iMechanica - Comments for "Textbook on linear algebra"
http://imechanica.org/node/15843
Comments for "Textbook on linear algebra"enLinear algebra done right
http://imechanica.org/comment/25396#comment-25396
<a id="comment-25396"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25296#comment-25296">Vector spaces and tensors</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, I have followed the linear algebra discussion on iMechanica with quite a bit of interest. Although I don't have anything meaningful to contribute beyond what has already been said, I would like to make a book recommendation.</p>
<p>
When I joined the University of Houston as a young assistant professor, Lewis Wheeler used to teach our required graduate math course sequence (--which I later taught as well). He is a big believer in teaching linear algebra in depth and also successfully convinced me of its importance for mechanicians. He refers to several books in the math course however, when I had asked him for an elementary text, he recommended the following; which he used heavily: "Linear Algebra Done Right": <br /><a href="http://www.amazon.com/Linear-Algebra-Right-Undergraduate-Mathematics/dp/0387982582">http://www.amazon.com/Linear-Algebra-Right-Undergraduate-Mathematics/dp/...</a>
</p>
<p>
I was amused to see on your post that there is also a book with precisely the opposite title! I had read the Linear Algebra Done many moons ago and liked it quite a bit. If I recall correctly, it does not quite get into tensors but lays out all the foundational material both rigorously and an easy to understand manner
</p>
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</ul>Wed, 01 Jan 2014 19:09:25 +0000Pradeep Sharmacomment 25396 at http://imechanica.orgWe are all applied mathematicians
http://imechanica.org/comment/25381#comment-25381
<a id="comment-25381"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25380#comment-25380">good memories</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>
My colleague, <a href="http://www.seas.harvard.edu/brenner/Home.html">Michael Brenner</a>, is spreading the wisdom that
</p>
<ol><li>The applied mathematician does not create mathematics just as the historian does not create history.</li>
<li>We are all applied mathematicians.</li>
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</ul>Tue, 31 Dec 2013 14:25:00 +0000Zhigang Suocomment 25381 at http://imechanica.orggood memories
http://imechanica.org/comment/25380#comment-25380
<a id="comment-25380"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25379#comment-25379">Re: why stress is a tensor</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>
Thank you for refreshing good memories :)
</p>
<p>
Happy New Year to you too!
</p>
<p>
Let us map old discussions into new year.
</p>
<p>
I am a big fan of iMechanica - I learn from it (not only your students do that). I have a feeling that you are becoming a mathematician. This is a dangerous process and I will try to slow it down in May :)
</p>
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</ul>Tue, 31 Dec 2013 14:12:00 +0000Konstantin Volokhcomment 25380 at http://imechanica.orgRe: why stress is a tensor
http://imechanica.org/comment/25379#comment-25379
<a id="comment-25379"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25374#comment-25374">why stress is a tensor</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 Kosta: Your notes on the<a href="http://imechanica.org/node/7521"> Mechanics of Soft Materials</a> are beautiful. Writing to you today brought back the memory of our first meeting, in August 2006, at Brown University, at the symposium celebrating the 60th birthday of Alan Needleman and Viggo Tvergaard. Or was that our first meeting? In any case, you were already deep in the mechanics of soft materials, and I was still a hardy. We were on separate recruiting missions. You were persuading me that the mechanics of soft materials was coming. I was persuading you that iMechanica was coming.
</p>
<p>
It turned out neither of us needed much persuasion. You became iMechanica User 134, and a very vocal one. <a href="http://imechanica.org/node/635">You gave me a hard time</a>, on iMechanica, all in good spirit, on my first paper on the mechanics of soft materials. I posted the paper on 1 January 2027, and you made your first comment on the paper on the second day. Your comments made great impression on me and my students. Thank you.
</p>
<p>
Now back to the question, Who gave the authority to Cauchy to define stress as a tensor, as we call it today?
</p>
<p>
The answer, it seems to me, is Euclid, Newton, and whoever invented the notion of tensor. The statement "stress is a tensor" results from a combination of geometry, mechanics, and algebra.
</p>
<p>
In <a href="/node/7521">your notes</a>, by force balance and geometry, you reach equation (3.17). The equation says that the force acting on an area is linear in the area vector.
</p>
<p>
The rest, as they say, is algebra. The area vector is an element in a vector space. The force is an element in another vector space. Equation (3.17) says that a linear map exists, mapping an area vector to the force acting on the area.
</p>
<p>
In linear algebra, we call a linear map between vector spaces a tensor.
</p>
<p>
In mechanics, we name this particular map the stress.
</p>
<p>
Incidentally, the existence of a linear map between the area vector and the force can be established without using any basis. I gave such a derivation in <a href="http://imechanica.org/node/538">my notes on finite deformation</a>. The derivation applies to both the nominal stress and the true stress.
