Some suggested time-pass (including ideas for Python scripts involving vectors and tensors)

Actually, I am busy writing down some notes on scalars, vectors and tensors, which I will share once they are complete. No, nothing great or very systematic; these are just a few notings here and there taken down mainly for myself. More like a formulae cheat-sheet, but the topic is complicated enough that it was necessary that I have them in one place. Once ready, I will share them. (They may get distributed as extra material on my upcoming FDP (faculty development program) on CFD, too.)

While I remain busy in this activity, and thus stay away from blogging, you can do a few things:


1.

Think about it: You can always build a unique tensor field from any given vector field, say by taking its gradient. (Or, you can build yet another unique tensor field, by taking the Kronecker product of the vector field variable with itself. Or, yet another one by taking the Kronecker product with some other vector field, even just the position field!). And, of course, as you know, you can always build a unique vector field from any scalar field, say by taking its gradient.

So, you can write a Python script to load a B&W image file (or load a color .PNG/.BMP/even .JPEG, and convert it into a gray-scale image). You can then interpret the gray-scale intensities of the individual pixels as the local scalar field values existing at the centers of cells of a structured (squares) mesh, and numerically compute the corresponding gradient vector and tensor fields.

Alternatively, you can also interpret the RGB (or HSL/HSV) values of a color image as the x-, y-, and z-components of a vector field, and then proceed to calculate the corresponding gradient tensor field.

Write the output in XML format.


2.

Think about it: You can always build a unique vector field from a given tensor field, say by taking its divergence. Similarly, you can always build a unique scalar field from a vector field, say by taking its divergence.

So, you can write a Python script to load a color image, and interpret the RGB (or HSL/HSV) values now as the xx-, xy-, and yy-components of a symmetrical 2D tensor, and go on to write the code to produce the corresponding vector and scalar fields.


Yes, as my resume shows, I was going to write a paper on a simple, interactive, pedagogical, software tool called “ToyDNS” (from Toy + Displacements, Strains, Stresses). I had written an extended abstract, and it had even got accepted in a renowned international conference. However, at that time, I was in an industrial job, and didn’t get the time to write the software or the paper. Even later on, the matter kept slipping.

I now plan to surely take this up on priority, as soon as I am done with (i) the notes currently in progress, and immediately thereafter, (ii) my upcoming stress-definition paper (see my last couple of posts here and the related discussion at iMechanica).

Anyway, the ideas in the points 1. and 2. above were, originally, a part of my planned “ToyDNS” paper.


3.

You can induce a “zen-like” state in you, or if not that, then at least a “TV-watching” state (actually, something better than that), simply by pursuing this URL [^], and pouring in all your valuable hours into it. … Or who knows, you might also turn into a closet meteorologist, just like me. [And don’t tell anyone, but what they show here is actually a vector field.]


4.

You can listen to this song in the next section…. It’s one of those flowy things which have come to us from that great old Grand-Master, viz., SD Burman himself! … Other songs falling in this same sub-sub-genre include, “yeh kisine geet chheDaa,” and “ThanDi hawaaein,” both of which I have run before. So, now, you go enjoy yet another one of the same kind—and quality. …


A Song I Like:

[It’s impossible to figure out whose contribution is greater here: SD’s, Sahir’s, or Lata’s. So, this is one of those happy circumstances in which the order of the listing of the credits is purely incidental … Also recommended is the video of this song. Mona Singh (aka Kalpana Kartik (i.e. Dev Anand’s wife, for the new generation)) is sooooo magical here, simply because she is so… natural here…]

(Hindi) “phailee huyi hai sapanon ki baahen”
Music: S. D. Burman
Lyrics: Sahir
Singer: Lata Mangeshkar


But don’t forget to write those Python scripts….

Take care, and bye for now…

 

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Exactly what does this script show?

Update on 02 March 2018, 15:34 IST: I have now added another, hopefully better, version of the script (but also kept the old one intact); see in the post below. The new script too comes without comments.


Here is a small little Python script which helps you visualize something about a state of stress in 2D.

If interested in understanding the concept of stress, then do run it, read it, try to understand what it does, and then, if still interested in the concept of stress, try to answer this “simple” little question:

Exactly what does this script show? Exactly what it is that you are visualizing, here?


