# A Very Small Slice of Pi

Rhubarb pie (Source: Wikipedia)

Some people know a suspiciously large number of the digits of pi. Perhaps you have met one of these people. They can typically be found hiding behind bushes and under the counters at pastry shops, just... waiting. At the slightest hint of a mention of pi, they will jump out and start reciting the digits like there's a prize at the end. After rattling off numbers for a few minutes they abruptly come to an end, grin like an idiot, and walk away. It is an unpleasant encounter. The sheer uselessness of this kind of thing has always bothered me, so I'd like to set a preliminary upper bound on the number of digits of pi that could ever possibly potentially kind of be useful (maybe). For those following along at home, now would be a good time to put on your numerology hats. Alright, so I hear this thing pi is fairly useful when dealing with circles. Let's say we want to make a really big circle and have its diameter only deviate by a very small amount from the correct value. To do this successfully, we will have to know pi fairly well. Let's take this to extremes now. Suppose I want to put a circle around the entire visible universe such that the uncertainty in the diameter is the size of a single proton. What would be the fractional uncertainty in the circumference in this case? If we know pi exactly, then we have that $$\delta C = \frac{\partial C}{\partial d} \delta d = \pi \delta d = C \frac{\delta d}{d},$$ where d is the diameter and C is the circumference. In other words, the fractional uncertainty in the circumference is just $$\frac{\delta C}{C} = \frac{\delta d}{d}.$$ Using a femtometer for the size of a proton and 90 billion light years for the size of the Universe [1], we get $$\frac{\delta C}{C} = \frac{\delta d}{d} = \frac{10^{-15}\mbox{m}}{(90\times10^9)(3\times10^7\mbox{s})(3\times10^8\mbox{m s}^{-1})} \sim\frac{10^{-15}}{10^{27}}\sim10^{-42}.$$ Alright, so how well do we need to know pi to get a similar fractional uncertainty? Well, we have that $$\frac{\delta \pi}{\pi} = \frac{\delta C}{C} = 10^{-42},$$ so we can afford an uncertainty in pi of $$\delta \pi = \pi \times 10^{-42}$$ and thus we'll need to know pi to about 42 digits. How's that for an answer? So if we have a giant circle the size of the entire visible universe, we can find its diameter to within the size of a single proton using pi to 42 digits. Therefore, I adopt this as the maximal number of digits that could ever prove useful in a physical sense (albeit under a somewhat bizzarre set of circumstances). If reciting hundreds of digits is what makes you happy, go for it. But 42 digits is more than enough pi for me.

[1] "But I thought the Universe was only 13.7 billion years! What voodoo is this!?" Yeah, I know. See here for a nice explanation.[back]

# Primes in Pi

Recently, I've been concerned with the fact that I don't know many large primes. Why? I don't know. This has led to a search for easy to remember prime numbers. I've found a few goods ones, namely

But then I remembered that I already know 50 digits of pi, memorized one boring day in grade school, so this got me wondering whether there were any primes among the digits of pi

Lo an behold, I wrote a little script, and found a few:

Found one, with 1 digits, is: 3 Found one, with 2 digits, is: 31 Found one, with 6 digits, is: 314159 Found a rounded one with 12 digits, is: 314159265359 Found one, with 38 digits, is: 31415926535897932384626433832795028841

I think it's usual for most science geeks to know pi to at least 3.14159, if you're one of those people, now you know a 6 digit prime! for free!

# F-91 Revisited

Farmer Uncle Sam...with a rifle. (Image Credit: Wikipedia)

Today was a sunny exception to the grey overcast rule of weather in Ithaca. I should be overjoyed by this anomaly, spending the day outside flying a kite or playing frisbee with a border collie in a bandanna. Unfortunately, today was also the beginning of Daylight Savings Time (DST) - my least favorite day of the year. For my colleagues unfamiliar with this temporal travesty (I'm looking at you Arizona), let me briefly explain DST. Once a year, the time lords steal a single hour from us and place it in an escrow account for future disbursement, presumably in some elaborate scheme to gain the favor of hat-throwing farmer-clock hybrids (see image left). The details are a bit murky, but the net result is that today I had one less hour to do my very favorite thing in the whole wide world - sleep. It also means that I have to set my watch, so I figured I'd check in and see how well my previous model for time-loss in my watch has held up. About a month ago, I looked at how my watch slowly deviated from the official time (the original post can be found here and a helpful clarification by Tom can be found here). Based on a little over 50 days worth of data, I found that my watch lost about 0.35 seconds per day against the official time. About 50 days have passed since my last measurement and today when I set the watch, so I thought it would be interesting to see how well my model fit the new data. The old data are presented in Figure 1 in blue, the old best fit line is in red, and the new data point (taken this morning) is in green. As always, click through the plot for a larger version.

