Making Discoveries in Reverse

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, $F(T)$. You can do this using by substituting the variable $u \equiv h \nu/kT$. This will allow you to split things into a temperature-dependent term, and a term comprising an integral over all frequencies. However, rather than solving for the integral, just set it equal to a constant, $\sigma$ , which is also known as the Stefan-Boltzmann constant. If you're really into calculus, go ahead and show that $\sigma \approx 5.7 \times 10^{-5}$ erg s$^{-1}$ cm$^{-2}$ K$^{-4}$. Otherwise, commit this number to memory.

This question is, in essence, a look at the math underlying my post from a couple weeks ago on the Ultraviolet Catastrophe, examining how the Planck function describes a blackbody. We're going to jump right in with that "frightening mass of algebra" from before, the equation for intensity of radiation emitted from a blackbody. $I_{\nu}(T)$.\[I_{\nu}(T) = \frac{2 \nu^2}{c^2} \frac{h \nu}{e^{\frac{h \nu}{kT}}-1} \equiv B_{\nu}(T)\: (a.1)\] It's good to have a sense of what this looks like graphically, so here's a quick refresher on what the Planck function actually looks like for a few different values of T.

This is a good image to have in the back of your head as we work through this. (Source)

Integrating this over a hemisphere centered on the blackbody turns intensity into the blackbody flux (energy per unit time per unit area) $F_{\nu}(T) = \pi I_{\nu}(T)$. Given \[F_{\nu}(T) = \pi \frac{2 \nu^2}{c^2} \frac{h \nu}{e^{\frac{h \nu}{kT}}-1}\: (a.2)\] we can make the substitution $u = \frac{h \nu}{kT}$, which we can solve for $\nu$ and $d\nu$ such that $\nu = \frac{ukT}{h}$ and $d\nu = \frac{kT}{h}du$. This gives us \[F_{\nu}(T) = 2\pi h c^{-2}(\frac{ukT}{h})^3(e^u-1)^{-1}\:(a.3)\] We now want to integrate this across all frequencies, from $\nu = (0, \infty)$. Upon setting up the integral we will pull out all constants not dependent on $\nu$  (note that because $u = \frac{h \nu}{kT}$, our bounds do not change).\[\frac{2\pi k^4 T^4}{h^3 c^2} \int_0^\infty \frac{u^3}{e^u-1}du\:(a.4)\] While it is true that $u$ is dependent on temperature, and so we have not, in the strictest sense, separated $T$ out into its own term, the integral actually works out in such a way that the dependence on $T$ is irrelevant when integrated from zero to infinity (note that if we were integrating between two specific frequencies this will not be the case). As it happens, the integral combined with all of its coefficients save for $T^4$ comes out to $\sigma$, leaving us with $F_{\nu}(T) = \sigma T^4$.


b) The Wien Displacement Law: Convert the units of the blackbody intensity from $B _{\nu}(T)$ to $B _{\lambda}(T)$ IMPORTANT: Remember that the amount of energy in a frequency interval $d\nu$ has to be exactly equal to the amount of energy in the corresponding wavelength interval $d\lambda$.

What we're doing here is deriving a crucially important equation, one which allow us to tell the wavelength of an object's peak emission based solely on its temperature. As the graph of the Planck function above shows, the warmer a blackbody is, the more the peak of its emission is shifted left (to shorter wavelengths). Since we can approximate pretty much any object in space as a blackbody, this paves the way for a pretty powerful mathematical relationship, which can tell you how hot an object is--whether it's a star, a planet, or something else--just by looking at where it gives off the most light.

A comparison of the peak emissions of the Sun and Earth. (Source)

This would at first seem like a straightforward problem, where we would need to apply the old, familiar relation $\nu \lambda = c$. However, the last part hints at an additional layer of complexity. Rather than simply plugging in $c/\lambda$ for every $\nu$ in $B_{\nu}(T)$ and calling it $B_{\lambda}(T)$, we must actually relate $B_{\nu}(T)d\nu$ and $B_{\lambda}(T)d\lambda$ in order to account for the fact that, because $\nu = c/\lambda$, $d\nu$ is actually equal to $-\frac{c}{\lambda^2}d\lambda$. Also worth noting is that, because an increase in frequency corresponds to a decrease in wavelength, $B_{\nu}(T)d\nu$ is actually equal to $-B_{\lambda}(T)d\lambda$.

We'll once again start with $B_{\nu}(T)$, this time tacking a $d\nu$ onto the end of it.\[B_{\nu}(T)d\nu = \frac{2 \nu^2}{c^2} \frac{h \nu}{e^{\frac{h \nu}{kT}}-1}d\nu \: (b.1)\] Now, taking into consideration the complications noted above, we'll substitute for $\nu$ and $d\nu$, giving us \[B_{\lambda}(T)d\lambda = -\frac{2 (\frac{c}{\lambda})^2}{c^2} \frac{h \frac{c}{\lambda}}{e^{\frac{hc}{\lambda kT}}-1}(-\frac{c}{\lambda^2}d\lambda)\:(b.2)\] That looks like a huge mess, but fortunately we can simplify it into something a little more useful. Combining like terms and canceling in the first part gives us \[B_{\lambda}(T)d\lambda = -\frac{2hc}{\lambda^3(e^{\frac{hc}{\lambda kT}}-1)}(-\frac{c}{\lambda^2}d\lambda)\:(b.3)\] Finally, we can distribute the negative signs and toss the $\frac{c}{\lambda^2}$ in with the rest, giving us \[B_{\lambda}(T)d\lambda = \frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)}d\lambda\:(b.4)\] Canceling out the $d\lambda$ on each side gives us an expression for $B_{\lambda}(T)$.


c) Derive an expression for the wavelength $\lambda_{max}$ corresponding to the peak of the intensity distribution at a given temperature $T$.  (HINT: How do you find the maximum of a function?) Once you do this, again substitute $u \equiv h\nu/kT$.  The expression you end up with will be transentental, but you can solve it easily to first order, which is good enough for this exercise.

