iMechanica - Comments for "On the nature of the Cauchy stress tensor"
http://imechanica.org/node/4356
Comments for "On the nature of the Cauchy stress tensor"enCovariant, contravariant and mixed tensors as such
http://imechanica.org/comment/18756#comment-18756
<a id="comment-18756"></a>
<p><em>In reply to <a href="http://imechanica.org/node/4356">On the nature of the Cauchy stress 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>One can look at the book By A I Borisenko "Vector and Tensor Analysis with Applications" page 91 within section "Covariant, contravariant and mixed tensors as such". Here it is said that in a physical problem, it is most natural that one starts with a particular set of components, say covariant components, then the tensor itself is said to be covariant, but the metrics can always be used to deduce all of its contravariant and mixed components. This should be applicable for a Riemannian manifold where a metric is well defined.</p>
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</ul>Wed, 21 Mar 2012 10:38:59 +0000chandrashekhar mukeshcomment 18756 at http://imechanica.org"Non-Riemannian" manifolds
http://imechanica.org/comment/9229#comment-9229
<a id="comment-9229"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/9218#comment-9218">Re: Non-Riemannian manifolds</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 Biswajit:</p>
<p>You're right about Riemannian manifolds. First you start with a differentiable manifold and then endow it with a metric tensor (that is positive-definite by definition) and that smoothly depends on position. "Non-Riemannian" is vague but it usually does not correspond to non-differentiability. </p>
<p>Having a general smooth manifold, there is no way to differentiate tensor fields intrinsically. For example, the partial derivative of a tensor is not a tensor, in general. One needs to add more structure by, for example, linearly connecting different tangent spaces. Doing this means that you introduce an "affine connection". This will allow you to define parallel transport of tensor fields and covariant differentiation of tensor fields.</p>
<p>Something amazing happens for Riemannian manifolds. Having a metric, there is a unique connection that is symmetric and its induced parallel transport preserves inner products (Fundamental Theorem of Riemannian Geometry). This unique connection is called the Levi-Civita connection.</p>
<p>Usually by "non-Riemannian" manifolds people mean a manifold with a non-symmetric affine connection, i.e. with non-vanishing torsion tensor (sometimes you see the name Riemann-Cartan manifolds). There are extensions of relativity to space-times with torsion. In the mechanics literature, "torsion" of a "material manifold" is believed to be related to the dislocation density tensor.</p>
<p>There are well-known examples of non-differentiable manifolds that have been of interest, namely simplicial complexes or the more general CW complexes. However, you would not define a tangent space there. There have been many attempts in defining discrete Riemannian manifolds. In the relativity literature, there is the so-called Regge Calculus that attempts to extend the usual Riemanian geometry concepts/quantities to simlicial complexes.</p>
<p>Regards,<br />
Arash</p>
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</ul>Tue, 25 Nov 2008 18:52:20 +0000Arash_Yavaricomment 9229 at http://imechanica.orgFoundations of Mechanics
http://imechanica.org/comment/9228#comment-9228
<a id="comment-9228"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/9226#comment-9226">Foundations of 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>Yes, I've seen this book. There is a recent second edition too.</p>
<p>If you read the reviews on <em>Mathematical Foundations of Elasticity</em>, one reviewer says "Turning simple problem into nightmare. How difficult can an elasticity problem be in engineering? But these guys just have a way to make 1+1=2 looks like the most mysterious problem mankind has ever come across. No wonder everyone hates engineering and physics nowadays." I don't think this book is the reason why many people avoid engineering these days. If you're a beginner in elasticity this may not be the right place to start but still it would be a good idea to know there are other ways of doing elasticity. It's like a new language, it can be very hard at the beginning but can open a new world of opportunities.</p>
<p>Arash</p>
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</ul>Tue, 25 Nov 2008 18:24:24 +0000Arash_Yavaricomment 9228 at http://imechanica.orgFoundations of Mechanics
http://imechanica.org/comment/9226#comment-9226
<a id="comment-9226"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/9225#comment-9225">vectors 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>
Kosta,
</p>
<p>
There's this book "Foundations of Mechanics" (FoM) by Ralph Abraham and J. E. Marsden can be found here for free:
</p>
<p>
<a href="http://caltechbook.library.caltech.edu/103/">http://caltechbook.library.caltech.edu/103/</a>
</p>
<p>
It describes tensors and differential forms in the context of manifolds. It uses language and notation very similar to that in the Marsden & Hughes elasticity book (some parts are taken from FoM). FoM, however, is a pretty difficult book - a reviewer on Amazon compares this book to "killing a cockroach with a bazooka". Arash must know about this book.
