The Virtuosi

The Stars Fell on Abe and Frederick

2012-01-02T20:02:00.001-05:00

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?

2012-01-01T20:46:00.000-05:00

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)

Conclusions and Caveats

Based on the simulations, the bootprint on the Moon would have about even odds of lasting at least 20 billion years if the primary means of destruction is through the formation of a crater from a meteorite. However, there are a few caveats that should be addressed. These deal with either the details of the simulation or the assumptions we have made.

In the simulation, we just took at 1 km square patch of the moon and scaled back the "crater flux" accordingly. However, this does not fully account for all possible craters that can form. For example, our simulation would miss an event that hit 50 km away from the target, but had a diameter of 100 km. Obviously this would hit the target, but we are only seeding craters in the 1 square km region. This would mean that the actual lifetime of the bootprint would be less than our 24 billion year figure. Re-running with a 10km by 10km square region, we find a lifetime of 18 billion years. Thus, an increase in area by a factor of 100 only reduces the age by 25%. Considering areas much larger than this makes the simulation prohibitively slow, but the order unity effect does not seem too significant.

Additionally, we have made a number of assumptions. The big one is that we have assumed that the craters currently seen on the Moon were formed uniformly in time. In fact, a large fraction of the craters may have been formed when the Moon was still very young (see Late Heavy Bombardment). If this were the case, we would have greatly overestimated the rate of crater formation and thus underestimated the time needed to hit the bootprint.

In spite of these caveats, let's take our value of 20 billion years to be accurate. What else can we say? Well, if we are right then we are wrong because the Moon may not last that long (and it's hard to have bootprints on the Moon without a Moon). Current estimates have that the Sun will expand into a red giant and (potentially) destroy the Earth (and the Moon) in about 5 billion years.

So a record of the Apollo astronauts' boot sizes could potentially last as long as the Moon [4]. Not bad.

Footnotes and Such

[1] Now with linked footnotes so Yariv doesn't have to scroll! [back]

[2] There was a fairly recent press release about Coronal Mass Ejections from the Sun "sandblasting" the lunar surface. For more info, check here, and note the acronymic acrobatics needed to make them the "DREAM team." But it's totally worth it. [back]

[3] A horribly forced pun. But it's totally worth it. [back]

[4] Also, Nixon [back]

Report from the Trenches: A CMS Grad Student's Take on the Higgs

2011-12-13T17:36:00.001-05:00

Hi folks. It's been an embarrassingly long time since I last posted, but today's news on the Higgs boson has brought me out of hiding. I want to share my thoughts on today's announcement from the CMS and ATLAS collaborations on their searches for the Higgs boson. I'm a member of the CMS collaboration, but these are my views and don't represent those of the collaboration.

The upshot is that ATLAS sees a 2.3 sigma signal for a Higgs boson at 126 GeV. CMS sees a 1.9 sigma excess around 124 GeV. CERN is being wishy-washy about whether or not this is actually a discovery. After all the media hype leading up to the announcement, this is somewhat disappointing, but maybe not too surprising.

First of all, what does a 2 sigma signal mean? The significance corresponds to the probability of seeing a signal as large or larger than the observed one given only background events. That is, what's the chance of seeing what we saw if there is no Higgs boson? You can think of the significance in terms of a Normal distribution. The probability of the observation corresponds to the integral of the tails of the Normal distribution from the significance to infinity. For those of you in the know, this is just 1 minus the CDF evaluated at the significance. For a 2 sigma observation, this corresponds to about 5%.

For both experiments, there was a 5% chance of observing the signal they observed or bigger if the Higgs boson doesn't exist. In medicine, this would be considered an unqualified success. So why is CERN being so cagey? In particle physics we require at least 3 sigma before we even consider something interesting, and 5 sigma to consider it an unambiguous discovery.

The reasons why the burden of proof is so much higher in particle physics than in other fields aren't entirely clear to me. I suspect is has to do with the relative ease of running the collider a little longer compared to recruiting more human test subjects, to use medicine as an example.

Given what I've just told you that we need a 3 sigma significance in particle physics, why is everyone so excited about a couple of 2 sigma results? Well, the first reason is that both results show bumps at approximately the same Higgs mass. Although it's not rigorous, you can get a rough idea of what the significance of the combined results are by adding the significances in quadrature. This gives us about 2.8 sigma. Higher, but still not up to the magic number of 3.

The explanation for the excitement that is most compelling brings us to Bayesian statistics. The paradigm of Bayesian statistics says that our belief in something given new information is the product of our prior beliefs and a term which updates them based on the new information. Physicists have long expected to find a Higgs boson with a mass around 120 GeV. So our prior degree of belief is pretty high. Thus, it doesn't take as much to convince us (or me anyway) that we have observed the Higgs boson. In contrast, consider the OPERA collaboration's measurement of neutrinos going faster than the speed of light. This claims to be a 6 sigma result, but no one expected to find superluminal neutrinos, so our (or at least my) prior for this is much lower. (Aside: If the OPERA result is wrong, it is likely due to a systematic effect rather than a statistical one. Nevertheless, I stand by my point.)

The final thing that excites me about this observation is that what we've seen is completely consistent with what we would expect to see from the Standard Model. Forgetting about significances for the moment, when the CMS experiment fits for the Higgs boson mass, they find a cross section that agrees very well with that predicted by the Standard Model. In the plot below, you're interested in the masses where the black line is near 1. The ATLAS experiment actually sees more signal than one would expect. This is likely just a statistical fluctuation, and explains why the ATLAS result has a higher significance.

In conclusion, while CERN is being non-committal, in my opinion, we have seen the first hints of the Higgs boson. This is mostly due to my high personal prior that there the Higgs boson exists around the observed mass. Unfortunately, Bayesian priors are for the most part a qualitative thing. Thus, ATLAS and CMS are sticking to the hard numbers, which say that what we have looks promising, but is not yet anything to get excited about.

I'll close by reminding you all to take this all with a grain of salt. There is every possibility that this is just a fluctuation. I'll remind you that at the end of last summer, CMS and ATLAS both showed a 3 sigma excess around 140 GeV, which went away just a month later at the next conference. So let's cross our fingers that next year's data will give us a definitive answer on this question.

By the way, if anyone wants to know more, fire away in the comments. I'll do my best.

Physics Challenge Award Show II

2011-11-06T23:22:00.001-05:00

Not a DeLorean. You're doing it wrong.

[Update: Prize Update / Added link to full solutions]

Welcome to the second Physics Challenge Award show!

[APPLAUSE]

Our judges have deliberated for several units of time and I now have in my hands the envelope holding our list of winners. I could easily just tell you who won right now and save everyone some time, but award shows need some suspense to work effectively, so let's first give some tedious background information!

[APPLAUSE]

You may recall that the winner of the first Physics Challenge contest won a CRC Handbook. We will not be giving out CRCs this time around. We felt that such a prize was far too ~~expensive~~ impersonal, so we have opted this year for something ~~much cheaper~~ from the heart. The following prizes will be awarded to our top three solutions:

First Prize: Our first prize winner will receive an actual back-of-an-envelope used in one of our posts (gasp!) signed by all the of the members of the Virtuosi that I can find at colloquium tomorrow. But that's not all! Alemi will also salute in your general direction.

Second Prize: For our second prize winner, we appear to have run out of envelopes... but Alemi will still salute in your general direction. You will not see him do this, but you will feel a major disturbance in the Awesome Force (mediated, of course, through the midi-chlorian boson).

Third Prize: You will receive no material prize, but on your deathbed you will receive total consciousness. So you've got that going for you, which is nice.

Let's first remind everyone what the Challenge problem was. The full text of the problem can be found here, but the gist is basically this: You've created a time machine and you're biggest fear is that you'll be stuck back in the past without any way to communicate to the future that your design worked and you deserve all kinds of Nobel prizes. The solution should be able to last long periods of time (who knows how far back in time you'll go?), should maximize the chances of modern people finding it, and be able to convince people that you have in fact gone back in time.

Alright, let's get to some solutions already!

First Place: The first place solution comes from Christian, who uses some biological wrangling to solve the time traveller conundrum. With some information from the announcement of "synthetic life" and some bio how-to from an entity known only as "steve," Christian plans to implant a message into the DNA of bacteria. The message will contain his name, identifying information, and the url of a website which will (presumably) contain a video of him with one hand outstretched saying "Nobel prize please."

Let's see how this solution satisfies our criteria for a successful solution. Does it work for an arbitrary amount of time? It appears to, so long as the bacteria manage to survive and the message doesn't become too garbled over time (perhaps some error-correction might be useful). Additionally, if one is worried about introducing non-native bacteria to the wild you could bring back a bunch of bacteria that were known to exist over wide periods of time and just release those alive at the time. Will modern humans find it? It seems that geneticists are decoding just about any genome they can get their hands on, so this is a strong possibility. Would it convince people that someone travelled in time? If the bacteria has dispersed enough, shows enough variation over geographic regions, and contains specific identifying information about a missing person who has allegedly created a time machine, I think that's pretty strong evidence. Neato, gang!

Second Place: The second place solution comes from Kyle, who offers a space-based answer. Kyle suggests etching detailed plans of the time travel mechanism (flux capacitor) onto a durable metal and putting that bad boy into space. He suggests that anyone capable of building a fully functional time machine should have no problem launching a small satellite. Fair enough. Additionally, the satellite would use some kind of solar power or the like to produce a low-power radio signal. In fact, this signal would only need to spit something out once every year or ten years or something. Since radio communication precedes space exploration, the detection of an artificial satellite sending a message would attract a fair deal of attention. The plans and successful reproduction of the time machine would then seal the deal.

Does this solution satisfy the necessary conditions? I think so. Assuming all goes according to plan, this would easily be detected by modern people and, assuming the time machine plans are accurate, would provide indisputable proof. My main concern would be that the satellite could be launched and survive to the present. Modern satellites need constant boosts to stay in orbit, without which they fall back onto Earth and burn up. One potential solution would be to put it on the Moon. This is technically much more difficult, but hey, you just created a time machine! Also, putting it on the moon then allows for a totally rad recreation of the Monolith scene in 2001: A Space Odyssey.

Third Place: The third place solution comes from Yariv. Though Yariv did not submit a solution through the proper channels (he follows no one's rules, not even his own), he was overheard to give a solution. While the Physics Challenge planning committee was discussing the problem over lunch, Yariv flippantly dismissed the entire premise as "trivial" and suggested a two-word solution: "radioactive paint." Personally, I like the idea of bewildered archaeologists finding a cave painting of Yariv riding a dinosaur done using a variety of radioactive paints which all date back 200 million years. For this amusement, I award Yariv the third place prize for this contest. As a member of the Virtuosi, however, Yariv is ineligible to receive a prize and instead receives 5 demerits on his record for his willful disregard of our institution's rules and excessive flippancy. One more slip-up and you'll lose your badge!

Full solutions are up on the Challenge website. Thanks for joining us for this episode of Physics Challenge Award Show, and thanks to everyone who submitted a response!

First and Second Prize Winners: We present the following in partial fulfillment of our prize offer.

For those who solve problems (he salutes you)

Betelgeuse, Betelgeuse, Betelgeuse!

2011-11-05T01:25:00.000-04:00

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 Hipparcos satellite 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)

Alright, so the moral of the story is that Betelgeuse is completely harmless to people on Earth. When it does explode, it will be a brilliant supernova that would likely be visible at least a little bit during the day. It will be the coolest thing that anyone alive (if there are people...) will ever see. Sadly, this explosion could take place at just about any time during the next million years. Assuming a uniform distribution over this time period and a human lifetime of order 100 years, there is something like a 1 in 10,000 chance you'll see this in your life.

Feel free to hope for a spectacular astronomical sight, but don't lose sleep worrying about being hurt by Betelgeuse!

Semi-excessive Footnotes:

[1] This has nothing to do with the Kessel Run. For a description of the actual distance unit see Wikipedia. For a circuitous retconning to correct for one throwaway line in Star Wars, see Wookieepedia.

[2] This is how anything heavier than helium gets distributed throughout the universe. The hydrogen and helium formed after the Big Bang gets fused into heavier elements in stars and then dispersed out through supernovae. In fact, most things heavier than Iron actually require supernovae to even exist. If you have any gold on you right now (I'm looking at you Mr. T), that only exists because a star exploded!

[3] Let's consider the case of turning 4 protons into a Helium nucleus. Helium-4 has a binding energy of about 28 MeV, which means that the total energy of a bound He-4 nucleus is 28 MeV less than its free protons and neutrons (in other words, we need to put in 28 MeV to break it up). So the process of turning 4 protons into a Helium nucleus gives off 28 MeV worth of energy. But we had a total of 4 times 1000 MeV worth of matter we could have turned into energy. Thus, the process was 28 MeV/4000 MeV ~ 0.7% efficient at turning matter into energy.

[4] Sometime last year, the United States disclosed that its nuclear arsenal as of Sept 2009 was something like 5000 warheads. Assume these to be Megaton warheads. A Megaton is about 4 * 10^15 J, so the total energy in the US arsenal is about 5000 * 4 * 10^15 J = 2 * 10^19 J.

[5] A fun fact about GRBs: They were discovered by a military satellite looking for illegal nuclear tests, which would emit some gamma-rays. Instead of seeing a signal on Earth, they saw bursts coming from space. I really really hope that someone's first thought was that the Russians were testing nukes on the Moon or something. We must not allow a moon-nuke gap!

[6] We here at the Virtuosi are contractually obligated to only destroy the Earth in our posts on Earth day. I apologize for any inconvenience this may cause.

