Imagine a 2-dimensional square of side-length 1. A diagonal line drawn between opposite corners has an easily calculable length using Pythagoras’ Theorem:

Of course there are other ways to get from one corner to the other… We could go halfway along one side, then take a ninety-degree turn, go halfway again (to the middle), then repeat to get to the far corner. Or we could take even more steps! Figure 1 shows a couple of these “staircase” options, in addition to the direct diagonal path.

The paradox then goes on that as the number of steps in the staircase (*n*) is increased it converges on the diagonal line. Therefore as *n* approaches infinity the staircase still has length 2, but it also converges infinitely close on the diagonal line.

**Does it Converge?**

From this, we can calculate the total area between the staircase and the diagonal line as:

And now applying limits we see that as the number of steps increases, the area of the difference between the two paths does actually converge on zero, while the distance traversed by each path remains different!

**Same Thing With Circles!**

The length of the *n*=2 red semicircle is:

If these calculations are repeated for *n*=4 and *n*=8:

At this point the paradox argument goes (like in the staircase above) that the semicircles eventually converge on the straight line. Since the straight line has a fixed length of 2 and the circles always have a length of *π*, that therefore *π* = 2!

In fact, the same area-based logic can be applied to this problem as well. First we find a general formula for the total area under the semicircles in terms of the number of semicircles along the line, *n*.

This is actually the same form of equation as in the staircase-case, except that the triangle coefficient in the numerator (^{1}/_{2}) is replaced by the circle coefficient (*π*)! Once again the limit shows that this area converges to zero as we increase the number of semicircles:

**So How Does This Resolve Anything?!**

It doesn’t, I’ve just made it more confusing… But Gödel *did* point out that paradoxes have to exist… For that matter, we can always prove that 0.9999[…] is exactly equal to 1:

The first piece of interest occurs due to the amazingly high accelerations. As everyone learns in high school physics, objects have kinetic energy as a result of their movement. This is

The key word here is of course approximation: this formula is only close to the true kinetic energy at the incredibly low speeds that we are used to on a day to day basis. Einstein’s relativity (specifically the special one) provides a much better calculation that holds as speeds approach the speed of light. It adds the Lorentz Factor to the equation (represented by the greek letter gamma: γ).

Interesting! As *v* approaches *c*(the speed of light in a vacuum), the denominator of gamma approaches zero, so gamma (and hence the kinetic energy) approaches infinity! In plot form, this looks like Figure 1. As you can see, the Newtonian equation is a pretty good approximation until 40-50% of *c* at which point the true kientic energy starts to diverge rapidly. This is also the reason that the speed of light is the ‘universe’s speed limit’: when *v* = *c*, the *v ^{2}*/

So what kind of energy does this really come out to? Lets use the space shuttle as our basic interstellar rocket ship – it may as well be used for something! According to The Only Information Source on the Internet, it has a weight of 2,030 tonnes, or 2,030,000 kg. Accelerating it to 50% of *c* would require 2.82×10^{22} joules of energy (about 28.2 zettajoules). By comparison the total planetwide energy consumption in 2008 was about 4.74×10^{20} joules (474 exajoules). So to accelerate one space shuttle to 50% of *c* we would have to devote every joule of energy used on the planet to the task for about 60 years! To get to 90% of *c* requires 2.36×10^{23} J, or 498 years of annual world energy use! There must be a better way…

**Energy Sources**

But what if we want to accelerate things to exceptionally high speeds? Say 99% of the speed of light, or 99.999% of *c*?! For the sake of clarity, lets re-design our speed-kinetic energy plot a little bit and use an inverse-logarithm plot. In this view, the first 90% of speed is crunched into one logarithmic division, the next 9% of speed in another division, the next 0.9% of speed into another and so on. This is useful as it shows the really high speeds clearly. The y-axis (energy) has also been changed to log-scale so that the massive changes in kinetic energy can be clearly visualized – a straight line on this log-log^{-1} plot is actually an exponential increase in energy. Figure 3 shows this plot with some reference points. The classical newtonian energy is almost flat in this view which is expected: the equation is of order two and thus a small change (percentage-wise) in the velocity results in twice that change in energy consumption. This isn’t a big change when the change in velocity is only 0.01% or 0.0001%. Conversely the relativistic energy shows massive exponential changes in energy for these small changes in velocity.

What if we were happy going 99% of the speed of light? That still gets us to Alpha Centauri (~4.4 ly) in 4 years and 5.3 months… For this feat we need about 1.1×10^{24} J, or about the amount of energy from the sun that reaches the earth in 2.4 months.

**What About The Enterprise?**

One last idea! Anyone whose anybody knows that the USS Enterprise from Star Trek runs on anti-matter. The ideal energy source since it can convert all of its mass into energy when it comes into contact with its regular-matter equivalent. Einstein’s most famous equation shows how much energy we can produce with a 100% efficient anti-matter reactor:

This means that is has an energy density of *c*^{2} J/g, or about 9×10^{16} J/g. That’s amazing! So if our 2,030,000 kg ship was outfitted with an anti-matter reactor and we wanted to accelerate it to, say 99% the speed of light so that it could transition into warp drive, how much mass is needed?

