# Science in Daily Life

### Coincidences and the stealthiness of the Calculus of Probabilities

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You know this story (or something similar) from your own life. I was walking from my parked car to the convenience store to purchase a couple of bottles of sparkling water. As I walked there, I noticed a car with the number 1966 – that’s the year I was born! This must be a coincidence – today must be a lucky day!

There are other coincidences, numerical or otherwise. Carl Sagan, in one of his books mentions a person that thought of his mother the very day she passed away in a different city. He (this person) was convinced this was proof of life after/before/during death.

There are others in the Natural World around us (I will be writing about the “Naturalness” idea in the future) – for eclipse aficionados, there is going to be a total solar eclipse over a third of the United States on the 21st of August 2017. It is a coincidence that the moon is exactly the right size to completely cover the sun (precisely – see eclipse photos from NASA below)

Isn’t is peculiar that the moon is exactly the right size? For instance, the moon has other properties – for instance, the face of the moon that we see is always the same face. Mercury does the same thing with the Sun – it also exhibits the same face to the Sun. This is well understood as a tidal effect of a small object in the gravitational field of a large neighbor. There’s an excellent Wikipedia article about this effect and I well explain it further in the future. But there is no simple explanation for why the moon is the right size for total eclipses. It is not believed to be anything but an astonishing coincidence. After all, we have 6000 odd other visible objects in the sky that aren’t exactly eclipsed by any other satellite, so why should this particular pair matter, except that they provide us much-needed heat and light?

The famous physicist Paul Dirac discovered an interesting numerical coincidence based on some other numerology that another scientist called Eddington was obsessed with. It turns out that a number of (somewhat carefully constructed) ratios are of the same order of magnitude – basically remember the number $10^{40}$!

• The ratio of the electrical and gravitational forces between the electron and the proton ($\frac{1}{4 \pi \epsilon_0} e^2$ vs. $G m_p m_e$) is approximately $10^{40}$
• The ratio of the size of the universe to the electron’s Compton wavelength, which is the de Broglie wavelength of a photon of the same energy as the electron – $10^{27}m \: vs \: 10^{-12} m \: \approx 10^{39}$

On the basis of this astonishing coincidence, Dirac made the startling observation that this could indicate (since the size of the universe is related to it’s age) the value of $G$ would fall with time (why not $e$ going up with time, or something else?). Whereas precision experiments in measuring the value of $G$ are beginning now, there would have been cosmological consequences if $G$ had indeed behaved as $1/t$ in the past! For this reason, people discount this “theory” these days.

I heard of another coincidence recently – the value of the quantity $\frac {c^2}{g}$, where $c$ is the speed of light and $g$ is the acceleration due to gravity at the surface of the earth ($9.81 \frac{m}{s^2}$), is very close to $1$ light year. This implies (if you put together the formulas for $g$, and $1$ light year), a relationship between the masses of the earth, the sun, the earth’s radius and the earth-moon distance!

The question with coincidences of any sort is that it is imperative to separate signal from noise. And this is why some simple examples of probability are useful to consider. Let’s understand this.

If you have two specific people in mind, the probability of both having the same birthday is $\frac{1}{365}$ – there are 365 possibilities for the second person’s birthday and only 1 way for it to match the first person’s birthday.

If, however, you have $N$ people and you ask for $any$ match of birthdays, there are $\frac {N(N-1)}{2}$ pairs of people to consider and you have a substantially higher probability of a match. In fact, the easy way to calculate this is to ask for the probability of $NO$ matches – that is $\frac {364}{365} \times \frac {363}{365} \times ... \frac {365 - (N-1)}{365}$, which is $\frac {364}{365}$ for the first non-match, then $\frac {363}{365}$ from the second non-match for the third person and so on. Then subtracting from 1 gives the probability of at least one match. Among other things, this implies that the chance of at least one match is over 50% (1 in 2) for a bunch of 23 un-connected people (no twins etc). And if you have 60 people, the probability is extremely close to 1.

I have a suggestion if you notice an unexplained coincidence in your life. Figure out if that same coincidence repeats itself in a bit – a week say. You have much stronger grounds for an argument with someone like me if you do! And then you still have to have a coherent theory why it was a real coincidence in the first place!

Addendum: Just to clarify, I write above “It is a coincidence that the moon is exactly the right size to completely cover the sun …” – this is from our point of view, of course. These objects would have radically different sizes when viewed from Jupiter, for instance.

