## You are here

# FRAME INDIFFERENCE

## Primary tabs

The attached notes are written for a course on plasticity. When I update the notes, I will post a link on my twitter account: https://twitter.com/zhigangsuo.

**Rheological behavior is independent of rigid-body motion**. The bumpy airplane makes us dizzy. We sense the acceleration of the airplane relative to the ground. By contrast, the rheological behavior of materials seems to be independent of rigid-body translation and rotation. The elastic modulus of the wing remains the same when the airplane rolls, yaws, and pitches. So does the viscosity of gasoline.

Here is the fundamental hypothesis: the rheological behavior of materials is unaffected by rigid-body motion of all kinds. We focus on the consequences of the hypothesis. We construct variables invariant with respect to rigid-body motion. Later we will use these variables to construct rheological models invariant with respect to rigid-body motion.

**Frame-indifference**. In the Euclidean space, a frame of reference consists of an origin and three base vectors. We can attach a frame to a spot on the ground, and attach another frame to a spot in the airplane. The two frames are rigid, but translate and rotate relative to each other. A variable independent of rigid-body motion is independent of the choice of the frame, and vice versa. Such a variable is known as a frame-indifferent variable.

**Separation is frame-indifferent**. The separation between two places in the Euclidean space is a frame-indifferent vector. The separation is the mother of all frame-indifferent variables. They have different fathers—time, energy, entropy, electric charge, as well as quantity of atoms, molecules and colloids of every species. They are scalars, and are frame-indifferent. We will watch the separation and the scalars breed other frame-indifferent variables.

**Relative velocity is frame-sensitive**. Even though the separation is frame-indifferent, its rate—the relative velocity—is frame-sensitive. In general, the rate of a frame-indifferent variable is frame-sensitive. This makes us question what we mean by rate, and generalize its definition to allow frame-indifferent rates.

Attachment | Size |
---|---|

frame indifference 2014 12 02.pdf | 5.31 MB |

- Zhigang Suo's blog
- Log in or register to post comments
- 24199 reads

## Comments

## Re: Frame Indifference and invariance under SRBM

Dear Zhigang - Thanks for pointing out this post. I have been meaning to find the time to read your notes in detail before responding, but you know how it goes…. I have looked at things very quickly though. Your caution about not confusing tensors with their components is something very worth paying attention too.

One remark in this regard - where you discuss the equation x’ = Qx + c – some may argue that invariance under superposed rigid body motions (Green and Rivlin) is a conceptually much easier way to think about the consequences of ‘frame-indifference’. Within that framework, x’ = Qx + c is a completely valid tensor equation, without any reference to components w.r.t. any basis. In fact, in all my work I prefer invariance under superposed rigid body motions as the basic postulate – and in my work I have done a fair bit of all this and it comes out right.

A second comment is that on page 28, where you talk about insignificance of objective rates, you are correct, but it is probably very important to emphasize that the function F cannot be kept fixed as the rates change – some, especially students, may interpret the ‘insignificance’ as saying that your corotational rate could be exchanged by any other objective rate in the statement without changing F. In fact, the whole controversy about objective rates for describing hypoelasticity can be ascribed to this desire to keep the function F fixed and corresponding to that finding the optimal objective rate.

A good example of this issue arises when one realizes that objective rates actually build in a nonlinearity through the coupling of the tensor of which a rate is being described with some function of the velocity gradient. As such, then it can have consequences with respect to stability of the evolution of the tensor, driven by the material deformation. Thus, for a given objective rate, the function F may be 0 in some circumstances. If one now changes the objective rate, this means that the corresponding function on the rhs should not be zero any further, transforming very precisely to keep the physical content of the first statement intact. That the choice of the objective rate can have an effect in predictions of bifurcation is mentioned in Rice and Rudnicki’s (1975) paper and Rice and Asaro’s (1977) JMPS papers, if memory serves me correctly.

A recent example came up in our work related to dislocation nucleation, pages 11- 14 of first paper at the following link – to appear in JMPS.

http://faculty.ce.cmu.edu/acharya/publications/

(actually this paper was also posted as a 2 part series on imechanica -

here - http://imechanica.org/node/16949

and here - http://imechanica.org/node/16950

since nowadays one cannot post a paper larger than 1Mb size and I had asked Akanksha to post a link to the whole paper but she probably did not find the time to get to it)

Also, in what you have written down, it is imprtant that the F function on the rhs of the corotational rate be an isotropic function of its arguments – that needs to be mentioned.

## Re: Frame Indifference

Dear Zhigang:

Thanks for sharing your notes. I agree that one should not confuse a vector (and, in general, any tensor) with its components. I also agree with you that traction is not a vector and that “traction vector” is an incorrect term. Another widely accepted incorrect term is “deformation gradient”, which is not a gradient as it does not depend on any metric; “deformation gradient” is simply the derivative of the deformation map.

