Flux and Luminosity

The objective of this problem was to introduce us to the concept of blackbodies and the method to finding one's flux and/or luminosity.

Problem 2(a) dealt with bolometric flux, which is energy per area per time.  To do this, one integrates the equation for blackbody flux, $F_\upsilon (T)$ (found in the last part of problem 1), over all frequencies.  Because the bolometric flux is found by adding up the flux at all frequencies from 0 to $\infty$, it is independent of frequency.

$F(T)=\int_{0}^{\infty}F_\upsilon (T)d\upsilon =\frac{2\pi h\upsilon ^3}{c^2(e^{\frac{h\upsilon }{kT}}-1)}d\upsilon $

$u=\frac{h\upsilon }{kT}\rightarrow \upsilon =\frac{ukT}{h}$

$d\upsilon =\frac{kT}{h}du$

$F(T)=T^4\int_{0}^{\infty}\left ( \frac{2\pi h}{c^2} \right )\left ( \frac{k}{h} \right )^4\left ( \frac{u^3}{e^u-1} \right )$

$\sigma =\int_{0}^{\infty}\left ( \frac{2\pi h}{c^2} \right )\left ( \frac{k}{h} \right )^4\left ( \frac{u^3}{e^u-1} \right )$

$F(T)=T^4\sigma$


In part (b), we were asked to convert the equation $B_\upsilon (T) =\frac{2\pi h\upsilon ^3}{c^2(e^{\frac{h\upsilon }{kT}}-1)}$ so that it was in terms of wavelength instead of frequency.  The tricky part was that this was not just a simple matter of substituting $\upsilon=\frac{c}{\lambda}$ into the equation.  Simple substitution would not conserve the amount of energy during a conversion.

$B_\upsilon (T) =\frac{2\pi h\upsilon ^3}{c^2(e^{\frac{h\upsilon }{kT}}-1)}$

$\upsilon=\frac{c}{\lambda}$

$d\upsilon=\frac{-c}{\lambda^2}d\lambda$

$B_\lambda(T) =\frac{2\pi h\left ( \frac{c}{\lambda } \right ) ^3c}{c^2(e^{\frac{hc}{\lambda kT}}-1)\lambda ^2}$
$B_\lambda(T)=\left ( \frac{2c^2h}{\lambda ^5} \right )\left ( \frac{1}{e^\frac{hc}{\lambda kT}-1} \right )$


Part (c) asked us derive an expression for $\lambda_{max}$, corresponding to the peak of the intensity distribution at a given temperature T.  To do this, we first derived the final equation we got for $B_\lambda(T)$ with respect to $\lambda$.  After many failed attempts at deriving this by myself and some help from WolframAlpha, I found that the derivative of $B_\lambda(T)$ is:

$B_\lambda'(T)=\left ( 2c^2h \right )\frac{\left ( \frac{hc}{kT} \right )\left ( e^\frac{hc}{\lambda kT} \right )-5\lambda \left ( e^\frac{hc}{\lambda kT}-1 \right )}{\lambda ^7\left ( e^\frac{hc}{\lambda kT}-1 \right )^2}$

To find $\lambda_{max}$, one has to set this derivative equal to 0 and solve for $\lambda$.

$5=\left ( \frac{hc}{\lambda kT} \right )\left ( \frac{e^\frac{hc}{\lambda kT}}{e^\frac{hc}{\lambda kT}-1} \right )$

Substituting $u=\frac{hc}{\lambda kT}$ leads to the following equation:

$5=\frac{ue^u}{e^u-1}$

I solved for u by graphing this equation on WolframAlpha.  I got the following graph,

and found a u-value of 4.965.

By solving $4.965=\frac{hc}{\lambda kT}\rightarrow \lambda T=\frac{hc}{4.965k}$ for $\lambda$, I found that $\lambda_{max}=2.85\times10^{-3}mK$

Part (d) of the problem asked us to find a simplified version of $B_{\lambda}(T)$ by using a first order Taylor expansion on $e^\frac{hc}{kT}$.

In a first order Taylor expansion, $e^x=1+x$.  Therefore, $e^\frac{hc}{kT}=1+\frac{hc}{kT}$.

The simplified version of  $B_{\lambda}(T)$ becomes $B_\lambda (T)= \frac{2h\upsilon ^3}{c^2\frac{h\upsilon }{kT}}=\frac{2kT\upsilon ^2}{c^2}$.

