# Category Archives: Fun

## The Physicist’s Proof II: Limits and the Monty Hall Problem

As an undergraduate, I was taught the concept of the “physicist’s proof”, a sort of silly concept that was a professor’s attempt to get us students to think a little harder about some problems. Here, I give you a “physicist’s proof” of the famous Monty Hall problem, which (to me!) is easier to think about than the typical Bayesian approach.

The Monty Hall problem, which was developed on a TV game show, goes something like this (if you already know the Monty Hall problem, you can skip the paragraphs in italics):

Suppose you are a contestant on a game show where there are three doors and a car behind one of them. You must select the correct door to win the car.

You therefore select one of the three doors. Now, the host of the show, who knows where the car is, opens a different door for you and shows you that there is no car behind that door.

There are two remaining unopened doors — the one you have chosen and one other. Now, before you find out whether or not you have guessed correctly, the host gives you the option to change your selection from the door you initially chose to the other remaining unopened door.

Should you switch or should you remain with you initial selection?

When I first heard this problem, I remember thinking, like most people, that there was a 50/50 chance of the car being behind either door. However, there is a way to convince yourself that this is not so. This is by taking the limit of a large number of doors. I’ll explain what I mean in a second, but let me just emphasize that taking limits is a common and important technique that physicists must master to think about problems in general.

In Einstein’s book, Relativity, he describes using this kind of thinking to point out absurd consequences of Galilean relativity. Einstein imagined himself running away from a clock at the speed of light: in this scenario, the light from the clock would be matching his pace and he would therefore observe the hands of the clock remaining stationary and time standing still. Were he able to run just a little bit faster than the light emanating from the clock, he would see the hands of the clock start to wind backwards. This would enable him to violate causality!  However, Einstein held causality to be a dearer physical truth than Newton’s laws. Special relativity was Einstein’s answer to this contradiction, a conclusion he reached by considering a physical limit.

Now, let us return to the Monty Hall problem. And this time, instead of three doors, let’s think about the limit of, say, a million doors. In this scenario, suppose that you have to choose one door from one million doors instead of just three. For the sake of argument, suppose you select door number 999,983. The host, who knows where the car is, opens all the other doors, apart from door number 17. Should you stick to door 999,983 or should you switch to door 17?

Let’s think about this for a second — there are two scenarios. Either you were correct on your first guess and the car is behind door 999,983 or you were incorrect on your first guess and the car is behind door 17. When you initially made your selection, the chance of you having made the right guess was 1/1,000,000! The probability of you having chosen the right door is almost zero! If you had chosen any other door apart from door 17, you would have been faced with the same option: the door you chose vs. door 17. And there are 999,999 doors for you to select and not win the car. In some sense, by opening all the other doors, the host is basically telling you that the car is behind door 17 (there is a 99.9999% chance!).

To me, at least, the million door scenario demonstrates quite emphatically that switching from your initial choice is more logical. For some reason, the three door case appears to be more psychologically challenging, and the probabilities are not as obvious. Taking the appropriate limit of the Monty Hall problem is therefore (at least to me) much more intuitive!

Especially for those who are soon to take the physics GRE — remember to take appropriate limits, this will often eliminate at least a couple answers!

For completeness, I show below the more rigorous Bayesian method for the three-door case:

Bayes theorem says that:

$P(A|B) = \frac{P(B|A) P(A)}{P(B)}$

For the sake of argument, suppose that you select door 3. The host then shows you that there is no car behind door 2. The calculation goes something like this. Below, “car3” translates to “the car was behind door 3” and “opened2” translates to “the host opened door 2”

$P(car3|opened2) = \frac{P(opened2 | car3) P(car3)}{P(opened2)}$

The probabilities in the numerator are easy to obtain $P(opened2 | car3) = 1/2$ and $P(car3) = 1/3$. However, the $P(opened2)$ is a little harder to calculate. It helps to enumerate all the scenarios. Given that you have chosen door three, if the car is behind door 1, then the probability that the host opens door two is 1. Given that you have chosen door three and are correct, the probability of the host choosing door 2 is 1/2. Obviously, the probability of the car being behind door 2 is zero. Therefore, considering that all doors have a 1/3 possibility of having the car behind them at the outset, the denominator becomes:

$P(opened2) = 1/3*(1/2 + 1 + 0) = 1/2$

and hence:

$P(car3|opened2) = \frac{1/2*1/3}{1/2} = 1/3$.