[7] I am not exactly sure why this is the case. It is certainly observed to be the case and I thought there was a straightforward explanation for why this was the case, but I don't really have a good explanation. Although, maybe there just isn't a good one yet.

[8] For comparison, there is about a 1 in 3000 chance you'll be struck by lightning in your lifetime.

Physics Challenge Update

2011-10-31T22:30:00.001-04:00

Marty McFly realizes he is running out
of time to submit his solution

Did you know that our Physics Challenge problem contest thingy is still up and going? It is! The contest will be open until the end of the day this Friday, November 4th. And, unlike last time, the winning solution will be chosen and posted by the end of the weekend. So even if you don't submit your own solution (though you totally should), check back here Monday morning for the winning entry.

Why should you submit a solution to our problem? Lots of reasons! The top ten reasons as decided by a random sample of me are given below the break.

Top Ten Reasons to Submit a Solution:

1) You will win super-awesome prizes! What kinds of prizes? Well how does a greeting card with kittens on the front and a collection of encouraging haikus from the entire Virtuosi staff written inside sound? It sounds awesome, awesome to the max.

2) You get to show off your totally rad physics skills!

3) You can put it on your resume/cv! The semi-annual Virtuosi Physics Challenge problem winners are held to the same esteem as Fields Medalists and Nobel Prize winners!

4) You will earn the respect and admiration of your peers! Winners of this contest develop an aura that all other people can see, fear, and respect.

5) You will become stronger, faster, and more productive than ever before! This is a scientifically proven fact perhaps.

6) You may gain the ability to talk to animals! Have you ever wanted to discuss espionage-related topics with a platypus (perhaps this platypus)? Of course you have! Winners can!

7) You will be in Presidential company! Did you know William Henry Harrison won the very first Virtuosi physics challenge contest shortly before becoming the ninth President of the United States? Yep! Tippecanoe and so can you!

8) Alemi will salute in your general direction. No questions asked.

9) You will give me work to do on the weekend! This will give my dull and uninteresting life a small glimmer of meaning! Hooray!

10) There's like a 50% chance that you will get three wishes from a magical genie named David Bowie (no relation). Seriously. Like 50%.

That should get you properly motivated! So check out the Challenge Problem website and when you have a solution, send it to our email address (given up in the sidebar).

Good luck!

The Linear Theory of Battleship

2011-10-03T00:48:00.002-04:00

Recently I set out to hold a Battleship programming tournament here among some of the undergraduates. Naturally, I myself wanted to win. So, I got to thinking about the game, and developed what I like to call "the linear theory of battleship".

A demonstration of the fruits of my efforts can be found here. Below, my aim is to guide you through how I developed this theory, as an exercise in using physics to solve an interesting unknown problem.

This is one of the things I really love about physics, the fact that obtaining an education in physics is essentially an education in reasoning and thinking through complicated problems, along with an honestly short list of tips and tricks that have proven successful for tackling a wide range of problems.

So, how do we develop the linear theory of battleship? First we need to quantify what we know, and what we want to know.

The Goal

So, how does one win Battleship? Since the game is about sinking your opponents ships before they can sink yours, it would seem that a good strategy would be to try to maximize your probability of getting a hit every turn. Or, if we knew the probabilities of there being a hit on every square, we could guess each square with that probability, to keep things a little random.

So, let's try to represent what we are after. We are after a whole set of numbers
\[ P_{i,\alpha} \]
where i ranges from 0 to 99 and denotes a particular square on the board, and alpha can take the values C,B,S,D,P for carrier, battleship, submarine, destroyer, and patrol boat respectively. This matrix should tell us the probability of there being the given ship on the given square. E.g.
\[ P_{53,B} \]
would be the probability of there being a battleship on the 53rd square.

If we had such a matrix, we could figure out the probability of there being at hit on every square by summing over all of the ships we have left, i.e.
\[ P_i = \sum_{\text{ships left}} P_{i, \alpha } \]

The Background

Alright, we seem to have a goal in mind, now we need to quantify what we have to work with. Minimally, we should try to measure the probabilities for the ships to be on each square given a random board configuration. Let's codify that information in another matrix
\[ B_{i,\alpha} \]
where B stands for 'background', i runs from 0 to 99, and alpha is either C,B,S,D, or P again, and stands for a ship. This matrix should tell us the probability of a particular ship being on a particular spot on the board assuming our opponent generated a completely random board.

This is something we can measure. In fact, I wrote a little code to generate random Battleship boards, and counted where each of the ships appeared. I did this billions of times to get good statistics, and what I ended up with is a little interesting. You can see the results for yourself over at my
results exploration page by changing the radio buttons for the ship you are interested in, but I have some screen caps below. Click on any of them to embiggen.

All

First of all, lets look at the sum of all of the ship probabilities, so that we have the probability of getting a hit on any square for any ship given a random board configuration, or in our new parlance
\[ B_i = \sum_{\alpha=\{C,B,S,D,P\} } B_{i,\alpha} \]
The results:

shouldn't be too surprising. Notice first that we can see that my statistics are fairly good, because our probabilities look more or less smooth, as they ought to be, and show nice left/right up/down symmetry, which it ought to have.

But as you'll notice, on the whole there is greater probability to get a hit near the center of the board than near the edges, an especially low probability of getting a hit in the corners. Why is that? Well, there are a lot more ways to lay down a ship such that there is a hit in a center square than there are ways to lay a ship so that it gives a hit in a corner. In fact, for a particular ship there are only two ways to lay it so that it registers a hit in the corner. But, for a particular square in the center, for the Carrier for example there are 5 different ways to lay it horizontally to register a hit, and 5 ways to lay it vertically, or 10 ways total. Neat. We see entropy in action.

Carrier

Next let's look just at the Carrier:

Woah. This time the center is very heavily favored versus the edges. This reflects the fact that the Carrier is a large ship, occupying 5 spaces, basically no matter how you lay it, it is going to have a part that lies near the center.

Battleship

Now for the Battleship:

This is interesting. This time, the most probable squares are not the center ones, but the not quite center ones. Why is that? Well, we saw that for the Carrier, the probability of finding it in the center was very large, and so respectfully, our battleship cannot be in the center as often, as a lot of the time it would collide with the Carrier. Now, this is not because I lay down the Carrier first, my board generation algorithm assigns all of the boards at once, and just weeds out invalid ones, this is a real entropic effect. So here we begin to see some interesting Ship-Ship interactions in our probability distributions. But notice again that on the whole, the battleship should also be found near the center as it is also a large ship.

Sub / Destroyer

Next let's look at the sub / destroyer. First thing to note is that our plot should be the same for both of these ships as they are both the same length.

Here we see an even more pronounced effect near the center. The Subs and Destroyers are 'pushed' out of the center because the Carriers and Battleships like to be there. This is a sort of entropic repulsion.

Patrol Boat

Finally, let's look at the patrol boat:

The patrol boat is a tiny ship. At only two squares long, it can fit in just about anywhere, and so we see it being strongly affected by the affection the other ships have for the center.

Neat stuff. So, we've experimentally measured where we are likely to find all of the battleship ships if we have a completely random board configuration. Already we could use this to make our game play a little more effective, but I think we can do better.

The Info

In fact, as a game of battleship unfolds, we learn a good deal of information about the board. In fact on every turn we get a great deal of information about a particular spot on the board, our guess. Can we incorporate this information into our theory of battleship? Of course we can, but first we need to come up with a good way to represent this information.

I suggest we invent another matrix! Let's call this one
\[ I_{j,\beta} \]
Where I is for 'information', j goes from 0 to 99 and beta marks the kind of information we have about a square, let's let it take the values M,H,C,B,S,D,P, where M means a miss, H means a hit, but we don't know which ship, and CBSDP mark a particular ship hit, which we would know once we sink a ship. This matrix will be a binary one, where for any particular value of j, the elements will all be 0 or 1, with only one 1 sitting at the spot marking our information about the square, if we have any.

That was confusing. What do I mean? Well, let's say its the start of the game and we don't know a darn thing about spot 34 on the board, then I would set
\[ I_{34,M}=I_{34,H}=I_{34,C}=I_{34,B}=I_{34,S}=I_{34,D}=I_{34,P}=0 \]
that is, all of the columns are zero because we don't have any information.
Now let's say we guess spot 34 and are told we missed, now that row of our matrix would be
\[ I_{34,M} = 1 \quad I_{34,H}=I_{34,C}=I_{34,B}=I_{34,S}=I_{34,D}=I_{34,P}=0 \]
so that we put a 1 in the column we know is right,
instead, if we were told it was a hit, but don't know which ship it was:
\[ I_{34,H} = 1 \quad I_{34,M}=I_{34,C}=I_{34,B}=I_{34,S}=I_{34,D}=I_{34,P}=0 \]
and finally, lets say a few turns later we sink our opponents sub, and we know that spot 34 was one of the spots the sub occupied, we would set:
\[ I_{34,S} = 1 \quad I_{34,M}=I_{34,H}=I_{34,C}=I_{34,B}=I_{34,D}=I_{34,P}=0 \]

This may seem like a silly way to codify the information, but I promise it will pay off.

As far as my Battleship Data Explorer goes, you don't have to worry about all this nonsense, instead you can just click on squares to set their information content. Note: shift-clicking will let you cycle through the particular ships, if you just regular click it will let you shuffle between no information, hit, and miss.

The Theory

Alright if we decide to go with my silly way of codifying the information, at this point we have two pieces of data,
\[ B_{i,\alpha} \]
our background probability matrix, and
\[ I_{j,\beta} \]
our information matrix, where what we want is
\[ P_{i,\alpha} \]
the probability matrix. Here is where the linear part comes in. Why don't we adopt the time honored tradition in science of saying that the relationship between all of these things is just a linear one? In matrix language that means we will choose our theory to be

\[ P_{i,\alpha} = B_{i,\alpha} + \sum_{j=[0,..,99],\beta=\{M,H,C,B,S,D,P\}} W_{i,\alpha,j,\beta} I_{j,\beta} \]

Whoa! What the heck is that!? Well, that is my linear theory of battleship. What the equation is trying to say is that I will try to predict the probability of a particular ship being in a particular square by (1) noting the background probability of that being true, and (2) adding up all of the information I have, weighting it by the appropriate factor.

So here, P is our probability matrix, B is our background info matrix, I is our information matrix, and W is our weight matrix, which is supposed to apply the appropriate weights.

That W guy seems like quite the monster. It has four indexes! It does, so let's try to walk through what they all mean. Here:
\[ W_{i,\alpha,j,\beta} \]
is supposed to tell us: "the extra probability of there being ship alpha at location i, given the fact that we have the situation beta going on at location j"

Read that sentence a few times. I'm sorry its confusing, but it is the best way I could come up with explaining W in english. Perhaps a visual would help.

Behold the following: (click to embiggen)

That is a picture of \[ W_{i,C,33,M} \]
that is, that is a picture of the extra probabilities for each square (i is all of them), of there being a carrier, (alpha=C) given that we got a miss (beta=M) on square 33, (j=33).

You'll notice that the fact that we saw a miss affects some of the squares nearby. In fact, knowing that there was a miss on square 33 means that the probability that the carrier will be found on the adjacent squares is a little lower (notice on the scale that the nearby values are negative), because there are now fewer ways the carrier could be on those squares without it overlapping over into square 33.

Let's try another:

That is a picture of
\[ W_{i,S,65,H} \]
that is, it's showing the extra probability of there being a submarine (alpha=S), at each square (i is all of them, since its a picture with 100 squares), given that we registered a hit (beta=H) on square 65 (j=65).

Here you'll notice that since we marked a hit on square 65, it is very likely that we will also get hits on the squares just next to this one, as we could have suspected.

In the end, by assuming our theory has this linear form, the benefit we gain is that by doing the same sort of simulations I did to generate the background information, I can back out what the proper values should be for this W matrix. By doing billions and billions of simulations, I can ask, for any particular set of information, I, what the probabilities are P, and solve for W. Given that the problem is linear, this solving step is particularly easy for me to do.

The Results

In the end, this is exactly what I did. I had my computer create billions of different battleship boards, and figure out what the proper values of B and W should be for every square of the matrix. I put all of those results together in a way that I hope is easy to explore up at the Fancy Battleship Results Page, where you are free to explore all of the results yourself. In fact, the way it's set up, you can even use the Superduper Results Page as a sort of Battleship Cheat Sheet. Have it open while you play a game of battleship, and it will show you the probabilities associated with all of the squares, helping you make your next guess.

I've used the page while playing a few games of battleship online, and have had some success, winning 9 of the 10 games I played against the computer player.

Of course, this linear theory isn't everything...

Why Linear isn't everything

But at the end of the day, we've made a pretty glaring assumption about the game of battleship, namely that all of the information on the board adds in a linear way.

Another way to say that is that in our theory of battleship, we have a principle of superposition.

Another way to say that is that in this theory, what you think is happening in a particular square is just the sum of the results from all of the squares, independent of one another.

Another way to say that is to show it with another picture. Consider the following:

Here, I've specified a bunch of misses, and am asking for the probability of there being a Carrier on all of the positions of the board. If you look in the center of that cluster of misses, especially in the inner left of the bunch, you'll see that the linear theory tells me that there is a small but finite chance that the Carrier is located on those squares. But if you stop to look at the board a little bit, you'll notice that I've arranged the misses such that there is a large swatch of squares in the center of the cluster where the Carrier is strictly forbidden. There is no way it can fit such that it touches a lot of those central squares.