### Arbitrage arguments in Finance and Physics

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Arbitrage refers to a somewhat peculiar and rare situation in the financial world. It is succinctly described as follows. Suppose you start with an initial situation – let’s say you have some money in an ultra-safe bank that earns interest at a certain basic rate $r$. Assume, also, that there is a infinitely liquid market in the world, where you can choose to invest the money in any way you choose. If you can end up with ${\bf {definite}}$ financial outcomes that are quite different, then you have an arbitrage between the two strategies. If so, the way to profit from the situation is to “short” one strategy (the one that makes less) and go “long” the other strategy (the one that makes more). An example of such a method would be to buy a cheaper class of shares and sell “short” an equivalent amount of an expensive class of shares for the same Company that has definitely committed to merge the two classes in a year.

An argument using arbitrage is hard to challenge except when basic assumptions about the market or initial conditions are violated. Hence, in the above example, suppose there was uncertainty about whether the merger of the two classes of shares in a year, the “arbitrage” wouldn’t really be one.

One of the best known arbitrage arguments was invented by Fischer Black, Myron Scholes and Robert Merton to deduce a price for Call and Put Options. Their argument is explained as follows. Suppose you have one interest rate for risk-free investments (the rate $r$ two paragraphs above). Additionally, consider if you, Dear Reader, own a Call Option, with strike price $\X$, on a stock price.  This is an instrument where at the end of (say) one year, you look at the market price of the stock and compute $\S - \X.$ Let’s say $X = \100$, while the stock price was initially $\76$. At the end of the year, suppose the stock price became $\110$, then the difference $\110 - \100 = \10$, so you, Dear Reader and Fortunate-Call-Option-Owner, would make $\10$. On the other hand, if the stock price unfortunately sank to $\55$, then the difference $\55 - \ 100 = - \45$ is negative. In this case, you, unfortunate Reader, would make nothing. A Call Option, therefore, is a way to speculate on the ascent of a stock price above the strike price.

Black-Scholes-Merton wanted to find a formula for the price that you should logically expect to pay for the option. The simplest assumption for the uncertainty in the stock price is to state that $\log S$ follows a random walk. A random walk is the walk of a drunkard that walks on a one-dimensional street and can take each successive step to the front or the back with equal probability. Why $\log S$ and not $S$? That’s because a random walker could end up walking backwards for a long time. If her walk was akin to a stock price, clearly the stock price couldn’t go below 0 – a more natural choice is $\log S$ which goes to $- \infty$ as $S \rightarrow 0$. A random walker is characterized by her step size. The larger the step size, the further she would be expected to be found relative to her starting point after $N$ steps. The step size is called the “volatility” of the stock price.

In addition to an assumption about volatility, B-S-M needed to figure out the “drift” of the stock price. The “drift”, in our example, is akin to a drunkard starting on a slope. In that case, there is an unconscious tendency to drift down-slope. One can model drift by assuming that there isn’t the same probability to move to the right, as to the left.

The problem is, while it is possible to deduce, from uncertainty measures in the market, the “volatility” of the stock, there is no natural reason to prefer one “drift” over the other. Roughly speaking, if you ask people in the market whether IBM will achieve a higher stock price after one year, half will say “Yes”, the other half will say “No”. In addition, the ones that say “Yes” will not agree on exactly by how much it will be up. The same for the “No”-sayers! What to do?

B-S-M came up with a phenomenal argument. It goes as follows. We know, intuitively, that a Call Option (for a stock in one year) should be worth more today if the stock price were higher today (for the same Strike Price) by, say $\1$. Can we find a portfolio that would decline by exactly the same amount if the stock price was up by $\1$. Yes, we can. We could simply “short” that amount of shares in the market. A “short” position is like a position in a negative number of shares. Such a position loses money if the market were to go up. And I could do the same thing every day till the Option expires. I will need to know, every day, from the Option Formula that I have yet to find, a “first-derivative” – how much the Option Value would change for a $\1$ increase in the stock price. But once I do this, I have a portfolio (Option plus this “short” position) that is ${\bf {insensitive}}$ to stock price changes (for small changes).

Now, B-S-M had the ingredients for an arbitrage argument. They said, if such a portfolio definitely could make more than the rate offered by a risk-less bank account, there would be an arbitrage. If the portfolio definitely made more, borrow (from this risk-free bank) the money to buy the option, run the strategy, wait to maturity, return the loan and clear a risk-free profit. If it definitely made less, sell this option, invest the money received in the bank, run the hedging strategy with the opposite sign, wait to maturity, pay off  the Option by withdrawing your bank funds, then pocket your risk-free difference.

This meant that they could assume that the portfolio described by the Option and the Hedge, run in that way, were forced to appreciate at the “risk-free” rate. This was hence a natural choice of the “drift” parameter to use. The price of the Option would actually not depend on it.

If you are a hard-headed options trader, though, the arguments just start here. After all, the running of the above strategy needs markets that are infinitely liquid with infinitesimal “friction” – ability to sell infinite amounts of stock at the same price as at which to buy them. All of these are violated to varying degrees in the real stock market, which is what makes the B-S-M formula of doubtful accuracy. In addition, there are other possible processes (not a simple random-walk) that the quantity $\log S$ might follow. All this contributes to a robust Options market.