</p>
<p>
Happy new year! May we have many more happy memories of talkig to each other in many more new years.
</p>
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</ul>Tue, 31 Dec 2013 13:45:00 +0000Zhigang Suocomment 25379 at http://imechanica.orgWho gives the authority to Cauchy?
http://imechanica.org/comment/25376#comment-25376
<a id="comment-25376"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25374#comment-25374">why stress is a tensor</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>Who gives the authority to Cauchy to define stress as a sum of dyads?</p>
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</ul>Tue, 31 Dec 2013 11:33:00 +0000Zhigang Suocomment 25376 at http://imechanica.orgwhy stress is a tensor
http://imechanica.org/comment/25374#comment-25374
<a id="comment-25374"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25370#comment-25370">Why is stress a tensor?</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, Cauchy stress is defined as a sum of three dyadic products of vectors - see Eq. (3.21) of my lecture notes on <a href="http://imechanica.org/node/7521">Mechanics of Soft Active Materials</a>.Thus, it is a tensor by construction.</p>
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</ul>Tue, 31 Dec 2013 10:37:00 +0000Konstantin Volokhcomment 25374 at http://imechanica.orgWhy is stress a tensor?
http://imechanica.org/comment/25370#comment-25370
<a id="comment-25370"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25369#comment-25369">maps and tensors</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 Kosta, Very well then. Please outline how you show that stress is a tensor.
</p>
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</ul>Tue, 31 Dec 2013 06:24:26 +0000Zhigang Suocomment 25370 at http://imechanica.orgmaps and tensors
http://imechanica.org/comment/25369#comment-25369
<a id="comment-25369"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25354#comment-25354">Re: Why map?</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>Not exactly. You define maps by their actions. Objects are defined by their properties. You do not need to map vector to vector in order to define tensor. You may define tensor by the transformation properties of its components (as Levi-Civita did). Or you may define tensor as a linear combination of dyads, tryads etc. They can be used for maps but they do not need maps for their definition.</span></p>
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</ul>Tue, 31 Dec 2013 06:00:00 +0000Konstantin Volokhcomment 25369 at http://imechanica.orgLet's move on to talk about scalars
http://imechanica.org/comment/25368#comment-25368
<a id="comment-25368"></a>
<p><em>In reply to <a href="http://imechanica.org/node/15843">Textbook on linear algebra</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>This thread has become very interesting and very long. I have another issue on which I'd like to get your opinion. It's about scalars. I have just posted a note titled <a href="http://imechanica.org/node/15857">Scalar Done Wrong.</a></p>
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</ul>Tue, 31 Dec 2013 04:41:00 +0000Zhigang Suocomment 25368 at http://imechanica.orgRe: Re: Re: Some thoughts on tensors
http://imechanica.org/comment/25367#comment-25367
<a id="comment-25367"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25359#comment-25359">Re: Re: Some thoughts on tensors</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 Arash,
</p>
<p>
I of course agree with what you say about a linear connection, but that takes us into calculus on manifolds while in this discussion everything is in some sense in 'one tangent space.'
</p>
<p>
In fact, in my opinion that is where the fuss with different linear spaces really starts to matter, and one has to take a stance. For instance, suppose we fuss a lot in the linear algebra and then when we have to do calculus we adopt balance of linear momentum and start adding traction vectors from different tangent spaces (the latter I am perfectly happy to do and have bought into, and I would not fuss with the linear algebra either), then what was the initial fuss while doing algebra about? I know of your work on covariantization of elasticity, so you seem to 'fuss consistently,' which I like - (this is a joke, take it lightly) .
</p>
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</ul>Tue, 31 Dec 2013 01:18:18 +0000Amit Acharyacomment 25367 at http://imechanica.orgdefinitions of tensors in mechanics textbooks
http://imechanica.org/comment/25366#comment-25366
<a id="comment-25366"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25362#comment-25362">Re: I have a dream--about tensors</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: Thank you very much. It is fun to make a list of definitions of tensors in textbooks of continuum mechanics. Of course, the authors may have more sophistictated ideas than they put in their books. We are doing a phenomenological search. What are in the textbooks are a reasonable proxy of what we teach to our students, I suppose.