I had written a few more notes and inline comments in the script, but have deliberately deleted most of them—or at least the ones which might have given you a clue towards answering the above question. I didn’t want to spoil your fun, that’s why.

Once you all finish giving it a try, I will then post another blog-entry here, giving my answer to that question (and in the process, bringing back all the deleted notes and comments).

Anyway, here is the script:


'''
A simple script to help visualize *something* about
a 2D stress tensor.

--Ajit R. Jadhav. Version: 01 March 2018, 21:39 HRS IST.
'''

import math
import numpy as np
import matplotlib.pyplot as plt

# Specifying the input stress
# Note:
# While plotting, we set the x- and y-limits to -150 to +150,
# and enforce the aspect ratio of 1. That is to say, we do not
# allow MatPlotLib to automatically scale the axes, because we
# want to appreciate the changes in the shapes as well sizes in
# the plot.
#
# Therefore, all the input stress-components should be kept
# to within the -100 to +100 (both inclusive) range.
#
# Specify the stress state in this order: xx, xy; yx, yy
# The commas and the semicolon are necessary.

sStress = "-100, 45; 90, 25"

axes = plt.axes()
axes.set_xlim((-150, 150))
axes.set_ylim((-150, 150))
plt.axes().set_aspect('equal', 'datalim')
plt.title(
    "A visualization of *something* about\n" \
    "the 2D stress-state [xx, xy; yx, yy] = [%s]" \
    % sStress)

mStress = np.matrix(sStress)
mStressT = np.transpose(mStress)

mUnitNormal = np.zeros((2, 1))
mTraction = np.zeros((2, 1))

nOrientations = 18
dIncrement = 360.0 / float(nOrientations)
for i in range(0, nOrientations):
    dThetaDegrees = float(i) * dIncrement
    dThetaRads = dThetaDegrees * math.pi / 180.0
    mUnitNormal = [round(math.cos(dThetaRads), 6), round(math.sin(dThetaRads), 6)]
    mTraction = mStressT.dot(mUnitNormal)
    if i == 0:
        plt.plot((0, mTraction[0, 0]), (0, mTraction[0, 1]), 'black', linewidth=1.0)
    else:
        plt.plot((0, mTraction[0, 0]), (0, mTraction[0, 1]), 'gray', linewidth=0.5)
    plt.plot(mTraction[0, 0], mTraction[0, 1], marker='.',
             markeredgecolor='gray', markerfacecolor='gray', markersize=5)
    plt.text(mTraction[0, 0], mTraction[0, 1], '%d' % dThetaDegrees)
    plt.pause(0.05)

plt.show()


Update on 02 March 2018, 15:34 IST:

Here is a second version of a script that does something similar (but continues to lack explanatory comments). One advantage with this version is that you can copy-paste the script to some file, say, MyScript.py, and invoke it from command line, giving the stress components and the number of orientations as command-line inputs, e.g.,

python MyScript.py "100, 0; 0, 50" 12

which makes it easier to try out different states of stress.

The revised code is here:


'''
A simple script to help visualize *something* about
a 2D stress tensor.

--Ajit R. Jadhav. 
History: 
06 March 2018, 10:43 IST: 
In computeTraction(), changed the mUnitNormal code to make it np.matrix() rather than python array
02 March 2018, 15:39 IST; Published the code
'''

import sys
import math
import numpy as np
import matplotlib.pyplot as plt

# Specifying the input stress
# Note:
# While plotting, we set the x- and y-limits to -150 to +150,
# and enforce the aspect ratio of 1. That is to say, we do not
# allow MatPlotLib to automatically scale the axes, because we
# want to appreciate the changes in the shapes as well sizes in
# the plot.
#
# Therefore, all the input stress-components should be kept
# to within the -100 to +100 (both inclusive) range.
#
# Specify the stress state in this order: xx, xy; yx, yy
# The commas and the semicolon are necessary.
# If you run the program from a command-line, you can also
# specify the input stress string in quotes as the first
# command-line argument, and no. of orientations, as the
# second. e.g.:
# python MyScript.py "100, 50; 50, 0" 12
##################################################

gsStress = "-100, 45; 90, 25"
gnOrientations = 18


##################################################

def plotArrow(vTraction, dThetaDegs, clr, axes):
    dx = round(vTraction[0], 6)
    dy = round(vTraction[1], 6)
    if not (math.fabs(dx) < 10e-6 and math.fabs(dy) < 10e-6):
        axes.arrow(0, 0, dx, dy, head_width=3, head_length=9.0, length_includes_head=True, fc=clr, ec=clr)
    axes.annotate(xy=(dx, dy), s='%d' % dThetaDegs, color=clr)


