A Posteriori

Attempts to grapple with and elucidate empirical knowledge

Mixes made equalNovember 9, 2009

Filed under: Uncategorized — Rāhul @ 16:38
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Today, I shall explore a simple puzzle which illustrates the advantages of extreme conditions in getting pointers towards the answer. Consider this scenario found at the Times’ Tierneylab blog a few weeks ago. Essentially, the question is-

If we have two vessels, X and Y, X containing a volume “a” of liquid A and Y containing  a volume “b” of liquid B which mixes with liquid A uniformly, and we pour a volume x of liquid A from X to Y first followed by pouring a volume x of the mixture of liquids in Y back to X, will there now be more of liquid A in Y than of liquid B in X? Assume that b is greater than a.

Before solving this algebraically, let’s think about the problem. Since the volume b is more than a, there is more of liquid B than liquid A. When x of b is poured into X, X will now have a+x amount of a mixture of A and B whereas Y will have b-x amount of just liquid B. When x amount of the mixture in X is poured back to Y, X will go back to having just a amount, but of the mixture of A and B, and Y will go back to having b amount, again of the mixture of A and B. There doesn’t seem to be any immediately compelling reason why the amount of B in X and A in Y should be the same or for the percentages to be the same either. So, let’s consider an extreme case for some help with the arguments.

Let us consider the case where x=0. Then, we can see that no liquid is exchanged between X and Y and so at the end the amount of B in X and A in Y are the same, which is zero.  Now consider the case where x=a. After the first exchange, we’ll have 2a amount of liquid in X, which is made of a amount of A and a amount of b, i.e. 50% of each. When x=a amount is poured back from X, this would involve a/2 amount each of A and B. Adding this to the b-a amount of B that was left in Y, we’ll have b-(a/2) amount of B and a/2 amount of A there. At the same time, X will have a/2 amount each of A and B. So, the amount of B in X and A in Y are the same- a/2.

In both the special cases we considered, the amount of B in X and amount of A in Y are the same. It is interesting to see that the concentration of B in X and of A in Y are not the same because the volumes a and b are not the same. Given these two special cases, which give the same result for very different values of x, it seems likely that the result might hold for all x. These examples do not prove so, but it does motivate us as to the answer to look for. So, while we solve it algebraically, not only will we have a higher confidence in the result if we prove it, it is likely that we are more likely to get to it since we know what we are looking for!

Getting to the algebra now, in the general case, after the first exchange of liquids, X will now have a-x amount of A and Y will have a total amount of b+x, of which there is b of B and x of A. When x amount of the liquid from Y is now poured back to X, since the liquids in Y mixed well, the amount poured back will involve [b/(b+x)]*x of B and [x/(b+x)]*x of A. In the proportion of their volumes, this now adds up to x amount poured back from Y to X.  So, the amount of A left in Y is x-[(x*x)/(b+x)] which is bx/(b+x). The amount of B now in X would just be the amount of B poured from Y to X, which again is bx/(b+x). Voila, we have the same number!

This means that the concentration of A in Y is much less than the concentration of B in X. So, if I gulp in a little sea water and spit out the same amount, there will be the same amount of sea water left in my mouth as the amount of my spit in the sea, but my mouth will taste much saltier than the sea will of my spit!

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A fair map of the worldNovember 1, 2009

Filed under: Public Policy — Rāhul @ 13:10
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Maps seek to represent the curved surface of the spherical earth on a flat surface. All methods we have of doing this introduce some distortion or another and involve trade-offs between distortions of shapes, relative distortion of areas at different latitudes, variations in distance between points on the map which are actually equidistant and so on. Despite these issues, it is clear that maps are a valuable resource in understanding geography and increasing awareness about the world. They are more compact and easier to carry than globes and are especially well suited for representation on flat computer screens as increasingly popular internet maps are. While there are infinitely many methods of map projection, of which a few are most popular, this post is about the considerations involved in choosing the relative orientations of maps- which part of the surface of the earth will they be centered on, which directions will form the horizontal and vertical parallels and which part of our wide world will be consigned to the far left and right, split partially between those two ends.