</p>
<p>
I have looked at the following books:
</p>
<ul><li><a href="http://rbowen.tamu.edu/">Bowen</a> (p.277) The same as Ogden.</li>
</ul><ul><li><a href="http://www.amazon.com/Mechanics-Thermodynamics-Continua-Morton-Gurtin/dp/052140598X">Gurtin-Fried-Anand</a> (p.9). "We use the tensor as a synonym for the phrase "linear transformation from V to V"." </li>
<li><a href="http://www.amazon.com/Nonlinear-Problems-Elasticity-Mathematical-Sciences/dp/0387208801/ref=sr_1_1?s=books&ie=UTF8&qid=1388450224&sr=1-1&keywords=Antman+elasticity">Antman</a> (p.400). The same as above.</li>
<li><a href="http://www.amazon.com/Non-Linear-Field-Theories-Mechanics/dp/3540027793/ref=sr_1_sc_1?s=books&ie=UTF8&qid=1388450408&sr=1-1-spell&keywords=Truesdel+Noll">Truesdel-Noll</a> (p.15). The same as above.</li>
<li><a href="http://www.amazon.com/Nonlinear-Solid-Mechanics-Continuum-Engineering/dp/0471823198/ref=sr_1_1?s=books&ie=UTF8&qid=1388450541&sr=1-1&keywords=holzapfel+nonlinear+solid+mechanics">Holzapfel</a> (p.9). The same as above.</li>
<li><a href="http://www.amazon.com/Continuum-Mechanics-Concise-Problems-Physics/dp/0486401804/ref=sr_1_1?s=books&ie=UTF8&qid=1388451979&sr=1-1&keywords=continuum+Chadwick">Chadwick</a> (p.16). The same as above </li>
</ul><ul><li><a href="http://www.amazon.com/Plasticity-Deformation-Heterogeneous-Inelastic-Monographs/dp/0521839793/ref=sr_1_1?s=books&ie=UTF8&qid=1388451457&sr=1-1&keywords=Nemat-Nasser">Nemat-Nasser</a> (p.14) defines a tensor through dyads.</li>
<li><a href="http://www.amazon.com/Nonlinear-Solid-Mechanics-Bifurcation-Instability/dp/1107025419/ref=sr_1_1?s=books&ie=UTF8&qid=1388451017&sr=1-1&keywords=bigoni">Bigoni</a> (p.95). The same as above</li>
<li><a href="http://www.amazon.com/Continuum-Mechanics-Dover-Books-Physics/dp/0486435946/ref=sr_1_1?s=books&ie=UTF8&qid=1388451627&sr=1-1&keywords=Spencer+continuum">Spencer</a> (p.20). The same as above</li>
<li><a href="http://www.amazon.com/Plasticity-Theory-Dover-Books-Engineering/dp/0486462900/ref=sr_1_sc_1?s=books&ie=UTF8&qid=1388451688&sr=1-1-spell&keywords=Lublinger+plasticity">Lublinger</a> (p.5), dyads, ect. </li>
</ul><p>Of all the books you and I have looked at so far, only <a href="http://www.amazon.com/Mathematical-Foundations-Elasticity-Mechanical-Engineering/dp/0486678652/ref=sr_1_1?s=books&ie=UTF8&qid=1388453551&sr=1-1&keywords=marsden+hughes">Marsden-Hughes</a> allows mappings between different vector spaces.</p>
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</ul>Tue, 31 Dec 2013 01:18:00 +0000Zhigang Suocomment 25366 at http://imechanica.orgRe: Re: A non-dimensional linear space as a canonical choice?
http://imechanica.org/comment/25365#comment-25365
<a id="comment-25365"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25360#comment-25360">Re: A non-dimensional linear space as a canonical choice?</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 Arash,
</p>
<p>
Yes, that's what I am saying. Of course, the main issue in this discussion that took a bit of thinking was the manner in which the copies should be constructed as there are infinitely many ways of doing so and one wants to find the way that is consistent with what we routinely do in continuum mechanics, the latter of course being sensible, given the weight of its results.
</p>
<p>
I find the the DND mappings interesting in that by dimensional consistency they link the various copies - they cannot be generated from completely arbitrary isomorphisms, say.
</p>
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</ul>Tue, 31 Dec 2013 01:04:16 +0000Amit Acharyacomment 25365 at http://imechanica.orgRe: Re: Higher order tensors, equivalent definitions
http://imechanica.org/comment/25364#comment-25364
<a id="comment-25364"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25358#comment-25358">Re: Re: Higher order tensors, equivalent definitions</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 Arash,
</p>
<p>
Here's how I see it:
</p>
<p>
The tangent map for either the reference or the deformed shell goes from 2-d to a 3-d space; thus the matrix representation is rectangular as you say.
</p>
<p>
The deformation gradient goes from a 2-d tangent space to a different 2-d tangent space (vectors in one in general do not lie within the linear span of a basis for the other). Thus its matrix is square 2x2, the 'canonical' representation in convected coordinates being the 2x2 identity matrix.
</p>
<p>
Did I miss anything?
</p>
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</ul>Tue, 31 Dec 2013 00:47:24 +0000Amit Acharyacomment 25364 at http://imechanica.orgrectangular matrices for tensor representation
http://imechanica.org/comment/25363#comment-25363
<a id="comment-25363"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25357#comment-25357">Re: Higher order tensors, equivalent definitions</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: This is in response to a statement you made.
