iMechanica - Comments for "Scalar done wrong"
https://imechanica.org/node/15857
Comments for "Scalar done wrong"enScalars
https://imechanica.org/comment/28572#comment-28572
<a id="comment-28572"></a>
<p><em>In reply to <a href="https://imechanica.org/node/15857">Scalar done wrong</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>I have just updated the <a href="http://imechanica.org/files/scalar%202016%2009%2004_1.pdf">notes on scalar</a>. The notes are now part of the <a href="http://imechanica.org/node/19709">notes on linear algebra</a>. </p>
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</ul>Mon, 05 Sep 2016 01:39:54 +0000Zhigang Suocomment 28572 at https://imechanica.orgNumbers and Scalars
https://imechanica.org/comment/26685#comment-26685
<a id="comment-26685"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/26680#comment-26680">Re: numbers and scalars</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>Dear Zhigang,</p>
<p>Many thanks for your comments. I'll read your lecture notes.</p>
<p>Regards,</p>
<p>Mohsen</p>
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</ul>Wed, 10 Dec 2014 16:00:29 +0000M. Jahanshahicomment 26685 at https://imechanica.orgRe: numbers and scalars
https://imechanica.org/comment/26680#comment-26680
<a id="comment-26680"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/26678#comment-26678">Numbers and Scalars</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p dir="ltr"><span>Dear Mohsen: In the </span><a href="http://imechanica.org/node/14164"><span>notes on tensors</span></a><span>, I have followed the terminology in algebra. In particular, the phrase “number field” means a specific algebraic structure: see the </span><a href="http://en.wikipedia.org/wiki/Field_(mathematics)"><span>Wikipedia entry on number field.</span></a><span> The word “field” in this context is unrelated to that in the phase “vector field”. My notes on tensor are entirely devoted to algebra, and do not talk about vector field (i.e., the variation of a vector in the Euclidean space).</span></p>
<p dir="ltr"><span>I list the attributes of “number field” because I wish to show that the number field is different from how we use energy, mass, and charge. I call the set of all values of energy a scalar set. Please see </span><a href="http://imechanica.org/node/14164"><span>my notes</span></a><span> for contrast between the number field and the scalar set.</span></p>
<p dir="ltr"><span>For our purpose, the number field is simply the set of real numbers.</span></p>
<p><span id="docs-internal-guid-fef76fda-31db-4365-3e6f-4548bd49b0a1"><span>Part of the </span><a href="http://imechanica.org/files/frame%20indifference%202014%2012%2002.pdf"><span>notes on frame indifference (pages 15-22)</span></a><span> describes the algebraic structure of the world. The notes show how all vectors and tensors in mechanics originate from a single vector space: the space of separation. </span></span></p>
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</ul>Wed, 10 Dec 2014 01:41:13 +0000Zhigang Suocomment 26680 at https://imechanica.orgNumbers and Scalars
https://imechanica.org/comment/26678#comment-26678
<a id="comment-26678"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/26676#comment-26676">Re: Scalars and Numbers</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>Dear Zhigang,</p>
<p>Actually that post goes back to a few months ago. That's ok to use number field for individual components of a tensor. But what for scalar fields such as temperature. The components of a second order tensor change under coordinate transformation but a scalar field does not. Can we still call a scalar field a number field? If we remove the constraint corresponding to scalar fields and assume that it is immaterial how numbers transform under coordinate transformation then we can call scalar fields, number fields as well. I figure out that this is the approach that is meant. We use number field regardless of how it transforms and we associate a unit to it right?</p>
<p>Regards</p>
<p>Mohsen</p>
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</ul>Tue, 09 Dec 2014 15:58:41 +0000M. Jahanshahicomment 26678 at https://imechanica.orgRe: Scalars and Numbers
https://imechanica.org/comment/26676#comment-26676
<a id="comment-26676"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25434#comment-25434">Scalars and Numbers</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p dir="ltr"><span>Dear Mohsen: In my </span><a href="http://imechanica.org/node/14164"><span>notes on tensors</span></a><span>, following the common practice in linear algebra, I defined the number field and vector space. I then defined scalar set as a one-dimensional vector space. Thus, a number field and a scalar set have distinct algebraic structures. The product of two numbers is another number, but the product of two energies is not another energy.</span></p>
<p><span id="docs-internal-guid-8acc545b-2f3f-bc01-1abc-38c8a3033c5e"><span>In linear algebra, a vector is a linear combination of base vectors. The coefficients are members in the number field. In using this idea in physics, however, it is convenient to make the base vectors unit vectors, and let the components carry the unit. This usage messes up terminology somewhat. But in terms of the algebraic structure, the components of a vector (or a tensor) are elements of the number field times a unit.</span></span></p>
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</ul>Tue, 09 Dec 2014 13:32:00 +0000Zhigang Suocomment 26676 at https://imechanica.orgFrom my understanding, when
https://imechanica.org/comment/26670#comment-26670
<a id="comment-26670"></a>
<p><em>In reply to <a href="https://imechanica.org/node/15857">Scalar done wrong</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>From my understanding, when saying numbers are scalar, we take the point of algegra. Here we only concern about "abstract numbers", don't care about its underlying physics or geometry. </p>
<p>When saying mass, density, temperature, energy and so on are scalars, we take the view of physics (or geometry). In linearized elasticity, ennergy density = stress*strain, stress is conjugate (dual) to strian. This has a geometrial meaning. </p>
<p> </p>
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</ul>Mon, 08 Dec 2014 21:34:56 +0000Junwei Xingcomment 26670 at https://imechanica.orgMother of all frame-indifferent variables, and their fathers
https://imechanica.org/comment/26613#comment-26613
<a id="comment-26613"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25403#comment-25403">The linear algebra of the world</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p><span id="docs-internal-guid-8acc545b-02c3-afec-7c0d-d493e4734924"><span>I have just developed this theme in </span><a href="http://imechanica.org/node/17565"><span>notes on frame-indifference</span></a><span>.</span></span></p>
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</ul>Sun, 30 Nov 2014 22:13:17 +0000Zhigang Suocomment 26613 at https://imechanica.orgRe^2: Answer to question posed + sundry remarks
https://imechanica.org/comment/25441#comment-25441
<a id="comment-25441"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25438#comment-25438">Re: Answer to question posed + sundry remarks</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
How does this sound? Strain is a troublesome variable. The strain in a steel specimen is different before the yield point and after it. The physics in the two intervals are different.
