An INALFO in my backyard

This being a Diwali day, I got up a bit earlier than usual this morning, and as I lazily shuffled out of my bed, as a routine first thing I do, I stumbled my way out to the smallish balcony/porch of our ground-floor flat [i.e. apartment].

While still being in that state of being half-hazy and half awake, in the dim light of the early morning, I spotted an Unidentified Object lying on the ground just 3–4 feet away from me, in the (very smallish) backyard of our house.

It was some 2–3 feet in size; the color was a pink.

Initially, I thought that it could be some piece of a baby clothing that got dropped from one of the flats above ours. Then I thought that it could be a smallish mosquito-net they use for infants. But then, soon enough, I realized that no one lives in the flat directly above ours, and with all the flower-trees and shrubs they have planted around in this backyard, the probability of a piece of clothing—a fairly heavy object—merely getting drifted away in the wind and landing precisely into that very smallish patch of the ground which lies in front of our flat, was very low. After all, there are a couple of 5–7 feet tall trees here, in our patch of the backyard. And, so, the question of how that piece of clothing got landed here was not a very easy one to figure out. … I yawned, decided to think about it later, and turned around to go inside, pick up my tooth-brush.

Some one or one-and-a-half hour later or so, I was in the balcony once again, and checked it out once again. … Whatever it was, I had to go out, pick it up, and may be alert my neighbours later in the day, I thought. The Sun had already been up, and the sky had got brightened up quite well by now.

More important, I had by now become fully awake.

In fact, a few minutes earlier, while sipping up my cup of coffee, while sitting in that same balcony, even “looking at” the same object, I had already thought of some small thing about QM, and so, I had by now picked up this book on QM by Eisberg and Resnick. I was trying to locate the thing I wanted to check out, its presentation in this book.

Yet, at the same time, the Unidentified Object lying out there wouldn’t let me go through the book right. That’s how, I had now decided to check that thing out there, first.

I had to get up from my chair and check it out. So, I leaned out of the balcony a bit, and had a good look at it. Under the brightened sky, and with me not just looking at it but also with my mental focus on it, I could now immediately recognize it for it was.

It was a very specific kind of a Flying Object.

Thus, the object had undergone a direct transformation from being (i) an Unidentified Object, to (i) a specific kind of a Flying Object. [Too bad, there never was an intermediate state of its formally being an Unidentified Flying Object.]

I anyway decided to take a snap of this INALFO (Identified, Not Any Longer Flying, Object). Here it is:

It sure was a Diwali-time balloon (with the hot air generated by a burning candle) of the kind they send it up in the sky at night.

OK. The UFO part of this post is now over.


I then slipped on my chappals and stepped out of our flat. As I picked this balloon up, I realized that something had been printed on the other side of it.

It turned out to be a message of love: there was a screen-printed outline of a rose flower, and also the words appearing in the capital letters: “I LOVE.” … The screen-printer had done a poor job of printing, and so, the “YOU” part had got only partially printed; it was almost completely unreadable.

It was funny, I thought. Right on the morning of the most important day of Diwali, I had happened to have received, literally, a message of love from the heavens. … There was this bit of that “family” / “Greeting Cards” / “Gift from the Heavens” / “TV serials” / “Hindi TV News Channels” / “Indians” sort of a drama built into it—the kind of a drama that everyone seems to so much love or at least appreciate, and, simultaneously, none has any actual use for.

Then, to let you gauge the size of the balloon, I placed my copy of the QM book over it (which I happened to be still carrying in my hand, absentmindedly, even in the act of stepping out). I then took another snap, which is here:

OK. Now… Taking a second look at this photograph…. I mean… if you take just this one photograph, and try to decipher the message contained in this photograph by reading it very straight-forwardly…

… Well, yes, I do LOVE QUANTUM PHYSICS, don’t I? Hadn’t the message arrived from the heavens only so I come to “learn” and “realize” and “appreciate” this part about myself? What do you think?


OK. The Mars-Man [^] [^] part of this post, too, is over.

But, have a Happy Diwali anyway!

And, as to me, well, let me now get back to the business of the accreditation-related work (which I’ve had to take home this Diwali season). Bye for now.

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Incidentals—Part 1

Yes, I have been working hard, very hard, and have been managing a responsibility, a very difficult and demanding and emotionally very draining a responsibility in a singular capacity, and yes, I have been having problems with people—their irrationalities. And, the irrational scripts they follow. [Mind you, the reference is to scripts and not to scriptures.]


But, just check this out this one, for instance (and, the people I have in mind in the above section wouldn’t do that, I am sure; they never do pursue links from such posts of mine, especially if they are just Indians—they are just self-confident, that’s all): [^].

But, leaving them aside—and I find it very, very easy do that at least in a moment like this—here is a suggestion: For tomorrow and the day after, and may be for a week or so, watch out the physics (esp. astronomy-related) Twitter-feeds, news-feeds, even blogs [complete with words like “kicking” people and all [Indians with ability to speak in English regard it as “unparliamentary,” together with words like “bloody”].


For obvious reasons, no “A Song I Like” section for this posts The news I am reporting about is exciting enough, all by itself.

Best,

–Ajit

Update on 21:35 IST the same day:

A couple of related posts are these: [^][^]

A Song I Like:

(Hindi) “meghaa chhaaye aadhee raat…”
Lyrics: Neeraj
Music: S. D. Burman
Singer: Lata Mangeshkar

 

 

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)


4. Comments on 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:

Start with Eq. 1 again:

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…]

Introducing the world at large to a new concept, viz., “Blog-Filling”—Part 1

I hereby introduce to the world at large, awaiting for it with a withheld breath, a new concept, viz. (which is read as “namely” and not “that is,” though the difference has been lost on the English Newspaper Editors of my current town, apparently, long ago; apparently, out of not only a very poor sense of English, but of equally poor sense of supervision descending here from the likes of Delhi and Mumbai—the two highly despicable towns of India).


The concept itself pertains to the idea of having to fill some column-centimeters (or, column-inches in that deprecated country, viz., USA), with whatever it is that you have to fill with.


The world (including the said USA) has been waiting precisely for such a new concept, and I am particularly glad at having not only announcing it, but also having had developed the requisite skills.


The concept in question may most aptly be named: “Blog-Filling.” Translated into a noun, it reads: a “blog filler.”

This post now is [in case you didn’t already guess] is The Blog Filler. [Guess I might have already announced its arrival, given my psycho-epistemological habits i.e. second natures.]


Ummm… In case you still are found wondering, may I repeat, this post really is a blog-filler.


OK. Honest. I will deliver on the promised count. So, here we go: I mean on the RD+Gulzar+Lata song I had had [and may be, also have had/had had/had have/etc.] promised…


A Song I Like:

(Hindi) “silli hawaa chhoo gayee, sillaa badan chhill_ gayaa”
Credits: Are you so dumb as not to be able to guess even these?
OK. I will tell you what? I will note these down, right here:
Lyrics: Gulzaar
Music: R. D. Burman
Singer: Lata Mangeshkar


A “Philanthropic” Assertion:

Even if you are so dumb, and, as usual, richer-than-me, or an Approved SPPU Mechanical Engineering Faculty (or of Any Other Indian University/AICTE/UGC), as not having been able to even guess it, or, in summary, if you are an American Citizen:

Don’t worry, even if you have not been able to guess it. … It was just a small simple game…

…Continuing on the same lines [which lines, people like me don’t need]: now, take care, and best, and good-bye; I mean it; etc.


Bye for now. Don’t bother me too much.