There seems to be a certain natural reasoning behind choosing the north and south poles as two opposite sides of a rectangular map. The earth spins on a north-south axis. So, it would seem logical to have the north-south axis either horizontal or vertical on the map. The north and south poles which are distorted the most by this projection are also the least populated areas in the world, which makes the map the most helpful for most people around the world. But, the question of which part of the world the map would center on doesn’t seem to follow from any easy utilitarian argument. The alignment I have most commonly encountered in the United States is shown below in Fig: 1. I have sourced all figures in this post from the ever helpful google maps. It has the United States right in the middle and cuts through Russia, China and countries in Southeast Asia.

Fig: 1 Americas in the center

Understandably, this is not very popular outside the Americas. Even as an objective analyst, it doesn’t strike as particularly helpful. The two biggest oceans of the world are close to the center whereas some of the most populous regions are consigned to the far corners or ever cut into parts. Another version of the map, which is more popular in much of Eurasia is shown below in Fig: 2.

Fig: 2 Greenwich in the middle

This map has the zero degree longitude at the center and splits at the international date line. This has the advantage of splitting the map in the pacific ocean which is the least populated stretch of the planet and which has no large land masses that are cut into two by the map. Although this map is a product of the Eurocentrism of the 19th century and reflects the primacy of the British Empire in global affairs in how the center of the map passes through London, it has the advantage of having much of the populated areas of the world towards the center of the map. A natural argument can also be made in favour of this alignment. Pangaea, the supercontinent which split to form all the current continents of the world would be left uncut by this map scheme. Pangaea was surrounded by what are today the Pacific, Arctic and Southern oceans, which this map scheme shows towards the corners, surrounding the inhabited world.

Fig: 3 The human story

Finally, Fig: 3 above shows another plausible map scheme which retains many of the advantages of Fig: 2, but adds a human element to the arguments. Here, the map is cut in the atlantic, which also avoids cutting through any large populated countries (except Greenland). Although the large pacific ocean is towards the center of the map, this alignment traces the migration of modern humans from Africa to populate the rest of the World. Humans crossed from Africa to the Middle East and onto Asia, Europe, Australia and across the Berring strait to the Americas. This map shows that entire route without break, which makes it an ideal candidate for a shared human map.

I do intend this only as a purely academic exercise. Maps are inherently political and are unlikely ever to be objectively drawn. Besides, with human progress I hope these issues fade away in importance rather than having to be renegotiated. Internet maps already are flexible enough to be centered at customisable places. But, that doesn’t take away from the fun of exploring what might be the pros and cons of different hypothetical possibilities!

Martin Gardner and the Jones sistersOctober 26, 2009

Filed under: Mathematics — Rāhul @ 20:04
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Today’s puzzle at New York Times’ Tierneylab blog was interesting enough to whet my appetite and yet not hard enough to elude me more than a tiny bit.  I used elementary probabiltity theory and algebra, and with a little help from ever-dependable OpenOffice Calc, I got results most amusing. So, let’s dive straight into it.

Quoting the first problem,

If you happen to meet two of the Jones sisters (this assumes that the two are random selections from the set of all Jones sisters), it is an exactly even-money bet that both girls will be blue-eyed. What is your best guess as to the total number of blue-eyed Jones sisters?

To attack this, I first realise that there isn’t one unique answer as the existence of one almost surely means the existence of others greater than it.  Assuming the total number of Jones sisters to be n and the subset of n that is blue-eyed to be m, we have- Probability of the two Jones sisters being blue eyed= (Probability of the first Jones sister being blue eyed) X (Probability of the second Jones sister being blue eyed). As the probability is 0.5, we have,

$(n/m) ((n-1)/(m-1))=0.5$

This is because first part follows because the first Jones sister could be blue eyed if any n of the m is chosen. But since 1 blue eyed sister has already been chosen as the first, the second could be blue eyed only if any one of the n-1 blue eyed sisters among the m-1 sisters remaining is chosen as the second. We are able to do this only because the choices of sisters are random. Attempting to solve this as a quadratic equation in n, we see that n has a different value for each value of m. Since the Jones sisters can only exist in whole numbers, the solutions should be the ones where both n and m are whole numbers. Do such solutions exist at all?

