# Fluxes, scalars, vectors, tensors…. and, running in circles about them!

0. This post is written for those who know something about Thermal Engineering (i.e., fluid dynamics, heat transfer, and transport phenomena) say up to the UG level at least. [A knowledge of Design Engineering, in particular, the tensors as they appear in solid mechanics, would be helpful to have but not necessary. After all, contrary to what many UGC and AICTE-approved (Full) Professors of Mechanical Engineering teaching ME (Mech – Design Engineering) courses in SPPU and other Indian universities believe, tensors not only appear also in fluid mechanics, but, in fact, the fluids phenomena make it (only so slightly) easier to understand this concept. [But all these cartoons characters, even if they don’t know even this plain and simple a fact, can always be fully relied (by anyone) about raising objections about my Metallurgy background, when it comes to my own approval, at any time! [Indians!!]]]

In this post, I write a bit about the following question:

Why is the flux $\vec{J}$ of a scalar $\phi$ a vector quantity, and not a mere number (which is aka a “scalar,” in certain contexts)? Why is it not a tensor—whatever the hell the term means, physically?

And, what is the best way to define a flux vector anyway?

1.

One easy answer is that if the flux is a vector, then we can establish a flux-gradient relationship. Such relationships happen to appear as statements of physical laws in all the disciplines wherever the idea of a continuum was found useful. So the scope of the applicability of the flux-gradient relationships is very vast.

The reason to define the flux as a vector, then, becomes: because the gradient of a scalar field is a vector field, that’s why.

But this answer only tells us about one of the end-purposes of the concept, viz., how it can be used. And then the answer provided is: for the formulation of a physical law. But this answer tells us nothing by way of the very meaning of the concept of flux itself.

2.

Another easy answer is that if it is a vector quantity, then it simplifies the maths involved. Instead of remembering having to take the right $\theta$ and then multiplying the relevant scalar quantity by the $\cos$ of this $\theta$, we can more succinctly write:

$q = \vec{J} \cdot \vec{S}$ (Eq. 1)

where $q$ is the quantity of $\phi$, an intensive scalar property of the fluid flowing across a given finite surface, $\vec{S}$, and $\vec{J}$ is the flux of $\Phi$, the extensive quantity corresponding to the intensive quantity $\phi$.

However, apart from being a mere convenience of notation—a useful shorthand—this answer once again touches only on the end-purpose, viz., the fact that the idea of flux can be used to calculate the amount $q$ of the transported property $\Phi$.

There also is another problem with this, second, answer.

Notice that in Eq. 1, $\vec{J}$ has not been defined independently of the “dotting” operation.

If you have an equation in which the very quantity to be defined itself has an operator acting on it on one side of an equation, and then, if a suitable anti- or inverse-operator is available, then you can apply the inverse operator on both sides of the equation, and thereby “free-up” the quantity to be defined itself. This way, the quantity to be defined becomes available all by itself, and so, its definition in terms of certain hierarchically preceding other quantities also becomes straight-forward.

OK, the description looks more complex than it is, so let me illustrate it with a concrete example.

Suppose you want to define some vector $\vec{T}$, but the only basic equation available to you is:

$\vec{R} = \int \text{d} x \vec{T}$, (Eq. 2)

assuming that $\vec{T}$ is a function of position $x$.

In Eq. 2, first, the integral operator must operate on $\vec{T}(x)$ so as to produce some other quantity, here, $\vec{R}$. Thus, Eq. 2 can be taken as a definition for $\vec{R}$, but not for $\vec{T}$.

However, fortunately, a suitable inverse operator is available here; the inverse of integration is differentiation. So, what we do is to apply this inverse operator on both sides. On the right hand-side, it acts to let $\vec{T}$ be free of any operator, to give you:

$\dfrac{\text{d}\vec{R}}{\text{d}x} = \vec{T}$ (Eq. 3)

It is the Eq. 3 which can now be used as a definition of $\vec{T}$.

