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Tensors and Tensors

  1. Jan 5, 2006 #1
    Greetings Gurus,

    I'm sure this has been asked before, so feel free to redirect me.

    What is the connection between tensors in physics and tensoring in algebra? That is, could anyone sketch the path between them?


    Thanks,

    Kevin
     
  2. jcsd
  3. Jan 5, 2006 #2

    George Jones

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    Let me proceed by example.

    Let [itex]V[/itex] be an n-dimensional real vector and [itex]V*[/itex] be its algebraic dual, and suppose [itex]T: V \times V \rightarrow \mathbb{R}[/itex] is a bilinear mapping, i.e., [itex]T[/itex] is in [itex]V* \otimes V*[/itex].

    Let [itex]\left\{ e_{1}, \dots, e_{n} \right\}[/itex] be a basis for [itex]V[/itex]. This basis can be used to define an array of numbers:

    [tex]T_{ij} = T \left( e_{i}, e_{j} \right).[/tex]

    Let [itex]L: V \rightarrow V[/itex] be an invertible linear transformation on [itex]V[/itex]. Then [itex]\left\{ e'_{1}, \dots, e'_{n} \right\}[/itex] with [itex]e'_{i} = Le_{i}[/itex] is also a basis for [itex]V[/itex] that can be used to define an array of numbers:

    [tex]T'_{ij} = T \left( e'_{i}, e'_{j} \right).[/tex]

    Each member of the primed basis can be written as linear combintion of elements of the unprimed basis:

    [tex]e'_{i} = L^{j} {}_{i} e_{j}.[/tex]

    (I have used (and will use repeatedly) the summation convention, i.e., the repeated index j is summed over.)

    This gives

    [tex]T'_{ij} = T \left( e'_{i}, e'_{j} \right) = T \left( L^{k} {}_{i} e_{k}, L^{l} {}_{j} e_{l} \right) = L^{k} {}_{i} L^{l} {}_{j} T \left( e_{k}, e_{l} \right).[/tex]

    Thus,

    [tex]T'_{ij} = L^{k} {}_{i} L^{l} {}_{j} T_{kl}.[/tex]

    In physics, this expression is often used as the definition of a tensor.

    Now consider an n-dimensional real differentiable manifold and let [itex]T[/itex] be a tensor (of the same type as above) field. Suppose further that the bases above are coordinate tangent vector fields that arise from 2 overlapping charts:

    [tex]e_{i} =\frac{\partial}{\partial x^{i}}[/tex] and [tex]e'_{i} =\frac{\partial}{\partial x'^{i}}.[/tex]

    Then, by the chain rule, the change of basis relation is

    [tex]e'_{i} = L^{j} {}_{i} e_{j} = \frac{\partial x^{j}}{\partial x'^{i}} e_{j},[/tex]

    and

    [tex]T'_{ij} = \frac{\partial x^{k}}{\partial x'^{i}} \frac{\partial x^{l}}{\partial x'^{j}} T_{kl}.[/tex]

    This express is also often used as the definition of a tensor in physics.

    A "metric" tensor gives a natural isomorphism between tangent spaces and cotangents spaces, their algebraic duals. In physics, this is used for raising and lowering of indices. See post #52 in the Foundations of Relativity Thread.

    Regards,
    George
     
    Last edited: Jan 5, 2006
  4. Jan 5, 2006 #3

    mathwonk

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    thanks George. that is very clear. but i am trying to see what the implication is for the answer to the OP's question. it seems to argue that physicists are people who prefer symbols to words, i.e. writing L upper i lower j times L upper s lower t, to just saying "bilinear".
    This puzzles me, as in my experience physicists do love to discuss things in words. Is it that they only like to discuss physics in words, and math gets relegated to symbols?::confused:
     
  5. Jan 5, 2006 #4

    George Jones

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    Issues of this sort interest me, and I have observed the inteactions (or lack thereof) between mathematicians and physicists at a number of institutions over the years. It would take more time than I have available right now to organize my thoughts, but, for now, let say that many physicists would find the term 'bilinear' to be too "highbrow" for their tastes. This might seem strange to you, as the concept of multilinearity is almost physical.

    I hope to give a more detailed answer on the weekend.

    Regards,
    George
     
  6. Jan 8, 2006 #5

    George Jones

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    I have decided not to write about about all my observations, but I will make a few more comments.

    By "highbrow" I mean too abstract. Physcists prefer their own "down to Earth" version of mathematics - sometimes with good reason. I can't imagine teaching quantum mechanics using a rigorous version of the spectral theorem for unbounded selfadjoint operators, and I have had disappointing results presenting tensors from the multilinear point of view to physics students in the relativistic portion of a senior electrodynamics course.

    Too much mathematical rigor can sometime obscure (but also sometimes illuminate!) physical concepts, but too little rigor can cause confusion. As Roger Penrose says in his book The Road to Reality, the amount of rigor appropriate for physics is not at all clear.

    This leads to another question: How should mathematics be presented to physics students? Standing something that Hilbert said on its head, many physcists take the point of view that the teaching of mathematics is too important to be left to the mathematicians. I do not (completely) agree, and I feel that many physicists take this point of view to far to an extreme.

    This has resulted in different mathematical "languages", so that now phycists and mathematicians often find it difficult to talk to each other.

    At times, I find this very depressing. However, there is hope - there are numerous examples of good mathematicians and physicists who speak both languages, and who can talk to both cultures.

    Regards,
    George
     
  7. Jan 8, 2006 #6

    gvk

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    George, mathwonk
    You raise a very interested question about relation between mathematical and physical languages, but the physical meaning of tensor, which George posted is wrong, unfortunately.
     
  8. Jan 8, 2006 #7

    Hurkyl

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    If you're going to assert someone's wrong, you ought to give an explanation. :grumpy:
     
  9. Jan 8, 2006 #8

    mathwonk

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    well, as a famous physicist supposedly said, what matters is not whether a hypothesis is right or wrong, but that it gives rise to useful work.
     
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