I disagree with your argument started on page 8 that there is some inconsistency in some treatments of frame indifference in the literature (I agree with Amit’s comment). This issue has been briefly discussed in the book by Marsden and Hughes in a more general context where you are formulating mechanics in a curved ambient space. There you would work with arbitrary smooth changes of coordinates. They explain that one can formulate covariance (frame indifference in non-Euclidean spaces) either actively or passively. What you are describing is the passive point of view and Amit’s approach is the active one. In the active point of view (as Amit mentioned) you start with a motion and then consider arbitrary superimposed rigid-body motions. There is no inconsistency and these two points of view are equivalent. I personally prefer the active point of view and that, for example, energy is invariant under superimposed rigid-body motions.

Regarding objective rates, as I discussed in the past in iMechanica all the well known objective rates are different components of the same tensor, which is a Lie derivative.

Regards,

Arash

## Objective Rates

Hi Arash - A couple of comments on your remarks on objective rates. I am not sure you really wanted to say what you did - "all the well-known objective rates are different components of the same tensor, which is a Lie derivative." This gives me the impression that you are saying all objective rates are actually the same tensor, differing only through a component representation. As I see it, the various objective rates of the same tensor, say Cauchy stress, are different tensors, being expressible with absolutely no reference to components or bases at all. It is true that the components of each objective rate of a particular tensor, with respect to very specifically convecting bases in time, may also be expressed as the time rate of change of the components of that tensor on the specific convecting basis chosen to define the objective rate. I.e. suppose e_i (t) = A(t) e_i^0, then OT = { d/dt ( e_i . T e_j) } e_i (0X) e_j , where e_i^0 is a fixed basis in time, A is an invertible tensor valued function of time, (OX) is the tensor product and OT is the objective rate of T (and I am considering T to be a tensor with both legs on the current configuration). Of course this seems magical that the right hand side should be a tensor expressible independently of any basis whatsoever, being dependent on T, A, Tdot, Adot, only, but this can be shown.

I guess a simple way of stating my main point is: Jaumann(T) = Tdot - WT + TW and upperconvected(T) = Tdot - LT -T L^t, where W is material spin and L is velocity gradient. Juamman(T) and upperconvected(T) are not the same tensors.

Now to the other point. Not all objective rates, even the well-known ones, are Lie derivatives - I've made this point many times, so once more cannot hurt. For Lie derivatives, the tensor A in the above has to be the deformation gradient (you can adjust the terminology to your liking, as you prefer) at a material point of some motion of a body. If you take the Jaumann rate for example, the Rotation tensor field arising from the material spin field at any instant need not correspond to the deformation gradient of any motion of the body - in fact the field of rotation tensors, if non-uniform on the body, cannot correspond, as a deformation gradient, to any deformation - this is a question of compatibility.

## Re: Objective Rates

Dear Amit:

Thanks for your comment. I should have said that all the well known objective stress rates are somewhat related to the Lie derivative of Cauchy stress with respect to velocity (or are different manifestations of it). Raising and lowering indices don’t commute with Lie derivative and for that reason you don’t get the same tensor; you are correct.

Regards,

Arash

## Re:Re: Frame Indifference

Dear Amit and Arash:

Thank you so much for all the comments. I agree with both of you that the two approaches are equivalent. They just write many formulas differently, and can be a source of confusion. Here I have adopted the two-frame approach. I should have been explicit about the two approaches. I will return to these notes and update them.

## good

i like this thank you

## Semantics

Thanks Zhigang, Arah, and Amit for a great discussion. I think we are wading deep into semantics, so I'd like to argue a few of my own.

While I understand that the set of all traction at a point as defined by Cauchy's theorem does not constitute a vector space, tangents are vectors because they exist in R3 and transform as vectors.

Arash's point about the deformation gradient not being a deformation gradient seems to depend on the definition of the gradient operator. If one were to define the gradient in the generalized setting using the contravariant bases vectors (reciprocal of the tangent bases vectors), then F can still be defined as the gradient of the deformation map.

Finally, for frame-indifference, I introduce both points of view in my class, but like Arash and Amit, very quickly run with the active point of view. It's conceptually much easier for students to intuit how a rigid body motion should not change the stress or strain energy density of an object.

## Re: Semantics

Dear Vicky:

Thanks for your message.

Deformation gradient being a gradient or not is not a matter of semantics. This is not really an important issue if you’re doing everything in R^3 with its standard structure. However, in a more general context a differential 1-form and gradient of a scalar are two very different objects. For example, you can intrinsically integrate a 1-form on an arbitrary curve but integrating a gradient over a curve does not make sense, in general. The relation between these two objects is established using a metric. Derivative (or exterior derivative) of a scalar does not need a metric while gradient explicitly depends on metric. In other words, equipping your space with different metrics you would have different gradients.