Straining back to first-semester calculus, we can recall that in order to find the maximum (or minimum) of a function, we find where its first derivative is equal to zero. We can ignore all the complicating factors that calculus teachers like to throw at you, since (having seen the Planck function before) we know that there is only one point on any curve where the derivative is zero--at its maximum.\[\frac{d}{d\lambda}B_{\lambda}(T) =  \frac{d}{d\lambda}\frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)} = \frac{d}{d\lambda}2hc^2\lambda^{-5}(e^{\frac{hc}{\lambda kT}} -1)^{-1}\:(c.1)\] The second step simply being a rearranging in order to make the equation a little more derivative-friendly. We're going to have to pull out all the stops here (power rule! product rule! chain rule!), so let's simplify things visually for the time being by letting $a \equiv hc/kT$. Taking derivatives gives us \[\frac{d}{d\lambda}B_{\lambda}(T) = 2hc^2[-\frac{5}{\lambda^6(e^{\frac{a}{\lambda}})} + \frac{ae^{\frac{a}{\lambda}}}{\lambda^7kT(e^{\frac{a}{\lambda}}-1)^2}]\:(c.2)\] We now set this equal to zero, divide by $dhc^2$, and add across, giving us \[\frac{5}{\lambda^6(e^{\frac{a}{\lambda}}-1)} = \frac{ae^{\frac{a}{\lambda}}}{\lambda^7(e^{\frac{a}{\lambda}}-1)^2}\:(c.3)\] We can now multiply both sides by $\lambda^6(e^{\frac{a}{\lambda}}-1)$, which simplifies things down to \[5 = \frac{ae^{\frac{a}{\lambda}}}{\lambda(e^{\frac{a}{\lambda}}-1)} \: (c.4)\] Finally, let's pull $\frac{hc}{\lambda kT}$ back out of $a$ so that our equation becomes \[5 = \frac{\frac{hc}{\lambda kT} e^{\frac{hc}{\lambda kT}}}{e^{\frac{hc}{\lambda kT}}-1} \: (c.5)\] That looks a whole lot messier, but we can now employ the substitution mentioned in the instructions, $x \equiv \frac{hc}{\lambda kT}$, which is dimensionless. That turns the ugly mess above into something a lot less frightening: \[5 = \frac{xe^x}{e^x-1} \: (c.6)\] That might be a lot prettier, but unfortunately it's not actually solvable, so let's turn to a Taylor approximation to give us something to go on. The Taylor expansion of $e^x$ is $1+x+x^2/2...$ but we're only going to bother with the first order approximation, and plug in $1+x$ for our two $e^x$ terms. That turns $(c.6)$ into \[5 \approx \frac{x(1+x)}{1+x-1} \approx {1+x} \: (c.7)\] Or, more simply, $x \approx 4$. From there, we can plug $\frac{hc}{\lambda kT}$ back in for $x$, then solve for $\lambda$ (which is really $\lambda_{max}$), giving us, at long last\[\lambda_{max} \approx \frac{4hc}{kT} \: (c.8)\] Having come all this way, it's worth making sure that at least our units make sense, so let's do a quick sanity check. Things will need to work out to units of m (length), since we solved for wavelength. On top, we have (erg $\times$ s $\times$ m $\times$ s$^{-1}$), while on the bottom we have (erg $\times$ T $\times$ T$^{-1}$). Ergs cancel, as does temperature (T) and time (s), leaving us with just the m term. Looks good!


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 $e^{\frac{h\nu}{kT}}$ to derive a simplified form of $B_{\nu}(T)$ in this low-energy regime.

Astonishingly, some people feel like the full expression for $B_{\nu}(T)$ is a little bit unwieldy, so it's not a bad idea to have some approximations on hand that we can use for specific regions of the Planck function. The Rayleigh-Jeans law attempted to use classical mechanics to arrive at the Planck function as a whole, but only succeeded in developing an approximation for low-energy photons--classical mechanics actually dictated that as wavelength dropped, energy would go to infinity (and, probably fortunately, we have yet to observe an infinite-energy photon). The explanation for the discrepancy between classical mechanics and observation (known, of course, as the ultraviolet catastrophe), was an early stage in the development of quantum mechanics.

The Very Large Array. Approximations based on the Rayleigh-Jeans law are still used for formulas used in radio astronomy. (Source)

We're thus working somewhat backwards in this case, but let's start with the full equation for $B_{\nu}(T)$ and try to break it down into a simpler expression for very low $\nu$. Remembering that \[B_{\nu}(T) = \frac{2\nu^2}{c^2}{h\nu}{e^{\frac{h\nu}{kT}}-1} \: (d.1)\] we use a Taylor expansion to simplify the troublesome $e^{\frac{h\nu}{kT}}$ term. This time we'll think of $\frac{h\nu}{kT}$ as $x$ in a Taylor expansion of $e^x$. This gives us the first-order Taylor expansion of $e^{\frac{h\nu}{kT}}$ as  $1+\frac{h\nu}{kT}$. Plugging this in for $e^{\frac{h\nu}{kT}}$ gives us \[B_{\nu}(T) \approx \frac{2\nu^2}{c^2}\frac{h\nu}{1+\frac{h\nu}{kT}-1}\: (d.2)\] Doing a bit of canceling then gives us \[B_{\nu}(T) \approx \frac{2kT\nu^2}{c^2} \: (d.3)\] Which, fortunately, is the Rayleigh-Jeans law in terms of frequency!


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

Let's pick up at that nasty integral in a), equation $(a.4)$, recalling that that equation is the blackbody flux of an entire object--its emissions integrated across all frequencies, measured in units of energy per time per unit area. What we need to do now is turn it into luminosity--the total output of the blackbody, irrespective of area. This is simpler than it sounds. We can assume our blackbody to be a sphere, which we know to be of radius $R$. Eliminating the unit area element of blackbody flux is a simple matter of multiplying it by the surface area of our sphere, so all we need to do is put $4\pi R^2$ out in front. Thus, \[L = 4\pi R^2 \times \frac{2\pi k^4 T^4}{h^3 c^2} \int_0^\infty \frac{u^3}{e^u-1}du = 4\pi R^2T^4\sigma\:(e.1)\] Once again employing our favorite method of double-checking (sorry, dimensional analysis), we can see that this is indeed the formula for the luminosity of a spherical blackbody.


f) You observe two gravitationally bound stars (a binary pair). One is blue and one is yellow. The yellow star is brighter than the blue star. Qualitatively compare their temperature and radii, i.e. which is hotter, which is smaller? Next, quantitatively compare their radii (to 1 significant  figure).

As we can see visually on the graph of the Planck function, and mathematically from Wien's Displacement Law, the hotter an object is, the shorter the wavelength of the peak of its emission curve. Blue light, at roughly $450$ nm $= 4.5 \times 10^{-7}$ m, has a shorter wavelength than yellow light, coming in at around $550$ nm $= 5.5 \times 10^{-7}$ m. Thus, we would expect the blue star to be hotter and brighter than the yellow one, as is typically the case when looking at main sequence stars.