In principle, you don’t have to go to Eq. 3. In principle, you could perhaps venture to use a bit of notation abuse (the way the good folks in the calculus of variations and integral transforms always did), and say that the Eq. 2 itself is fully acceptable as a definition of $\vec{T}$. IMO, despite the appeal to “principles”, it still is an abuse of notation. However, I can see that the argument does have at least some point about it.

But the real trouble with using Eq. 1 (reproduced below)

$q = \vec{J} \cdot \vec{S}$ (Eq. 1)

as a definition for $\vec{J}$ is that no suitable inverse operator exists when it comes to the dot operator.

3.

Let’s try another way to attempt defining the flux vector, and see what it leads to. This approach goes via the following equation:

$\vec{J} \equiv \dfrac{q}{|\vec{S}|} \hat{n}$ (Eq. 4)

where $\hat{n}$ is the unit normal to the surface $\vec{S}$, defined thus:

$\hat{n} \equiv \dfrac{\vec{S}}{|\vec{S}|}$ (Eq. 5)

Then, as the crucial next step, we introduce one more equation for $q$, one that is independent of $\vec{J}$. For phenomena involving fluid flows, this extra equation is quite simple to find:

$q = \phi \rho \dfrac{\Omega_{\text{traced}}}{\Delta t}$ (Eq. 6)

where $\phi$ is the mass-density of $\Phi$ (the scalar field whose flux we want to define), $\rho$ is the volume-density of mass itself, and $\Omega_{\text{traced}}$ is the volume that is imaginarily traced by that specific portion of fluid which has imaginarily flowed across the surface $\vec{S}$ in an arbitrary but small interval of time $\Delta t$. Notice that $\Phi$ is the extensive scalar property being transported via the fluid flow across the given surface, whereas $\phi$ is the corresponding intensive quantity.

Now express $\Omega_{\text{traced}}$ in terms of the imagined maximum normal distance from the plane $\vec{S}$ up to which the forward moving front is found extended after $\Delta t$. Thus,

$\Omega_{\text{traced}} = \xi |\vec{S}|$ (Eq. 7)

where $\xi$ is the traced distance (measured in a direction normal to $\vec{S}$). Now, using the geometric property for the area of parallelograms, we have that:

$\xi = \delta \cos\theta$ (Eq. 8)

where $\delta$ is the traced distance in the direction of the flow, and $\theta$ is the angle between the unit normal to the plane $\hat{n}$ and the flow velocity vector $\vec{U}$. Using vector notation, Eq. 8 can be expressed as:

$\xi = \vec{\delta} \cdot \hat{n}$ (Eq. 9)

Now, by definition of $\vec{U}$:

$\vec{\delta} = \vec{U} \Delta t$, (Eq. 10)

Substituting Eq. 10 into Eq. 9, we get:

$\xi = \vec{U} \Delta t \cdot \hat{n}$ (Eq. 11)

Substituting Eq. 11 into Eq. 7, we get:

$\Omega_{\text{traced}} = \vec{U} \Delta t \cdot \hat{n} |\vec{S}|$ (Eq. 12)

Substituting Eq. 12 into Eq. 6, we get:

$q = \phi \rho \dfrac{\vec{U} \Delta t \cdot \hat{n} |\vec{S}|}{\Delta t}$ (Eq. 13)

Cancelling out the $\Delta t$, Eq. 13 becomes:

$q = \phi \rho \vec{U} \cdot \hat{n} |\vec{S}|$ (Eq. 14)

Having got an expression for $q$ that is independent of $\vec{J}$, we can now use it in order to define $\vec{J}$. Thus, substituting Eq. 14 into Eq. 4:

$\vec{J} \equiv \dfrac{q}{|\vec{S}|} \hat{n} = \dfrac{\phi \rho \vec{U} \cdot \hat{n} |\vec{S}|}{|\vec{S}|} \hat{n}$ (Eq. 16)

Cancelling out the two $|\vec{S}|$s (because it’s a scalar—you can always divide any term by a scalar (or even  by a complex number) but not by a vector), we finally get:

$\vec{J} \equiv \phi \rho \vec{U} \cdot \hat{n} \hat{n}$ (Eq. 17)

In Eq. 17, there is this curious sequence: $\hat{n} \hat{n}$.