Deformation gradient is the derivative of deformation mapping and is always the same object no matter what metric you choose in your reference configuration. In classical continuum mechanics you assume your reference configuration (stress free) is the Euclidean space but this is not always the case and in the presence of residual stresses you may want to work with a non-trivial material metric.

Regards,

Arash

## 1-forms, gradients, and tangent maps

Arash,

Clearly 1-forms and (tangent) vectors are different beasts in general. However I think missing from this discussion is a precise definition of what you firstly mean by gradient and secondly by what you mean by "deformation gradient". Without those definitions one can not make precise statements.

For example, Marsden & Hughes as well as Abrahamson, Marsden & Ratiu consider the gradient of a function f to be the object grad(f) such that <grad(f), v> = df(v) for all vectors v, where < , > is an inner product pairing on the tangent space, and df is the differential 1-form. In this sense grad(f) is a vector. However, this is far from universal. Schutz, for example, defines the gradient of a function f to be the differential 1-form df, as do many other authors. Thus it would be helpful to know what your definition of gradient is. Likewise what precisely do you consider to be the definition of the "deformation gradient".

Knowing these two points is required for a cogent discussion.

-sanjay

## 1-forms, gradients and tangent maps

Dear Sanjay,

According to Schutz the gradient of a scalar function f is a covector df (1-form) when operating on a vector v gives the directional derivative of f in the direction v i.e. <df, v>. But here in the case of continuum mechanics if we assume that Φ(X,t) is a mapping from reference configuration to spatial configuration then F=∂Φ/∂X is a mapping from reference configuration to the tangent space in spatial configuration. In other words a map from the neighborhood of P in reference configuration to the space tangent to p in spatial configuration.

Mohsen

## Re: 1-forms, gradients, and tangent maps

Dear Sanjay:

Thanks for your comment. By gradient of a function I mean exactly what you quoted from those two books. Changing the inner product (metric) you would get different gradients. I also agree that many of these definitions are not universally accepted.

Regarding deformation gradient, I define it exactly how Marsden and Hughes define it: given the coordinate charts {X^A} and {x^a} for the reference and current configurations, F^a_A = \partial x^a / \partial X^A. You can think of deformation gradient as a vector-valued 1-form. What I mean by not being a gradient is the 1-form part (the leg in the reference configuration). This object (deformation gradient) will not be affected by the choice of metric

Gin the reference configuration. So, I think “deformation derivative” would be more appropriate. However, I still go with the accepted term and call it deformation gradient knowing that changing the material metricFwill not be affected.Regards,

Arash

## Dear Arash,

Dear Arash,

Here we are talking about the gradient of a vector field such as Φ(X, t) and not of course the gradient of a scalar function such as f in previous post. These are completely different right? In the latter case we have the directional derivative of f along v, but in the former one we are dealing with [Φ,v] or the covariant derivative of Φ (grad_v Φ) and these are all different from F=∂Φ/∂X. Then the question is how we can connect these together?

Mohsen

## Dear Mohsen:

Dear Mohsen:

Deformation gradient is a two-point tensor. The part in the ambient space is a vector and we’re not worried about its “gradient” here. But you are right that its spatial “gradient” should be defined in terms of a covariant derivative (if needed). To define deformation gradient we are simply taking the derivative of the deformation map. This map acts on a vector in the reference configuration and gives you a vector in the ambient space and in this sense is a vector-valued 1-form. You can make the referential leg (the 1-form part) a vector by using a metric and similar to a scalar define a (referential) gradient (that was the point I was trying to make). The important point is: Deformation gradient does not depend on any metric (referential or spatial) or connection. It is simply the derivative of the deformation map.

Regards,

Arash

## How about tangent map

Arash,

If you go with the Marsden and Hughes definition, then I agree that deformation gradient is not an appropriate bit of terminology.

Fmaps vectors between two tangent spaces and thus needs to be a ${1 \choose 1}$ tensor, which by the Marsden definition of gradient can not possibly be a deformation gradient (given the tensor legs involved). However, if you accept the other common definition, then it works.-sanjay

## I meant to argue in the first

I meant to argue in the first point that traction are vectors. Typos.

"While I understand that the set of all traction at a point as defined by Cauchy's theorem does not constitute a vector space, tractions are vectors because they exist in R3 and transform as vectors. "

Arash, I will have to digest your explanations for a while. Thanks for the discussion.