</p>
<p>
A third-order tensor on a 3-d vector space V_3 in the direct formulation of continuum mechanics (a la Carlson, say) is a linear transformation from 3-d space to the 3x3 = 9 dimensional space of second order tensors on V_3. Thus its representation on bases yields a rectangular 9 by 3 matrix. Similarly the transpose of this tensor is a linear transformation from the 9 dimensional space of second order tensors to the 3-d space - that yields a 3 x 9 matrix.
</p>
<p>
So what we do in continuum mechanics very much results in rectangular matrices when required and quite naturally (unlike what you claimed to be the case), but obviously these are trivial details not to be fussed over.
</p>
<p>
For some serious actual use of 3rd order tensors in continuum mechanics with the direct formulation definition:
</p>
<p>
<a href="http://www.imechanica.org/node/15585">http://www.imechanica.org/node/15585</a>
</p>
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</ul>Tue, 31 Dec 2013 00:39:00 +0000Amit Acharyacomment 25363 at http://imechanica.orgRe: I have a dream--about tensors
http://imechanica.org/comment/25362#comment-25362
<a id="comment-25362"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25361#comment-25361">I have a dream--about tensors</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>The book by R.W. Ogden (Non-Linear Elastic Deformations) defines tensors very similarly to what has been done in Carlson's notes (page 43). Let me also say that this is an excellent book that focuses on finding exact solutions in nonlinear elasticity.</p>
<p>Marsden and Hughes (Mathematical Foundations of Elasticity) differentiate between tensors (page 65) and two-point tensors (page 70). I think they did not define anything more general as they didn't find much use for it in the context of elasticity.</p>
<p>Green and Zerna (Theoretical Elasticity) define tensors in terms of coordinate transformations and how tensor components change under such transformations (Amit already mentioned this).</p>
<p>Regards,<br />
Arash</p>
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</ul>Tue, 31 Dec 2013 00:03:47 +0000Arash_Yavaricomment 25362 at http://imechanica.orgI have a dream--about tensors
http://imechanica.org/comment/25361#comment-25361
<a id="comment-25361"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25358#comment-25358">Re: Re: Higher order tensors, equivalent definitions</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: Thank you so much for the comment. I believe that you and I are both happy with defining tensor as the following
</p>
<ol><li>V1, V2,..., Vp, V are vector spaces over a field F.</li>
<li>A tensor is a <a href="http://en.wikipedia.org/wiki/Multilinear_map">multilinear map</a> that maps elements in V1, V2,...,Vp to an element in V.</li>
</ol><p>
<span>I don't believe that this definition is commonly adopted among mechanictians.</span>
</p>
<p>
<span>Now this statement is a socialogical one, not a scientific one. The statement can be amply confirmed by a look at existing textbooks. For example </span><a href="http://www.amazon.com/Continuum-Mechanics-Thermodynamics-Fundamental-Governing/dp/1107008263/ref=sr_1_1?s=books&ie=UTF8&qid=1388442348&sr=1-1&keywords=tadmor+continuum">Tadmor-Miller-Elliot</a> <span>(p.36) restrict V1, V2,...Vp to be vector spaces of the same dimension. Their definition is equivalent to that in the </span><a href="http://imechanica.org/node/15845">Carlson notes</a> <span>(Definition A4.1.1). One can go on to list many textbooks in mechanics that give definition equivalent to that in Tadmor-Miller-Elliot. </span>
</p>
<p>
<span>Are you aware of any textbook in mechanics that adopts the definition above (Point 1 and 2)? They must be a very small minority. That is what I mean that this definition is not commonly adopted by mechanicians.</span>
</p>
<p>
<span>Is this definition </span><span>(Point 1 and 2 above)</span><span> too general for our basic needs? That is debatable. But this much is sure. </span><span>As you and I discussed before, the definition comfortably accomodates the two most basic tensors in continuum mechanics: stress and deformation gradient. Let me list our steps once more (for people who wish to see more details, please take a look at my </span><a href="http://imechanica.org/node/538">notes on finite deformation</a><span>): </span>
</p>
<p>
<span><strong>A special case of the definition (Point 1 and 2)</strong>. A linear map that maps an element in one vector space to an element in another vector space is a tensor.</span>
</p>
<p>
<span><strong>Stress</strong>. By balancing forces, we find a linear map that maps an area vector to the force acting on the area. The area vector is an element in one vector space, and the force is an element in another vector space. We call this linear map the stress, and it fits our definition of tensor. </span>
</p>
<p>
<span>(Incidentally, here I use area vector, rather than unit vector normal to the area, because the latter does not form a vector space: a linear combination of two unit vectors is in general not another unit vector.) </span>
</p>
<p>
<span><strong>Deformation gradient</strong>. By the geometry of homogeneous deformation, we find a linear map that maps a segment of a straight line of material particles from the reference state to the current state. The segment in the reference state is an element in one vector space, and the segment in the current state is an element in another vector space. We call this linear map the deformation gradient, and it fits our definition of tensor.</span>
</p>
<p>
<span>By contrast, we have to do some dances to fit the two objects to the conventional definition of tensor. Why should we? Again, this is socialogical question, not a scientific one. </span>
</p>
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</ul>Mon, 30 Dec 2013 23:23:00 +0000Zhigang Suocomment 25361 at http://imechanica.orgRe: A non-dimensional linear space as a canonical choice?
