# See, how hard I am trying to become an Approved (Full) Professor of Mechanical Engineering in SPPU?—4

In this post, I provide my answer to the question which I had raised last time, viz., about the differences between the $\Delta$, the $\text{d}$, and the $\delta$ (the first two, of the usual calculus, and the last one, of the calculus of variations).

Some pre-requisite ideas:

A system is some physical object chosen (or isolated) for study. For continua, it is convenient to select a region of space for study, in which case that region of space (holding some physical continuum) may also be regarded as a system. The system boundary is an abstraction.

A state of a system denotes a physically unique and reproducible condition of that system. State properties are the properties or attributes that together uniquely and fully characterize a state of a system, for the chosen purposes. The state is an axiom, and state properties are its corollary.

State properties for continua are typically expressed as functions of space and time. For instance, pressure, temperature, volume, energy, etc. of a fluid are all state properties. Since state properties uniquely define the condition of a system, they represent definite points in an appropriate, abstract, (possibly) higher-dimensional state space. For this reason, state properties are also called point functions.

A process (synonymous to system evolution) is a succession of states. In classical physics, the succession (or progression) is taken to be continuous. In quantum mechanics, there is no notion of a process; see later in this post.

A process is often represented as a path in a state space that connects the two end-points of the staring and ending states. A parametric function defined over the length of a path is called a path function.

A cyclic process is one that has the same start and end points.

During a cyclic process, a state function returns to its initial value. However, a path function does not necessarily return to the same value over every cyclic change—it depends on which particular path is chosen. For instance, if you take a round trip from point $A$ to point $B$ and back, you may spend some amount of money $m$ if you take one route but another amount $n$ if you take another route. In both cases you do return to the same point viz. $A$, but the amount you spend is different for each route. Your position is a state function, and the amount you spend is a path function.

[I may make the above description a bit more rigorous later on (by consulting a certain book which I don’t have handy right away (and my notes of last year are gone in the HDD crash)).]

The $\Delta$, the $\text{d}$, and the $\delta$:

The $\Delta$ denotes a sufficiently small but finite, and locally existing difference in different parts of a system. Typically, since state properties are defined as (continuous) functions of space and time, what the $\Delta$ represents is a finite change in some state property function that exists across two different but adjacent points in space (or two nearby instants in times), for a given system.

The $\Delta$ is a local quantity, because it is defined and evaluated around a specific point of space and/or time. In other words, an instance of $\Delta$ is evaluated at a fixed $x$ or $t$. The $\Delta x$ simply denotes a change of position; it may or may not mean a displacement.

The $\text{d}$ (i.e. the infinitesimal) is nothing but the $\Delta$ taken in some appropriate limiting process to the vanishingly small limit.

Since $\Delta$ is locally defined, so is the infinitesimal (i.e. $\text{d}$).

The $\delta$ of CoV is completely different from the above two concepts.

The $\delta$ is a sufficiently small but global difference between the states (or paths) of two different, abstract, but otherwise identical views of the same physically existing system.

Considering the fact that an abstract view of a system is itself a system, $\delta$ also may be regarded as a difference between two systems.

Though differences in paths are not only possible but also routinely used in CoV, in this post, to keep matters simple, we will mostly consider differences in the states of the two systems.

In CoV, the two states (of the two systems) are so chosen as to satisfy the same Dirichlet (i.e. field) boundary conditions separately in each system.

The state function may be defined over an abstract space. In this post, we shall not pursue this line of thought. Thus, the state function will always be a function of the physical, ambient space (defined in reference to the extensions and locations of concretely existing physical objects).

Since a state of a system of nonzero size can only be defined by specifying its values for all parts of a system (of which it is a state), a difference between states (of the two systems involved in the variation $\delta$) is necessarily global.

In defining $\delta$, both the systems are considered only abstractly; it is presumed that at most one of them may correspond to an actual state of a physical system (i.e. a system existing in the physical reality).

The idea of a process, i.e. the very idea of a system evolution, necessarily applies only to a single system.

What the $\delta$ represents is not an evolution because it does not represent a change in a system, in the first place. The variation, to repeat, represents a difference between two systems satisfying the same field boundary conditions. Hence, there is no evolution to speak of. When compressed air is passed into a rubber balloon, its size increases. This change occurs over certain time, and is an instance of an evolution. However, two rubber balloons already inflated to different sizes share no evolutionary relation with each other; there is no common physical process connecting the two; hence no change occurring over time can possibly enter their comparative description.

Thus, the “change” denoted by $\delta$ is incapable of representing a process or a system evolution. In fact, the word “change” itself is something of a misnomer here.

Text-books often stupidly try to capture the aforementioned idea by saying that $\delta$ represents a small and possibly finite change that occurs without any elapse of time. Apart from the mind-numbing idea of a finite change occurring over no time (or equally stupefying ideas which it suggests, viz., a change existing at literally the same instant of time, or, alternatively, a process of change that somehow occurs to a given system but “outside” of any time), what they, in a way, continue to suggest also is the erroneous idea that we are working with only a single, concretely physical system, here.

But that is not the idea behind $\delta$ at all.

To complicate the matters further, no separate symbol is used when the variation $\delta$ is made vanishingly small.

In the primary sense of the term variation (or $\delta$), the difference it represents is finite in nature. The variation is basically a function of space (and time), and at every value of $x$ (and $t$), the value of $\delta$ is finite, in the primary sense of the word. Yes, these values can be made vanishingly small, though the idea of the limits applied in this context is different. (Hint: Expand each of the two state functions in a power series and relate each of the corresponding power terms via a separate parameter. Then, put the difference in each parameter through a limiting process to vanish. You may also use the Fourier expansion.))

The difference represented by $\delta$ is between two abstract views of a system. The two systems are related only in an abstract view, i.e., only in (the mathematical) thought. In the CoV, they are supposed as connected, but the connection between them is not concretely physical because there are no two separate physical systems concretely existing, in the first place. Both the systems here are mathematical abstractions—they first have been abstracted away from the real, physical system actually existing out there (of which there is only a single instance).

