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Insights What Is a Tensor? - Comments

  1. Jun 18, 2017 #1

    fresh_42

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  2. jcsd
  3. Jun 20, 2017 #2
    If I interpret correctly this sentence says the underlying set of U⊗V is U×V. This is not correct. This works for the direct sum U⊕V but the tensor product is a "bigger" set than U×V. One usual way to encode it is to quotient ##\mathbb{R}^{(U \times V)}## (the set of finitely-supported functions from U×V to ##\mathbb{R}##) by the appropriate equivalence relation.
     
  4. Jun 20, 2017 #3

    lavinia

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    - Students of General Relativity learn about tensors from their transformation properties. Tensors are arrays of number assigned to each coordinate system that transform according to certain rules. Arrays that do not transform according to these rules are not tensors. I think that it would be helpful to connect this General Relativity approach to the mathematical approach that you have explained in this Insight.

    Also these students need to understand how a metric allows one to pass back and forth between covariant and contra-variant tensors. One might show how this is the same as passing between a vector space and its dual.

    - Students of Quantum Mechanics learn about tensors to describe the states of several particles e.g. two entangled electrons. In this case, the mathematical definition is more like the Quantum Mechanics definition but for the Quantum Mechanics student it is also important to understand how linear operators act on tensor products of vector spaces.

    - If one wants to discuss tensor products purely mathematically, then one might show how they are defined when the scalars are not in a field but in a commutative ring - or even a non-commutative ring. The formal properties do not depend on a field per se.
     
    Last edited: Jun 20, 2017
  5. Jun 20, 2017 #4

    lavinia

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    The tensor product of two 1 dimensional vector spaces is 1 dimensional so it is smaller not bigger than the direct sum. The tensor product of two 2 dimensional vector spaces is 4 dimensional so this is the the same size as the direct sum not bigger.
     
    Last edited: Jun 20, 2017
  6. Jun 20, 2017 #5
    This is correct but missing the relevant point: that the presentation contains a false statement. The fact that you can indeed find counter examples where the direct sum is bigger than the tensor product does not makes the insight presentation any more correct.
     
  7. Jun 20, 2017 #6

    fresh_42

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    No, this interpretation was of course not intended, rather a quotient of the free linear span of the set ##U \times V##.
    I added an explanation to close this trapdoor. Thank you.
     
  8. Jun 20, 2017 #7
    Thank you
     
  9. Jun 20, 2017 #8

    fresh_42

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    I know, or at least assumed all this. And I was tempted to explain a lot of these aspects. However, as I recognized that this would lead to at least three or four parts, I concentrated on my initial purpose again, which was to explain what kind of object tensors are, rather than to cover all aspects of their applications. It was meant to answer this basic question which occasionally comes up on PF and I got bored retyping the same stuff over and over again. That's why I've chosen Strassen's algorithm as an example, because it uses linear functionals as well as vectors to form a tensor product on a very basic level, which could easily be followed.
     
  10. Jun 20, 2017 #9
  11. Jun 21, 2017 #10

    Orodruin

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    Let me first say that I think that the Insight is well written in general. However, I must say that I have had a lot of experience with students not grasping what tensors are based on them being introduced as multidimensional arrays. Sure, you can represent a tensor by a multidimensional array, but this does not mean that a tensor is a multidimensional array or that a multidimensional array is a tensor. Let us take the case of tensors in ##V\otimes V## for definiteness. A basis change in ##V## can be described by a matrix that will tell you how the tensor components transform, but in itself this matrix is not a tensor.

    Furthermore, you can represent a tensor of any rank with a row or column vector - or (in the case of rank > 1) a matrix for that matter (just choose suitable bases). This may even be more natural if you consider tensors as multilinear maps. An example of a rank 4 tensor being used in solid mechanics is the compliance/stiffness tensors that give a linear relation between the stress tensor and the strain tensor (both symmetric rank 2 tensors). This is often represented as a 6x6 matrix using the basis ##\vec e_1 \otimes \vec e_1##, ##\vec e_2 \otimes \vec e_2##, ##\vec e_3 \otimes \vec e_3##, ##\vec e_{\{1} \otimes \vec e_{2\}}##, ##\vec e_{\{1} \otimes \vec e_{3\}}##, ##\vec e_{\{2} \otimes \vec e_{3\}}## for the symmetric rank 2 tensors. In the same language, the stress and strain tensors are described as column matrices with 6 elements.
     