This is an example of the failure of the linear model. All the linear model knows is that in the spots nearby misses there is a lower probability of the ship being there, but what it doesn't know to do is look at the arrangement of misses and check to see whether there is any possible way the ship can fit. This is a nonlinear effect, involving information at more than one square at a time.

It is these kinds of effects that this theory will miss, but as you'll notice, it still does pretty well. Even though it reports a finite positive probability of the Carrier being inside the cluster, the value it reports is a very small one, about 1 percent at most. So the linear theory will have corrections at the 1 percent level or so, but that's pretty good if you ask me.

Summary

And so it is. I've tried to develop a linear theory for the game Battleship, and display the results in a Handy Dandy Data Explorer. I encourage you to play around with the website, use it to win games of Battleship, and in the comments, point out interesting effects, things you think I've missed, or ideas for how to come up with linear theories of other things.

Physics Challenge II: Marty McPhysics

2011-09-05T14:16:00.001-04:00

Doc Brown didn't have a time-travel backup plan.

In light of the incredible success of our last Physics Challenge Problem (we received several responses), we here at the Virtuosi have decided to reinstate what was nominally a "monthly" contest. In addition to a brand-spankin'-new problem (with "prizes", see [1]), we have also tried to make a nice collaborative environment for discussing interesting physics problems and posting your own solutions. So I will discuss our new problem and then I'll throw it over to Alemi to discuss the goal of our new Physics Challenge webpage.

Our last challenge problem (here, with winning solutions) was deemed a bit "unrealistic" by many because it assumes that you were shipwrecked on a desert island with a CRC handbook! What a ridiculous situation! Fear not, dear readers, we have heard your complaints and have adjusted accordingly. Our new problem relies on a far less ridiculous assumption.

Assume you have created a time machine out of a DeLorean [2]. You are sure it will send you back in time and, causality-be-damned, you are ready to give it a test drive. But you are less confident in the machine's ability to take you exactly when you want. Additionally, the positive tachyonic chrono-coupler is a bit finicky - meaning, of course, that there is a non-negligible chance that once you go back in time, you may never be able to come Back to the Present. Thus, there is a chance that your time machine will have worked but no one will ever know!

As an unabashed narcissist, you find this totally unacceptable! But first and foremost, you are a scientist. As a scientist, your challenge is to design a strategy that would allow you to convince your present-day scientific peers that you did travel back in the time even if you are stuck in the past. What could you build/make/leave in the past that would be most likely to convince modern scientists that it had to be from a time-traveler and not just some hoaxer? How long would your strategy be effective? Would it work if you were stuck ten thousand years in the past? What about a million years in the past? What about 100 million?

Though this is an inherently silly problem, we feel that the idea behind it is a very interesting one. What could one motivated person do/build/make that would survive a long time into the future? Also, the winning solution will be submitted as a policy White Paper for all future time-travelers.

So that's the new problem. You can find the "official" posting here. I now cede the floor and my remaining time to the gentleman from Wisconsin.

A sense of community...

So, we've had this blog going for a while now, and we get a decent number of comments on our posts, but felt as though something has been lacking. A certain sense of community. Over the past year, we've done some neat things here, and now it's your turn. Step right up, because we've just launched a google groups: TheVirtuosi where we hope people from all corners of the globe will gather to discuss their kooky physics projects.

Have you recently explored the physics of cats? Launch a discussion. Do you have an idea but need some help investigating it? Create a discussion and query us and all of our fantastic readers. Have an idea for a blog post you want to see? Let us know. Want to take a stab at solving one of the newly minted Physics Challenge problems in public? Go for it!

I know the site is sparse now, but we'd really like to build a little bit of a community here, so please, post away. You all have already bookmarked it I'm sure, but a permanent link has been added to the top of the sidebar as well. That is all.

[1] These "prizes" will no longer be as exciting as the CRC from last time, for several reasons. First of all, we are all poor grad students and have no money. Secondly, the winners from last time lived as far away from us as physically possible and it's expensive to ship heavy things. So what can you expect to win? Well, you could win a "certificate of accomplishment" which would just be drawn in crayon by one of this. But it will be made with love! Or you could win a floppy disk with a recording of my very own version of the Jurassic Park theme. The surprise is part of the fun (I'm told).

[2] Quoth the most famous chrononaut of our time, Dr. Emmett Brown: "The way I see it, if you're gonna build a time machine into a car, why not do it with some style?"

A Tweet is Worth (at least) 140 Words

2011-08-30T23:49:00.001-04:00

So, I recently read An Introduction to Information Theory: Symbols, Signals and Noise.

It is a very nice popular introduction to Information Theory, a modern scientific pursuit to quantify information started by Claude Shannon in 1948.

This got me thinking. Increasingly, people try to hold conversations on Twitter, where posts are limited to 140 characters. Just how much information could you convey in 140 characters?

After some coding and investigation, I created this, an experimental twitter English compression algorithm capable of compressing around 140 words into 140 characters.

So, what's the story? Warning: It's a bit of a story, the juicy bits are at the end.

UPDATE: Tomo in the comments below made a chrome extension for the algorithm

Entropy

Ultimately, we need some way to assess how much information is contained in a signal. What does it mean for a signal to contain information anyway? Is 'this is a test of twitter compression.' more meaningful than '歒堙丁顜善咮旮呂'? The first is understandable by any english speaker, and requires 38 characters. You might think the second is meaningful to a speaker of chinese, but I'm fairly certain it is gibberish, and takes 8 characters. But, the thing is if you put those 8 characters into the bottom form here, you'll recover the first. So, in some sense to the messages are equivalent. They contain the same amount of information.

Shannon tried to quantify how we could estimate just how much information any message contains. Of course it would be very hard to try to track down every intelligent being in the universe and ask them if any particular message had any meaning to them. Instead, Shannon reserved himself to trying to quantify how much information was contained in a message produced by a random source. In this regard, the question of how much information a message contains becomes a more tractable question: How unlike is a particular message from all other messages produced by the same random source?

This question might sound a little familiar. It is similar to a question that comes up a lot in Statistical Physics, where we are interested in just how unlike a particular configuration of a system is from all possible configurations of a system. In Statistical physics, the quantity that helps us answer questions like this is the Entropy, where the entropy is defined as
\[ S = -\sum_i p_i \log p_i \]
where p_i stands for the probability of a particular configuration, and we are supposed to sum over all possible configurations of the system.

Similarly, for our random message source, we can define the entropy in exactly the same way, but for convenience, let's replace the logarithm with the logarithm base 2.

\[ S = -\sum_i p_i \log_2 p_i \]
At this point, the Shannon Entropy, or Information Entropy takes on a real quantitative meaning. It reflects how many bits of information the message source produces per character.

The result of all of this aligns quite well with intuition. If we have a source that outputs two symbols 0 or 1 randomly, each with probability 1/2. The shannon entropy comes out to be 1, meaning each of the symbols of our source is worth one bit, which we already new. If instead of two symbols, our source can output 16 symbols, 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F say, the shannon entropy comes out to be 4 bits per symbol, which again we should have suspected since with four bits we can count up to the number 16 in base 2 (e.g. 0000 - 0, 0001 - 1, 0010 - 2 , etc ).

Where it begins to get interesting is when all of our symbols don't occur with equal probability. To get a sense of this situation, I'll show 5 example outputs:

'000001000100000000010000010000'
'000000000010000000000001000000'
'010100000000000000000000111000'
'010100000000000000000000111000'
'000000000100000000110000000010'

looking at these examples, it begins to become clear that since we have a lot more zeros than ones, each of these messages contain less information than the case when 0 and 1 occur with equal probability.

In fact, in this case, if 0 occurs 90% of the time, and 1 occurs 10% of the time, the shannon entropy comes out to be 0.47. Meaning each symbol is worth just less than half a bit. We should expect our messages in this case to have to be about twice as long to encode the same amount of information.

In an extreme example, imagine you were trying to transmit a message to someone in binary, but for some reason, your device had a sticky 0 key so that every time you pushed 0, it transmitted 0 10 times in a row. It should be clear in this case that as far as the receiver is concerned, this is not a very efficient transmission scheme.

English

What does this have to do with anything? Well, all of that and I really only wanted to build up a fact you already know. The fact is, the English language is not very efficient on a per symbol basis. For example, I'm sure everyone knows exactly what word will come at the end of this _______. There you go, I was able to express exactly the same thought with at least 8 fewer characters. n fct, w cn d _ lt bttr [in fact, we can do a lot better], using 22 characters to express a thought that normally takes 31 characters.

In fact, Shannon has a nice paper where he attempted to measure the entropy of the english language itself. Using more sophisticated methods, he concludes that english has an information entropy of between 0.6 and 1.3 bits per character, let's call it 1 bit per character. Whereas, if each of the 27 symbols (26 letters + space) we commonly use each showed up equally frequently, we would have 4.75 bits per character possible.

Of course, from a practical communication standpoint, having redundancies in human language can be a useful thing, as it allows us to still understand one another even over noisy phone lines and with very bad handwriting.

But, with modern computers and faithful transmission of information, we really ought to be able to do better.

Twitter

This brings me back to twitter. If you are unaware, twitter allows users to post short, 140 character messages for the rest of the world to enjoy. 140 characters is not a lot to go on. Assuming 4.5 characters per word, this means that in traditionally written english you're lucky to fit 25 words in a standard tweet.

But, we know now that we can do better. In fact, if we could come up with some kind of crazy scheme to compress english in such a way as to use each of the 27 usual characters so that each of those characters appeared with roughly equal probability, we've seen that we could get 4.75 bits per character, with 140 characters and 5.5 symbols per word, this would allow us to fit not 25 words in a tweet but 120 words. A factor of 4.8 improvement.

Of course we would have to discover this miraculous encryption transformation. Which to my knowledge remains undiscovered.

But, we can do better. It turns out that twitter allows you to use Unicode characters in your tweets. Beyond enabling you to talk about Lagrangians (ℒ) and play cards (♣), it enables international communication, by including foreign alphabets.

So, in fact we don't need to limit ourselves to the 27 commonly used English symbols. We could use a much larger alphabet, Chinese say. I choose Chinese because there are over 20,900 Chinese alphabet symbols in Unicode. Using all of these characters, we could theoretically encode 14.3 bits of information per character, with 140 characters, and 1 bit per English character, and 5.5 symbols per English word, we could theoretically fit over 365 English words in a single tweet. But alas, we would have to discover some magical encoding algorithm that could map typed English to the Chinese alphabet such that each of the Chinese symbols occurred with equal probability.

I wasn't able to do that well, but I did make an attempt.

My Attempt

So, I tried to compress the English language, an design an effective mapping from written English to the Chinese character set of Unicode. We know that we aim to have each of these Chinese characters occur with equal probability, so my algorithm was quite simple. Let's just look at a bunch of English and see which pair of characters occur with the highest probability, and map these to the first Chinese character in the Unicode set. Replace their occurring in the text, rinse, and repeat. This technique is guaranteed to reduce the probability at which the most common character occurs at every step, by taking some if its occurrences and replacing them, so it at least aims to achieve our ultimate goal. That's it. Of course, I tried to bootstrap the algorithm a little bit by first mapping the most common 1500 words to their own symbols.

For example, consider the first stanza of the Raven by Edger Allen Poe

Once upon a midnight dreary, while I pondered, weak and weary,
Over many a quaint and curious volume of forgotten lore--
While I nodded, nearly napping, suddenly there came a tapping,
As of some one gently rapping, rapping at my chamber door.
"'Tis some visiter," I muttered, "tapping at my chamber door--
                     Only this and nothing more."

The most common character is ' ' (the space). The most common pair is 'e ' (e followed by space), so let's replace 'e ' with the first Chinese Unicode character '一' we obtain:

Onc一upon a midnight dreary, whil一I pondered, weak and weary,
Over many a quaint and curious volum一of forgotten lore--
Whil一I nodded, nearly napping, suddenly ther一cam一a tapping,
As of som一on一gently rapping, rapping at my chamber door.
"'Tis som一visiter," I muttered, "tapping at my chamber door--
                     Only this and nothing more.'

So we've reduced the number of spaces a bit. Doing one more step, now the most common pair of characters is 'in', which we replace by '丁' obtaining:

Onc一upon a midnight dreary, whil一I pondered, weak and weary,
Over many a qua丁t and curious volum一of forgotten lore--
Whil一I nodded, nearly napp丁g, suddenly ther一cam一a tapp丁g,
As of som一on一gently rapp丁g, rapp丁g at my chamber door.
"'Tis som一visiter," I muttered, "tapp丁g at my chamber door--
                     Only this and noth丁g more.'

etc.

The end results of the effort are demo-ed here. Feel free to play around with it. For the most part, typing some standard English, I seem to be able to get compression ratios around 5 or so. Let me know how it does for you.

I'll leave you with this final message:

儌咹乺悃巄格丌凣亥乄叜

Python code that I used to do the heavy lifting is available as a gist.

Futurama Physics

2011-08-22T23:44:00.000-04:00

The rotting corpses of sunbeams cause global warming.

Good news, everyone! While rummaging through all my old stuff at home, I found my long-lost copy of Toto IV. Huzzah for me! This is entirely unrelated to what I wanted to talk about, but I have it on good authority that Toto's Africa syncs up really well with this post [1]. I'll tell you when to press play.