Starting from n=1 and counting all natural numbers, we see that $n=3$ and $m=4$ is one combination that agrees. In fact, that is the only combination of n and m under 10. So, the answer must be that there are 4 Jones sisters of whom 3 are blue eyed. Then, the problem goes on to say-

A simplified version of eye color inheritance, long held to be true, went as follows. Whether we are blue or brown-eyed is decided by a single pair of eye-color genes, one of which we get from our mother and the other from our father. For a person to have blue eyes, both genes must be for blue eyes; if either gene is for brown eyes, the person’s eye color is brown. The brown eye-color gene is dominant over blue. Each parent randomly passes on either one of his or her two eye-color genes to any given child. Thus if Mr. Jones has one blue and one brown eye color gene, he will be brown-eyed, and any given child will get either the blue or brown eye-color gene from him. We now know that human eye-color inheritance is much more complex than this, but for the purposes of the next question, assume that this simple model is true.

Based on the solution of problem 3, what would you guess are the colors of Mr. and Mrs. Jones’s eyes? What is your best guess about the eye-color genes that Mr. and Mrs. Jones have? How confident can you be about this, assuming that the only information you have is the colors of their daughters’ eyes?

Since the Jones’ have 3 blue-eyed and one brown-eyed daughter, the 4 eye-genes they have are not all blue or all brown. If one of them has both genes brown, then all the daughters will be brown eyed. So, that is ruled out too. So, at least one of the parents needs to have one blue and one brown gene. The other parent may have either both blue genes or one brown and one blue genes. Let us consider the former case first. The probability of each daughter being blue eyed is 0.5, that is when she gets blue-eye genes from both parents. Using binomial probability distribution (which I won’t elaborate here), the probability then of having 3 blue eyed and one brown eyed daughter is 0.25. In the latter case, the probability of each daughter being blue eyed is 0.25, when she gets blue eyed genes from both parents which in this case happens only in one case out of four. Using binomial probability distribution again, the probability now of having 3 blue eyed and 1 brown eyed daughter is 0.046875. Comparing the two probabilities, 0.25 and 0.046875, we can be reasonably confident that one of the parents is blue-eyed and the other is brown-eyed with one brown-eye gene and one blue-eye gene. Our confidence, in my opinion will be the ratio of 0.25 to the combined probability, which is close to 84.21%. Not, the widely accepted 95%, but very high.

Then, the puzzle goes on-

In the same neighborhood there are a whole bunch of animal couples who want to keep up with the Joneses and produce offspring that have some prized trait that we will metaphorically call being “blue-eyed.” The animals (and the range of eggs they produce) are: snakes (10-40), tortoises (50-150), frogs (500-800), beetles (1000-5000), lobsters (5000-50,000), bees (100,000-500,000), termites (500,000-1 million), codfish (1 million-8 million) and oysters (10 million-80 million).

It turns out that all these animal couples produced a bunch of eggs within their expected range, such that (as with the Jones sisters) if you picked any two eggs from the bunch randomly, there is an exactly even chance that they would both be “blue-eyed.” How many “blue-eyed eggs” and total eggs did each of these animal couples produce? Can you further extend this series of numbers using some analogy other than animal reproduction (which we may have exhausted, notwithstanding nature’s amazing profligacy)?