http://imechanica.org/comment/25360#comment-25360
<a id="comment-25360"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25350#comment-25350">A non-dimensional linear space as a canonical choice?</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 Amit,</p>
<p>If I understand you correctly, pretty much you're saying let's look at different copies of the same linear space (using your DND mappings) and each quantity of interest lives in one copy. What do you think?</p>
<p>Regards,<br />
Arash</p>
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</ul>Mon, 30 Dec 2013 22:07:56 +0000Arash_Yavaricomment 25360 at http://imechanica.orgRe: Re: Some thoughts on tensors
http://imechanica.org/comment/25359#comment-25359
<a id="comment-25359"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25346#comment-25346">Re: Some thoughts on tensors</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:</p>
<p>Thanks for your remarks. I think the main one I'd like to understand is the first remark (I pretty much agree with your other comments). I'm not sure if this is the solution but one may be able to think of two copies of the same linear space. In other words, position vector and velocity vector live on two copies of the same linear space. This would mean that you can't even mathematically add vectors from the two copies.</p>
<p>This idea of different copies of the same linear space you clearly see in differential geometry. Given an n-manifold, tangent space at any point is an n-dimensional linear space. You have different copies of the same linear space at different points. These linear spaces are all isomorphic but you cannot add vectors from different copies of the same linear space. The extra structure that would allow you to do so is "connection". Just some thoughts of course.</p>
<p>Regards,<br />
Arash</p>
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</ul>Mon, 30 Dec 2013 21:49:45 +0000Arash_Yavaricomment 25359 at http://imechanica.orgRe: Re: Higher order tensors, equivalent definitions
http://imechanica.org/comment/25358#comment-25358
<a id="comment-25358"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25357#comment-25357">Re: Higher order tensors, equivalent definitions</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>I haven't looked at this book but I don't think we're restricted to work with linear spaces of the same dimension the way we define tensors. For example, deformation gradient for a shell is a linear map from a two-dimensional vector space to a three-dimensional vector space. So, it is a two-point tensor and its matrix representation is not a square matrix.</p>
<p>I agree with Amit that one shouldn't worry too much about what definition is better and/or more general. I would say one should use what is appropriate for the application of interest.</p>
<p>Regards,<br />
Arash</p>
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</ul>Mon, 30 Dec 2013 21:32:44 +0000Arash_Yavaricomment 25358 at http://imechanica.orgRe: Higher order tensors, equivalent definitions
http://imechanica.org/comment/25357#comment-25357
<a id="comment-25357"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25355#comment-25355">Higher order tensors, equivalent definitions</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: Sorry for being unclear. When I wrote, "We essentially stop at 2", I meant to say stop at Point 2 in the definition listed in the comment. That is, we essentially stop at the following point: tensor is a multiliear map that maps elements in V1, V2,..., Vp to an element in V. This definition, of course, let us go beyond second-order tensor.
</p>
<p>
I believe that the definition of tensor in Treil's book is broader than what we normally use. For example, this definition will allow a tensor with compoenents forming a rectangular matrix, instead of being limited to a squre matrix. That is, this definition is not equivalent to what we commonly adopt.
</p>
<p>
I like your spirit: everything goes, so long as we do good mechanics.