An arbitrage argument is akin to an argument by contradiction.

Arguments of the above sort, abound in Physics. Here’s a cute one, due to Hermann Bondi. He was able to use it to deduce that clocks should run slower in a gravitational field. Here goes (this paraphrases a description by the incomparable T. Padmanabhan from his book on General Relativity).

Bondi considered the following sort of apparatus (I have really constructed my own example, but the concept is his).

One photon rushes from the bottom of the apparatus to the top. Let’s assume it has a frequency $\nu_{bottom}$ at the bottom of the apparatus and a frequency $\nu_{top}$ at the top. In our current unenlightened state of mind, we think these will be the same frequency. Once the photon reaches the top, it strikes a target and undergoes pair production (photon swerves close to a nucleus and spontaneously produces an electron-positron pair – the nucleus recoils, not in horror, but in order to conserve energy and momentum). Let’s assume the photon is rather close to the mass of the electron-positron pair, so the pair are rather slow moving afterwards.

Once the electron and positron are produced (each with momentum of magnitude $p_{top}$), they experience a strong magnetic field (in the picture, it points out of the paper). The law that describes the interaction between a charge and a magnetic field is called the Lorentz Force Law. It causes the (positively charged) positron to curve to the right, the (negatively charged) electron to curve to the left. The two then separately propagate down the apparatus (acquiring a momentum $p_{bottom}$) where they are forced to recombine, into a photon, of exactly the right frequency, which continues the cycle. In particular, writing the energy of the photons in each case.

$h \nu_{top} = 2 \sqrt{(m_e c^2)^2+p_{top}^2 c^2} \approx 2 m_e c^2$

$h \nu_{bottom} = 2 \sqrt{(m_e c^2)^2+p_{bottom}^2 c^2} \approx 2 m_e c^2 + 2 m_e g L$

In the above, $p_{bottom} > p_{top}$, the electrons have slightly higher speed at the bottom than at the top.

We know from the usual descriptions of potential energy and kinetic energy (from high school, hopefully), that the electron and positron pick up energy $m_e g L$ (each) on their path down to the bottom of the apparatus. Now, if the photon doesn’t experience a corresponding loss of energy as it travels from the bottom to the top of the apparatus, we have an arbitrage. We could use this apparatus to generate free energy (read “risk-less profit”) forever. This can’t be – this is nature, not a man-made market! So the change of energy of the photon will be

$h \nu_{bottom} - h \nu_{top} =2 m_e g L \approx h \nu_{top} \frac{g L}{c^2}$

indeed, the frequency of the photon is higher at the bottom of the apparatus than at the top. As photons “climb” out of the depths of the gravitational field, they get red-shifted – their wavelength lengthens/frequency reduces. This formula implies

$\nu_{bottom} \approx \nu_{top} (1 + \frac{g L}{c^2})$

writing this in terms of the gravitational potential due to the earth (mass $M$) at a distance $R$ from its center

$\Phi(R) = - \frac {G M}{R}$

$\nu_{bottom} \approx \nu_{top} (1 + \frac{\Phi(top) - \Phi(bottom)}{c^2})$

so , for a weak gravitational field,

$\nu_{bottom} (1 + \frac{ \Phi(bottom)}{c^2}) \approx \nu_{top} (1 + \frac{\Phi(top)}{c^2})$

On the other time intervals are related to inverse frequencies (we consider the time between successive wave fronts)

$\frac {1}{\Delta t_{bottom} } (1 + \frac{ \Phi(bottom)}{c^2}) \approx \frac {1}{\Delta t_{top}} (1 + \frac{\Phi(top)}{c^2})$

so comparing the time intervals between successive ticks of a clock at the surface of the earth, versus at a point infinitely far away, where the gravitational potential is zero,

$\frac {1}{\Delta t_{R} } (1 + \frac{ \Phi(R)}{c^2}) \approx \frac {1}{\Delta t_{\infty}}$

which means

$\Delta t_{R} = \Delta t_{\infty} (1 + \frac{ \Phi(R)}{c^2})$

The conclusion is that the time between successive ticks of the clock is measured to be much smaller on the surface of the earth vs. far away. Note that $\Phi(R)$ is negative, and the gravitational potential is usually assumed to be zero at infinity. This is the phenomenon of time dilation due to gravity. As an example, the GPS systems are run off clocks on satellites orbiting the earth at a distance of $20,700$ km. The clocks on the earth run slower than clocks on the satellites. In addition, as a smaller effect, the satellites are travelling at a high speed, so special relativity causes their clocks to run a little slower compared to those on the earth. The two effects act in opposite directions. This is the subject of a future post, but the effect, which has been precisely checked, is about 38 $\mu$seconds per day. If we didn’t correct for relativity, our planes would land at incorrect airports etc and we would experience total chaos in transportation.