</p>
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</ul>Tue, 25 Nov 2008 13:45:55 +0000kwlimcomment 9226 at http://imechanica.orgvectors and tensors
http://imechanica.org/comment/9225#comment-9225
<a id="comment-9225"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/9214#comment-9214">covariancy and contravariancy in a more general context</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 Malik,
</p>
<p>
I did not find Marsden&Hughes on the Internet for free. However, you can go to the webpage of <a href="http://repositories.tamu.edu/handle/1969.1/2500">Professor Bowen</a> where you can download a comprehensive treatise on vectors and tensors for free.
</p>
<p>
Best regards, Kosta
</p>
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</ul>Tue, 25 Nov 2008 08:49:00 +0000Konstantin Volokhcomment 9225 at http://imechanica.orgre
http://imechanica.org/comment/9223#comment-9223
<a id="comment-9223"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/9213#comment-9213">Contravariant and Covariant COMPONENTS of 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 Prof. Nemat-Nasser: I agree with what you say only for continua deforming in the Euclidean space where a Cartesian structure can be enforced. This is not always the case. The intermediate configurations in multiplicative theories of plasticity, for example, are not in the Euclidean space. Another example of physical theories in Non-Euclidean space is the general relativity, of course. <span>If the space is </span>Non-Euclidean then the co- contra- and mixed- variant quantities designate different objects rather than different representations of the same object. Kosta
</p>
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</ul>Tue, 25 Nov 2008 06:27:00 +0000Konstantin Volokhcomment 9223 at http://imechanica.orgStress == (1,1) tensor
http://imechanica.org/comment/9220#comment-9220
<a id="comment-9220"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/9215#comment-9215">covariant vs contravariant 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>
Let me second Sia's remark that stress is neither covariant nor contravariant. Coordinates are simple artifices that we introduce for convenience.
</p>
<p>
I think that Arash's comment is a much better way of thinking about stress; i.e. one should think of stress in an operational sense. In this regard, I like to think of stress as a (1,1) tensor. Why? Operationally stress takes in a surface normal (which geometrically is a one-form) and returns the force per unit area to the surface. Force itself operationally takes in a velocity and returns a scalar, power. Thus stress can be nicely thought of as acting on a surface normal (a one-form) and a velocity (a tangent-vector) and returning a scalar the power. Thus from an operational perspective stress is nicely thought of as (1,1) tensor. Note that how you think of stress depends upon your operational definition. For example, in Marsden and Hughes <em>Mathematical Foundations of Elasticity</em>, they state that stress should be a (2,0) tensor (as does Schutz). However this depends upon how once wishes to treat forces -- as one-forms or tangent-vectors. I like to think about the possibility of integrating traction on a surface and thus it makes more sense to me to think of traction as a one-form. Burke for example agrees with treating forces as one-forms.
</p>
<p>
For further reading on such concepts, I like the book of Bernard Schutz <em>Geometrical methods of mathematical physics </em>and the book of William L. Burke <em>Applied Differential Geometry.</em>
</p>
<p>
</p>
<p>
Prof. Dr. Sanjay Govindjee<br />
University of California, Berkeley
</p>
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</ul>Mon, 24 Nov 2008 23:02:49 +0000Sanjay Govindjeecomment 9220 at http://imechanica.orgRe: non Euclidean spaces
http://imechanica.org/comment/9219#comment-9219
<a id="comment-9219"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/9216#comment-9216">non-Euclidian spaces</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>
Arash says: " Can someone give an example which the conditions of problem, justifies working with non-euclidian spaces?"
</p>
<p>
Shells, for instance. As an example you can see
</p>
<p>
Simo and Fox, " <strong>On stress resultant geometrically exact shell model. Part I: formulation and optimal parametrization", </strong><span class="mediumb-text">Computer Methods in Applied Mechanics and Engineering, </span><a href="http://portal.acm.org/toc.cfm?id=J148&type=periodical&coll=GUIDE&dl=GUIDE&CFID=12041540&CFTOKEN=13796652" target="_self" class="small-link-text"></a><span class="small-text">Volume 72 , Issue 3 (March 1989), </span><br />
Pages: 267 - 304.
