User login

You are here

I am [still] confused about gradients, vectors, deformation gradient, etc.

I am creating this blog entry to have my confusions about gradients, vectors, and deformation gradient, etc., straightened out once (and hopefully for all time!) ... My confusions got exposed (even to me) while commenting on a thread started by Prof. Suo here [^]. In particular, I realized my confusions after writing this comment [^] there.

So, let me note down my confusions in the form of a few questions; the order may not be exactly from the simplest/basic to the complex:

  1. What is a vector space?
  2. What is a vector field?
  3. (Perhaps a repetition) What is the difference between (a) a function that is defined on a scalar field and outputs a vector-valued "field," and (b) a vector space?
  4. Is the set of traction vectors directed along a given (local) direction, a field? Does it span a vector space? which one?
  5. What exactly is a gradient of a scalar field? is it a vector field?
  6. Is there a gradient of a vector field? What is the nature of its output?
  7. Is the nabla operator always a vector? why or why not? examples?
  8. Is there any technical difference between the terms: displacement gradient and deformation gradient?
  9. Is the deformation gradient a gradient?
  10. Is the deformation gradient a vector?

May be, I will add more questions as they strike me, but this should be enough to get going.

Thanks in advance for clarifying these matters.

Best,

--Ajit

[E&OE]

 

Comments

Zhigang Suo's picture

Dear Ajit:  Thank you for this post.  I eagerly await illumination by experts on geometry.  Here is a try on your Question 1.  I wrote some notes on linear algebra.  Vector space is defined on page 4.  I just follow the standard usage of this phrase "vector space" in linear algebra.

Dear Ajit: If you really want to go deep, then please check

http://en.wikipedia.org/wiki/Discrete_exterior_calculus

and the references. 

Dear Zhigang and all,

Thanks for pointing out your notes. I will go through them.

Also, I now realize that these questions are too broad. So, here is what I propose to do to get my concepts straightened out.

I would go through Prof. Abeyaratne's notes [^], Prof. Suo's notes (a/a), as well as some other standard material once again. (My copy of Schaum's series on continuum mechanics is in Pune, and not with me in Mumbai right now. I will pick it up the next time I visit Pune.)

Also, to all: feel free to leave the pointers to materials, esp. notes and books freely available on the 'net.

Then, I would myself write down a quick set of answers to these questions for context. I would write my answers as if I were an UG who had gone through the preliminary materials. On the basis of these answers, I would raise further specific queries.

This can be the only practicable way; otherwise the scope is too big.

Please give me a week or so to come back. I promise I would like to finish this thing off (even if only as an on-and-off thing) before this year ends, i.e. during these "vacation" times (even if, frankly, I would be too busy checking some 200-300 university final exam. answer-books over the upcoming weeks, and so, would be able to pursue everything else---my research, programming, and this activity---mostly only in the evenings/nights or so).

Best,

--Ajit

[E&OE]

 

Yes it is, if placed in the appropriate vector space (and hence it has components, like any other vector). The essence here is the duality pairing between a vector space and its dual. Thus:

A vector space is an additive group whose elements can be (conveniently) multiplied by numbers (e.g., real, complex). The archetype is the familiar set of "tied" vectors of the 3D space. When talking about derivatives and, in general, tensors/operators, there's another important example: given a vector space V, its
dual V' is the set of all linear functionals f:V->R (for real spaces). V' too is a vector space. The important fact to be mentioned here is that a basis in V induces a basis in V'; and that a change of basis in V induces a change of basis in V' with a different coordinate/component transformation rule: This is the basis for "contravariant" versus "covariant". If V is actually Euclidean (thus equipped with the usual scalar product, denoted here by <.,.>), this distinction disappears, because the action of every linear functional can be represented as a dot product: there exists and is unique a vector G in V such that f(v) = <G,v> for every v in V; V and V' have become one and the same (via the identification G->f).

A vector field is a function assigning a vector to each element of a set. An illustrative example: to each element "x" of a subset of a manifold we assign an element of the tangent space at "x". Down to Earth: the wind velocity over Europe is a vector field (defined over a subset of a sphere).

Finally, before "gradient", one needs to talk about the derivative of a function; function which may be defined over quite abstract objects, e.g., manifolds. However, it is sufficient to consider the case of a function defined over some open subset of a normed space (a vector space equipped with a norm |.|, which, for the purpose of taking limits, should be complete), for then everything extends easily (by composing with coordinate maps - charts). Then we have the classic: given E and F normed spaces and an open subset M of E, a function f:M->F is differentiable at x in M iff there exists a linear operator Df(x):E->F such that f(y)=f(x)+Df(x):(y-x)+o(|y-x|^2). When F is R, the field of real numbers, we get a functional, or a "scalar" function/field. Thus Df is an element of E' (the dual of E). This is the nabla operator. Now if the normed space is actually a Euclidean space (the norm being generated by a scalar product <.,.>), then any linear functional can be uniquely represented as Df:v = <G,v>, for any v in E. We call G the gradient of f; it is an element of E, thus a vector (recall the identification of V and V' at point 1.). When F is R^n, the gradient becomes a tensor (still a "vector", but with more indices).