Figure 1

The new data point appears to be in fair agreement with the old best-fit model, but it's a little hard to see here. Zooming in a bit, though, we see that the model lies outside the error bar of the new data point.

Figure 2

So is this a big deal? Not really. But if it will help you sleep at night, we can redo the fitting with the new data point included to see how much things change. The the plots with the updated model to include all data points are provided below with the old data in blue and the new point in green.

Figure 3

The new model looks a whole lot like the old one, except the best fit line now appears to go through the new data point. Zooming in a little, we see that it does indeed fall within the error bars of our new point.

Figure 4

Alright, so the new model fits with our new point, but how much did the model have to change? Well, the fit to just the old data gave an offset of 0.36 seconds and a loss rate of 0.35 seconds per day. The new model has an offset of 0.40 seconds and a loss rate of 0.348 seconds per day. Overall, not a significant change. It looks as though I may continue to not worry about the accuracy of my watch. I have set it to match the official time and have no intention of fiddling with it until I have to set it again at the end of Daylight Savings Time - my favorite day of the year.

# Time Keeps On Slippin'

This is picture of a watch. (Source: Wikipedia)

A couple of months ago, the Virtuosi Action Team (VAT) assembled for lunch and the discussion quickly and unexpectedly turned to watches. As Nic and Alemi argued over the finer parts of fancy-dancy watch ownership, I looked down at my watch: the lowly Casio F-91W. Though it certainly isn't fancy, it is inexpensive, durable, and could potentially win me an all-expense paid trip to the Caribbean. But how good of a watch is it? To find out, I decided to time it against the official U.S. time for a couple of months. Incidentally, about half-way in I found out that Chad over at Uncertain Principles had done essentially the same thing already. No matter, science is still fun even if you aren't the first person to do it. So here's my "new-to-me" analysis. Alright, so how do we go about quantifying how "good" a watch is? Well, there seem to be two main things we can test. The first of these is accuracy. That is, how close does this watch come to the actual time (according to some time system)? If the official time is 3:00 pm and my watch claims it is 5:00 am, then it is not very accurate. The second measure of "good-ness" is precision or, in watch parlance, stability. This is essentially a measure of the consistency of the watch. If I have a watch that is consistently off by 5 minutes from the official time, then it is not accurate but it is still stable. In essence, a very consistent watch would be just as good as an accurate one, because we can always just subtract off the known offset. To test any of the above measures of how "good" my cheap watch is, we will need to know the actual time. We will adopt the official U.S. time as provided on the NIST website. This time is determined and maintained by a collection of really impressive atomic clocks. One of these is in Colorado and the other is secretly guarded by an ever-vigilant Time Lord (see Figure 1).

Figure 1: Flavor Flav, Keeper of the Time

At 9:00:00 am EST on November 30th, I synchronized my watch with the time displayed on the NIST website. For the next 54 days, I kept track of the difference between my watch an the NIST time. On the 55th day, I forgot to check the time and the experiment promptly ended. The results are plotted below in Figure 2 (and, as with all plots, click through for a larger version).