</p>
<p>
The absolute zero itself would be the sought "origin," even if temperature can't be modeled as a vector. But, yes, it should be possible to think of addition of temperatures. I addressed this issue in an earlier comment somewhere in this thread. And, that way, you never add electric potentials, anyway; you only displace charges from one point to another point. Yet, we always regard potentials as quantities that can be added to each other. If so, why not temperature?
</p>
<p>
The third law only prohibits <em>going</em> from a non-zero temperature to the absolute zero temperature; it does not prohibit some thing/some region somewhere in the universe possibly <em>already</em> being <em>at</em> the absolute zero---the law leaves that possibility open. The issue is similar to c. Just because massive objects cannot be accelerated from v < c to v = c does not mean that objects cannot move at c---photons do. Of course, to my knowledge, we haven't yet actually run into something which already was at absolute zero. But no known law prohibits our possibly running into it. If a prohibitive principle is found, it will be a new knowledge; a new law, unknown as of today.
</p>
<p>
There were certain reports of negative (absolute) temperatures being reached in experiments. These were gimmicks (like certain other reports of light going faster than c). Barring these, an object at absolute zero simply means that it cannot give up its energy; that it can only absorb some. Known laws and empirical data do not rule out such a possibility.
</p>
<p>
Of course, to define mathematical object, it's only necessary to know that there is no physical constraint to at all have 0K on the scale. So, the scale can indeed be defined, and used for additions/subtractions of temperatures.
</p>
<p>
See you sometime tomorrow (IST).
</p>
<p>
--Ajit
</p>
<p>
- - - - - <br />
[E&OE]
</p>
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</ul>Sat, 04 Jan 2014 18:59:10 +0000Ajit R. Jadhavcomment 25441 at https://imechanica.orgRe:^2 Answer to question posed + sundry remarks
https://imechanica.org/comment/25440#comment-25440
<a id="comment-25440"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25438#comment-25438">Re: Answer to question posed + sundry remarks</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Hi Zhigang: The DND map from say the traction vector space to the non-dimensional 3d vector space would be generated as follows:
</p>
<p>
One would have already (arbitrarily) chosen some fundamental system of units for say mass, time, and length, and this system would be common to all the 'DND maps' from the various dimensional 3d vector spaces to the non-dimensional one. Now simply nondimensionalize each of the traction vectors in the dimensional vector space. Then I think each dimensional traction vector corresponds to an element of the canonical nondimensional vector space chosen. Now one can do the same thing for say the vector space of position vectors and velocities etc. It is amusing to note that in effect a set of these DND maps (parametrized by a choices of the fundamental unit set) allows one to associate the velocity vector representing say 30deg NE of magnitude 2 m/s with some traction vector (not to be construed as anything physical of course).
</p>
<p>
Then the stress tensor, which we understand in the dimensional context as a mapping of a certain area vector to a certain traction vector can be translated to a linear mapping from V_3 to V_3 where I now think of V_3 being the canonical nondimensional space just as we want in continuum mechanics...... and so on and so forth for other tensors....
</p>
<p>
Of course, all this requires careful checking about preservation of linearity, and whether these are really 1-1 maps etc.
</p>
<p>
</p>
<p>
And yes, I agree with you, with these things the deeper you dig the murkier it gets. There's probably a philosophical reason behind mathematics (and physics) not being decidable if they are logically consistent.....
</p>
<p>
In this context, I sometimes really like David Mermin's comment - "shut up and calculate" about quantum mechanics.
</p>
<p>
There's also been talk of Strang here - a great saying of his is "adept at abstraction, inept at computation" I am very happy to be an engineer, so that we come up with some answers, however imperfect....
</p>
<p>
And while I am at it with random comments, here is a plug for one of Strang's books that I think is absolutely terrific - it is "Introduction to Applied Mathematics"
</p>
<p>
</p>
<p>
</p>
<p>
</p>
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</ul>Sat, 04 Jan 2014 18:55:00 +0000Amit Acharyacomment 25440 at https://imechanica.orgRe:^7 How to think of force linear functional as force vector?
https://imechanica.org/comment/25439#comment-25439
<a id="comment-25439"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25424#comment-25424">Re:^5 How to think of force linear functional as force vector?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>Hi Zhigang: I don't see why my previous question is relevant in some crucial way to this discussion.</p>
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</ul>Sat, 04 Jan 2014 18:28:36 +0000Amit Acharyacomment 25439 at https://imechanica.orgRe: Answer to question posed + sundry remarks
https://imechanica.org/comment/25438#comment-25438
<a id="comment-25438"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25437#comment-25437">Answer to question posed + sundry remarks</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Amit: Thank you. I just came back from shoveling snow.