But, yes, there is a sense in which we can say that $\delta$ does have a physical meaning: it carries the same physical units as for the state functions of the two abstract systems.

An example from biology:

Here is an example of the differences between two different paths (rather than two different states).

Plot the height $h(t)$ of a growing sapling at different times, and connect the dots to yield a continuous graph of the height as a function of time. The difference in the heights of the sapling at two different instants is $\Delta h$. But if you consider two different saplings planted at the same time, and assuming that they grow to the same final height at the end of some definite time period (just pick some moment where their graphs cross each other), and then, abstractly regarding them as some sort of imaginary plants, if you plot the difference between the two graphs, that is the variation or $\delta h(t)$ in the height-function of either. The variation itself is a function (here of time); it has the units, of course, of m.

Summary:

The $\Delta$ is a local change inside a single system, and $\text{d}$ is its limiting value, whereas the $\delta$ is a difference across two abstract systems differing in their global states (or global paths), and there is no separate symbol to capture this object in the vanishingly small limit.

Exercises:

Consider one period of the function $y = A \sin(x)$, say over the interval $[0,2\pi]$; $A = a$ is a small, real-valued, constant. Now, set $A = 1.1a$. Is the change/difference here a $\delta$ or a $\Delta$? Why or why not?

Now, take the derivative, i.e., $y' = A \cos(x)$, with $A = a$ once again. Is the change/difference here a $\delta$ or a $\Delta$? Why or why not?

Which one of the above two is a bigger change/difference?

Also consider this angle: Taking the derivative did affect the whole function. If so, why is it that we said that $\text{d}$ was necessarily a local change?

An important and special note:

The above exercises, I am sure, many (though not all) of the Officially Approved Full Professors of Mechanical Engineering at the Savitribai Phule Pune University and COEP would be able to do correctly. But the question I posed last time was: Would it be therefore possible for them to spell out the physical meaning of the variation i.e. $\delta$? I continue to think not. And, importantly, even among those who do solve the above exercises successfully, they wouldn’t be too sure about their own answers. Upon just a little deeper probing, they would just throw up their hands. [Ditto, for many American physicists.] Even if a conceptual clarity is required in applications.

(I am ever willing and ready to change my mind about it, but doing so would need some actual evidence—just the way my (continuing) position had been derived, in the first place, from actual observations of them.)

The reason I made this special note was because I continue to go jobless, and nearly bank balance-less (and also, nearly cashless). And it all is basically because of folks like these (and the Indians like the SPPU authorities). It is their fault. (And, no, you can’t try to lift what is properly their moral responsibility off their shoulders and then, in fact, go even further, and attempt to place it on mine. Don’t attempt doing that.)

A Song I Like:

[May be I have run this song before. If yes, I will replace it with some other song tomorrow or so. No I had not.]

Hindi: “Thandi hawaa, yeh chaandani suhaani…”
Music and Singer: Kishore Kumar
Lyrics: Majrooh Sultanpuri

[A quick ‘net search on plagiarism tells me that the tune of this song was lifted from Julius La Rosa’s 1955 song “Domani.” I heard that song for the first time only today. I think that the lyrics of the Hindi song are better. As to renditions, I like Kishor Kumar’s version better.]

[Minor editing may be done later on and the typos may be corrected, but the essentials of my positions won’t be. Mostly done right today, i.e., on 06th January, 2017.]

[E&OE]

# See, how hard I am trying to become an Approved (Full) Professor of Mechanical Engineering in SPPU?—3

I was looking for a certain book on heat transfer which I had (as usual) misplaced somewhere, and while searching for that book at home, I accidentally ran into another book I had—the one on Classical Mechanics by Rana and Joag [^].

After dusting this book a bit, I spent some time in one typical way, viz. by going over some fond memories associated with a suddenly re-found book…. The memories of how enthusiastic I once was when I had bought that book; how I had decided to finish that book right within weeks of buying it several years ago; the number of times I might have picked it up, and soon later on, kept it back aside somewhere, etc.  …

Yes, that’s right. I have not yet managed to finish this book. Why, I have not even managed to begin reading this book the way it should be read—with a paper and pencil at hand to work through the equations and the problems. That was the reason why, I now felt a bit guilty. … It just so happened that it was just the other day (or so) when I was happily mentioning the Poisson brackets on Prof. Scott Aaronson’s blog, at this thread [^]. … To remove (at least some part of) my sense of guilt, I then decided to browse at least through this part (viz., Poisson’s brackets) in this book. … Then, reading a little through this chapter, I decided to browse through the preceding chapters from the Lagrangian mechanics on which it depends, and then, in general, also on the calculus of variations.

It was at this point that I suddenly happened to remember the reason why I had never been able to finish (even the portions relevant to engineering from) this book.

The thing was, the explanation of the $\delta$—the delta of the variational calculus.

The explanation of what the $\delta$ basically means, I had found right back then (many, many years ago), was not satisfactorily given in this book. The book did talk of all those things like the holonomic constraints vs. the nonholonomic constraints, the functionals, integration by parts, etc. etc. etc. But without ever really telling me, in a forth-right and explicit manner, what the hell this $\delta$ was basically supposed to mean! How this $\delta y$ was different from the finite changes ($\Delta y$) and the infinitesimal changes ($\text{d}y$) of the usual calculus, for instance. In terms of its physical meaning, that is. (Hell, this book was supposed to be on physics, wasn’t it?)

Here, I of course fully realize that describing Rana and Joag’s book as “unsatisfactory” is making a rather bold statement, a very courageous one, in fact. This book is extraordinarily well-written. And yet, there I was, many, many years ago, trying to understand the delta, and not getting anywhere, not even with this book in my hand. (OK, a confession. The current copy which I have is not all that old. My old copy is gone by now (i.e., permanently misplaced or so), and so, the current copy is the one which I had bought once again, in 2009. As to my old copy, I think, I had bought it sometime in the mid-1990s.)