Uh-oh! So the amount of mass needed to produce our accelerating energy is more than the weight of the ship itself! And the situation is made even worse by the fact that an equal amount of regular matter is needed to react with the anti-matter. So 24,720,000 kg of mass are required to accelerate a 2,030,000 kg ship to 99% of the speed of light… doesn’t seem promising!

]]>The population density of Tripoli is given by Wonderful Wikipedia as 4500 people/km^{2} [2]. If we assume that the cross-sectional area of an average human being from above is roughly 0.5 m^{2} and knowing that one square kilometer is composed of 1,000,000 m^{2}, then:

This means that looking straight down on a square kilometer of Tripoli, we would expect 0.225% of the total area to be occupied by bits of “people”. This also means that a random bullet falling down on our imagined piece of Tripoli has a 0.225% chance of hitting part of a person. That is all well and good but a far more interesting question is how many bullets do we have to shoot to have a 50% chance of hitting somebody? or a 90% chance of hitting somebody?

A Bit of Probability

A person needs to roll a six (on a six-sided die) to win a game. How many times would they have to roll the die in order to have a 50% chance of getting a 6? This is analogous to our bullet scenario, but a bit easier to visualize. It turns out that the best way to calculate this is the combine the probabilities of **not** rolling a 6 on each throw. If *P(A)* is the probability of rolling a 6 in *n* throws, and *P(A)’* is the probability of **not** rolling a 6 in *n* throws, then we know that *P(A) = 1 – P(A)’*. We are interested in the number of throws of the dice at which *P(A)’* becomes less than 0.50, where *P(A)’* is the probability of not throwing a 6 on each toss (an independent event):

Then the probability “break-even” point can be calculated as follows:

So this means that 4 rolls are required to have a > 50% chance of rolling one six. We can apply the same reasoning to our bullet problem. The chance (*P(A)*) of a bullet hitting someone is 0.00225, so the individual, independent chance of a bullet **not** hitting someone is 0.99775:

To have a 90% chance of hitting someone (by accident of course!) we would have to shoot off:

While 1,022 is a lot of bullets, if you had a sizeable crowd celebrating, that wouldn’t take long at all! Of course with all this shooting going on, it is quite likely that most people are indoors, but even if 75% of people were hiding indoors (and presumably the bullets were stopped by their roofs), that’s still only 1,232 bullets to have a 50% chance of a hit, or 4,093 bullets for a 90% chance. There is still a lot of room for error in this approximation, but it seems reasonable that there were some accidental hits – I would be very surprised if it came anywhere near the death toll of the conflict though!

*(For those interested, Libya has an average population density of 3.6 per km ^{2}, again from our trusty Wikipedia [3]. This means that if the gunfire and people were spread out over the entire country, the 50% level would be 385,082 bullets and the 90% level would take an astonishing 1,279,213 bullets. The moral of the story is, of course: don’t live in cities!)*

The Quick Physics/Medicine Digression

So now that we’ve established that some unlucky souls have a reasonable chance of having a thoughtlessly fired bullet coming down on them: will it be worse than a bump on the head? The unfortunate answer is yes. Basic physics tells us that a bullet shot up will return to earth at the same speed it left, in the absence of any air resistance. Air resistance however means that the bullet will come down slower, but how much slower? Amazingly, this research has been undertaken and the result ends up being a roughly 90 m/s terminal velocity for a .30 caliber bullet (the size used in an AK-47) [4]. This is more than fast enough to penetrate not only human skin, but also the human skull [5]: a target that is rather unfortunately positioned to intercept bullets falling from the sky.

In fact a case report in *The Annals of Thoracic Surgery* published in 2007 describes a case report of a patient presenting to a Michigan ED with severe injuries from celebratory gunfire [6]. The bullet entered the patient’s chest at the 5th intercostal space about 2 cm left of the sternal margin (see figure at left). The bullet then passed downwards through the right ventricle of the heart, through the diaphragm, and through the lesser curvature of the stomach. The bullet then came to rest inside the stomach but not before impacting the greater curvature of the stomach and injuring the splenic hilum (requiring a splenectomy). The bullet is clearly visible in the left upper quadrant in the pyelogram shown below.

So What!?

So there you have it! While it is unlikely that celebratory gunfire associated with the death of Colonel Gadhafi caused deaths to rival the death toll of the conflict (estimates in excess of 10,000 people are being bandied about), it is absolutely likely that some unfortunate innocent bystanders were hit and seriously injured or killed. While the terminal velocity of a bullet falling from the sky is much lower than the muzzle velocity of a typical assault rifle, the speed is more than enough to maim or kill.

References

[1] Martin Kennedy, “Letter: Careless shooting may have marred Libyan celebration“, The Calgary Herald. October 22, 2011.

[2] “Tripoli“, Wikipedia. Retrieved October 22, 2011.

[3] “Libya“, Wikipedia. Retrieved October 22, 2011.

[4] Julian Hatcher, “Hatcher’s Notebook”, p. 154, Stackpole Books. 1962.

[5] Michael J Stewart, “Head, Face and Neck Trauma: Comprehensive Management”, p. 189, Thieme Medical Publishers. 2005.

[6] Angelo Incorvaia MD, Despina Poulos MPH, Robert Jones MD, James Tschirhart MD, “Can a Falling Bullet Be Lethal at Terminal Velocity? Cardiac Injury Caused by a Celebratory Bullet“, The Annals of Thoracic Surgery, Vol. 83 No. 1, p. 283-284. January 2007.