What Are Tensors and Why Should You Care?

[uiulic]

Any interested in this topic?

 

[Chris Hillman]

What topic precisely? Sure, something about "tensors", but what about them?

 

[uiulic]

Thank you for the link, Chris.

Let’s limit the discussion to the tensor used in physics esecially in solid mechanics first. I have a series of questions and will ask them separately in the future.

Question1: Why is force a vector?

Question2: In continuum mechanics (Malvern 1969), a second order tensor (stress tensor) is defined as a linear vector function with both the argument and output of the function are vectors. In my mind, tensor is a quantity. Why is the function is an quantity?

Question3: In terms of linear vector function. Let’s assume is a linear vector function b=f(a), a and b are vectors. Then f is said to be a second order tensor. Why don’t you say b (b has been defined as a vector already) is a second order tensor? Since we can understand b=f(a),the linear function relation, as b(a)=f(a) , and b is then a second order tensor. Where does the problem occur for this explanation?

 

[Chris Hillman]

Can you clarify the questions?

Question1: Why is force a vector?

When someone asks a why question like this, sometimes it can be difficult to guess what kind of answer they want. Mathematical? Physical? Philosophical?

Since no-one else jumped in here, I guess others are equally puzzled.

Question2: In continuum mechanics (Malvern 1969), a second order tensor (stress tensor) is defined as a linear vector function with both the argument and output of the function are vectors. In my mind, tensor is a quantity. Why is the function is an quantity?

Do you mean this book?

Lawrence E Malvern,
Introduction to the mechanics of a continuous medium
Englewood Cliffs, N.J., Prentice-Hall,1969.

If so, I am not familiar with it, but as it happens I have been discussing the stress tensor in the context of continuum mechanics in another thread, "What is the Theory of Elasticity?". That might provide some physical insight into why we define the stress tensor in terms of the strain tensor the way we do.

Somewhere in there you might be asking about linear operators, which take vectors to vectors. Bundling this, we obtain:
A^k \rightarrow {T^k}_m \, A^m [/itex]<br /> where T_{km} is a second rank tensor. <br /> <br /> You might also be asking about quantities versus intensities.<br /> <br /> <blockquote data-attributes="" data-quote="uiulic" data-source="post: 1340965" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> uiulic said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Question3: In terms of linear vector function. </div> </div> </blockquote><br /> I am not sure what you mean by &quot;linear vector function&quot;, but I guess you might mean what I wrote down above.<br /> <br /> <blockquote data-attributes="" data-quote="uiulic" data-source="post: 1340965" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> uiulic said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Let’s assume is a linear vector function b=f(a), a and b are vectors. Then f is said to be a second order tensor. </div> </div> </blockquote><br /> Second rank?<br /> <br /> <blockquote data-attributes="" data-quote="uiulic" data-source="post: 1340965" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> uiulic said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Why don’t you say b (b has been defined as a vector already) is a second order tensor? Since we can understand b=f(a),the linear function relation, as b(a)=f(a) , and b is then a second order tensor. Where does the problem occur for this explanation? </div> </div> </blockquote><br /> Huh?<br /> <br /> Your terminology seems a bit odd to me--- does Malvern use all the terms you mentioned? If so, does he define them?

 

[uiulic]

Sorry,
I have deleted the one with wrong sign.

 

[mathwonk]

Chris, we're just glad to have new blood to deal with these questions. we've answered this question so many hundreds of times we're dizzy, as a search of our old posts may reveal.

 

[Chris Hillman]

Thanks, mathwonk! Unfortunately, inspiration has yet to strike. That is to say, I still haven't thought of an interesting way of making sense of the OP's questions (and then, of course, answering them).

uiulic, just to make sure I am not misunderstood: I suspect there are ways of reformulating your questions so that interesting answers can be given, so I am not saying they are bad questions, just that at the moment I can't think of interesting/useful answers, because at the moment I can't think up a suitable restatement of them.

 

[pmb_phy]

Any interested in this topic?