Here, we have to find values of n and m beyond 3 and 4. For this, I sought the help of MS excel and using the relationship connecting n and m and iterating n through increasing natural numbers while searching for whole number m I got the following list.

 n m (n+1)/n (m+1)/m) m/n 3 4 1.3333333333 15 21 5.0000000000 5.2500000000 1.4000000000 85 120 5.6666666667 5.7142857143 1.4117647059 493 697 5.8000000000 5.8083333333 1.4137931034 2871 4060 5.8235294118 5.8249641320 1.4141414141 16731 23661 5.8275862069 5.8278325123 1.4142011834 97513 137904 5.8282828283 5.8283250919 1.4142114385

Although, I didn’t go into the millions as the question demands, we can see patterns emerginng already. The ratio of any value of n or m to its preceeding value increases quickly at first and then plateaus at some point a little above 5.8. Also, the ratio of m to n also increases first and then plateaus near 1.414. The puzzle goes on to explore these numbers.

Not surprisingly, the ratio between “blue-eyed” eggs to the total number of eggs approaches a finite limit as we go towards infinity. But, unexpectedly, so does the ratio between consecutive numbers of eggs in the series, both blue-eyed and total. Why should this be? Can you figure out what these limits are and come up with a general formula that will yield all possible numbers of even-probability blue-eyed babies?

The last question asks for the general formula for all possible values of n, at which my limited math skills seem stumped. But, to find the finite limit for the ratio of m to n, let’s go back to the first equation. As n and m tend to infinity, 1 can be neglected in comparison to n or m. So, n-1 and m-1 tend to n and m respectively. Thus, the equation reduces to-

$(n/m) (n/m)=0.5$

which can be solved to get

$(m/n) = sqrt{2} = 1.41421...$

which is indeed what we saw it plateauing at. After some thought, I am still at a loss why the ratio of consecutive values of n and m also converge to a number as n and m tend to infinity. I suppose I’ll have to wait till other Tierney lab readers solve that! But, since the ratio of n and m converges to a values, if the ratio of the consecutive values of n converges to a number, we can see that the ratio of consecutive values of m also converges to the same number.

Relativity and the Electromagnetic fieldOctober 24, 2009

Filed under: General Physics — Rāhul @ 10:50
Tags: , ,

This week, I shall continue on the theme of the magnetic field and using a thought experiment prove the complete  interdependence of the electric and magnetic field, which we’ll see to be two ways of looking at the same thing. I first came across this fascinating line of reasoning in the pages of the second volume of the Feynman lectures on Physics, within which I have sought the most revealing wisdom on nature since my undergraduate days.

Consider a wire carrying current I and a charge at a distance l from it. The current will produce a magnetic field around the wire in the right-hand direction. The charge q is in the presence of the magnetic field, as shown in Fig: 1 below.

Fig: 1 Wire at rest, Charge moving down

Let us consider now that q is moving in a downward direction (note that current is in the upward direction) at a velocity equal to the drift velocity of electrons in the wire carrying current. Now, we shall analyse the forces on the system from two different inertial frames of reference and consider whether the observable results are the same, a requirement that is critical if we are to have any confidence in our methods of analyses. If our physical theory predicts different realities for observers in different inertial frames of reference, that would be a significant shortcoming indeed in the theory.

The first frame of reference is the one in which the wire is at rest. The current I is moving upwards and hence the drift electrons are moving downwards, say with a velocity v. The charge outside the wire, in its magnetic field is also moving down with the same velocity v. Since the wire is neutral, there is no electric force between it and the charge. But, since the charge moves in a magnetic field with a component of its velocity perpendicular to the field, there is a magnetic force on the charge. In this case, using the left hand rule, the force is towards the right if q is a positive charge and towards the left if q is negative. Assuming a positive q, observers in this frame of reference see the charge repelled from the wire. So, its velocity which was initially downwards, would now diverge away from the wire.

Let us now consider another frame of reference, this one moving at the same velocity as the charge and the drift electrons in the wire, as shown in Fig: 2 below.

Fig: 2 Wire moving up, Charge at rest

This would be like imagining the observer in a vessel moving downwards parallel to the motion of the external charge and the wire. In this frame of reference, the drift electrons are at rest and so there is no current in the wire. Hence, there is no magnetic field around the wire and no magnetic force on the charge. Does this mean the charge is at rest in this frame of reference? If so, in the previous frame, the charge should be seen as moving down at a constant velocity v. But, we had seen that the charge diverges away from the wire as it moves down. Since the observations in the two inertial frames have to agree with each other, there has to be something wrong with our reasoning in at least one frame.

