a career in industry when I know my dream job is to be a professor. <o:p></o:p></p> <p class="MsoNormal">Pursuing this goal, I have learned from this project and the rest of term, will involve facing rejection many times. Not only do I have to apply to grad school and perhaps fellowships, but I will eventually apply to postdocs and eventually faculty positions. I have also realized how important other people are. I would never have learned any of the things I learned this term if I had been in a regular class, because I would have spent my time doing problem sets instead of meeting interesting people and going interesting places (Palomar, the Solar laboratory here, seeing the adaptive optics in the basement). There is nothing wrong with asking questions and there is nothing wrong with having fun.<o:p></o:p></p> <p class="MsoNormal">From the interview with Annelia Sargent, I learned that it is possible to have a family and pursue a career in astronomy simultaneously. She also spoke to us about all the different committees and projects she has been on. It was cool when I read an article of Science a few days later and saw her mention in concert with yet another project that she works on, but the fact that she considers all the extra things she does as “community service” made me realize that it is very difficult to put limits on what a professor does, as every position is different and a lot of what they do is what they choose to do. <o:p></o:p></p> <p class="MsoNormal">Reading posts online, I am struck by the sheer volume of posts by older people in the field giving advice to younger ones seeking to attain their positions. Grad students advise undergrads with seeking guidance from postdocs, and professors offer advice to all. Everyone in the field seems to realize that it is a tough path, and they want to make it easier for those attempting the arduous journey. Astronomy seems, at least as seen through the eyes of the blogosphere, as a cooperative and friendly field. This certainly corroborates my experience this term here at Caltech.<o:p></o:p></p> <p class="MsoNormal">There is also a lot of discussion online about the problems with the current system. Many people lament that the ‘problem set/test/grade’ status quo is alienating to some minds that would be great in astronomy, and theorists have an easier time than people who could be great coders or instrumentationists. I would agree with this – although here at Caltech, we have theory pounded into our brains until it is all we can think, qualities that make a great astronomer cannot necessarily be measured by checking someone’s ability to do a physics problem. <o:p></o:p></p> <p class="MsoNormal">This semester has given me a lot to think about, and changed my habits forever. I really do enjoy talking to others in the field, and now I know how to approach them. One of the goals Professor Johnson gave at the beginning of term was to be able to talk to an astronomer, be it a professor or someone else. He wanted us to see someone that we think is way “above” us, and just be able to talk to them. The me of a year ago could not have done this, and the me of this summer didn’t know to try, but ever since the day I met up with Dr. Kirby on Professor Johnson’s suggestion, I have realized that approaching other people isn’t so scary as it seems. I look forward to continuing the habits of this class, and pursuing the career path of being a professional astronomer. <o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com1tag:blogger.com,1999:blog-5849210431402234073.post-90329639130237583272011-12-04T13:34:00.001-08:002011-12-04T20:52:50.960-08:00Habitable Zone<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjRUHw7V8iB7HuKb2llSWGLlSIwoQ3DVtfgk_2Ro3JP_4Qxn0DwgE07Kmp_AEUsMseHzuiDYMcOgs7OYMSRbQSHYJUjzUrBY9d0ZndeGS7rADztGJ-dlmtPepBl2YKMvdpdopwaxHkPP25B/s1600/habit+1.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img style="float:right; margin:0 0 10px 10px;cursor:pointer; cursor:hand;width: 320px; height: 281px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjRUHw7V8iB7HuKb2llSWGLlSIwoQ3DVtfgk_2Ro3JP_4Qxn0DwgE07Kmp_AEUsMseHzuiDYMcOgs7OYMSRbQSHYJUjzUrBY9d0ZndeGS7rADztGJ-dlmtPepBl2YKMvdpdopwaxHkPP25B/s320/habit+1.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5682389802779751650" /></a>The so-called “habitable zone” around a star is the range of semimajor axes corresponding to<br />planetary surface temperatures warm enough for liquid water. How does the location of the<br />center of the habitable zone scale with stellar mass? (Recall the scaling relationships from our activities on stellar structure.<br />We will have liquid water from 0 degrees Celsius to 100 degrees Celsius. This is equivalent to 273K to 373K.<br />We know from our activities on stellar structure that mass scales with radius. We also know relationships for luminosity, flux, temperature and radius:<br /><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20F\sim%20T^{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20F\sim%20T^{4}" title="\bg_black F\sim T^{4}" /></a><br /><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20L\sim%20R^{2}F" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20L\sim%20R^{2}F" title="\bg_black L\sim R^{2}F" /></a><br /><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20L\sim%20R^{2}T^{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20L\sim%20R^{2}T^{4}" title="\bg_black L\sim R^{2}T^{4}" /></a><br /><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20a{_{}}^{2}T_{a}^{4}\sim%20R{_{*}}^{2}T_{*}^{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20a{_{}}^{2}T_{a}^{4}\sim%20R{_{*}}^{2}T_{*}^{4}" title="\bg_black a{_{}}^{2}T_{a}^{4}\sim R{_{*}}^{2}T_{*}^{4}" /></a><br />Note that Ta is NOT the value of the temperature on the planet. It is the effective temperature, due to the sun, at a distance (a) from the sun. Luminosity remains constant, thus the further away from the star you get, the smaller the flux is and the smaller temperature is.<br /><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20{a}%20\sim%20\sqrt{\frac{R{_{*}}^{2}T_{*}^{4}}{T_{p}^{4}}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20{a}%20\sim%20\sqrt{\frac{R{_{*}}^{2}T_{*}^{4}}{T_{p}^{4}}}" title="\bg_black {a} \sim \sqrt{\frac{R{_{*}}^{2}T_{*}^{4}}{T_{p}^{4}}}" /></a><div><br /></div><div><b>MORE</b>!!! After Professor Johnson's comment... I didn't go quite far enough!!</div><div><br /></div><div>I want an expression for a in terms of M. This means that if I use the current expression but find scaling relations for R, T* and Tp in terms of a and M, I should have a better answer. </div><div><br /></div><div>M scales with R. </div><div><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20\frac{dP}{dr}=\frac{GMp}{{r}^{2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\frac{dP}{dr}=\frac{GMp}{{r}^{2}}" title="\bg_black \frac{dP}{dr}=\frac{GMp}{{r}^{2}}" /></a></div><div><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20P=-\frac{GMp}{{r}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20P=-\frac{GMp}{{r}}" title="\bg_black P=-\frac{GMp}{{r}}" /></a></div><div>This is the energy due to pressure. Balance this with thermal energy:</div><div><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20P=-\frac{GMp}{{r}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20P=-\frac{GMp}{{r}}" title="\bg_black P=-\frac{GMp}{{r}}" /></a></div><div><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20\frac{3}{2}kT=\frac{GM(\frac{M}{V})}{{r}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\frac{3}{2}kT=\frac{GM(\frac{M}{V})}{{r}}" title="\bg_black \frac{3}{2}kT=\frac{GM(\frac{M}{V})}{{r}}" /></a></div><div><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20\frac{3}{2}kT=\frac{GM(\frac{M}{\frac{4}{3}\pi%20r^{3}})}{{r}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\frac{3}{2}kT=\frac{GM(\frac{M}{\frac{4}{3}\pi%20r^{3}})}{{r}}" title="\bg_black \frac{3}{2}kT=\frac{GM(\frac{M}{\frac{4}{3}\pi r^{3}})}{{r}}" /></a></div><div><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20\frac{3}{2}kT=\frac{3GM^2}{{4\pi%20r^{4}}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\frac{3}{2}kT=\frac{3GM^2}{{4\pi%20r^{4}}}" title="\bg_black \frac{3}{2}kT=\frac{3GM^2}{{4\pi r^{4}}}" /></a></div><div><br /></div><div><br /></div><div>For temperature, we can just use flux, since we have a direct relationship between flux and temperature. flux is dependent on the distance from the star. </div><div><br /></div><div><br /></div><div>(more coming!!! sorry!!)</div><div><br /></div>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com2tag:blogger.com,1999:blog-5849210431402234073.post-43164079337361556422011-12-03T22:03:00.001-08:002011-12-03T22:18:09.175-08:00Aladin Lab #3<div style="text-align: left;">Aladin is a visual FITS file viewer software. It is available online: <a href="http://aladin.u-strasbg.fr/" target="_blank" title="This external link will open in a new window" style="text-align: left; "><span style="mso-bidi-font-family: Calibri;mso-bidi-theme-font:minor-latin;background:black">http://aladin.u-strasbg.fr/</span></a><span class="Apple-style-span" style="text-align: left; "> It is easy to install – no Cygwin necessary for Windows computers (phew!). In this blog post, I will show step-by-step what I did… not all of which was right! I headed down dead ends at some points! But hopefully seeing my mistakes will make it easier for everyone else to learn how to use this software </span><span style="text-align: left; font-family: Wingdings; ">J</span></div><p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal">With this software, you can analyze fits images, found online or stored locally on your computer. For this lab, I used the file found on Professor Johnson’s website with a binary system. Loaded into the Aladin viewer, a FITS file is represented visually as a field of stars:</p><p class="MsoNormal"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhn9_7FrkOFnBFXwm4Ay5sW06IuCmwLh5eZuqvYgx74D4jrUjnmyJIVd4iuO8vsJGXT7TMtkM9HP1GUsWKfW3Wj-ykQ-Ydq5-krHrJ5HW63tAOPmNt6OCwUsCP6BSBGYPPLO2W-Rpw9hVY2/s320/1.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5682150096564059602" style="color: rgb(0, 0, 238); text-decoration: underline; display: block; margin-top: 0px; margin-right: auto; margin-bottom: 10px; margin-left: auto; text-align: center; cursor: pointer; width: 320px; height: 297px; " /></p><div>What is a FITS file? A FITS file is a format of storing data specific to astronomy. A FITS file is not simply an image, but it also comes with information about the image (such as ascension and declination of the targets, field of view, photon counts for each pixel, etc).</div><p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal">From this specific FITS file, we can determine the location of the targets: from Aladin, I found the following for each source:<o:p></o:p></p> <div align="center"> <table class="MsoTableGrid" border="1" cellspacing="0" cellpadding="0" style="border-collapse:collapse;border:none;mso-border-alt:solid windowtext .5pt; mso-yfti-tbllook:1184;mso-padding-alt:0in 5.4pt 0in 5.4pt"> <tbody><tr> <td width="85" valign="top" style="width:63.9pt;border:solid windowtext 1.0pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; "><b>Source<o:p></o:p></b></p> </td> <td width="132" valign="top" style="width:99.0pt;border:solid windowtext 1.0pt; border-left:none;mso-border-left-alt:solid windowtext .5pt;mso-border-alt: solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; "><b>RA<o:p></o:p></b></p> </td> <td width="120" valign="top" style="width:1.25in;border:solid windowtext 1.0pt; border-left:none;mso-border-left-alt:solid windowtext .5pt;mso-border-alt: solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; "><b>DEC<o:p></o:p></b></p> </td> <td width="120" valign="top" style="width:1.25in;border:solid windowtext 1.0pt; border-left:none;mso-border-left-alt:solid windowtext .5pt;mso-border-alt: solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; "><b>RA (degrees)<o:p></o:p></b></p> </td> <td width="120" valign="top" style="width:1.25in;border:solid windowtext 1.0pt; border-left:none;mso-border-left-alt:solid windowtext .5pt;mso-border-alt: solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; "><b>DEC (degrees)<o:p></o:p></b></p> </td> </tr> <tr> <td width="85" valign="top" style="width:63.9pt;border:solid windowtext 1.0pt; border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt; padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; "><i>A<o:p></o:p></i></p> </td> <td width="132" valign="top" style="width:99.0pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; ">18:05:27.73<o:p></o:p></p> </td> <td width="120" valign="top" style="width:1.25in;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; ">+2:29:48.0<o:p></o:p></p> </td> <td width="120" valign="top" style="width:1.25in;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; ">270.5455<o:p></o:p></p> </td> <td width="120" valign="top" style="width:1.25in;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; ">2.4967<o:p></o:p></p> </td> </tr> <tr> <td width="85" valign="top" style="width:63.9pt;border:solid windowtext 1.0pt; border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt; padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; "><i>B<o:p></o:p></i></p> </td> <td width="132" valign="top" style="width:99.0pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; ">18:05:27.18<o:p></o:p></p> </td> <td width="120" valign="top" style="width:1.25in;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; ">+2:29:44.4<o:p></o:p></p> </td> <td width="120" valign="top" style="width:1.25in;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; ">270.5363<o:p></o:p></p> </td> <td width="120" valign="top" style="width:1.25in;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; ">2.4956<o:p></o:p></p> </td> </tr> </tbody></table> </div> <p class="MsoNormal" style="margin-bottom: 0.0001pt; background-image: initial; background-attachment: initial; background-origin: initial; background-clip: initial; background-color: black; background-position: initial initial; background-repeat: initial initial; ">We have found the angular separation of the binaries to be 0.009 degrees. This distance was found using the distance tool on Aladin (just click on your two sources, and the program computes the exact distance between them!). Also, it can be found manually using the source positions. <o:p></o:p></p> <p class="MsoNormal" style="margin-bottom: 0.0001pt; background-image: initial; background-attachment: initial; background-origin: initial; background-clip: initial; background-color: black; background-position: initial initial; background-repeat: initial initial; "><o:p> </o:p></p> <p class="MsoNormal" style="margin-bottom: 0.0001pt; background-image: initial; background-attachment: initial; background-origin: initial; background-clip: initial; background-color: black; background-position: initial initial; background-repeat: initial initial; ">Given that the angular resolution is based on the ‘airy’ pattern – it would be useful to know, what is an airy pattern? The airy pattern refers to the lighter and darker regions, where different strengths of intensities of light are visible.