Figure 2: Best-fit to time difference

As you can see from Figure 2, the amount of time the watch lost over the timing period appears to be fairly linear. There does appear to be a jagged-ness to the data, though. This is mainly caused by the fact that both the watch and the NIST website only report times to the nearest second. As a result, the finest time resolution I was willing to report was about half a second. Adopting an uncertainty of half a second, I did a least-squares fit of a straight line to the data and found that the watch loses about 0.35 seconds per day. As far as accuracy goes, that's not bad! No matter what, I'll have to set my watch at least twice a year to appease the Daylight Savings Gods. The longest stretch between resetting is about 8 months. If I synchronize my watch with the NIST time to "spring forward" in March, it will only lose about $$t_{loss} = 8\~\mbox{months} \times 30\frac{\mbox{days}}{\mbox{month}} \times 0.35 \frac{\mbox{sec}}{\mbox{day}} = 84\~\mbox{sec}$$ before I have to re-synchronize to "fall back" in November. Assuming the loss rate is constant, I'll never be more than about a minute and a half off the "actual" time. That's good enough for me. Furthermore, if the watch is consistently losing 0.35 seconds per day and I know how long ago I last synchronized, I can always correct for the offset. In this case, I can always know the official time to within a second (assuming I can add). But is the watch consistent? That's a good question. The simplest means of finding the stability of the watch would be to look at the timing residuals between the data and the model. That is, we will consider how "off" each point is from our constant rate-loss model. A plot of the results is shown below in Figure 3.

Figure 3: Timing residuals

From Figure 3, we see that the data fit the model pretty well. There's a little bit of a wiggle going on there and we see some strong short-term correlations (the latter is an artifact of the fact that I could only get times to the nearest second). To get some sense of the timing stability from the residuals, we can calculate the standard deviation, which will give us a figure for how "off" the data typically are from the model. The standard deviation of the residuals is $$\sigma_{res} = 0.19\~\mbox{sec}.$$ A good guess at the fractional stability of the watch would then just be the standard deviation divided by the sampling interval, $$\frac{\sigma_{res}}{T} = 0.19\~\mbox{sec} \times \frac{1}{1\~\mbox{day}} \times \frac{1\~\mbox{day}}{24\times3600\~\mbox{sec}} \approx 2\times10^{-6}.$$ In words, this means that each "tick" of the watch is consistent with the average "tick" value to about 2 parts in a million. That's nice...but isn't there something fancier we could be doing? Well, I have been wanting to learn about Allan variance for some time now, so let's try that. The Allan variance (refs: original paper and a review) can be used to find the fractional frequency stability of an oscillator over a wide range of time scales. Roughly speaking, the Allan variance tells us how averaging our residuals over different chunks of time affects the stability of our data. The square root of the Allan variance, essentially the "Allan standard deviation," is plotted against various averaging times for our data below in Figure 4.

Figure 4: Allan variance of our residuals

From Figure 4, we see that as we increase the averaging time from one day to ten days, the Allan deviation decreases. That is, the averaging reduces the amount of variation in the frequency of the data, making it more stable. However, at around 10 days of averaging time it seems as though we hit a floor in how low we can go. Since the error bars get really big here, this may not be a real effect. If it is real, though, this would be indicative of some low-frequency noise in our oscillator. For those who prefer colors, this would be "red" noise. Since the Allan deviation gives the fractional frequency stability of the oscillator, we have that $$\sigma_A = \frac{\delta f}{f} = \frac{\delta(1/t)}{1/t} = \frac{\delta t}{t}.$$ Looking at the plot, we see that with an averaging time of one day, the fractional time stability of the watch is $$\frac{\delta t}{t} \approx 2\times10^{-6},$$ which corresponds nicely to our previously calculated value. If we average over chunks that are ten days long instead, we get a fractional stability of $$\frac{\delta t}{t} \approx 10^{-7},$$ which would correspond to a deviation from our model of about 0.008 seconds. Not bad. The initial question that started this whole ordeal was "How good is my watch?" and I think we can safely answer that with "as good as I'll ever need it to be." Hooray for cheap and effective electronics!

# The Stars Fell on Abe and Frederick

The 1833 Leonids (Source: Wikipedia)

Word on the street is there's a meteor shower set for late Tuesday night, peaking at 2 am EST on January 4th [1]. The meteors in question are the Quadrantids, which often go unnoticed for two good reasons. Reason the first: apparently [2], they are usually pretty awful. Unlike the "good" meteor showers, the Quadrantids are bright and pretty for only a few hours (instead of a few days). This means that a lot of the time, we just miss them. Reason the second: they have a lame name [3]. But this year, they should be pretty good if the weather is right. Now, there's lots of neat physics to talk about with meteors, but that's not why I bring it up. This has all just been flimsy pretext so I could share a historical anecdote about a meteor shower. Trickery, indeed. Those who feel cheated are free to leave now with heads held high. Those still around (Hi, Mom!) will hear about the night in 1833 when the stars fell on Alabama (and the rest of the country, too). The Leonids typically put on a pretty good show, but their showing in 1833 was so dramatic that the term "meteor shower" was coined to describe what was happening. The 1833 Leonids were truly one for the ages and made such an impression that people were often able to remember when events happened by their relation to the night when "the stars fell." It was in this use as a "calendar anchor" that I first heard of this particular meteor shower. While home for the holiday I was reading Life and Times of Frederick Douglass, one of the later autobiographies written by the former slave and noted abolitionist. Recounting when he was moved from Baltimore to a plantation on the Eastern Shore of Maryland, Douglass writes:

I went to St. Michaels to live in March, 1833. I know the year, because it was the one succeeding the first cholera in Baltimore, and was also the year of that strange phenomenon when the heavens seemed about to part with their starry train. I witnessed this gorgeous spectacle, and was awe-struck. The air seemed filled with bright descending messengers from the sky. It was about daybreak when I saw this sublime scene. I was not without the suggestion, at the moment, that it might be the harbinger of the coming of the Son of Man; and in my then state of mind I was prepared to hail Him as my friend and deliverer. I had read that the "stars shall fall from heaven," and they were now falling. I was suffering very much in my mind. It did seem that every time the young tendrils of my affection became attached they were rudely broken by some unnatural outside power; and I was looking away to heaven for the rest denied me on earth.

Douglass wrote these words almost 50 years after the fact and it is evident that the meteor shower clearly had an effect on him. By this time (at age 15), Douglass had already made up his mind to escape from slavery. Three years later, he made a failed attempt. Two years after that, in 1838, Frederick Douglass escaped to the North and became an influential abolitionist. After reading the above passage from Douglass, I wondered who else may have seen the 1833 Leonids. After a bit of research, I found a paper by Olson & Jasinski (1999) which provides an excerpt from Walt Whitman recounting a story told by Abraham Lincoln. Whitman writes:

In the gloomiest period of the war, he [Lincoln] had a call from a large delegation of bank presidents. In the talk after business was settled, one of the big Dons asked Mr. Lincoln if his conﬁdence in the permanency of the Union was not beginning to be shaken — whereupon the homely President told a little story. “When I was a young man in Illinois,” said he, “I boarded for a time with a Deacon of the Presbyterian church. One night I was roused from my sleep by a rap at the door, & I heard the Deacon’s voice exclaiming ‘Arise, Abraham, the day of judgment has come!’ I sprang from my bed & rushed to the window, and saw the stars falling in great showers! But looking back of them in the heavens I saw all the grand old constellations with which I was so well acquainted, ﬁxed and true in their places. Gentlemen, the world did not come to an end then, nor will the Union now."

Abraham Lincoln witnessed the 1833 meteor shower and was still telling stories about it 30 years later.

So what's the point of this whole story? Is there any significance to the fact that the man who escaped slavery to tell the world of its evils and "The Great Emancipator" both saw the same meteor shower? Probably not. Tons of people saw it.

Regardless, it is interesting to think about. Though these men would cross paths several times over the next 30 years, the earliest memory they shared was of a night in 1833, when a 15 year old slave in Maryland and a 24 year old boarder in Illinois watched the stars fall from the sky.

[1] I use "Tuesday night" here to mean, of course, "Wednesday morning." [back]

[2] I say "apparently" because I have never heard of these guys before, so this is all Wikipedia, baby! [back]

[3] Like other meteor showers, the Quadrantids take their name from the constellation from which the meteors seem to emerge. In this case, Quadrans Mural: The Mural Quadrant. Unfortunately for Quadrans Mural, the constellations dumped it like the planets dumped Pluto. [back]

# How Long Will a Bootprint Last on the Moon?

Buzz Aldrin's bootprint (source: Wikipedia)