Anyway, what I really wanted to talk about was a fairly well-posed problem in Futurama. In the episode "Crimes of the Hot," all of the Earth's robots vent their various "exhausts" into the sky at the same time, using the thrust to push the Earth into an orbit slightly further away from the sun. As a result of this new orbit, the year is made longer by "exactly one week." Anything that quantitative is pretty much asking to be analyzed. Let's explore this problem a bit more then, why not?

[ Those wishing to get the full aural experience of this post should press play on their cassette players ... now ]

First, a little background. In this episode, it is learned that all the robots (especially Hedonism Bot) emit the greenhouse gases responsible for Global Warming. The previous solution (detailed here) is no longer viable, so it is decided that all robots must be destroyed (especially Hedonism Bot). The disembodied head of Richard Nixon rounds up all the world's robots on the Galapagos [2] to have a "party" so that they may be destroyed by a giant space-based electromagnetic pulse cannon. In a last-ditch effort to save the robots, Professor Farnsworth has all the robots blast their exhausts into the sky, using the thrust to push the Earth into an orbit further away from the sun, thus solving the problem of global warming once and for all. As a result of changing the Earth's orbit, the year is "exactly one week longer."

First Pass Through

Ok, so what can we say about the new orbit if all we know is that its orbital period is exactly one week longer? Well, we know from our good buddy Kepler that the square of the period of a bound orbit is proportional to the cube of its semi-major axis [3], so

\[\tau^2 \propto a^3.\]

We already know the Earth's period (1 year) and semi-major axis (1 AU) before the robo-boost, so we can get rid of the proportionality by writing things in terms of the initial values. In other words,

\[ \left(\frac{\tau}{1~\mbox{yr}}\right)^2=\left(\frac{a}{1~\mbox{AU}}\right)^3.\]

Alright, so we know that our new orbital period is 1 year + 1 week, or since there are 52 weeks in a year, 53/52 years. So our new semi-major axis is

\[ a = \left(\frac{1 + 1/52~\mbox{yr}}{1~\mbox{yr}}\right)^{2/3}\mbox{AU}\approx1.013~\mbox{AU},\]

or a little over 1% larger than it is currently. Fair enough. So would this fix Global Warming for ever and ever? Let's see.

The solar flux at some distance d is given by

\[ S = \frac{L_{\odot}}{4\pi d^2}, \]

where L is the luminosity of the sun. So the ratio of the flux at the new semi-major axis [4] to that before the orbit was changed is

\[ \frac{S}{S_0}=\left(\frac{a_0}{a}\right)^2=\left(\frac{a}{1~\mbox{yr}}\right)^{-2}.\]

OK, so we have the flux, but how do we relate this to temperature? Well, we know that the power radiated by a blackbody of temperature T is given by

\[ P = \sigma A T^4, \]

where sigma is the Stefan-Boltzmann constant, A is the area of the emitting region and T is the temperature. For a blackbody in equilibrium, the power coming in is going to be equal to the power going out. The power coming in is just the solar flux times the cross-section area of Earth, given by

\[ P_{in} = S\times\pi R^2_{\oplus} \]

and the power going out is just that radiated by the Earth as a blackbody

\[ P_{out} =\sigma \times 4\pi R^2_{\oplus}\times T^4. \]

Equating the power in to the power out gives

\[ T = \left( \frac{S}{4\sigma}\right)^{1/4}. \]

Now we can find the ratio of the new average Earth temperature to the temperature before the orbital move. We have

\[ \frac{T}{T_0} =\left( \frac{S}{S_0}\right)^{1/4}=\left( \frac{a_0}{a}\right)^{1/2}\approx0.994\]

If we take the initial average Earth temperature to be something like T = 300 K, then we find a new temperature of T = 298 K. Huzzah, a whole 2 degrees cooler!

That may not sound like a lot, but remember, that's the mean global temperature. Apparently, it only takes a few degrees increase in global average temperatures to make things a bit uncomfortable for people. The IPCC indicates that the average global surface temperature on Earth in the next century is likely to rise by about 1 to 2 degrees under optimistic scenarios or about 3 to 6 degrees under pessimistic scenarios. So the robo-boost option is at least in the right ballpark here. Neat!

How Big Is The Push?

So how big of a push did the robots need to give the Earth to boost it out to this new orbit? Is this possible? Let's find out!

First we need to find out how the Earth's velocity has changed. We can do this by finding the change in energy. The total energy of a bound orbit is given by

\[ E = -\frac{k}{2a}\]

where a is the semi-major axis of the orbit and k = GMm. So the difference in Earth's energy before and after the robo-boost is

\[ E_f - E_0 =-\frac{k}{2a_f}-\left(-\frac{k}{2a_0}\right)=\frac{k}{2a_0}\left(1-\frac{a_0}{a_f}\right).\]

But we also know that Earth's energy in the orbit is given by

\[ E = -\frac{k}{r} + \frac{1}{2}mv^2, \]

so the difference in energy before and after is

\[ E_f - E_0 =\frac{1}{2}m\left(v^2_f-v^2_0\right), \]

where we have found the energies immediately before and immediately after the boost so Earth is pretty much at the same distance from the sun, so the potential energy terms cancel.

Combining our expressions for the change in energy and solving for the final velocity, we find

\[ v_f = \left[\frac{GM_{\odot}}{a_0}\left(1-\frac{a_0}{a_f}\right)+v^2_0\right]^{1/2}.\]

Taking the initial orbital velocity of the Earth to be 30 km/s, we find that the final velocity of the Earth immediately after the robo-boost is

\[ v_f = 30.2~\mbox{km/s}\]

So the robots just need to give a "little" 200 m/s boost to the Earth, right? Well, we are adding velocity vectors here, so it depends on which direction the robots are pushing. The magnitude of the final velocity is given by

\[v_f = \sqrt{\left({\bf v_0}+{\bf \Delta v} \right)^2}=\sqrt{v^2_0+\Delta v^2-2v_0\Delta v \cos{\beta}},\]

where delta v is the boost in velocity caused by the robots and beta is the angle between the initial orbital motion of the Earth and the velocity boost from the robots.

If they wanted to make it slightly easier on themselves, the robots would have boosted the Earth in the direction it was already moving. That would make the cosine beta term equal to one and thus minimize the necessary boost. However, in the show the robots appear to point their exhaust right at the sun (see figure below). This is essentially at a 90 degree angle to the Earth's orbital motion, so the cosine beta term goes to zero in our expression above.

Plugging it all in we find that the magnitude of the robo-boost is

\[\Delta v \approx 3.5~\mbox{km/s}.\]

We see that this is a fair bit larger than the 0.2 km/s needed for a boost parallel to Earth's initial velocity.

Robots blasting from the Galapagos (which now appear to be in China...)

Alright, so how much effort would it take to give that kind of boost to the Earth? We can quantify this effort in terms of a force or in terms of the energy difference. Let's do both.

For the force, we have

\[ F = \frac{\Delta p}{\Delta t}=\frac{M_{\oplus}\Delta v}{\Delta t}.\]

Here we are a little stuck unless we can figure out the duration of the robo-boost. Watching the episode again, the robots are blasting up exhaust for about a minute but then the show cuts to commercial. So we don't really know how long they were pushing. Let's just say an hour, but we'll leave the time in if we want to fiddle with that.

Plugging in numbers, the total force is

\[ F = 6\times10^{24}~\mbox{N}\left(\frac{\Delta t}{1~\mbox{hr}}\right)^{-1}.\]

If this force is spread evenly over the billion robots present [5], then each robot would be applying a force of

\[ F = 6\times10^{15}~\mbox{N},\]

which is roughly equivalent to the force it would take to lift up Mount Everest [6].

That wasn't terribly helpful. Let's look at the work then instead. The total work done by the robots to move the Earth is

\[ W = \frac{1}{2}M_{\oplus}\left(v^2_f - v^2_0\right) \approx 4\times10^{31}~\mbox{J}.\]

Well, that's a large number. Could a billion robots feasibly do that much work? If the robots are each 100 kg, then if the mass of all billion robots were directly converted to energy, we would get

\[ E = mc^2 = 10^9\times10^2~\mbox{kg}\times\left(3\times10^8~\mbox{m/s}\right)^2 \approx 10^{28}~\mbox{J},\]

or less than a thousandth the total energy needed. So it looks unlikely that the robots would be able to push the Earth, but that was to be expected.

Changes to The Orbit

Let's take a look at how the robo-boost affects the entirety of the Earth's new orbit. In our first pass through the problem, we ignored the fact that the shape of the orbit changed and only focused on the new semi-major axis. To see why we must also consider the changes to the "shape" of the orbit, take a look at the figure below.

In the figure, the initial orbit is plotted (black dashed line) as well as two new orbits that each have the appropriate semi-major axis so that the period of revolution is one year and one week. The difference between the two final orbits comes from the robots pushing in different directions. In the blue orbit, the boost was made in a direction radially outward from the Sun (that is, perpendicular to the orbital velocity of the Earth). This is the case shown in the Futurama episode. In the red orbit, the boost was made parallel to the orbital velocity of the Earth. In each case, the boost was applied at the point labeled with an "X."

One thing that jumps out from this figure is that the Earth is always further away from the sun on the red orbit than it was on the initial (dashed black) orbit. But on the blue orbit, the Earth is further away from the Sun than it was initially for only half the orbit. On the other half, the blue orbit would actually make the Earth's temperature higher than it was on the old orbit!

The temperature calculation we made earlier should hold pretty well for the red orbit, since it is essentially a circle. It would be a little more tricky for the blue orbit, as one would need to get a time-averaged value of the flux over the course of the whole orbit. A hundred Quatloos to anyone that does the calculation.

Wrap-Up

So what have we found out here? Well, it seems that there are certain scenarios in which boosting the Earth out to a new orbit with period of 1 year + 1 week could cool the Earth by a few degrees. Granted, we have made some simplifications (the Earth is not a blackbody), but the general idea of the thing should still hold.

I had some fun playing around with this problem and I thought it was neat that there was a good deal of information to get started with from the episode. The Futurama people gave an exact period and at least a visual representation of the direction the robots apply their push. So 600 Quatloos for the writers!

Not Quite As Useless as Usual Footnotes

[1] At least I think it says so if you play this song backwards.

[2] According to the Wikipedia page for the episode, the location of the Galapagos for the party was chosen because the writers felt that it would be most convenient to push the Earth near the equator.

[3] The semi-major axis of an ellipse is half of the longest line cutting through the center of the ellipse. Likewise, the semi-minor axis is half of the shortest line drawn through the center of the ellipse. Check this out for some more fun stuff on ellipses.

[4] This is technically incorrect, since the semi-major axis is measured from the center of the ellipse, but the sun is located on one of the foci. However, this requires information on the eccentricity of the orbit, which we are currently glossing over right now. Our method is then approximate, but becomes exact in the case where both orbits are circles. The effect, however, is minor. At worst, it is semi-minor. Zing!

[5] My source for this billion robots number comes from Professor Farnsworth himself. When it looks like the robots will all be destroyed the Professor says "A billion robot lives are about to be extinguished. Oh, the Jedis are going to feel this one!"

[6] Well, sort of. The height of Everest is ~10^4 m, so this gives a volume of ~10^12 m^3. The density of most metals is around ~10 g/cm^3, which is ~10^4 kg/m^3. This gives a mass of ~10^16 kg. The weight is then ~10^17 N. Each robot exerts a force of ~10^16 N. So not quite, but hey it was the first thing I thought of and it almost worked out so I'm sticking with it!

Fun with an iPhone Accelerometer

2011-08-03T03:24:00.019-04:00

The iPhone 3GS has a built-in accelerometer, the LIS302DL, which is primarily used for detecting device orientation. I wanted to come up with something interesting to do with it, but first I had to see how it did on some basic tests. It turns out that the tests gave really interesting results themselves! A drop test gave clean results and a spring test gave fantastic data; however a pendulum test gave some problems.

You might guess the accelerometer would give a reading of 0 in all axes when the device is sitting on a desk. However, this accelerometer measures "proper acceleration," which essentially is a measure of acceleration relative to free-fall. So the device will read -1 in the z direction (in units where 1 corresponds to 9.8 m/s^2, the acceleration due to gravity at the surface of Earth).

Armed with this knowledge, let's take a look at the drop test:

To perform this test, I stood on the couch which was in my office (before it was taken away from us!), and dropped my phone hopefully into the hands of my officemate. I suspected that the device would read magnitude 1 before dropping, 0 during the drop, and a large spike for the large deceleration when the phone was caught.

As you can see, the results were basically as expected. The purple line shows the magnitude of the acceleration relative to free-fall. Before the drop, the magnitude bounces around 1, which is due to my inability to hold something steadily. The drop occurred near time 12.6, but I wasn't able to move my hand arbitrarily quickly so there's not a sharp drop to 0 magnitude. The phone fell for around 0.4 seconds corresponding to

\[y = \frac{1}{2} g t^2 = \frac{1}{2} (9.8 \frac{m}{s^2})*(0.4 s)^2 = 0.784 m = 2.57 feet \]
As for the spike at 13 seconds, the raw data shows that the catch occurs in
\[ t = 0.02 \pm 0.01 s \]. In order for the device to come to rest in such a short amount of time, there needs to be a large deceleration provided by my officemate's hands.

Now the pendulum test consisted of taping my phone to the bottom of a 20 foot pendulum.

I didn't think enough about this, but the period of a pendulum, assuming we have a small amplitude, is given by:
\[T = 2 \pi \sqrt{\frac{L}{g}}\] which is about 5 seconds. With a relatively small amplitude, the acceleration in the x direction will be small. Basically I'm reaching the limit of the resolution of the acceleration device. It appears that the smallest increment the device can measure is 0.0178 g. This happens to match the specifications from the spec sheet I linked at the top of the page, where they specify a minimum of 0.0162 g, and a typical sensitivity of 0.018 g!