</p>
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</ul>Mon, 30 Dec 2013 18:58:00 +0000Zhigang Suocomment 25357 at http://imechanica.orgOnline lecture notes on linear algebra
http://imechanica.org/comment/25356#comment-25356
<a id="comment-25356"></a>
<p><em>In reply to <a href="http://imechanica.org/node/15843">Textbook on linear algebra</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>All, thanks for a fruitful cascade of discussion, which I follow with great interest. I've updated the <a href="http://imechanica.org/node/1551">Lecture Notes</a> page of iMechanica to include downloadable lecture notes on linear algebra alluded in the discussions under the "Mathematics for Mechanics" section.</p>
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</ul>Mon, 30 Dec 2013 18:03:00 +0000Teng Licomment 25356 at http://imechanica.orgHigher order tensors, equivalent definitions
http://imechanica.org/comment/25355#comment-25355
<a id="comment-25355"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25351#comment-25351">Linear Algebra 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>
Hi Zhigang: I haven't had the time to take a look at the book you mention, but here is a guess as to why it makes a lot of sense to define whatever one needs (e.g. 2nd order tensors, higher order tensors), especially when doing mechanics - it is to ultimately to do the mechanics one wants to do. If 2nd order tensors are all one needs, one does that. If higher order tensors are needed (as from the time of the Brothers Cosserat, at least, in continuum mechanics, through Green-Rivlin, Noll, Truesdell etc.; of course linear elasticity came even before that), one uses a logically satisfactory definition for that, and then gets on with life. And, based on these examples, I wouldn't say mechanicians stop at order 2.
</p>
<p>
As for "routinely "creating"/"discovering" new tensors" - I feel it is important to add to your statement that while any mechanician with an iota of basic training and good sense appreciates equivalences between disparate definitions as a deeper understanding of his/her subject, none in this class think that coming up with such definitions is an achievement in its own right or that one particular definition among equivalent definitions is somehow better. Moreover, they are also completely agnostic and mercenary in their approach to such definitions - they learn all of them as well as they can to deploy (all of) them effectively as particular situations demand.
</p>
<p>
And, a final note - let's not forget the purely component-transformation-rule based approach for introducing tensors that we have not discussed at all. Again, that is an equivalent definition, and Einstein would be very mad with us - or think we were really ignorant - if we were to tell him that the multilinear approach of introducing tensors is really the way to think about tensors and their applications to mechanics.
</p>
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</ul>Mon, 30 Dec 2013 17:25:00 +0000Amit Acharyacomment 25355 at http://imechanica.orgRe: Why map?
http://imechanica.org/comment/25354#comment-25354
<a id="comment-25354"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25348#comment-25348">why map?</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>
Maps are objects as well. In linear algebra, tensors are defined as linear maps. See a comment below: <a href="http://imechanica.org/node/15843#comment-25351">linear algebra done wrong</a>.
</p>
<p>
More importantly, we mechanicians have been using linear maps to define tensors all the time. Here are familiar examples:
</p>
<ul><li>Stress is a linear map that maps an area vector to the force acting on the area.</li>
<li>Deformation gradient is a linear map that maps a segment of staight line of material particles from the reference state to the current state.</li>
</ul><p>
Maybe you and I use different words, but we mean the same things. I have descibed these definitions in my notes on <a href="http://imechanica.org/node/538">finite deformaton</a>.
</p>
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</ul>Mon, 30 Dec 2013 16:10:00 +0000Zhigang Suocomment 25354 at http://imechanica.orgLinear Algebra Done Wrong
http://imechanica.org/comment/25351#comment-25351
<a id="comment-25351"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25347#comment-25347">Linear Algebra Done Right-Sheldon Axler, Springer</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>
This is the title of the book by the mathematician Sergei Treil, of Bowen University.
</p>
<p>
He noted that the book is for a first course on linear algebra for mathematically advanced students. He posted the book online. For some time I have had the 2004 edition, and have enjoyed the book: the beautiful layout, careful sequencing of ideas, and masterful pedagogy. It is a short book, ~200 pages.
</p>
<p>
This morning I went to <a href="http://www.math.brown.edu/~treil/papers/LADW/LADW.html">his website</a>, and found a 2009 edition! In this new edition, he has added a chapter on dual spaces and tensors.
</p>
<p>
Here is how he defines tensors:
</p>
<ol><li>V1, V2,..., Vp, V are vector spaces over a field F.</li>
<li>A multilinear map that maps elements in V1, V2,...,Vp to an element in V.</li>
<li>The multilinear map is called a tensor if V is a scalar over the field F. </li>
</ol><p>
Not sure why he does not stop at 2, and simply defines tensor as a multilinear map from vector spaces to yet another vector space. Noneheless a few pages later in the book, he did show that a linear map from one vector space to another vector is also a tensor.
</p>
<p>
I believe that mechanicians have been routinely "discovering" or "creating" new tensors from known tensors, as I commented elsewhere: <a href="http://imechanica.org/node/15329#comment-25335">recipe to make new tensors</a>. We essentially stop at 2.
</p>
<p>
Please also see the discussion above among Arash, Amit, and me.
</p>
<p>
What do you think?
</p>
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</ul>Mon, 30 Dec 2013 14:51:00 +0000Zhigang Suocomment 25351 at http://imechanica.orgA non-dimensional linear space as a canonical choice?
http://imechanica.org/comment/25350#comment-25350
<a id="comment-25350"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25334#comment-25334">why conventional mechanics gets away with it?</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 and Arash:
</p>
<p>
Here is a thought that I have not completely checked through for logical consistency:
</p>
<p>
One takes a linear space, of say, dimension 3, that contains objects that have no physical dimensions.