The Hertzsprung-Russell Diagram: Temperature increases (and wavelength decreases) to the left, while brightness increases moving upward. The main sequence is the line going diagonally across. (Source)

However, we know that in this case, the yellow star is actually brighter than the blue one. If we're assuming that they're the same temperature, that would be impossible (unless Planck, and most of modern astronomy, were way off the mark). However, we can account for this by assuming that the yellow star is significantly larger than the blue one--thus, its temperature might be lower, and thus its blackbody curve shifted to the right of the blue star's, but it manages to be brighter just by pumping out a huge number of lower-energy photons. While this wouldn't really work for main sequence stars, you can see looking at the H-R diagram that this situation, while unusual, is not totally out of the question if we were to be dealing with a dimmer blue star and a yellow giant (or supergiant) star.

A yellow supergiant: Polaris A, an unusually interesting star--as well as being the north star it's a triple star system and the closest Cepheid Variable star to Earth (Source)

The blue star is still hotter--there's no getting around the Planck function--but if we beef up its yellow companion so that it's significantly bigger, it's still possible for it to be brighter.

Getting Sidetracked

I started this post with what I thought was a rather clever idea (that I may revisit in another post) that involved a bit of research into the final stages of the sun's life. Apparently it wasn't clever enough to keep my attention, because I ran into a few articles about the helium flash and got a little bit distracted.

The sun isn't always going to stay its current friendly size. In five billion years or so, it will begin to run out of hydrogen to fuel nuclear fusion and begin to fuse heavier and heavier elements (at this point it'll also have about double its present-day luminosity). As it does so, it will expand to many times its current size, careening off the main sequence and ballooning into a red giant. Initially it's dramatic but slow--a billion years spent growing to 200 times its current size and thousands of times its current luminosity. Then things get crazy.

The sun's core is now made up entirely of helium, which is added onto by helium "ash" that sinks into the core as a product of the last of its hydrogen fusing in an outer shell. For stars the size of our sun (and up to around twice its size), this core doesn't get hot enough to begin burning helium. As a result, there is no thermal pressure to counteract gravity, and the star begins to collapse in on itself, and the core gets so densely packed that the helium in the very center actually turns into degenerate matter. Degenerate helium heats up, but cannot expand to compensate as a normal gas would. Eventually, it hits a temperature of around $10^8$ K, and the helium in the core suddenly begins to fuse in a runaway chain reaction--as helium ignites it heats the core up even faster, which simply leads to the ignition of even more helium. In a few rather dramatic moments (and this really is on the timescale of seconds), 6% of the core's helium turns into carbon, and the sun's core is producing as much energy as our entire galaxy.

I wanted to do a clever scale-relating analogy like I did in this post, but there is literally no way to scale down something that insane. So here's the Milky Way. Imagine packing it into something around the same size as Jupiter. (Source)

And yet, despite that patently insane fact, that's not even the most astounding part. You would expect that a galaxy of luminosity packed into the center of a solar system would fry every rock, gas, and dwarf planet all the way out to Quaoar--but it doesn't. The thermal pressure given off by the helium flash is finally enough to counteract the crushing gravity of the rest of the overblown star pushing down on it, puffing our beleaguered sun back out and returning the core to its previous nondegenerate state. But doing so requires so much energy that the entire explosion is absorbed into the core that gives rise to it--despite the prodigious amount of energy being released, the process is unobservable from outside the star.

It's over in seconds, but it's really the death knell for the sun. Ten billion years old, it's life is 99.99% over. It will become increasingly unstable over the next few million years, fusing what helium it has left and blowing out mass in the solar wind. Meanwhile, a series of massive pulses will briefly but dramatically increase its size and luminosity. After the last of these, the sun will finally release its outermost shell, which will expand out into a planetary nebula. The core that saw so much dramatic action will be left as a white dwarf.

The downside? We'll all be fried. The upside? Aside from being ridiculously cool, our solar system will be left looking something like this:

Mosaic of the Helix Nebula, composed of Hubble and Kitt Peak National Observatory images. The white dwarf is visible at the very center. A much larger (as in 16,000 x 16,000) version is available here. (Source)

At least we'll be going out in style.

Ganymede

When I was younger, my favorite author--hands down--was Robert A. Heinlein. One of the "Big Three" science fiction writers (along with Arthur C. Clarke and Isaac Asimov), most of Heinlein's early work was geared toward teenagers, telling rousing adventure stories in space. Later, his writing got more adult-oriented, more overtly politically charged, and to be completely honest, worse. But I still enjoy picking up one of the books I read and reread as a kid and reading through it in an afternoon, largely because Heinlein put a lot of effort into writing "hard" scifi that was as grounded as possible in actual science.

One of my favorite books of his was called Farmer in the Sky, and tells the story of a family which emigrates to the Galilean moon of Ganymede in order to participate in its terraforming and escape a crushingly overcrowded Earth (frequent themes in Heinlein's writing). The book devotes a great deal of time to explaining (in a way that, at least to me, managed to be consistently interesting) how the colonists would go about seeding volcanic dust and crushed rock with organic material in order to convert it into arable soil, as well as how pumping oxygen and greenhouse gases into the moon's atmosphere could give it breathable--and warm enough--air. Later, near the climax of the book, a syzygy of Ganymede, Europa, Io, and Jupiter* causes enormous tidal strains that cause a catastrophic earthquake (moonquake? Ganymedequake?).

Ganymede, as seen by the Galileo probe (Source)

Despite being a pretty smart guy, Heinlein was an author, not a scientist, and he got a few things wrong. But I don't mind, because looking into the science behind the book led me to some pretty interesting discoveries.

First off, that pseudo-syzygy: Heinlein goes to great pains in the book to emphasize how infrequent such an event would be, given the orbits of the three moons involved. But what he didn't realize is that Ganymede, Europa, and Io lining up in front of Jupiter wouldn't just be infrequent--it's actually impossible. The three inner Galilean moons are actually in a 4:2:1 orbital resonance, where it takes Europa twice as long to orbit as it does Io, and it takes Ganymede four times as long. As a result, the three can actually never line up all at once. If you don't believe me, try watching this great little animation for a while:

(Source)

As I looked into things further, I started to wonder how plausible growing anything on Ganymede would really be (assuming an amenable atmosphere and range of temperatures). The book is very aware of how far away Jupiter is--a large section of it is devoted entirely to the trip there--but once the characters arrive on Ganymede, this is never really mentioned again. However, plants require sun to grow, and there's a whole lot less sunlight illuminating Ganymede as there is powering photosynthesis on Earth. So to figure it out, I decided to bust out the old inverse-square law (to his credit, my first exposure to that term actually did come from reading Heinlein) and figure out exactly how much energy is reaching Ganymede.\[F=\frac{L_{sun}}{4 \pi D_G^2}\] We know the luminosity of the sun pretty well as about $4 \times 10^{33}$ erg/sec, and Ganymede is on average the same distance from the sun as Jupiter--around $8 \times 10^{11}$ meters. This gives us a flux of about $5 \times 10^8$ ergs per second per square meter. What does that mean? Well, an astronomical unit away from the Sun, the Earth is pulling in a flux of about $1.5 \times 10^{10}$ ergs per second per square meter. To be fair, many plants can grow in the shade on Earth, where they may be receiving 1%-20% of the solar energy they would receive in full sunlight. Ganymede, receiving 3% of the flux that Earth gets, would only be able to host the very hardiest of these--true agriculture would be impossible. Wheat, a vital crop, struggles to grow even in very mild shade. To drive the point home, here's a great illustration of how the Sun would look from various planets in the solar system. 