## Traction is a vector!

Dear Vicky:

I’m glad you posted this comment. I went back and read Zhigang’s notes more carefully. It is true that given let’s say uniform tractions on the faces of a tetrahedron one cannot simply look at the vector sum of tractions; one must multiply each by its corresponding area. On page 18 of his notes Zhigang considers a planar region and defines traction to be the force acting on this surface divided by its area. This is not the correct definition of traction; this is simply the average traction acting on the surface. Given a point one starts with a small region containing this point and shrinking this surface to the point of interest traction is the limit of force divided by area. In a body traction field depends on both position and

N(unit normal vector) that belongs to the unit sphere.To be precise force is a 1-form (when paired with velocity it gives you power, a scalar) but using the Euclidean metric one can consider the corresponding vector. Now fixing a point and a plane passing through it (i.e. fixing the unit normal

N), the set of all tractions at this point is a vector space. Why? Because all properties of a vector space are satisfied. In particular, given two tractions acting at the same point their sum is a traction at that point as well.What is the source of confusion here? We are used to moving vectors from one point to another and adding them. One learns such operation the first time in undergraduate statics in equilibrium of rigid bodies (sum of all the forces acting on a rigid body in static equilibrium must be the zero vector). Note that this has nothing to do with force being a vector or not. Fixing a point on the body the set of all forces acting at that point under scalar multiplication and vector addition is a vector space.

Now for adding two vectors acting at two different points one needs more structure. One may consider parallel transporting one vector along a curve. This operation is, in general, path dependent, and hence not well defined. However, in a flat space (zero curvature) it becomes well defined. This is what we do in classical mechanics; we use the standard parallel transport of Euclidean space and move all the forces to one point and then add them up.

Regarding your comment, “… the set of all traction at a point as defined by Cauchy's theorem does not constitute a vector space…”, 1) Traction is not defined through Cauchy’s theorem; Cauchy’s theorem tells us that the dependence of traction vector on the unit normal

Nis linear, i.e. there exists a second-order tensor (Cauchy stressσ) such thatt=σ.N. 2) As I just mentioned, the set of tractions at a fixed point is indeed a vector space.In summary: 1) I should take back what I said earlier: “traction vector” is a perfectly fine term (but “deformation gradient” is not). 2) What Zhigang mentions in his notes regarding traction is not correct. However, it should be emphasized that one should not think of adding tractions acting at different points and in that sense force and traction are very different.

Regards,

Arash

## Tractions violate the rule of vectors

Dear Vicky and Arash: Thank you very much for your comments. I assume that you are both looking at the draft dated 2014 12 02. Here is the text I wrote on pages 18 and 19:

“Traction. Consider a planar region of area a, normal to a unit vector

n.Acting on the planar region is a forcef. In general, the force has components normal and tangential to the plane. Define the traction t by the force acting on the planar region divided by the area of the region:t=f/a.The body is in a homogeneous state. The traction is independent of the shape and the area of the region, but depends on the direction

nof the region. For instance, for the chewing gum in tension, a plane not normal to the axial direction will have both normal and shear traction.Traction is commonly called a vector. This designation is wrong. Tractions acting on various planar regions form a set. This set, however, is not a vector space. Tractions do not obey the rule of vectors: the sum of tractions acting on two planar regions, in general, does not give traction acting on another planar region.”

Dear Arash: About your comments “This is not the correct definition of traction; this is simply the average traction acting on the surface.” As I wrote, the body is in a homogeneous state. This definition of traction is correct. For a body in a homogeneous state, the traction only depends on the orientation of the planar region, but not on the location of the planar region.

Dear Vicky and Arash: You both seem to object the word “wrong” in the sentence “This designation is wrong”. But you both agree with the following statements: “Tractions acting on various planar regions form a set. This set, however, is not a vector space. Tractions do not obey the rule of vectors: the sum of tractions acting on two planar regions, in general, does not give traction acting on another planar region.” These statements say what I feel is wrong with designating traction as a vector.

In the notes, I went on to define the stress

Tas a function that maps the area vector a of a planar region to the force acting on the planar region:f=T(a).The set of area vectors of all planar regions form a vector space. The balance of forces shows that the set of forces acting on all planar regions constitute another vector space. The notes show that the function T is a linear map from one vector space to the other vector space. Thus, the stress T is a tensor.

Dividing the above equation by the area a, we get

t=T(n).This equation is correct for planar regions of all directions. However, the set of unit vectors normal to various planar regions do not form a vector space. Neither do the set of tractions acting on various planar regions. Thus, the equation

t=T(n) is not a linear map between two vector spaces.By contrast, the equation

f=T(a) is a linear map between two vector spaces. I like this equation. Do you?Perhaps I should simply stick to the fact: the set of tractions acting on all planar regions do not form a vector space. Calling

ta vector will be OK if you agree that it violates the basic rule of vector operation. Indeed, we callna vector, even though the set of unit vectors do not form a vector space.Arash: Of course I agree with your statement “Now fixing a point and a plane passing through it (i.e. fixing the unit normal N), the set of all tractions at this point is a vector space.” But we do not restrict ourselves to pannar region in a fixed direction in using the equation

t=T(n). The equation applies to planar regions of all directions.## Re: Tractions violate the rule of vectors

Dear Zhigang:

I agree that if you assume that the body is in a uniform state of stress then traction would only depend on the unit normal

Nand you can simply divide the force by area. But why do you have to restrict yourself to such a special case? What about an arbitrary body in an arbitrary state of stress? But I think this is not the main issue.“Tractions acting on various planar regions form a set. This set, however, is not a vector space. Tractions do not obey the rule of vectors: the sum of tractions acting on two planar regions, in general, does not give traction acting on another planar region.” Here you first define a set and then argue that it’s not a linear space. Fine. But the set you’re defining has no physical significance and as far as I know no where in the literature there is any mention of traction being a vector in the sense that you are defining it.

