# Is something like a re-discovery of the same thing by the same person possible?

Yes, we continue to remain very busy.

However, in spite of all that busy-ness, in whatever spare time I have [in the evenings, sometimes at nights, why, even on early mornings [which is quite unlike me, come to think of it!]], I cannot help but “think” in a bit “relaxed” [actually, abstract] manner [and by “thinking,” I mean: musing, surmising, etc.] about… about what else but: QM!

So, I’ve been doing that. Sort of like, relaxed distant wonderings about QM…

Idle musings like that are very helpful. But they also carry a certain danger: it is easy to begin to believe your own story, even if the story itself is not being borne by well-established equations (i.e. by physic-al evidence).

But keeping that part aside, and thus coming to the title question: Is it possible that the same person makes the same discovery twice?

It may be difficult to believe so, but I… I seemed to have managed to have pulled precisely such a trick.

Of course, the “discovery” in question is, relatively speaking, only a part of of the whole story, and not the whole story itself. Still, I do think that I had discovered a certain important part of a conclusion about QM a while ago, and then, later on, had completely forgotten about it, and then, in a slow, patient process, I seem now to have worked inch-by-inch to reach precisely the same old conclusion.

In short, I have re-discovered my own (unpublished) conclusion. The original discovery was may be in the first half of this calendar year. (I might have even made a hand-written note about it, I need to look up my hand-written notes.)

Now, about the conclusion itself. … I don’t know how to put it best, but I seem to have reached the conclusion that the postulates of quantum mechanics [^], say as stated by Dirac and von Neumann [^], have been conceptualized inconsistently.

Please note the issue and the statement I am making, carefully. As you know, more than 9 interpretations of QM [^][^][^] have been acknowledged right in the mainstream studies of QM [read: University courses] themselves. Yet, none of these interpretations, as far as I know, goes on to actually challenge the quantum mechanical formalism itself. They all do accept the postulates just as presented (say by Dirac and von Neumann, the two “mathematicians” among the physicists).

Coming to me, my positions: I, too, used to say exactly the same thing. I used to say that I agree with the quantum postulates themselves. My position was that the conceptual aspects of the theory—at least all of them— are missing, and so, these need to be supplied, and if the need be, these also need to be expanded.

But, as far as the postulates themselves go, mine used to be the same position as that in the mainstream.

Until this morning.

Then, this morning, I came to realize that I have “re-discovered,” (i.e. independently discovered for the second time), that I actually should not be buying into the quantum postulates just as stated; that I should be saying that there are theoretical/conceptual errors/misconceptions/misrepresentations woven-in right in the very process of formalization which produced these postulates.

Since I think that I should be saying so, consider that, with this blog post, I have said so.

Just one more thing: the above doesn’t mean that I don’t accept Schrodinger’s equation. I do. In fact, I now seem to embrace Schrodinger’s equation with even more enthusiasm than I have ever done before. I think it’s a very ingenious and a very beautiful equation.

A Song I Like:

(Hindi) “tum jo hue mere humsafar”
Music: O. P. Nayyar
Singers: Geeta Dutt and Mohammad Rafi
Lyrics: Majrooh Sultanpuri

Update on 2017.10.14 23:57 IST: Streamlined a bit, as usual.

# A prediction. Also, a couple of wishes…

The Prediction:

While the week of the Nobel prizes always has a way to generate a sense of suspense, of excitement, and even of wonderment, as far as I am concerned, the one prize that does that in the real sense to me is, of course, the Physics Nobel. … Nothing compares to it. Chemistry can come close, but not always. [And, Mr. Nobel was a good guy; he instituted no prize for maths! [LOL!]]. …

The Physics Nobel is the King of all awards in all fields, as far as I am concerned.

That’s why, this year, I have this feeling of missing something. … The reason is, this year’s Physics Nobel is already “known”; it will go to Kip Thorne and pals.

[I will not eat crow even if they don’t get it. [… Unless, of course, you know a delicious recipe or two for the same, and also demonstrate it to me, complete with you sampling it first.]]

But yes, Kip Thorne richly deserves it, and he will get it. That’s the prediction. I wanted to slip it in even if only few hours before the announcement arrives.

I will update this post later right today/tonight, after the Physics Nobel is actually announced.