</p>
<p>
What is the "DND mapping" that you mentioned in your <a href="http://imechanica.org/node/15843#comment-25350">previous comment?</a>
</p>
<p>
Indeed, temperature is a troublesome variable. This fact has troubled me for some years. The temperature from 50K to 70K is different from 100K to 120K. The physics in the two intervals are different.
</p>
<p>
We'd love to have a additive quantity, but temperature is not. I agree with you: it is still troublesome even if we choose an origin. Then we need to say how far away this origin is from the absolute zero. Thus, temperature does not have the "translational symmetry", unlike time and spatial point.
</p>
<p>
Entropy may also have no translational symmetry.
</p>
<p>
Why does energy have translational symmetry? Perhaps energy does not have this "translational symmetry", at some fundamental level. We are just operating at a point so far away from its "absolute zero", whatever that may be.
</p>
<p>
Random questions for which I have no answers now.
</p>
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</ul>Sat, 04 Jan 2014 18:25:36 +0000Zhigang Suocomment 25438 at https://imechanica.orgAnswer to question posed + sundry remarks
https://imechanica.org/comment/25437#comment-25437
<a id="comment-25437"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25424#comment-25424">Re:^5 How to think of force linear functional as force vector?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
1) Since I do not want to have the dubious distinction of "raiser of questions without trying to answer", my answer to the question I posed appeared some ways down in that long thread, <a href="http://imechanica.org/node/15843#comment-25350">so here is a link to it</a>.
</p>
<p>
2) Somewhere in these threads I read about the importance of linearity. I just want to point out, may be only as a side remark, that while an object like a derivative Df(x) of a scalar potential f: V --> R is by definition a linear map on the underlying vector space
</p>
<p>
i.e. it acts like Df(x)[au + bv] = aDf(x)[u] + bDf(x)[v], for all a,b in R and u,v in V and (to keep things simple) say x in V
</p>
<p>
Df's dependence on x can be completely nonlinear. Therefore the gradient of f at x can also be completely nonlinear in x.
</p>
<p>
3) My feeling is linear PDEs of physics should not play a big role in our discussions. As my friend Luc Tartar likes to think, may be at the smallest scales our physical world is at best a semilinear system of hyperbolic PDE, but the smallest scale does not physics make...
</p>
<p>
In particular, just based on rotational invariance, we know our beloved Elasticity cannot be a linear theory.
</p>
<p>
4) Discussion on temperature - much like a point space and its translation space, the space of temperatures may be considered something close to a vector space by introducing an 'origin' - so differences of temperatures make sense (otherwise we would have to say bye-bye to the heat equation, e.g.). However, it is only 'close to a vector space' because even temperature differences would not be closed under addition (there is a physical meaning to absolute zero)?
</p>
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</ul>Sat, 04 Jan 2014 15:51:00 +0000Amit Acharyacomment 25437 at https://imechanica.orgRe: Multiplication of particle masses (electric charges)
https://imechanica.org/comment/25436#comment-25436
<a id="comment-25436"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25385#comment-25385">Multiplication of particle masses (electric charges)</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Arash: Excellent question. In the case of <a href="http://en.wikipedia.org/wiki/Coulomb's_law">Coulomb's law</a>, we can derive the law from the Maxwell equations. Of course, many textbooks will start with Coulomb's law as an empirical observation, and then formulate the Maxwell equations to be consistent with it. The approach follows historical development of electrostatics. But we have the Maxwell equations now, and we can choose to start with them, and derive Coulomb's law. This is the apporach I will take here.
</p>
<p>
The Mawell equations are linear. This fact itself is interesting, but here I will take it as given. Once you solve the boundary value problem and find that the electric field is linear in charge, you need to calculate force. Most people will use the <a href="http://en.wikipedia.org/wiki/Lorentz_force">Lorentz force</a>, where force equals charge times electric field. Thus, force involves charge times charge.
</p>
<p>
But Lorentz force looks mysterious. You don't need his equation. You can find this quadratic dependence in charge from calculating energy. In the following review article, I describe a simple approach in Section 3.
</p>
<p>
<span>Zhigang Suo. </span><a href="http://www.seas.harvard.edu/suo/papers/243.pdf">Theory of dielectric elastomers.</a><span> Acta Mechanica Solida Sinica 23, 549-578 (2010).</span>
</p>
<p>
To sum up, Maxwell equations are linear. The force is qudratic in charge because energy is involved.
</p>
<p>
Perhaps the case of gravity can be rationalized in a similar way, but I am not sure at this point.
</p>
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</ul>Sat, 04 Jan 2014 11:32:44 +0000Zhigang Suocomment 25436 at https://imechanica.orgRe: vector space vs. scalar set
https://imechanica.org/comment/25435#comment-25435
<a id="comment-25435"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25431#comment-25431">vector space vs. scalar set</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Zhigang,
</p>
<p>
Happy new year to you, too! (Now, continuing in the jocular vein, let me say: apologies not accepted <img src="http://www.imechanica.org/modules/tinymce/includes/jscripts/tiny_mce/plugins/emotions/images/smiley-laughing.gif" border="0" alt="Laughing" title="Laughing" />. More seriously: An apology here was neither necessary nor expected. )
</p>
<p>
With your noting that calling S a vector space is confusing, guess we essentially are in agreement.