##################################################

def computeTraction(mStressT, dThetaRads):
    vUnitNormal = [round(math.cos(dThetaRads), 6), round(math.sin(dThetaRads), 6)]
    mUnitNormal = np.reshape(vUnitNormal, (2,1))
    mTraction = mStressT.dot(mUnitNormal)
    vTraction = np.squeeze(np.asarray(mTraction))
    return vTraction


##################################################

def main():
    axes = plt.axes()
    axes.set_label("label")
    axes.set_xlim((-150, 150))
    axes.set_ylim((-150, 150))
    axes.set_aspect('equal', 'datalim')
    plt.title(
        "A visualization of *something* about\n" \
        "the 2D stress-state [xx, xy; yx, yy] = [%s]" \
        % gsStress)

    mStress = np.matrix(gsStress)
    mStressT = np.transpose(mStress)
    vTraction = computeTraction(mStressT, 0)
    plotArrow(vTraction, 0, 'red', axes)
    dIncrement = 360.0 / float(gnOrientations)
    for i in range(1, gnOrientations):
        dThetaDegrees = float(i) * dIncrement
        dThetaRads = dThetaDegrees * math.pi / 180.0
        vTraction = computeTraction(mStressT, dThetaRads)
        plotArrow(vTraction, dThetaDegrees, 'gray', axes)
        plt.pause(0.05)
    plt.show()


##################################################

if __name__ == "__main__":
    nArgs = len(sys.argv)
    if nArgs > 1:
        gsStress = sys.argv[1]
    if nArgs > 2:
        gnOrientations = int(sys.argv[2])
    main()



OK, have fun, and if you care to, let me know your answers, guess-works, etc…..


Oh, BTW, I have already taken a version of my last post also to iMechanica, which led to a bit of an interaction there too… However, I had to abruptly cut short all the discussions on the topic because I unexpectedly got way too busy in the affiliation- and accreditation-related work. It was only today that I’ve got a bit of a breather, and so could write this script and this post. Anyway, if you are interested in the concept of stress—issues like what it actually means and all that—then do check out my post at iMechanica, too, here [^].


… Happy Holi, take care to use only safe colors—and also take care not to bother those people who do not want to be bothered by you—by your “play”, esp. the complete strangers…

OK, take care and bye for now. ….


A Song I Like:

(Marathi [Am I right?]) “rang he nave nave…”
Music: Aditya Bedekar
Singer: Shasha Tirupati
Lyrics: Yogesh Damle

 

Stress is defined as the quantity equal to … what?

Update on 01 March 2018, 21:27, IST: I had posted a version of this post also at iMechanica, which led to a bit of a very interesting interaction there [^] too. Check it out, if you want… Also see my today’s post concerning the idea of stress, here [^].


In this post, I am going to note a bit from my personal learning history. I am going to note what had happened when a clueless young engineering student that was me, was trying hard to understand the idea of tensors, during my UG years, and then for quite some time even after my UG days. May be for a decade or even more….

There certainly were, and are likely to be even today, many students like [the past] me. So, in the further description, I will use the term “we.” Obviously, the “we” here is the collegial “we,” perhaps even the pedagogical “we,” but certainly neither the pedestrian nor the royal “we.”


What we would like to understand is the idea of tensors; the question of what these beasts are really, really like.

As with developing an understanding of any new concept, we first go over some usage examples involving that idea, some instances of that concept.

Here, there is not much of a problem; our mind easily picks up the stress as a “simple” and familiar example of a tensor. So, we try to understand the idea of tensors via the example of the stress tensor. [Turns out that it becomes far more difficult this way… But read on, anyway!]

Not a bad decision, we think.