It was many years later, guess some time while teaching FEM to the undergraduates in Mumbai, that the concept had finally become clear enough to me. Most especially, while I was going through P. Seshu’s and J. N. Reddy’s books. [Reflected Glory Alert! Professor P. Seshu was my class-mate for a few courses at IIT Madras!] However, even then, even at that time, I remember, I still had this odd feeling that the physical meaning was still not clear to me—not as as clear as it should be. The matter eventually became “fully” clear to me only later on, while musing about the differences between the perspective of Thermodynamics on the one hand and that of Heat Transfer on the other. That was some time last year, while teaching Thermodynamics to the PG students here in Pune.

Thermodynamics deals with systems at equilibria, primarily. Yes, its methods can be extended to handle also the non-equilibrium situations. However, even then, the basis of the approach summarily lies only in the equilibrium states. Heat Transfer, on the other hand, necessarily deals with the non-equilibrium situations. Remove the temperature gradient, and there is no more heat left to speak of. There does remain the thermal energy (as a form of the internal energy), but not heat. (Remember, heat is the thermal energy in transit that appears on a system boundary.) Heat transfer necessarily requires an absence of thermal equilibrium. … Anyway, it was while teaching thermodynamics last year, and only incidentally pondering about its differences from heat transfer, that the idea of the variations (of Cov) had finally become (conceptually) clear to me. (No, CoV does not necessarily deal only with the equilibrium states; it’s just that it was while thinking about the equilibrium vs. the transient that the matter about CoV had suddenly “clicked” to me.)

In this post, let me now note down something on the concept of the variation, i.e., towards understanding the physical meaning of the symbol $\delta$.

Please note, I have made an inline update on 26th December 2016. It makes the presentation of the calculus of variations a bit less dumbed down. The updated portion is clearly marked as such, in the text.

The Problem Description:

The concept of variations is abstract. We would be better off considering a simple, concrete, physical situation first, and only then try to understand the meaning of this abstract concept.

Accordingly, consider a certain idealized system. See its schematic diagram below:

There is a long, rigid cylinder made from some transparent material like glass. The left hand-side end of the cylinder is hermetically sealed with a rigid seal. At the other end of the cylinder, there is a friction-less piston which can be driven by some external means.

Further, there also are a couple of thin, circular, piston-like disks ($D_1$ and $D_2$) placed inside the cylinder, at some $x_1$ and $x_2$ positions along its length. These disks thus divide the cylindrical cavity into three distinct compartments. The disks are assumed to be impermeable, and fitting snugly, they in general permit no movement of gas across their plane. However, they also are assumed to be able to move without any friction.

Initially, all the three compartments are filled with a compressible fluid to the same pressure in each compartment, say 1 atm. Since all the three compartments are at the same pressure, the disks stay stationary.

Then, suppose that the piston on the extreme right end is moved, say from position $P_1$ to $P_2$. The final position $P_2$ may be to the left or to the right of the initial position $P_1$; it doesn’t matter. For the current description, however, let’s suppose that the position $P_2$ is to the left of $P_1$. The effect of the piston movement thus is to increase the pressure inside the system.

The problem is to determine the nature of the resulting displacements that the two disks undergo as measured from their respective initial positions.

There are essentially two entirely different paradigms for conducting an analysis of this problem.

The first paradigm is based on an approach that was put to use so successfully by Newton. Usually, it is called the paradigm of vector analysis.

In this paradigm, we focus on the fact that the forced displacement of the piston with time, $x(t)$, may be described using some function of time that is defined over the interval lying between two instants $t_i$ and $t_f$.

For example, suppose the function is:
$x(t) = x_0 + v t$,
where $v$ is a constant. In other words, the motion of the piston is steady, with a constant velocity, between the initial and final instants. Since the velocity is constant, there is no acceleration over the open interval $(t_i, t_f)$.

However, notice that before the instant $t_i$, the piston velocity was zero. Then, the velocity suddenly became a finite (constant) value. Therefore, if you extend the interval to include the end-instants as well, i.e., if you consider the semi-closed interval $[t_i, t_f)$, then there is an acceleration at the instant $t_i$. Similarly, since the piston comes to a position of rest at $t = t_f$, there also is another acceleration, equal in magnitude and opposite in direction, which appears at the instant $t_f$.

The existence of these two instantaneous accelerations implies that jerks or pressure waves are sent through the system. We may model them as vector quantities, as impulses. [Side Exercise: Work out what happens if we consider only the open interval $(t_i, t_f)$.]

We can now apply Newton’s 3 laws, based on the idea that shock-waves must have begun at the piston at the instant $t = t_i$. They must have got transmitted through the gas kept under pressure, and they must have affected the disk $D_1$ lying closest to the piston, thereby setting this disk into motion. This motion must have passed through the gas in the middle compartment of the system as another pulse in the pressure (generated at the disk $D_1$), thereby setting also the disk $D_2$ in a state of motion a little while later. Finally, the pulse must have got bounced off the seal on the left hand side, and in turn, come back to affect the motion of the disk $D_2$, and then of the disk $D_1$. Continuing their travels to and fro, the pulses, and hence the disks, would thus be put in a back and forth motion.

After a while, these transients would move forth and back, superpose, and some of their constituent frequencies would get cancelled out, leaving only those frequencies operative such that the three compartments are put under some kind of stationary states.

In case the gas is not ideal, there would be damping anyway, and after a sufficiently long while, the disks would move through such small displacements that we could easily ignore the ever-decreasing displacements in a limiting argument.

Thus, assume that, after an elapse of a sufficiently long time, the disks become stationary. Of course, their new positions are not the same as their original positions.

The problem thus can be modeled as basically a transient one. The state of the new equilibrium state is thus primarily seen as an effect or an end-result of a couple of transient processes which occur in the forward and backward directions. The equilibrium is seen as not a primarily existing state, but as a result of two equal and opposite transient causes.