Hard to imagine a dot 1/5 the diameter of the Sun on Earth keeping any plants alive. (Source)

One last mistake of Heinlein's can be forgiven, since the information wasn't available at the time. However, his characterization of Ganymede as a dusty, dormant-volcanic ball was pretty far off the mark. As it turns out, the moon's surface is extremely rich with water ice. It does have an iron core, and its convection makes Ganymede the only moon in our solar system known to have a magnetosphere. However, salts on the surface suggest that 200 kilometers below lies a vast saltwater ocean--in fact, after growing up fascinated by Europa's subterranean (subeuropean?) oceans and the possibility of life there, I was surprised to find that Ganymede and Callisto are both thought to harbor deep oceans as well (though the conditions are much less favorable for life on the outer moons, which are not heated by tidal forces to the same degree as Europa). Volcanism on Ganymede is virtually nonexistent.

So what prompted me to start pondering Ganymede? Well, a few days ago JPL and the USGS put out a pretty remarkable map of the moon using data collected by the Pioneer and Voyager missions along with the Galileo probe. I'm including an image here, but its worth following through to the source and really examining it, because it's incredibly detailed. The color, obviously, is false, but it represents an incredibly chaotic landscape, with jarring divisions between old and new terrain shaped by three distinct phases of the moon's geologic evolution, from a stretch of frequent impacts to one of tectonic upheaval into a final period of relative calm.

Ganymede (you should really click through to the Source)

Doesn't look like there are any farms down there.

*Actually, this technically isn't a syzygy--a syzygy is defined as a straight-line alignment of exactly three celestial bodies (I imagine because the term was coined to describe the conditions that result in a solar eclipse on Earth). There isn't a word that I'm aware of to describe the straight-line alignment of four celestial bodies, so I went with syzygy anyway. Also because it's just a fun word (and a good-but-not-great episode of The X-Files--wow, this is turning into a nerdy post).

Small Moves

Hey, remember this post? In it, we estimated a few quantities that are going to be useful here--the size of the Earth and Moon, along with the distance between the two. We were off a bit, so here are the actual figures:

Diameter of the Earth = $D_+ = 1.3 \times 10^7$ meters

Diameter of the Moon = $D_m = 3.5 \times 10^6$ meters

Distance from the Earth to the Moon = $R_{+M} = 4 \times 10^8$ meters

As with all big numbers where the zeroes have started to pile up, that can be hard to visualize. Most people imagine the Earth and Moon looking something like this:

(Source)

But actually, a scale diagram of the two is a little more extreme-looking:

Earth is the blue one.

The distance between the two is about 31 times the diameter of the earth--so on this scale, where Earth is about a centimeter wide, the moon is nearly 1/3 of a meter away. That is the farthest from home that a human being has ever traveled.

But what about what's past there? The nearest another planet ever comes to Earth is Venus, when it's at its closest approach (the term for this is "opposition"). In the case of Venus, this is about $3.8*10^{10}$ meters.

That's 96 times the distance to the Moon. I was going to add Venus to the scale drawing above, but apparently Google Docs drawings don't handle 32-meter long lines very well. WolframAlpha has helpfully suggested that this is approximately the length of a blue whale, so if you want to imagine the distance between the two, you could try envisioning two dots the width of your pinky at opposite ends of this guy.

The guy on the right, specifically (Source)

Or you could stick to the more formulaic analogy of one dot in a football field's end zone and another on the 35-yard line. Mars is a little trickier, since its orbital eccentricity--at 1%--is fairly high. In theory, it's possible for Mars to come as close as $5.4\times 10^{10}$ meters (which would put Mars 1.4 blue whales away, or almost exactly on the 50-yard line), but this hasn't actually happened in human memory. The 2004 opposition between the planets, where they reached $5.6 \times 10^{10}$ meters (1.5 blue whales, or halfway between the 51- and 52-yard lines), was the closest we've been to the red planet in 50,000 years.

But this is still just a third of the way from the Earth to the Sun, clocking in at $1.5 \times 10^{11}$ meters. Rather than measuring that in blue whales, it might make a bit more sense to apply that to something a little more familiar. Let's put the Earth, still a centimeter across, right here:

It's a better sport than football anyway.

If the centimeter-Earth is one one end of the rubber, the moon is sitting at the other. Venus, at opposition, is sitting in shallow center, just past the edge of the infield dirt. Mars at its 2004 opposition would be just at the bottom of the "B" in center field. The sun, meanwhile, makes it over the wall, landing in the fifth row of bleachers in straightaway center. At a meter across, it's also taking up two seats.

No human being has been this far, but our robotic probes have made it substantially further. Saturn is currently $1.5 \times 10^{12}$ meters away, with Cassini circling it. On our scale, that works out to about $1.1 \times 10^3$ meters, putting Saturn right on the Boston side of the BU bridge:

Interestingly, Saturn also could go near the mound of MIT's baseball field (top). But that's a bad analogy because no one knows where that is. (Source: Google Earth)

Let's go a bit further, to Pioneer 10--the first spacecraft to achieve escape velocity from the solar system. It was launched in 1972 and performed a flyby of Jupiter 21 months later. Contact with Pioneer 10 was lost just eleven years ago, when it was a healthy 12 billion kilometers (sorry, $1.2 \times 10^{13}$ meters) from Earth. Judging from its trajectory at the time, that should put it about $1.7 \times 10^{13}$ meters away today. On our scale, that would mean Pioneer 10 would be stuck in traffic at the intersection of the Pike (I-90 for non-natives) and 128 (I-95) in Waltham, MA.