It’s a sequence of two vectors, but the vectors apparently are not connected by any of the operators that are taught in the Engineering Maths courses on vector algebra and calculus—there is neither the dot ($\cdot$) operator nor the cross $\times$ operator appearing in between the two $\hat{n}$s.

But, for the time being, let’s not get too much perturbed by the weird-looking sequence. For the time being, you can mentally insert parentheses like these:

$\vec{J} \equiv \left[ \left( \phi \rho \vec{U} \right) \cdot \left( \hat{n} \right) \right] \hat{n}$ (Eq. 18)

and see that each of the two terms within the parentheses is a vector, and that these two vectors are connected by a dot operator so that the terms within the square brackets all evaluate to a scalar. According to Eq. 18, the scalar magnitude of the flux vector is:

$|\vec{J}| = \left( \phi \rho \vec{U}\right) \cdot \left( \hat{n} \right)$ (Eq. 19)

and its direction is given by: $\hat{n}$ (the second one, i.e., the one which appears in Eq. 18 but not in Eq. 19).

5.

We explained away our difficulty about Eq. 17 by inserting parentheses at suitable places. But this procedure of inserting mere parentheses looks, by itself, conceptually very attractive, doesn’t it?

If by not changing any of the quantities or the order in which they appear, and if by just inserting parentheses, an equation somehow begins to make perfect sense (i.e., if it seems to acquire a good physical meaning), then we have to wonder:

Since it is possible to insert parentheses in Eq. 17 in some other way, in some other places—to group the quantities in some other way—what physical meaning would such an alternative grouping have?

That’s a delectable possibility, potentially opening new vistas of physico-mathematical reasonings for us. So, let’s pursue it a bit.

What if the parentheses were to be inserted the following way?:

$\vec{J} \equiv \left( \hat{n} \hat{n} \right) \cdot \left( \phi \rho \vec{U} \right)$ (Eq. 20)

On the right hand-side, the terms in the second set of parentheses evaluate to a vector, as usual. However, the terms in the first set of parentheses are special.

The fact of the matter is, there is an implicit operator connecting the two vectors, and if it is made explicit, Eq. 20 would rather be written as:

$\vec{J} \equiv \left( \hat{n} \otimes \hat{n} \right) \cdot \left( \phi \rho \vec{U} \right)$ (Eq. 21)

The $\otimes$ operator, as it so happens, is a binary operator that operates on two vectors (which in general need not necessarily be one and the same vector as is the case here, and whose order with respect to the operator does matter). It produces a new mathematical object called the tensor.

The general form of Eq. 21 is like the following:

$\vec{V} = \vec{\vec{T}} \cdot \vec{U}$ (Eq. 22)

where we have put two arrows on the top of the tensor, to bring out the idea that it has something to do with two vectors (in a certain order). Eq. 22 may be read as the following: Begin with an input vector $\vec{U}$. When it is multiplied by the tensor $\vec{\vec{T}}$, we get another vector, the output vector: $\vec{V}$. The tensor quantity $\vec{\vec{T}}$ is thus a mapping between an arbitrary input vector and its uniquely corresponding output vector. It also may be thought of as a unary operator which accepts a vector on its right hand-side as an input, and transforms it into the corresponding output vector.

6. “Where am I?…”

Now is the time to take a pause and ponder about a few things. Let me begin doing that, by raising a few questions for you:

Q. 6.1:

What kind of a bargain have we ended up with? We wanted to show how the flux of a scalar field $\Phi$ must be a vector. However, in the process, we seem to have adopted an approach which says that the only way the flux—a vector—can at all be defined is in reference to a tensor—a more advanced concept.