Notice that throughout this process, Newton’s laws can be applied directly. The nature of the analysis is such that the quantities in question—viz. the displacements of the disks—always are real, i.e., they correspond to what actually is supposed to exist in the reality out there.

The (values of) displacements are real in the sense that the mathematical analysis procedure itself involves only those (values of) displacements which can actually occur in reality. The analysis does not concern itself with some other displacements that might have been possible but don’t actually occur. The analysis begins with the forced displacement condition, translates it into pressure waves, which in turn are used in order to derive the predicted displacements in the gas in the system, at each instant. Thus, at any arbitrary instant of time $t > t_i$ (in fact, the analysis here runs for times $t \gg t_f$), the analysis remains concerned only with those displacements that are actually taking place at that instant.

The Method of Calculus of Variations:

The second paradigm follows the energetics program. This program was initiated by Newton himself as well as by Leibnitz. However, it was pursued vigorously not by Newton but rather by Leibnitz, and then by a series of gifted mathematicians-physicists: the Bernoulli brothers, Euler, Lagrange, Hamilton, and others. This paradigm is essentially based on the calculus of variations. The idea here is something like the following.

We do not care for a local description at all. Thus, we do not analyze the situation in terms of the local pressure pulses, their momenta/forces, etc. All that we focus on are just two sets of quantities: the initial positions of the disks, and their final positions.

For instance, focus on the disk $D_1$. It initially is at the position $x_{1_i}$. It is found, after a long elapse of time (i.e., at the next equilibrium state), to have moved to $x_{1_f}$. The question is: how to relate this change in $x_1$ on the one hand, to the displacement that the piston itself undergoes from $P_{x_i}$ to $P_{x_f}$.

To analyze this question, the energetics program (i.e., the calculus of variations) adopts a seemingly strange methodology.

It begins by saying that there is nothing unique to the specific value of the position $x_{1_f}$ as assumed by the disk $D_1$. The disk could have come to a halt at any other (nearby) position, e.g., at some other point $x_{1_1}$, or $x_{1_2}$, or $x_{1_3}$, … etc. In fact, since there are an infinity of points lying in a finite segment of line, there could have been an infinity of positions where the disk could have come to a rest, when the new equilibrium was reached.

Of course, in reality, the disk $D_1$ comes to a halt at none of these other positions; it comes to a halt only at $x_{1_f}$.

Yet, the theory says, we need to be “all-inclusive,” in a way. We need not, just for the aforementioned reason, deny a place in our analysis to these other positions. The analysis must include all such possible positions—even if they be purely hypothetical, imaginary, or unreal. What we do in the analysis, this paradigm says, is to initially include these merely hypothetical, unrealistic positions too on exactly the same footing as that enjoyed by that one position which is realistic, which is given by $x_{1_f}$.

Thus, we take a set of all possible positions for each disk. Then, for each such a position, we calculate the “impact” it would make on the energy of the system taken as a whole.

The energy of the system can be additively decomposed into the energies carried by each of its sub-parts. Thus, focusing on disk $D_1$, for each one of its possible (hypothetical) final position, we should calculate the energies carried by both its adjacent compartments. Since a change in $D_1$‘s position does not affect the compartment 3, we need not include it. However, for the disk $D_1$, we do need to include the energies carried by both the compartments 1 and 2. Similarly, for each of the possible positions occupied by the disk $D_2$, it should include the energies of the compartments 2 and 3, but not of 1.

At this point, to bring simplicity (and thereby better) clarity to this entire procedure, let us further assume that the possible positions of each disk forms a finite set. For instance, each disk can occupy only one of the positions that is some $-5, -4, -3, -2, -1, 0, +1, +2, +3, +4$ or $+5$ distance-units away from its initial position. Thus, a disk is not allowed to come to a rest at, say, $2.3$ units; it must do so either at $2$ or at $3$ units. (We will thus perform the initial analysis in terms of only the integer positions, and only later on extend it to any real-valued positions.) (If you are a mechanical engineering student, suggest a suitable mechanism that can ensure only integer relative displacements.)

The change in energy $E$ of a compartment is given by
$\Delta E = P A \Delta x$,
where $P$ is the pressure, $A$ is the cross-sectional area of the cylinder, and $\Delta x$ is the change in the length of the compartment.

Now, observe that the energy of the middle compartment depends on the relative distance between the two disks lying on its sides. Yet, for the same reason, the energy of the middle compartment does depend on both these positions. Hence, we must take a Cartesian product of the relative displacements undergone by both the disks, and only then calculate the system energy for each such a permutation (i.e. the ordered pair) of their positions. Let us go over the details of the Cartesian product.

The Cartesian product of the two positions may be stated as a row-by-row listing of ordered pairs of the relative positions of $D_1$ and $D_2$, e.g., as follows: the ordered pair $(-5, +2)$ means that the disk $D_1$ is $5$ units to the left of its initial position, and the disk $D_2$ is $+2$ units to the right of its initial position. Since each of the two positions forming an ordered pair can range over any of the above-mentioned $11$ number of different values, there are, in all, $11 \times 11 = 121$ number of such possible ordered pairs in the Cartesian product.

For each one of these $121$ different pairs, we use the above-given formula to determine what the energy of each compartment is like. Then, we add the three energies (of the three compartments) together to get the value of the energy of the system as a whole.

In short, we get a set of $121$ possible values for the energy of the system.

You must have noticed that we have admitted every possible permutation into analysis—all the $121$ number of them.

Of course, out of all these $121$ number of permutations of positions, it should turn out that $120$ number of them have to be discarded because they would be merely hypothetical, i.e. unreal. That, in turn, is because, the relative positions of the disks contained in one and only one ordered pair would actually correspond to the final, equilibrium position. After all, if you conduct this experiment in reality, you would always get a very definite pair of the disk-positions, and it this same pair of relative positions that would be observed every time you conducted the experiment (for the same piston displacement). Real experiments are reproducible, and give rise to the same, unique result. (Even if the system were to be probabilistic, it would have to give rise to an exactly identical probability distribution function.) It can’t be this result today and that result tomorrow, or this result in this lab and that result in some other lab. That simply isn’t science.