  12. Jun 21, 2017 #11

    fresh_42

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    You can consider every matrix as a tensor (defining the matrix rank by tensors) or write a tensor in columns, as it is a vector (element of a vector space) in the end. Personally I like to view a tensor product as the solution of a couniversal mapping problem. As I said, I was tempted to write more about the aspect of "How to use a tensor" instead of "What is a tensor" but this would have led to several chapters and the problem "Where to draw the line" were still an open one. Therefore I simply wanted to take away the fears of the term and answer what it is, as I did before in a few threads, where the basic question was about multilinearity and linear algebra and the constituencies of tensors. The intro with the numbers should show that the degree of complexity depends on the complexity of purpose. I simply wanted to shortcut future answers to threads rather than write a book about tensor calculus. That was the main reason for the examples, which can be understood on a very basic level. Otherwise I would have written about the Ricci tensor and tensor fields which I find far more exciting. And I would have started with rings and modules and not with vector spaces. Thus I only mentioned them, because I wanted to keep it short and to keep it easy: an answer for a thread. Nobody on an "A" and probably as well on an "I" level reads a text about what a tensor is.
     
  13. Jun 21, 2017 #12

    Orodruin

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    Perhaps I was mislead regarding the intended audience from the beginning. I am pretty sure most engineering students will not remember what a homomorphism is without looking it up. Certainly a person at B-level cannot be expected to know this?

    In the end, I suspect we would give different answers to the question in the title based on our backgrounds and the expected audience. My students would (generally) not prefer me to give them the mathematical explanation, but instead the physical application and interpretation, more to the effect of how I think you would interpret "how can you use tensors in physics?" or "how do I interpret the meaning of a tensor?"

    This must mean I am B-level. :oops::eek::frown:
     
  14. Jun 21, 2017 #13

    fresh_42

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    Yes, you are right. My goal was really to say "Hey look, a tensor is nothing to be afraid of." and that's why I wrote
    And to be honest, I'm bad at basis changes, i.e. frame changes and this whole rising and lowering indices is mathematically completely boring stuff. I first wanted to touch all these questions but I saw, that would need a lot of more space. So I decided to write a simple answer and leave the "several parts" article about tensors for the future. Do you want to know where I gave it up? I tried to get my head around the covariant and contravariant parts. Of course I know what this means in general, but what does it mean here? How is it related? Is there a natural way how the ##V's## come up contravariant and the ## V^{*'}s## covariant? Without coordinate transformations? In a categorial sense, it is again a different situation. And as I've found a source where it was just the other way around, I labeled it "deliberate". Which makes sense, as you can always switch between a vector space and its dual - mathematically. I guess it depends on whether one considers ##\operatorname{Hom}(V,V^*)## or ##\operatorname{Hom}(V^*,V)##. But if you know a good answer, I really like to hear it.
    Well, your motivation can't have been to learn what a tensor is. That's for sure. :cool: Maybe you have been curious about another point of view. As I started, I found there are so many of them, that it would be carrying me away more and more (and thus couldn't be used as a short answer anymore). It is as if you start an article "What is a matrix?" by the sentence: "The Killing form is used to classify all simple Lie Groups, which are classical matrix groups. There is nothing special about it, all we need is the natural representation and traces ... etc." Could be done this way, why not.

    This is the skeleton I originally planned:

    \subsection*{Covariance and Contravariance}
    \subsection*{To Rise and to Lower Indices}
    \subsection*{Natural Isomorphisms and Representations}
    \subsection*{Tensor Algebra}
    \section*{Stress Energy Tensor}
    \section*{Cauchy Stress Tensor}
    \section*{Metric Tensor}
    \section*{Curvature Tensor}
    \section*{The Co-Universal Property}
    \subsection*{Graßmann Algebras}
    \subsection*{Clifford Algebras}
    \subsection*{Lie Algebras}
    \section*{Tensor Fields}
     
  15. Jun 21, 2017 #14

    fresh_42

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    Corrected. Thanks.
     
  16. Jun 21, 2017 #15

    WWGD

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    I thought it would be nice to have a good understanding of what a singleton ## a \otimes b ## represents in a tensor product. It is one of these things that I have understood and then forgotten many times over.
     
  17. Jun 21, 2017 #16

    WWGD

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    I think this is done before the moding out and arranging into equivalence classes is done.
     
  18. Jun 21, 2017 #17

    fresh_42

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    It's the freely generated vector space (module) on the set ##U \times V##. The factorization indeed guarantees the multilinearity and the finiteness of sums which could as well be formulated as conditions to hold.
     
  19. Jun 21, 2017 #18

    WWGD

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    I was replying to someone else's post.
     
  20. Jun 21, 2017 #19

    fresh_42

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    Sorry, was a bit in "defensive mode".
     
  21. Jun 21, 2017 #20

    WWGD

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    No problem.
     
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