<o:p></o:p></p> <p class="MsoNormal" style="margin-bottom: 0.0001pt; background-image: initial; background-attachment: initial; background-origin: initial; background-clip: initial; background-color: black; background-position: initial initial; background-repeat: initial initial; "><o:p> </o:p></p> <p class="MsoNormal" style="margin-bottom: 0.0001pt; background-image: initial; background-attachment: initial; background-origin: initial; background-clip: initial; background-color: black; background-position: initial initial; background-repeat: initial initial; "><b><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:Calibri; mso-bidi-theme-font:minor-latin;color:red">[I start to go down the wrong road here… skip ahead if you want!]<o:p></o:p></span></b></p> <p class="MsoNormal" style="margin-bottom: 0.0001pt; background-image: initial; background-attachment: initial; background-origin: initial; background-clip: initial; background-color: black; background-position: initial initial; background-repeat: initial initial; "><o:p> </o:p></p> <p class="MsoNormal" style="margin-bottom: 0.0001pt; background-image: initial; background-attachment: initial; background-origin: initial; background-clip: initial; background-color: black; background-position: initial initial; background-repeat: initial initial; ">The sinc function, which is the Fourier transform of the box function, is defined by:<o:p></o:p></p> <p class="MsoNormal" style="margin-bottom: 0.0001pt; background-image: initial; background-attachment: initial; background-origin: initial; background-clip: initial; background-color: black; background-position: initial initial; background-repeat: initial initial; "><a href="http://www.blogger.com/<a%20href=" com="" latex="\bg_black"" target="_blank">http://www.codecogs.com/eqnedit.php?latex=\bg_black</a>\textup{sinc}(x)=\textup{sin}(πx)/(πx) \textup{ if } x ≠0, \textup{and sinc}(0) = 1" target="_blank"><imgsrc="<a href="http://latex.codecogs.com/gif.latex?\bg_black" target="_blank">http://latex.codecogs.com/gif.latex?\bg_black \textup{sinc}(x) =\textup{sin}(πx)/(πx) \textup{ if } x ≠ 0, \textup{and sinc}(0) = 1" title="\bg_black \textup{sinc}(x) =\textup{sin}(πx)/(πx) \textup{ if } x ≠ 0, \textup{and sinc}(0) = 1 "/><o:p></o:p></imgsrc="<a></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">You can load the Simbad database into Aladin as well. Also, you can select individual sources and see an image of each source. If you do this with this data file, you get what looks like a giant region of exposure. </p><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiWpN3ZfwfNfmWsVNseDZND0c6FnsX9TT9uopKU12f1NmJwEjHJjchNFOubvUQOuHAE0aLifznfCtcwFyl5Jf2a4uTBgMKpB1HNYm9IWPks4k-wO7YQHmvUk0bZ05-bvS5ATelLfRpDA2f1/s1600/5.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiWpN3ZfwfNfmWsVNseDZND0c6FnsX9TT9uopKU12f1NmJwEjHJjchNFOubvUQOuHAE0aLifznfCtcwFyl5Jf2a4uTBgMKpB1HNYm9IWPks4k-wO7YQHmvUk0bZ05-bvS5ATelLfRpDA2f1/s320/5.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5682150563263764498" style="display: block; margin-top: 0px; margin-right: auto; margin-bottom: 10px; margin-left: auto; text-align: center; cursor: pointer; width: 320px; height: 297px; " /></a><p class="MsoNormal">What I wanted to do next was to find the sinc function of the light from these images. Theoretically, you could do this by extracting a spectrum of the region containing the point source, and plotting how intensity oscillates as you move further away from the center (maximum). I played around with Aladin for a ridiculously long amount of time until I found a feature that gives you the full width at half maximum in each direction. To find this, I used the “Phot”[ometry] command to select a circular region containing the point source, then examined the table of values to find the full width values. This was harder for me than it should have been, because I wasn’t sure exactly what I was looking for. I only vaguely remembered that “full width at half maximum” can refer to a sinc function.</p><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg1YffpSDdmtlaZ5KKCYT88ItQ0PSyl8tA-zZwbJXB_2AoqNpTRcbLFv0H5dUCvLvhUr4-cwumuiczDsEuAxBDZxRRPws_gK6mZtMRT7e_GBsPajXvYSFWG2OMjVBSGsOiUUEn_ca4eG0Be/s1600/6.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg1YffpSDdmtlaZ5KKCYT88ItQ0PSyl8tA-zZwbJXB_2AoqNpTRcbLFv0H5dUCvLvhUr4-cwumuiczDsEuAxBDZxRRPws_gK6mZtMRT7e_GBsPajXvYSFWG2OMjVBSGsOiUUEn_ca4eG0Be/s320/6.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5682151356463095218" style="text-align: left;display: block; margin-top: 0px; margin-right: auto; margin-bottom: 10px; margin-left: auto; cursor: pointer; width: 320px; height: 318px; " /></a><p class="MsoNormal" style="text-align: left;"> The values I found for full width at half maximum for each source were:<o:p></o:p></p> <div align="center"> <table class="MsoTableGrid" border="1" cellspacing="0" cellpadding="0" style="border-collapse:collapse;border:none;mso-border-alt:solid windowtext .5pt; mso-yfti-tbllook:1184;mso-padding-alt:0in 5.4pt 0in 5.4pt"> <tbody><tr> <td width="85" valign="top" style="width:63.9pt;border:solid windowtext 1.0pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; "><b>Source<o:p></o:p></b></p> </td> <td width="132" valign="top" style="width:99.0pt;border:solid windowtext 1.0pt; border-left:none;mso-border-left-alt:solid windowtext .5pt;mso-border-alt: solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; "><b>FWHM (appx)<o:p></o:p></b></p> </td> </tr> <tr> <td width="85" valign="top" style="width:63.9pt;border:solid windowtext 1.0pt; border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt; padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; "><i>A<o:p></o:p></i></p> </td> <td width="132" valign="top" style="width:99.0pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; ">13<o:p></o:p></p> </td> </tr> <tr> <td width="85" valign="top" style="width:63.9pt;border:solid windowtext 1.0pt; border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt; padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; "><i>B<o:p></o:p></i></p> </td> <td width="132" valign="top" style="width:99.0pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom: 0.0001pt; text-align: center; ">11<o:p></o:p></p> </td> </tr> </tbody></table> </div> <p class="MsoNormal">We want to see how the full width at half maximum (FWHM) relates to other qualities of the telescope. <o:p></o:p></p> <p class="MsoNormal"><a href="http://www.codecogs.com/eqnedit.php?latex=\textup{FWHM}%20=%20\frac{\lambda%20}{D}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{FWHM}%20=%20\frac{\lambda%20}{D}" title="\textup{FWHM} = \frac{\lambda }{D}" /></a><o:p></o:p></p> <p class="MsoNormal"><b><span style="mso-bidi-font-family:Calibri;mso-bidi-theme-font:minor-latin;color:red">WAIT</span></b><span style="mso-bidi-font-family:Calibri;mso-bidi-theme-font:minor-latin;color:red">! </span>At this point, I asked Professor Johnson if I was going about this project the right way, and he suggested a much easier way to do it. Instead of using the FWHM (which I am still unclear if this would actually work), he suggested I change the display range and measure the airy rings.<o:p></o:p></p> <p class="MsoNormal">I was having some trouble finding the way to change the display range (sad, I know) until I found that there is an undergraduate mode for the Aladin software! It looks like the undergraduate mode if just the regular version with a bunch of features locked out, but it’s a little less overwhelming to start with.<o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">If you zoom in, you can see what look like airy rings. These are the distances at which there appears to be a dark spot. We can see the cause of these if we look at the function of light reflected on a screen. The airy ring, therefore, occurs at this first node. On the image, we can measure how far it is between the center (max) and this first mode (min). The image gives a value of approximately 0.1 degrees.</p><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3H6VBmAPolgMCUi_yXhrsWLa60cfB82NLOfNEH51v697NThP51kQdycKj8IG70afbZFOSCBaMx67yDTvfDRDxWmQdFl0kWnd4qOzRNj7y_CYnnfhyphenhyphenBstpWNbdslFxfFjUn5OaPmFTLjj7/s1600/7.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3H6VBmAPolgMCUi_yXhrsWLa60cfB82NLOfNEH51v697NThP51kQdycKj8IG70afbZFOSCBaMx67yDTvfDRDxWmQdFl0kWnd4qOzRNj7y_CYnnfhyphenhyphenBstpWNbdslFxfFjUn5OaPmFTLjj7/s320/7.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5682152460179286722" style="float: right; margin-top: 0px; margin-right: 0px; margin-bottom: 10px; margin-left: 10px; cursor: pointer; width: 320px; height: 175px; " /></a><p class="MsoNormal">How can we use this to find the angular resolution of the image? It is given that the image is in K-band. This means that the frequency of light is between 20 and 40 GHz. This corresponds to 7.5*10^-1 to 1.5 cm for the wavelength. Using the average-ish value, we can say that lamba equals 1.15 cm. Using the formula:<o:p></o:p></p> <p class="MsoNormal"><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20sin(\Theta%20)%20=%201.22\frac{\lambda%20}{D}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20sin(\Theta%20)%20=%201.22\frac{\lambda%20}{D}" title="\bg_black sin(\Theta ) = 1.22\frac{\lambda }{D}" /></a><o:p></o:p></p> <p class="MsoNormal">Substituting theta with the value found by the airy ring method. Note that these values are in radians and nanometers. Also, the small angle approximation is used here.<o:p></o:p></p> <p class="MsoNormal"><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20sin(.0017)=1.22{\frac{\1.5}{D}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20sin(.0017)=1.22{\frac{\1.5}{D}}" title="\bg_black sin(.0017)=1.22{\frac{\1.5}{D}}" /></a><o:p></o:p></p> <p class="MsoNormal">So, <o:p></o:p></p> <p class="MsoNormal"><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20D\approx%2010.48%20\textup{%20meters}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20D\approx%2010.48%20\textup{%20meters}" title="\bg_black D\approx 10.48 \textup{ meters}" /></a><o:p></o:p></p> <p class="MsoNormal">This implies that the diameter of the telescope is roughly 10 meters. Which telescopes do we know of that have an aperture of 10 meters? The first one that springs to mind for me is Keck. However, how could this be? The K-band is infrared. Now, there are two ways to figure out from which telescope this file was observed. One way is to look at the declination of the pair.<o:p></o:p></p> <p class="MsoNormal">The source is at about +2:30. This means that the object is just slightly north of the celestial equator. Also, I found the observation date in the image – 4/27/2010, at 11:28 – so we can check if the pair would have been visible from Keck at that time. I will assume the time given is PM, rather than military time (though it is possible there is some mix-up with times zones)? 11:30 pm is probably an acceptable approximation, however). The right ascension of the image is around 18 hours. At the vernal equinox, the RA is 0 hours at noon. So, we must calculate how much that changes in a month.<o:p></o:p></p> <p class="MsoNormal">Each sidereal day is four minutes shorter than a solar day. So, the sidereal time will be at about 3:00 at noon on April 27 (this is a rough estimation). At 18:00, sidereal time, it will be roughly be 3:00 AM April 28. This is the time that the source would be directly overhead. It would totally be possible to see this source from Keck at 11:28 PM, April 27. <o:p></o:p></p> <p class="MsoNormal">If you look at this paper: <a href="http://www.astro.cornell.edu/academics/courses/astro233/symp06/symp06.pdf">http://www.astro.cornell.edu/academics/courses/astro233/symp06/symp06.pdf</a>, you will see that there is actually a way for Keck to observe in infrared! The NIRC is a near-infrared camera that was on Keck at one point. NIRC2 would have been operational on April 27, 2010, when this data was taken. <o:p></o:p></p><p class="MsoNormal"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg4mjAsQBs20j0d05Vco3QRfqlr4hlyhwGEo-UdoDXWTREGBL88KqVjAyTJWT7QIniJ3nTNsKgyUOAeU_TCvdeoU_6AsA8jhCQGC1OJUbKUpKhT2nGbXFAzVX6HxkWIFbAC-YCrei9o-f4v/s1600/10.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg4mjAsQBs20j0d05Vco3QRfqlr4hlyhwGEo-UdoDXWTREGBL88KqVjAyTJWT7QIniJ3nTNsKgyUOAeU_TCvdeoU_6AsA8jhCQGC1OJUbKUpKhT2nGbXFAzVX6HxkWIFbAC-YCrei9o-f4v/s320/10.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5682153024550367154" style="display: block; margin-top: 0px; margin-right: auto; margin-bottom: 10px; margin-left: auto; text-align: center; cursor: pointer; width: 320px; height: 290px; " /></a></p><div>The second way to figure out from which telescope the image came involves some marginally sketchy stalking… I looked up the observers, who were named in the FITS header, and saw that the file mentioned Cornell IR. NIRC is associated with Cornell, as well. Also, I looked at a paper published by these observers, using data taken over the nights centered around April 27, 2010, and the paper said that Keck was used. Although none of these people were actually named in the list of people who were allocated that night at Keck, it does say on the telescope time allocation site that that night, NIRC 2 was being used. (<a href="http://www2.keck.hawaii.edu/observing/schedule/index.php?sched=Both+Tels&year=2010&month=4&go=GO">http://www2.keck.hawaii.edu/observing/schedule/index.php?sched=Both+Tels&year=2010&month=4&go=GO#</a>) I am a little skeptical, since the people listed in the FITS file as observers were not listed as observing at Keck that night, but I would say that it is likely that this image was taken with NIRC2 at Keck2.</div><p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com3tag:blogger.com,1999:blog-5849210431402234073.post-23339492937359848862011-12-03T19:46:00.000-08:002011-12-03T19:47:22.266-08:00The Exoplanet App. Also... the floodgates lift<p class="MsoNormal">Thus begins my backlog of posts, all of which have been in the “almost done” stage of being for varying amounts of time (2 weeks-today). <o:p></o:p></p> <p class="MsoNormal">First, I wanted to draw everyone’s attention to this cool Iphone app that Professor Johnson showed me! It’s called “Exoplanet,” Hanno Rein, and it is basically a visual catalog of exoplanets! You can zoom in on systems, see the orbit, see light curves and other cool stuff! There were a few systems that I have found that are particularly cool:<o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><b>HD 98649</b> – Gas giant, 7 Jupiter masses, orbital period of 10400 days (which is around 30 years!)