</p>
<p>
Yes, the change of basis for vectors (and tensors) involves rotation whereas speaking of rotation is not meaningful for scalars because there is no direction associated with them in the first place.
</p>
<p>
--Ajit
</p>
<p>
- - - - - <br />
[E&OE]
</p>
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</ul>Sat, 04 Jan 2014 11:19:09 +0000Ajit R. Jadhavcomment 25435 at https://imechanica.orgScalars and Numbers
https://imechanica.org/comment/25434#comment-25434
<a id="comment-25434"></a>
<p><em>In reply to <a href="https://imechanica.org/node/15857">Scalar done wrong</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Zhigang,
</p>
<p>
If we are going to use numbers for scalars then what can we call<br />
individual elements of tensors they are not scalars because they change under<br />
coordinate trasformation we cannot call them numbers as well because we are<br />
going to substitute them for scalars.
</p>
<p>
Regards
</p>
<p>
Mohsen
</p>
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</ul>Sat, 04 Jan 2014 07:26:54 +0000M. Jahanshahicomment 25434 at https://imechanica.orgNot quite clear where the problem is....
https://imechanica.org/comment/25432#comment-25432
<a id="comment-25432"></a>
<p><em>In reply to <a href="https://imechanica.org/node/15857">Scalar done wrong</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
For "mass", for example, is not a "scalar"; mass, volume, etc, are measures defined on a measurable space (the real line, a body in 3D, etc).
</p>
<p>
Happy New Year !
</p>
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</ul>Sat, 04 Jan 2014 04:26:29 +0000Stefan C. Soarecomment 25432 at https://imechanica.orgvector space vs. scalar set
https://imechanica.org/comment/25431#comment-25431
<a id="comment-25431"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25430#comment-25430">Re: A scalar set is defined as a 1D vector space over real numbe</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Ajit: Happy new year! I apologize for misspelling your name.
</p>
<p>
I am trying to construct a linear map between two vector spaces. f: V-->S. Here V is a 3D vector space, S is a 1D vector space, both over the same number field R. A linear combination of elements in V is an element in V. All elements in S scale with each other. f is a linear map. Being linear is crucial to address <a href="http://imechanica.org/node/15857#comment-25424">an issue raised by Amit</a>.
</p>
<p>
I also dislike the term one-dimensional vector space, although the object S does have the same structure as the one-dimensional vector space. The phrase is too long. Calling a scalar a 1D vector is confusing.
</p>
<p>
I have proposed to call S a scalar set over the field of real numbers. Thus, we have the parallel between a vector space and scalar set:
</p>
<ul><li>vector space, scalar set</li>
<li>a vector is an element in the vector spae, a scaler is an element in the scalar set</li>
<li>a basis is a set of three linearly independent vectors in V, a unit is a non-zero scalar in S</li>
<li>components of a vector, magnitude of a scalar</li>
<li>change of basis, change of unit </li>
</ul><p>The last pair of parallel terms are worth commenting. Because the 3D vector space V is an inner-product space, the chosen basis is usually orthonormal. A change of basis invove a charge in the orinetation of the basis, not the length of the each base vector. In changing the unit of a scalar set, however, we change the unit form one scalar to a scalar of a different magnitude. </p>
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</ul>Sat, 04 Jan 2014 03:13:00 +0000Zhigang Suocomment 25431 at https://imechanica.orgRe: A scalar set is defined as a 1D vector space over real numbe
https://imechanica.org/comment/25430#comment-25430
<a id="comment-25430"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25425#comment-25425">A scalar set is defined as a 1D vector space over real numbers</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Zhigang,
</p>
<p>
0. Paraphrasing Ian Fleming [<a href="http://en.wikipedia.org/wiki/Ian_Fleming" target="_blank">^</a>] (and changing the order of the (extended) operands of the operator phrase in question), the name is Ajit---Ajit R. Jadhav.
</p>
<p>
1. No, the definitions are different. The object that both Alecsander and I agree on would be a composite object that is neither a vector space nor derived from one.
</p>
<p>
2. Your fourth operation is an <em>external</em> binary map [<a href="http://en.wikipedia.org/wiki/Binary_operation#External_binary_operations" target="_blank">^</a>]. This is analogous to the reason why we would like to make it a <em>composite</em> object.
</p>
<p>
3. The terms do <em>parallel</em> to those used in the vector spaces, but that still does not mean that it should be called a <em>vector</em> space.
</p>
<p>
Let me suggest a new name: <em>physical unit</em> space, or, if mathematicians feel repelled by it, the PU space. (The moniker "unit space" would make it too general---even the real number field could arguably be called that.)
</p>
<p>
4. Making it a composite object <em>seems</em> to also automatically take care of the idea that there should be multiple instances of F. I have to think more on this aspect.
</p>
<p>
5. Indeed, as I was writing my above replies in this thread, though I didn't mention it, I had also thought that also the vector and tensor spaces would have multiple instances---same abstract mathematical structure but different objects, instantiated according to their physical units.</p>
<p>Re. your recent reply to Amit on this thread here [<a href="http://www.imechanica.org/node/15857#comment-25424" target="_blank">^</a>]. I did not know that Amit had mentioned this point (concerning different instances of the same structure) before me, because I had not read the previous thread. Thus, I got this point later than him, but, sure, also independently of him. (I came from the OO programming and ADT perspective.) I also got the point regarding the law of the dimensional homogeneity independently of him---I now realize that he does mention dimensional analysis, too. So, let me first go through the previous thread.