In the second frame, since the charge is not moving, there is no magnetic force on it.  Maybe there is another force equivalent to the magnetic force in the first frame that will produce the same effect? One difference between the frames that we have ignored till now is that the wire is moving upward in the second frame and the electrons are at rest, while in the first frame, the wire is at rest and the electrons are moving down. When there is no current through the wire, the electrons and rest of the wire are at rest in the same frame and the charges balance each other. When there is a current and the electrons move downwards and wire is at rest, as in the first frame, according to special relativity, the charge of the electrons is unaffected by the velocity. But, the volume they occupy shrinks because the length of this space along the direction of the velocity shrinks according to the famous Lorentz transformation. Hence the charge density of electrons as seen in the first frame is higher than that in the second frame. By using a similar argument, we can see that the charge density of the positive charges in the wire is higher by the same amount, but in the second frame. So, we see that the second frame sees the wire as charged more positively than the first frame. In the first frame, we know that the wire is neutral. Otherwise, a wire will spontaneously attract or repel charges in its neighbourhood when the current through it is switched on. We see enough current carrying wires in common life to know this is not true. Hence, in the second frame, the wire is positively charged. As the charge q is also positive, it will be repelled from the wire. So, when we take the electric force also into consideration, the charge in the second frame repels from the wire, which when combined with the relative velocity of the second frame downwards with respect to the first, agrees with the observation in the first frame that the charge moved downwards in a divergent manner from the wire.

Similar observations in terms of velocity is arrived at  in both the frames of reference when we consider the effects of the electric and magnetic fields together. This adds to our understanding that they are but two facets of the general electromagnetic field. In different inertial frames, the electromagnetic field splits into different portions of electric and magneric fields and hence forces due to them. In our example, by choosing extremes of frames in which either the wire or the electrons were at rest, we could split the field into just the electric or just the magnetic field. But, many times it is a combination of both. By sacrificing a mathematical treatment of the problem, we have missed out on much elegance for the sake of simplicity. We also don’t know whether the velocities of the electron as seen in both the frames will agree in magnitude as well as in direction. They actually do, and they do this by having forces different in each frame in such a way as to have the velocities the same! I shall explore a mathematical treatment of the problem, again derived from Feynman’s lectures (Section 13-6), in a later post.

Why the left hand rule?October 11, 2009

Filed under: General Physics — Rāhul @ 22:57
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Today’s post is about the magnetic field and what it really means in the context of the force felt by a charge moving through it. From high school physics, we know that a moving charge (0r current) in a magnetic field feels a certain force on it which is perpendicular both to the direction of motion and the direction of the magnetic field. To find out the direction of this force, we can either follow through with the vector cross product of the current and the magnetic field or use one of the many hand tricks like the left hand rule below.

Fig: 1 Left hand rule

The first question that comes to my mind really is, what is magnetic field? A moving charge as well as the effects of force in motion are both observables. But, how do we observe the magnetic field? If we don’t, how do we know it is there? The magnetic field, like other fields, is a creation of science, not nature. We use it to help explain observables like current and acceleration.

Let’s assume there are magnets as shown in Fig:1, between which is a conductor carrying current in the direction shown. Then, we observe a force on the conductor in the upward direction. We observe that when the magnets are removed from the picture, there is no longer a force on the conductor. Also, when one of the magnets is flipped so that they are repelling, there is no force on the conductor. But, when both the magnets are flipped so that they attract again, we see that the force on the conductor is now of the same intensity as the first case, but downwards. If we are to define two poles for a magnet- north and south and imaginary lines emanating from the north pole and sinking into the south pole, we see that there are such lines crossing the conductor only when there is a north pole and a south pole on either side of it and this, we might hypothesise, encapsulates the effect of the magnet on the conductor. So, the idea of the magnetic field (it it works) replaces the whole structure of the magnets with one space and time variant vector, as far as the current carrying conductor is concerned.

