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On the nature of the Cauchy stress tensor

Sia Nemat-Nasser's picture

Checking the iMechanica web, I notice some discussion about the Cauchy stress tensor, whether it is covariant or contravariant.  Well, a tensor is neither covariant nor contravariant, while it can be expressed by its covariant, contravariant, or mixed *components* with respect to any arbitrary coordinate system. See the attached short explanation.

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Hi Dr Nasser,

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?

Sia Nemat-Nasser's picture

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

Thanks Pfr Nasser,

Indeed the notion on covariancy and contravariancy of tensors depend on the representation  that you choose.

Malik

Konstantin Volokh's picture

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.

Sia Nemat-Nasser's picture

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

Konstantin Volokh's picture

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. If the space is Non-Euclidean then the co- contra- and mixed- variant quantities designate different objects rather than different representations of the same object. Kosta

Professor Volokh,

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.

 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 ?

 

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). 

 

I am a little bit confused but i believe that these two notions might be different in the case of complicated manifolds.

 

In the case of non-riemanian manifolds is it correct and does it make sense to say that covariant and contravariant tensors are different? 

 

Regards,

 

Malik

 

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?

Also, can a tangent space be defined uniquely at a point on a non-differentiable manifold?

-- Biswajit 

Arash_Yavari's picture

Dear Biswajit:

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.

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.

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.

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.

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.

Regards,
Arash

Konstantin Volokh's picture

Dear Malik,

I did not find Marsden&Hughes on the Internet for free. However, you can go to the webpage of Professor Bowen where you can download a comprehensive treatise on vectors and tensors for free.

Best regards, Kosta

Kosta,

There's this book "Foundations of Mechanics" (FoM) by Ralph Abraham and J. E. Marsden can be found here for free:

http://caltechbook.library.caltech.edu/103/

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. 

Arash_Yavari's picture

Yes, I've seen this book. There is a recent second edition too.

If you read the reviews on Mathematical Foundations of Elasticity, 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.

Arash

Arash_Yavari's picture

Dear Sia,

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.

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.

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.

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.

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.

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.

Regards,
Arash

Let me second Sia's remark that stress is neither covariant nor contravariant.  Coordinates are simple artifices that we introduce for convenience.

 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 Mathematical Foundations of Elasticity, 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.

 For further reading on such concepts, I like the book of Bernard Schutz Geometrical methods of mathematical physics and the book of William L. Burke Applied Differential Geometry.

 

Prof. Dr. Sanjay Govindjee
University of California, Berkeley

Dear all,

i'm new to tensors. Can someone give an example which the conditions of problem, justifies working with non-euclidian spaces?

Thanks.

Arash says: " Can someone give an example which the conditions of problem, justifies working with non-euclidian spaces?"

Shells, for instance.  As an example you can see

Simo and Fox, " On stress resultant geometrically exact shell model. Part I: formulation and optimal parametrization", Computer Methods in Applied Mechanics and Engineering, Volume 72 ,  Issue 3  (March 1989),
Pages: 267 - 304.

-- Biswajit 

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.

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