</p>
<p>
</p>
<p>
--Ajit
</p>
<p>
- - - - -
</p>
<p>
[E&OE]
</p>
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</ul>Sat, 04 Jan 2014 02:31:12 +0000Ajit R. Jadhavcomment 25430 at https://imechanica.orgRe:^6 How to think of force linear functional as force vector?
https://imechanica.org/comment/25427#comment-25427
<a id="comment-25427"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25421#comment-25421">Re:^4 How to think of force linear functional as force vector?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Amit: I have just looked again at the book <a href="http://books.google.com/books?id=bE-9tUH2J2wC&printsec=frontcover&dq=landau+lifshitz&hl=en&sa=X&ei=AfTGUo3PJKa0sQTvroDADQ&ved=0CEEQ6AEwAw#v=onepage&q=landau%20lifshitz&f=false">Mechanics by Landau and Lifshitz</a>. Chapter 1 describes the principle of least action, basis-free. The book defines force by the gradient of energy.
</p>
<p>
At some point the isotropy of the space is used. Also the magnitude of velocity enters. These would rely on an inner product of the position space.
</p>
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</ul>Fri, 03 Jan 2014 17:30:00 +0000Zhigang Suocomment 25427 at https://imechanica.orgA scalar set is defined as a 1D vector space over real numbers
https://imechanica.org/comment/25425#comment-25425
<a id="comment-25425"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25418#comment-25418">Dear Ajid and others,
I'm</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Aleksander and Ajid: In the initial post, I defined a scalar set as follows
</p>
<p>
<strong>Definition</strong>. A scalar set is a one-dimensional vector space, S, over the field of real numbers, R.
</p>
<p>
I followed the <a href="http://en.wikipedia.org/wiki/Field_(mathematics)#Definition_and_illustration">definition of number field</a> and the <a href="http://en.wikipedia.org/wiki/Vector_space#Definition">defintion of vector space</a>. This definition involves two sets S and R, and four binary maps:
</p>
<ul><li>Adding two elements in R gives an element in R</li>
<li>Multiplying two elements in R gives an element in R</li>
<li>Additing two elements in S gives an element in S</li>
<li>Multiplying an element in R and an element in S gives an element in S </li>
</ul><p>
These binary maps follow the familar rules, as listed in the definitions of vector space and number field. Each element in S is called a scalar. Each element in R is called a number.
</p>
<p>
Let u be a nonzero element in S. Each element s in S scales with u by a real number r, namely, s = ru. We call u a unit of S, and r the magnitude of s relative to the unit u. These terms parallel those in a higher dimensional vector space: scalar~vector, unit~basis, magnitude~component.
</p>
<p>
The definition is abstract, but is intended to model objects like energy, mass, charge, entropy. For example, for the scalar set of masses, 1.7kg means a mass, which is 1,7 times the mass of a <a href="http://en.wikipedia.org/wiki/Kilogram">particular block of metal</a>. The mass of the particular block of metal is an element in the scalar set of masses and is chosen as a unit. 1.7kg is another element in the set.
</p>
<p>
The purpose of this definition is to understand the linear algebra of our physical world, as noted in <a href="http://imechanica.org/node/15857#comment-25403">another comment</a>.
</p>
<p>
Is this definition the same as yours?
</p>
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</ul>Fri, 03 Jan 2014 12:11:00 +0000Zhigang Suocomment 25425 at https://imechanica.orgRe:^5 How to think of force linear functional as force vector?
https://imechanica.org/comment/25424#comment-25424
<a id="comment-25424"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25421#comment-25421">Re:^4 How to think of force linear functional as force vector?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Amit: A <a href="http://imechanica.org/node/15843#comment-25328">question you posted in an earlier thread</a> gives a focus to this discussion. Here is your question again:
</p>
<p>
"...why conventional mechanics gets away with treating position vectors, displacements, velocities, area vectors, tractions etc. as all belonging to the same vector space (when physically we cannot really add displacements and velocities, e..g.)..."
</p>
<p>
The answer, it seems to me, is that all these objects are generated from a single 3D vector space: the space of position vectors.
</p>
<p>
As I noted in a <a href="http://imechanica.org/node/15857#comment-25403">previous comment</a> , the linear algebra of our world has the following basic building blocks:
</p>
<ul><li>The field of real numbers, R,</li>
<li>3D vector space of position vectors, V</li>
<li>many scalar sets, time, entropy, energy, mass, charge</li>
</ul><p>
We can then use these basic parts to generate objects like velocity, force, area vector, traction, etc, by using linear maps among V and various scalar sets. We can then map the maps to generate even more objects; for example, stress is a linear map that maps area vector to force. And then the set of all stresses is a vector space, ready to be mapped.
</p>
<p>
In the above list of the basic building blocks, only a single 3D vector space V appears. We use linear maps to generate other objects. Thus, a change of basis in V will change the components of all the objects so generated, according to the rules of tensor.
</p>
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</ul>Fri, 03 Jan 2014 11:28:00 +0000Zhigang Suocomment 25424 at https://imechanica.orgRe: Alecsander's suggestion
https://imechanica.org/comment/25423#comment-25423
<a id="comment-25423"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25418#comment-25418">Dear Ajid and others,
I'm</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Hi Alec (hope this way of calling you is OK by you), and others:
</p>
<p>
Re. Your suggestion of S = F + PU: Yes, I think that's a very neat suggestion. (And you are right in noting that a constraint also needs to be put on more operations like addition.) There is no reason why a mathematical object or a structure cannot be a composite. And, yes, the physical units is an essential part of the consideration here.