As myriad experiments prove the utility of the magnetic field idea and validates our left hand rule, we still don’t know if this is the best method to reduce the effects of the magnets. We see that this definition of magnetic field is perpendicular both to the current and the direction of force, the two directions where there are observable effects. Why not define the magnetic field in the direction of the force like the electric field or gravitational field is? To explore why, let us use the method of contradictions. Assuming the direction of magnetic field is defined along the force on the conductor, we see that the current is perpendicular to both the magnetic field and the force which lie along the same axis. If the current is now flipped, the force, according to the previous rule will still be along the same direction as it remains perpendicular to current and parallel to the magnetic field. But, we observe that when the direction of current is flipped, so is the direction of the force on the conductor. So, we cannot define magnetic field in the direction of the force. We can also rule out defining it in the direction of the current because then, we do not have any easy way to judge the direction of force, apart from restricting it to a plane perpendicular to the current and field. But, if we do choose a direction orthogonal to both the current and force, we see that simple rules like the left hand rule can be developed to find one observable from the other in a way that agrees with experiment.

While all fields are human creations, I have always found the idea of the magnetic field the most artificial. I don’t have any rational arguments for this feeling, but it has to do with how it is defined in a direction which is apart from either of the observable vectors involved. But, as we see, this is the only way we see how to define the magnetic field in a way that facilitates accurate predictions between current and force.

Replacement and Population MomentumOctober 4, 2009

Filed under: Public Policy — Rāhul @ 17:53

China implemented the one-child policy back in 1979, under which most of the country’s couples were encouraged, and some would say compelled, to have only one child. Even though many Chinese parents, as decided by geography or ethnic minority status, were allowed to have more than 1 child, on the average, Chinese families had less than 2 children per woman. Indeed by 2005, the fertility rate of mainland Chinese women has fallen to about 1.75. But, China’s population continues to increase even today, as can be seen from Fig: 1 below, courtesy Wikipedia.

Fig: 1 Growth of China's population

Replacement fertility rate of a group is defined as the average number of children a newborn girl needs to give birth to in her life so that the population of her group remains constant. At first glance, this number seems like it should be around 2, so that the children born would replace (at the risk of sounding rather inhumane) their parents. But, unfortunately, not all baby girls live long enough to be of childbearing age, which means the Replacement fertiliy rate has to be a little higher than 2. Also, as the sex ratio at birth in most places in the world is skewed towards boys, less than half of the children born are girls who will give birth to the next generation. So, to have the replacement number of girls, the replacement fertility rate has to be a little higher still. Obviously, since infant/child mortality rates and sex ratio at birth differ widely across the world, the amount by which the Replacement fertility rate is above 2 is different in different countries. According to Wikipedia, while the number is around 2.1 for most developed countries, it ranges from 2.5 to 3.3 in developing countries.  So, when fertility rate in China is so low, how does population still increase?

To understand why changes in fertility don’t immediately translate to changes in population, let’s see what is the immediate reason for changes in population. The net-population growth is the difference between the total of births and immigration and the total of deaths and emigration. Migrations, being independent of fertility, obviously do not affect how fertility rate affects population. Then, the question is whether replacement fertility during a particular year guarantees equality between total births and deaths. It is easy to see that it doesn’t. Even after a few years of replacement fertility, the society only guarantees that the number of births is equal to the number of births X years ago, where X is the average childbearing age of women in that society. But, the number of deaths is different! It will be the number of births Y years ago, where Y is the life span of the society. If the society had a growing population in the near past, then X would be greater than Y and so the population will continue to increase despite the achievement of fertility at or below replacement levels. Similarly, population will continue to fall for a few years in countries like Russia, which has a declining population now, even if they achieve Replacement fertility. This whole phenomenon where Population lags behind fertility rate trends is called Population Momentum.