Likewise, the probability that the car is behind door 1 is:

$P(car1|opened2) = \frac{P(opened2 | car1) P(car1)}{P(opened2)}$

which can similarly be calculated:

$P(car1|opened2) = \frac{1*1/3}{1/2} = 2/3$.

It is a bizarre answer, but Bayesian results often are.

## Spot the Difference

A little while ago, I wrote a blog post concerning autostereograms, more commonly referred to as Magic Eye images. These are images that, at first sight, seem to possess nothing but a random-seeming pattern. However, looked at in a certain way, a three-dimensional image can actually be made visible. Below is an example of a such an image (taken from Wikipedia):

Autostereogram of a shark

In my previous post about these stereograms, I pointed out that the best way to understand what is going on is to look at a two-image stereogram (see below). Here, the left eye looks at the left image while the right eye looks at the right image, and the brain is tricked into triangulating a distance because the two images are almost the same. The only difference is that part of the image has been displaced horizontally, which makes that part appear like it is at a different depth. This is explained at the bottom of this page, and an example is shown below:

Boring old square

In this post, however, I would like to point out that this visual technique can be used to solve a different kind of puzzle. When I was in middle school, one of the most popular games to play was called Photo-Hunt, essentially a spot-the-difference puzzle. You probably know what I’m referring to, but here is an example just in case you don’t:

The bizarre thing about these images is that if you look at them like you would a Magic Eye image, the differences between the two images essentially “pop out” (or rather they flicker noticeably). Because each of your eyes is looking at each image separately, your brain is tricked into thinking there is a single image at a certain depth. Therefore, the differences reveal themselves, because while the parts of the image that are identical are viewed with a particular depth of view, the differences don’t have the same effect. Your eyes cannot triangulate the differences, and they appear to flicker. I wish I had learned this trick in middle school, when this game was all the rage.

While this may all seem a little silly, I noticed recently while zoning out during a rather dry seminar, that I could notice very minute defects in TEM images using this technique. Here is an example of an image of a bubble raft (there are some really cool videos of bubble rafts online — see here for instance), where the defects immediately emerge when viewed stereoscopically (i.e. like a Magic-Eye):

Bubble raft image taken from here

I won’t tell you where the defects are, but just to let you know that there are three quite major ones, which are the ones I’m referring to in the image. They’re quite obvious even if not viewed stereoscopically.

Because so many concepts in solid state physics depend on crystal symmetries and periodicity, I can foresee entertaining myself during many more dry seminars in the future, be it a seminar with tons of TEM images or a wealth of diffraction data. I have even started viewing my own data this way to see if anything immediately jumps out, without any luck so far, but I suspect it is only a matter of time before I see something useful.

## An Excellent Intro To Physical Science

On a recent plane ride, I was able to catch an episode of the new PBS series Genius by Stephen Hawking. I was surprised by the quality of the show and in particular, its emphasis on experiment. Usually, shows like this fall into the trap of giving one the facts (or speculations) without an adequate explanation of how scientists come to such conclusions. However, this one is a little different and there is a large emphasis on experiment, which, at least to me, is much more inspirational.

Here is the episode I watched on the plane:

## Holiday Puzzle

During the holidays, I’ve spent some time on puzzles like Sudoku, Kakuro and crosswords. In this spirit, I made a condensed matter themed crossword puzzle for you to enjoy (click to enlarge and print). Happy holidays!

