R.S.I.

I got that, again, for the 4th time in my life, so far. Taking rest. Giving rest to my RHS wrist. Have to. No “A Song I Like” section.

 

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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…]

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.)]