Part (e) asked us to find an equation for the bolometric luminosity, L, for a blackbody with radius, R.  This quantity should have units of energy per time.  We were told to start with the equation for bolometric flux, $F(T)=T^4\sigma$.  From there, because we know that blackbodies emit blackbody radiation isotropically, we can simply multiply the flux by the surface area of the blackbody.  

$L=4\pi \sigma R^2T^4$

This works out units-wise because bolometric flux is in units of $\frac{energy}{area\cdot time}$.  multiplying by area will give units of energy per time.  

The final part of the problem posed a situation in which two stars are gravitationally bound.  One is blue and one is yellow; the yellow one is a lot brighter than the blue one.  We were first asked to qualitatively compare the temperatures and radii of the two stars.  The blue star is hotter, because blue corresponds to higher energy, which corresponds to higher temperature.  The yellow star has a bigger radius because it appears to give off more light than the blue star.  The only way this would happen (assuming the two stars are roughly the same distance from Earth) is if the yellow star were just a lot bigger than the blue star.  

We were then asked to quantitatively compare the radii of the two stars.  To answer this question, I randomly assigned relative values to the luminosities and temperatures of the two stars.  I said that the blue star was twice as hot as the yellow star, which was twice as luminous.  

$8\pi \sigma R_B^2T_B^4 = 4\pi \sigma R_Y^2T_Y^4$

$R_B^2T_B^4 = R_Y^2\left ( \frac{1}{2}T_B \right )^4$

$\left ( \frac{R_B}{R_Y} \right )^2=\frac{1}{32}$

$\frac{R_B}{R_Y}=.2$

Mayan Astronomy

Unlike the ancient Celts who measured time with respect to the night, the ancient Maya were more interested in studying the sun.  This is probably a consequence of their beliefs, which include a sun god, Kinich Ahau, but no lunar deity.  That's not to say that they completely ignored the moon, because they eventually started tracking the lunar cycle, but it wasn't their primary concern.

Predating the Coligny Calendar by a few thousand years, the (in)famous "Mayan Calendar" is actually a system of calendars.  We know about their calendars because before the fall of Mayan civilization, they, again unlike the ancient Celts, left behind a lot of written pieces documenting their work.  There are four such pieces, known as the Dresden, Madrid, Paris, and Grolier Codices.

The Dresden Codex is in the best condition of all the codices.  Its most well-known contents are the accurate calculations of the lunar cycle, including predictions of eclipses, and descriptions of the motion of the planet Venus.  One of their many calendars was based on the movements of Venus.  It was 260 days long, which is the amount of time Venus is seen in the morning and, 50 days later, in the evening.  This Venus calendar ran concurrently with the 365-day calendar marking the solar year and the lunar calendar that kept track of the lunar cycle.

From the Wikipedia page on the Dresden Codex

The Madrid and Paris Codices mostly contain information about the Maya religious deities and rituals.  They document the constellations and what they meant to the Maya, describing the ceremonies done by priests at different times of the year.  The existence of the fourth codex is under some scrutiny because it's the least intact and contains simple descriptions of the Venus motion described in the Dresden Codex.  

The ancient Maya were able to to find such accurate measurements not because they had advanced technology*, but because they recorded careful observations over hundreds of years.  That was the method used by all ancient civilizations interested in astronomy.  

*They did have a really cool observation setup that resembled our modern inferometry.  They spaced out their temples in such a way that they could observe objects in the sky from different angles and get a more complete view of the same celestial object.  

From authenticmaya.com

Black Body Radiation and Chemical Composition

This post is going to make a lot of assumptions about the "average human," but astronomers and physicists base much of their work on assumptions, so I don't think it will matter too much.


Blackbodies are physical structures that absorb all electromagnetic radiation and emit what scientists call blackbody radiation.  They are ideal emitters, so at a given temperature, they emit as much or more energy in all frequencies than other bodies of the same temperature.  The energy they emit is diffused isotropically, so it's the same in every direction.   Every other physical body's ability to emit radiation is measured in relation to a blackbody.  That measurement is called emissivity, and blackbodies have an emissivity coefficient of 1.