Given a point

xand a unit normalN, I think we both agree that traction depends on both, i.e.t=t(x,n). Now can we addt(x,n1) andt(x,n2)? The answer is no and I completely agree with you. It’s completely meaningless to add these two quantities. However, if you fix bothxandn, then the set {t(x,n)} is a vector space. This is what people mean by traction being a vector. Cauchy’s theorem sayst(x,n)=σ(x).nand thatσ(x) is a second-order tensor. You are correct that the set of all unit vectors (the unit sphere) is not a vector field. However, Cauchy stress is a linear map and in the equationt(x,n)=σ(x).n,σ(x) as a map is restricted to the unit sphere.Regards,

Arash

## Re. Tractions

Hi Arash,

You say in your above reply to Zhigang: "if you fix both

xandn, then the set {t(x,n)} is a vector space."I don't quite get it. Suppose that the state of stress is fixed---a particular static (time-independent) stress field is assumed to exist. Fixing

xmeans fixing the point of interest. Fixingnmeans fixing the orientation of the infinitesimal mathematical cut. Now, if the state of stress at this point is fixed, then obviouslyt(x,n) is also fixed. If so, how does it make for an entire space? ... What am I missing? Or do you mean to say that the set has only one element?Best,

--Ajit

[E&OE]

## Thank you, Ajit, for this

Thank you, Ajit, for this clear question. I will use it to teach the ideas.

## Re: Ajit's question

Dear Ajit:

Thanks for your excellent question. Fixing

xandn,t(x,n) is an element of a vector space (3-dimesnional for us). It doesn’t mean the underlying vector space has only one element (also note that traction depends on time as well, in general). Perhaps the following example will make this clearer. Let’s consider motion of a body and the velocity field. We all agree that at every material point velocity is a vector (at a fixed pointxvelocityvis an element of the tangent space at that point, which is a 3-dimensional vector space). Now following Zhigang’s line of argument one may say velocity is not a vector because if you add the velocity of two different material points it doesn’t give you velocity of another material point. That is true but adding velocities of different points is as meaningless as trying to add tractions acting on different directions. What is meaningful in the Euclidean space is adding linear momentum vectors. You consider a small volume element dv multiply it by velocityvand mass density ρ and then adding all these vectors ρvdv (integrating) you have the total linear momentum vector. Now this vector must be equal to the total force acting on the body (balance of linear momentum).Also note that when you consider a superimposed riding body motion (or a change of coordinates as Zhigang has in his notes) traction for fixed

xandntransforms like a vector (perhaps this is what Vicky was referring to).Regards,

Arash

## Reply to Arash about the traction vector

Hi Arash,

... Well... What I meant here is that the very act of talking about the two vectors

xandnitself presumes that there is an embedding vector space spanned by some basis vectors, say the three Cartesian basis vectors, and then, for a given state of stress, the tractiontis a vector because, simply put, itscomponentsobey the cosine law under rotation of the embedding space, even iftitself, as the mathematical object, remains the same (and, of coursse, as you point out, it also obeys the translational invariance).Now, for a given fixed frame defining the embedding space, and for a given stress state, the vector

tis unique, and hence, the set notation isn't called for. That's what I meant. The embedding vector space, under the continuum assumption, does contain an infinity of traction vectors (in different directions at different points) and so the set notation may be used for thatembedding space. But fixing thexandnfixes the value of the vector functiont, and therefore you can't say: "if you fix bothxandn, then the set {t(x,n)} is a vector space." I think there was a small typo in there to what you wrote above. Anyway, the matter is settled now.On another note, sure, a slight clarification is called for in Zhigang's notes to distinguish planar regions (or mathematical cuts) at a certain point fixed in an embedding space vs. those at different points in that same space.

BTW, if adding linear momenta (at a material point) is meaningful, then so should be velocities (at the same material point).

Incidentally, apart from time (which you consider), yet another case should help close the issue completely: You

canadd two different traction vectors---provided bothxandnare fixed. Think of a 1D rod put under tension to, say, 1 MPa axial stress and then adding another 10 MPa tension to it. Now, it's meaningful to say:t(x1,n1) =t1(x1,n1)+t2(x1,n1).BTW, I have much appreciated your clarifications in this thread on the 1-form, gradient, and why deformation gradient is not a vector.