Thus, out of all those $121$ different ordered-pairs, one and only one ordered-pair would actually correspond to reality; the rest all would be merely hypothetical.

The question now is, which particular pair corresponds to reality, and which ones are unreal. How to tell the real from the unreal. That is the question.

Here, the variational principle says that the pair of relative positions that actually occurs in reality carries a certain definite, distinguishing attribute.

The system-energy calculated for this pair (of relative displacements) happens to carry the lowest magnitude from among all possible $121$ number of pairs. In other words, any hypothetical or unreal pair has a higher amount of system energy associated with it. (If two pairs give rise to the same lowest value, both would be equally likely to occur. However, that is not what provably happens in the current example, so let us leave this kind of a “degeneracy” aside for the purposes of this post.)

(The update on 26 December 2016 begins here:)

Actually, the description  given in the immediately preceding paragraph was a bit too dumbed down. The variational principle is more subtle than that. Explaining it makes this post even longer, but let me give it a shot anyway, at least today.

To follow the actual idea of the variational principle (in a not dumbed-down manner), the procedure you have to follow is this.

First, make a table of all possible relative-position pairs, and their associated energies. The table has the following columns: a relative-position pair, the associated energy $E$ as calculated above, and one more column which for the time being would be empty. The table may look something like what the following (partial) listing shows:

(0,0) -> say, 115 Joules
(-1,0) -> say, 101 Joules
(-2,0) -> say, 110 Joules

(2,2) -> say, 102 Joules
(2,3) -> say, 100 Joules
(2,4) -> say, 101 Joules
(2,5) -> say, 120 Joules

(5,0) -> say, 135 Joules

(5,5) -> say 117 Joules.

Having created this table (of $121$ rows), you then pick each row one by and one, and for the picked up $n$-th row, you ask a question: What all other row(s) from this table have their relative distance pairs such that these pairs lie closest to the relative distance pair of this given row. Let me illustrate this question with a concrete example. Consider the row which has the relative-distance pair given as (2,3). Then, the relative distance pairs closest to this one would be obtained by adding or subtracting a distance of 1 to each in the pair. Thus, the relative distance pairs closest to this one would be: (3,3), (1,3), (2,4), and (2,2). So, you have to pick up those rows which have these four entries in the relative-distance pairs column. Each of these four pairs represents a variation $\delta$ on the chosen state, viz. the state (2,3).

In symbolic terms, suppose for the $n$-th row being considered, the rows closest to it in terms of the differences in their relative distance pairs, are the $a$-th, $b$-th, $c$-th and $d$-th rows. (Notice that the rows which are closest to a given row in this sense, would not necessarily be found listed just above or below that given row, because the scheme followed while creating the list or the vector that is the table would not necessarily honor the closest-lying criterion (which necessarily involves two numbers)—not at least for all rows in the table.

OK. Then, in the next step, you find the differences in the energies of the $n$-th row from each of these closest rows, viz., the $a$-th, $b$-th, $c$-th and $c$-th rows. That is to say, you find the absolute magnitudes of the energy differences. Let us denote these magnitudes as: $\delta E_{na} = |E_n - E_a|$$\delta E_{nb} = |E_n - E_b|$$\delta E_{nc} = |E_n - E_c|$ and $\delta E_{nd} = |E_n - E_d|$.  Suppose the minimum among these values is $\delta E_{nc}$. So, against the $n$-th row, in the last column of the table, you write the value $\delta E_{nc}$.

Having done this exercise separately for each row in the table, you then ask: Which row has the smallest entry in the last column (the one for $\delta E$), and you pick that up. That is the distinguished (or the physically occurring) state.

In other words, the variational principle asks you to select not the row with the lowest absolute value of energy, but that row which shows the smallest difference of energy from one of its closest neighbours—and these closest neighbours are to be selected according to the differences in each number appearing in the relative-distance pair, and not according to the vertical place of rows in the tabular listing. (It so turns out that in this example, the row thus selected following both criteria—lowest energy as well as lowest variation in energy—are identical, though it would not necessarily always be the case. In short, we can’t always get away with the first, too dumbed down, version.)

Thus, the variational principle is about that change in the relative positions for which the corresponding change in the energy vanishes (or has the minimum possible absolute magnitude, in case the positions form a discretely varying, finite set).

(The update on 26th December 2016 gets over here.)

And, it turns out that this approach, too, is indeed able to perfectly predict the final disk-positions—precisely as they actually are observed in reality.

If you allow a continuum of positions (instead of the discrete set of only the $11$ number of different final positions for one disk, or $121$ number of ordered pairs), then instead of taking a Cartesian product of positions, what you have to do is take into account a tensor product of the position functions. The maths involved is a little more advanced, but the underlying algebraic structure—and the predictive principle which is fundamentally involved in the procedure—remains essentially the same. This principle—the variational principle—says:

Among all possible variations in the system configurations, that system configuration corresponds to reality which has the least variation in energy associated with it.

(This is a very rough statement, but it will do for this post and for a general audience. In particular, we don’t look into the issues of what constitute the kinematically admissible constraints, why the configurations must satisfy the field boundary conditions, the idea of the stationarity vs. of a minimum or a maximum, i.e., the issue of convexity-vs.-concavity, etc. The purpose of this post—and our example here—are both simple enough that we need not get into the whole she-bang of the variational theory as such.)

Notice that in this second paradigm, (i) we did not restrict the analysis to only those quantities that are actually taking place in reality; we also included a host (possibly an infinity) of purely hypothetical combinations of quantities too; (ii) we worked with energy, a scalar quantity, rather than with momentum, a vector quantity; and finally, (iii) in the variational method, we didn’t bother about the local details. We took into account the displacements of the disks, but not any displacement at any other point, say in the gas. We did not look into presence or absence of a pulse at one point in the gas as contrasted from any other point in it. In short, we did not discuss the details local to the system either in space or in time. We did not follow the system evolution, at all—not at least in a detailed, local way. If we were to do that, we would be concerned about what happens in the system at the instants and at spatial points other than the initial and final disk positions. Instead, we looked only at a global property—viz. the energy—whether at the sub-system level of the individual compartments, or at the level of the overall system.