Instead of simplifying things, we seem to have ended up complicating the matters. … Have we? really? …Can we keep the physical essentials of the approach all the same and yet, in our definition of the flux vector, don’t have to make a reference to the tensor concept? exactly how?

(Hint: Look at the above development very carefully once again!)

Q. 6.2:

In Eq. 20, we put the parentheses in this way:

$\vec{J} \equiv \left( \hat{n} \hat{n} \right) \cdot \left( \phi \rho \vec{U} \right)$ (Eq. 20, reproduced)

What would happen if we were to group the same quantities, but alter the order of the operands for the dot operator?  After all, the dot product is commutative, right? So, we could have easily written Eq. 20 rather as:

$\vec{J} \equiv \left( \phi \rho \vec{U} \right) \cdot \left( \hat{n} \hat{n} \right)$ (Eq. 21)

What could be the reason why in writing Eq. 20, we might have made the choice we did?

Q. 6.3:

We wanted to define the flux vector for all fluid-mechanical flow phenomena. But in Eq. 21, reproduced below, what we ended up having was the following:

$\vec{J} \equiv \left( \phi \rho \vec{U} \right) \cdot \left( \hat{n} \otimes \hat{n} \right)$ (Eq. 21, reproduced)

Now, from our knowledge of fluid dynamics, we know that Eq. 21 seemingly stands only for one kind of a flux, namely, the convective flux. But what about the diffusive flux? (To know the difference between the two, consult any good book/course-notes on CFD using FVM, e.g. Jayathi Murthy’s notes at Purdue, or Versteeg and Malasekara’s text.)

Q. 6.4:

Try to pursue this line of thought a bit:

$q = \vec{J} \cdot \vec{S}$ (Eq. 1, reproduced)

Express $\vec{S}$ as a product of its magnitude and direction:

$q = \vec{J} \cdot |\vec{S}| \hat{n}$ (Eq. 23)

Divide both sides of Eq. 23 by $|\vec{S}|$:

$\dfrac{q}{|\vec{S}|} = \vec{J} \cdot \hat{n}$ (Eq. 24)

“Multiply” both sides of Eq. 24 by $\hat{n}$:

$\dfrac{q} {|\vec{S}|} \hat{n} = \vec{J} \cdot \hat{n} \hat{n}$ (Eq. 25)

We seem to have ended up with a tensor once again! (and more rapidly than in the development in section 4. above).

Now, looking at what kind of a change the left hand-side of Eq. 24 undergoes when we “multiply” it by a vector (which is: $\hat{n}$), can you guess something about what the “multiplication” on the right hand-side by $\hat{n}$ might mean? Here is a hint:

To multiply a scalar by a vector is meaningless, really speaking. First, you need to have a vector space, and then, you are allowed to take any arbitrary vector from that space, and scale it up (without changing its direction) by multiplying it with a number that acts as a scalar. The result at least looks the same as “multiplying” a scalar by a vector.

What then might be happening on the right hand side?

Q.6.5:

Recall your knowledge (i) that vectors can be expressed as single-column or single-row matrices, and (ii) how matrices can be algebraically manipulated, esp. the rules for their multiplications.

Try to put the above developments using an explicit matrix notation.

In particular, pay particular attention to the matrix-algebraic notation for the dot product between a row- or column-vector and a square matrix, and the effect it has on your answer to question Q.6.2. above. [Hint: Try to use the transpose operator if you reach what looks like a dead-end.]

Q.6.6.

Suppose I introduce the following definitions: All single-column matrices are “primary” vectors (whatever the hell it may mean), and all single-row matrices are “dual” vectors (once again, whatever the hell it may mean).

Given these definitions, you can see that any primary vector can be turned into its corresponding dual vector simply by applying the transpose operator to it. Taking the logic to full generality, the entirety of a given primary vector-space can then be transformed into a certain corresponding vector space, called the dual space.

Now, using these definitions, and in reference to the definition of the flux vector via a tensor (Eq. 21), but with the equation now re-cast into the language of matrices, try to identify the physical meaning the concept of “dual” space. [If you fail to, I will sure provide a hint.]