</p>
<p>
But, no, this is not a <em>philosophical</em> discussion. It's about the logical part of <em>mathematics</em>, and about <em>foundational</em> issues of <em>mechanics</em>. You could call it philosophy of science/mechanics, if you wish. But inasmuch as science, including engineering science, requires precision in not only data but also <em>concepts</em>, there is a definite place for such discussions, right in the realm of science itself---not of philosophy.
</p>
<p>
Yes, we engineers do usually rely on mathematicians to supply us with the most precise definitions, but if their offering falls short, it sure is our responsibility to point out our requirements to them. At least I see it that way. So, it's very much a discussion in science. (Think what a Kant, a Hume, or, for that matter, a Budhist will say by way of a discussion or an answer to this same question, and you can immediately see that this is not philosophy per say.)
</p>
<p>
Re. temperature: Epistemologically, it sure is true that concepts are formed in reference to two or more instances/objects. (A notable exception is the concept of existence, or in physics, of the physical universe taken as a whole: what it refers to is a single object, in a way.) But once a concept like temperature is formed, I think that it also is epistemologically proper to regard it as a property or attribute of just single bodies taken one at a time. A <em>change</em> of temperature would sure take two bodies, but right there we first assume that each body does have that attribute of temperature in the first place.
</p>
<p>
It is heat that means energy in transit, and thus requires at least two bodies, not temperature.
</p>
<p>
Inasmuch as temperature means the level of the energy content of a (single) body, I think that we can think of adding (or subtracting) temperatures. Can you add two heights? Yes. Two potentials? Yes. If so, then on similar lines, adding (or subtracting) two temperatures also <em>should</em> make sense. To understand temperature as a level, it is useful to make reference to the kinetic theory (as a measure of the kinetic energy/phononic energy), but I also think that this procedure is not entirely necessary---you can define temperature "purely" in continuum terms, too, and even if defined this way, it means a level of a kind.
</p>
<p>
Coming back to your suggestion, let me share some further thoughts:
</p>
<p>
Mathematicians have never cared for physical units; indeed, progressing further in their abstractions, they often don't care even for constants, ratios, or sometimes even limits (in fractal theory) or measures like lengths/areas (in topology). And, their pursuit of ideas like these certainly have been useful, too; I won't deny that. But that does not mean that they should <em>never</em> care for physical units. It indeed is their task to ponder over the <em>theoretical</em> necessity to have physical units, and then suggest proper logico-mathematical ways to handle it.
</p>
<p>
As an aside: In earlier, better times, e.g. around 19th century, mathematics was not divorced from physical reality, and innovations like dimensional analysis were possible. The law of dimensional homogeneity (first put forth, I think, by Fourier) is usually taken as a physical law, but inasmuch as it applies to <em>equations</em>---<em>any</em> physically valid equation---there also is an essential <em>mathematical</em> aspect to it; it also is, I think, a mathematical law of a kind. (But, of course, the nature of the relation between physics and mathematics is way out of scope here.) In a similar vein, mathematicians should not shirk from tackling the kind of issues that have been raised in this thread in general, and of physical units or dimensions in particular---suggesting proper mathematical structures/forms/objects to encapsulate these concerns.
</p>
<p>
--Ajit
</p>
<p>
- - - - - <br />
[E&OE]
</p>
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</ul>Fri, 03 Jan 2014 10:11:38 +0000Ajit R. Jadhavcomment 25423 at https://imechanica.orgRe:^4 How to think of force linear functional as force vector?
https://imechanica.org/comment/25421#comment-25421
<a id="comment-25421"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25415#comment-25415">Re:^3 How to think of force linear functional as force vector?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Hi Zhigang: Everything I am saying has to do with algebra (the special example from particle mechanics involved a little calculus and I made a few side remarks about functional analysis, a subject that may be viewed as infinite dimensional linear algebra). As to the question you ask, being able to allow an algebraic rule for 'multiplying' elements of S and V would help I think in the problem I was considering, but I am not sure that by itself will solve the problem and it needs more work than I did for a clean answer. Moreover, similarly, it is not clear to me that there are not other important questions where the fact that your number field is actually an algebra (roughly, vector space that allows 'multiplication' of elements) becomes important too. I simply don't know. I think it needs a real expert with deep knowledge of the ramifications of the interaction of the number field and the vector space......
</p>
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</ul>Fri, 03 Jan 2014 04:12:40 +0000Amit Acharyacomment 25421 at https://imechanica.orgDear Ajid and others,
I'm
https://imechanica.org/comment/25418#comment-25418
<a id="comment-25418"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25407#comment-25407">Re: Scalars. I was being wrong (or, may be, half right!)</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Ajid and others,
</p>
<p>
I'm not familiar with abstract algebra (still on my to-do list!) but I totally see the Your point. I agree that if field of physical quantities is not closed under multiplication then other structure should be formulated. One could think of physical quantities as a special structure (S) created in following way:
</p>
<p>
S consists of two elements (so it's similar to vectors) but with one element being formulated over real number field (F) and the other over physical units (let's call it PU). Now we can postulate that addition on S is defined only if Si(2)=Sj(2) -> second terms must match, then Si+Sj = [Si(1)+Sj(1), Si(2)]. while under multiplication terms are multiplied respectively. I don't know if that makes sense but that's the way I see it.
</p>
<p>
PS1. Wikipedia differentiate between scalars in mathematics and physics.