A couple of months ago, I stumbled across a bunch of pictures of Apollo landing sites taken by one of the cameras onboard the Lunar Reconnaissance Orbiter. The images have a resolution high enough that you can resolve features on the surface down to about a meter. Looking at the Apollo 17 landing site, you can see the trails of both astronauts and a moon buggy. It's pretty cool. It also got me thinking about how long the landing sites would be preserved. More specifically, I want to know how long Buzz Aldrin's right bootprint (shown, incidentally, to the left) will last on the Moon. Since the Moon has no atmosphere, the wind and rain that would weather away a similar bootprint here on Earth are not present and it seems as though the print would last a really long time. But how long? Let's try to quantify it [1]. Pick Your Poison Before we get going, we need to figure out what physical process would be most important in erasing a bootprint from the Moon. Although the Moon lacks the conventional "weathering" we experience on Earth (due to wind, rain, etc), it does experience something called "space weathering." Space weathering is the changing of the lunar surface due to cosmic rays, micrometeorite collisions, regular meteorite collisions, and the solar wind [2]. Of these phenomena, the most apparent and well-studied would be the meteorites which have covered the Moon in craters. We adopt the meteorite impact as our primary means of wiping out a bootprint and restate our question as follows: "How long would it take for a meteorite to hit the Moon such that the resulting crater wipes out Aldrin's right bootprint?" Background As it is currently stated, we can answer our question if we knew the rate of formation and size distribution of the craters on the Moon. We could count up all the craters on the Moon (or a particular region of interest) and tabulate their sizes. This would give us the size distribution. It would also give us a headache and potentially drive us to lunacy [3]. Luckily, someone has beat us to it. Cross (1966) used images from the Ranger 7 and 8 missions to count craters and determine the size distribution of craters in three regions of the Moon. The data for the crater distribution in the Sea of Tranquility (where Apollo 11 landed) are given in the figure below. Cross found that in the Sea of Tranquility, the number of craters with diameters greater than X meters (per million square kilometers) is given by: $$N(d>X) = 10^{10}\left(\frac{X}{1\~\mbox{m}}\right)^{-2},$$ which holds for craters with diameters between 1 meter and 10 kilometers (see figure below).

Figure 2 from Cross (1966)

We can also estimate the rate at which craters are formed from this data. If we assume that the craters formed at a constant rate over the age of the Moon (about 4 billion years), then we get about 2.5 craters with diameters above 1 meter formed in a million square kilometer area every year. This is a "crater flux" for the Moon. Written another way, the crater flux in the Sea of Tranquility is $$F \approx 1\~{\mbox{km}}^{-2} \frac{1}{4\times10^5\~\mbox{yr}},$$ so we get that roughly one crater with diameter greater than 1 meter is formed on a square kilometer of the Moon once every 400,000 years or so. We now have enough information to do some simulations. Simulation I wrote up a code that simulates craters being formed on a 1 square kilometer patch of the Moon. A crater is randomly placed in the 1 square kilometer region with a diameter pulled from the above distribution. The bootprint is placed at the center of the grid and craters are formed until we get a "hit." At that point, the time is recorded and the run stops. As a sanity check, I thought it would be fun to just let the simulation run without caring if the boot was hit or not. By simulating the craters in this way for 4 billion years, I should get something that looks like the Moon at the present day. Here's a 200 m square from my simulation: and here's a picture of the same-sized region on the surface of the Moon:

Cropped from this image (Source: LRO)

Just eyeballing it, things look pretty good. Now it's time for the actual simulation. I ran the simulation 10,000 times and tabulated the amount of time needed before the bootprint was hit. The figure below gives the CDF for the hit times in the simulation. That is, for each time T, we find the fraction of simulations in which the bootprint got hit in a time less than or equal to T. The dashed lines in the plot indicate the amount of time needed to pass for half of the simulations to have recorded a hit. This time turns out to be about 24 billion years.

(Click for larger, actually readable version)

# Physics Challenge Award Show II

Not a DeLorean. You're doing it wrong.

For those who solve problems (he salutes you)