The Two Paradigms Contrasted from Each Other:

If we were to follow Newton’s method, it would be impossible—impossible in principle—to be able to predict the final disk positions unless all their motions over all the intermediate transient dynamics (occurring over each moment of time and at each place of the system) were not be traced. Newton’s (or vectorial) method would require us to follow all the details of the entire evolution of all parts of the system at each point on its evolution path. In the variational approach, the latter is not of any primary concern.

Yet, in following the energetics program, we are able to predict the final disk positions. We are able to do that without worrying about what all happened before the equilibrium gets established. We remain concerned only with certain global quantities (here, system-energy) at each of the hypothetical positions.

The upside of the energetics program, as just noted, is that we don’t have to look into every detail at every stage of the entire transient dynamics.

Its downside is that we are able to talk only of the differences between certain isolated (hypothetical) configurations or states. The formalism is unable to say anything at all about any of the intermediate states—even if these do actually occur in reality. This is a very, very important point to keep in mind.

The Question:

Now, the question with which we began this post. Namely, what does the delta of the variational calculus mean?

Referring to the above discussion, note that the delta of the variational calculus is, here, nothing but a change in the position-pair, and also the corresponding change in the energy.

Thus, in the above example, the difference of the state (2,3) from the other close states such as (3,3), (1,3), (2,4), and (2,2) represents a variation in the system configuration (or state), and for each such a variation in the system configuration (or state), there is a corresponding variation in the energy $\delta E_{ni}$ of the system. That is what the delta refers to, in this example.

Now, with all this discussion and clarification, would it be possible for you to clearly state what the physical meaning of the delta is? To what precisely does the concept refer? How does the variation in energy $\delta E$ differ from both the finite changes ($\Delta E$) as well as the infinitesimal changes ($\text{d}E$) of the usual calculus?

Note, the question is conceptual in nature. And, no, not a single one of the very best books on classical mechanics manages to give a very succinct and accurate answer to it. Not even Rana and Joag (or Goldstein, or Feynman, or…)

I will give my answer in my next post, next year. I will also try to apply it to a couple of more interesting (and somewhat more complicated) physical situations—one from engineering sciences, and another from quantum mechanics!

In the meanwhile, think about it—the delta—the concept itself, its (conceptual) meaning. (If you already know the calculus of variations, note that in my above write-up, I have already supplied the answer, in a way. You just have to think a bit about it, that’s all!)

An Important Note: Do bring this post to the notice of the Officially Approved Full Professors of Mechanical Engineering in SPPU, and the SPPU authorities. I would like to know if the former would be able to state the meaning—at least now that I have already given the necessary context in such great detail.

Ditto, to the Officially Approved Full Professors of Mechanical Engineering at COEP, esp. D. W. Pande, and others like them.

After all, this topic—Lagrangian mechanics—is at the core of Mechanical Engineering, even they would agree. In fact, it comes from a subject that is not taught to the metallurgical engineers, viz., the topic of Theory of Machines. But it is taught to the Mechanical Engineers. That’s why, they should be able to crack it, in no time.

(Let me continue to be honest. I do not expect them to be able to crack it. But I do wish to know if they are able at least to give a try that is good enough!)

Even though I am jobless (and also nearly bank balance-less, and also cashless), what the hell! …

…Season’s greetings and best wishes for a happy new year!

A Song I Like:

[With jobless-ness and all, my mood isn’t likely to stay this upbeat, but anyway, while it lasts, listen to this song… And, yes, this song is like, it’s like, slightly more than 60 years old!]

(Hindi) “yeh raat bhigee bhigee”
Music: Shankar-Jaikishan
Singers: Manna De and Lata Mangeshkar
Lyrics: Shailendra

[E&OE]

# The Infosys Prizes, 2015

I realized that it was the end of November the other day, and it somehow struck me that I should check out if there has been any news on the Infosys prizes for this year. I vaguely recalled that they make the yearly announcements sometime in the last quarter of a year.

Turns out that, although academic bloggers whose blogs I usually check out had not highlighted this news, the prizes had already been announced right in mid-November [^].

It also turns out also that, yes, I “know”—i.e., have in-person chatted (exactly once) with—one of the recipients. I mean Professor Dr. Umesh Waghmare, who received this year’s award for Engineering Sciences [^]. I had run into him in an informal conference once, and have written about it in a recent post, here [^].

Dr. Waghmare is a very good choice, if you ask me. His work is very neat—I mean both the ideas which he picks out to work on, and the execution on them.

I still remember his presentation at that informal conference (where I chatted with him). He had talked about a (seemingly) very simple idea, related to graphene [^]—its buckling.

Here is my highly dumbed down version of that work by Waghmare and co-authors. (It’s dumbed down a lot—Waghmare et al’s work was on buckling, not bending. But it’s OK; this is just a blog, and guess I have a pretty general sort of a “general readership” here.)

Bending, in general, sets up a combination of tensile and compressive stresses, which results in the setting up of a bending moment within a beam or a plate. All engineers (except possibly for the “soft” branches like CS and IT) study bending quite early in their undergraduate program, typically in the second year. So, I need not explain its analysis in detail. In fact, in this post, I will write only a common-sense level description of the issue. For technical details, look up the Wiki articles on bending [^] and buckling [^] or Prof. Bower’s book [^].

Assuming you are not an engineer, you can always take a longish rubber eraser, hold it so that its longest edge is horizontal, and then bend it with a twist of your fingers. If the bent shape is like an inverted ‘U’, then, the inner (bottom) surface has got compressed, and the outer (top) surface has got stretched. Since compression and tension are opposite in nature, and since the eraser is a continuous body of a finite height, it is easy to see that there has to be a continuous surface within the volume of the eraser, some half-way through its height, where there can be no stresses. That’s because, the stresses change sign in going from the compressive stress at the bottom surface to the tensile stresses on the top surface. For simplicity of mathematics, this problem is modeled as a 1D (line) element, and therefore, in elasticity theory, this actual 2D surface is referred to as the neutral axis (i.e. a line).