Blackbodies don't actually exist in nature (though there are objects that come close enough as to be called blackbodies by scientists), but the physical description is a hole in a box.

Picture of blackbody representation from Wikipedia website on blackbodies

Light can go in at any wavelength, but because the walls are opaque to radiation and the hole is small, the light most likely won't get out.  Any radiation the box emits will be a function of temperature and will radiate from all sides of the box.  

Given this description of a blackbody, can the average human be called one? Of course not (technically)!  We're visible, for one, which means we don't absorb all kinds of electromagnetic radiation.  We aren't spherical, so any energy we do emit isn't emitted isotropically.  Still, astronomers like to assume that a lot of things are blackbodies.  Is the emissivity coefficient of the human body close enough to 1 to be considered a blackbody?   

The interesting question is whether or not our chemical composition would lead one to believe that humans are blackbodies.  All solid substances are considered grey bodies, meaning their emissivity coefficients are between 0 and 1.  I'm curious to see if an object's emissivity coefficient can be found by adding up all of the emissivities of its components (taking into account percent composition).

The average human body is mostly oxygen, carbon, hydrogen, and nitrogen (in order of decreasing presence).  

From Wikipedia's page on the composition of the human body

Some of those elements, because they're not metals, don't actually have emissivity coefficients.  Thinking I had reached a dead end, I almost gave up.  But then I realized that the internal composition doesn't matter anywhere near as much as the external composition.  In other words, I figured that I should actually be looking at the chemical composition of human skin.  

According to a study done in 1927 on the composition of human skin (which, despite its age, was the most straightforward data I could find), the average 30-year-old human's skin is composed of water, calcium, magnesium, each making up about 1% of the skin.  

A quick google search told me that the emissivity coefficient of water is .95 (from infrared-thermography.com).  Given this information and the (rough) chemical composition of human skin, it's safe to say that the emissivity coefficient of human skin is very close to 1.  

Therefore, the human body is, in fact, a very close approximation of a blackbody.  

Celtic Astronomy (a little more archaeoastronomy)

Until the late 1800s, little was known about the ancient Celts' knowledge of or interest in astronomy because archaeologists just couldn't find any artifacts to indicate it.  For centuries, most of the knowledge held by the Celtic peoples was concentrated in a small group of wise people they called Draoi, or Druids, and they passed their knowledge from generation to generation orally.  In 1897, archaeologists found a calendar that dated back to the Gauls in modern-day France.  This tablet, called the Coligny Calendar, was the first evidence of ancient Celtic interest in astronomy.

Taken from the Coligny Calendar page on Wikipedia

Measuring about 1.5 x 1 m, the calendar is a lunisolar calendar thought by anthropologists to be a rebellion against the implementation of the Julian calendar in the 2$^{_nd}$ century AD.  The Gauls measured time starting with night, so night came before day and dark seasons(what we call winter) came before light seasons (summer).  

The ancient Gauls had no astronomical tools to observe the sky.  They made their calculations based on years--even generations--of naked-eye observations.  And (as far as we know) they didn't write any of it down!  Druids used the modern statistical method of Stochastic Estimation to make their predictions, and though their work wasn't as accurate or as far-reaching as the famous Mayan Calendar, it was accurate enough.  

They were able to discern that the phases of the moon were periodic, and even figured out the geocentric relationship between the Earth and the moon.  Their work in predicting eclipses was particularly accurate, discerning an 18-year-long cyclical pattern, which they called the Saros Cylcle.  

By pinpointing different spots in the sky and keeping track of how long it took the moon to reach those spots time after time, the Druids developed an intricate system of cycles.  When combined, the different cycles allowed the Druids to accurately track time, including corrections similar to our leap year to make up for the difference between solar and sidereal times.  

Based on the evidence provided by the Coligny Calendar that the ancient Celts were interested in studying the moon, archaeoastronomers are convinced that they also would have studied the stars.  Now it's just a matter of finding the artifacts to prove it.  

Betelgeuse

The constellation Orion has always been special to me.  It was the first constellation I could name by sight, and when I was younger, my mom told me he was watching over me.  Orion's right shoulder (left, when we look at him) is the star Betelgeuse.