Pioneer 10 should've taken Route 16 and skipped the traffic. (Source: Google Earth)

Pioneer 11, launched the following year, is a comparable distance away, but headed in the opposite direciton. So now, finally, let's indulge the Voyager fanbase out there (and yes, it is a fanbase--you can tell that much just from the fact that, while I had to do a bit of math to figure out just how far away the Pioneer probes are, the Voyager 1 Wikipedia page was updated to reflect its distance from Earth as of this morning). Courtesy of a very avid Wikipedia updater, I know that Voyager 1 is $1.9 \times 10^{13}$ meters away right now. While that would, finally, get us out of Greater Boston on our scale, we'd have to be going due west. If not, well, it's almost exactly the distance from Fenway to the intersection of 128 and route 2 a little farther north...

(Source: Google Earth)

So that's the farthest manmade object from Earth. It's been traveling for 36 years, 5 months, and 12 days as of today (thanks, Wikipedians!), and on the centimeter-Earth scale, it doesn't even make it into another area code.

Of course, I could scale this however I wanted to make that distance seem insignificant ("If the Earth were a grain of sand on the mound, Voyager would barely have made it to the warning track! Take that, NASA!"). So let's zoom out a bit more. This is Proxima Centauri:

Hello. (Source)

It's a pretty interesting star. A seventh the diameter of the sun, it's so faint it wasn't discovered until 1915--all despite the fact that,  4.2 light years distant, it's our nearest stellar neighbor. Traded in for slightly more useful units, those 4.2 light years come out to just about $4.0 \times 10^{16}$ meters. Plug it into the super-complicated math I've been doing to these distances (AKA multiplying them by $1 \: cm / D_+$), that works out to $3.1 \times 10^7$ meters. Where does this land Proxima Centauri? Well, if you were to travel along the 42nd parallel (just south of Boston) this would leave you out in the Atlantic, halfway between London and Boston--if you were to go west. At 31,000 kilometers, Proxima Centauri is a long way away. In fact, the Earth's semicircumference is 20,000 kilometers, which means that no two locations on the planet are farther apart than that (if you're planning on traveling 31,000 kilometers to get anywhere, you might want to consider turning around--it'd be a whole lot faster).

I personally see the Apollo Program as one of the greatest achievements in the history of our species, and I think rightly so. The idea that fellow humans have set foot on another world, however small, should quietly blow the mind of anyone looking up at the night sky. But it's worth keeping in mind that, even by the standards of our stellar neighborhood, we've just barely started to move.

Baby steps. (Source)

Another Sense of Fourier Transforms

So we've talked a lot about how telescopes (and, in fact, other optical systems) perform Fourier transforms of oncoming light waves. Interferometers function because the information in images tends to be concentrated in low-frequency/high-wavelength scales, so we can form a relatively complete image even if we aren't sampling every possible frequency. As it turns out, we can do pretty much the same thing to sound--and that's basically how mp3s work.

When you create a CD-quality recording of an audio track, each minute is about 10 megabytes of data. That may not sound like a lot now that everyone is packing terabyte hard drives, but it would still put a pretty serious strain on the tubes if everyone were downloading CD-quality music all the time. To allow us to compress audio files, the process of creating mp3s uses Fourier transforms to split a track up into 576 frequency bands. Within each of these bands, everything that is inaudible to humans is thrown away.

Fourier transforms allow us to store sound simply by tracking the intensities of different frequencies (Source).

Combined with other techniques, Fourier transforms allow us to take those 10 megabytes/minute of CD-quality audio and compress it down to 1 megabyte. Without them, the digital music industry probably just wouldn't exist.

That atmosphere

(Source)

What you're looking at there is two different images of the same binary system, LP 359-186. The two stars are a mere 0.31 arcseconds apart--substantially less than the 1 arcsecond scale at which the atmosphere is stable. The left-hand image was produced using speckle imaging, while the one on the right is the result of a longer exposure. The comparison pretty clearly shows the effect atmospheric diffraction can have on the resolving power of a telescope. Angular resolution might be $\lambda/D$, but only up to the point that the atmosphere allows. Unless you're compensating for atmospheric diffraction with a technique like speckle imaging or using adaptive optics, getting down to a resolution like .31 arcseconds just isn't possible from the ground.

Lambda over D

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? What's "J-band"?

The MMT (Multiple Mirror Telescope). It only has one primary mirror.

The key to this problem is a very simple equation, one that draws on the Airy disk (which I discussed the origins of in my previous post) to determine angular resolution. Essentially, in order to resolve two nearby objects as being truly separate, the center of one needs to be far enough from the center of the other that it is located at the first "null," or minimum, of the other object's Airy disk. This gives us a crucially important formula, known as the Rayleigh Criterion, or the diffraction limit:\[\theta = 1.22\lambda/D\] Where $\theta$ is the smallest possible angle a telescope can resolve, $D$ is the telescope's diameter, and $\lambda$ is, as always, the wavelength of incoming light (the 1.22 coefficient comes from calculating the first zero of a Bessel function).

So, having looked up the infrared J-band and knowing that it's centered at $\lambda = 1250$ nanometers (confusingly, there is also a J-band in the radio, ranging rom $\lambda = 1.5$ to $3$ centimeters), here is all the information we need:

$\lambda_{MMT} = 1.25 \times 10^{-6}$ meters

$\lambda_{CCAT} = 8.5 \times 10^{-4}$ meters

$D_{MMT} = 6.5$ meters

$D_{CCAT} = 2.5 \times 10$ meters

Thus, $\theta_{MMT} = \frac{1.25 \times 10^{-6}}{6.5} = 2 \times 10^{-7}$ radians, and $\theta_{CCAT} = \frac{8.5 \times 10^{-4}}{2.5 \times 10} = 3 \times 10^{-5}$ radians. So, despite being considerably smaller, the MMT is actually giving better angular resolution than the CCAT when observing in much shorter wavelengths. There are caveats to this, of course--it might have better angular resolution, but the MMT's light-collecting area is always going to be much smaller than the CCAT's. But this relationship between wavelength and diameter is still very important, and is actually the reason why the folks at the CfA are able to do meaningful radio astronomy research using a 1.2-meter telescope right here in Cambridge.