# Betelgeuse, Betelgeuse, Betelgeuse!

A very cold person points out Betelgeuse

Betelgeuse is a massive star at the very end of its life and could explode any second now! Every time I hear that I get really really excited. Like a kid in a candy store that's about to see a star blow up like nobody's business. This giddiness will last for a solid minute before I realize that "any second now" is taken on astronomical timescales and roughly translates to "sometime in the next million years maybe possibly." Then I feel sad. But you know what always cheers me up? Calculating things! Hooray! So let's take a look at the ways Betelgeuse could end its life (even if it's not going to happen tomorrow) and how these would affect Earth. First, a little background. Betelgeuse is the bright orangey-red star that sits at the head/armpit of Orion. It is one of the brightest stars in the night sky. Its distance has been measured by the Hipparcossatellite to be about 200 parsecs [1] from Earth (about 600 light years). Betelgeuse is at least 10 times as massive as our Sun and has a diameter that would easily accomodate the orbit of Mars. In fact, the star is big enough and close enough that it can actually be spatially resolved by the Hubble Space Telescope! Being so big and bright, Betelgeuse is destined to die young, going out with a bang as a core-collapse supernova. This massive explosion ejects a good deal of "star stuff" into interstellar space [2] and leaves behind either a neutron star or a black hole. Alright, now that we're all caught up, let's turn our focus on this "massive explosion" bit. What kind of energy scale are we talking about if Betelgeuse blows up? Well, a pretty good upper bound would be if all of the star's mass (10 solar masses worth!) were converted directly to energy, so $$E_{max} = mc^2 = 10M_{\odot}\times\left(\frac{2\times10^{30}\~\mbox{kg}}{1\~M_{\odot}}\right)\times \left(3\times10^8\~\mbox{m/s}\right)^2$$ which is about $$E_{max} \sim 10^{48}\~\mbox{J}$$ and that's nothing to shake a stick at. But remember, this is if the entire star were converted directly to energy, and that would be hard to do. Typical fusion efficiencies are about \~1% [3], so let's say a reasonable estimate for the total nuclear energy available is $$E_{nuc} \sim \eta_{f} \times E_{max} \sim 10^{-2} \times 10^{48}\~\mbox{J} \sim 10^{46}\~\mbox{J}.$$ This is the total energy released by a typical supernova. As it turns out though, 99% of this energy is carried away in the form of neutrinos and only about 1% is carried away in photons. Since we are mainly concerned with how this explosion will affect Earth, and the neutrinos will just pass on by, we will only consider the 1% of energy released in photons that would reasonably interact with Earth. That gives us $$E_{ph} \sim 0.01 \times E_{nuc} \sim 10^{44}\~\mbox{J}.$$ Neato, so that's the total amount of energy released in a supernova in the form of photons. How much of this energy would be deposited at the Earth if Betelgeuse exploded? Well, if the energy is deposited isotropically (that is, the same in all directions), then the fluence (or time integrated energy flux) is given by $$F_{ph} = \frac{E_{ph}}{4\pi d^2}.$$ All this is saying is that the total energy release by the supernova spreads out uniformly over a sphere of radius d, so the fluence will give us the amount of energy deposited in each square meter of that sphere (the units of fluence here are J/m^2). The total energy deposited on Earth is then $$E_{\oplus} = F_{ph} \times \pi R^2_{\oplus}.$$ Hot dog! Let's plug in some numbers, already. The total energy deposited on the Earth by a symmetrically exploding Betelgeuse at a distance of d = 200 pc (where 1 pc = 3 10^16 m) is $$E_{\oplus}=\frac{E_{ph}}{4\pi d^2}\times\pi R^2_{\oplus}\sim 10^{19}\~\mbox{J}\left(\frac{E_{ph}}{10^{44}\~\mbox{J}}\right)\left(\frac{d}{200\~\mbox{pc}}\right)^{-2}.$$ Well, 10^19 J certainly seems* like a lot of energy. In fact, it is roughly the amount of energy contained in the entire nuclear arsenal of the United States [4]. But it is spread over the entire atmosphere. Is there a way to gauge how this would affect life on Earth? We could see how much it would heat up the atmosphere using specific heats: $$E = m_{atm}c_{air}\Delta T$$ where c is the specific heat of air (\~10^3 J per kg per K). Oops, looks like we need to know the mass of the atmosphere. But we can figure this out, the answer is pushing right down on our heads! We know the pressure at the surface of the Earth (1 atm = 101 kPa) and that pressure is just the result of the weight of the atmosphere pushing down on us. Since pressure is just force / area, we have $$P = F/A = m_{atm}g / A_{\oplus}$$ So $$m_{atm} = \frac{P\times4\pi R^2_{\oplus}}{g}=\frac{10^5\~\mbox{Pa}\times4\pi (6\times10^6\~\mbox{m})^2}{9.8\~\mbox{m/s}^2}\approx4\times10^{18}\~\mbox{kg}.$$ Neato, gang. So we could see a temperature rise of about $$\Delta T = \frac{E_{ph}}{m_{atm}c_{air}}=\frac{10^{19}\~\mbox{J}}{4\times10^{18}\~\mbox{kg}\times10^3\~\mbox{J/ kg K}}\approx0.003\~\mbox{K},$$ or three one-thousandths of a degree. Remember, too, that this will be an upper bound since we are assuming that all this energy is deposited into the atmosphere before it has a chance to cool. In fact, if the energy is deposited over the course of hours or days, this value will be much less. So it looks like we've wrapped this thing up: Betelgeuse exploding will most certainly not put the Earth in any danger. Or did we? We have considered the case of a symmetric supernova, but there's more than one way to blow up a star. Massive stars can also end their lives in a fantastic explosion called a gamma-ray burst (GRBs to the hep cats that study them, some fun facts relegated to [5]). GRBs are still an intense area of current study, but the current picture (for one type of GRB, at least) is that they are the result of a star blowing up with the energy of the explosion focussed into two narrow beams (see picture below). Since the flux isn't distributed over the whole sphere, GRBs can be seen at much greater distances than a typical supernova.

