Week 3: WS5 #2

2. a) In astronomy, it is often useful to deal with something called the "bolometric flux," or the energy per area per time, independent of frequency. Integrate the blackbody flux \(F_\nu(T)\) over all frequencies to obtain the bolometric flux emitted from a blackbody, F(T). 

We were given that \(F_\nu(T)=\frac{4\pi\nu^2}{c^2}\frac{h\nu}{e^{h\nu/kT}-1}\), which we then have to integrate.
\(\frac{4\pi}{c^2}\int^\infty_0\frac{h\nu^3}{e^{h\nu/kT}-1}d\nu\)
We define \(u=h\nu/kT\).
\(\frac{4\pi}{c^2}\int^\infty_0\frac{h(ukT/h)^3}{e^u-1}(kT/h)du\)
\(\frac{4\pi k^4T^4}{c^2h^3}\int^\infty_0\frac{u^3}{e^u-1}du\)
The integral is defined as \(\frac{\pi^4}{15}\), so we can pull that out and define \(\frac{4\pi^5k^4}{15c^2h^3}=\sigma\) (a constant).
The final equation is \(F(T)=\sigma T^4\)

b) The Wein Displacement Law: Convert the units of the blackbody intensity from \(B_\nu(T)\) to \(B_\lambda(T)\). 

The relation between the blackbody intensity in terms of frequency and the intensity in terms of wavelength is defined such that the integral of the two equations are equal.
\(\int^\infty_0B_\nu d\nu=\int^0_\infty B_\lambda d\lambda=-\int^\infty_0B_\lambda d\lambda\)
We can get rid of the integral such that \(B_\nu d\nu=-B_\lambda d\lambda\).
Since we know that \(B_\nu(T)=\frac{2\nu^2}{c^2}\frac{h\nu}{e^{h\nu/kT}-1}\) and that \(\nu=\frac{c}{\lambda}\) (and therefore that \(d\nu=\frac{-c}{\lambda^2}d\lambda\)), we can simply plug in the values to do this conversion.
\(B_\lambda(T)d\lambda=\frac{-2(c/\lambda)^2}{c^2}\frac{h(c/\lambda)}{e^{hc/\lambda kT}-1}\frac{-c}{\lambda^2}d\lambda\)
\(B_\lambda(T)d\lambda=\frac{2}{\lambda^5}\frac{hc^2}{e^{hc/\lambda kT}-1}d\lambda\)
We can cancel out \(d\lambda\) to get the final answer:
\(B_\lambda(T)=\frac{2}{\lambda^5}\frac{hc^2}{e^{hc/\lambda kT}-1}\)

c) Derive an expression for the wavelength \(\lambda_{max}\) corresponding to the peak of the intensity distribution at a given temperature T. 

In order to do this, we want to take the partial derivative in terms of wavelength of the equation we derived in part (b) and set it equal to zero to find the maximum wavelength.
\(\frac{\partial}{\partial\lambda}[\frac{2}{\lambda^5}\frac{hc^2}{e^{hc/\lambda kT}-1}]\)
By u-substitution (\(u=\frac{hc}{\lambda kT}\)), we can condense this expression:
\(\frac{\partial}{\partial u}[\frac{2}{(hc/\lambda kT)^5}\frac{hc^2}{e^u-1}]\)
After a bunch of calculus, we have found the derivative and can set it to zero.
\(\frac{2(kT)^5(hc^2)}{(hc)^5}\frac{(e^u-1)(5u^4)-(e^u)(u^5)}{(e^u-1)^2}=0\)
Solving this equation gives us \(5=\frac{ue^u}{e^u-1}\).
If we assume that u is very large, we can simplify to \(5=\frac{ue^u}{e^u}\) yielding the final result that \(u=\frac{hc}{\lambda kT}\approx5\). Solving for \(\lambda_{max}\) and plugging in the constants, we end with the equation \(\lambda_{max}=\frac{0.283}{T}\).

d) The Rayleigh-Jeans Tail: Next, let's consider photon energies that are much smaller than the thermal energy. Use a first-order Taylor expansion on the term \(e^{h\nu/kT}\) to derive a simplified form of \(B_\nu(T)\) in this low-energy regime. 

If we assume that \(\frac{h\nu}{kT}\) is very small, we can assume that \(e^{h\nu/kT}\approx1+\frac{h\nu}{kT}\).
\(B_\nu(T)=\frac{2}{c^2}\frac{h\nu^3}{e{h\nu/kT}-1}\)
\(B_\nu(T)=\frac{2}{c^2}\frac{h\nu^3}{1+\frac{h\nu}{kT}-1}\)
\(B_\nu(T)=\frac{2\nu^2kT}{c^2}\)

e) Write an expression for the total power output of a blackbody with radius R, starting with the expression for \(F_\nu\). This total energy output per unit time is also known as the bolometric luminosity, L. 

We've already figured out that \(F(T)=\sigma T^4\). Since the luminosity is simply all of the flux over the entire surface area of the sphere of expanding radiation, we can simply multiply the flux by the surface area of a sphere to yield \(L=\sigma T^44\pi R^2\).