The CfA 1.2-meter

(Fourier) Transformers

To understand the basics of astronomical instrumentation, I find it useful to go back to the classic Young's double slit experiment. Draw a double slit setup as a 1-D diagram on the board. Draw it big, use straight lines, and label features clearly. The two slits are separated by a distance D, and each slit is w wide, where w << D such that its transmission function is basically a delta function. There is a phosphorescent screen placed a distance L away from the slits, where L >> D. We'll be thinking of light as plane-parallel waves incident on the slit-plane, with a propagation direction perpendicular to the slit plane. Further, the light is monochromatic with a wavelength $\lambda$.

a) Convince yourself that the brightness pattern of light on the screen is a cosine function. (HINT: Think about the conditions for constructive and destructive interference of the light waves emerging from each slit).

b) Now imagine a second set of slits placed just inward of the rst set. How does the second set of slits modify the brightness pattern on the screen?

c) Imagine a continuous set of slit pairs with ever decreasing separation. What is the resulting brightness pattern?

d) Notice that this continuous set of slits forms a "top hat" transmission function. What is the Fourier transform of a top hat, and how does this compare to your sum from the previous step?

e) For the top hat functions FT, what is the relationship between the distance between the first nulls and the width of the transmission function (HINT: it involves the wavelength of light and the width of the aperture)?

f) Take a step back and think about what I'm trying to teach you with this activity, and how it relates to a telescope primary mirror.

The two-slit experience is a deceptively simple one. We've got a single-wavelength beam of light shining on a screen with two minuscule slits in it. For now, we're assuming that these waves are coming in at exactly 90 degrees to the screen, and are perfectly straight and parallel (a star, for instance, is emitting light in spherical shells in every direction, but after traveling light-years to get to us, the shells are essentially flat planes--or lines if simplified into two dimensions). When it hits them, waves of light spread out of each one, then interfere constructively and destructively with each other, leaving an interference pattern on another screen in front of it. I've drawn this setup about a thousand times over the past week and a half, so rather than subjecting everyone to my somewhat questionable artistic skills yet again, here's a much better illustration.

(Source)

As you can see, the light is emitted isotropically from each slit, and each set of waves interferes with the other. Wherever crests line up (along the blue lines), we end up with a bright spot on the right-hand screen (constructive interference). Directly in between is where the two lines are exactly half a wavelength out of phase with each other, giving us dark spots (destructive interference), with the intensity of the light waxing and waning in between to make a sinusoidal pattern--a cosine rather than a sine, since we have a maximum directly in the center at the "origin," with brightness on the y axis and distance from the central maximum on the x-axis.

So what happens when we add a second slit in between? To figure that out, I'm afraid I'm going to have to subject you to my drawing skills, so we can work out a mathematical relationship between D, the distance between the slits, and the spacing between the bright spots on the detector screen.

This is a pretty confusing diagram, so let's work out what's going on here. Light is coming in from the left, just like in the last image, and passing through the slits, which are $D$ apart. $x_1$ is the distance from each slit to the central maximum. The next maximum is $d$ away. It is also $x_2$ away from the top slit and $x_2 + y$ away from the bottom slit.

However, since maxima will only occur an when crests line up, light has to be traveling $x_2+\lambda$ from the bottom slit to the first maximum (if the next maximum is $x_3$ away from the top slit, it will be $x_2+2\lambda$ from the bottom slit, and so on), so we can tell that $y = \lambda$. We can also work out some angle equalities as well--we know that the two angles labeled $\phi$ are equal by the vertical angle theorem, which gives us $\phi + \theta = 90$. Bearing in mind that $L$, the distance between the two screens, is much larger than $D$ (yeah, my drawing isn't exactly to scale), we can also approximate that the two lines $x_2$ and $x_2 + y$ are parallel, and that the bottom line $x_1$ and $L$ are parallel, giving us the two right angles shown (and the "rectangle" redrawn below, which tells us that $x_1 \approx L$). This means that the angle between the line $x_2 + y$ and the left-hand screen is also $\theta$. The topmost angle of that right triangle (also redrawn below), then, is $\phi$.

Phew. That was a lot of geometry just to get everything labeled. But if we apply some trig to it, it can tell us what we need to know (and give us the answer to part (e) for good measure). That small triangle on the left tells us that $cos(\theta) = \lambda/D$.  Meanwhile, we also know from the triangle made by $d$, $x_2 + y$, and $x_1$ that $cos(\theta) = d/x_1 = d/L$. This allows us to construct a key equality.\[cos(\theta) = \lambda/D = d/L\] We can rearrange this to give us a function for $d$ in terms of $D$: $d(D) = \frac{\lambda L}{D}$. Thus, the closer together our slits get (the smaller $D$ gets, the bigger $d$ will be--in other words, closer slits will give us fatter cosine functions. So if $D=1$ gives us cosines that look like this:

Then $D=2$ will give us cosines that look like this:

And $D=3$ will give us cosines that look like this:

We could keep doing this with steadily decreasing values for $D$, until the limit where $D=0$, we have only one slit, and there is no interference pattern, just an even distribution of light. However, more interesting than looking at all of these individual cases is looking what happens when we combine all of them, superimposing every possible cosine and giving us an entire aperture rather than a pair of slits. That gives us what's called a sinc function:

As you can see, superimposing all the slits gives us a huge, bright central maximum, with much smaller maxima at regular intervals off to the sides. Since the y-axis is brightness and the x-axis is distance from a central point, that translates to an image with a bright center surrounded by evenly spaced dimmer rings:

(Source)

Anyone who has looked through a telescope will recognize this as the Airy disk one sees when looking at a bright point source. This is really cool--we've taken the results of the two-slit experiment and generalized them to apply not to two slits, but to an entire aperture. 

And if we go deeper, there's even another layer to all of this. Here's the transmission function for our first set of slits:

Where the x-axis is position and the y-axis is the amount of light passing through (zero meaning none, one meaning all). The two lines are the locations of our two slits, $D$ apart. Thus, the transmission function for a set of two closer-in slits just has the two lines moved slightly closer:

And if we move the two slits even closer together, the lines follow suit:

Thus, if we take every possible spacing of slits from our first set all the way down to zero, we end up adding every possible pair of lines until they all run together into a rectangle:

That's called a top hat function, and it models the transmission of an aperture. It also just so happens to have a Fourier transform that looks a little familiar:

What this means is that whenever you point your telescope at an object in the sky, it's actually taking the Fourier transform of the light it's receiving across its aperture, going from planar waves of light spread across a mirror or lens to a single, defined image. So when you look through a telescope, you're not just seeing photons that have traveled through space for years to reach you, you're seeing some pretty complicated (and elegant) math, too.

Catastrophe!

Let me present for your consideration a rather frightening mass of algebra:\[B_{\lambda}(T) = \frac{2hc^2/\lambda^5}{e^{hc/\lambda kT}-1}\] That's the Planck function, which describes the radiation curve of a perfect blackbody as a function of its temperature--the constants in it are wavelength ($\lambda$), the speed of light ($c$, as always), the Boltzmann constant $k$, and Planck's constant $h$. This equation was the result of centuries of head-scratching on the part of physicists, who had noticed fairly early on that a blackbody would emit light at different temperatures depending on how hot it was. The first real attempts to model this phenomenon led to the Wien Approximation, which accurately described short-wavelength emission, and Rayleigh-Jeans Law, which accurately described long-wavelength emission. However, under a contemporary understanding of physics, Rayleigh-Jeans also implied that a blackbody contained an infinite amount of energy due to the presence of an infinite number of infinitesimally short wavelengths. Scientists don't always name things particularly creatively...