### The earth is flat – in Cleveland

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I stopped following basketball after Michael Jordan stopped playing for the Bulls – believe it or not, the sport appears to have become the place to believe and practice outlandish theories that might be described (in comparison to the Bulls) as bull****.

There’s a basketball star, that plays for the Cleveland Cavaliers. His name is Kyrie Irving. He believes that the earth is flat. He wishes to leave the Cleveland Cavaliers – but not go away too far, since he might fall off the side of the earth. However, he has inspired a large number of middle-schoolers (none of whom I have had the pleasure of meeting, but apparently they exist) that the earth is flat and that the “round-earthers” are government-conspiracy-inspired, pointy-headed, Russian spies – read this article if you want background. In fact, there is a club called the Flat Earth Society, that has members around the globe, that all believe the earth is flat as a pancake.

It would be really interesting, I thought, if, like my favorite detective – Sherlock Holmes – I decided to write the “Intelligent Person’s Guide to Why the Earth is Round”. I would ask you, dear Skeptical Reader, to use no more than tools readily available, some believable friends who possess phones with cameras and the ability to send and receive pictures by mail or text, as well as not being in the pay of the FSB  (or the North Koreans, who decidedly are trying very hard to check the flat earth theory by sending out ICBMs at increasing distances).

I live  in south New Jersey. At my location, the sun rose today at 5:57 am (you could figure this out by typing it out on Google search or just wake up in time to look for the sun). I have two friends, that live in Denver (Colorado) and Cheyenne (Wyoming). Their sunrises occur at 6:00 am and 5:53 am (their time) – averages to 5:56:30 am roughly. I realize that Denver is a mile high, which is also roughly Cheyenne’s height, but hey, you don’t pick your friends. I also live at an elevation of roughly 98′, which isn’t much and I ignore it. They sent me pictures of when the sun rose and I was able to prove they weren’t lying to me or part of a government conspiracy.

The distance from my town to these places is 1766 miles (to Denver) and 1613 miles (to Cheyenne). I used Google to calculate these, but you could schlep yourself there too. Based on just these facts, I should conclude that the earth curves between New Jersey and those places. To my mind, this should clinch the question of whether the earth is round. Since the  roughly 1700 mile separation equals 2 hours of time difference (in sunrises), a 24-hour time difference corresponds to 20,400 miles. This is roughly equal to 24,000  miles times the cosine of 40 degrees, which is the latitude of both New York City, Denver and Cheyenne (which is the circumference for radius $r$, rather than Earth’s radius $R$). This means $2 \pi R$ (which is the earth’s equatorial circumference) is roughly 26,000 miles, which is close to the correct figure to within $4%$. The extreme height at Denver and Cheyenne has something to do with it! The sun ${\bf should}$ have risen later in Denver and Cheyenne if they had been at lower elevations, so 1700 miles would ${\bf really}$ have corresponded to a few minutes more than 2 hours, which would have meant a lower estimate for the earth’s equatorial circumference.

By the way, I picked Cheyenne because of its auditory resemblance to the town that Eratosthenes picked for his diameter-of-Earth measurement, Syrene in present-day Libya. Yes, the first person to measure the Earth’s diameter was Libyan!

Some objections to these entirely reasonable calculations include – if the earth is actually rotating, why doesn’t it move under you when you go up in a balloon. Sorry, this has been thought of already! When I was young, I was consumed by Yakov Perelman’s “Astronomy for Entertainment” – a book written by the tragically short-lived Soviet popularizer of science who died during the siege of Leningrad (St. Petersburg) in 1942. Perelman wrote about a young, enterprising, French advertising executive/scammer at the turn of the 19th century that dreamed up a new scheme to separate people from their money. He advertised balloon flights that would take you to different parts of the world without moving – just go up in a balloon and stay aloft till your favorite country comes up beneath you. It doesn’t happen because all the stuff around you is moving with you. Why? Its the same reason why the rain drops don’t fly off your side-windows even when you are driving on the road at high speed in the rain – forgetting for a second about the gravitational force that pulls things towards the earth’s center. There is a boundary layer of material that rotates or moves as fast as a moving object – its a consequence of the mechanics of fluids and we live with it in various places. For instance, it is one reason why icing occurs on airplane wings – if there was a terrible force of wind all the time, ice wouldn’t form.

So, if you are willing to listen to reason, no reason to restrict yourself to Cleveland. The world is invitingly round.

Addendum : a rather insightful friend of mine just told me that Kyrie Irving was actually born in Australia on the other side of the Flat Earth. If so, I doubt that even my robust arguments would convince him to globalize his views.