As a part of this exercise, you will also be able to figure out which of the two $\hat{n}$s forms the “primary” vector space and which $\hat{n}$ forms the dual space, if the tensor product $\hat{n}\otimes\hat{n}$ itself appears (i) before the dot operator or (ii) after the dot operator, in the definition of the flux vector. Knowing the physical meaning for the concept of the dual space of a given vector space, you can then see what the physical meaning of the tensor product of the unit normal vectors ($\hat{n}$s) is, here.

Over to you. [And also to the UGC/AICTE-Approved Full Professors of Mechanical Engineering in SPPU and in other similar Indian universities. [Indians!!]]

A Song I Like:

[TBD, after I make sure all LaTeX entries have come out right, which may very well be tomorrow or the day after…]

# Machine “Learning”—An Entertainment [Industry] Edition

Yes, “Machine ‘Learning’,” too, has been one of my “research” interests for some time by now. … Machine learning, esp. ANN (Artificial Neural Networks), esp. Deep Learning. …

Yesterday, I wrote a comment about it at iMechanica. Though it was made in a certain technical context, today I thought that the comment could, perhaps, make sense to many of my general readers, too, if I supply a bit of context to it. So, let me report it here (after a bit of editing). But before coming to my comment, let me first give you the context in which it was made:

Context for my iMechanica comment:

It all began with a fellow iMechanician, one Mingchuan Wang, writing a post of the title “Is machine learning a research priority now in mechanics?” at iMechanica [^]. Biswajit Banerjee responded by pointing out that

“Machine learning includes a large set of techniques that can be summarized as curve fitting in high dimensional spaces. [snip] The usefulness of the new techniques [in machine learning] should not be underestimated.” [Emphasis mine.]

Then Biswajit had pointed out an arXiv paper [^] in which machine learning was reported as having produced some good DFT-like simulations for quantum mechanical simulations, too.

A word about DFT for those who (still) don’t know about it:

DFT, i.e. Density Functional Theory, is “formally exact description of a many-body quantum system through the density alone. In practice, approximations are necessary” [^]. DFT thus is a computational technique; it is used for simulating the electronic structure in quantum mechanical systems involving several hundreds of electrons (i.e. hundreds of atoms). Here is the obligatory link to the Wiki [^], though a better introduction perhaps appears here [(.PDF) ^]. Here is a StackExchange on its limitations [^].

Trivia: Kohn and Sham received a Physics Nobel for inventing DFT. It was a very, very rare instance of a Physics Nobel being awarded for an invention—not a discovery. But the Nobel committee, once again, turned out to have put old Nobel’s money in the right place. Even if the work itself was only an invention, it did directly led to a lot of discoveries in condensed matter physics! That was because DFT was fast—it was fast enough that it could bring the physics of the larger quantum systems within the scope of (any) study at all!

And now, it seems, Machine Learning has advanced enough to be able to produce results that are similar to DFT, but without using any QM theory at all! The computer does have to “learn” its “art” (i.e. “skill”), but it does so from the results of previous DFT-based simulations, not from the theory at the base of DFT. But once the computer does that—“learning”—and the paper shows that it is possible for computer to do that—it is able to compute very similar-looking simulations much, much faster than even the rather fast technique of DFT itself.

OK. Context over. Now here in the next section is my yesterday’s comment at iMechanica. (Also note that the previous exchange on this thread at iMechanica had occurred almost a year ago.) Since it has been edited quite a bit, I will not format it using a quotation block.

[An edited version of my comment begins]

A very late comment, but still, just because something struck me only this late… May as well share it….

I think that, as Biswajit points out, it’s a question of matching a technique to an application area where it is likely to be of “good enough” a fit.

I mean to say, consider fluid dynamics, and contrast it to QM.