Credit: The inimitable Bill Watterson

... but they managed to come up with a pretty dramatic-sounding one for the implications of the Rayleigh-Jeans law at shorter wavelengths: the ultraviolet catastrophe. This inconsistency between mathematics and observations remained until 1900, when Max Planck came along with a critical realization: there weren't actually infinite possible standing waves that could exist within a blackbody; rather, the only possibilities were integral multiples of a minimum possible wave energy, or quantum--this shows up as the $hc/\lambda$ in the Planck Function. As a result, the model no longer predicted energy going to infinity at short wavelengths, and the Wien Approximation and Rayleigh-Jeans Law had been tied together rather prettily in an extremely accurate model.

Planck's Function predicts the colored radiation curves, while Rayleigh-Jeans predicts the black curve. (Source)

The Story of AY Sixteenus: Part II

a) Delineate the band corresponding to nighttime as a function of season, on a graph with hour on the y-axis (5pm to 7am); and months on the x-axis (January to December). The line corresponding to sunrise should be at the top of the graph, and sunset at the bottom.

b) On the same graph, sketch a qualitative diagram illustrating the times during which AY Sixteenus is observable from a northern latitude like Mt. Hopkins as a function of time of year (x-axis). This should appear as a band in your diagram.

c) Consider what happens to that band if you observe from di erent sites located at latitudes of -10 and +60 degrees.

d) At what declination (higher or lower) than AY Six, are stars more frequently observable from a northern site like Mt. Hopkins?

So here we are with a bunch more questions about our favorite definitely-not-made-up star, AY Sixteenus (located, as you'll recall, at RA = 18:00, Declination = +32). We'll start by figuring out when the sun will be down at our observatory, located at Mount Hopkins, AZ (conveniently located at latitude = +32). As we know, days get longer in winter and shorter in summer, but by how much? To figure out the exact times of sunrise and sunset at +32 degrees latitude, I turned to my trusty friend the online Nautical Almanac, which gives the sunrise and sunset times in UT for any given latitude and longitude (set longitude to zero so that UT would be the same as local time), then plotted it in Excel and fitted sinusoidal trendlines to them--the blue line is sunset, the red one sunrise, and the duration of night is the time between the two lines.

The times on the left look pretty confusing at first, which is understandable. Excel can be finicky about time (it "thinks" in units of days, with all smaller units being expressed as fractions of a day), so the best way to visually express it with midnight in the middle was to just measure time in minutes before and after midnight. So 1 AM would be 60, 6 AM would be 360, and noon would be 720. Looking backward, 11 PM would be -60, 6 PM would be -360, and the preceding noon would be -720. Months are measured numerically along the x-axis (January = 1, February = 2, etc.) and repeat in order to show how times change periodically (so 13 is the following January, 14 the following February, etc.). As you can see, the nighttime gets noticeably longer in midwinter and shorter in midsummer (though less so than they do here in Cambridge, ten degrees farther north).

Now we need to figure out when AY Sixteenus will be visible. There are two things that need to be the case for us to be able to see it, both of them fairly obvious. The first is that the star must be above the horizon--we're going to have a hard time seeing the star if our planet is in the way. The second is that the sun needs to be down--it's usually pretty difficult to see more than one star during the day (though if you count exploding stars, it it's not always impossible).

Since the declination of AY Sixteenus is equal to our latitude, we know that it's going to reach a maximum altitude of 90 degrees--in astronomy-speak, it will be at the zenith when it crosses our meridian. Every object moves 360 degrees each day (not accounting for variation due to precession and the Earth's movement around the sun, which has negligible effects for objects outside of the solar system). Some of that angular movement is altitudinal and some azimuthal, depending on the relationship between the object's declination and the observer's latitude. At the poles, for instance, objects experience no daily change in altitude--they simply circle at the same height in the sky. However, for an object to reach the zenith, its declination must equal the observers latitude, and it must rise straight up in the east and set straight down in the west, moving exactly 180 degrees. This means that it is completing exactly half of its daily 360 degree movement while above the horizon, and will be up for exactly half of the day. For the more mathematically inclined out there, the relationship looks like this (letting $d$ = the angular distance covered on the sky, and $t$ = the time an object spends above teh horizon):\[\frac{d}{360} = \frac{t}{24} \: (1)\] This is a very useful equation, which we'll come back to later, and it tells us that $t$ = 12 hours. Thus, it will rise six hours before it crosses the meridian, and set six hours later.

Now that we know how long before and after meridian passage AY Sixteenus will spend above the horizon, we need to figure out when meridian passage will occur each month. To do this, we just need to figure out when our LST will be 18:00 over the course of the year. LST = 18:00 is noon on the meridian at noon on the northern hemisphere's winter solstice (when the RA of the sun is also 18:00). Knowing that the RA of the sun changes by two hours every month, we know that the time we're looking for will also change by two hours every month--all we need to know is whether it's going up or down. As it turns out, since LST = 18:00 will occur at 6 AM on the vernal equinox, it's going down. We can extrapolate this for the entire year, giving us this neat improvement on the graph from earlier, where the shaded region represents the time AY Sixteenus will be above the horizon:

Thus, the star will be visible in the parts of the shaded region that fall between sunset and sunrise.

But what if we aren't in the neat scenario where our latitude is the same as AY Sixteenus' declination? Let's fly down to -10 degrees and see how things look there. The sunrise and sunset times will be different, for one--we're much closer to the equator, so the variation is much less significant (once you're on the equator itself, the length of days doesn't really change over the course of the year). It's also the opposite hemisphere, so days will be longer from September to March than they are from March to September. However, we can pull the times from the Nautical Almanac, so once we know that it'll be different, we can figure out exactly how different pretty easily. What's trickier is gauging how long the star will be up for. For that, I'm going to introduce a scary-looking, but incredibly helpful equation which ties together an object's declination $\delta$, its right ascension $\alpha$, its altitude $h$, your latitude $\theta$, and another value which I'll explain in a moment, $a$. Here it is:\[sin(h) = sin(\delta)sin(\theta)+cos(\delta)cos(\theta)cos(a) \: (2)\]The mystery variable, $a$, is something called the semi-diurnal arc--the angular distance an object travels from rising to meridian passage, and again from meridian passage down to setting (the diurnal arc itself is the full angular distance covered between rising and setting). You can probably see why this is a nice value to know--we knew that the semi-diurnal arc when observed from Mt. Hopkins was 90 degrees, and we used that to figure out how long it was above the horizon. Since that's all we're looking for, we can simplify the equation a bit by setting $h = 0$ to put the object on the horizon and solving for $cos(a)$, giving us a much more manageable version.\[cos(a) = tan(\delta)tan(\theta)\] Solving again for $a$ gives us 83.67 degrees. We can plug that all the way back into equation $(1)$ and solve again for $t$ to find that, at our new position, AY Sixteenus will spend 11:09 above the horizon every day. Since meridian passage will occur at the same time as at Mt. Hopkins (we've only changed our latitude, not our longitude), all we need to do is calculate the rise and set times (5:35 before and after meridian passage), plot when AY Sixteenus will be up over a plot of sunset and sunrise times, and we're set.