Greetings uiulic and welcome to the forum. I'm currently in the process of creating a branch on my website to describe geometrical methods of physics. There is a page there on an intro to tensors. It gives the basics. Its located at

http://www.geocities.com/physics_world/gr_math/geo_tensor.htm

As far as why force is a vector; If you are referring to a 3-vector then force is a vector because it has both a magnitude and direction per the definition of 3-vector (which can also be defined by how its components transform under a rotation of the coordinate system). If you're referring to a 4-vector then this is something which proved useful to defined for SR and is a generalization of force into a covariant form.

Hope that helps.

Pete

 

[robphy]

Question1: Why is force a vector?

I would answer this by saying that physical forces (at a point) add and scalar-multiply like mathematical vectors. More loosely speaking, forces superpose according to the parallelogram rule. (The magnitude of a vector is based on a choice of metric, which is an additional structure... and which is not part of the standard textbook definition of a vector.)

 

[pmb_phy]

(The magnitude of a vector is based on a choice of metric, which is an additional structure... and which is not part of the standard textbook definition of a vector.)

Hi Rob

I'm curious about this magnitude thing. While I do not disagree with you on this part, I do have a Dover text called Differential Geometry, by Erwin Kreyszig, in which the author defines a vector according to the simple definition I gave above. I gave the correct definition in parentheses for those more informed on the matter. In your opinion, what would be your guess as to why the author would define a vector in this way with no mention of a metric??

I this instance I chose that definition so as not to bring in higher order of mathematics that the OP may not understand.

Thanks

Pete

 

[uiulic]

1) To Chris,
Thank you for bringing my questions to popular attention. But I must say that I answered each of your questions in much detail in this forum at the cost of 3-5 A4 word-document pages, but I deleted them because of a confusing signature and your were unhappy with the confusing signature. Some information related to your questions will maybe be induced in the future.

2) To mathwonk,
Thank you for your attention for this topic and thank you for your ever efforts in answering questions. But your reply did seem a complaint, in my eyes. If there is a link for this old topic, could you give a link? I am really a newcomer here. What’s more, some old topic like “force is a vector or not” as an example, if you do not consider it as a problem at all, then you really do not need to think about it. If you think it is, then it is.

3) To pmb_phy in your first reply
Thank you for the link although the topic related there is a little beyond my current knowledge. In your reply, you mentioned 3-vector, 4-vector, I take my understanding simply as a vector which is in 3D or 4D space (3 or 4 vector components). Your answer because it has both a magnitude and direction per the definition of 3-vector is a little different from what I read in normal textbooks. What’s more, your “per” means “each of the three” or “each of all the possible”, because per direction can really mean each possible direction in the 3D (take 3D as an example) space. Normally, a vector is said to be have a fixed direction and magnitude (0 vector excluded of course), but this definition does not seem to be preferred by mathematicians or physicists since it is hard for them to extend the use of vector in a broader sense. But can “magnitude and direction” define a vector at all or not, since mathematicians and physicists can really make a broader sense of definitions for magnitude and direction?

4) To robphy,
Thank you for your reply. Your consideration of the definition of the vector is based on that by some mathematicians. But this definition is the same (consistent with? Broader than? Less broad than?) as that given in the subject of tensor analysis? In tensor analysis, it is given by some transformation rules on the vector (tensor) components under different coordination systems. Your complementary reply (The magnitude of a vector is based on a choice of metric, which is an additional structure... and which is not part of the standard textbook definition of a vector.) does arise a similar question “ who is talking about the vector”.