In (C)FD, the nonlinearity present in the advective term is a major headache. As far as I can gather, this nonlinearity has all but been “proved” as the basic cause behind the phenomenon of turbulence. If so, using machine learning in CFD would be, by the simple-minded “analysis”, a basically hopeless endeavour. The very idea of using a potential presupposes differential linearity. Therefore, machine learning may be thought as viable in computational Quantum Mechanics (viz. DFT), but not in the more mundane, classical mechanical, CFD.

But then, consider the role of the BCs and the ICs in any simulation. It is true that if you don’t handle nonlinearities right, then as the simulation time progresses, errors are soon enough going to multiply (sort of), and lead to a blowup—or at least a dramatic departure from a realistic simulation.

But then, also notice that there still is some small but nonzero interval of time which has to pass before a really bad amplification of the errors actually begins to occur. Now what if a new “BC-IC” gets imposed right within that time-interval—the one which does show “good enough” an accuracy? In this case, you can expect the simulation to remain “sufficiently” realistic-looking for a long, very long time!

Something like that seems to have been the line of thought implicit in the results reported by this paper: [(.PDF) ^].

Machine learning seems to work even in CFD, because in an interactive session, a new “modified BC-IC” is every now and then is manually being introduced by none other than the end-user himself! And, the location of the modification is precisely the region from where the flow in the rest of the domain would get most dominantly affected during the subsequent, small, time evolution.

It’s somewhat like an electron rushing through a cloud chamber. By the uncertainty principle, the electron “path” sure begins to get hazy immediately after it is “measured” (i.e. absorbed and re-emitted) by a vapor molecule at a definite point in space. The uncertainty in the position grows quite rapidly. However, what actually happens in a cloud chamber is that, before this cone of haziness becomes too big, comes along another vapor molecule, and “zaps” i.e. “measures” the electron back on to a classical position. … After a rapid succession of such going-hazy-getting-zapped process, the end result turns out to be a very, very classical-looking (line-like) path—as if the electron always were only a particle, never a wave.

Conclusion? Be realistic about how smart the “dumb” “curve-fitting” involved in machine learning can at all get. Yet, at the same time, also remain open to all the application areas where it can be made it work—even including those areas where, “intuitively”, you wouldn’t expect it to have any chance to work!

[An edited version of my comment is over. Original here at iMechanica [^]]

“Boy, we seem to have covered a lot of STEM territory here… Mechanics, DFT, QM, CFD, nonlinearity. … But where is either the entertainment or the industry you had promised us in the title?”

You might be saying that….

Well, the CFD paper I cited above was about the entertainment industry. It was, in particular, about the computer games industry. Go check out SoHyeon Jeong’s Web site for more cool videos and graphics [^], all using machine learning.

And, here is another instance connected with entertainment, even though now I am going to make it (mostly) explanation-free.

Check out the following piece of art—a watercolor landscape of a monsoon-time but placid sea-side, in fact. Let me just say that a certain famous artist produced it; in any case, the style is plain unmistakable. … Can you name the artist simply by looking at it? See the picture below:

A sea beach in the monsoons. Watercolor.

If you are unable to name the artist, then check out this story here [^], and a previous story here [^].

A Song I Like:

And finally, to those who have always loved Beatles’ songs…

Here is one song which, I am sure, most of you had never heard before. In any case, it came to be distributed only recently. When and where was it recorded? For both the song and its recording details, check out this site: [^]. Here is another story about it: [^]. And, if you liked what you read (and heard), here is some more stuff of the same kind [^].

Endgame:

I am of the Opinion that 99% of the “modern” “artists” and “music composers” ought to be replaced by computers/robots/machines. Whaddya think?

[Credits: “Endgame” used to be the way Mukul Sharma would end his weekly Mindsport column in the yesteryears’ Sunday Times of India. (The column perhaps also used to appear in The Illustrated Weekly of India before ToI began running it; at least I have a vague recollection of something of that sort, though can’t be quite sure. … I would be a school-boy back then, when the Weekly perhaps ran it.)]

# Micro-level water-resources engineering—8: Measure that water evaporation! Right now!!

It’s past the middle of May—the hottest time of the year in India.