<o:p></o:p></p> <p class="MsoNormal">This planet is so big it makes the sun move! It is worth looking at on the Exoplanet app because not only are its light curves interestingly-shaped, but the animation shows the sun moving! Both planet and sun are orbiting around their mutual center of mass. The eccentricity of this sun’s sole planet is 0.860 (a highly elliptical orbit). It is surprising that not all parameters for this system are posted - maybe the system is so new that no papers have been published on it yet? I want to study this system, but its HUGE period probably makes it impractical to study.<o:p></o:p></p> <p class="MsoNormal"><b>HD 20794 system: b, c and d</b> – each around 2-4 Earth masses, orbital periods of 18, 40, and 90 days, respectively.<o:p></o:p></p> <p class="MsoNormal">This system exists unfortunately right inside the habitable zone of their star. The star is 0.7 times the mass of our sun, and the orbits are nearly circular. I love systems with many planets – three may not be “many,” but it’s more than usual for sure!<o:p></o:p></p> <p class="MsoNormal"><b>PSR 1257 12 b</b> – 0.02 Earth masses, orbital period of 25 days. <o:p></o:p></p> <p class="MsoNormal">How is this possible? How can a planet be so tiny? This seems more like an asteroid than a planet… it is roughly ten times smaller than Mercury. Other planets in this system are 3 and 4 times earth mass, but all the planets orbit too close to be in the habitable zone. <o:p></o:p></p><p class="MsoNormal"><br /></p> <p class="MsoNormal">Honestly, there is no limit to cool systems. Each one has something new and interesting – “how is a planet that big orbiting so fast?” – “how did such a skewed system form?” – “why hasn’t that planet collapsed into the sun?”. This app is a good diving board for new research projects!<o:p></o:p></p>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com0tag:blogger.com,1999:blog-5849210431402234073.post-88468592597045462252011-11-18T22:10:00.000-08:002011-11-18T22:15:00.542-08:00And again...<a href="http://inspirehep.net/record/928153/files/arXiv:1109.4897.pdf?version=2">http://inspirehep.net/record/928153/files/arXiv:1109.4897.pdf?version=2</a><div><br /></div><div>I'm still skeptical.... </div>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com0tag:blogger.com,1999:blog-5849210431402234073.post-37018215537214647462011-11-18T12:23:00.000-08:002011-11-18T12:24:02.504-08:00What does it mean to be a professional astronomer?<p class="MsoNormal">When I think of being a professional astronomer, I can’t help but keep thinking about how amazingly awesome it would be to go observing! After our class trip to Palomar, I want nothing more than to have a project where I can go observing. Which, of course, requires that I think up a project and get telescope time. But after that…… <span style="font-family: Wingdings; ">J</span></p><p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal">The process to become a professional astronomer, as far as I know, is basically constrained to academia. Professors and staff scientists of universities/ organizations such as CfA are the only people who can actually participate in this process, as it helps tremendously to be associated with a telescope-owning institution to study the sky. In this vein, to become a professional astronomer, one would need to follow the path of being an undergraduate -> grad student -> postdoc -> assistant professor -> professor. Each step of this transition needs something new. To become a grad student, I need to do research as an undergrad to learn research skills, and get good grades and test scores so I can get into a grad school. To become a postdoc, I would need to win a fellowship. I am unclear if this is the only way to become a postdoc. I’m also unsure how one goes about becoming a professor – I assume that if you are a really good postdoc, then you can apply for jobs and if you’re lucky, you’ll get one?<o:p></o:p></p> <p class="MsoNormal">Also, your job can limit the science you can do. As a undergrad, I cannot apply for telescope time, unless someone qualified submits the proposal for me and I am listed only as a coauthor. Something I have learned in that last few months is how extremely important knowing people is. Who you know is almost more important than what you know. Working with people (or, in my case, learning from people) is one of the most important ways to grow as scientist. No matter how many papers on adaptive optics I read, I could never have gotten the same emotional fascination with them nor rough understanding of them that I did from listening to people explain it. <o:p></o:p></p> <p class="MsoNormal">There also seems to be a lot of luck in becoming a professor – if the system you are studying happens to contain some crazy new structure, it could boost your career hugely. However, this also requires you have enough skill to recognize an opportunity when you see it. I have met a few postdocs who are doing such interesting projects, and being so successful with them, that I can’t believe they aren’t professors yet. This shows how high the standards are in the field. I think that rather than it always being the researcher who is the BEST who become a professor, there is an element of luck to it. There is not a definite measurement of being the “best” researcher, after all. <o:p></o:p></p> <p class="MsoNormal">Like any field, I think the most important things to becoming a professional astronomer are connections and luck, with a necessary amount of skill to take advantage of the luck you have. More skill can offset a lack of luck, but luck cannot offset a lack of skill. <o:p></o:p></p>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com1tag:blogger.com,1999:blog-5849210431402234073.post-21835689559155248812011-11-06T20:51:00.000-08:002011-11-06T22:29:28.038-08:00The Formation of Stars<div style="text-align: center;"><b style="text-align: left; "><br /></b></div><div style="text-align: center;"><b style="text-align: left; ">Introduction</b></div><div> <p class="MsoNormal">Here we consider the time scale of star formation and stability. We do this by considering the time it would take a cloud of particles to collapse into a star, and in this process derive an approximation for the Jean’s Length. <b><o:p></o:p></b></p><p class="MsoNormal"></p><div><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjsEgHt2nldLJTwb1iIhtMN3XZc0dCLiVluf-AiLNvwk7Q7Flbzntw2PPPrlavLziRUjgYZJadlx-fd17hirsz4H47q_VEbYiTZbQnF8ZHG4Gw6li4x5Bot2ubv-95Ix8WTFxCnrbYr0B9n/s1600/ay8.png" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjsEgHt2nldLJTwb1iIhtMN3XZc0dCLiVluf-AiLNvwk7Q7Flbzntw2PPPrlavLziRUjgYZJadlx-fd17hirsz4H47q_VEbYiTZbQnF8ZHG4Gw6li4x5Bot2ubv-95Ix8WTFxCnrbYr0B9n/s320/ay8.png" border="0" alt="" id="BLOGGER_PHOTO_ID_5672132243754636882" style="display: block; margin-top: 0px; margin-right: auto; margin-bottom: 10px; margin-left: auto; text-align: center; cursor: pointer; width: 320px; height: 256px; " /></a></div><div>This is the system we must consider. The grey cloud is all the dust orbiting the point mass. The test particle is at the far right edge of the red line. The red line is the semimajor axis of the ellipse. The test particle must travel along the red line to reach the center. The time it takes to travel this distance is the free fall time.</div><div><br /></div><p></p> <p class="MsoNormal"><b>Solution<o:p></o:p></b></p> <p class="MsoListParagraphCxSpFirst" style="text-indent:-.25in;mso-list:l0 level1 lfo1"><!--[if !supportLists]-->a.<span style="font:7.0pt "Times New Roman""> </span><!--[endif]--><i>Consider a test particle in an e=1 “orbit” around a point mass.</i><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">An orbit with an eccentricity of 1 would be a straight line. That is, it would mean that that the planet would orbit back and forth straight across the point mass:<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><o:p> </o:p></p><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi4wlRdF-Tnti7601C6REsAOKMFrE4c7E_5ZEA9RxbwgZQP2O1qo7fkJpSTExQ4zhmOz24TokLmf9oObKA3OblZVxYZffLdXIfS17j0zK5b4ZRfYkj8bjWcQ5GVWWAARV8TjVwQQMJP96Xf/s1600/ay7.png" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi4wlRdF-Tnti7601C6REsAOKMFrE4c7E_5ZEA9RxbwgZQP2O1qo7fkJpSTExQ4zhmOz24TokLmf9oObKA3OblZVxYZffLdXIfS17j0zK5b4ZRfYkj8bjWcQ5GVWWAARV8TjVwQQMJP96Xf/s320/ay7.png" border="0" alt="" id="BLOGGER_PHOTO_ID_5672132090666251922" style="display: block; margin-top: 0px; margin-right: auto; margin-bottom: 10px; margin-left: auto; text-align: center; cursor: pointer; width: 320px; height: 135px; " /></a><p class="MsoListParagraphCxSpMiddle">If the test particle starts at the edge of the orbit - that is, r away from the point mass, then the distance it will have to fall is equal to r (the semimajor axis of the whole clous - see image above!)</p><p class="MsoListParagraphCxSpMiddle">Kepler’s third law, from memory, is: <ahref="http: com="" latex="\bg_black" dm="p(r)4\pi" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\bg_black%20dM%20=%20p(r)4\pi%20r^2dr" title="\bg_black dM = p(r)4\pi r^2dr" /><o:p></o:p></ahref="http:></p> <p class="MsoListParagraphCxSpMiddle">The expression for mass of a sphere, in terms of its density, is: <a href="http://www.codecogs.com/eqnedit.php?latex=M%20=%20\frac{4\pi%20r^{3}\bar{p}}{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20M%20=%20\frac{4\pi%20r^{3}\bar{p}}{3}" title="M = \frac{4\pi r^{3}\bar{p}}{3}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">Combining these two expressions, we can solve for the free fall time by considering the period. Since the period, p, is the time it would take the test particle to “cross” over the point mass twice and return to its initial position, one half of the period will equal the free-fall time. Also, we can substitute 2<b><span style="color:#4BACC6;mso-themecolor:accent5">a</span></b> for <b><span style="color:red">r</span></b> (see the diagram above). <o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\frac{a^{3}4\pi%20^{2}}{Gp^{2}}%20=%20\frac{4\pi%20r^{3}\bar{p}}{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\frac{a^{3}4\pi%20^{2}}{Gp^{2}}%20=%20\frac{4\pi%20r^{3}\bar{p}}{3}" title="\frac{a^{3}4\pi ^{2}}{Gp^{2}} = \frac{4\pi r^{3}\bar{p}}{3}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">Solve for p (period. Note that this is different from the symbol for density, which is p-bar):<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\sqrt{\frac{3\pi%20a^{3}}{G%20r^{3}\bar{p}}}%20=%20{p}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\sqrt{\frac{3\pi%20a^{3}}{G%20r^{3}\bar{p}}}%20=%20{p}" title="\sqrt{\frac{3\pi a^{3}}{G r^{3}\bar{p}}} = {p}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">We know the relationships between t<sub>ff </sub>and p, as well as r and a:<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\frac{p}{2}%20=%20T_{ff}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\frac{p}{2}%20=%20T_{ff}" title="\frac{p}{2} = T_{ff}" /></a> <a href="http://www.codecogs.com/eqnedit.php?latex=2a%20=%20r" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%202a%20=%20r" title="2a = r" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\sqrt{\frac{3\pi%20a^{3}}{G%20(2a)^{3}\bar{p}}}%20=%20{\frac{p}{2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\sqrt{\frac{3\pi%20a^{3}}{G%20(2a)^{3}\bar{p}}}%20=%20{\frac{p}{2}}" title="\sqrt{\frac{3\pi a^{3}}{G (2a)^{3}\bar{p}}} = {\frac{p}{2}}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\sqrt{\frac{3\pi%20a^{3}}{8G%20a^{3}\bar{p}}}%20=%20{\frac{p}{2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\sqrt{\frac{3\pi%20a^{3}}{8G%20a^{3}\bar{p}}}%20=%20{\frac{p}{2}}" title="\sqrt{\frac{3\pi a^{3}}{8G a^{3}\bar{p}}} = {p}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\frac{1}{2}\sqrt{\frac{3\pi}{8G\bar{p}}}%20=%202T_{ff}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\frac{1}{2}\sqrt{\frac{3\pi}{8G\bar{p}}}%20=%20T_{ff}" title="\frac{1}{2}\sqrt{\frac{3\pi}{8G\bar{p}}} = T_{ff}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\sqrt{\frac{3\pi}{32G\bar{p}}}%20=%20T_{ff}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\sqrt{\frac{3\pi}{32G\bar{p}}}%20=%20T_{ff}" title="\sqrt{\frac{3\pi}{32G\bar{p}}} = T_{ff}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">This expression describes the free fall time of a particle on the edge of the cloud, and this the time it would take the cloud of dust to collapse into a star. We are assuming that the test particle experiences exactly one half a period of motion. <o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><o:p> </o:p></p> <p class="MsoListParagraphCxSpMiddle" style="text-indent:-.25in;mso-list:l0 level1 lfo1"><!--[if !supportLists]-->b.<span style="font:7.0pt "Times New Roman""> </span><!--[endif]--><i>Determine the dynamical time for a sound wave to cross this same distance. </i><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">Defining the speed of sound as c<sub>s</sub>, we can define the time it would take a sound wave to cross the same distance as a test particle as;<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=t_{s}%20=%20\frac{2a}{c_{s}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20t_{s}%20=%20\frac{2a}{c_{s}}" title="t_{s} = \frac{2a}{c_{s}} " /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">We can solve for the length variable – that is, <b><span style="color:#00B0F0">a</span></b>.<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=t_{s}%20=%20t_{ff}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20t_{s}%20=%20t_{ff}" title="t_{s} = t_{ff}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\frac{2a}{c_{s}}%20=%20\sqrt{\frac{3%20\pi}{32G\bar{p}}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\frac{2a}{c_{s}}%20=%20\sqrt{\frac{3%20\pi}{32G\bar{p}}}" title="\frac{2a}{c_{s}} = \sqrt{\frac{3 \pi}{32G\bar{p}}}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=a%20=%20\frac{c_{s}}2{}\sqrt{\frac{3%20\pi}{32G\bar{p}}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20a%20=%20\frac{c_{s}}2{}\sqrt{\frac{3%20\pi}{32G\bar{p}}}" title="a = \frac{c_{s}}2{}\sqrt{\frac{3 \pi}{32G\bar{p}}}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">Which is equivalent to:<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=R_{j}%20=%20\sqrt{\frac{3%20\pi%20c_{s}^{2}}{128G\bar{p}}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20R_{j}%20=%20\sqrt{\frac{3%20\pi%20c_{s}^{2}}{128G\bar{p}}}" title="R_{j} = \sqrt{\frac{3 \pi c_{s}^{2}}{128G\bar{p}}}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><o:p> </o:p></p> <p class="MsoListParagraphCxSpLast">This length signifies the distance at which the forces of gravity and pressure are at an equilibrium. The gas cloud will neither expand nor contract at this size. <o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoListParagraphCxSpFirst"><o:p> </o:p></p> <p class="MsoListParagraphCxSpMiddle" style="text-indent:-.25in;mso-list:l0 level1 lfo1"><!