Week 3: Gravitational waves

Yesterday, a breakthrough was announced in astrophysics: the first detection of gravitational waves. Gravitational waves were initially predicted by Einstein's theory of general relativity, but they hadn't been detected until September 2015. Their detection provides strong support for general relativity, as well as providing an avenue for studying the universe that is fundamentally different from just using electromagnetic radiation. I think this is probably the aspect of this discovery that I am most excited about--just last week I was wondering what telescopes would be like if they functioned less like the human eye, and using gravity as an observational tool could play that role. (Was I in on the discovery? Was it just coincidence? The mystery continues...)

Essentially what happened was that two black holes, one 36 times and one 29 times the mass of the Sun, collided and released 3 times the Sun's mass in pure energy. Remember, \(E=mc^2\) (another shout-out to Einstein), so this is a lot of energy--50 times the power of everything else in the visible universe, in fact. This energy was released largely in the form of gravitational waves, which is why LIGO (Laser Interferometer Gravitational Wave Observatory) was able to detect them, even though the event occurred 1.3 billion light years away (and therefore 1.3 billion years ago). The video below is a visualization of the two black holes orbiting each other and finally colliding. This video is in fact in slow motion--the actual event happened in just tenths of a second, with the black holes reaching half the speed of light in their orbit.

While this has been the most impressive signal received so far, there were a couple of smaller signals received during the 5 month detecting period. This is pretty cool because it means that the technology is working and that events are happening in the universe that are within our threshold of detection. It'll be really exciting to see how this can be used to advance our understanding of gravity, general relativity, and the universe as a whole.




Sources
https://www.ligo.caltech.edu/video/ligo20160211v3
http://gizmodo.com/physicists-are-freaking-out-about-gravitational-waves-a-1758605484
http://www.nytimes.com/2016/02/12/science/ligo-gravitational-waves-black-holes-einstein.html?_r=0
https://en.wikipedia.org/wiki/General_relativity#Gravitational_wave_astronomy
http://apod.nasa.gov/apod/ap160212.html

Week 2: Astrophysics in a Nutshell Chapter 1

Chapter 1 of Astrophysics in a Nutshell essentially focused on how telescopes worked, but the most interesting part for me was the parallels drawn between telescopes and the human eye (again, I'm much more a biologist than a physicist). Like telescopes, the human eye has a both a "camera" and a "detector"--the lens in the front of the eye focuses the light onto the retina in the back of the eye in the same way telescopes may use mirrors to focus light onto a CCD or other type of detector.

Many vision problems arise from misalignments of the focused light in the eye, and for centuries, concave and convex lenses have been used as vision correctors. On top of that, people have been staring at the sky for pretty much as long as people have been around to look at anything. It's not surprising, then, that in some ways the human eye served as inspiration for telescopes, and that the first telescopes used lenses to magnify the light.

The lens focuses light onto the retina, although the image is flipped--a problem which is corrected by the brain. 

However, telescopes soon surpassed eyes in almost every regard (which would be sad if it weren't kind of the point of telescopes in the first place). First of all, modern telescopes can increase their integration time. This allows for the detection of a larger number of photons, which strengthens the signal of faint sources. Our eye has a very short, fixed integration time of only a fraction of a second, which isn't conducive to astronomy, but also prevents us from being blinded by the Sun, so it's probably a net positive. Our eyes are also pretty small, which again isn't conducive to gathering tons of photons. If you've ever had your eyes dilated, that increases the pupil size, allowing more photons to enter. This is why the ophthalmologist's office always gives you those super cheap sunglasses after every appointment. 

Swag.

You never see telescopes wearing those cool glasses though, because it's to an astronomer's advantage to gather as many photons as possible, and this is facilitated by having a larger area over which to gather photons. Finally, the eye is limited in that it can only detect electromagnetic radiation in the visible spectrum (hence the name "visible spectrum"). Telescopes, on the other hand, can be set to detect wavelengths in different parts of the spectrum. This is helpful since a lot of astronomy occurs outside of the visible light range, so we'd never be able to be able to detect all that information without them, even if our eyes were otherwise comparable to telescopes. 

One interesting thing to think about is whether there are alternative forms that telescopes could take that are less based on the human eye, and what the advantages and disadvantages of something like that would be. 

Sources
https://en.wikipedia.org/wiki/History_of_the_telescope
http://www.antiquetelescopes.org/history.html
https://cnx.org/resources/465ef69789306dc609b2650a92697be3e852a6e7/Figure_27_01_02.jpg
http://www.easyeyes.com/images/pedsc.jpg

Week 2: WS4 #2

2. CCAT is a 25-meter telescope that will detect light with wavelengths up to 850 microns. How does the angular resolution of this huge telescope compare to the angular resolution of the much smaller MMT 6.5-meter telescope observing in the infrared J-band? 