</p>
<p>
PS2. This is more philosophy discussion than engineering one (again, my opinion)
</p>
<p>
PS3. Addition of temperatures is meaningless as temperature only indicates direction of energy flow. Temperature has no meaning if there aren't at least 2 bodies. So what would be adding temperature mean? If you join two bodies they won't have added temperature altogether, but rather their difference will determine heat flux (I guess). On the other hand difference is subtraction (so we got addition here) and this quantity is used for example in Newton's Law of Cooling.
</p>
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</ul>Fri, 03 Jan 2014 00:22:29 +0000Aleksander Marekcomment 25418 at https://imechanica.orgRe:^3 How to think of force linear functional as force vector?
https://imechanica.org/comment/25415#comment-25415
<a id="comment-25415"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25414#comment-25414">Re:^2 How to think of force linear functional as force vector?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Amit: I have not been thinking about analysis. Just algebra at the moment.
</p>
<p>
By definition, a scalar set S is a one-dimensional vector space over the field of real number. Thus, once we choose any one non-zero element u in S, all elements in S take the form au, where a is a real number.
</p>
<p>
In linear algebra, u is a basis of S
</p>
<p>
In physics, u is a unit for the scalar in S.
</p>
<p>
Thus, in terms of analysis, S has the same structure as the field of real number. Also the position space V is still an Euclidean space.
</p>
<p>
Does this solve your issue?
</p>
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</ul>Thu, 02 Jan 2014 20:53:00 +0000Zhigang Suocomment 25415 at https://imechanica.orgRe:^2 How to think of force linear functional as force vector?
https://imechanica.org/comment/25414#comment-25414
<a id="comment-25414"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25405#comment-25405">Re: How to think of force linear functional as force vector?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Zhigang: Thanks for trying to explain and the link to the Wiki page. There is a subtlety here.</p>
<p>I will call your number field R instead of F (I'll need F for Newtonian force below).</p>
<p>The dual space of V w.r.t. S that you are defining is simply the space of all linear transformations, say L(V,S), between V and S. I agree that it can be proved (and it does require a proof) that L(V,S) is again a vector space, but just from its definition it is also clear that that vector space does not contain elements of V.</p>
<p>Just for the moment, think that R = S. The whole job of the representation theorem for linear functionals in linear algebra (and the Riesz representation theorem in Functional Analysis) is to show that corresponding to each element f of L(V,S) (in functional analysis, L(V,S) would have to be the set of bounded linear functionals) there exists a unique element A(f) in V such that the action of f belonging to L(V,S) on *any* u in V is given by the inner-product (in V) of A(f) and u. Thus A(f) is the representation in V of f in L(V,S). In mechanics, this (A) would be the force vector corresponding to the force linear functional (f). And it is this force vector that I was trying to construct.</p>
<p>Why is such a construction important? Let's say I want to solve a problem in classical particle mechanics F = ma. Let's say it is also mentioned that F is conservative and so generated from the negative spatial gradient of a potential. The potential field is specified as a given function, say psi, of position in ambient Euclidean 3-d space. What is the meaning of the gradient? Clearly the inertial force and the acceleration gets its direction from what we specify for F; F better be a force vector, a member of V or a close cousin that looks and smells like V with the right physical dimensions of force. What CANNOT suffice in this context is to say that F is a member of L(V,S), even though L(V,S) has the right dimension (both vector space meaning and physical units meaning), and setting it equal to ma which is a member of V (or the close 'force' cousin of V). Alternatively, you would have to raise ma to be a member of L(V,S) somehow.</p>
<p>Now given psi as a function of position, do its derivative. What is the derivative of psi at any given point of space? It is a member of L(V,S), by definition. Is the gradient the derivative? No. The gradient is the representation in V of the derivative in L(V,S), when S = R. If you wrote psi as a function say of rectangular Cartesian coordinates, and just did what we would think of doing for a 'derivative', what do you get? One gets the components of the gradient vector of psi on the orthonormal basis corresponding to the Rectangular Cartesian coordinates chosen (all this can be done in arbitrary coordinates). Now I ask, give me a similar 'formula' for the derivative. One will be hard-pressed to come up with something that is not contrived, even though most would accept that the derivative exists. It is in this sense I find the algebraic dual space of a vector space an 'abstract' object.
</p>
<p>
<br />
So, my point is in applications, the representation of linear functionals in the vector space matters a great deal and in what has been spelled out until now, how this can work out is not clear. This is what I was trying to do in what I described. When S = R (your F) all this works out fine, but when S is not R it seems problematic.</p>
<p>My feeling is that some notion of multiplication between elements of S and elements in V is required at the very least (to set up the 'close cousin' vector space of forces), but a satisfactory resolution seems to require some careful thinking. Making S separate from R is a good idea, but coming up with a satisfactory alternate framework requires more than the definitions we have seen on imechanica until now (keep in mind we have poked with just one example as yet). It would be good if we could get the help of some people good at both algebra and mechanics.
</p>
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</ul>Thu, 02 Jan 2014 18:49:00 +0000Amit Acharyacomment 25414 at https://imechanica.orgRe: The trend for the development of rational mechanics
https://imechanica.org/comment/25411#comment-25411
<a id="comment-25411"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25408#comment-25408">The trend for the development of rational mechanics</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>Dear Jing: The adoption of scalar sets cleans up our practice, but it does not introduce any new mechancs. We are looking into the use of Latex on iMechanica.</p>
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</ul>Thu, 02 Jan 2014 14:53:00 +0000Zhigang Suocomment 25411 at https://imechanica.orgThe trend for the development of rational mechanics
https://imechanica.org/comment/25408#comment-25408
<a id="comment-25408"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25403#comment-25403">The linear algebra of the world</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Prof. Suo,
</p>
<p>
If this is the case, the physical world could have a rigorous and beautiful mathematical structure. We could use the revised definitions to analyze and redefine classical mechanical concepts. Hence, in this aspect, it seems to deserve a lot of work??