I had just mentioned this part in my undergraduate class on FEM last year, but without being able to offer any explanation on the fly---after all, it was an introductory course on linear FEA, and so, I myself was not prepared about this part. It just so happened that the issue struck me in the flow of the lecture at that time, though I couldn't on the fly recall the exact logic for it. With your clear answers, I think I should be able to do much better---I mean in an UG class---this year. Last year, I had just told them that I had stopped applying \vec{} on every \nabla wherever it appeared in equations, that's all, and then had moved on to the syllabus portion to be finished :). ... Anyway, I would love it if you could spare some time to write a small note/paper (to, say, ASEE) clarifying this matter at a level understandable to the typical engineering UGs...

Best,

--Ajit

[E&OE]

## Reply to Arash, continued...

Hi Arash,

No, my above reply (mainly the last two paragraphs) make it clear that I am still confused about these matters.

Let me own up my confusions, and request you to respond to questions at a new post I created specifically to address the issue of vectors, gradients, deformation gradients, etc., here [^].

[To admins: BTW, due to a posting mistake (faulty 'net connections, page reloads, etc.), I ended up posting this reply at the bottom of the thread, but the confused reply I had in mind is the above one. Sorry for the repetition.]

Best,

--Ajit

[E&OE]

## A state of stress maps one vector space to another.What vectors?

The significance of homogeneous state. I follow the familiar division of labor in continuum mechanics. We wish to study a body undergoing inhomogeneous deformation. We divide the body into many small pieces. Each small piece undergoes a sequence of homogeneous deformation in time. A rheological model answers the following question: given a history of stress, how do we predict the history of strain, or the other way around? Different pieces in the body communicate through the compatibility of deformation, the balance of forces, and the conservation of energy and matter. These notes focus on homogeneous state because commonly used rheological models rely on homogeneous state. See page 1 of the beginning notes of the course for division of labor.Tractions violate the rule of vectors. You and I agree on the basic fact: the sum of tractions on two planar regions does not give a traction on another planar region. Thus, the equationt=T(n)does not map a vector space to another vector space.

This fact is unremarkable by itself. But we know that stress

Tis a tensor, which should be a linear map between two vector spaces. What are these two vector spaces? Writef=T(a).A state of stress

Tmaps an area vector of a planar region to the force acting on the planar region. This equation says that a state of stressTis a map between two spaces. One vector space is the set of area vectors of various planar regions, and the other vector space is the set of forces acting on various planar regions.In the notes I show that the map

f=T(a) is a linear map. See the triangle on page 20 of the notes.Mother of all vectors and tensors, and their fathers.A larger purpose of these notes is to show that the vector space of separations is the mother of all vectors and tensors used in rheology. The notes also identify their fathers.## 'Deep thoughts by AA'

(The title of this comment is meant to be funny. For those not familiar, try out some old episodes of Saturday Night Live which has a portion 'Deep thoughts by Jack Handey')

Hi Zhigang: Admittedly I have not read the section of your notes on this traction business, so strictly speaking I don't know what I am talking about - but I have gathered a good idea about what the issue is in this discussion on 'tractions not being a vector space.' So, let me set up a potential straw man. We have also just finished classes here, so I am in a good mood and I won't mind it being shot down - that's the purpose. And everything I say below is to be taken in a light vein - we are getting too serious and the Holiday season is coming, so it's time for some levity (more on this after my 'theorem.')

Theorem: Tractions in continuum mechanics form a vector space.

Remark 1 - My theorem is so deep, it is unpublishable.

Remark 2 - I do understand why Zhigang says that the sum of tractions on two planar regions does not give a traction on another planar region. I believe that the vector space of tractions that continuum mechanics talks about is not restricted to one point and one normal in one body for one specific distribution of contact forces. It is the set of all tractions that can be generated for all possible points, for all normals, for all possible bodies subjected to all possible contact force distributions.

Proof: Take a traction vector at point x_1, and for normal n_1 through x_1. This is an arrow with a direction and magnitude, call it t_1. Take another traction vector at point x_2, and for normal n_2 through x_2. This is another arrow with a direction and magnitude, call it t_2. Now we add these two arrows as we do to add vectors. Now I take the resultant arrow, call it t, take any point x_3 (could be x_1, or x_2) choose an arbitrary unit vector and call it n (could be n_1 or n_2), set up a body with a plane surface going through x_3 with normal in direction n, with area A, put a force distribution on this face with resultant (vectorial) force At. So I have constructed a planar region on which the traction is the sum of the traction vectors t_1 and t_2.

It is in this sense that one can also understand Arash's example of a vector space of tractions when the point x and normal n are fixed. Consider tractions generated on this face at the point x by all possible contact force distributions that it can be subjected to.

I don't see anything wrong with this argument. Do you?

