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Problem with definition of tensor

  1. May 18, 2015 #1

    ShayanJ

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    In textbooks, a tensor is usually defined in terms of its transformation properties. But this definition actually seems vague when it comes to checking a set of quantities to see whether they form a tensor or not. Imagine I have four functions and want to see whether they form a 2d 2nd rank tensor together or not. Then if I want to use the transformation properties, I should check whether ## f_{ij}=J(i,k)J(j,l)f'_{kl} ##. But that's weird because that needs us to know what is ## f'_{kl} ## which means we should have an accepted method of transformation. But what is that method? Surely just replacing the coordinates doesn't work. Another way I thought of was through seeing tensors as operators. Then we would want the tranformed operator to do the same thing as the old operator, to the transformed vectors. But this method seems too subjective and doesn't seem general enough.
    So I wanna know, how can I check whether a bunch of functions form a tensor or not?
    Thanks
     
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  3. May 18, 2015 #2

    Orodruin

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    You do not necessarily need a transformation rule for this, what you need to know is what the expression for ##f'_{k'l'}## is. For example, take the permutation symbol ##\varepsilon_{ijk}## and ask yourself whether the object that takes values 1 if the indices are an even permutation and -1 if they are an odd permutation of 123 is a tensor. There is no transformation rule involved here, just an assumption on how the components look in different frames.

    That being said, I am not a big fan of introducing tensors by defining the transformation properties. Instead, it is more elegant to define them as linear maps or linear combinations of outer products. The transformation properties follow from these definitions and you can still check whether a given set of functions fulfil the transformation rules the components of a tensor would have to fulfil.
     
  4. May 18, 2015 #3

    ShayanJ

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    OK, so let's consider your favorite definitions. Then given a set of functions, can you check that whether there is a tensor with those components or not?
     
  5. May 18, 2015 #4

    Orodruin

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    As long as you are given the components in all frames, yes. It should only be a matter of checking whether the transformation rules are obeyed, i.e., take your knowledge about the expression of the functions and insert into the transformation rules.
     
  6. May 18, 2015 #5

    ShayanJ

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    So information about the functions in only one frame is insufficient for that. Knowing this, now it seems really a bad idea to me to define tensors in terms of their transformation properties.
    Thanks man
     
  7. May 18, 2015 #6

    Orodruin

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    Well, it does have its merits. However, I think that it is a way that has had many students utterly confused about what a tensor is and what they are good for in physics. I strongly believe that introducing tensors to students with the approach "a tensor has the following transformation properties, just eat it" is not a good way to have them learn.
     
  8. May 18, 2015 #7

    ShayanJ

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    Yeah, strongly agree with that. I've seen many students confused with that definition. But the problem is, other definitions seem to require the student to be able to make sense of other parts of mathematics at least a bit.
     
  9. May 20, 2015 #8

    HallsofIvy

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    If you know the functions in one frame and know that it is a tensor then it is easy to use those transformation properties to determine them in any other frame. If you know the functions in one frame and want to prove from this that it is a tensor, of course, you cannot do it. There exist an infinite number of "objects" that have those specific functions in one frame, other values in other frames, some of which are tensors and some of which are not.
     
  10. May 21, 2015 #9

    martinbn

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    That is precisely why "definitions" like this are rubbish. And when you say in textbooks, you mean in physics textbooks. In good ones i.e. maths textbooks tensors are defined properly and without ambiguities.
     
  11. May 21, 2015 #10

    ShayanJ

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    Some physicists do that to reduce the amount of math the student has to struggle with. Some students actually appreciate that but I don't like it because it takes some years for that student to understand the subject at a level that he could reach much sooner if they committed themselves to learning the concept in a more proper way.
    But I can still point out some physicists that write in a way you may like. For instance I can point to my favorite SR book. I think you'll like its treatment of tensors, or at least won't hate it!(Actually all of the material in the book is as nice as the treatment of tensors

    EDIT:
    or even nicer!):biggrin:
     
    Last edited: May 21, 2015
  12. May 21, 2015 #11

    Fredrik

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    Right, a 4-tuple of functions can't possibly be a tensor. The tensor would be something like a map that associates a 4-tuple of functions with each coordinate system. Then you can consider two arbitrary coordinate systems x and x', and the corresponding 4-tuples ##(f_1,f_2,f_3,f_4)## and ##(f_1',f_2',f_3',f_4')##.

    This is handled much more elegantly in math books (e.g. "Linear algebra done wrong" by Sergei Treil or "Introduction to smooth manifolds" by Lee) and a few physics books (e.g. "A first course in general relativity" by Schutz, chapter 3 I think). For example, a tensor of type (1,2) over a vector space V would be a multilinear map ##T:V^*\times V\times V\to\mathbb R##, where V* is the dual space of V (the set of all linear maps from V to ℝ). Given an ordered basis ##(e_1,\dots,e_n)## for V, you can define an ordered basis for V* by ##e^i(e_j)=\delta^i_j##. The components of T are the numbers ##T^i{}_{jk}## defined by ##T^i{}_{jk}=T(e^i,e_j,e_k)##. The tensor transformation law describes how the components change when you change the ordered basis for V (as well as the associated ordered basis for V*).

    In differential geometry, V is the tangent space of a manifold at some point p. Its elements are called tangent vectors at p. The elements of V* are called cotangent vectors at p. A coordinate system is a map from a subset of the manifold into ##\mathbb R^n##. It can be shown that every coordinate system with p in its domain can be used to define an ordered basis for the tangent space at p. So a change of coordinates induces a change of ordered basis.
     
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