The day-time is still lengthening. And it will continue doing so well up to the summer solstice in the late June, though once monsoon arrives some time in the first half of June, the solar flux in this part of the world would get reduced due to the cloud cover, and so, any further lengthening of the day would not matter.

In the place where I these days live, the day-time temperature easily goes up to 42–44 deg. C. This high a temperature is, that way, not at all unusual for most parts of Maharashtra; sometimes Pune, which is supposed to be a city of a pretty temperate climate (mainly because of the nearby Sahyaadris), also registers the max. temperatures in the early 40s. But what makes the region where I currently live worse than Pune are these two factors: (i) the minimum temperature too stays as high as 30–32 deg. C here whereas in Pune it could easily be falling to 27–26 deg. C even during May, and (ii) the fall of the temperatures at night-time proceeds very gradually here. On a hot day, it can easily be as high as 38 deg C. even after the sunset, and even 36–37 deg. C right by the time it’s the mid-night; the drop below 35 deg. C occurs only for the 3–4 hours in the early morning, between 4 to 7 AM. In comparison, Pune is way cooler. The max. temperatures Pune registers may be similar, but the evening- and the night-time temperatures fall down much more rapidly there.

There is a lesson for the media here. Media obsesses over the max. temperature (and its record, etc.). That’s because the journos mostly are BAs. (LOL!) But anyone who has studied physics and calculus knows that it’s the integral of temperature with respect to time that really matters, because it is this quantity which scales with the total thermal energy transferred to a body. So, the usual experience common people report is correct. Despite similar max. temperatures, this place is hotter, much hotter than Pune.

And, speaking of my own personal constitution, I can handle a cold weather way better than I can handle—if at all I can handle—a hot weather. [Yes, in short, I’ve been in a bad shape for the past month or more. Lethargic. Lackadaisical. Enervated. You get the idea.]

But why is it that the temperature does not matter as much as the thermal energy does?

Consider a body, say a cube of metal. Think of some hypothetical apparatus that keeps this body at the same cool temperature at all times, say, at 20 deg. C.  Here, choose the target temperature to be lower than the minimum temperature in the day. Assume that the atmospheric temperature at two different places varies between the same limits, say, 42 to 30 deg. C. Since the target temperature is lower than the minimum ambient temperature, you would have to take heat out of the cube at all times.

The question is, at which of the two places the apparatus has to work harder. To answer that question, you have to calculate the total thermal energy that has be drained out of the cube over a single day. To answer this second question, you would need the data of not just the lower and upper limits of the temperature but also how it varies with time between two limits.

The humidity too is lower here as compared to in Pune (and, of course, in Mumbai). So, it feels comparatively much more drier. It only adds to the real feel of a real hot weather.

One does not realize it, but the existence of a prolonged high temperature makes the atmosphere here imperceptibly slowly but also absolutely insurmountably, dehydrating.

Unlike in Mumbai, one does not notice much perspiration here, and that’s because the air is so dry that any perspiration that does occur also dries up very fast. Shirts getting drenched by perspiration is not a very common sight here. Overall, desiccating would be the right word to describe this kind of an air.

So, yes, it’s bad, but you can always take precautions. Make sure to drink a couple of glasses of cool water (better still, fresh lemonade) before you step out—whether you are thirsty or not. And take an onion with you when you go out; if you begin to feel too much of heat, you can always crush the onion with hand and apply the juice onto the top of your head. [Addendum: A colleague just informed me that it’s even better to actually cut the onion and keep its cut portion touching to your body, say inside your shirt. He has spent summers in eastern Maharashtra, where temperatures can reach 47 deg. C. … Oh well!]

Also, eat a lot more onions than you normally do.

And, once you return home, make sure not to drink water immediately. Wait for 5–10 minutes. Otherwise, the body goes into a shock, and the ensuing transient spikes in your biological metabolism can, at times, even trigger the sun-stroke—which can even be fatal. A simple precaution helps avoid it.