Now we come to the most exciting test, the spring test! Setup: I taped my phone to the end of a spring and let it go. Ok.

Here is the actual acceleration data:

The first thing I see is that the oscillation frequency looks constant, as it should be for a simple harmonic oscillator. There is also a decay which looks exponential! Let's see how well the data fits if we have a frictional term proportional to the velocity of the phone. This gives is a differential equation which looks like this:

\[ m \ddot{x} + F\dot{x} + k x = 0 \]
Now we can plug in an ansatz (educated guess) to solve this equation:

\[ x(t) = A*e^{i b t} \]
\[-b^2 mx(t) + i b Fx(t) + kx(t) = 0\]
\[-m b^2+iFb+k = 0\]
We can solve this equation for b with the quadratic equation:

\[ b = \frac{\sqrt{4km - F^2}}{2m} + i\frac{F}{2m} \equiv \omega + i \gamma \]

where I defined two new constants here. So we see that our ansatz does solve the differential equation. Now we want acceleration, which is the second time derivative of position with respect to time.

\[a(t) \equiv \ddot{x} = -b^2 A e^{ibt} \]

Now are only interested in the real part of this solution, which gives us (adding in a couple of constants to make the solution more general):

\[a(t) = -(\omega^2 - \gamma^2) A e^{-\gamma t} cos(\omega t + \phi) + C \]
Let's redefine the coefficient of this acceleration to make things a little cleaner!

\[a(t) = B e^{-\gamma t} cos(\omega t + \phi) + C \]
Ok, with that math out of the way (for now), we can try to fit this data. I actually used Excel to fit this data using a not-so-well-known tool called Solver. This allows you to maximize or minimize one cell while Excel varies other cells. In this case, I defined a cell which is the Residual Sum of Squares of my fit versus the actual data, and I tell Excel to vary the 5 constants which make the fit! The values jump around for a little while then it gives up when it thinks it converged to a solution. Using this you can fit arbitrary functions, neato!

With this, I come up with the following plot:

\[B = 0.633740943\]
\[\gamma = 0.012097581 \]
\[\omega = 8.599670376 \]
\[\phi = 0.693075811 \]
\[C =-1.004454967 \]
with an R^2 value of 0.968!

At this point it should be noted that if I discretize my smooth fit to have the same resolution (0.0178 g) as the accelerometer, then see what the error is comparing the smooth fit to its own discretization, I get an R^2 of 0.967! This means that there is a decent amount of built-in error to these fits due to discretization on the order of the error we're seeing for our actual fits.

Immediately we can recognize that C should be -1, since this is just a factor relating "free-fall" acceleration to actual iPhone acceleration. If we wanted, we could solve for the ratio of the spring constant to the mass, but I'll leave that as an exercise for the reader.

If you look closely, you can see that the frequency appears to match very well. The two lines don't go out of phase. One problem with the fit is the decay. The beginning and the end of the data are too high compared to the fit, which is a problem. This implies that there is some other kind of friction at work.

Some larger objects or faster moving objects tend to experience a frictional force proportional to the square of the velocity. I don't think my iPhone is large or fast (compared to a plane for example), but I'll try it anyway.

The differential equation is:

\[ m \ddot{x} + F\dot{x}^2 + k x = 0 \]
yikes. This is a tough one because of the velocity squared term. One trick I found here attempts a general solution for a similar equation. They make an approximation in order to solve it, but the approximation is pretty good in our case. Take a look at the paper if you're interested.

The basic idea is to note that the friction term is the only one that affects the energy. So, assuming that the energy losses are small in a cycle, we can look at a small change in energy with respect to a small change in time due to this force term. This gives us an equation which can let us solve for the amplitude as a function of time approximately! Really interesting idea.

So I plugged the following equation into the Excel Solver:

\[a(t) = \frac{A cos(\omega t + \phi)}{\gamma t + 1} + B\]

Here's the fit:

Which uses these values:
\[A = 0.772773705 \]
\[\gamma = 0.029745368 \]
\[\omega = 8.600177692 \]
\[\phi = 0.688610161 \]
\[B = -1.004530009 \]
with an R^2 value of 0.964!

This fit seems to have the opposite effect. The middle of the data is too high compared to the fit, while the beginning and end of the data seems too low. This makes me think that the actual friction terms involved in this problem are possibly a sum of a linear term and a squared term. I don't know how to make progress on that differential equation, so I wasn't able to fit anything. If you try the same trick I mentioned earlier, you run into a problem where you can't separate some variables which you need to separate in the derivation unfortunately.

So there you have it, I wanted to find something neat to do, and I got really cool data from just testing the accelerometer.

Stay tuned for an interesting challenge involving some physical data from my accelerometer!

Physics in Sports: The Fosbury Flop

2011-08-01T01:30:00.023-04:00

Physics has greatly influenced the progress of most sports. There have been continual improvements in equipment for safety or performance as well as improvements in technique. I'd like to talk about some physics in sports over a series of posts. Here I'll talk about a technique improvement in High Jumping, the Fosbury Flop.

The Fosbury Flop came into the High Jumping scene in the 1968 Olympics, where Dick Fosbury used the technique to win the gold medal. The biggest difference between the Flop and previous methods is that the jumper goes over the bar upside down (facing the sky). This allows the jumper to bend their back so that their arms and legs drape below the bar which lowers the center of mass (See the picture above). Here is a video of the Fosbury Flop executed very well.

Let’s assume Dick Fosbury is shaped like a semi-circle as he moves over the bar. The bar is indicated as a red circle, as this is a side view. From this diagram, we can guess his center of mass is probably near the marked 'x', since most of his mass is below the bar.

It is important to recall the definition of center of mass, which is the average location of all of the mass in an object.

\[ \vec{R} = \frac{1}{M} \int \vec{r} dm \]
Note that this is a vector equation, and the integral should be over all of the mass elements. This integral gets easier because I'm going to assume that Dick Fosbury is a constant density semi-circle. This means that

\[ M = C*h \]
where C is a constant equal to the ratio of the mass to the height, and

\[dm = C * dh \].
This is a vector equation, so in principle we need to solve the x integral and the y integral; however, due to the symmetry about the y-axis, the x integral is zero. Finally we'll convert to polar coordinates, leaving us with:

\[ y = \frac{1}{C \pi R} \int_0^\pi R\sin{\theta} C R d\theta = \frac{1}{C \pi R} R (-\cos{\theta}) C R \bigg|_0^\pi = \frac{2R}{\pi} \]
Ok, so this is the y-coordinate of the center of mass of our jumper relative to the bottom of the semi-circle. Now we need to calculate relative to the top of the bar, which is roughly the location of the top of the circle. We just need to subtract from R:

\[ R - \frac{2}{\pi} R = R * (1 - \frac{2}{\pi}) = \frac{h}{\pi} * (1 - \frac{2}{\pi}) \]
Now Dick Fosbury was 1.95m tall, which gives us a distance of 22.6 cm BELOW the bar! Of course he's not a semi-circle, but this isn't a terrible approximation, as you can see from the video linked above. Further, wikipedia mentions that some proficient jumpers can get their center of mass 20 cm below the bar, which matches pretty well with our guess.

A nifty technique in physics is looking at the point-particle system, which allows us to see the underlying motion of a system. If you’re not familiar with this method, you collect any given number of objects and replace them with a single point at the center of mass of the object. We can use energy conservation now for our point-mass instead of the entire body of the jumper.^note

In this case, we can simply deal with the center of mass motion of the jumper. All of my kinetic energy will be converted to gravitational potential energy. Again this is an approximation because some energy is spent on forward motion, as well as the slight twisting motion which I'll ignore.

\[E = \frac{1}{2} mv^2 = mgh\]
Now let’s look at some data. Here is a plot of each world record in the high jump. The blue data show jumps before the Flop, and the red data show records after the Flop. **Note: In 1978, the straddle technique broke the world record, being the only non-flop technique to do so since 1968. Thanks Janne!** The Flop was revealed in 1968, so I’ll assume that all jumps before this year used a method where the center of mass of the jumper was roughly even with the bar, while all jumps after this year used the flop (see the previous note). Clearly something happened just before the Flop came out, and this is something called the Straddle technique.

I want to know the percent difference in the initial energies required, so I will calculate

\[ 100\% * \frac{E_0-E_f}{E_0} = 100\% * \frac{mgh_0-mgh_f}{mgh_0} = 100\% * \frac{h_0-h_f}{h_0} \]
where
\[E_0\] is the initial energy without the force, err the flop, and
\[ E_f \] is the initial energy using the flop.

Since we are using the point-particle system, the gravitational potential energy only cares about the center of mass of the flopper, and we need to know the height of the center of mass for a 2.45m flop, which is the current world record.

This corresponds to a flop center of mass height of 2.25m, which gives us an 8.2% decrease in energy using the flop (versus a method where the center of mass is even with the bar)! The current world record is roughly 20 cm higher than it was when the flop came out. This could be due to athletes getting stronger, but this physics tells us that some of the height increase could have been from the technique change.

To sum up, the high jump competition, along with many other sports, is being exploited by physics!

[note] Here we're relying on the center of mass being equal to something called the center of gravity of the jumper. The center of mass is as defined above. The center of gravity is the average location of the gravitational force on the body. This happens to be the same as the center of mass if you assume we are in a uniform gravitational field, which is essentially true on the surface of the Earth.

Grains of Sand

2011-07-18T11:33:00.000-04:00

Have you ever sat on a beach and wondered how many grains of sand there were? I have, but I may be a special case. Today we're going to take that a step further, and figure out how many grains of sand there are on the entire earth. (Caveat: I'm only going to consider sand above the water level, since I don't have any idea what the composition of the ocean floor is).

I'm going to start by figuring out how much beach there is in the world. If you look at a map of the world, there are four main coasts that run, essentially, a half circumference of the world. We'll say the total length of coast the world has is roughly two circumferences. As an order of magnitude, I would say that the average beach width is 100 m, and the average depth is 10 m. This gives a total beach volume of

\[ (100 m)(10 m)(4 \pi (6500 km) )= 82 km^3\]

That's not a whole lot of volume. Let's think about deserts. The Sahara desert is by far the largest sandy desert in the world. Just as a guess, we'll assume that the rest of the sandy deserts amount to 20% (arbitrary number picked staring at a map) as much area as the Sahara. According to wikipedia the area of the Sahara is 9.4 million km^2. We'll take, to an order of magnitude, that the sand is 100 m deep. 10 m seems to little, and 1 km too much. That amounts to ~1 million km^3 of sand.

We're going to assume that a grain of sand is about 1 mm in radius The volume occupied by a grain of sand is then 1 mm^3. Putting that together with our previous number for the occupied volume gives

\[ \frac{1\cdot 10^6 km^3}{1 mm^3}=\frac{1 \cdot 10^{15}}{1\cdot 10^{-9}}=1\cdot 10^{24}\]

That's a lot of grains of sand.

Addendum:
Carl Sagan is quoted as saying

"The total number of stars in the Universe is larger than all the grains of sand on all the beaches of the planet Earth"

If we just use our beach volume, that gives a total number of grains of sand as ~1*10^20, which is large, but not as large as what we found above. Is that less than the number of stars in in the universe? Well, that's a question for another day (or google), but the answer is, to our best estimate/count, yes.

Lifetime of Liquid Water

2011-07-10T22:07:00.000-04:00

Apologies for the hiatus recently, it's been a busy time (when isn't it). I hope to get back to talking about experiments soon, but for now I wanted to write up a quick problem I thought up a while back. The question is this: how long does a molecule of H2O on earth remain in the liquid state, on average?

I'm going to treat this purely as an order of magnitude problem. I'm also going to have to start with one assumption that is almost certainly inaccurate, but makes things a lot easier. I'm going to assume perfect mixing of all of the water on earth.

Given that assumption, I really only need to figure out two things. The first is how much liquid water there is on earth. The second is now much liquid water leaves the liquid phase each year. Let's start with the total amount of liquid water on earth. This is relatively easy to estimate. I happen to know that about 70% of the earth's surface is covered in water. Most all of that is ocean. To an order of magnitude, the average depth of the ocean must be 1 km, as it is certainly not 100 m or 10 km [1]. For a thin spherically shell, the volume of the shell is roughly

\[ 4 \pi r_e^2 \Delta r \]

where r_e is the radius of the earth. Thus, the total volume of water on the earth is

\[.7*4 \pi r_e^2 (1 km)\]

Now, we need to figure out how much H20 leaves the liquid phase every year. To an order of magnitude, it rains 1 m everywhere on earth each year, it's not .1 m or 10 m [2]. I'm going to ignore any freezing/melting in the ice caps, assuming that is small fraction of the water that leaves the liquid phase each year. Since we have a closed system, all the water that rains must have left the liquid phase. So, on average, the total volume of water that leaves the liquid phase is

\[4 \pi r_e^2 (1 m) \]

Thus, the fraction of liquid water that changes phase per year is

\[ \frac{4 \pi r_e^2 (1m)}{.7*4\pi r_e^2 (1 km)} = .0014 \]

This means that, given my assumption of perfect mixing, in somewhere around 1/.0014 = 700 yr all of the water on earth will have cycled through the vapor phase. Since we're only operating to an order of magnitude, I'll call this 1000 years. This is the answer to our question if every molecule has been in the vapor phase once in 1000 years, then we expect a molecule to stay in the liquid phase for 1000 years

[1] According to wikipedia, this is really about 4 km, so we're underestimating a bit.