The difference isn't incredibly dramatic (though you can easily see how much flatter the sunset and sunrise lines are). However, since AY Sixteenus is rising half an hour earlier and setting half an hour later, the band is a bit more narrow here than it was at +32 degrees.

Moving back north, up to +60 degrees latitude, yields some pretty interesting changes. At that latitude it's possible to see objects from declinations of -30 up to +90 degrees--you can figure this out just by adding and subtracting 90 to your latitude to determine the declinations of objects on your northern and southern horizons. Obviously +90 isn't 60+90, but since declination doesn't go past +/- 90, you end up actually seeing "over" the North Pole and looking at objects from +90 down to +30 degrees on the far side of the Earth. Since AY Sixteenus' declination is +32 degrees, it never actually drops below the horizon. Instead, it'll circle the North Celestial Pole in the sky, still 2 degrees above the northern horizon at its lowest point.

AY Sixteenus on the far side of the Earth--it will cross the meridian twice, once at an altitude of 58 degrees (in the South) and once at an altitude of 2 degrees (in the North)

This means that AY Sixteenus is, for an observer at +60 degrees latitude, a circumpolar star. This gives us a pretty useful formula:\[|\delta_{min}| = 90-|l|\:(3)\] Give $\delta_{min}$ the same sign as your latitude and it tells you the minimum declination for a star to be circumpolar. Give it the opposite sign as your latitude and it tells you what stars will never rise above your horizon.

The Story of AY Sixteenus: Part I

Sketch the elevation of the star AY Sixteenus that is located at RA = 18 hours and Dec = +32 degrees along the meridian as a function of time (month) throughout the year as viewed from Mt. Hopkins, Arizona. Your sketch should be qualitative, yet clear, as if you were showing it to a class you are teaching. But don't be one of those teachers with horrible handwriting/drawing skills: neatness counts. Label each monthly point with the UT (Universal Time) that the star is on the meridian. What is the corresponding LST at each point?

Right Ascension (RA) and Declination (Dec) are the celestial sphere's analog of longitude and latitude, respectively. Declination is by far the simpler of the two--it's just latitudes on Earth projected out onto the sky. A star that orbits directly above the equator has zero declination, a star whose orbit takes it directly above us in Cambridge (42 degrees longitude) has a declination of 42 degrees, a star that sits directly above the North Pole (90 degrees latitude) has a declination of 90 degrees... You get the point. 

RA (which is measured in hours and minutes) is a bit more complex, since (just like longitude) it has to be measured from an arbitrary position. In the same way that people chose the Royal Observatory in Greenwich as zero degrees longitude, astronomers have picked the "first point of Aries," or the place in the sky where the sun sits at the Vernal equinox, as zero degrees RA. You can also think of this as being the one of the intersections of the ecliptic (the plane in which most of the planets orbit) and the celestial equator, which you can imagine these both as great circles arcing over your head. 

A section of the sky, showing the intersection of the ecliptic and horizon (Source)

RA is measured as the distance around the equator in a circle starting at Aries. Since every star has a permanent RA coordinate, the range of RA you can see is constantly changing as the Earth rotates. However, since the Earth is also circling the sun along its orbit, each day's rotation doesn't translate to 24 hours of RA--at this time tomorrow, I'll be looking at a slightly different patch of sky than I am right now (about four minutes of longitudinal difference). This also means that, while the right ascension of stars doesn't change, the sun's does, making a full circle every year. On March 21st at the vernal equinox the sun's RA is 00:00, three months later at the northern hemisphere's summer solstice it's 06:00, at the autumnal equinox it's 12:00, at the winter solstice it's 18:00, and by the time March 21st rolls around again it's back to 00:00 and we've made a full orbit. 

Local Sidereal Time (LST) is a related concept, and thankfully it's a simpler one. It essentially refers to whatever RA is crossing your meridian at a given time. So if it's March 21st and the sun is on my meridian, my LST is 00:00. Likewise, if a star should happen to have an RA of 18:00 (remember, the RA of stars other than the sun doesn't change), my LST will be 18:00 when it crosses my meridian. As I mentioned above, the sun's RA is 18:00 on the Winter Solstice, so on that day our LST will be 18:00 at noon. Time is determined by the sun, so a month (or two hours or RA) later, our LST will be 18:00 two hours before noon--extrapolating throughout the year, we can tell that LST = 18:00 will occur at 06:00 on the vernal equinox, midnight on the summer solstice, 18:00 on the autumnal equinox, and will be back at noon come late December.

All right, so let's come back to AY Sixteenus. What we know is this:

-Its declination, 32 degrees North (or +32 degrees, for those of you who find +/- more intuitive than N/S)

-Its RA, 18:00 hours

-Our own latitude, 32 degrees North

And that's all we need! If you've been paying attention, you should have already realized that the question is actually a very straightforward one--and that you're in store for the two least interesting graphs of all time. I'm feeling artistic today, so here's a diagram of what the setup looks like when the star is crossing our meridian.

Since stars' declinations don't change, that means that they trace the exact same path in the sky every day--you just sometimes can't see them because the sun occasionally happens to be in the way. Since AY Sixteenus has the same declination as our latitude, it's going to be directly overhead whenever it does this, regardless of what time of year it is. So here's boring graph #1:

Not Pictured: A Dependent Variable

Figuring our LST at the star's meridian passage is going to end similarly. Since LST is defined as the RA of all the objects on your meridian, and AY Sixteenus' RA is 18:00, well...

We can figure out what time it will be for each of these meridian passages by using the calculations we did for the time of LST = 18:00 earlier. Knowing that it'll occur at noon in late December and move two hours earlier every month, it's easy to put it together into a neat little table.