For the same reason, take care to sit down in the shade of a tree for a few minutes before you eat that slice of water-melon. Water-melon is nothing but more than 95% water, thrown with a little sugar, some fiber, and a good measure of minerals. All in all, good for your body because even if the perspiration is imperceptible in the hot and dry regions, it is still occurring, and with it, the body is being drained of the necessary electrolytes and minerals. … Lemonades and water-melons supply the electrolytes and the minerals. People do take care not to drink lemonade in the Sun, but they don’t always take the same precaution for water-melon. Yet, precisely because a water-melon has so much water, you should take care not to expose your body to a shock. [And, oh, BTW, just in case you didn’t know already, the doctor-recommended alternative to Electral powder is: your humble lemonade! Works exactly equivalently!!]

Also, the very low levels of humidity also imply that in places like this, the desert-cooler is effective, very effective. The city shops are full of them. Some of these air-coolers sport a very bare-bones design. Nothing fancy like the Symphony Diet cooler (which I did buy last year in Pune!). The air-coolers locally made here can be as simple as just an open tray at the bottom to hold the water, a cube made of a coarse wire-mesh which is padded with the khus/wood sheathings curtain, and a robust fan operating [[very] noisily]. But it works wonderfully. And these local-made air-coolers also are very inexpensive. You can get one for just Rs. 2,500 or 3,000. I mean the ones which have a capacity to keep at least 3–4 people cool.(Branded coolers like the one I bought in Pune—and it does work even in Pune—often go above Rs. 10,000. [I bought that cooler last year because I didn’t have a job, thanks to the Mechanical Engineering Professors in the Savitribai Phule Pune University.])

That way, I also try to think of the better things this kind of an air brings. How the table salt stays so smoothly flowing, how the instant coffee powder or Bournvita never turns into a glue, how an opened packet of potato chips stays so crisp for days, how washed clothes dry up in no time…

Which, incidentally, brings me to the topic of this post.

The middle—or the second half—of May also is the most ideal time to conduct evaporation experiments.

If you are looking for a summer project, here is one: to determine the evaporation rate in your locality.

Take a couple of transparent plastic jars of uniform cross section. The evaporation rate is not very highly sensitive to the cross-sectional area, but it does help to take a vessel or a jar of sizeable diameter.

Affix a mm scale on the outside of each jar, say using cello-tape. Fill the plastic jars to some level almost to the full.

Keep one jar out in the open (exposed to the Sun), and another one, inside your home, in the shade. For the jar kept outside, make sure that birds don’t come and drink the water, thereby messing up with your measurements. For this purpose, you may surround the jar with an enclosure having a coarse mesh. The mesh must be coarse; else it will reduce the solar flux. The “reduction in the solar flux” is just a fancy [mechanical [thermal] engineering] term for saying that the mesh, if too fine, might cast too significant a shadow.

Take measurements of the heights of the water daily at a fixed time of the day, say at 6:00 PM. Conduct the experiment for a week or 10 days.

Then, plot a graph of the daily water level vs. the time elapsed, for each jar.

Realize, the rate of evaporation is measured in terms of the fall in the height, and not in terms of the volume of water lost. That’s because once the exposed area is bigger than some limit, the evaporation rate (the loss in height) is more or less independent of the cross-sectional area.

Now figure out:

Does the evaporation rate stay the same every day? If there is any significant departure from a straight-line graph, how do you explain it? Was there a measurement error? Was there an unusually strong wind on a certain day? a cloud cover?

Repeat the experiment next winter (around the new year), and determine the rate of evaporation at that time.

Later on, also make some calculations. If you are building a check-dam or a farm-pond, how much would be the evaporation loss over the five months from January to May-end? Is the height of your water storage system enough to make it practically useful? economically viable?

A Song I Like:

(Hindi) “mausam aayegaa, jaayegaa, pyaar sadaa muskuraayegaa…”
Music: Manas Mukherjee
Singers: Manna Dey and Asha Bhosale
Lyrics: Vithalbhai Patel

# 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]

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