After all, even if the tensor algebra (and tensor calculus) was an achievement wrought only in the closing decade(s) of the 19th century, Cauchy was already been up and running with the essential idea of the stress tensor right by 1822—i.e., more than half a century earlier. We come to know of this fact, say via James Rice’s article on the history of solid mechanics. Given this bit of history, we become confident that we are on the right track. After all, if the stress tensor could not only be conceived of, but even a divergence theorem for it could be spelt out, and the theorem even used in applications of engineering importance, all some half a century before any other tensors were even conceived of, then developing a good understanding of the stress tensor ought to provide a sound pathway to understanding tensors in general.

So, we begin with the stress tensor, and try [very hard] to understand it.


We recall what we have already been taught: stress is defined as force per unit area. In symbolic terms, read for the very first time in our XI standard physics texts, the equation reads:

\sigma \equiv \dfrac{F}{A}               … Eq. (1)

Admittedly, we had been made aware, that Eq. (1) holds only for the 1D case.

But given this way of putting things as the starting point, the only direction which we could at all possibly be pursuing, would be nothing but the following:

The 3D representation ought to be just a simple generalization of Eq. (1), i.e., it must look something like this:

\overline{\overline{\sigma}} = \dfrac{\vec{F}}{\vec{A}}                … Eq. (2)

where the two overlines over \sigma represents the idea that it is to be taken as a tensor quantity.

But obviously, there is some trouble with the Eq. (2). This way of putting things can only be wrong, we suspect.

The reason behind our suspicion, well-founded in our knowledge, is this: The operation of a division by a vector is not well-defined, at least, it is not at all noted in the UG vector-algebra texts. [And, our UG maths teachers would happily fail us in examinations if we tried an expression of that sort in our answer-books.]

For that matter, from what we already know, even the idea of “multiplication” of two vectors is not uniquely defined: We have at least two “product”s: the dot product [or the inner product], and the cross product [a case of the outer or the tensor product]. The absence of divisions and unique multiplications is what distinguishes vectors from complex numbers (including phasors, which are often noted as “vectors” in the EE texts).

Now, even if you attempt to “generalize” the idea of divisions, just the way you have “generalized” the idea of multiplications, it still doesn’t help a lot.

[To speak of a tensor object as representing the result of a division is nothing but to make an indirect reference to the very operation [viz. that of taking a tensor product], and the very mathematical structure [viz. the tensor structure] which itself is the object we are trying to understand. … “Circles in the sand, round and round… .” In any case, the student is just as clueless about divisions by vectors, as he is about tensor products.]

But, still being under the spell of what had been taught to us during our XI-XII physics courses, and later on, also in the UG engineering courses— their line and method of developing these concepts—we then make the following valiant attempt. We courageously rearrange the same equation, obtain the following, and try to base our “thinking” in reference to the rearrangement it represents:

\overline{\overline{\sigma}} \vec{A} = \vec{F}                  … Eq (3)

It takes a bit of time and energy, but then, very soon, we come to suspect that this too could be a wrong way of understanding the stress tensor. How can a mere rearrangement lead from an invalid equation to a valid equation? That’s for the starters.

But a more important consideration is this one: Any quantity must be definable via an equation that follows the following format:

the quantiy being defined, and nothing else but that quantity, as appearing on the left hand-side
=
some expression involving some other quantities, as appearing on the right hand-side.

Let’s call this format Eq. (4).

Clearly, Eq. (3) does not follow the format of Eq. (4).

So, despite the rearrangement from Eq. (2) to Eq. (3), the question remains:

How can we define the stress tensor (or for that matter, any tensors of similar kind, say the second-order tensors of strain, conductivity, etc.) such that its defining expression follows the format given in Eq. (4)?


Can you answer the above question?

If yes, I would love to hear from you… If not, I will post the answer by way of an update/reply/another blog post, after some time. …

Happy thinking…


A Song I Like:
(Hindi) “ye bholaa bhaalaa man meraa kahin re…”
Singers: Kishore Kumar, Asha Bhosale
Music: Kishore Kumar
Lyrics: Majrooh Sultanpuri


[I should also be posting this question at iMechanica, though I don’t expect that they would be interested too much in it… Who knows, someone, say some student somewhere, may be interested in knowing more about it, just may be…

Anyway, take care, and bye for now…]

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…]