Picture of Betelgeuse from the Hubble site taken with the Hubble Space Telescope

Betelgeuse is about 1000 times bigger and 100,000 times more luminous than our sun.  It's one of the brightest stars in the night sky, despite the fact that it's almost 200 parsecs away.  To us, it appears to be slightly red, indicating a low energy and cooler temperature relative to our sun (in astronomy, red means cold and blue means hot).  Because of its color and size, Betelgeuse is classified as a red giant, so it's further along in its life cycle than our sun is by about 3 billion years.  Eventually, because it's so massive, Betelgeuse will most likely explode into a supernova.  

Found on an astronomy wiki page

http://nothingnerdy.wikispaces.com/E5+STELLAR+PROCESSES+AND+STELLAR+EVOLUTION

Fourier Transforms in Oceans

I had trouble thinking of an application of Fourier Transforms.  At first, I thought I might look into acoustics and how sound waves propagate through the air, but I discarded that idea quickly because I didn't find it all that interesting.  It seemed too similar to the work we had done with light waves.  I sat for a while, trying to think of other things that came in waves.  Eventually (I'll spare you the details of my mind's twisty journey) I remembered myself sitting on the beach over winter break, looking up at the moon and hearing the tide come in.

I knew, then, that tides were the result of the gravitational pull from the moon and sun and the Earth's rotation.  I also knew that people had had the knowledge to predict the tides for years (though I had never given much though to how long this knowledge had been around).  Now that I know about Fourier transforms, I can (mostly) understand the theory behind tidal prediction.

Picture of ocean currents around the world from an Indiana University geology class site

http://www.indiana.edu/~geol105/1425chap4.htm

Just like the light waves in Young's double slit experiment, the currents in the above picture interfere with each other destructively and constructively.  When they interfere constructively, they create waves (or tsunamis when they interfere constructively and nature's in a bad mood).

The exact equations vary slightly from source to source, but the general idea is that tidal predictors do their work by relating Fourier transforms of the currents and average current information obtained from months or years of data collection.  In 1966, Munk and Cartwright developed the Response Method$^{_1}$ (which is the easiest method to explain) for predicting shallow water tides.  It uses the equation Z(f)=$\frac{G(f)}{H(f)}$ where G(f) and H(f) are Fourier transforms of the tide's potential and the data gathered, respectively and f is frequency.

$^{_1}$oceanworld.tamu.edu

Comparing Telescopes

In this problem, we were told about two telescopes that observe at different wavelengths and asked to compare their angular resolutions.  The CCAT is a 25 m telescope that observes at a wavelength of 850 microns.  The MMT is only a 6.5 m telescope that observes in the J-band.  A quick Google search taught me that the J-band is a range of infrared frequencies centered around wavelengths of 1.25 microns.

In class, we learned that angular resolution, $\theta = \frac{\lambda }{D}$.

$\theta_{CCAT}=\frac{850 \mu m}{25 m}\left ( \frac{1 m}{1\times 10^{-6}\mu m} \right )= 3.4\times 10^{-5} rad$

$\theta_{MMT}=\frac{1.25 \mu m}{6.5 m}\left ( \frac{1 m}{1\times 10^{-6}\mu m} \right )= 1.9\times 10^{-7} rad$

The angular resolution of the MMT is smaller, and therefore better than that of the CCAT.  This might be slightly surprising because the CCAT has a larger diameter, which, as a very loose rule of thumb, is usually better than smaller telescopes because they can collect more light.  It's still necessary to have CCAT, though, because the MMT isn't big enough to capture wavelengths as big as those captured by CCAT.

Angular Resolution

In the first part of this problem, we were asked to reconstruct Young's double slit experiment.  We drew this picture, with the curves representing the light waves coming from behind the slits.

We were told to convince ourselves that the monochromatic light hits the screen in a cosine function.  We drew this picture:

which represented a situation in which the light waves are interfering in one spot on the screen.  We figured that in order for the waves to interfere constructively, the line labeled $\lambda$ would have to be equal to the wavelength multiplied by an integer. That way, the two waves would be in phase together. The waves are interfering constructively in the picture, creating a peak at a distance, d, from the center of the screen.  If the two waves aren't in phase, they will interfere each other destructively.  We knew that the waves hit the screen in a periodic wave function, so we were able to convince ourselves it was a cosine function that looked kind of like this:

Because we were told that L>>D, the lengths of all of the waves going to the screen from the two slits are almost equal and the angle formed where the line L meets the screen is a right angle.  With this information, we made the following equation:

$\frac{d}{L}=\frac{n\lambda}{D}\rightarrow d=\frac{nL\lambda }{D}$

The second part of the problem asked us to imagine what would happen to the brightness pattern on the screen if there were two more slits closer together than the already existing two.  The distance, D, between these two slits would be less than in the first part of the problem.  According to the equation above D and d are inversely proportional, so as D decreases, d increases.  An increase in d results in a brightness pattern on the screen with more space between each peak where the waves interfere constructively.  The light coming in through the four slits would result in a more intense peak in the middle of the screen where the two waves interfere constructively and lower amplitudes elsewhere on the screen.  The resulting wave function, relative to the first, would look like this:

In the third part of the problem, we were asked to find the brightness pattern on the screen if there were a continuous set of slits with ever increasing D.  So, essentially we were asked to find the brightness pattern on the screen if there were a giant gap in front of the light source instead of a wall with slits.  Again, using the equation above, as D decreases, d increases, thus increasing the distance between peaks in the brightness pattern.  As more and more slits are added, the peak in the middle of the screen will get higher because of the constructive interference of all of the light waves. The resulting brightness pattern, again relative to the first two, will look like this:

In the next part of the problem, we were asked to compare the double slit experiment to the top hat Fourier transform function.  The top hat transform is a function that shows a frequency spike at 0 on the x-axis.  I find it easier to picture the graphical representation of the top hat function from an aerial view, which looks like a ripple in water with a big splash in the middle and diminishing waves radiating out.

http://www.radiantzemax.com/content_images/apodization/ap2.jpg

The final part of the problem asked us to relate what we had done to a telescope primary mirror.  Working with the slit experiment and convincing myself that a brightness pattern left by light on a surface will be a cosine function.  All of the light waves interfere constructively and destructively in such a way that there is a peak of information in the center of the primary mirror.  

   

The Great Wall

I was looking through the Center for Astrophysics' website, trying to find an interesting research project for my fifth blog post.  Clicking through the different tabs, I saw a link to the cosmology page and, remembering my fondness for the subject from AY 17, I decided to take a look.  I didn't get too far, though, because I saw a picture labeled "Great Wall" and excited about learning about this part of the universe that never came up in my class on extragalactic astronomy.

The Hercules-Corona Borealis Great Wall is the largest structure in our observable universe, larger even than superclusters by far. It was discovered in 2013 while scientists were mapping gamma ray bursts at high redshifts.  The Great Wall is at a redshift >1, so it's about 3x10$^{_9}$ pc away. 

Before last year, the largest structure in the universe was believed to be the CfA2 Great Wall, located about 6.1x10$^{_7}$ pc away.  Scientists believe its volume to be about 2x10$^{_60}$ cm$^{_3}$.  They can't be certain, though, because part of the Great Wall is near the galactic plane.  Objects near the galactic plane are obscured by the dust and gas of the milky way. 

The existence of Great Walls--these super, super, super, Superclusters-- is such a big deal because it seems to go against the cosmological principle, which says that the universe is homogeneous and isotropic.  In laymen's terms, objects, temperature, chemical composition, and all other characteristics are all evenly distributed throughout the universe.  The fact that there's such a large collection of objects in one place is pretty antithetical to that even distribution.  

They also complicate scientists' ideas of the evolution of the universe.  The Hurcules-Corona Borealis Great Wall is 10 billion light years away, which suggests that the way we see it from Earth now is the way it looked 10 billion years ago, really close to the Big Bang.  Such a large and complicated structure shouldn't have had enough time to form.  

Archaeoastronomy

Coined in 1973 by Elizabeth Baity$^{_1}$, the term archaeoastronomy is relatively new, but the idea of using astronomical data to solve mysteries of the past has existed for centuries. Over the past 40 years, archaeoastronomers have worked towards establishing the field as the anthropology of astronomy, comparing current astronomical knowledge to that of ancient civilizations to learn about their cultures.  They prefer to make a clear distinction between their work and the study of the history of astronomy, which is more like the study of when discoveries were made.  Astronomers have been around for thousands of years, but before there were archaeologists, there were antiquarians.  People who chose to combine the two fields did things like find relationships between stars and church locations in England and point out the interesting orientation of the Great Pyramids.