[2] According to wikipedia, this is spot on (.99 m on average).

Coriolis Effect on a Home Run

2011-07-03T11:52:00.001-04:00

Citizen's Bank Park

I like baseball. Well, technically, I like ~~laying~~[3] lying on the couch for three hours half-awake eating potato chips and mumbling obscenities at the television. But let's not split hairs here.

Anyway, out of curiosity and in partial atonement for the sins of my past [1] I would now like to do a quick calculation to see how much effect the Coriolis force has on a home-run ball.

The Coriolis force is one of the artificial forces we have to put in if we are going to pretend the Earth is not rotating. For a nice intuitive explanation of the Coriolis force see this post over at Dot Physics.

Let's now consider the following problem. Citizen's Bank Park (home to the Philadelphia Phillies) is oriented such that the line from home plate to the foul pole in left field runs essentially South-North. Imagine now that Ryan Howard hits a hard shot down the third base line (that is, he hits the ball due North). Assuming it is long enough to be a home run, how with the Coriolis force effect the ball's trajectory?

This is a well-posed problem and we could solve it as exactly as we wanted. But please don't make me. It's icky and messy and I don't feel like it. So let's do some dimensional analysis! Hooray for that!

So what are the relevant physical quantities in this problem? Well, we'll certainly need the angular velocity of the Earth and the speed of the baseball. We'll also need the acceleration due to gravity. Alright, so what do we want to get out of this? Well, ideally we'd like to find the distance the ball is displaced from its current trajectory. So is there any way we can combine an angular velocity, linear velocity and acceleration to get a displacement?

Let's see. We can write out the dimensions of each in terms of some length, L, and some time, T. So:

\[ \left[ \Omega \right] = \frac{1}{T} \]

\[ \left[ v \right] = \frac{L}{T} \]

\[ \left[ g \right] = \frac{L}{T^2} \]

where we have used the notation that [some quantity] = units of that quantity. Combining these in a general way gives:

\[ L = \left[ v^{\alpha} \Omega^{\beta} g^{\gamma} \right] = \left( \frac{L}{T}\right)^{\alpha}\left( \frac{1}{T}\right)^{\beta}\left( \frac{L}{T^2}\right)^{\gamma} = L^{\alpha+\gamma} T^{-(\alpha+\beta+2\gamma)}\]
Since we want just want a length scale here, we need:

\[\alpha+\gamma = 1~~~\mbox{and}~~~\alpha+\beta+2\gamma = 0. \]

We can fiddle around with the above two equations to get two new equations that are both functions of alpha. This gives:

\[\beta = \alpha - 2~~~\mbox{and}~~~\gamma = 1 - \alpha. \]

Unfortunately, we have two equations and three unknowns, so we have an infinite number of solutions. I've listed a few of these in the Table below.

Ways of getting a length

At this point, we have taken Math as far as we can. We'll now have to use some physical intuition to narrow down our infinite number of solutions to one. Hot dog!

One way we can choose from these expressions is to see which ones have the correct dependencies on each variable. So let's consider what we would expect to happen to the deflection of our baseball by the Coriolis force if we changed each variable.

What happens if we were to "turn up" the gravity and make g larger? If we make g much larger, then a baseball hit at a given velocity will not be in the air as long. If the ball isn't in the air as long, then it won't have as much time to be deflected. So we would expect the deflection to decrease if we were to increase g. This suggests that g should be in the denominator of our final expression.

What happens if we turn up the velocity of the baseball? If we hit the ball harder, then it will be in the air longer and thus we would expect it to have more time to be deflected. Since increasing the velocity would increase the deflection, we would expect v to be in the numerator.

What happens if we turn up the rotation of the Earth? Well, if the Earth is spinning faster, it's able to rotate more while the ball is in the air. This would result in a greater deflection in the baseball's path. Thus, we would expect this term to be in the numerator.

So, using the above criteria, we have eliminated everything on that table with alpha less than 3 based on physical intuition. Unfortunately, we still have an infinite number of solutions to choose from (i.e. all those with alpha greater than or equal to 3). But, we DO have a candidate for the "simplest" solution available, namely the case where alpha = 3. Since we have exhausted are means of winnowing down our solutions, let's just go with the alpha = 3 case.

Our dimensional analysis expression for the deflection of a baseball is then

\[ \Delta x \sim \frac{v^3 \Omega}{g^2} \]

Plugging in typical values of

\[ v = 50~\mbox{m/s}~~~(110~\mbox{mi/hr}) \]
\[ \Omega = 7 \times 10^{-5}~\mbox{rad/s} \]
\[ g = 9.8~\mbox{m/s}^2 \]
we get
\[ \Delta x \approx 0.1~\mbox{m} = 10~\mbox{cm}. \]

That's all fine and good, but which way does the ball get deflected? Is it fair or foul? Well, remembering that the Coriolis force is given by:

\[ {\bf F} = -2m{\bf \Omega} \times {\bf v} \]

and utilizing Ye Olde Right Hand Rule, we see that a ball hit due north will be deflected to the East. In the case of Citizen's Bank Park, that is fair territory.

But how good is our estimate? Well, I did the full calculation (which you can find here) and found that the deflection due to the Coriolis force is given by

\[ \Delta x =-\frac{4}{3}\frac{\Omega v^3_0}{g^2} \cos \phi \sin^3 \alpha \left[1 -3 \tan \phi \cot \alpha \right] \]

where phi is the latitude and alpha is the launch angle of the ball. We see that this is essentially what we found by dimensional analysis up to that factor of 4/3 and some geometrical terms. Not bad!

Plugging in the same numbers we used before, along with the appropriate latitude and a 45 degree launch angle we find that the ball is deflected by:

\[ \Delta x = 5~\mbox{cm}. \]

For comparison, we note that the diameter of a baseball is 7.5 cm. So in the grand scheme of things, this effect is essentially negligible. [2]

That wraps up the calculation, but I'm certain that many of you are still a little wary of this voodoo calculating style. And you should be! Although dimensional analysis will give you a result with the proper units and will often give you approximately the right scale, it is not perfect.

But, it can be formalized and made rigorous. The rigorous demonstration for dimensional analysis is due to Buckingham and his famous pi-theorem. The original paper can be found behind a pay-wall here and a really nice set of notes can be found here. It's a pretty neat idea and I highly recommend you check it out!

Unnecessary Footnotes:

[1] Once in college I argued with a meteorologist named Dr. Thunder over the direction of the Coriolis force on a golf ball for the better half of the front nine at Penn State's golf course. I was wrong. Moral of the story: don't play golf with meteorologists.

[2] For a counterargument, see Fisk et al. (1975)

[3] Text has been corrected to illustrate our enlightenment by a former English major as to the difference between 'lay' and 'lie' through the following story: 'Once in a college psych class, a young student said "It's too hot. Let's lay down." A mature student, a journalist, asked, "Who's Down?" '

Counting Critters

2011-06-28T00:19:00.000-04:00

This picture allows us to set a lower bound on the number
of creatures that ever lived of ~4.

We recently had a big book sale [1] here in town where books were being sold for about a quarter. Needless to say, I bought far more than I'll probably ever need or read. One of the books I bought was called General Paleontology by A. Brouwer [2].

Anyways, I didn't really make it too far in the book. In fact, I only made it to the first sentence of the second paragraph of the first chapter, when I encountered this line:

"The number of individuals which has populated the Earth since life began is beyond estimation."

Horse feathers, I say! Horse Feathers!

The number of things that ever lived may very well be unknowable, but it's certainly not beyond estimation. So below, Alemi and I each provide an estimate for the total number of creatures that have ever lived on Earth.

We flipped a coin and I lost, so I guess I'll go first. My estimation will be a genuine guess-y kind of estimate that doesn't draw too heavily on too many physical considerations. Instead, I will formulate a series of assumptions and base my final answer on that. So assuming my assumptions are valid, the answer should give a reasonable estimate to the total number of creatures that have ever lived.

My assumptions are as follows:

(1) The number of individuals that have ever lived will be almost entirely dominated by the number of bacteria that have ever lived on Earth. So to leading order, all the life that has ever lived on Earth is bacteria (or something similar).

(2) Life began at some time, T, in the past and immediately spread to all places on Earth. Hey, man, it's the power of geometric progression!

(3) The majority of life is found within h = 1 m from the surface of water. I picked this number since it's roughly (order of magnitude) how far down I can see in really clear water. Most of the life will be photosynthetic and thus need a fair amount of sunlight.

(4) The number density of organisms in water is n ~ 10^5. I have no real justification for this.

(5) The average lifetime of an organism is t ~ 1 hr.

Alright, so if these assumptions are valid (a big if [3]), then the following prediction should be fairly accurate. So the total volume in which these creatures may live is just the shell of the Earth down to about a meter:

\[ V = 4 \pi R_{\oplus}^2 h \]

where R = 6 * 10^6 m is the radius of the Earth. Alright, so the number of creatures at any given moment will be the volume times the number density which I will take to be n ~ 10^5. That will give us the total number of creatures at any given moment. But we want it for all the moments. So I will take the total number of "generations" to be the time life has been around divided by the average lifetime of a given organism. Putting this all together gives

\[N = 4 \pi R_{\oplus}^2 h \times \frac{T}{t} \]

Plugging in some numbers I get:

\[ N \sim 10^{39} \left(\frac{h}{1~\mbox{m}}\right)\left(\frac{n}{10^5~\mbox{cm}^{-3}}\right)\left(\frac{T}{3\times10^9~\mbox{yrs}} \right)\left(\frac{t}{1~\mbox{hr}} \right)^{-1}\]

So for the nominal values I've plugged in, I'll get that about 10^39 creatures have ever lived on Earth [4]. I've left my equation in a dimensionless form above, so if you think my individual estimates are garbage, you can easily plug in your own estimates to see how things change.

Except for the completely arbitrary factor for the average number density of organisms per cubic centimeter of water, I feel alright about this estimate. And I'm fairly confident that the number density will not be off more than about 3 orders of magnitude either high or low. So my final estimate is:

\[ N \sim 10^{39 \pm 3} \]

I promised that there would be two estimates, so I present below in picture form, Alemi's back-of-the-wrapper estimate.

Click for the full Jimmy Johns experience

To explain these scribbles, I now cede the floor (and the mic) to Alemi.

[ SEAMLESS TRANSITION ]

So, when Corky posed this question to me while we ate some tasty sandwiches, another approach came to mind.

Namely, I wanted to try to estimate the number of critters that have ever lived by putting some kind of energy bound on the number.

Ultimately, all critters come from the sun. That is, all life on Earth is only able to exist in so much as it consumes energy, and for almost all life [ignoring the under ocean heat vent guys], the energy they consume, one way or another comes from the sun.

So, let's estimate the number of critters in three parts (1) We need the energy the Earth recieves from the sun. (2) We need to estimate the energy density of life (3) These two things, combined with a characteristic length or volume scale for a critter would enable us to figure out the rate at which the Earth could produce critters. (4) Assuming this rate and a time scale for how long life has been around on the Earth would give us a total number of critters.

Let's begin

(1) Energy from the sun.

Corky and I happen to know that the solar flux on the sun is roughly 1000 W/m^2. Multiplying this by half the surface area of the earth gives us a rough total solar flux

\[ (1000 \text{ W/m}^2) ( 2\pi R_{\oplus}^2) \sim 2 \times 10^{17} \text{ W} \]

(2) Energy density of critters

For this we used the bag of potato chips we had on hand, assuming that all life matter has roughly the same energy content. The bag of chips was 150 calories in a serving size of 28 grams. This and assuming that life forms are the density of water gives us a life energy density

\[ \left( \frac{ 150 \text{ kcal} }{ 28 \text{ g}} \right)\left( \frac{ 1 \text{ g}}{ \text{ cm}^3} \right) \sim 2 \times 10^4 \text{ J/cm}^3 \]

(3) Length scale of critters

We assumed that bacteria are the most abundant life form, so we chose a length scale of 100 microns.

Putting these pieces together gives

\[ \frac{ 2 \times 10^{17} \text{ W} }{ \left( 2 \times 10^4 \text{ J/cm}^3 \right) \left( 100\ \mu\text{m} \right)^3 } \sim 10^{19} \text{ 1/s} \]

Which is our estimated critter creation rate

(4) Time scale for life generation

Finally, we estimate that life creation has been chugging along on earth for about 3 billion years, this gives us our final estimate for the number of critters that have every lived

\[ \left( 10^{19} \text{ 1/s} \right) \left( 3 \times 10^9 \text{ years} \right) \sim 10^{36} \]

So, there we have it. If the Earth was 100% efficient at converting solar energy into life, and that life is characteristically the energy density of a potato chip and the size of a bacteria, we should have had 1 billion billion billion billion critters ever. To make us feel a little better we would like to tack on a 10% efficiency, since we don't actually expect the Earth to be 100% efficient, and because 10% seems to be the rule of thumb efficiency estimation used when it comes to food chains and the like, so our final estimate, motivated purely by physics is

\[ \boxed{ \text{ Total number of critters ever } = 10^{35\pm 3} } \]

This number seems pretty good, and is in general agreement with Corky's earlier method. Notice that the only parameter we are a little worried about the is the length scale, especially because our final answer depends on the inverse cube of this number, so, our error is probably something around 3 orders of magnitude, as before, since an order of magnitude goof in the size would cause 3 orders of magnitude error in the final estimate.