It is evident that if fertility rate continues to be low, the Chinese population eventually has to fall. When will this happen? When the one-child policy has continued for long enough that the total number of births in the current year (in the future) is less than than the number of births Y years before current, the population will begin to fall. As births are falling each year and the number of deaths Y years ago increasing each year, that day will come soon. Of course, Chinese policymakers will find it prudent to reverse the policy, at least partially, much before that day as the dependency ratio in China is rapidly increasing, threatening to curtail economic growth as the average working age Chinese will have to support more and more retired seniors.

Policy-making is intensely intricate and dependent of good data collection and analysis. Leaving aside for another venue the morality of an overly planned state, one can’t but admire how intelligent data analysis can lead to policies that can quickly uplift hundreds of millions from poverty, as seen in China today.

Jump not to conclusionsSeptember 30, 2009

Filed under: General Physics — Rāhul @ 11:52
Tags: ,

Today’s post is on why it isn’t wise in science to jump to conclusions, at least not ones which we’re too hasty to take as facts. Yesterday, I was in the cleanroom, cleaning some wafer pieces for SIMS analysis. My samples were being cleaned at high temperature in a water bath, when I noticed something interesting. I whisked out my phone and got some pictures to illustrate my points.

I had 3 small beakers in the water bath, almost evenly spaced out, as shown in Fig: 1 below.

Fig: 1 Beakers immersed in water bath at even spacing

After about 10 minutes, I noticed that beakers had migrated within the water bath to cluster together at one end, as shown below in Fig: 2.

Fig: 2 Beakers clustered together in the water bath

The water was at about 80 degrees centigrade and the various stresses in it was creating many eddies and waves, but the question was why there was a net. effect towards clustering the beakers. I wondered if there was some reason why the forces on the beaker surface would vary in magnitude based on the amount of water beyond it. This might mean that the beakers will tend to cluster closer together as the greater force from the larger quantity of water outside the cluster than the force from inside the cluster would push them together. This would also explain why in Fig: 2, all three beakers didn’t move to the center of the water bath. There water-force proportionality argument works to keep the beakers closer to the edge of the water bath too.

I started exploring reasons why the quantity of water in the neighbourhood might affect the force on the beakers and cause clustering, but could not come up with any satisfactory answer. After all, the bombardment of high energy water molecules on the surface of the beaker (pressure) is just a function of the local temperature. Although temperature might vary radially in the water bath, that wouldn’t explain the behaviour. So, my hypothesis was proving hard to substantiate, even with the limited physics I employed. Then, I thought that maybe there is a gradient along the water bath towards the far side so that beakers move that way because gravity can overcome friction on the beakers when there is random agitation of water. To test this, I replaced the beakers as shown in Fig: 3 below.

Fig: 3 Beakers placed together on the near side

If the reason for the 3 beakers in Figs 1 and 2 clustering on the far side was an effective force bringing them together, then, the beakers should now remain clustered. But, after a short while, the beakers moved as shown in Fig: 4 below.

Fig: 4 Beakers moving to the far side

The glass beakers moved over to the far side of the water bath while the Teflon beaker remains steady. We don’t know if this difference is related to the material or weight of the beakers or just due to their relative positions. But, from this observation, we can indeed reject any tendency to clusterise as the main reason why the beakers in Figs 1 and 2 moved over to the far side. There does seem to be a gravitational (or other) gradient favouring the far side of the water bath.  Even if there is a force favouring clustering, the gravitational gradient overcomes it, as seen in Figs 3 and 4. The Teflon beaker might not have moved to the far side in Fig: 4 because of a local energy minimum there. But, I am not certain because it was time for me to remove my samples from the water bath, and hence my experiments with the beakers had to cease!

This post just intends to illustrate why in Science we should remain ever-vigilant in seeking new information and never too dogmatic to re-evaluate the theories we might find interesting. My former theory sounded more fun to me, but further evidence pointed to something more prosaic. Prosaic, yet agreeing with evidence!