The deformation of the eraser is elastic, which means that it remains in the bent state only so long as you are applying a bending “force” to it (actually, it’s a moment of a force).

The classical theory of bending allows you to relate the curvature of the beam, and the bending moment applied to it. Thus, knowing bending moment (or the applied forces), you can tell how much the eraser should bend. Or, knowing how much the eraser has curved, you can tell how big a pair of fforces would have to be applied to its ends. The theory works pretty well; it forms of the basis of how most buildings are designed anyway.

So far, so good. What happens if you bend, not an eraser, but a graphene sheet?

The peculiarity of graphene is that it is a single atom-thick sheet of carbon atoms. Your usual eraser contains billions and billions of layers of atoms through its thickness. In contrast, the thickness of a graphene sheet is entirely accounted for by the finite size of the single layer of atoms. And, it is found that unlike thin paper, the graphen sheet, even if it is the the most extreme case of a thin sheet, actually does offer a good resistance to bending. How do you explain that?

The naive expectation is that something related to the interatomic bonding within this single layer must, somehow, produce both the compressive and tensile stresses—and the systematic variation from the locally tensile to the locally compressive state as we go through this thickness.

Now, at the scale of single atoms, quantum mechanical effects obviously are dominant. Thus, you have to consider those electronic orbitals setting up the bond. A shift in the density of the single layer of orbitals should correspond to the stresses and strains in the classical mechanics of beams and plates.

What Waghmare related at that conference was a very interesting bit.

He calculated the stresses as predicted by (in my words) the changed local density of the orbitals, and found that the forces predicted this way are way smaller than the experimentally reported values for graphene sheets. In other words, the actual graphene is much stiffer than what the naive quantum mechanics-based model shows—even if the model considers those electronic orbitals. What is the source of this additional stiffness?

He then showed a more detailed calculation (i.e. a simulation), and found that the additional stiffness comes from a quantum-mechanical interaction between the portions of the atomic orbitals that go off transverse to the plane of the graphene sheet.

Thus, suppose a graphene sheet is initially held horizontally, and then bent to form an inverted U-like curvature. According to Waghmare and co-authros, you now have to consider not just the orbital cloud between the atoms (i.e. the cloud lying in the same plane as the graphene sheet) but also the orbital “petals” that shoot vertically off the plane of the graphene. Such petals are attached to nucleus of each C atom; they are a part of the electronic (or orbital) structure of the carbon atoms in the graphene sheet.

In other words, the simplest engineering sketch for the graphene sheet, as drawn in the front view, wouldn’t look like a thin horizontal line; it would also have these small vertical “pins” at the site of each carbon atom, overall giving it an appearance rather like a fish-bone.

What happens when you bend the graphene sheet is that on the compression side, the orbital clouds for these vertical petals run into each other. Now, you know that an orbital cloud can be loosely taken as the electronic charge density, and that the like charges (e.g. the negatively charged electrons) repel each other. This inter-electronic repulsive force tends to oppose the bending action. Thus, it is the petals’ contribution which accounts for the additional stiffness of the graphene sheet.

I don’t know whether this result was already known to the scientific community back then in 2010 or not, but in any case, it was a very early analysis of bending of graphene. Further, as far as I could tell, the quality of Waghmare’s calculations and simulations was very definitely superlative. … You work in a field (say computational modeling) for some time, and you just develop a “nose” of sorts, that allows you to “smell” a superlative calculation from an average one. Particularly so, if your own skills on the calculations side are rather on the average, as happens to be the case with me. (My strengths are in conceptual and computational sides, but not on the mathematical side.) …

So, all in all, it’s a very well deserved prize. Congratulations, Dr. Waghmare!

A Song I Like:

(The so-called “fusion” music) “Jaisalmer”
Artists: Rahul Sharma (Santoor) and Richard Clayderman (Piano)
Album: Confluence

[As usual, may be one more editing pass…]

[E&OE]

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# The most economic particles model of a[n utterly] fake fluid—part 1

Real fluids are viscous.

Newton was the first to formulate a law of viscosity; his law forms an essential part of the engineering fluid mechanics even today.

The way the concept of viscosity is usually presented to undergraduates is in reference to a fluid moving over a horizontal solid surface, e.g., water flowing over a flat river-bed. The river-bed itself is, of course, stationary. The students are then asked to imagine a laminar flow in which the horizontal layers of fluid go slipping past each other with different velocities. The viscous forces between the fluid layers tend to retard their relative motion. Now, under the assumption that the layer adjacent to the stationary solid surface has zero velocity, and that the flow is laminar, a simple parabolic profile is obtained for the velocity profile. The velocity progressively increases from 0 at the solid surface to some finite mainstream value as you go up and away from the horizontal solid surface. Newton’s law is then introduced via the equation:

$\tau \propto \dfrac{dU}{dy}$

where $\tau$ is the shear stress between the fluid layers slipping past each other, and $\frac{dU}{dy}$ is the velocity gradient along the vertical direction. The constant of proportionality is viscosity, $\mu$:

$\mu \equiv \dfrac{\tau}{\left(\dfrac{dU}{dy}\right)}$

This picture of layers of fluids slipping past with progressively greater velocities, as in a deck of card given a gentle horizontal push, is easy to visualize; it helps people visualize what otherwise is not available to direct perception.

That’s quite fine, but then, as it happens, sometimes, concrete pictures also tend to over-concretize the abstract ideas. The above mentioned picture of viscosity is one of these.

You see, the trouble is, people tend to associate viscosity to be operative only in this shear mode. They can’t readily appreciate the fact that viscous forces also arise in the normal direction. The reason is, they can’t as easily imagine velocity gradients along the flow direction. No engineering (or physics) text-book ever shows a diagram illustrating the action of viscosity along the direction of the flow.