To conclude, when an ambient scalar product is present, we can safely identify the derivative of a function with its gradient. In this context the classical terminology, "deformation gradient", makes perfect sense and is very convenient ("The derivative of the configuration of a body is called the deformation gradient", Marsden and Hughes, Mathematical foundations of elasticity, Dover ed, pp 47).

 

 

Hi Stefan,

Thanks for an amzingly nifty reply.

Though, of course, I didn't understand it fully. Which is an understatement. I should have said that I more or less stopped following you whenever you came to the real core of the theory: functionals (are these the same animals as in the calculus of variation?), dual spaces (I would get it if I get what the term functional means here), and then, the "completeness" (of a normed space), the ":" operator (or whatever mathematical object it is), etc.

I predict that if you should have begun to feel frustrated by now.

But if you therefore feel too frustrated, please don't. (i) A major reason is that you will feel very similarly far too often in life, simply because there are so many engineers out there. (ii) A minor reason is that I actually did get some poetic sort of a "feel" here and there for some of the things you wrote. For instance when you came to what you call the "classic," what I got is that it states something like: y is differentiable if there exists an m such that the condition y = c + m x + higher order terms in the Taylor series expansion, is fulfilled, and if it is, then m represents the differentiation of y with respect to x. ... But of course, my understanding being firmly rooted only in some poetic sort of grounds [no sarcasm], I couldn't possibly have got why you have that (y-x) in place of the simple x [which, in fact, is the case].

Never mind. Let me show you a way to uplift yourself out of your present state of semi-/full-frustration---the one in which I must have pushed you. ...

Please recommend us the simplest-to-an-engineer book that covers all (or at least most) of this theory. ... Ok. Make it two books. Or three, at the most.

And, if possible, please mention at least one killer app (from engineering, not from physics/mathematical physics) for this theory. It will be the greatest motivator.

Personally, I think the theory is intriguing, and could even be understandable (even if I am sure none would have worked out many of the unforeseen consequences in going so abstract)... Indeed, this theory could even be beautiful. But the way I see it, it has not been sold right. There are no visualizations. More important: There is no killer app. Not at least from the usual core engineering viewpoint (i.e. the one as is covered in the UG/master's studies).

But, still, the faint outlines of the theory do seem interesting... They are interesting enough to make someone like me wonder aloud, here.

Thanks, once again, for your helpful reply, and please see if you can come back about the books and all.

Best,

--Ajit

[E&OE]

 

 

You are a formidable (mind)reader. So, to the point:

The "killer app" of this abstraction is, recalling where part of this discussion was initiated, a better appreciation/understandig of the concepts involved in the description of the motion of a continuum, particularly its "deformation gradient".

Functional is any function defined over a vector space and taking values in the field of real numbers (or, the field of the multipliers of the vector space). Of common interest in continuum mechanics (and in particular in engineering) is the case when the vector space is finite dimensional, but the definitions and the
theory extend to infinite dimensions. I used the symbol ":" to denote the application of a linear function (can be operator, can be functional) to its argument, or to indicate the domain of definition of a function.

Unintentionally, it did come out like a Taylor expansion formula.
So, for the sake of rigor:

A function f:M->F defined over an open set in the normed space E, taking values in the normed space F (one may take E and F as R^n and R^m, respectively, with n and m natural numbers), is differentiable at point x in M iff there exists a linear function, denoted Df(x), and a function g:M->F with the property that lim(y->x)g(y) = 0, such that, for any y in M the function f admits the representation

f(y)=f(x)+Df(x):(y-x)+g(y)|y-x|,

where |y-x| is the distance between "points" x and y.
Df(x) is the differential/derivative of f at x; it is a linear application from E to F (we write this as Df(x):E->F). Df(x):(y-x) is the application of Df(x) to the vector y-x.

What is known as the Jacobian of the function f (the matrix of partial derivatives) is nothing but the coordinate expression of the tensor Df(x); in addition, in the most general case, two coordinate systems may be in use (one in E, the other in F, from where the "two point tensor" terminology).

One can find good textbooks at the following links:

http://rbowen.tamu.edu/

http://www.cds.caltech.edu/~marsden/books/

I would particularly mention "Manifolds, Tensor Analysis and Applications", by Abraham, Marsden and Ratiu (a quick browse should find several pdf's online) and, of course, the introductory chapters from "An introduction to continuum mechanics" by Gurtin. Further references can be found in these works.

Best,

SS

Hi Stefan,

Thanks for pointing out the books. After initial browsing, I think these will require quite a concentrated reading... Let me see if I can find time for the same.

In the meanwhile, since your replies seem to have addressed most of my above questions, I am cancelling the idea of writing down my set of answers so as to get the discussion going.

Thanks, once again, for the replies and the pointers. Hope you would make yourself available if I need some help in this direction in future.

Best,

--Ajit

[E&OE]

 

Subscribe to Comments for "I am [still] confused about gradients, vectors, deformation gradient, etc."

Recent comments

More comments

Syndicate

Subscribe to Syndicate