</p>
<p>
Suppose we now take a fundamental set of physical units. Then we nondimensionalize each of the elements of each of the dimension 3 physical vector spaces of interest - space of position vectors, velocities, area vectors, tractions, etc. - appropriately. For each of these physical vector spaces, this act of nondimensionalization induces a mapping between them and the non-dimensional vector space (say, DND mapping).
</p>
<p>
We then do linear algebra and mechanics on the non-dimensional space, defining tensors etc. based on it. For interpretation of results, we go back, using the inverse of the appropriate DND mapping to the relevant vector space.
</p>
<p>
The DND mappings may be isomprhisms - I have not checked carefully. But even if they are not, it seems like this is the right physical choice. Moreover, there should be an invariance of implied mechanics over all possible choice of fundamental sets of physical units chosen to induce the DND mappings, simply by the invariance of the laws of mechanics over choices of fundamental units.
</p>
<p>
May be this is what is implied in mechanics when we lump position vectors, velocities, tractions etc. in the same vector space.
</p>
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</ul>Mon, 30 Dec 2013 13:48:08 +0000Amit Acharyacomment 25350 at http://imechanica.orgLinear Algebra Done Right-Sheldon Axler, Springer
http://imechanica.org/comment/25347#comment-25347
<a id="comment-25347"></a>
<p><em>In reply to <a href="http://imechanica.org/node/15843">Textbook on linear algebra</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>
<font size="3">Dear Prof. Suo, </font>
</p>
<p>
<font size="3">As the best of my knowledge, this book has a wide audience for beginners from math and physics departments in our university. </font>
</p>
<p>
<font size="3">Most linear algebra textbook written for engineering students may neglect the mathematical aspects of linear space. Although the book "Linear Algebra Done Right" is only an introductory book, it covers and describes all the basic knowledge of finite vector space completely with abract formulation and concrete examples.</font>
</p>
<p>
<font size="3">The elegant arrangement of the book, namely, abract formulation, proof in conjunction with concrete examples make it accessible even to students with only knowledge of basic caculus. The electronic copy could be downloaded by the following link.</font>
</p>
<p>
<font size="2"> <a href="http://ishare.iask.sina.com.cn/f/6443278.html?sudaref=www.baidu.com&retcode=0">http://ishare.iask.sina.com.cn/f/6443278.html?sudaref=www.baidu.com&retcode=0</a></font>
</p>
<p>
<font size="3"> Sincerely </font>
</p>
<p>
</p>
<p>
</p>
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</ul>Mon, 30 Dec 2013 11:44:51 +0000Jing_Yangcomment 25347 at http://imechanica.orgwhy map?
http://imechanica.org/comment/25348#comment-25348
<a id="comment-25348"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25297#comment-25297">linear map and tensor</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>
Map is an action. Nobody has to to define tensor as an action. It can be defined as an object - analogously to vector. No maps no problems :)
</p>
<p>
Kosta
</p>
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</ul>Mon, 30 Dec 2013 10:50:00 +0000Konstantin Volokhcomment 25348 at http://imechanica.orgRe: Some thoughts on tensors
http://imechanica.org/comment/25346#comment-25346
<a id="comment-25346"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25344#comment-25344">Re: Some thoughts on tensors</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 Arash: A few remarks on your comments:
</p>
<p>
1. "We think of both the position vector and velocity as vectors living in<br />
the same linear space. But with the understanding that we cannot add a<br />
position vector and a velocity vector as they have different physical<br />
dimensions."
</p>
<p>
This is the crux of the problem, as it sounds like a contradiction in terms. Ideally, we would like to be able to say that if two objects live in the same linear space then they absolutely should be'addable' and if not, then they do not belong to the same linear space.
</p>
<p>
2. "If this particle moves in a curved space then position vector is not even defined."
</p>
<p>
I get the point you are making and the context, but just to play devil's advocate, as long there is a good, physical embedding space, it all works out just fine. E.g a 2-d shell in 3-d space - of course, we do very good mechanics using position vectors on the shell.
</p>
<p>
3. "Given two linear spaces V and W an isomorphism is a one-to-one and onto<br />
linear map. If you specify the map on a basis for V then the map is<br />
uniquely defined. The two vector spaces being isomorphic doesn't mean<br />
you can add a vector in V to a vector in W. In this sense, identifying<br />
the linear space of forces with that of velocities doesn't mean you can<br />
add the two."
</p>
<p>
For any linear transformation (on a finite dimensional space at least), whether an isomorphism or not, knowing its action on a basis completely determines it, obviously. However, this property of all linear transformations (on fd spaces) obviously does not uniquely pick any one of them.