One reason for that, in turn, is that while in solids stress depends on the extent of deformation, in fluids, it depends on the rate of deformation. Indeed the extent of possible deformation, in fluids, is theoretically undefined (or infinite, if you wish). A fluid will continually go on changing its shape so long as a shear stress is applied to it… It’s easily possible to pour water from a tap onto a tilted plate, and then, from that plate onto the bottom of a kitchen sink, without any additional stress coming into picture as the water continuously goes on deforming in the act of pouring. The fact that water has already suffered deformation while being poured from the tap to the plate, does not hinder the additional deformation that it further suffers while falling off from the plate. And, all this deformation, inasmuch as it involves a change of the initial shape, involves only shear. And, as to the stress, when it comes to fluids, the extent of deformation does not matter; the rate of deformation—or the velocity gradient—does. Stress in fluids is related to the velocity gradient, not to the deformation gradient (as in solids).

Another, related, reason for the difficulty in visualizing viscosity appearing in the normal direction is that, in our usual imagination, we can’t visualize fluids being gripped from its ends and pulled apart, the way solids (e.g. rubber band) can be. The trouble is not in the stretching part of it; the trouble is in the “being held” part of it: you can’t grab of a piece of a fluid in exactly the same way as you can, say, a bite of food. There is no bite of water, only a gulp of it. But the practical impossibility of holding fast onto an end of a fluid also carries over when it comes to imagining fluids being stretched purely along the normal direction, i.e., without involving shear.

Of course, as far as exerting a normal force to a fluid is concerned, people have no difficulty imagining that. You can always exert a compressive normal force on a fluid, by applying a pressure. But then, that is only a compressive force, and, a static situation. You don’t have to have spatially varying velocities to arrive at the concept of pressure—indeed, you don’t make any reference to the very idea of velocity, in that concept. Pressure refers to static forces.

Now, when people try to visualize velocity gradients in the normal direction, they unwittingly tend to take the visualization on the lines parallel to the viscosity-defining picture. So, they take, say, a 10 m/s velocity vector at origin, an 8 m/s velocity vector at the point x = 1, a 6 m/s vector at x = 2, and so on. Soon, they end up imagining having a zero magnitude velocity vector.

But this is a poorly imagined situation because it can never be realized in one-dimension—quantitatively, it violates the mass conservation principle, i.e. the continuity equation (at least for the incompressible 1D flow without sources/sinks, it does).

Now, when pushed further, people do end up imagining an L’ kind of bend in a pipe (or a fluid bifurcating at aT’ joint), i.e., taking velocity vectors to be just x-components of a 2D/3D velocity field.

But, speaking in general terms, at least in my observation, people still can’t easily imagine viscosity being defined in reference to velocity gradients along the direction of the flow. Many engineers in fact express a definite surprise at such a definition of viscosity. The only picture ever presented to them refers to the shear deformation, and given the peculiar nature of fluids, velocity gradients in the normal direction (i.e. along the flow) are not as easy to visualize unless you are willing to break continuity.

Recently when Prof. Suo wrote an iMechanica post about viscosity (in reference to a course he is currently teaching at Harvard), the above-mentioned observations came rushing to my mind, and that’s how I had a bit of discussion with him on this topic, here [^].

As mentioned in that discussion, to help people visualize the normal viscosity, I then thought of introducing a particles model of fluid, specifically, the Lennard-Jones (LJ) fluid [^]. It also goes well with my research interests concerning the particles approaches to fluids.

But then, of course, I have been too busy just doing the class-room teaching this semester, and find absolutely no time to pursue anything other than that—class-room teaching, or preparation for the same, or follow-up activities concerning the same (e.g. designing assignments, unit tests, etc.). But no time at all is left for research, blogging, or why, even just building a few software toys at home. (As a matter of fact, I find myself hard-pressed to find time even for just grading of unit-test answer-books.)

Therefore, writing some quick-n-simple illustrative software (actually, completing writing this software—something which I had began last summer) was out of the question. Still, I wanted to steal some time, to think about this question.

I therefore decided to drastically simplify the matters. I would work on the problem, but only to the extent that I can work on it off my head (i.e. without using even paper and pencil, let alone a computer or a software)—that’s what I decided.

So, instead of taking the $(1/r)^{12} - (1/r)^6$ potential, I began wondering what if I take a simple $1/r$ attractive potential (as in Newtonian gravity). After all, most every one knows about the inverse-square law, and so, it would be easier for people to make the conceptual connections if a fluid could be built also out of the plain inverse-distance potential.

So, the question was: (Q1) if I take a few particles with (only attractive) gravitational interactions among them—would they create a fluid out of them, just the way the LJ potential does? And if the answer is yes, then would these particles also create a solid out of them, too, just the way the LJ potential does?

Before you rush into an affirmative answer, realize here that the LJ potential carries both attractive and repulsive terms, whereas the gravitational interaction is always only attractive.

But, still, suppose such a hypothetical fluid is possible, then, (Q2) what would distinguish this hypothetical fluid from its corresponding solid? How precisely would the phase transition between the solid and fluid occur? For instance, how would the fluid consisting of only gravity-interacting particles, melt or solidify?

And, (Q3) what is the minimum number of such particles that must be present before they can create a solid? a fluid? a liquid? a gas?

Of course, answering these questions is not a big deal (neither is thinking up these questions). The point is, I had some fun thinking along these lines, in whatever time I could still find.

However, since this post is already more than a thousand words-long, let me stop here, and ask you to think about the above mentioned questions. In my next post, I will give my answers to them. In the meanwhile, think about it, have fun, and if you think you have got an answer that you could share with me, feel free to drop a comment or an email.

* * * * *   * * * * *   * * * * *

A Song I Like:
(Marathi) “waaT ithe swapnaatil sampali jaNu…”
Singer: Suman Kalyanpur
Music: Ashok Patki
Lyrics: Ashok Paranjape

[E&OE]

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