--[if !supportLists]-->c.<span style="font:7.0pt "Times New Roman""> </span><!--[endif]--><i>What is the significance of Jean’s Length?</i><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">Jean’s Length shows the distance at which a cloud of dust would not be stable, and thus susceptible to collapse. The force of gravity attempts to pull the dust into the center, while the force of pressure tries to push it back out. If a sound wave takes longer to cross the distance than the actual particle of dust, then the force of pressure won’t be strong enough to prevent the dust cloud from collapsing. So, if a dust cloud is larger than Jean’s Length, that means that it cannot sustain itself, and will collapse into its center, resulting in stars!<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><o:p> </o:p></p> <p class="MsoListParagraphCxSpMiddle" style="text-indent:-.25in;mso-list:l0 level1 lfo1"><!--[if !supportLists]-->d.<span style="font:7.0pt "Times New Roman""> </span><!--[endif]--><i>Consider a collapsing cloud with a radius equal to Jean’s Length: when the cloud reaches a radius equal to one half of its initial radius, by what fractional amount had R<sub>J </sub>changed?</i><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">To see the ratio of the Jeans’ Length for a cloud of equal mass but different radii, we must consider how the density changes from one size to another. So, first, we can find the densities of the two clouds, assuming a constant mass. <o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\bar{p_{o}}=\frac{M}{\frac{4}{3}%20\pi%20R_{o}^{3}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\bar{p_{o}}=\frac{M}{\frac{4}{3}%20\pi%20R_{o}^{3}}" title="\bar{p_{o}}=\frac{M}{\frac{4}{3} \pi R_{o}^{3}}" /></a> <a href="http://www.codecogs.com/eqnedit.php?latex=\bar{p_{J}}=\frac{M}{\frac{4}{3}%20\pi%20R_{J}^{3}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\bar{p_{J}}=\frac{M}{\frac{4}{3}%20\pi%20R_{J}^{3}}" title="\bar{p_{J}}=\frac{M}{\frac{4}{3} \pi R_{J}^{3}}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">So:<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\bar{p_{J}}\frac{4}{3}%20\pi%20R_{J}^{3}%20=%20\bar{p_{o}}\frac{4}{3}%20\pi%20R_{o}^{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\bar{p_{J}}\frac{4}{3}%20\pi%20R_{J}^{3}%20=%20\bar{p_{o}}\frac{4}{3}%20\pi%20R_{o}^{3}" title="\bar{p_{J}}\frac{4}{3} \pi R_{J}^{3} = \bar{p_{o}}\frac{4}{3} \pi R_{o}^{3}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\bar{p_{J}}R_{J}^{3}%20=%20\bar{p_{o}}%20R_{o}^{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\bar{p_{J}}R_{J}^{3}%20=%20\bar{p_{o}}%20R_{o}^{3}" title="\bar{p_{J}}R_{J}^{3} = \bar{p_{o}} R_{o}^{3}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">We can plug in 0.5R<sub>O</sub> and see how R<sub>J </sub>changes. As R<sub>o</sub> shrinks, the density p must grow by a balancing amount. <o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=X\bar{p_{o}}(0.5R_{o})^{3}%20=%20\bar{p_{o}}%20R_{o}^{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20X\bar{p_{o}}(0.5R_{o})^{3}%20=%20\bar{p_{o}}%20R_{o}^{3}" title="X\bar{p_{o}}(0.5R_{o})^{3} = \bar{p_{o}} R_{o}^{3}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=X\bar{p_{o}}(0.5R_{o})^{3}%20=%20\bar{p_{o}}%20R_{o}^{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20X\bar{p_{o}}(0.5R_{o})^{3}%20=%20\bar{p_{o}}%20R_{o}^{3}" title="X\bar{p_{o}}(0.5R_{o})^{3} = \bar{p_{o}} R_{o}^{3}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=X\bar{p_{o}}0.125R_{o}^{3}%20=%20\bar{p_{o}}%20R_{o}^{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20X\bar{p_{o}}0.125R_{o}^{3}%20=%20\bar{p_{o}}%20R_{o}^{3}" title="X\bar{p_{o}}0.125R_{o}^{3} = \bar{p_{o}} R_{o}^{3}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=(8)\bar{p_{o}}0.125R_{o}^{3}%20=%20\bar{p_{o}}%20R_{o}^{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20(8)\bar{p_{o}}0.125R_{o}^{3}%20=%20\bar{p_{o}}%20R_{o}^{3}" title="(8)\bar{p_{o}}0.125R_{o}^{3} = \bar{p_{o}} R_{o}^{3}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">So the density increases by a factor of 8 if the radius decreases by a factor of 2. <o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">Thus, <o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=R_{J}%20=\sqrt{\frac{\pi%20c_{s}^{2}}{G%20\bar{p}}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20R_{J}%20=\sqrt{\frac{\pi%20c_{s}^{2}}{G%20\bar{p}}}" title="R_{J} =\sqrt{\frac{\pi c_{s}^{2}}{G \bar{p}}}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">If R becomes 0.5R<sub>o</sub> then the density becomes 8p:<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=R_{J}%20=\sqrt{\frac{\pi%20c_{s}^{2}}{G%20\bar{8p}}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20R_{J}%20=\sqrt{\frac{\pi%20c_{s}^{2}}{G%20\bar{8p}}}" title="R_{J} =\sqrt{\frac{\pi c_{s}^{2}}{G \bar{8p}}}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=R_{J}%20=\frac{1}{^{\sqrt{8}}}\sqrt{\frac{\pi%20c_{s}^{2}}{G%20\bar{p}}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20R_{J}%20=\frac{1}{^{\sqrt{8}}}\sqrt{\frac{\pi%20c_{s}^{2}}{G%20\bar{p}}}" title="R_{J} =\frac{1}{^{\sqrt{8}}}\sqrt{\frac{\pi c_{s}^{2}}{G \bar{p}}}" /></a><br /><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">So, R<sub>j </sub>decreases by a factor of 0.35. This is fractional change of about 65%; the new value of R<sub>J</sub> will be around 65% of the old one.<sub><o:p></o:p></sub></p> <p class="MsoListParagraphCxSpLast"><o:p> </o:p></p> <p class="MsoNormal"><b>Conclusion<o:p></o:p></b></p> <p class="MsoNormal">Jeans’ Length, the point of balance between the forces of pressure and gravity, can help you determine the behavior of a cloud of dust. By knowing the radius and mass, or by knowing just the density, you can calculate whether the cloud will expand or contract. If the cloud contracts, then it will likely become a star. Decreasing the radius of a cloud by 50% decreasing the Jeans’ Length by 35%, so the new length is 65% of the old one. <o:p></o:p></p> <p class="MsoNormal"><b>Acknowledgements<o:p></o:p></b></p> <p class="MsoNormal">Thanks to Iryna and Monica! We had a lot of fun with this problem :)<o:p></o:p></p></div>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com2tag:blogger.com,1999:blog-5849210431402234073.post-31377142482364375342011-11-06T17:08:00.001-08:002011-11-06T17:17:13.193-08:00Look what I found!!<span class="Apple-style-span">So, as I attempt to do the write-up for the most recent worksheet, I instead find myself procrastinating by reading Astrobites... and look what I found!!</span><div><span class="Apple-style-span"><br /></span></div><div><a href="http://astrobites.com/2011/08/15/qa-with-john-johnson-what-should-you-do-in-graduate-school/"><span class="Apple-style-span">http://astrobites.com/2011/08/15/qa-with-john-johnson-what-should-you-do-in-graduate-school/</span></a></div><div><span class="Apple-style-span"><br /></span></div><div><span class="Apple-style-span">:) There are lots of other articles about grad school, both about getting in and what to do once you have gotten in, but I thought you guys you especially like this one!! For obvious reasons...</span></div><div><span class="Apple-style-span"><br /></span></div><div><span class="Apple-style-span">Here is one comment that he made regarding classes in graduate school, one that I think applies to us undergrads at Caltech:<span class="Apple-style-span"> Professor Johnson explains that it was a bad thing that "<span class="Apple-style-span" style="line-height: 19px; background-color: rgb(0, 0, 0); ">I was just concentrating on getting the current week’s problem sets finished so I could get to the next one and then finish the course with a good grade." This is something I have been thinking about a lot the last few months, mostly inspired by the way Ay20 is run. I think I lot of people get through Caltech this way, but we don't have to!</span></span></span></div><div><span class="Apple-style-span"><span class="Apple-style-span" style="line-height: 19px;"><br /></span></span></div><div><span class="Apple-style-span"><span class="Apple-style-span" style="line-height: 19px;">Astrobites is actually pretty cool! There are a lot of contributors, resulting in very diverse reading! .... and many hours worth of procrastination... I started my write-up at three.... oops....</span></span></div>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com0tag:blogger.com,1999:blog-5849210431402234073.post-9913811924536130182011-10-27T23:22:00.000-07:002011-10-27T23:32:13.886-07:00Hydrostatic Equilibrium and the Virial Theorem<p class="MsoNormal"><b>Introduction<o:p></o:p></b></p> <p class="MsoNormal">Here we consider how to show the equation for hydrostatic equilibrium. The methodology used is a force balance followed by some algebra. <span> </span><b><o:p></o:p></b></p> <p class="MsoNormal"><b>Solutions<o:p></o:p></b></p> <p class="MsoNormal"><b><i>Question</i></b>: <i>Find the differential equation describing hydrostatic equilibrium. <span> </span></i><o:p></o:p></p> <p class="MsoListParagraphCxSpFirst" style="text-indent:-.25in;mso-list:l0 level1 lfo1"><!--[if !supportLists]--><span><span>a.<span style="font:7.0pt "Times New Roman""> </span></span></span><!--[endif]--><i>Write an expression for the differential element dM contained in the cell.</i><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">Considering a mass shell with thickness dr and density p(r), we can write the equation for the differential element contained in this cell by finding the volume of the shell and multiplying this by the density, as follows:<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><o:p> </o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20dM%20=%20p(r)4\pi%20r^2dr" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20dM%20=%20p(r)4\pi%20r^2dr" title="\bg_black dM = p(r)4\pi r^2dr" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><o:p> </o:p></p> <p class="MsoListParagraphCxSpMiddle" style="text-indent:-.25in;mso-list:l0 level1 lfo1"><!--[if !supportLists]--><span><span>b.<span style="font:7.0pt "Times New Roman""> </span></span></span><!--[endif]--><i>What is the gravitational force, F<sub>g</sub>, acting on the shell?</i><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">We know the differential element for the mass (that is, the mass of an infinitely thin shell). We can use the general equation for gravitational force:<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20F_{g}=-\frac{{GMm}}{r^2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20F_{g}=-\frac{{GMm}}{r^2}" title="\bg_black F_{g}=-\frac{{GMm}}{r^2}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">When M and m describe the two masses. However! We do not have two masses here. We have a mass, M, and the differential element dM that represents the mass of the shell. So, we shold rewrite the expression for F<sub>g </sub>as:<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20F_{g}=-\frac{{GM(r)dm}}{r^2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20F_{g}=-\frac{{GM(r)dm}}{r^2}" title="\bg_black F_{g}=-\frac{{GM(r)dm}}{r^2}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">When M(r) describes the mass of the sun up to a radius r. <o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><o:p> </o:p></p> <p class="MsoListParagraphCxSpMiddle" style="text-indent:-.25in;mso-list:l0 level1 lfo1"><!--[if !supportLists]--><span><span>c.<span style="font:7.0pt "Times New Roman""> </span></span></span><!--[endif]--><i>Assuming there is a pressure P(r) acting to balance F<sub>g</sub>, write the net force equation in terms of the relevant qualities.</i><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">We know:<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20P(r)=\frac{force}{area}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20P(r)=\frac{force}{area}" title="\bg_black P(r)=\frac{force}{area}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">We want to show that the forces balance, so we can set this equal to the expression we found for F<sub>g </sub>in part b. Solving for the F<sub>p</sub>, force of pressure, and substitute this expression in:<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20P(r){4\pi%20r^{2}}=F_{p}%20=%20F_{g}%20=%20-G\frac{M(r)dm}{r^{2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20P(r){4\pi%20r^{2}}=F_{p}%20=%20F_{g}%20=%20-G\frac{M(r)dm}{r^{2}}" title="\bg_black P(r){4\pi r^{2}}=F_{p} = F_{g} = -G\frac{M(r)p(r)dm}{r^{2}}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">It is correct that the signs are opposite, because the forces are acting in opposite directions. <o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><o:p> </o:p></p> <p class="MsoListParagraphCxSpMiddle" style="text-indent:-.25in;mso-list:l0 level1 lfo1"><!--[if !supportLists]--><span><span>d.<span style="font:7.0pt "Times New Roman""> </span></span></span><!--[endif]--><i>Show the equation of hydrostatic equilibrium.</i><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">To show that this equation fits the form of the equation of hydrostatic equilibrium, we can substitute in the value of dm that we found in part a, and then manipulate the equation algebraically:<o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20P(r){4\pi%20r^{2}}%20=%20-G\frac{M(r))dm}{r^{2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20P(r){4\pi%20r^{2}}%20=%20-G\frac{M(r))dm}{r^{2}}" title="\bg_black P(r){4\pi r^{2}} = -G\frac{M(r)dm}{r^{2}}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20P(r){4\pi%20r^{2}}%20=%20-G\frac{M(r))p(r)4\pi%20r^2)}{r^{2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20P(r){4\pi%20r^{2}}%20=%20-G\frac{M(r))p(r)4\pi%20r^2)}{r^{2}}" title="\bg_black P(r){4\pi r^{2}} = -G\frac{M(r))p(r)4\pi r^{2}dr)}{r^{2}}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20P(r)%20=%20-G\frac{M(r)p(r)dr}{r^{2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20P(r)%20=%20-G\frac{M(r)p(r)dr}{r^{2}}" title="\bg_black P(r) = -G\frac{M(r)p(r)dr}{r^{2}}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle"><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20\frac{P(r)}{dr}%20=%20-G\frac{M(r)p(r)}{r^{2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\frac{P(r)}{dr}%20=%20-G\frac{M(r)p(r)}{r^{2}}" title="\bg_black \frac{P(r)}{dr} = -G\frac{M(r)p(r)}{r^{2}}" /></a><o:p></o:p></p> <p class="MsoListParagraphCxSpMiddle">This is the equation of hydrostatic equilibrium!!