Finally, Zhigang - It seems you turn into a hard taskmaster as the winter holidays approach. If I recall correctly, last year too around this time you started this high-falutin vector space, shmector space discussion involving vectors, co-vectors, gradients and derivatives etc which I enjoyed and participated in, but I did have to think, because you question the fundamentals which is a good thing for all of us to understand. Ideally, during this break, I just want to work with my students on things that we have finished thinking about and need to write up, spend some time with near and dear ones, and generally vegetate. I don't want to get worried about whether linear algebra, Cauchy and Euler were wrong - at least during this time.

So I propose a vote on imechanica - from Dec 15 to Jan 5, there will be no questioning the fundamentals.

As you can tell, I have gone completely nuts, so I will sign off here.

## Dear Arash,

Dear Arash,

Thanks for your comments.

Regards,

Mohsen

## A few additional comments.

"The bumpy airplane makes us dizzy. We sense the acceleration of the airplane relative to the ground". Yet the degree of sensibility is certainly dependent on how resistent or how accommodated we are - on our constitutive system. Acceleration has triggered a response. As in a rotating spring. The mathematical

expression of this, according to the Newtonian determinism, is that accelerations are determined by positions and velocities - Newton's law. This extends from particles to continuum bodies, where acceleration enters the equations as a force field, and stresses are determined by relative positions and/or relative velocities (that is, by strains and/or deformation rates): a uniform acceleration field does not stress a falling ball, whereas gravity does deform a resting ball (via boundary conditions). Thus rheology subtracts only motions that do not induce relative changes - rigid body motions.

Regarding Lie derivatives: every objective rate is a push-forward of the time-rate of a pullback transformation of the Cauchy stress. Hence all objective rates are Lie derivatives. Depending on what frame is used for observation, the pull-back and push-forward transformations may be associated with the motion of the body (as in the case of convective rates, associated with convective frames) or with the motion of an abstract frame (as in the case of Jaumann rate, where the pull-back generates the rotated stress).

Regarding the gradient: indeed, what is usually called deformation gradient is actually the derivative of the motion. The derivative is the linear part of the variation of a function - a linear operator. In other words, defining derivatives requires only the structure of a vector space and a limiting process, a topology. When this topology is generated by a metric associated with a scalar product, linear operators/functionals acquire representations in terms of the dot product with a characteristic vector - the gradient.

## Regarding the difference

Regarding the difference between a generalized Convected derivative (w.r.t. an arbitrary, invertible, time-dependent two point tensor function, say A(t) (not just F(t) or F^-1(t)), of time that convects frames or co-frames) and the Lie derivative - Perhaps you miss the point.The Lie derivative, as a matter of definition, requires a flow of a manifold to begin with. The convected derivative does not, and needs only the fn. A (it is a pointwise operation). Not all fields of A on a body can be compatible with some flow of the body. Thus all Lie derivatives are generalized convected derivatives but not vice-versa.

One could say, so what. Given an A(.) function at a given material point, say X, I can always construct an abstract homogeneous motion for the body such that A(.) is the spatially homogeneous deformation gradient for the motion. So then w.r.t this motion the generalized convected derivative is a Lie derivative. That would be almost there except for the following two points to consider.

1. For any motion of a body, any Lie derivative FIELD on the body is a generalized convected derivative field. Any generalized convected derivative field is not necessarily a Lie derivative field w.r.t ONE specific motion of the body (or flow of the manifold).

2. The more important point - on a manifold, say a shell, talking about a homogeneous deformation makes no sense, as doformation gradients from two different base points cannot be compared - their domains are different. Thus constructing the abstract motion with a 'uniform in time homogeneous deformation gradient' as I did earlier would not be an operational idea.

In summary, the issue is that generalized convected derivatives are pointwise notions on a manifold. A Lie derivative is almost that, except for the flow that appears in its definition.

I think enough electronic ink (i.e. energy) has been spent on this Lie derivative business for all of us (especially students who may be reading this) to form our own opinions on it. I am signing off from it.

## Combination of components of different temporal origin

<p>Dear Zhigang,<br /><br />Thank you very much for this interesting work. I very much liked that you pointed out the confusion between tensors and their components.</p>

<p><br />Additionally, I would also like to mention a second point that often was an origin of confusion in the classes of the chair I am working for. This is: frame-indifference of combined parameters measured at different times. For example, this may be interesting for the definition you used in your equation</p>

<p><br />“ […] A’_{ab} = Q_{ai} Q_{bj} A_{ji}<br />They both mean that the tensor is frame-indifferent. […]”<br /><br />In such definitions, Q generally applies to the total A but what if A consists of components, which were determined at different times? Such an example can be the deformation gradient (independent of its definition as discussed above):<br />Assume two observers measure the length l_1 of a steel bar at t_1. Afterwards, the bar is stretched and both observers measure the new length l_2 at time t_2. The lengths and the ratio l_2/l_1 will be the same for both observers and they will be frame-indifferent (surely assuming velocities << speed of light as you also stated in your introduction). <br />Assume now, that we replace l_1 and l_2 with the limits of d_x and d_X, i.e. the spatial differentials of the “actual” configuration at time t_2 and of a reference configuration at time t_1, both defined with respect to the 3-dimensional space. The tensor d_x/d_X then requires the following relationship to be frame-indifferent:</p>