</p>
<p>
The goal in utilizing an isomorphism would not be to add vectors from the different spaces. The goal would be to work with only one space and hopefully be able to do mechanics relevant to the other through this space, utilizing the identification afforded by the isomorphism.
</p>
<p>
4. "However, again, I agree with Zhigang that for mechanics applications one<br />
should work with multilinear maps on a collection of linear spaces that<br />
are not necessarily identical. The obvious examples are deformation<br />
gradient and the first Piola-Kirchhoff stress."
</p>
<p>
Since this is just an opinion, let me share mine from practical experience. I have followed beautiful, nonlinear shell theory (of the Antman, Fox-Simo, etc. variety), done some of that theory, and also numerically implemented some without ever thinking of multilinear maps. Anyone who as done shell theory will immediately realize that you have to deal with the deformation gradient on a shell by considering the tangent spaces at corresponding points on reference and target shell as fundamentally different, but I do not see this as any reason for making the multilinear map point of view as essential for mechanics application......It is what it is - just a point of view.
</p>
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</ul>Mon, 30 Dec 2013 06:43:01 +0000Amit Acharyacomment 25346 at http://imechanica.orgRe: Some thoughts on tensors
http://imechanica.org/comment/25344#comment-25344
<a id="comment-25344"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/25340#comment-25340">Re: why conventional mechanics gets away with it?</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 Amit:</p>
<p>Thanks for this interesting discussion. Just some thoughts:</p>
<p>Let us consider a very simple example. A particle moving in the Euclidean three space. Position vector <strong>x</strong>(t) at any instant of time t can be written with respect to the canonical Cartesian basis vectors {<strong>i</strong>,<strong>j</strong>,<strong>k</strong>}. If you now differentiate <strong>x</strong>(t) with respect to time you get the velocity <strong>v</strong>(t) that has components with respect to the same basis {<strong>i</strong>,<strong>j</strong>,<strong>k</strong>}. We think of both the position vector and velocity as vectors living in the same linear space. But with the understanding that we cannot add a position vector and a velocity vector as they have different physical dimensions.</p>
<p>Note also that when curvilinear coordinates are used the natural bases would have different physical dimensions and consequently components of a vector may not have the right physical dimensions, e.g. in spherical coordinates. Here, one would have to use the so-called physical components.</p>
<p>If this particle moves in a curved space then position vector is not even defined. However, velocity is defined and at any given time t it lies in the tangent space to the ambient space at the point <strong>x</strong>(t). So, in this case one cannot even pose the question of adding "position" and velocity vectors.</p>
<p>Let us call the linear space of velocities at a point <strong>x</strong>, V. Using rate of change of energy or power, velocity and force are dual to each other (similarly force and displacement are dual to each other as Zhigang mentioned). Forces live in the dual space of V, i.e. V*. This is the space of one-forms or covectors. Choosing a basis for V, one has a natural dual basis for V*. These, in general, have different physical dimensions. There is a natural pairing between elements of V and V* and this natural pairing does not require an inner product (or metric). An inner product (metric) induces an isomorphism between V and V*, i.e. one can identify the two linear spaces. Note that, in general, this isometry changes from point to point. </p>
<p>Given two linear spaces V and W an isomorphism is a one-to-one and onto linear map. If you specify the map on a basis for V then the map is uniquely defined. The two vector spaces being isomorphic doesn't mean you can add a vector in V to a vector in W. In this sense, identifying the linear space of forces with that of velocities doesn't mean you can add the two.</p>
<p>I think Cauchy stress is a tensor in the sense of tensors in the notes of Prof. Carlson using the isomorphy between vectors and covectors. One doesn't even need to worry about identifying the linear spaces of forces and velocities. Cauchy stress can be simply represented using a basis of covectors (and is obviously symmetric). However, again, I agree with Zhigang that for mechanics applications one should work with multilinear maps on a collection of linear spaces that are not necessarily identical. The obvious examples are deformation gradient and the first Piola-Kirchhoff stress.</p>
<p>Regards,<br />
Arash</p>
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</ul>Mon, 30 Dec 2013 05:00:41 +0000Arash_Yavaricomment 25344 at http://imechanica.orghow to create new tensors
http://imechanica.org/comment/25342#comment-25342
<a id="comment-25342"></a>
<p><em>In reply to <a href="http://imechanica.org/node/15843">Textbook on linear algebra</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 added a comment titled "<a href="http://imechanica.org/node/15329#comment-25335">recipe to make new tensors</a>". I suppose this recipe is how mechanicians routinely "discover" or "create" new tensors.</p>
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</ul>Sun, 29 Dec 2013 23:35:42 +0000Zhigang Suocomment 25342 at http://imechanica.org