<o:p></o:p></p> <p class="MsoListParagraphCxSpLast"><o:p> </o:p></p> <p class="MsoNormal"><b>Conclusion<o:p></o:p></b></p> <p class="MsoNormal">We have shown the derivation for the equation of hydrostatic equilibrium, using only simple physics and algebra. Pretty cool!!<o:p></o:p></p> <p class="MsoNormal"><b>Acknowledgements<o:p></o:p></b></p> <p class="MsoNormal">Thanks to Iryna and Monica for working with me on this worksheet – see their blogs for the discussions of other questions!<o:p></o:p></p><p class="MsoNormal">Also, I would like to thank my classmates whose blogs I have been reading - they gave me the idea to use an online Latek editor instead of typing equations in plain text. It looks way better this way! Thanks guys!</p>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com0tag:blogger.com,1999:blog-5849210431402234073.post-88699598009850510022011-10-22T17:28:00.000-07:002011-10-22T17:45:21.401-07:00Stellar Properties from Afar<div style="text-align: left;"><b>Introduction</b></div> <p class="MsoNormal">Here we consider a series of questions related to the sun and its properties. Topics for consideration include the luminosity of the sun and furthest distance at which a similar star could be seen, and what this distance is in various units of measurement. We are addressing questions 5-8 from the worksheet. </p> <p class="MsoNormal"><b>Solutions<o:p></o:p></b></p> <p class="MsoNormal"><b><i>Question</i></b>: <i>Given the that eye must receive 10 photons in order to see something, how far away would a solar twin have to be for it to be barely visible from Earth?</i><o:p></o:p></p> <p class="MsoNormal">We will need to make a few assumptions for this problem:</p><p class="MsoNormal"><span style="font-family:Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family: Symbol"><span><span class="Apple-tab-span" style="white-space:pre"> </span>·<span style="font:7.0pt "Times New Roman""> </span></span></span>The “read rate” of the eye (in what length of time the photons must arrive to be visible) is 1/30 second (this is the frame rate for most movies)</p><p class="MsoNormal"><span style="font-family:Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family: Symbol"><span><span class="Apple-tab-span" style="white-space:pre"><span class="Apple-tab-span" style="white-space:pre"> </span> </span>·<span style="font:7.0pt "Times New Roman""> </span></span></span>The sun emits light at 500 nm.</p><p class="MsoNormal"><span style="font-family:Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family: Symbol"><span><span class="Apple-tab-span" style="white-space:pre"> <span class="Apple-tab-span" style="white-space:pre"> </span></span>·<span style="font:7.0pt "Times New Roman""> </span></span></span>The human eye has a surface area of .0625 square centimeters that can receive photons.</p><p class="MsoNormal"><span style="font-family:Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family: Symbol"><span><span class="Apple-tab-span" style="white-space:pre"> <span class="Apple-tab-span" style="white-space:pre"> </span></span>·<span style="font:7.0pt "Times New Roman""> </span></span></span>The sun’s luminosity is 4*10<sup>33</sup> ergs.</p><p class="MsoListParagraphCxSpLast" style="text-indent:-.25in;mso-list:l0 level1 lfo1"><span class="Apple-tab-span" style="white-space:pre"> <span class="Apple-tab-span" style="white-space:pre"> </span></span>Using this assumption, we can calculate the minimum flux needed to see something. If the sun emits light at 500 nm, we can calculate the energy of 10 photons, and then find what this energy implies for the flux needed to see something.</p><p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>E=hv and c=v<span style="font-family:Symbol;mso-ascii-font-family: Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;mso-char-type:symbol;mso-symbol-font-family: Symbol"><span>l</span></span><o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>3*10<sup>10</sup>= v*500*10<sup>-7<o:p></o:p></sup></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>v=6*10<sup>14 </sup><o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>E=(6.6*10<sup>-27</sup>)( 6*10<sup>14</sup>)<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>E=3.96*10<sup>-12 </sup><o:p></o:p></p> <p class="MsoNormal">This is the energy for one photon. So, the energy for ten photons is 3.96*10<sup>-11 </sup><span> </span>ergs. Also, if we want to find the minimum flux that would bring this energy to the eye, we must consider the surface area of the eye.<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>Minimum flux = (3.96*10<sup>-11 </sup><span> </span>ergs)/(.0625 cm<sup>2</sup>)<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>Flux = 6.336*10<sup>-10 </sup><span> </span>ergs/ cm<sup>2</sup><o:p></o:p></p> <p class="MsoNormal">Now, we have the minimum flux that the eye needs to see something. If we find the distance such that a star with the sun’s luminosity will have this flux at Earth, we will have our answer. <o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>Flux=(luminosity)/(area)<o:p></o:p></p> <p class="MsoNormal">What we can take this equation to mean for this problem is that given the luminosity of the sun, if we divide it by the surface area of the sphere with radius r, we will get the flux at some point r away from the sun. We can do this because photons are emitted isotropically from the sun. Now, in a previous problem on this worksheet (go see Iryna’s blog for the solution to that problem!), we found the luminosity of the sun to be roughly 4*10<sup>33</sup> ergs. So:<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>6.336*10<sup>-10 </sup><span> </span>ergs/ cm<sup>2 </sup>= (4*10<sup>33</sup> ergs)/(4<span style="font-family:Symbol;mso-ascii-font-family:Calibri;mso-ascii-theme-font: minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin; mso-char-type:symbol;mso-symbol-font-family:Symbol"><span>p</span></span>(distance in cm)<sup>2</sup>)<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>Distance = 7.087*10<sup>20 </sup>cm <o:p></o:p></p> <p class="MsoNormal">We can find this in AU (again, this is based on the answer to a previous problem! See Iryna’s post for details…): <o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>7.087*10<sup>20 </sup>cm * (1 AU)/(1.5*10<sup>13</sup>cm) = x AU<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>X = 4.7*10<sup>7 </sup>AU<o:p></o:p></p> <p class="MsoNormal">For fun, let’s convert to lightyears. <o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>4.7*10<sup>7 </sup>AU * (8 lightminutes/1 AU)*(1 hour/60 minutes)*(1 day/24 hours)*(1 year/365 <span class="Apple-tab-span" style="white-space:pre"> </span>days) =715 lightyears<o:p></o:p></p> <p class="MsoNormal"><span style="color:red">So, the farthest star we could see is 4.7*10<sup>7 </sup>AU away. This is about 715 lightyears.<o:p></o:p></span></p> <p class="MsoNormal"><b><i>Question</i></b>: <i>The assumption in the last problem that all light from the sun is received at 500 nm is incorrect. In fact, light is processed by the eye at a “passband” of 450-600nm. Given this, how does the farthest distance at which a sun-strength star can be seen change?</i><o:p></o:p></p> <p class="MsoNormal">Knowledge from earlier in the problem set and assumptions that we will use for this problem include:</p><p class="MsoNormal"><span style="font-family:Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family: Symbol"><span><span class="Apple-tab-span" style="white-space:pre"> <span class="Apple-tab-span" style="white-space:pre"> </span></span>·<span style="font:7.0pt "Times New Roman""> </span></span></span>The “read rate” of the eye (in what length of time the photons must arrive to be visible) is <span class="Apple-tab-span" style="white-space:pre"> </span>1/30 second (this is the frame rate for most movies)</p><p class="MsoNormal"><span style="font-family:Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family: Symbol"><span><span class="Apple-tab-span" style="white-space:pre"> <span class="Apple-tab-span" style="white-space:pre"> </span></span>·<span style="font:7.0pt "Times New Roman""> </span></span></span>The sun emits light from 450 to 600 nm – however, not necessarily equally at all <span class="Apple-tab-span" style="white-space:pre"> </span>wavelengths.</p><p class="MsoNormal"><span style="font-family:Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family: Symbol"><span><span class="Apple-tab-span" style="white-space:pre"> <span class="Apple-tab-span" style="white-space:pre"> </span></span>·<span style="font:7.0pt "Times New Roman""> </span></span></span>The human eye has a surface area of .0625 square centimeters that can receive photons.</p><p class="MsoNormal"><span style="font-family:Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family: Symbol"><span><span class="Apple-tab-span" style="white-space:pre"> <span class="Apple-tab-span" style="white-space:pre"> </span></span>·<span style="font:7.0pt "Times New Roman""> </span></span></span>The sun’s luminosity is 4*10<sup>33</sup> ergs.</p><p class="MsoListParagraphCxSpLast" style="text-indent:-.25in;mso-list:l0 level1 lfo1"><o:p></o:p></p> <p class="MsoNormal">No longer can we use the equations used in the previous problem, since now we have a band of wavelengths to consider. Instead, we can consider the equation for energy per wavelength, which was derived in a previous problem set:<o:p></o:p></p> <p class="MsoNormal"><a href="http://www.codecogs.com/eqnedit.php?latex=\bg_black%20\textup{B}_{\lambda%20}(T)=\frac{2hc^{^{2}}}{\lambda%20^{5}}\frac{1}{{e^{\frac{hc}{\lambda%20kT}-1}}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\textup{B}_{\lambda%20}(T)=\frac{2hc^{^{2}}}{\lambda%20^{5}}\frac{1}{{e^{\frac{hc}{\lambda%20kT}-1}}}" title="\bg_black \textup{B}_{\lambda }(T)=\frac{2hc^{^{2}}}{\lambda ^{5}}\frac{1}{{e^{\frac{hc}{\lambda kT}-1}}}" /></a><o:p></o:p></p> <p class="MsoNormal">To make this solvable, we will use a Taylor expansion around our e term. If we integrate this expression, from 450nm to 600nm: <o:p></o:p></p> <p class="MsoNormal"><a href="http://www.codecogs.com/eqnedit.php?latex=%20\bg_black%20\int_{450}^{600}\textup{B}_{\lambda%20}(T)=\frac{2hc^{^{2}}}{\lambda%20^{5}}\frac{1}{{\frac{hc}{\lambda%20kT}}}dT" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\bg_black%20\int_{450}^{600}\textup{B}_{\lambda%20}(T)=\frac{2hc^{^{2}}}{\lambda%20^{5}}\frac{1}{{\frac{hc}{\lambda%20kT}}}dT" title="\bg_black \int_{450}^{600}\textup{B}_{\lambda }(T)=\frac{2hc^{^{2}}}{\lambda ^{5}}\frac{1}{{\frac{hc}{\lambda kT}}}dT" /></a><o:p></o:p></p> <p class="MsoNormal">The answer to this integral is 1.01*10<sup>-10</sup>. We can use as the total energy for all wavelengths. Then, we divide this by 150 to get the energy for one photon – an average photon. This value is 6.7*10<sup>-13</sup>. Then, we can multiply this by 10 to get the energy for ten photons, which is 6.7*10<sup>-12</sup>. Then, we can use this to find the distance the star would need to be. <o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>Minimum flux = (6.7*10<sup>-12 </sup><span> </span>ergs)/(.0625 cm<sup>2</sup>)</p><p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>Flux = 1.077*10<sup>-10 </sup><span> </span>ergs/ cm<sup>2</sup></p><p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal">Using this flux, we can use the same method as in the previous problem. <o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>Flux=(luminosity)/(area)<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>1.077*10<sup>-10 </sup><span> </span>ergs/ cm<sup>2 </sup>= (4*10<sup>33</sup> ergs)/(4<span style="font-family:Symbol;mso-ascii-font-family:Calibri;mso-ascii-theme-font: minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin; mso-char-type:symbol;mso-symbol-font-family:Symbol"><span>p</span></span>(distance in cm)<sup>2</sup>)<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>Distance = 1.72*10<sup>21 </sup>cm <o:p></o:p></p> <p class="MsoNormal">We can find this in AU:<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>1.72*10<sup>21 </sup>cm * (1 AU)/(1.5*10<sup>13</sup>cm) = x AU<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>X = 1.14*10<sup>8 </sup>AU<o:p></o:p></p> <p class="MsoNormal">Again, for fun:<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>1.14*10<sup>8 </sup>AU * (8 lightminutes/1 AU)*(1 hour/60 minutes)*(1 day/24 hours)*(1 year/365 <span class="Apple-tab-span" style="white-space:pre"> </span>days) =1735 lightyears<o:p></o:p></p> <p class="MsoNormal"><span style="color:red">So, a more accurate measurement of the farthest star we could see is 1.14*10<sup>8 </sup>AU away. This is 1735 lightyears.<o:p></o:p></span></p> <p class="MsoNormal"><b><i>Question</i></b>: <i>How far away is this star in centimeters?</i><o:p></o:p></p> <p class="MsoNormal">We actually found this figure mid-way through our calculation in the previous problem, so lifting that answer gives us:<o:p></o:p></p> <p class="MsoNormal"><span style="color:red">Distance farthest star we can see is from Earth: 1.72*10<sup>21 </sup>cm <o:p></o:p></span></p> <p class="MsoNormal"><b><i>Question</i></b>: <i>How far is this is parsecs?</i><o:p></o:p></p> <p class="MsoNormal">Knowledge from earlier in the problem set and assumptions that we will use for this problem include:</p><p class="MsoNormal"><span style="font-family:Symbol;mso-fareast-font-family:Symbol;mso-bidi-font-family: Symbol"><span><span class="Apple-tab-span" style="white-space:pre"> <span class="Apple-tab-span" style="white-space:pre"> </span></span>·<span style="font:7.0pt "Times New Roman""> </span></span></span>The farthest a star of sun luminosity could be from the Earth for us to see it is 1.14*10<sup>8 </sup>AU</p><p class="MsoListParagraph" style="text-indent:-.25in;mso-list:l1 level1 lfo2"><o:p></o:p></p> <p class="MsoNormal">First, we should find the distance of a parsec. A parsec is, as the problem explains, the distance from Earth a star must be in order to have a parallax of 1 arcsecond (recall, 1 arcsecond = 1/3600 degrees) from the sun. <o:p></o:p></p><p class="MsoNormal"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixeeI7mS7YDa29l4f5-EeTCqTNs5cSt22oeLLQ67CBPrX1bYlBosQKaBv7EFrRwe12QD02a3R1jJ-f2uHLh42z6UmHWcHJ6NWzoI1y4qkjK1BXaz7fJ2xhUItGaYXKDucTZGy8LGY9SVg6/s1600/ay5.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixeeI7mS7YDa29l4f5-EeTCqTNs5cSt22oeLLQ67CBPrX1bYlBosQKaBv7EFrRwe12QD02a3R1jJ-f2uHLh42z6UmHWcHJ6NWzoI1y4qkjK1BXaz7fJ2xhUItGaYXKDucTZGy8LGY9SVg6/s320/ay5.