<p><br />(d_x/d_X)’ = Q_ai(t_1) Q_(bj)(t_2) d_x/d_X.</p>

<p><br />Here, we have to consider that parts of our object of investigation were determined at different times (d_x at t_2, and d_X at t_1). Consequently, we need to verify that they were frame-indifferently measured at those individual times. A time-independent relationship via a general Q is not sufficient for this criterion. The criterion has to be applied for each individual component with respect to the actual time of “measurement” or “determination”.</p>

<p><br />More generally, if we have an expression E, which depends on n vectors, a_1, …, a_n, measured at n different times, t_1, …, t_n, the criterion for frame-indifference can be modified to (neglecting indices)<br />E(a’_1, …, a’_n) = E(Q(t_1) * a_1, …, Q(t_n) * a_n).<br />Similar expressions apply to E for tensors of higher rank, which were measured at the same time (Q(t_1) * A_1 * Q(t_1)^T, …).<br /><br />Still, there does not seem to be an agreement in literature, as some textbooks treat this topic very differently, using different approaches/definitions of objectivity or frame-indifference, respectively. <br />The following works consider time-dependence<br /><br />I.-S. Liu. On the transformation property of the deformation gradient<br />under a change of frame. J. Elasticity, 71:73–80, 2003.<br /><br />P. Haupt. Continuum mechanics and theory of materials. Springer, 2002,<br />whereas many others neglect this for their definitions (especially textbooks, which can make it even more difficult for teaching purposes).</p>

<p><br />Thank you again for this detailed treatment of the topic and I wish you interesting discussions in your classes<br /><br />Patrick</p>

## (plain text, sorry for double posting)

Dear Zhigang,

Thank you very much for this interesting work. I very much liked that you pointed out the confusion between tensors and their components.

Additionally, I would also like to mention a second point that often was an origin of confusion in the classes of the chair I am working for. This is: frame-indifference of combined parameters measured at different times. For example, this may be interesting for the definition you used in your equation

“ […] A’_{ab} = Q_{ai} Q_{bj} A_{ji}

They both mean that the tensor is frame-indifferent. […]”

In such definitions, Q generally applies to the total A but what if A consists of components, which were determined at different times? Such an example can be the deformation gradient (independent of its definition as discussed above):

Assume two observers measure the length l_1 of a steel bar at t_1. Afterwards, the bar is stretched and both observers measure the new length l_2 at time t_2. The lengths and the ratio l_2/l_1 will be the same for both observers and they will be frame-indifferent (surely assuming velocities << speed of light as you also stated in your introduction).

Assume now, that we replace l_1 and l_2 with the limits of d_x and d_X, i.e. the spatial differentials of the “actual” configuration at time t_2 and of a reference configuration at time t_1, both defined with respect to the 3-dimensional space. The tensor d_x/d_X then requires the following relationship to be frame-indifferent:

(d_x/d_X)’ = Q_ai(t_1) Q_(bj)(t_2) d_x/d_X.

Here, we have to consider that parts of our object of investigation were determined at different times (d_x at t_2, and d_X at t_1). Consequently, we need to verify that they were frame-indifferently measured at those individual times. A time-independent relationship via a general Q is not sufficient for this criterion. The criterion has to be applied for each individual component with respect to the actual time of “measurement” or “determination”.

More generally, if we have an expression E, which depends on n vectors, a_1, …, a_n, measured at n different times, t_1, …, t_n, the criterion for frame-indifference can be modified to (neglecting indices)

E(a’_1, …, a’_n) = E(Q(t_1) * a_1, …, Q(t_n) * a_n).

Similar expressions apply to E for tensors of higher rank, which were measured at the same time (Q(t_1) * A_1 * Q(t_1)^T, …).

Still, there does not seem to be an agreement in literature, as some textbooks treat this topic very differently, using different approaches/definitions of objectivity or frame-indifference, respectively.

The following works consider time-dependence

I.-S. Liu. On the transformation property of the deformation gradient

under a change of frame. J. Elasticity, 71:73–80, 2003.

P. Haupt. Continuum mechanics and theory of materials. Springer, 2002,

whereas many others neglect this for their definitions (especially textbooks, which can make it even more difficult for teaching purposes).

Thank you again for this detailed treatment of the topic and I wish you interesting discussions in your classes

Patrick

PS: I am sorry for the double posting, the editor converted the html-coding incorrectly the first time. Unfortunately, I was no able to correct it by myself and perhaps an administrator can delete the first version.