## Too Close to Home

I haven’t been blogging much recently because I just moved from Chicago to Boston. Also, I don’t currently have access to internet in my new apartment. As always, there’s an XKCD comic to capture this scenario:

Hopefully, I’ll be back and posting more often soon!

## Lunar Eclipse and the 22 Degree Halo

The beautiful thing about atmospheric optics is that (almost) everyone can look up at the sky and see stunning optical phenomena from the sun, moon or some other celestial object. In this post I’ll focus on two particularly striking phenomena where the physical essence can be captured with relatively simple explanations.

The 22 degree halo is a ring around the sun or moon, which is often observed on cold days. Here are a couple images of the 22 degree halo around the sun and moon respectively:

22 degree halo around the sun

22 degree halo around the moon

Note that the 22 degree halo is distinct from the coronae, which occur due to different reasons. While the coronae arise due to the presence of water droplets, the 22 degree halo arises specifically due to the presence of hexagonal ice crystals in the earth’s atmosphere. So why 22 degrees? Well, it turns out that one can answer the question using rather simple undergraduate-level physics. One of the most famous questions in undergraduate optics is that of light refraction through a prism, illustrated below:

Fig. 1: The Snell’s Law Prism Problem

But if there were hexagonal ice crystals in the atmosphere, the problem is exactly the same, as one can see below. This is so because a hexagon is just an equilateral triangle with its ends chopped off. So as long as the light enters and exits on two sides of the hexagon that are spaced one side apart, the analysis is the same as for the triangle.

Equilateral triangle with ends chopped off, making a hexagon

It turns out that $\theta_4$ in Fig. 1 can be solved as a function of $\theta_1$ with Snell’s law and some simple trigonometry to yield (under the assumption that $n_1 =1$):

$\theta_4 = \textrm{sin}^{-1}(n_2 \times \textrm{sin}(60-\textrm{sin}^{-1}(\textrm{sin}(\theta_1)/n_2)))$

It is then pretty straightforward to obtain $\delta$, the difference in angle between the incident and refracted beam as a function of $\theta_1$. I have plotted this below for the index of refraction of ice crystals for three different colors of light, red, green and blue ($n_2 =$ 1.306, 1.311 and 1.317 respectively):

The important thing to note in the plot above is that there is a minimum angle below which there is no refracted beam, and this angle is precisely 21.54, 21.92 and 22.37 degrees for red, green and blue light respectively. Because there is no refracted beam below 22 degrees, this region appears darker, and then there is a sudden appearance of the refracted beam at the angles listed above. This is what gives rise to the 22 degree halo and also to the reddish hue on the inside rim of the halo.

Another rather spectacular celestial occurrence is the lunar eclipse, where the earth completely obscures the moon from direct sunlight. This is the geometry for the lunar eclipse:

Geometry of the lunar eclipse

The question I wanted to address is the reddish hue of the moon, despite it lying in the earth’s shadow. It would naively seem like the moon should not be observable at all. However, there is a similar effect occurring here as with the halo. In this case, the earth’s atmosphere is the refracting medium. So just as light incident on the prism was going upward and then exited going downward, the sun’s rays similarly enter the atmosphere on a trajectory that would miss the moon, but then are bent towards the moon after interacting with the earth’s atmosphere.

But why red? Well, this has the same origins as the reddish hue of the sunset. Because light scatters from atmospheric particles as $1/\lambda^4$, blue light gets scattered away much more easily than red light. Hence, the only color of light left by the time the light reaches the moon is primarily of red color.

It is interesting to imagine what the earth looks like from the moon during a lunar eclipse — it likely looks completely dark apart from a spectacular red halo around the earth. Anyway, one should realize that Snell’s law was first formulated in 984 by Arab scientist Ibn Sahl, and so it was possible to come to these conclusions more than a thousand years ago. Nothing new here!