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5666481657864665090" style="display: block; margin-top: 0px; margin-right: auto; margin-bottom: 10px; margin-left: auto; text-align: center; cursor: pointer; width: 320px; height: 98px; " /></a></p><div>Using the small angle approximation, and converting our 1/3600 into radians (<span style="font-family:Symbol;mso-ascii-font-family: Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;mso-char-type:symbol;mso-symbol-font-family: Symbol"><span>p</span></span>/648000)</div><p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>tan(<span style="font-family:Symbol;mso-ascii-font-family: Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;mso-char-type:symbol;mso-symbol-font-family: Symbol"><span>p</span></span>/648000) = 1 AU / (x)<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>1/(<span style="font-family:Symbol;mso-ascii-font-family: Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;mso-char-type:symbol;mso-symbol-font-family: Symbol"><span>p</span></span>/648000)= 1/x<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>1 parsec = x =1 / (<span style="font-family:Symbol; mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family: Calibri;mso-hansi-theme-font:minor-latin;mso-char-type:symbol;mso-symbol-font-family: Symbol"><span>p</span></span>/648000) = 2.06*10<sup>5 </sup>AU<o:p></o:p></p> <p class="MsoNormal">Now we know how many AU there are in a parsec, we can calculate how far the farthest star we can see is. Using the value for distance we previously calculated, <o:p></o:p></p> <p class="MsoNormal"><span class="Apple-tab-span" style="white-space:pre"> </span>1.14*10<sup>8 </sup>AU * (1 parsec/ 2.06*10<sup>5 </sup>AU) = 5.52*10<sup>2 </sup>parsecs = <span style="color:red">552 parsecs</span><o:p></o:p></p> <p class="MsoNormal"><b>Conclusion<o:p></o:p></b></p> <p class="MsoNormal">We have found that the farthest distance a star of sun luminosity could be from the Earth for us to still be able to see in with our naked eye is 1.14*10<sup>8 </sup>AU. This is equivalent to 1.72*10<sup>21 </sup>cm, or 552 parsecs, or 1735 lightyears. Our preliminary estimate was 4.7*10<sup>7 </sup>AU (which is about 715 lightyears), and we have also found the distance of a parsec (2.06*10<sup>5 </sup>AU).<o:p></o:p></p> <p class="MsoNormal"><b>Acknowledgements<o:p></o:p></b></p> <p class="MsoNormal">Thanks to Iryna Butsky for significant help on these problems! See <a href="http://ibutsky.blogspot.com/">her blog</a> for problems 1-4.<o:p></o:p></p>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com1tag:blogger.com,1999:blog-5849210431402234073.post-65821441109645174762011-10-21T13:53:00.001-07:002011-10-23T11:36:42.796-07:00How far to the earth?<div style="text-align: left;"><b>Abstract</b></div><div> <p class="MsoNoSpacing"><o:p> </o:p></p> <p class="MsoNoSpacing">By taking visual data from the TRACE satellite, which tracked the position of Mercury as measured against the sun, we will determine the distance between earth and the sun (the AU) and other qualities about the sun, Earth, Mercury. <o:p></o:p></p> <p class="MsoNoSpacing"><o:p> </o:p></p> <p class="MsoNoSpacing"><b>Introduction <o:p></o:p></b></p> <p class="MsoNoSpacing"><o:p> </o:p></p> <p class="MsoNoSpacing">For the purposes of this problem, we will consider the angles from the satellite (which we assume is at a distance equal to Earth’s radius from the center of the planet) to some point on the sun (the lines defining these angles are black on the diagram). We will also consider the shadow that Mercury casts on the sun (shown as the blue lines on the diagram). The angle these two lines make from the earth is ‘a’, while the angle from the centroid of earth to the centroid of the sun to the satellite is ‘b’.</p><p class="MsoNoSpacing"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhaqjOPAigJhwJ8ieJ95Py9x79VQvDZv3LvTqKdbxx-P65fMuFA8dnUrZa9l8miIhyphenhyphenMgNa3WwLKKjn0Xvy_H3DzT0idV7DiOSW7OslVCxyPjCW6FeqxJqlkI5d2JEjSXKdOcPR4ej4FM83Y/s1600/im1.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhaqjOPAigJhwJ8ieJ95Py9x79VQvDZv3LvTqKdbxx-P65fMuFA8dnUrZa9l8miIhyphenhyphenMgNa3WwLKKjn0Xvy_H3DzT0idV7DiOSW7OslVCxyPjCW6FeqxJqlkI5d2JEjSXKdOcPR4ej4FM83Y/s320/im1.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5666052194310139250" style="display: block; margin-top: 0px; margin-right: auto; margin-bottom: 10px; margin-left: auto; text-align: center; cursor: pointer; width: 320px; height: 251px; " /></a></p><div><b>Solution</b></div><div><p class="MsoNoSpacing"><o:p></o:p></p> <p class="MsoNoSpacing"><b><o:p> </o:p></b></p> <p class="MsoNoSpacing"><o:p> </o:p></p> <p class="MsoNoSpacing">We know that the angle ‘a’ refers to one-half the amplitude of Mercury’s oscillations as seen transposed against the sun. TRACE’s orbit is what causes this oscillation, so the angle it makes with the sun is based on the radius of its orbit (which we defined as equal to the radius of the earth).<o:p></o:p></p> <p class="MsoNoSpacing">We define theta, <span style="font-family:Symbol; mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family: Calibri;mso-hansi-theme-font:minor-latin;mso-char-type:symbol;mso-symbol-font-family: Symbol"><span>q</span></span>, as seen on the diagram.<o:p></o:p></p> <p class="MsoNoSpacing">Consider this section of the diagram, bringing special attention to the triangle outlined in blue:<o:p></o:p></p><p class="MsoNoSpacing"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjo2Lis1fYDWbuLxap4QRZ8i2Jk3XzG96WTu_R6TqDJZzQnfe0zDiFP9J4uRUhte-Ikc6RaCvs9GkaDK5HLccROF2zjBkcg4d8VZi3xSKAhB51wHQRQRgFUne7kczEs5ecvIk_NNbU_dL1b/s1600/im2.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjo2Lis1fYDWbuLxap4QRZ8i2Jk3XzG96WTu_R6TqDJZzQnfe0zDiFP9J4uRUhte-Ikc6RaCvs9GkaDK5HLccROF2zjBkcg4d8VZi3xSKAhB51wHQRQRgFUne7kczEs5ecvIk_NNbU_dL1b/s320/im2.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5666052342836962322" style="display: block; margin-top: 0px; margin-right: auto; margin-bottom: 10px; margin-left: auto; text-align: center; cursor: pointer; width: 320px; height: 110px; " /></a></p><div>The angles in this triangle must sum to <span style="font-family:Symbol;mso-ascii-font-family:Calibri;mso-ascii-theme-font: minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin; mso-char-type:symbol;mso-symbol-font-family:Symbol"><span>p</span></span>. So, we can write the expression:</div><p class="MsoNoSpacing"><o:p></o:p></p> <p class="MsoNoSpacing" style="text-indent:.5in"><span style="font-family:Symbol; mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family: Calibri;mso-hansi-theme-font:minor-latin;mso-char-type:symbol;mso-symbol-font-family: Symbol"><span>p</span></span>=a+b+<span style="font-family:Symbol;mso-ascii-font-family:Calibri;mso-ascii-theme-font: minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin; mso-char-type:symbol;mso-symbol-font-family:Symbol"><span>p</span></span>-<span style="font-family: Symbol;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin; mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin;mso-char-type: symbol;mso-symbol-font-family:Symbol"><span>q</span></span>/2<o:p></o:p></p> <p class="MsoNoSpacing">Which simplifies to:<o:p></o:p></p> <p class="MsoNoSpacing" style="text-indent:.5in">a+b=<span style="font-family: Symbol;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin; mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin;mso-char-type: symbol;mso-symbol-font-family:Symbol"><span>q</span></span>/2<o:p></o:p></p> <p class="MsoNoSpacing">This expression relates one-half the amplitude of Mercury’s oscillations (a) to angles b and <span style="font-family:Symbol; mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family: Calibri;mso-hansi-theme-font:minor-latin;mso-char-type:symbol;mso-symbol-font-family: Symbol"><span>q</span></span>/2. We can measure the oscillations using the image given in class. On his image, we measure the amplitude of the oscillations as about .4 cm (on the paper). Meanwhile, the scale of the sun on the paper is about 11 cm across. We know the angular diameter of the sun from earth is about 0.5<span>°</span> (the same as the moon), so this means that the amplitude of the orbit, as an angle, is 3*10<sup>-5</sup> (.4/11*.05/360*2*<span style="font-family:Symbol;mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font: minor-latin;mso-char-type:symbol;mso-symbol-font-family:Symbol"><span>p</span></span>). So a = 3*10<sup>-5</sup><o:p></o:p></p><p class="MsoNoSpacing"><sup><span class="Apple-style-span" style="font-size: 16px; "><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjo2Lis1fYDWbuLxap4QRZ8i2Jk3XzG96WTu_R6TqDJZzQnfe0zDiFP9J4uRUhte-Ikc6RaCvs9GkaDK5HLccROF2zjBkcg4d8VZi3xSKAhB51wHQRQRgFUne7kczEs5ecvIk_NNbU_dL1b/s1600/im2.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><span class="Apple-style-span" style="color: rgb(0, 0, 0); -webkit-text-decorations-in-effect: none; "></span></a><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiYZJfSzC1VZBMTVxubAcMa6h9P5YJ4Ft2SSNTVwBuNqWeiiaIncLUbdAoXr7g6RLk2rhAPGORrdPdzWyAc7LpAeS0x-f9Scrr2DCVJO27ADcClgln4OecOG4R91FH1UmYrOgBXAWHCwpBs/s1600/im3.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0mnWV2UdGScs3bIIfXq32sVleQ2hZikmcnROGhyphenhyphenmvyBzgT7lRViwoNin1IfbEAhxggwGEzmDJ7nG-ITz2NujKW0sZ-VDL6jQMmT7OoId3_2MOM7hlAXLvuLJJApbHmbrhjPit2xe5s46B/s320/im4.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5666052654864643714" style="display: block; margin-top: 0px; margin-right: auto; margin-bottom: 10px; margin-left: auto; text-align: center; cursor: pointer; width: 320px; height: 83px; " /></a></span></sup></p><div></div><p></p><p class="MsoNoSpacing">Now, we can look at yet another sub-triangle of the Mercury-Earth-Sun system.</p><p class="MsoNoSpacing"><o:p></o:p></p> <p class="MsoNoSpacing"><o:p> </o:p></p> <p class="MsoNoSpacing">Here, we consider the right triangle that has as its legs the radius of the earth and the distance from earth to mercury. We can set up this triangle as follows:<o:p></o:p></p><p class="MsoNoSpacing"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiYZJfSzC1VZBMTVxubAcMa6h9P5YJ4Ft2SSNTVwBuNqWeiiaIncLUbdAoXr7g6RLk2rhAPGORrdPdzWyAc7LpAeS0x-f9Scrr2DCVJO27ADcClgln4OecOG4R91FH1UmYrOgBXAWHCwpBs/s1600/im3.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiYZJfSzC1VZBMTVxubAcMa6h9P5YJ4Ft2SSNTVwBuNqWeiiaIncLUbdAoXr7g6RLk2rhAPGORrdPdzWyAc7LpAeS0x-f9Scrr2DCVJO27ADcClgln4OecOG4R91FH1UmYrOgBXAWHCwpBs/s320/im3.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5666052470237646034" style="display: block; margin-top: 0px; margin-right: auto; margin-bottom: 10px; margin-left: auto; text-align: center; cursor: pointer; width: 320px; height: 121px; " /></a></p><p class="MsoNoSpacing" style="text-indent:.5in">tan(<span style="font-family: Symbol;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin; mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin;mso-char-type: symbol;mso-symbol-font-family:Symbol"><span>q</span></span>/2) <span> </span>= R<sub>earth</sub> / R<sub>earth-mercury<o:p></o:p></sub></p> <p class="MsoNoSpacing">By a small angle approximation:<o:p></o:p></p> <p class="MsoNoSpacing" style="text-indent:.5in"><span> </span>(<span style="font-family:Symbol;mso-ascii-font-family: Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;mso-char-type:symbol;mso-symbol-font-family: Symbol"><span>q</span></span>/2) <span> </span>= R<sub>earth</sub> / R<sub>earth-mercury<o:p></o:p></sub></p> <p class="MsoNoSpacing">We earlier solved for <span style="font-family:Symbol; mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family: Calibri;mso-hansi-theme-font:minor-latin;mso-char-type:symbol;mso-symbol-font-family: Symbol"><span>q</span></span>/2! So, we can substitute, as follows:<o:p></o:p></p> <p class="MsoNoSpacing" style="text-indent:.5in">a+b = R<sub>earth</sub> / R<sub>earth-mercury</sub><o:p></o:p></p> <p class="MsoNoSpacing">We found a, we know R<sub>earth. </sub>Based again on our triangles, we can define b as follows:<o:p></o:p></p> <p class="MsoNoSpacing" style="text-indent:.5in">tan(b) = R<sub>earth </sub>/ AU<o:p></o:p></p> <p class="MsoNoSpacing">Again, by the small angle approximation:<o:p></o:p></p> <p class="MsoNoSpacing" style="text-indent:.5in">(b) = R<sub>earth </sub>/ AU<o:p></o:p></p> <p class="MsoNoSpacing">Substitute!<o:p></o:p></p> <p class="MsoNoSpacing" style="text-indent:.5in">a +R<sub>earth </sub>/ AU = R<sub>earth</sub> / R<sub>earth-mercury<o:p></o:p></sub></p> <p class="MsoNoSpacing" style="text-indent:.5in">R<sub>earth</sub>/( a +R<sub>earth </sub>/ AU )= R<sub>earth-mercury<o:p></o:p></sub></p> <p class="MsoNoSpacing" style="text-indent:.5in">AU - R<sub>earth</sub>/( a +R<sub>earth </sub>/ AU ) = R<sub>mercury<o:p></o:p></sub></p> <p class="MsoNoSpacing"><span> </span>(P<sub>mercury</sub>)<sup>2</sup>/(P<sub>earth</sub>)<sup>2</sup> <span> </span>= (A<sub>mercury</sub>)<sup>3</sup>/(A<sub>earth</sub>)<sup>3</sup> <o:p></o:p></p> <p class="MsoNoSpacing">We know P<sub>earth</sub>=365 days and P<sub>mercury</sub>=87 days. Also, we solved for R <sub>mercury </sub>=A<sub>mercury </sub>earlier. So:<o:p></o:p></p> <p class="MsoNoSpacing" style="text-indent:.5in">(P<sub>mercury</sub>)<sup>2</sup>/(P<sub>earth</sub>)<sup>2</sup> <span> </span>= (AU - R<sub>earth</sub>/( a +R<sub>earth </sub>/ AU ) = R<sub>mercury</sub>)<sup>3</sup>/(A<sub>earth</sub>)<sup>3<o:p></o:p></sup></p> <p class="MsoNoSpacing" style="text-indent:.5in">(87)<sup>2</sup>/(365)<sup>2</sup> <span> </span>= (AU - 6378/( 3*10<sup>-5</sup> +6378<sub> </sub>/ AU ))<sup>3</sup>/(AU)<sup>3</sup><sub><o:p></o:p></sub></p> <p class="MsoNoSpacing">What an equation! Plugging this into Wolfram Alpha (or solving for an approximate solution) give the value:<o:p></o:p></p> <p class="MsoNoSpacing"><span> </span>AU = 1.33 * 10<sup>8 </sup>km<o:p></o:p></p> <p class="MsoNoSpacing">The real value for an AU is (a quick Google search can give us an approximate value):<o:p></o:p></p> <p class="MsoNoSpacing"><span> </span>AU = 1.496 *10<sup>8 km</sup><o:p></o:p></p> <p class="MsoNoSpacing">Not only are these values on the same order of magnitude, they are actually pretty close – so you can see that even rough measurements such as measuring the period of Mercury’s projections function using a ruler and a print-out image are sufficient to estimate the AU fairly accurately. <o:p></o:p></p> <p class="MsoNoSpacing">For a final answer, we can convert this value to cm, as the lab asked:<o:p></o:p></p> <p class="MsoNoSpacing"><b><span style="color:red"><span> </span>One AU = 1.33 * 10<sup>13 </sup>cm<o:p></o:p></span></b></p> <p class="MsoNoSpacing">We can find more values using these same few equations, as well!