Now let me come to the couple of wishes, as mentioned in the title. I will try to be brief. [Have been too busy these days… OK. Will let you know. We are going in for accreditation, and so, it’s been all heavy documentation-related work for the past few months. Despite all that hard-work, we still have managed to slip a bit on the progress, and so, currently, we are working on all week-ends and on most public holidays, too. [Yes, we came to work yesterday.] So, it’s only somehow that I manage to find some time to slip in this post—which is written absolutely on the fly, with no second thoughts or re-reading before posting. … So excuse me if there is a bit of lack of balance in the presentation, and of course, typos etc.]

Wish # 1:

The first wish is that a Physics Nobel should go, in a combined way, to what actually are two separate, but very intimately related, and two most significant advances in the physical understanding of man: (i) chaos theory (including fractals) and (ii)catastrophe theory.

If you don’t like the idea of two ideas being given a single Nobel, then, well, let me put it this way: the Nobel should be given for achieving the most significant advancements in the field of the differential nonlinearities, for a very substantial progress in the physical understanding of the behaviour of nonlinear physical systems, forging pathways for predictive capacity.

Let me emphasize, this has been one of the most significant advances in physics in the last century. No, saying so is emphatically not a hyperbole.

And, yes, it’s an advance in physics, primarily, and then, also in maths—but only secondarily.

… It’s unfortunate that an advancement which has been this remarkable never did register as such with most of the S&T “manpower”, esp., engineers and practical designers. It’s also unfortunate that the twin advancement arrived on the scene at the time of bad cultural (even epistemological) trends, and so, the advancements got embedded in a fabric of hyperbole, even nonsense.

But regardless of the cultural tones in which the popular presentations of these advancements (esp. of the chaos theory) got couched, taken as a science, the studies of nonlinearity in the physical systems has been a very, very, original, and a very, very creative, advancement. It needs to be recognized as such.

That way, I don’t much care for what it helped produce on the maths side of it. But yes, even a not very extraordinarily talented undergraduate in CS (one with a special interest in deterministic methods in cryptography) would be able to tell you how much light got shone on their discipline because of the catastrophe and chaos theories.

The catastrophe theory has been simply marvellous in one crucial aspect: it actually pushed the boundaries of what is understood by the term: mathematics. The theory has been daring enough to propose, literally for the first time in the entire history of mankind, a well-refined qualitative approach to an infinity of quantitative processes taken as a group.

The distinction between the qualitative and the quantitative had kept philosophers (and laymen) pre-occupied for millenia. But the nonlinear theory has been the first theoretical approach that tells you how to spot and isolate the objective bases for distinguishing what we consider as the qualitative changes.

Remove the understanding given by the nonlinear theory—by the catastrophe-theoretical approach—and, once in the domain of the linear theory, the differences in kind immediately begin to appear as more or less completely arbitrary. There is no place in theory for them—the qualitative distinctions are external to the theory because a linear system always behaves exactly the same with any quantitative changes made, at any scale, to any of the controlling parameters. Since in the linear theory the qualitative changes are not produced from within the theory itself, such distinctions must be imported into it out of some considerations that are in principle external to the theory.

People often confuse such imports with “applications.” No, when it comes to the linear theory, it’s not the considerations of applications which can be said to be driving any divisions of qualitative changes. The qualitative distinctions are basically arbitrary in a linear theory. It is important to realize that that usual question: “Now where do we draw the line?” is basically absolutely superfluous once you are within the domain of the linear systems. There are no objective grounds on the basis of which such distinctions can be made.

Studies of the nonlinear phenomena sure do precede the catastrophe and the chaos theories. Even in the times before these two theories came on the scene, applied physicists would think of certain ideas such as differences of regimes, esp. in the areas like fluid dynamics.

But to understand the illuminating power of the nonlinear theory, just catch hold of an industrial CFD guy (or a good professor of fluid dynamics from a good university [not, you know, from SPPU or similar universities]), and ask him whether there can be any deeper theoretical significance to the procedure of the Buckingham Pi Theorem, to the necessity, in his art (or science) of having to use so many dimensionless numbers. (Every mechanical/allied engineering undergraduate has at least once in life cursed the sheer number of them.) The competent CFD guy (or the good professor) would easily be at a loss. Then, toss a good book on the Catastrophe Theory to him, leave him alone for a couple of weeks or may be a month, return, and raise the same question again. He now may or may not have a very good, “flowy” sort of a verbal answer ready for you. But one look at his face would tell you that it has now begun to reflect a qualitatively different depth of physical understanding even as he tries to tackle that question in his own way. That difference arises only because of the Catastrophe Theory.

As to the Chaos Theory (and I club the fractal theory right in it), more number of people are likely to know about it, and so, I don’t have to wax a lot (whether eloquently or incompetently). But let me tell you one thing.