5) To pmb_phy in your second reply,
I am an engineering student, who has no experience in advanced mathematics and physics. But you can still discuss it, because this topic is open to all others as well. I read many different definitions on tensors (also vectors), and I find that there is a big gap for me and it hard to catch what some authors say in their texts. As to what is the different definitions in the eyes of mathematicians, physicists, engineers, this is my originally intentioned purpose. But I proposed a topic like that, and nobody replied and therefore I later changed my topic to the current topic as it is now. We can limit the discussion why force is a vector this topic and later in the future extend the discussion in due time. You can expand the topic and talk about the vector space and dual space and a tensor defined on these two. But I definitely will keep silence when there you talk about that. But there may be induced interesting questions and others will give a comment. My question related to this in the future will maybe be “ is your definition of vector in terms of vector space or your probable definition of tensor in terms of vector space and its dual space can cover all possible definitions in the literature?”

I have a unmature reply as to why the author would define a vector in this way with no mention of a metric??. Because we often talk about vectors in 3D Eulid, and the metric is given by Pythagoras’s rule, and this is tacitly assumed at least in engineering sense without causing confusion.

 

[robphy]

Hi Rob

I'm curious about this magnitude thing. While I do not disagree with you on this part, I do have a Dover text called Differential Geometry, by Erwin Kreyszig, in which the author defines a vector according to the simple definition I gave above. I gave the correct definition in parentheses for those more informed on the matter. In your opinion, what would be your guess as to why the author would define a vector in this way with no mention of a metric??

I this instance I chose that definition so as not to bring in higher order of mathematics that the OP may not understand.

Thanks

Pete

Classic tensor calculus usually emphasized geometrical quantities by their invariance under transformations, probably in the spirit of Felix Klein's Erlangen Program: "every system of geometry deals only with such relations of space as remain unchanged by the transformations of its group." Upon specifying the group of transformations (e.g., Euclidean, Lorentz, conformal, projective, etc..), you may be implicitly defining a specific metric structure.

---
I don't have a copy of Kreyszig's Differential Geometry handy, although I did use it as my supplementary text for the first differential geometry course I took. I do have Kreyszig's Advanced Engineering Mathematics (5th ed, 1983), which possibly has a similar philosophy on this issue. A.E.M., Chapter 6 is called "Linear Algebra I: Vectors".

From (A.E.M., 6.1 "Scalars and Vectors"),
"A directed line segment is called a vector. Its length is called the length of the vector and its direction is called the direction of the vector. Two vectors are equal if and only if they have the same length and direction.

The length of a vector is also called the Euclidean norm or magnitude of the vector."
---

Here, the emphasis on the Euclidean length of a vector is probably based on familiarity with what is generally taught at the elementary level. However, as I usually argue [see below], the most important feature of a vector is the parallelogram rule. A quantity can have a magnitude and a direction... but without the parallelogram rule, it's not a vector.


Here is how I use the directed-line-segment concept to describe vectors.

I draw a directed-line-segment (an "arrow") and refer to its size (rather than "magnitude" which is already claimed by standard definitions and which involves additional structure) and direction. At this point, there are no numbers assigned to either the size or the direction. Indeed, without additional structure, there is no way to compare two directed-line-segments in the plane... unless the direction of the two directed-segments coincide [in which case, one can express one as a scalar multiple of the other]. In spite of not having any additional structure [and independent of any structure that might be imposed], one can, however, still do the parallelogram rule. The parallelogram rule is fundamental to the notion of a vector.

If one now introduces a metric, one can now form numbers from vectors. Indeed, coordinate-free vector notation [including the abstract index notation] practically enforces this idea: given v^a, one cannot form a scalar without additional structure. A metric g_{ab} allows one to form a scalar g_{ab}v^av^b. (Graphically, a metric can be introduced by specifying (for example) a circle centered at the base-point.)

Rather than introducing the metric directly, one can introduce a group of transformations-of-coordinate-systems which can preserve the scalar nature of g_{ab}v^av^b, then possibly determine what that g_{ab} must be to be compatible with that transformation. Thus, by specifying that the group is the "ordinary rotation group", you implicitly introduce the Euclidean metric... "Lorentz transformations"... Minkowski metric, etc... (Note, however, that there are groups of transformations that will not introduce a metric.)

 

[matheinste]

Hello robphy.

Thanks for a simple and clear explanation of the metric relating to a vector.

Matheinste.