<o:p></o:p></p> <p class="MsoNoSpacing" style="text-indent:.5in">(P<span class="Apple-style-span" >earth</span>)<sup>2</sup>/( A<sub>earth</sub>)<sup>3</sup> <span> </span>= 4<span style="font-family:Symbol;mso-ascii-font-family:Calibri;mso-ascii-theme-font: minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin; mso-char-type:symbol;mso-symbol-font-family:Symbol"><span>p</span></span><sup>2</sup>/(MG)<sup><o:p></o:p></sup></p> <p class="MsoNoSpacing" style="text-indent:.5in">(365*24*60*60)<sup>2</sup>/( 1.33 * 10<sup>13</sup>)<sup>3</sup> <span> </span>= 4<span style="font-family:Symbol;mso-ascii-font-family:Calibri;mso-ascii-theme-font: minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin; mso-char-type:symbol;mso-symbol-font-family:Symbol"><span>p</span></span><sup>2</sup>/(M*6.7*10<sup>-8</sup>)<o:p></o:p></p> <p class="MsoNoSpacing" style="text-indent:.5in">So, M = 1.39*10<sup>33 </sup>grams<span> </span>(this figure is a little off…)<o:p></o:p></p> <p class="MsoNoSpacing">Actual value: 5.97*10<sup>27 </sup>grams<span> </span><o:p></o:p></p> <p class="MsoNoSpacing">Can this be attributed to error? It seems like way too big of an error… could someone else who did the problem tell me what I did wrong here? Thanks!<o:p></o:p></p> <p class="MsoNoSpacing"><o:p> </o:p></p> <p class="MsoNoSpacing"><b>Epilogue<o:p></o:p></b></p> <p class="MsoNoSpacing"><o:p> </o:p></p> <p class="MsoNoSpacing">We have determined one AU to be <b><span style="color:red">1.33 * 10<sup>13 </sup>cm. </span></b>Additionally, we have found the mass of the sun to be 1.39*10<sup>33 </sup>grams. The orbital period of mercury is roughly 87 days. <o:p></o:p></p> <p class="MsoNoSpacing"><o:p> </o:p></p> <p class="MsoNoSpacing"><b>Acknowledgements<o:p></o:p></b></p> <p class="MsoNoSpacing"><b><o:p> </o:p></b></p> <p class="MsoNoSpacing">Thanks to Iryna and Monica for working with me on this problem! <o:p></o:p></p></div></div>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com4tag:blogger.com,1999:blog-5849210431402234073.post-68959856098197721912011-10-17T14:51:00.000-07:002011-10-17T14:52:10.768-07:00A swift dinner...<p class="MsoNormal">Yesterday at dinner, I sat with two scientists from Oxford, who were at Palomar working on their instrument, SWIFT. When they mentioned that SWIFT uses Integral imagery, I immediately started thinking about x-rays (Swift and Integral are two x-ray telescopes), but the naming similarities were just a coincidence. SWIFT is an instrument created specifically for the Hale 200-inch telescope. There are many different types of instruments, each for a different purpose. Installing an instrument onto the telescope alters the results of the observation. For example, one instrument that had been used in the past is PHARO, which is an adaptive optics instrument. (However, the two scientists told me that adaptive optics is no longer being used at the Hale 200 inch! This was very surprising…)<o:p></o:p></p> <p class="MsoNormal">What does SWIFT do? Best I can understand it, SWIFT is an integral field spectrograph. This means that it records not only the location (in the sky) of an event, but it also records the spectra at this point. This is useful because for larger sources, you can see the differences in spectra from one spot to another (still part of the same source). This technique is used for extended sources; it is not necessary for point sources, where a single spectrum is sufficient. An integral field spectrograph uses a different configuration of slits than a classic long-slit spectrograph (which took observations for the same purpose, seeing spectra over area). In long-slit observations, an observation is made through a (as the name implies) long slit aperture, and then the light is split by a prism. An integral field spectrograph speeds up this process. The result of using an instrument such as SWIFT is that an astronomer can study extended or grouped sources.<o:p></o:p></p> <p class="MsoNormal">While explaining their project to me, the two scientists also ended up explaining a lot of science – I learned a lot! For example, I never knew that Keck wasn’t a monolithic mirror (I didn’t know what monolithic meant, either – it means that the mirror was one solid piece). Instead, Keck is many smaller mirrors aligned into a plane. The Extremely Large Telescope, which is a British telescope not yet built, will be 40 meters in diameter (!!!!) but will similarly be composed of many smaller mirrors instead of one single monolithic mirror. Also, apparently the problem they are having with building the ELT is not that they do not have enough money, but that they “can’t spend it fast enough” – that is, they cannot find enough qualified people to hire! Manufacturing the mirrors is a long process, and apparently not very efficient… car companies are more efficient than the people who build the telescopes!<o:p></o:p></p> <p class="MsoNormal">I was very surprised at the degree to which the SWIFT scientists were aware and even bitter at the difference between the astronomers and the people who build the instruments. They both (although considering themselves scientists) lamented the fact that scientists never respect the people who build the instruments. I never thought about it before, but without the people who build the instruments, the astronomers who take and analyze data could never do their science. All modern science hinges on the fact that someone in history decided it was worth it to build a telescope – visiting the Hale 200-inch and seeing its battleship-like construction made me aware that telescopes are not built in a day, and no astronomer is self-sufficient… behind every paper are thousands of people whose labors have made it possible!<o:p></o:p></p>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com1tag:blogger.com,1999:blog-5849210431402234073.post-6585361282023976182011-10-16T11:27:00.001-07:002011-10-16T11:27:42.477-07:00Telescope time!<p class="MsoNormal">I met with Dr. Evan Kirby last week (<a href="http://www.astro.caltech.edu/~enk/">http://www.astro.caltech.edu/~enk/</a>), and he very nicely spent an hour explaining to me various facets of the telescope time acquisition process! Here are some highlights of what I learned from him:<o:p></o:p></p> <p class="MsoNormal">When Keck telescope was built, the costs were divided between the UC system and Caltech. Caltech paid the bulk of the initial costs, but the UCs were to pay for maintenance. Over the course of many, many years, this will eventually even out, so each institution was initially given a half stake in the telescope. The University of Hawaii also received a stake (10%) in the telescope, because the telescope was built on Hawaiian land. <o:p></o:p></p> <p class="MsoNormal">What a ‘stake’ in a telescope means is that out of all the nights of the year, that institution will receive that many nights for their scientists. So, a higher stake means that the institution will have more observing time – which is a good thing! For Keck, this meant that the UC system and Caltech each had equal shares. However, you might notice that Caltech has FAR fewer people vying for this time than the entire UC system – indeed, this means that it is far easier for Caltech scientists to get telescope time. There are also some Keck nights each year available to anyone in the country. Getting time on these nights is extremely competitive, since there are professors and professional scientists from across the country who perhaps have no other way to get the telescope time they need. <o:p></o:p></p> <p class="MsoNormal">Caltech has stakes in other telescopes as well. It operates Palomar telescope here in California. The access that Caltech scientists have to it is actually quite amazing – did you know that grad students can apply for time on Palomar?? That’s pretty awesome. <o:p></o:p></p> <p class="MsoNormal">Wait – so who exactly can apply for time to telescopes? And how do you apply?<o:p></o:p></p> <p class="MsoNormal">To get telescope time, applicants must write a proposal, which is reviewed by a panel of scientists (this panel changes – Dr. Kirby has been on this panel before). Proposals are due each semester, for the following semester of observing time. These proposals take a lot of work, and they must be submitted by someone who has access to the telescope. For example, an undergraduate cannot write and submit a proposal. However, an undergraduate CAN write a proposal and convince someone (a professor, postdoc) with access to the telescope to submit it for them. <o:p></o:p></p> <p class="MsoNormal">At most institutions, there are strict limits to who can apply for telescope time. For example, grad students are not generally allowed to apply for telescope time, and even postdocs cannot apply in most places. At Caltech, there are certainly such limits, although they are less strict than at other places. Either postdocs or Professors can apply for Keck time here, but grad students, postdocs and Professors all can apply for Palomar time. <o:p></o:p></p> <p class="MsoNormal">The national ratio for proposals submitted to proposals accepted is roughly 10 to 1. The UC system’s average is roughly 5:1, and Caltech’s is much, much lower than that!<o:p></o:p></p> <p class="MsoNormal">I never knew just how good Caltech’s access to telescopes is. It is inspiring to know that if you had science you wanted to do, at Caltech you’d be able to actually do that science! It makes me want to write my own proposal…<o:p></o:p></p>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com1tag:blogger.com,1999:blog-5849210431402234073.post-48558400867238193462011-10-10T14:20:00.000-07:002011-10-13T18:42:05.362-07:00More on the purposes of varying apertures…<p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;mso-line-height-alt:13.5pt"></p><p class="MsoNormal" align="center" style="text-align: left;margin-bottom: 0.0001pt; "></p><p class="MsoNormal"><b>Abstract</b></p> <p class="MsoNormal"><o:p><span class="Apple-style-span"> </span></o:p></p> <p class="MsoNormal"><span class="Apple-style-span">The size of a telescope affects what its ideal observation targets are. Larger telescopes will have higher angular resolution for greater wavelengths of light, while telescopes with smaller apertures will have better angular resolution for shorter wavelengths. However, this dependence is only valid to an order of magnitude: increasing the order of magnitude for the wavelength but not the aperture will result in a worse angular resolution. <o:p></o:p></span></p> <p class="MsoNormal"><o:p><span class="Apple-style-span"> </span></o:p></p> <p class="MsoNormal"><o:p><span class="Apple-style-span"> </span></o:p></p> <p class="MsoNormal"><span class="Apple-style-span"><b>Introduction</b><o:p></o:p></span></p> <p class="MsoNormal"><o:p><span class="Apple-style-span"> </span></o:p></p> <p class="MsoNormal"><span class="Apple-style-span">J-band refers to a specific range of wavelengths, from 1.1 to 1.4 microns (a micrometer is equal to 10-6 meters). This value means that light in the J-band spectrum falls into the infrared. Considering CCAT and Keck, two telescopes with different apertures (25 meters and 10 meters respectively), how does the angular resolution compare? The CCAT will detect wavelengths up to 850 microns, while Keck observes in J-band (1.1-1.4 microns). This question elaborates on my earlier blog post concerning the differences between large and small telescopes: it is a practical example of why you might want a telescope with a smaller aperture in some cases.<o:p></o:p></span></p> <p class="MsoNormal"><o:p><span class="Apple-style-span"> </span></o:p></p> <p class="MsoNormal"><span class="Apple-style-span"><b>Solution</b><o:p></o:p></span></p> <p class="MsoNormal"><o:p><span class="Apple-style-span"> </span></o:p></p> <p class="MsoNormal"><span class="Apple-style-span">The relationship between aperture, wavelength and angular resolution can be defined as follows:<o:p></o:p></span></p> <p class="MsoNormal"><span class="Apple-style-span">sin </span><span class="Apple-style-span" style="font-family: Symbol; ">q</span><span class="Apple-style-span"> = (1.220)(</span><span class="Apple-style-span" style="font-family: Symbol; ">l /</span>D)</p><p class="MsoNormal"><o:p></o:p></p><p></p><p class="MsoNormal"><span class="Apple-style-span">Taking care to convert units (using meters as the standard unit), we can calculate the values for each telescope.<o:p></o:p></span></p> <p class="MsoNormal"><span class="Apple-style-span">For CCAT, sin </span><span class="Apple-style-span" style="font-family: Symbol; ">q</span><span class="Apple-style-span"> = (1.220)(850*10-6 /25) = 4.15*10-5 and using the small angle approximation, we can say that </span><span class="Apple-style-span" style="font-family: Symbol; ">q</span><span class="Apple-style-span"> = 4.15*10-5. This value is given in radians, so a quick conversion to degrees gives us .0024° as the angular resolution of CCAT at this wavelength. Since there are 3600 arcseconds in a degree, this is an angular resolution of 8.64 arcseconds. <o:p></o:p></span></p> <p class="MsoNormal"><span class="Apple-style-span">Keck, on the other hand, is observing at a much smaller frequency, at about 1 micron, with a slightly smaller aperture as well. So, sin </span><span class="Apple-style-span" style="font-family: Symbol; ">q</span><span class="Apple-style-span"> = (1.220)(1*10-6 /10) = 1.22*10-7. This is equivalent to 0.000007°, which is equal to 0.03 arcseconds. This is two orders of magnitude smaller than the resolution for CCAT, which makes sense since the wavelength Keck is studying here is two orders of magnitude smaller as well. <o:p></o:p></span></p> <p class="MsoNormal"><o:p><span class="Apple-style-span"> </span></o:p></p> <p class="MsoNormal"><span class="Apple-style-span"><b>Epilogue</b><o:p></o:p></span></p> <p class="MsoNormal"><o:p><span class="Apple-style-span"> </span></o:p></p> <p class="MsoNormal"><span class="Apple-style-span">The angular resolution for CCAT was better for 850 micrometer wavelengths than Keck’s resolution was for 1 micrometer waves: we can compare CCAT’s angular resolution of 8.64 arcseconds to Keck’s resolution of 0.03 arcseconds. However, if we were to study different wavelengths of light, then the angular resolution for each telescope would also change – this one measurement of angular resolution cannot be considered the standard by which we judge the telescope.<span> </span><o:p></o:p></span></p> <p class="MsoNormal"><o:p><span class="Apple-style-span"> </span></o:p></p> <p class="MsoNormal"><span class="Apple-style-span"><b>Questions for Further Study (next blog post?)</b><o:p></o:p></span></p> <p class="MsoNormal"><span class="Apple-style-span">Is it true that the formula for angular resolution can only work for ground-based telescopes? That is, how broadly can the term ‘aperture’ be defined? Can the resolution of x-ray telescopes be defined this way – do x-ray telescopes even have an angular resolution?<o:p></o:p></span></p> <p class="MsoNormal"><span class="Apple-style-span"><b>Acknowledgements</b><o:p></o:p></span></p> <p class="MsoNormal"><o:p><span class="Apple-style-span"> </span></o:p></p> <p class="MsoNormal"><span class="Apple-style-span">Thanks to Iryna Butsky and Monica He for their help!