Feigenbaum’s discovery of the universal constant remains, to my mind, one of the most ingenious advancements in the entire history of physics, even of science. Especially, given the experimental equipment with which he made that discovery—a handheld HP Calculator (not a computer) in the seventies (or may be in the sixties)! … And yes, getting to that universal constant was, if you ask me, an act of discovery, and not of invention. (Invention was very intimately involved in the process; but the overall act and the end-product was one of discovery.)

So, here is a wish that these fundamental studies of the nonlinear systems get their due—the recognition they so well deserve—in the form of a Physics Nobel.

…And, as always, the sooner the better!

Wish # 2:

The second wish I want to put up here is this: I wish there was some commercial/applied artist, well-conversant with the “art” of supplying illustrations for a physics book, who also was available for a long-term project I have in mind.

To share a bit: Years ago (actually, almost two decades ago, in 1998 to be precise), I had made a suggestion that novels by Ayn Rand be put in the form of comics. As far as I was concerned, the idea was novel (i.e. new). I didn’t know at that time that a comics-book version of The Fountainhead had already been conceived of by none other than Ayn Rand herself, and it, in fact, had also been executed. In short, there was a comics-book version of The Fountainhead. … These days, I gather, they are doing something similar for Atlas Shrugged.

If you think about it, my idea was not at all a leap of imagination. Newspapers (even those in India) have been carrying comic strips for decades (right since before my own childhood), and Amar Chitrakatha was coming of age just when I was. (It was founded in 1967 by Mr. Pai.)

Similarly, conceiving of a comics-like book for physics is not at all a very creative act of imagination. In fact, it is not even original. Everyone knows those books by that Japanese linguistics group, the books on topics like the Fourier theory.

So, no claim of originality here.

It’s just that for my new theory of QM, I find that the format of a comics-book would be most suitable. (And what the hell if physicists don’t take me seriously because I put it in this form first. Who cares what they think anyway!)

Indeed, I would even like to write/produce some comics books on maths topics, too. Topics like grads, divs, curls, tensors, etc., eventually. … Guess I will save that part for keeping me preoccupied during my retirement. BTW, my retirement is not all that far away; it’s going to be here pretty soon, right within just five years from now. (Do one thing: Check out what I was writing, say in 2012 on this blog.)

But the one thing I would like write/produce right in the more immediate future is: the comics book on QM, putting forth my new approach.

So, in the closing, here is a request. If you know some artist (or an engineer/physicist with fairly good sketching/computer-drawing skills), and has time at hand, and has the capacity to stay put in a sizeable project, and won’t ask money for it (a fair share in the royalty is a given—provided we manage to find a publisher first, that is), then please do bring this post to his notice.

A Song I Like:

And, finally, here is the Marathi song I had promised you the last time round. It’s a fusion of what to my mind is one of the best tunes Shrinivas Khale ever produced, and the best justice to the words and the tunes by the singer. Imagine any one else in her place, and you will immediately come to know what I mean. … Pushpa Pagdhare easily takes this song to the levels of the very best by the best, including Lata Mangeshkar. [Oh yes, BTW, congrats are due to the selection committe of this year’s Lata Mangeshkar award, for selecting Pushpa Pagdhare.]

(Marathi) “yeuni swapnaat maajhyaa…”
Singer: Pushpa Pagdhare
Music: Shrinivas Khale
Lyrics: Devakinandan Saraswat

[PS: Note: I am going to come back and add an update once this year’s Physics Nobel is announced. At that time (or tonight) I will also try to streamline this post.

Then, I will be gone off the blogging for yet another couple of weeks or so—unless it’s a small little “kutty” post of the “Blog-Filler” kind or two.]

# Blog-Filling—Part 2

I don’t know if I ran this song before or not… Most probably not. Never mind; this is just a blog-filler… .

A Song I Like:

[Once again, as an exception, I make reference to also the video of this song—in particular, those frames in which Dilip Kumar is not present, and in most particular, those frames in which Waheedaa is.

[For the sake of completeness: As to the frames in which both are present, well, why even raise that question? Haven’t you mastered the art of applying a suitable selective visual filter yet?]]

(Hindi) “kal ke sapane aaj bhee aanaa…”
Singer: Lata Mangeshkar

OK, bye for now, and work hard and do well and take care of yourself and bye [really] for now…

Oh, before I forget, let me make a note of it: I have wanted to run a certain Marathi song, and I will do it the next time, sure. [OK. Now, repeat the above para.]

# 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:

$q = \vec{J} \cdot \vec{S}$ (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!!]]

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