So there you have it. Two not-egregiously-horrible estimates for the total number of critters that have ever lived. All in all, I think that book was a quarter well spent!

Unnecessary footnotes:

[1] This is a bit misleading. They actually sold books of all sizes.

[2] I like to read about paleontology and such just in case I'm ever sent back in time. This way, I'll know what dinosaurs are safe to eat.

[3] Here's a bigger if: if

[4] FUN FACT: The total number of atoms in all the people on Earth is roughly 10^39. A proof of this is left as an exercise for the reader.

Day in the life of Clicky

2011-06-01T16:34:00.030-04:00

Remember when you first learned about other planets and their many fun facts? You were probably bombarded by such truisms as: "Jupiter is approximately the mass of 318 Earths, has a orbital period that is 4,300 Earth days, is made out of pure love, and is mostly transparent." Well, I was curious about what Clicky's day was like in terms of Earth days. When did Clicky get to sleep, when did he eat dinner, what is his orbital radius and eccentricity? To do this, I looked at only the 3rd column of the data that you may or may not have downloaded yesterday from here.

Starting out, it should be noted that this is not an ordinary time series where you have some value measured at given intervals. It is actually a series of time points at which a click was entered. I thought it would be most natural to go ahead and bins these to give a sense of the access to Clicky over the past two months. It looks something like this:

Well, that wasn't as informative as I had hoped. We see some spikes here and there with a general trend toward neglect and abandonment towards the end of the second month. What if we take these days and bin them into one single day worth of traffic? We get this:

Again, not so informative yet, but we are definitely starting to see some structure in the day. In particular, there is a lot of activity in the afternoon and evening with a definite lull around 5pm. There is also a distinct minima around 5am. It appears that Clicky is on the average most active between noon and 9pm, getting a break through most of the night.

Next, let's look at the autocorrelation of the times. For standard time series, the autocorrelation is defined to be

\[ C_{ss}(\tau) = \int_{-\infty}^{\infty}s(t) * \bar s(t-\tau) dt \]

which measures the amount of similarity in a signal as a function of the time separation between two points. Again, we don't have a continuous signal, so our autocorrelation function is instead a histogram of the all-pairs differences in the data points that we do have. That is, start at the first time point and subtract it from all of the other time points in the series. Then, move to the next data point and subtract it from all the subsequent times while keeping tracking of all of these differences. This is our autocorrelation in real space.

We can also zoom in and smooth the data a bit

Ah, now there we go. We see a distinct over-abundance of time differences around 1 day, 2 days, etc. What is the primary oscillation we see in the correlation? To see that, let's look at the frequency space autocorrelation and plot its power, or square.

Finally, we get the primary component of the variation of Clicky's visits throughout the 2 month period that he was in operation. It occurs, to within error, at 1 day. No very surprising at all. I'm sure we could look more closely at the peak and its width, but I am satisfied to say that Clicky's day is defined by the Earth day to within a few percent.

*In response to more messages in Clicky, we agree that it is "So slow." Rest assured, management is looking into the problem.

Clicky Data v1.0

2011-05-31T11:36:00.012-04:00

As we sift through the Clicky data Corky presented yesterday, let us not forget the Clicky that came before.

Let us not forget the Great Server Move of 2011 and the great pains we felt while waiting for Clicky to come back to us: Corky sat in his room crying into his pillow, Alex was pressing his arrow keys longingly, and I just couldn't seem to get up in the morning. Let us not forget the many hours in which untold numbers of anonymous internet users tried in vain to spell simple words using only discrete steps on a finite lattice. Let us not forget the Great Server Crash that caused the physics department to be charged extra for data usage. Let us not forget Clicky v1.0.

So today we remember him, our beloved Clicky v1.0, by releasing his data as well. It can be found close to the other Clicky data dump from yesterday at this location:

Download Clicky v1.0

Be warned, this was actually available to the anonymous internet and there are a few things that may surprise you. That being said, the last half of this data set is actually comprised of multiple users interacting with Clicky at once so it should have different statistics than yesterday's data, though we have yet to sort through that too. More of our analysis is to follow..

Also, props to whoever made the dinosaur.

Clickin' the Night Away

2011-05-31T00:07:00.001-04:00

Hey, everybody! Do you remember Clicky [1]? Well, we finally got around to analyzing data, so here goes. But first, a brief summary.

Matt, Alemi and I came up with the idea for Clicky in the beginning of April. Perhaps "idea" is a bit too generous... it was really just a passing thought: "Hey wouldn't it be cool if we had an internet Ouija board?" It was just a stupid lunch-time discussion that wouldn't have gone anywhere had Alemi and Matt not taken it as some sort of challenge. So after a few hours that night we had Clicky.

To say we had some goal with Clicky would be an overstatement. But, if anything, we were kind of hoping to see some sort of Brownian motion. We figured if we had lots of people pulling on the same dot, some kind of Brownian walk would show up. This was grossly overestimating how many people actually view this blog and it turned out that most of the time Clicky was moved by one person at a time. Anyway, what we did end up finding was more interesting than just a Brownian random walk...

Behold, in its full 133,000 point glory, Clicky!

Far View of Clicky. Click to super-size.

Well, I guess that isn't that impressive. But you can click on it for a larger view or download the data and plot it yourself from here. Alternatively, you can view a super-duper large version of the above picture that will almost certainly make your browser sad (seriously, it's big) at his website here.

We note that each step taken by Clicky is 1 unit long, and the above image goes about 2500 on the y-axis and about 5000 in the x-axis. Though we make no explicit comparison between our humble traveler and the great men of lore, we do note that Clicky's long and tortuous path both begins and ends in Ithaca [2].

Now the big picture is all well and good for some folks, but let's zoom in a bit. We'll now zoom into a portion that is about 1000 by 1000 and is located about in the middle of the Clicky map.

Mid-level view of Clicky. Click for a more cromulent view.

So this view is pretty neat. Whereas the previous view appeared largely random, we start to seem some structure here. We can see that some brave soul has made a spiral that, at its biggest, goes for about a hundred squares (remember, you could only see ten squares at a time!). We can also see that most of the steps are small and tend to cluster, but every now and again there is a large jump to uncharted territory.

Neat! Let's zoom in a bit more. Now we will zoom down to about a 100 by 100 square.

Near view. Note the primitive form of communication.

So this is neat. We see some very non-random structures. We see spelled out the phrase "IM IN FIVE-TEN" (Phys 510 is the required graduate physics lab here). This was actually not uncommon. There are lots of little communications that go on throughout the Clicky map. Most are just people marking their territory, but there are some fun ones. If you find anything neat, let us know! (MILD WARNING: As this was open to The Internet, we make no claims that everything written is appropriate, but the worst thing I've seen so far is "butts lol." So I think you're safe).

Now dedication to write something is fine, but how about some real dedication. I found this little Italian gem here, although its means of creation are suspect, to say the least...

It's a Mario!

Hot dog. So is there anything quantitative we can say about the path of Clicky? Sure. Let's take a look at the distribution of step sizes. By step sizes I mean the lengths of continuously straight paths. So if you go right for 5 clicks in a row, the length will be five. Unfortunately, this will not include the lengths of diagonal paths. Anyway, here's what I get:

Power-law fit to Click step-size distribution

The red line here is a maximum likelihood fit to a power law distribution of the form:

\[ p(x) \propto x^{-\mu}. \]

(For an outstanding reference guide to fitting power law distributions see this preprint.)

So it appears as though we have a power law distribution here (but see the paper above!). Well what does that mean? Well it seems we have a roughly random walk path where the step sizes are pulled from a power law distribution. This type of random walk is called a Levy flight (a nice tutorial here) and shows up (or at least appears to) in all kinds foraging patterns in animals (for example, sharks).

To test this, we can simulate a Levy flight on a grid like Clicky. Doing this with the power law found in the above fit gives:

Impostor Clicky!

Not exactly the same, but still looks pretty close!

So that's all for this installment of Virtuosi Theatre, but there's still a whole lot to be analyzed with Clicky. With that much data, you're bound to find something (whether it's there or not!). So if you find something neat, let us know. (Remember the data can be downloaded as a txt file here).

---------------------------------------------

Superfluous Footnotes:

[1] Yes, I know you loved him as Mr. Dottington. I did too! But apparently "the man" thought that was a "lame" name and made it all "commercial" with the buzzname Clicky. So it goes.

[2] Although if Clicky is Odysseus, then I guess that makes you Homer. D'oh! [3]

[3] All my knowledge of "culture" comes from The Simpsons.

Anatomy of an Experiment I - The Question

2011-05-05T21:06:00.000-04:00

Warning: picture has little or no relation to this post.

I realized the other day that I've seen a lot of people talk about research results, but it is much more rare that I see someone talk about how we do research. I think that may be because, to us as scientists, the process is second nature. We've been doing it for years. Other folks may be less familiar with the process though. With this in mind, I'm going to do a short series of posts focused on how we do an experiment. Not the results, not so much the physics, but the process that we go through to create, setup, and carry out an experiment. As my example I'll use a short little experiment that I built from the ground up in the last few weeks, that I'm currently in the process of (hopefully) wrapping up. Today I'll talk about the driving force behind almost any experiment: The Question.

It might be argued that there are two types of experiments. There are those that set out to answer a specific question, and those that set out to explore what happens under certain conditions (explore some part of phase space). An example of the first type that comes immediately to mind is the recently announced results from Gravity Probe B (GP-B). This was an experiment designed with one goal in mind, to test the validity of einstein's theory of general relativity, specifically geodesic precession and frame dragging. They asked the question, built the apparatus, and then got results. Here's a spoiler from the article: Einstein was right, to remarkable precision.

I'm going to mostly ignore the second type of experiment. While very important, I argue that those exploratory experiments are (almost?) always done on experimental apparatus that was built for another experiment. You don't spend the time, money, and energy to build an experimental apparatus without having good evidence that it's worth doing, that is, without expecting to see something.

This brings me to The Question. The name might be misleading, the motivation for an experiment might not be a question (though it can usually be phrased as one). One common motivation is to test theoretical predictions, as was the case with GP-B. Theory without experimental verification is empty. It may sound nice, but we can't trust it unless we've tested it against what nature actually does. Sometimes theory develops because of experimental results, for example the knowledge of the quantization of light came out of anomalous experimental effects of the photoelectric effect and blackbody radiation (among others). Other times, experiment develops to test theory, the GP-B and the Large Hadron Collider for example.

Another common motivation is a question based on a physical observation, for example: how does a fly fly? That question is, as these things go, very simply stated. For an idea of how complicated they can get, just take a look at any recent collection of articles from any physics journal, wherein we find things like the form and source of 'itinerant magnetism in FeAs' (grabbed from a recent Phys. Rev. B article).

I classify a third type of question, one that is more process based: "How can we do X?". This third category is where the question that motivates (at least in the broad sense) the experiment I'm going to describe comes from. I can phrase it as: "How can we successfully cryopreserve biological samples?" For those unfamiliar with biological cryopreservation, this is something I discussed almost a year ago. From there, we get into smaller questions, most of those are type two, based on physical observations.

This particular small experiment has grown out of my work on cryopreservation, and has more to do with the structure of water on freezing. Over the past year, one of the projects I've been working on has been to measure the so called critical warming rate of aqueous solutions. This is the rate at which you have to warm vitreous aqueous solutions (see my earlier cryopreservation post for more details) to prevent ice formation on warming. The question that has grown out of this work is: how does the cooling history of my vitreous sample affect the critical warming rate?

Having arrived at the question, we'll next discuss the apparatus.

End of the Earth VI: Nanobot destruction

2011-04-22T13:11:00.000-04:00

Let's destroy the earth with technology.

A while ago, I read the novel Postsingular by Rudy Rucker, and in the first chapter the Earth gets destroyed, and then undestroyed, and then the novel unfolds and the Earth's likelihood is threatened again, and it looks like the Earth will be destroyed, but it isn't.

How does all of this craziness happen you might ask: nanobots! The story revolves around little self-replicating robots. The story explores what it would be like to live in a world where every surface on Earth was coated in little computers, all of which were networked together. It's certainly a neat idea, but whenever you have self-replicating things, you need to worry a bit about what might happen if they get out of control.

So, let's assume we, evil scientists that we are, have managed to create a little self-replicating nanobot. This little guy can scurry around, running off something ubiquitous, probably some combination of solar, and some kind of infrared photovoltaics. This little guy, call him Bob, his only mission in life is to create a friend. He scurries around collecting the various ingredients necessary, and using his little robot arms, he slices and dices up the pieces and welds them together to create another copy of himself, Rob. Not satisfied with his work; Bob found Rob quite the bore, and honestly Rob didn't too much like Bob either, both of them part ways and try to fashion a new friend.

How long until Bob and Rob and their cohorts manage to chew through all of the material on Earth?

What we have here is the setup to a problem in Exponential Growth.

Exponential Growth

Let's simplify things a bit and assume that the nanobots always take a fixed amount to time to make a new copy of themselves, call that time T. We'll start with one guy, so we know that at t =0, we have 1 bot
\[ N(t =0 ) = 1 \]
And we know that after T seconds we should have 2
\[ N(T) = 2 \]
and after 2T seconds, we've managed to double twice and get 4
\[ N(2T) = 4 \]
after 3T seconds we'll double again to 8, etc. In fact, after nT seconds, so m repetitions we should have doubled m times
\[ N(nT) = 2^n \]

So if we want to describe all times, we need only ask how many doublings can fit into t seconds
\[ t = n T \]
which gives us
\[ N(t) = 2^{t/T} \]

At this point you might object, as this formula doesn't always give an integer, so we could ask things like how many bots are there after 0.5T seconds? We know the true answer is still 1, Bob hasn't finished Rob yet, but our formula tells us the answer is 1.414...