</span><b><o:p></o:p></b></p><p></p><p></p><p></p>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com3tag:blogger.com,1999:blog-5849210431402234073.post-83243066795996442522011-10-08T13:49:00.000-07:002011-10-09T17:50:04.305-07:00The ideal size for a telescope...<p class="MsoNormal">At the end of class Friday, Professor Johnson made a comment about there being telescopes on earth with much, much larger apertures than the 10-25 meters discussed on the problem set (I still think these are pretty big…) I didn’t really think that he was serious, partially because it’s so hard to imagine a telescope mirror being so large, but in fact I was thinking about it wrong: the creatively named Very Large Array, which consists of 27 radio antennas in New Mexico, is capable of making very high resolution measurements. On a side note - If you think its name is unoriginal, then just consider for comparison the Very Large Telescope (VLT) and Overwhelmingly Large Telescope (OLT).</p><p class="MsoNormal"><br /><img src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgvB5ZsIzkzPQAIZVnn_bx2s6MJuP03AwCwciwlC_G28ufPmKoTXDiU-5v0omLWlI5IvdG5Nm_MXhfH8gOUqbytF25ZrG5fNE568z9njk5cY2vT4RPzJWzWnu60o7pe8bCCGYP8LvMtKrC4/s320/blog+3.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5661253731555486386" style="float: right; margin-top: 0px; margin-right: 0px; margin-bottom: 10px; margin-left: 10px; cursor: pointer; width: 320px; height: 185px; " />The VLA looks small compared to the Very Large Baseline Array (VLBA), however. VLBA is a system of similar radio antennas, also 25 meters in diameter, which are spread across the whole United States. To the left, the locations of the dishes are shown - image from the NRAO website (you can also go there for more information): <a href="http://www.vlba.nrao.edu/">http://www.vlba.nrao.edu/</a> <o:p></o:p></p> <p class="MsoNormal"><o:p></o:p></p><p class="MsoNormal"><br /></p> <p class="MsoNormal">Why is this useful? The relationship between aperture (D), wavelength (<span style="font-family:Symbol;mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font: minor-latin;mso-char-type:symbol;mso-symbol-font-family:Symbol"><span>l</span></span>), and angular resolution (<span style="font-family:Symbol;mso-ascii-font-family: Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;mso-char-type:symbol;mso-symbol-font-family: Symbol"><span>q</span></span>, measured in radians) is as follows for single telescopes:<o:p></o:p></p> <p class="MsoNormal">sin(<span style="font-family:Symbol;mso-ascii-font-family: Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;mso-char-type:symbol;mso-symbol-font-family: Symbol"><span>q</span></span>)=(<span class="Apple-style-span"><span class="Apple-style-span" style="font-size: 15px; line-height: 17px;">1.220)</span></span>(<span style="font-family: Symbol; "><span>l/</span></span>D<span class="Apple-style-span" style="font-family: Symbol; ">)</span></p><!--[if gte msEquation 12]><m:omathpara><m:omath><m:f><m:fpr><span style="'font-family:"><m:ctrlpr></m:ctrlPr></span></m:fPr><m:num><span style="'font-size:11.0pt;line-height:115%;font-family:Symbol;mso-ascii-font-family:"><span style="'mso-char-type:symbol;mso-symbol-font-family:Symbol'">l</span></span></m:num><m:den><i style="'mso-bidi-font-style:normal'"><span style="'font-size:11.0pt;"><m:r>D</m:r></span></i></m:den></m:f></m:oMath></m:oMathPara><![endif]--><!--[if !msEquation]--><!--[endif]--> <p class="MsoNormal">From this, we can see that aperture of a telescope affects what wavelengths of light it can observe at an optimal resolution. Resolution is important because if a source is larger than the resolution of the telescope observing it, then the observation will not contain the entire source. This is called an extended source, as opposed to a point source, which is smaller than the resolution of the telescope observing it. Also, sources that are too close together (that is, the distance between them is smaller than the angular resolution) cannot be accurately studied, as astronomers would not be able to see where one point source ended and the other began! It would just look like one larger, blurrier point source…<o:p></o:p></p> <p class="MsoNormal">HOWEVER – this equation does not hold for the VLBA, because the VLBA is an array, not a single telescope. Instead, for arrays, the relationship depends upon the separation between radio dishes (B) and wavelength of the source (<span style="font-family:Symbol;mso-ascii-font-family: Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;mso-char-type:symbol;mso-symbol-font-family: Symbol"><span>l</span></span>). Again, the angular resolution is measured in radians (<span style="font-family: Symbol;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin; mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin;mso-char-type: symbol;mso-symbol-font-family:Symbol"><span>q</span></span>):<o:p></o:p></p> <p class="MsoNormal"><span class="Apple-style-span" style="font-family: Symbol; ">q</span>=<span class="Apple-style-span"><span class="Apple-style-span" style="font-size: 15px; line-height: 17px;">(1.220)(</span></span><span class="Apple-style-span" style="font-family: Symbol; ">l</span><span class="Apple-style-span" style="font-family: Calibri, sans-serif; font-size: 15px; line-height: 17px; ">/B)</span></p><p class="MsoNormal"><o:p></o:p></p> <p class="MsoNormal">Regardless, a larger telescope can observe SOME frequencies of light with higher resolution. Longer wavelengths require a larger aperture, which is why the largest receivers on earth study radio waves, which have a higher wavelength than visible light. <o:p></o:p></p> <table class="MsoTableGrid" border="1" cellspacing="0" cellpadding="0" style="border-collapse:collapse;border:none;mso-border-alt:solid windowtext .5pt; mso-yfti-tbllook:1184;mso-padding-alt:0in 5.4pt 0in 5.4pt"> <tbody><tr> <td width="97" valign="top" style="width:72.9pt;border:solid windowtext 1.0pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">Type of light studied<o:p></o:p></p> </td> <td width="126" valign="top" style="width:94.5pt;border:solid windowtext 1.0pt; border-left:none;mso-border-left-alt:solid windowtext .5pt;mso-border-alt: solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">(Approximate) Wavelength<o:p></o:p></p> </td> <td width="160" valign="top" style="width:119.85pt;border:solid windowtext 1.0pt; border-left:none;mso-border-left-alt:solid windowtext .5pt;mso-border-alt: solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">Approximate aperture width to study this light from Earth<o:p></o:p></p> </td> <td width="254" valign="top" style="width:190.65pt;border:solid windowtext 1.0pt; border-left:none;mso-border-left-alt:solid windowtext .5pt;mso-border-alt: solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">Example of a telescope operating at this wavelength<o:p></o:p></p> </td> </tr> <tr> <td width="97" valign="top" style="width:72.9pt;border:solid windowtext 1.0pt; border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt; padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">Radio waves<o:p></o:p></p> </td> <td width="126" valign="top" style="width:94.5pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">1 cm – 100 m<o:p></o:p></p> </td> <td width="160" valign="top" style="width:119.85pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">Antennas are spaced 8000 km across north America<o:p></o:p></p> </td> <td width="254" valign="top" style="width:190.65pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">VLBA<o:p></o:p></p> </td> </tr> <tr> <td width="97" valign="top" style="width:72.9pt;border:solid windowtext 1.0pt; border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt; padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">Infrared<o:p></o:p></p> </td> <td width="126" valign="top" style="width:94.5pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">1000 nm – 0.1 cm<o:p></o:p></p> </td> <td width="160" valign="top" style="width:119.85pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">*<o:p></o:p></p> </td> <td width="254" valign="top" style="width:190.65pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">Spitzer<o:p></o:p></p> </td> </tr> <tr> <td width="97" valign="top" style="width:72.9pt;border:solid windowtext 1.0pt; border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt; padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">Visible<o:p></o:p></p> </td> <td width="126" valign="top" style="width:94.5pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">400 nm – 800 nm<o:p></o:p></p> </td> <td width="160" valign="top" style="width:119.85pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">10 meters<o:p></o:p></p> </td> <td width="254" valign="top" style="width:190.65pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">Keck<o:p></o:p></p> </td> </tr> <tr> <td width="97" valign="top" style="width:72.9pt;border:solid windowtext 1.0pt; border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt; padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">UV<o:p></o:p></p> </td> <td width="126" valign="top" style="width:94.5pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">10 nm – 200 nm<o:p></o:p></p> </td> <td width="160" valign="top" style="width:119.85pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">*<o:p></o:p></p> </td> <td width="254" valign="top" style="width:190.65pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">Extreme Ultraviolet Imaging Telescope (EIT)<o:p></o:p></p> </td> </tr> <tr> <td width="97" valign="top" style="width:72.9pt;border:solid windowtext 1.0pt; border-top:none;mso-border-top-alt:solid windowtext .5pt;mso-border-alt:solid windowtext .5pt; padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">X-rays<o:p></o:p></p> </td> <td width="126" valign="top" style="width:94.5pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">0.01 nm – 10 nm<o:p></o:p></p> </td> <td width="160" valign="top" style="width:119.85pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">*<o:p></o:p></p> </td> <td width="254" valign="top" style="width:190.65pt;border-top:none;border-left: none;border-bottom:solid windowtext 1.0pt;border-right:solid windowtext 1.0pt; mso-border-top-alt:solid windowtext .5pt;mso-border-left-alt:solid windowtext .5pt; mso-border-alt:solid windowtext .5pt;padding:0in 5.4pt 0in 5.4pt"> <p class="MsoNormal" align="center" style="margin-bottom:0in;margin-bottom:.0001pt; text-align:center;line-height:normal">Chandra<o:p></o:p></p> </td> </tr> </tbody></table> <p class="MsoNormal">*not ground-based telescopes<o:p></o:p></p> <p class="MsoNormal">So – larger telescopes do exist, but not in the capacity that I was imagining. According to a quick Google search, the largest single telescope in the world is the Gran Telescipo Canarias (don’t think you need to be fluent to translate that one…) in the Canary Islands (<a href="http://www.universetoday.com/17652/largest-telescope/">http://www.universetoday.com/17652/largest-telescope/</a>). It has a 10.4 meter aperture. No single telescope would actually be 1 km or more across, but what Professor Johnson was referring to was the necessity for much larger values of B to study higher wavelength waves (such as radio waves!) which is why the VLBA was built across such a large distance!<o:p></o:p></p> <p class="MsoNormal">Most important thing I took away from this was that depending on what wavelength of light you are studying, different sized telescopes will have different levels of effectiveness. There is no ideal size for a telescope because what you need depends on what you are studying. Bigger isn’t always better!<o:p></o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><span> </span><o:p></o:p></p>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com2tag:blogger.com,1999:blog-5849210431402234073.post-25497092423982323432011-10-03T09:31:00.000-07:002011-10-03T11:46:47.979-07:00Lab at the beach<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjUzdQ3l7b0_3SsoCJp5AwWhyuoiVKRzHlQRGZrxK9d1zszTVh8ppHR2V_utizzq-ymBiTD2xl4gpfrPpOm1FwRzz4DGZ7bcEN7LaBVD9ilxSkS4nUlUBbycnplDu6TObTJoTrMKM9uqggZ/s1600/IMG_8330.JPG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;width: 320px; height: 240px;" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjUzdQ3l7b0_3SsoCJp5AwWhyuoiVKRzHlQRGZrxK9d1zszTVh8ppHR2V_utizzq-ymBiTD2xl4gpfrPpOm1FwRzz4DGZ7bcEN7LaBVD9ilxSkS4nUlUBbycnplDu6TObTJoTrMKM9uqggZ/s320/IMG_8330.JPG" border="0" alt="" id="BLOGGER_PHOTO_ID_5659332425386496962" /></a>In what will unfortunately probably be my only trip to the beach for the rest of the year, our Ay class enjoyed our first lab at the Santa Monica beach, measuring the time it takes the sun to set from different altitudes. Next we have to calculate the radius of the Earth using this data collected at the beach.<div><br /></div><div>This lab reminds me of the fact that telescopes are often in fun, fun places - such as Hawaii! To do an Ay lab, we go to the beach. To take measurements for a real project, an astronomer might go to Hawaii. Correlation does not prove causation, but I think I see a theme here... </div><div><br /></div><div>On the drive back from Santa Monica, Professor Johnson explained the importance of the telescope in astronomy projects. As I understand it, part of his motivation for accepting a professorship where he did was that he could continue to use the same telescope that he had been using previously. This is interesting because it shows how important tools are to an astronomer's work - indeed, I asked Professor Johnson how important telescope time was to success as an astrophysicist, and he said that a lot of telescope time could let anyone publish a decent paper, but that brilliant scientists can get by with less. </div><div><br /></div><div>At Caltech, these two categories are combined as all the brilliant scientists have a plenthora of telescope time! Caltech has a lot of shares in many telescopes: take a look at the Caltech Astro department's summary, <a href="http://www.astro.caltech.edu/observatories/">http://www.astro.caltech.edu/observatories/</a>. Also, Caltech has a lot of very good astronomers/astrophysicists publishing papers using data from those telescopes! It seems to me almost like a monopoly of astronomy: Caltech and other leading institutions have all the telescope time, which makes these institutions even better as their scientists can use all this time to publish papers!</div><div><br /></div><div>I would love to hear more about how the telescope time system works from people familiar with it!</div><div><br /></div><div><br /></div>Juliette Beckerhttp://www.blogger.com/profile/15496311467108019749noreply@blogger.com2