Example of a gamma-ray burst, with the explosion in two beams.

So how will this change our answer? Well, it's going to change the fluence we calculated above. Instead of spreading the energy out over the whole sphere, it's only going to go to some fraction of the 4pi steradians. So we get $$F_{ph} = \frac{E_{ph}}{4\pi f_{\Omega} d^2},$$ where f_{omega} is called the "beaming fraction" and tells us what fraction of the sphere the energy goes through. Typical GRB beams range from 1 to 10 degrees in radius. Converting this to radians, we can find the beaming fraction as $$f_{\Omega} = \frac{2 \times \pi \theta^2}{4\pi} \approx 10^{-4}\left(\frac{\theta}{1^\circ}\right)^2,$$ so the beaming fraction is 10^-4 and 10^-2 for a beam angle of 1 degree and 10 degrees, respectively. Alright, so now we can redo the calculations we did for the supernova case, but keeping this beaming fraction around. The total amount of energy that would hit Earth is then about $$E_{\oplus}=\frac{E_{ph}}{4\pi f_{\Omega} d^2}\times\pi R^2_{\oplus}\sim 10^{23}\~\mbox{J}\left(\frac{E_{ph}}{10^{44}\~\mbox{J}}\right)\left(\frac{d}{200\~\mbox{pc}}\right)^{-2}\left(\frac{\theta}{1^\circ}\right)^{-2}.$$ Holy sixth-of-a-moley! Continuing as we did above, we find that this could potentially heat up the atmosphere by $$\Delta T = \frac{E_{ph}}{m_{atm}c_{air}}=\frac{10^{23}\~\mbox{J}}{4\times10^{18}\~\mbox{kg}\times10^3\~\mbox{J/ kg K}}\approx3\~\mbox{K}\left(\frac{\theta}{1^\circ}\right)^{-2},$$ which is certainly non-negligible. Now, this won't destroy the planet [6], but it could make things really uncomfortable. This will be especially true when you realize that a fair amount of the energy carried away from a gamma-ray burst is in the form of (wait for it...) gamma-rays, which will wreck havoc on your DNA. Remember, though, that this is an absolute worst-case scenario since we have assumed the smallest beaming angle. But this may still make us a little nervous, so is there anyway to figure out if Betelgeuse could, in fact, beam a gamma-ray burst towards Earth? Yes, yes there is. Jets and beams like those in GRBs typically point along the rotation axis of the star [7]. If we could determine the rotational axis of Betelgeuse, then we could say whether or not there's a chance it's pointed towards us. It just so happens that Betelgeuse is the only star (aside from our Sun) that is spatially resolved. If you could measure spectra along the star, you could look for Doppler shifting of absorption lines and say something about the velocity at the surface of the star. Luckily, this has already been done for us (see, for example Uitenbroek et al. 1998). These measurements are hard to do since the star is only a few pixels wide, but it appears as though the rotation axis is inclined to the line-of-sight by about 20 degrees (see figure below). That means this would require a beam with at least a 20 degree radius to hit the Earth. This appears to be outside the typical ranges observed. So even if Betelgeuse were to explode in a gamma-ray burst, the beam would miss Earth and hit some dumb other planet nobody cares about.

Figure reproduced from Uitenbroek et al. (1998)