There are two general fields of archaeoastronomy, Brown and Green, which tend to be separated not by tendency towards social or physical science, but by availability of information.  Brown archaeoastronomers , for example, tend to live in newer civilizations where they don't have direct access to the specific site they're studying.  Because of this, they rely more on ethnographic writings and historical records of the cultures they study.  Green archaeoastronomers, on the other hand, usually live in the Old World and are known for emphasizing alignment, orientation, and statistical data in areas where there might not be as much historical data.

I chose to write about archaeoastronomy because many of my future posts will probably be acrhaeoastronomical pieces about different cultures.  Since declaring a joint concentration in Astrophysics and Folklore/Mythology, I've been interested in studying the astronomy of ancient cultures, but I didn't have a word for that interest until I read the introductory chapter in the textbook.


$^{_1}$ Wikipedia page on Archaeoastronomy

Local Sidereal Time

Local Sidereal Time was explained in this problem as the right ascension at the local meridian at that moment.  At noon on the vernal equinox, when the Earth is pointing directly at the sun, the right ascension is 0.00 hours.  Every month, the right ascension increases by two hours, a result of the difference between a 24-hour solar day and a slightly longer stellar day.  We then drew the following picture to help us understand right ascension.

The first part of the problem asked what the LST would be at midnight on the vernal equinox.  My initial thought was that it would be 12:00, but then I realized that I had failed to take into account the four-minute difference between a solar day and a stellar day.  Since midnight is half a day after noon, it would be twelve hours and 2 minutes after 0:00, making it 12:02.  

The second part asked what the LST would be 24 hours after the time from part (a).  Following the same thought pattern as in the first part of the problem, I found that it would be 24 hours and the additional four minutes after 12:02 hours.  The LST would be 12:06. 

In the third part of the problem, we were asked to find the LST at that moment.  When we solved the problem, it was around 3:00 pm on February 5.  That date is about a month and a half before the vernal equinox, meaning The RA is 21 hours after 0:00.  We figured that, if noon on that date is 21:00, then 3 hours after noon is 0:00.  The next part of the problem asked us to find the LST at midnight on that same date.  To find this, I just added nine because midnight is nine hours after 3:00, making the LST 9:00.  

The final problem asked me to find the LST at sunset on my birthday.  My birthday is in the beginning of January, and because I've always lived in the northern hemisphere where days are shorter in the winter, sunset is usually around 6.  The beginning of January is about two and a half months before the vernal equinox, meaning the LST at noon is about 19 hours after 0:00.  Six hours after 19:00 would be 25:00, but since LST resets at 0:00, the LST at sunset on my birthday is 1:00.  

AY Sixteenus (cont.)

The first part of this problem asked me to create a graph of night time as a function of season.  My first step was to set up the axes of the graph with hours from 5 p.m. to 7 a.m. on the y-axis and months of the year on the x-axis.  My next step was to separate the x-axis into the four seasons.

I approached the problem with the knowledge that day and night are equal at the equator, using that knowledge to set up a basis for comparison of night and day lengths for the rest of Earth.  The foundation graph looked like this:

The next step was to label the sunrise and sunset times for the vernal and autumnal equinoxes because that's when night and day are both 12 hours long.  Then I used my experience of the length of day and night during the different to seasons to plot the sunrise and sunset times at the solstices (relative to the base lines in the graph above).  I have many winter memories of walking home from practice in the dark at 5:00, so I knew sunset happened earlier in the winter.  I also have memories of riding to school at 7:00 in the morning when it was still dark, so sunrise in the winter is later.  My experiences also told me that summer is exactly the opposite, which led me to the following graph.  

The second part of the problem asked me to sketch a diagram illustrating the times that astronomers can observe AY Sixteenus from the Mt. Hopkins observatory.  We* started with a review of right ascension which led to the following picture.  

Looking at this picture, and operating under the infinite distance idea described in the last post, we realized that on the vernal equinox, AY Sixteenus comes into the field of view at midnight.  Before that, the observatory is facing the wrong half of the night sky to see the star.  Twelve hours later, at noon the next day, the observatory is once again facing the wrong side of the sky and the star sets.  

