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.