We determined in a previous problem that the resolution \(\theta\) is equal to \(\frac{\lambda}{D}\), where \(\lambda\) is the wavelength of the light observed and D is the diameter of the telescope. If we say the J-band is 1.2 microns, we can simply calculate and compare the resolutions.
\(\theta_{CCAT}=\frac{850\times10^{-6}m}{25m}=3.4\times10^{-5}\)
\(\theta_{MMT}=\frac{1.2\times10^{-6}m}{6.5m}=1.8\times10^{-7}\)
Since smaller resolution is better, we know that MMT has the better resolution.


Corrected 2/14/16


Sources
https://en.wikipedia.org/wiki/J_band

Week 2: WS3 #2

2. The Local Sidereal Time (LST) is the right ascension that is at the meridian right now. LST = 0:00 is at noon on the Vernal Equinox (the time when the Sun is on the meridian March 20th, for 2013 and 2014). 
a) What is the LST at midnight on the Vernal Equinox? 

Sidereal time is based on the time it takes for the Earth to complete exactly one rotation, rather than how long it takes for the same for the same point to realign with the Sun. As a result of the Earth's rotation around the Sun, these times differ slightly--LST picks up 4 minutes per day over "regular" time. Therefore, at midnight (12 hours after noon) on the Vernal Equinox, the LST would be 12:02.

b) What is the LST 24 hours later (after midnight in part 'a')? 

An extra 24 hours would mean an extra 4 minutes; 12:06.

c) What is the LST right now (to the nearest hour)? 

During our class time (2:30-4pm) on Tuesday, February 2, we calculated that the LST would be approximately 1:00--1 month and 18 days before the Vernal Equinox translates to a loss of about 3 hours. At 4pm, this would make the LST about 1:00.

d) What will the LST be tonight at midnight (to the nearest hour)? 

Approximately 8 hours after class time would make the LST about 9:00.

e) What LST will it be at sunset on your birthday? 

My birthday is July 27. If sunset is at 8pm, that makes this 4 months, 7 days, and 8 hours after noon on March 20th. This translates to a gain of 8:29 and an LST of 16:29.

Week 2: The Pillars of Creation

"Pillars of Creation" is one of my favorite photographs of all time--in fact, it's actually my desktop background, which hopefully contributes to my astro street cred without being too nerdy. First of all, its colors and form are beautiful. The picture was taken in both visible and infrared spectra, which allows the capture of the densest regions of the structure. The glow around the pillars (which might be my favorite part of the picture) is the vaporization of the material on the periphery. Unfortunately, this means that this incredible structure is gradually disappearing. 

Second, this thing is huge--the pillar on the left is about four light years long. It's composed of molecular hydrogen and is an active star forming region. Thinking about the sheer size and the vast amount of hydrogen, the lightest element, that must be present to form structures as large and dense as stars is mind-boggling. In fact, the stars that this thing has created provide a large part of the energy that is destroying the pillars. Kind of sad, isn't it? 

A final interesting fact about this photo is that it was taken in 2014 as kind of a tribute to a famous picture taken by the Hubble telescope in 1995--an awesome image in its own right: 

Sources

http://www.astronomy.com/-/media/Images/News%20and%20Observing/News/2015/01/Pillars2014.jpg?mw=600

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

https://upload.wikimedia.org/wikipedia/commons/b/b2/Eagle_nebula_pillars.jpg

Week 2: Planet X

As probably everyone who is remotely interested in astronomy has heard, a new planet was recently discovered. What's notable about this planet is that it orbits our own Sun, which makes it a candidate for inclusion in The Official Planet Roster Of Our Solar System--a list from which Pluto was crossed off in 2006.

While Planet X (whose name for some reason just reminds me of Chemical X from the Powerpuff Girls) has been detected due to its gravitational effects on other objects in our solar system, including the minor planet Sedna, it hasn't been visually detected. Part of the reason for this is because of how far away it is. Estimates for its closest approach are around 200 AU--almost 7 times that of Neptune, the most distal planet in our solar system (since Pluto's demotion at least)--and its farthest distance could be as large as 1,200 AU. At that distance, it would take light from the Sun almost a week to reach the planet. This makes it pretty difficult to find with a telescope.

In an ironic twist, one of the scientists who published results supporting the existence of Planet X was Mike Brown, a Caltech scientist whose previous work directly led to Pluto's demotion. On discovering Planet X, he said, "Killing Pluto was fun, but this is head and shoulders above everything else." Harsh.

Sources: 
https://www.centralmaine.com/2016/01/27/the-discovery-of-planet-x/
http://ih0.redbubble.net/image.77569687.6291/raf,220x200,075,f,black.u5.jpg
https://en.wikipedia.org/wiki/Pluto
https://en.wikipedia.org/wiki/90377_Sedna
http://www.sciencemag.org/news/2016/01/feature-astronomers-say-neptune-sized-planet-lurks-unseen-solar-system
https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEim9temp3WPCmNNxpr03SbtAAKjeBqRM6ZpyowVuvYbQtxMNYOsXCcvExzwsHez5aPwTM_fhOsjAQ1ntIIVV9SgRLKBPnRRfmYVF6P7_FzefLwGhGpxMVaImToYCHMhl6sUErQ3MsEfNv81/s1600/sad+pluto4.jpg