What we've done is made a continuous approximation to a discrete function. Certainly, we've paid a price, in that our new formula doesn't get answers right in fractions of T, but its a small price to pay for the mathematical simplicity afforded by the nice continuous function, and as long as we don't really care about time scales smaller than T in the long run, we haven't done any real harm.

These kinds of approximations show up all over the place in physics, and going both ways too. Sometimes it is advantageous to treat some discrete quantity as continuous, and sometimes it might be beneficial to treat some continuous quantity as discrete. These kinds of approximations are more than adequate, provided you don't really take the answers they give you in the cases where your approximation starts to break too seriously.

In this case, as long as we don't try to seriously predict the number of nanobots to an exact count in time scales less than a fraction of their doubling time, we will have a nice prediction of the number of bots running around.

Earth Destruction

As promised, we wanted to calculate the time it would take the nanobots to devour the earth. For this we need a little bit more to our model. How will the nanobots eat the earth, I reckon it will be through using up its mass. Assuming the bots are made out of elements that are rich enough, something like iron, they ought to have a field day on Earth, seeing as it's composed of about 5% iron on the surface, and with an interior that is probably about 32% iron overall [ref].

So, we need to estimate the mass of a single nanobot. Let's say the nanobot is roughly a 1 micron sized cube, made out of iron. This gives us a nanobot mass of
\[ m = (\text{ density of iron }) * (\text{ 1 micron} )^3 = \rho_{\text{Fe}} L^3 \sim 8 \text{ picograms} \]

From here we can estimate the time it would take to chew through the earth, as the time for the nanobots to be as massive as the earth.

\[ \frac{M_{\oplus}}{\rho_{\text{Fe}} L^3 } = N(t) = 2^{t/T} \]
Solving for t we obtain
\[ t = T \log_2 \frac{ M_{\oplus}}{ \rho_{\text{Fe}} L^3 } \]

Solution

Let's say it takes Bob one month to make Rob, which I don't think is a completely unrealistic time for nanobot replication, assuming Bob and Rob and all of their cohorts are 1 micron in size, I calculate that in 10 years time they would chew through the Earth.

The power of exponential growth! Even with a 1 month gestation, if left unabashed, the self-replicating robots would eat the entire earth in 10 years time. They could eat through Mars in about 2. In fact in Postsingular this is what the humans planned. They wanted a Dyson sphere, so they sent some self-replicating robots to Mars, let them chew through it a couple years, and they had 10^37 little robots to do their bidding. That is of course until the nants set their sites on Earth as their next target...

In order to let you play around with the doubling time and bot size, I've created a Wolfram Alpha widget that solves the above equation, feel free to play around with the parameters and see how long Earth would survive.

The widget should be right above this text. If it isn't working for some reason, here's a link

Earth Day Special: Post-Apocalyptic Literature

2011-04-22T10:30:00.007-04:00

At some point in elementary school I got into the habit of reading Franz Kafka's The Metamorphosis every time that I got sick. I found it strangely comforting to be reminded that while I might have scarlet fever and be intermittently hallucinating about Mickey Mouse, at least I had not been (spoiler alert!) turned into a giant cockroach and disowned by my family.

Today is Earth Day! The earth has seen better days, and I got too depressed googling various environmental problems to even come up with a suitable list of examples. However, look on the bright side: things could be much, much worse. To explore how much worse it could be, here's a few of my favorite works of post-apocalyptic fiction - perfect reading for Earth Day. Skip past the cut to check them out.

In no particular order, here's some of my favorite post-apocalyptic fiction. Many of these are aimed more toward young adults, and since this is a science blog, I've also tried to score them arbitrarily on their scientific plausibility (0-10). Check out the associated amazon pages for better descriptions and reviews.

Z for Zachariah - Robert C. O'Brien - Where's the best place to be when a nuclear war goes down? In a isolated valley in upstate New York, apparently! Z for Zachariah follows a 16 year old girl who is left to fend for herself after the bombs go off, until a Cornell chemistry postdoc shows up in a radiation suit. 7/10
The Postman - David Brin - Another in the post-nuclear war sub-genre. The story gets bogged down in weird survivalist themes in the second half, but paints a rather believable portrait of the aftermath of a nuclear winter. 5/10
A Canticle for Liebowitz - Walter Miller - Have you ever been stuck in a waiting room at the dentist's and the only thing to read is a Reader's Digest from 1983? In the future, it's like that, only way worse. 6/10
Childhood's End - Aurthur C. Clark - Sometimes the end of the world is surprisingly zen. 3/10
A Gift Upon the Shore - M. K. Wren - Rural Oregon also turns out to be a decent place to ride out a nuclear winter. Everything is great, unless your only surviving neighbors are fundamentalists. 7/10
Emergence - David R. Palmer - This is probably my favorite work in this genre. Emergence is the diary of a very plucky Candidia Smith-Foster, who, along with a pet parrot, has survived a communist bio attack. Things get a bit nutty in the end, but overall a very enjoyable read. Despite great reviews it's currently out of print, although a movie may be in the works. 8/10
I am Legend - It turns out that Will Smith was actually playing an older white dude. Who knew? It shares strange religious overtones with the movie, but much better written and with a totally different ending. 4/10
The Pesthouse - Jim Crace - Society has gone a long way backwards, but they hear everything is better in Europe. 5/10
The Road - Cormac McCarthy - I can't say I'm a huge fan of his writing style, but the world that Cormac McCarthy creates here is very compelling, although mind-numbingly depressing. 9/10

This is by no means an exhaustive list - there's a lot of classics that I haven't gotten around to reading yet such as On the Beach. I also have high hopes for Kim Stanley Robinson's The Wild Shore since I enjoyed his Mars series. Anyone else have anything to add? Happy Earth Day!

End of the Earth V: There Goes the Sun

2011-04-22T09:42:00.000-04:00

The Sun [photo courtesy of NASA]

People that know me well know that I have a lot in common with Robert Frost. We both were born in March and we both employ rural New England settings to explore complex social and philosophical themes in our poetry. We also like the same rap groups.

In honor of my literary doppelganger, I will now, having already had the world end in fire, try my hand at ice.

Let's try to answer the question: "If the sun blinks out of existence this instant, what is the temperature of the Earth as a function of time?"

The Sun, in addition to being the King of Planets, is also what keeps us all warm and toasty and alive. What happens if we turn that off? Well, the Earth will cool by radiating its heat away into space. To see how long this would take, let's make some assumptions.

Let's model the surface of the Earth as an ocean 1 km deep and let's pretend that all the heat is stored in this ocean. Let's take the ocean to be liquid water at T = 0 degrees Celsius. How long will it take this ocean to freeze into ice at 0 degrees Celsius?

Well, the amount of energy released from the oceans as the water freezes is given by

\[ Q = L_{w} M_{ocean} \]

where L is the "latent heat of fusion" and M is the mass of the water. The "latent heat of fusion" is a fancy way of saying "the amount of energy released per unit mass as water turns to ice at constant temperature." For water, we have that

\[ L_{w} = 3.3 \times 10^5 \mbox{J/kg} \]

And for the mass of the ocean, it will be convenient later to write it as

\[ M_{ocean} = 4\pi {R_{\oplus}}^2 \Delta R \rho \]

Alright, so now we've got enough to say how much heat energy we have, so how fast do we lose it? We can take the Earth to be a blackbody radiator, so the power lost in such a case is:

\[ P =4\pi \sigma R_{\oplus}^2 T^4 \]

Since Power is just Energy per unit Time, we now have all we need to get the time for total freezing of all the oceans. We have:

\[ t = \frac{Q}{P} = \frac{4\pi {R_{\oplus}}^2 \Delta R \rho L_{w}}{4\pi \sigma R_{\oplus}^2 T^4} \]

Simplifying the above expression a bit, we get

\[ t =\frac{\Delta R \rho L_{w}}{\sigma T^4} \]

Now we can plug in some numbers,

\[ t =\frac{\left(10^3 \mbox{ m}\right) \times \left(10^3 \frac{kg}{m^3}\right) \times \left(3.3 \times 10^5 \mbox{J/kg}\right)}{\left( 5.67 \times 10^{-8} J s^{-1} m^{-2} K^{-4}\right) \times \left( 273 K\right)^4} \]

where we have made sure to put our temperatures in Kelvin. Crunching the numbers with the calculator we "borrowed" from Nic three months ago gives:

\[ t = 10^9 \mbox{ s} \]

Remembering that a year is very nearly

\[ 1 \mbox{ year} = \pi \times 10^7 \mbox{ s}, \]

we find that the time for the oceans to freeze after the sun disappears is about 30 years. Hooray!

Now this model was very simple. First of all, I assumed that the ocean temperatures were already at 0 degrees, but they are a bit warmer. If the oceans are about 300 K (ie 80 degrees in not-Yariv units), then we get another 30 years to cool down to freezing temperatures. Secondly, I have completely neglected the heat stored in the Earth. Will this change my answer by an embarrassingly large factor? Lastly, I have ignored all internal heating mechanisms (ie, radioactive decay) that will also heat up the Earth. But ignoring all that....

So is there a way for anyone to survive this? Well, for the most part it will mean the end of life on Earth. There could potentially be a few exceptions, like by geothermal vents and such. But for the most part, it's one quick cold spiral down to eternal nothingness. But what about a few people, could they survive for a bit even if the human race is doomed?

I'm glad you asked! You see, I have this plan involving mine shafts. Hunkering down underground with a nuclear power plant and all the canned food food we can stomach should allow us to at least ride the rest of our lives. Details can be found here.

End of the Earth IV - Shocking Destruction

2011-04-22T07:54:00.002-04:00

Earth day is upon us once more. So many other namby-pamby bloggers out there (don't hurt me!) are writing about how wonderful the earth is and how great earth day is. We here at The Virtuosi take a more hardline approach. Today I'm going to tell you how to destroy the earth. Completely and totally. Unlike last year's methods, this one should work.

In fact, this method is so simple that I can tell you what to do right now. Just tweak the charge on the electron so it is a bit out of balance with the charge on the proton. Just a little bit. How little a bit, you might ask? A very little bit. Really, this doesn't sound hard. I mean, sure, you have to do it for all of the electrons in the earth, but we're talking about a very very small percentage change. Not convinced? Let me show you just how small a change we're talking.

If there is a charge imbalance in the electron and the proton, this will give the earth a net charge, throughout it's volume. I've got to make a few assumptions about the earth here, so hold on. I'm going to assume that the earth is a uniform density everywhere, and I'm going to assume that the earth is made entirely of iron.* Now, the net charge of any iron atom will be
\[ (q_e-q_p)Z=(q_e-q_p)26\]
where Z is the atomic number of iron, the number of protons (and electrons) the atom has. The net charge of the earth, Q, is the number of iron atoms, N, times this charge,
\[Q=(q_e-q_p)ZN\]
I've previously estimated that N is about 3*10^50 atoms. Now, the electric potential energy of a sphere of radius r with charge q uniformly distributed throughout it's volume is
\[U_e=\frac{3kq^2}{5r}\]
where k is the coulomb constant.

Dissolution of the earth will occur when the electrostatic energy of the earth equals the gravitational potential energy of the earth. The gravitational bound energy of the earth is given by
\[U_g=\frac{3GM^2}{5R}\]
Where M is the mass of the earth, G is Newton's gravitational constant, and R is the radius of the earth. Setting this equal to the electrostatic energy of the earth,
\[\frac{3GM^2}{5R}=\frac{3kQ^2}{5R}\]
\[Q^2=\frac{G}{k}M^2\]
so
\[(q_e-q_p)ZN=\left(\frac{G}{k}\right)^{1/2}M\]
Now, N is given, in our approximations, by
\[N=\frac{M_{earth}}{m_{iron}}\]
so
\[q_e-q_p=\left(\frac{G}{k}\right)^{1/2}\frac{m_{iron}}{Z_{iron}}\]

Now we can plug in some numbers. G=6.7*10^-11 m^3*kg^-1*s^-2, k= 9*10^9 m^3*kg*s^-2*C^-2, m_iron=9*10^-26 kg, Z=26. Thus,

\[q_e-q_p=3*10^{-37} C\]

To put this in perspective, the charge on the electron is 1.6*10^-19 C, so this is roughly 10^18 times less than that charge. Put another way, if the charge on the electron was imbalanced from that of the proton by roughly 1 part in 10^18, the earth would cease to exist due to electrostatic repulsion.**

As I told you at the beginning, you only have to change the charge by a very small amount! So get working. There are only about 1000000000000000000000000000000000000000000000000000 electrons you need to modify!

*According to the internet, the density of the earth, on average, is roughly 5.5 g/cm^3. The density of iron is 7.9 g/cm^3 at room temperature, and the density of water is 1 g/cm^3 at room temperature. So, while the earth is not entirely iron (of course), it is a better approximation to assume the earth is iron than the earth is water. And those, of course, were really our only two choices.

** It turns out that this is a good argument for the charge balance of the electron and the proton.

Matt Raises Clicky From the Dead

2011-04-08T13:52:00.001-04:00

After languishing for four days in the digital hereafter, Clicky has been successfully raised from the dead (thanks to Matt). He can be found at his new (and bandwidth-unrestricted) home here.

For the initial post about Clicky, see this earlier post.

Enjoy!