</p>
<p>
-- Biswajit
</p>
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</ul>Mon, 24 Nov 2008 21:57:20 +0000Biswajit Banerjeecomment 9219 at http://imechanica.orgRe: Non-Riemannian manifolds
http://imechanica.org/comment/9218#comment-9218
<a id="comment-9218"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/9214#comment-9214">covariancy and contravariancy in a more general context</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 believe Riemannian manifolds are structures consisting of a differentiable manifold with a positive definite metric tensor. Non-Riemannian manifolds may either be non-differentiable or have either no metric tensor or a non pos-def metric. Intutively, I don't think it's very fruitful to think of physical tensor quantities in the a general non-Riemannian context. Could someone provide me with an example that counters my intuition?
</p>
<p>
Also, can a tangent space be defined uniquely at a point on a non-differentiable manifold?
</p>
<p>
-- Biswajit
</p>
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</ul>Mon, 24 Nov 2008 21:48:49 +0000Biswajit Banerjeecomment 9218 at http://imechanica.orgnon-Euclidian spaces
http://imechanica.org/comment/9216#comment-9216
<a id="comment-9216"></a>
<p><em>In reply to <a href="http://imechanica.org/node/4356">On the nature of the Cauchy stress 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 all,
</p>
<p>
i'm new to tensors. Can someone give an example which the conditions of problem, justifies working with non-euclidian spaces?
</p>
<p>
Thanks.
</p>
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</ul>Mon, 24 Nov 2008 20:44:00 +0000Arash Erfaniancomment 9216 at http://imechanica.orgcovariant vs contravariant tensors
http://imechanica.org/comment/9215#comment-9215
<a id="comment-9215"></a>
<p><em>In reply to <a href="http://imechanica.org/node/4356">On the nature of the Cauchy stress 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 Sia,</p>
<p>Thanks for the notes and your comment. What you pointed out is perfectly fine when you're working in Euclidean spaces. of course you can use Cartesian or any non-orthogonal curvilinear coordinates even for an intrinsically flat space like an Euclidean space. </p>
<p>One important thing to note is that tensors are multi-linear maps that take a bunch of vectors and co-vectors and give you a real number. Given a linear space V, its dual V* (space of all linear functionals on V) is not "equal" to V. So, a map from V to V is not "equal" to a map from V* to V* or a map from V to V*. This can all be semantics if you're working in Euclidean spaces but not in general manifolds.</p>
<p>This can be explained more clearly in terms of vectors (contravariant vectors) and co-vectors (covariant vectors or differential forms). Given a vector in a linear space, the corresponding dual vectors or forms can always be defined even if you don't have any metric structure. One interesting distinction between vectors and co-vectors (1-forms) is the fact that you cannot integrate a vector field intrinsically on a general manifold but differential forms can always be integrated intrinsically. </p>
<p>However, as soon as you endow you manifold with a metric then there is a way to identify a tangent space with its corresponding cotangent space and this is what happens in Riemannian manifolds.</p>
<p>To be precise, given a manifold M, a tensor S of type (m,n) at a point X is a multilinear map from Cartesian product of m copies of T*M (cotangent space at X) and n copies of TM (tangent space at X) to real numbers. In this case, S is contravariant of rank m and covariant of rank n. This definition does not depend on the local chart (coordinate system) chosen. </p>
<p>What I think can be a rigorous interpretation of what you mentioned in your comment is that Cauchy stress is a physical quantity that can be described by different associated tensors.</p>
<p>Regards,<br />
Arash</p>
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</ul>Mon, 24 Nov 2008 16:46:02 +0000Arash_Yavaricomment 9215 at http://imechanica.orgcovariancy and contravariancy in a more general context
http://imechanica.org/comment/9214#comment-9214
<a id="comment-9214"></a>
<p><em>In reply to <a href="http://imechanica.org/node/4356">On the nature of the Cauchy stress 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>
Professor Volokh,
</p>
<p>
I got your point. I believe what you mean is that there might be some cases where co- and contravariant tensors are different regarding the structure in continuum. Your point about the "multiplicative decomposition of the deformation gradient" reminds us of some configutations in Continuum Mechanics that cannot be assigned a cartesian structure.