We know that one stellar day is equal to four minutes more than exactly 24 hours.  Four minutes every day for three months (the length of one season) gives a difference of about six hours.  Knowing this, we knew that at every change of season, the starrise and starset would be six hours earlier than they were at the previous change.  This idea led to the following graph:

The final parts of the problem asked us to consider how the above graph would change in different situations.  If the observatory were at a latitude of -10$^{\circ}$, The lines indicating sunrise and sunset would be closer together in the winter (beginning of the graph) and further apart in the summer. If the observatory were further north than in is at Mt. Hopkins, at a latitude of 60$^{\circ}$, the lines indicating sunrise and sunset would be further apart in the winter and closer together in the summer.  

The final part asked whether a higher or lower declination than +32$^{\circ}$ would be better for observation.  A higher declination would be better because an observer would have to look through more atmosohere to see anything much lower.  

*"We" are Dennis Lee, Delfina Martinez-Pandiani, and me.

AY Sixteenus

In this problem, we were told that there was a star, AY Sixteenus, located at a R.A. of 18 hours and a declination of +32$^{\circ}$.  We were asked to draw a graph of elevation vs. time of the star as seen from the meridian at the Mt. Hopkins observatory in Arizona.

The first step in solving this problem was to figure out if, and if so how, the elevation would change throughout the year.  Intuition told us* it would change because as we observe throughout the night, the star appears to rise and fall in the sky.  Further thought, however, reminded us that we were only graphing the view of the star as seen through the meridian.  When the star passes over the meridian every night, it's at the same elevation, which makes sense because a star's declination remains constant throughout the year.

The next obstacle was trying to find a constant value for the the star's elevation.  Originally, we thought it should be equal to the star's declination.  But declination is the angle from the equator to the star and elevation is the angle of the star above the observatory's horizon.  The Mt. Hopkins observatory is approximately 32$^{\circ}$ above the equator.  Using the information that the star is infinitely far away, we determined that the angle of view from Mt. Hopkins was basically the same as that from the equator.

When viewed from the ground through the meridian, the star is at zenith, meaning it's directly overhead.  That led us to the conclusion that AY Sixteenus, when viewed through the meridian, is at a constant elevation of 90$^{\circ}$.     

*By us, I mean our group made up of Dennis Lee, Delfina Martinez-Pandiani, and me.

Dimensional Analysis

Before I came to college, I had never heard of dimensional analysis.  Once I learned what it was, I realized that it was just another example of scientists giving simple things scary names.  Dimensional analysis is extremely useful when you don't know an equation, but know the units.

The first dimensional analysis problem asked me to find the relationship between pressure, density, and the speed of sound in a gas. I knew that the speed of sound had units of m/s and density had units of g/cm^3.  I had always heard of pressure in units of lbs/in^2, and it took me a few minutes of thought to recognize that that translated into science units of N/m^2.  (To get to this point, I first had to remember that pounds were actually a unit of force achieved by multiplying mass and the gravity at earth's surface.)  With this knowledge, I was able to rearrange the units of pressure and density to become m/s in the following way:

Cs^2 = gm/(m^2 s^2 ) * m^3/g

         = g/(ms^2 ) * m^3/g

         = m^2/s^2 

Cs = √(m^2/s^2 )

     = √(P/ρ) 

"Yep, I see that."

In this problem, we were asked to find the power output of a light bulb in a large, dark cave.  We were told that the bulb is 1 kilometer away and that the human eye has to receive 10 photons to register a source of light.  We were also given the equation for finding the energy of a proton, E = hv, where h is Plank's constant in erg*s and v is the frequency of the light.

Knowing the relationship between frequency, wavelength, and the speed of light to be c = vλ, I rearranged the Energy equation to read E = hc/λ.  I then solved to find the energy in ergs using the wavelength λ=580 nm, the average wavelength in the visual spectrum.  I multiplied this energy by 10, the number of photons reaching the eye if the light is barely visible.

Power is a quantity in units of energy per time.  Flux is in units of power per unit of area.  To find the power output, I first found the flux of the light by dividing the energy of 10 photons by the area of of the light's "surface" from 1 kilometer away, which is 1 x 10^10 square centimeters.  This left me with a flux of 3.3 x 10^-22 erg/(s*square cm).  To convert flux to luminosity, the astronomer's measure of power, I used the Inverse Square Law: L = F4πd^2.  With this equation, I found the luminosity to be 4 x 10^-11 ergs/s.