</p>
<p>
It's pity that at present I don't have direct access to you comments on twitter since I'm in Zhejiang University, China. Just as I suggested in the e-mail that I sent to you, I think imechanica could consider to incoporate latex in order to enable its users to edit math formulas directly. Technically,it seems not to hard for a website to be compatible with Latex.
</p>
<p>
With the help of compatibility with latex , maybe we can describe the applications of the linear maps in detail.
</p>
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</ul>Thu, 02 Jan 2014 14:32:00 +0000Jing_Yangcomment 25408 at https://imechanica.orgRe: Scalars. I was being wrong (or, may be, half right!)
https://imechanica.org/comment/25407#comment-25407
<a id="comment-25407"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25406#comment-25406">Re:Re:Re: Is scalar a synonym of number?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Zhigang, and others,
</p>
<p>
Guess I got the problem <em>now</em> (!)---or so I think (!!)
</p>
<p>
So, I only now realize that, no, when it comes to scalars, even the ordinary "real number field" wouldn't be suitable either. The real number field by definition <em>is</em> closed under multiplication and we need a structure that isn't so. (I appreciate this part better only now!!)
</p>
<p>
But then, for the reasons mentioned above, it won't be a vector space either---whether 1D or otherwise.
</p>
<p>
So, at least for scalars like mass and energy, it would be neither a vector space nor a real number field, but a new structure that is like the real number field except for a further constraint that it isn't closed under multiplication. Suppose we call this new structure S. Now, inasmuch as the physical units of mass, mass^2 and mass^4 do differ (not to mention mass^3), the idea of there being different instances of this new structure each for the <em>operands</em> and the <em>results</em> of the multiplication operation, might still hold. (I would be half-right only in this sense.)
</p>
<p>
And, I don't know what to think of adding two temperatures. Shouldn't addition be a valid operation if the <em>absolute</em> temperature scale is used? I tend to think so, but now am no longer sure of anything (LOL!)
</p>
<p>
(Ok, now, see you tomorrow, really!)
</p>
<p>
--Ajit
</p>
<p>
- - - - - <br />
[E&OE]
</p>
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</ul>Thu, 02 Jan 2014 14:04:10 +0000Ajit R. Jadhavcomment 25407 at https://imechanica.orgRe:Re:Re: Is scalar a synonym of number?
https://imechanica.org/comment/25406#comment-25406
<a id="comment-25406"></a>
<p><em>In reply to <a href="https://imechanica.org/comment/25404#comment-25404">Re:Re: Is scalar a synonym of number?</a></em></p>
<div class="field field-name-comment-body field-type-text-long field-label-hidden"><div class="field-items"><div class="field-item even"><p>
Dear Zhigang,
</p>
<p>
1. Looks like I used a wrong term. I meant to keep a scalar quantity only to a simple real number field [<a href="http://en.wikipedia.org/wiki/Real_number" target="_blank">^</a>], whereas a vector quantity would properly belong to a vector space [<a href="http://en.wikipedia.org/wiki/Vector_space" target="_blank">^</a>].
</p>
<p>
So, please substitute "real number field" wherever I have used the term "number field" in my above reply.
</p>
<p>
2. The main point I then made is that the real number field for mass would not be the same as the real number field for the quantity: mass X mass. <em>Both</em> would be real number fields, but since their physical units differ, they would belong to two different <em>instances</em> of real number fields. Neither would be a vector space.
</p>
<p>
3. The structure of a vector space, even if it be only a one-dimensional vector space, brings in the additional baggage of a basis set.
</p>
<p>
A basis set is nothing but, what else, just a minimal set of <em>vectors</em> that can be used to span that entire space. For 1D space, the basis would consist of just one vector, but it would still be a <em>vector</em>! Yet, when we talk of a scalar like mass, we don't think in terms of some real number multiplied by a "unit vector of a mass," where the latter notion supposedly might mean going from an origin point of a zero mass to a point of a unit mass---the very ideas of a "from" and "to" don't apply for a scalar quantity like mass.
</p>
<p>
This extra conceptual baggage that the basis set introduces is neither necessary for scalar quantities nor does it make any sense for them.
</p>
<p>
4. Yes, it is true that a vector space is not closed under multiplication. But the converse is not true (I am not sure if "converse" is the word I want here). The idea of a vector space is not necessarily an answer for every operation that is not closed under multiplication.
</p>
<p>
As I pointed out, the answer is: <em>multiple</em> instances, identified by physical units, of a certain "scalar-like" structure. In my first reply above, I called it the number field. Looks like I should have called it the "real number field." The key idea I introduced is, however, the <em>multiple</em> instances of the <em>same</em> structure, not an introduction of <em>another</em> structure.
</p>
<p>
I would like to know what other mechanicians think of these points, too.
</p>
<p>
(PS: Signing off for now, will be back tomorrow morning IST, i.e. after about 15 hours.)
</p>
<p>
--Ajit
</p>
<p>
- - - - - <br />
[E&OE]
</p>
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</ul>Thu, 02 Jan 2014 13:27:07 +0000Ajit R. Jadhavcomment 25406 at https://imechanica.org