</p>
<p>
However since i have not read the book that your cited in your comment i would like you to give us some explanations about the case that you mentioned. I mean in what sense are the co- and contra-variant different ?
</p>
<p>
</p>
<p>
Indeed Pfr. Nasser makes the point that we speak of covariancy and contravariancy only for the components of a tensor. I believe he is considering Riemanian manifolds where a metric tensor can be defined (i am refering to Pfr. Arash Yavari's comment in the previous discussion).
</p>
<p>
</p>
<p>
I am a little bit confused but i believe that these two notions might be different in the case of complicated manifolds.
</p>
<p>
</p>
<p>
In the case of non-riemanian manifolds is it correct and does it make sense to say that covariant and contravariant tensors are different?
</p>
<p>
</p>
<p>
Regards,
</p>
<p>
</p>
<p>Malik </p>
<p>
</p>
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</ul>Mon, 24 Nov 2008 16:36:12 +0000Malik Ait-Bachircomment 9214 at http://imechanica.orgContravariant and Covariant COMPONENTS of Tensors
http://imechanica.org/comment/9213#comment-9213
<a id="comment-9213"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/9212#comment-9212">contra- and co-variant tensors can be different</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>Tensors are neither COVARIANT nor CONTRAVARIANT. These terms refer to the COMPONENTS of a tensor. With respect to a general curvilinear coordinate system, the matrices of these components are generally not the same. Thus, we should not refer to tensors as either covariant or contravariant. These terms are for their components once a specific coordinate system is chosen. A tensor, e.g., a vector, in general, has a physical significant independently of any coordinate system that one may choose to represent its covariant or contravariant components. SNN</p>
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</ul>Mon, 24 Nov 2008 14:59:00 +0000Sia Nemat-Nassercomment 9213 at http://imechanica.orgcontra- and co-variant tensors can be different
http://imechanica.org/comment/9212#comment-9212
<a id="comment-9212"></a>
<p><em>In reply to <a href="http://imechanica.org/node/4356">On the nature of the Cauchy stress 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>Putting aside the Cauchy stress specifically, I should notice that the contra- and co-variant tensors can be different if there is no Cartesian structure in continuum (Marsden and Hughes: “Mathematical foundations of elasticity”). In the classical continuum mechanics we almost always have the Cartesian structure except for the theories based on the multiplicative decomposition of the deformation gradient which imply the existence of incompatible intermediate configurations. Such configurations are not physical continua where a global Cartesian structure can be introduced.</p>
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</ul>Mon, 24 Nov 2008 13:17:33 +0000Konstantin Volokhcomment 9212 at http://imechanica.orgTensor algebra
http://imechanica.org/comment/9209#comment-9209
<a id="comment-9209"></a>
<p><em>In reply to <a href="http://imechanica.org/node/4356">On the nature of the Cauchy stress 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>
Thanks Pfr Nasser,
</p>
<p>
Indeed the notion on covariancy and contravariancy of tensors depend on the representation that you choose.
</p>
<p>
Malik
</p>
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</ul>Mon, 24 Nov 2008 08:46:02 +0000Malik Ait-Bachircomment 9209 at http://imechanica.orgTensors & their components
http://imechanica.org/comment/9205#comment-9205
<a id="comment-9205"></a>
<p><em>In reply to <a href="http://imechanica.org/comment/9204#comment-9204">Regarding 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>Yes, a vector is a first-order tensor. Any tensor can be represented by its covariant, contravariant, or mixed components with respect to any suitable coordinate system. Please read the sections cited in the referense at the bottom of my original comments. SNN</p>
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</ul>Mon, 24 Nov 2008 02:17:23 +0000Sia Nemat-Nassercomment 9205 at http://imechanica.orgRegarding tensors
http://imechanica.org/comment/9204#comment-9204
<a id="comment-9204"></a>
<p><em>In reply to <a href="http://imechanica.org/node/4356">On the nature of the Cauchy stress 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>
Hi Dr Nasser,
</p>
<p>
thanks for the clarification. so it all depends on what kind of basis vectors are selected (covariant or contravariant basis). then every tensor,regardless of it's type (Velocity vector, couchy stree tensor,etc ) can be expressed and transformed covariantly or contravariantly.correct?
</p>
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</ul>Sun, 23 Nov 2008 19:43:57 +0000Arash Erfaniancomment 9204 at http://imechanica.org