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New Member question about Tensors

  1. Nov 13, 2006 #1

    No doubt this question has been asked by many before. I am new to the study of Relativity. I have spent several months reading this and that and generally getting very confused.

    One thing that I just can't get passed is understanding even what the heck tensors are. Every single book or article that I have found that calls itself an introduction starts with a definition which just reads to my limited understanding as complete double dutch. Can anyone help explain in completely non mathematical terms what tensors are?


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  3. Nov 13, 2006 #2


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  4. Nov 13, 2006 #3


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    But tensors are mathematical objects. How do you explain them in completely non-mathematical terms?

    Do you know what scalars, vectors and matrices are?
  5. Nov 13, 2006 #4


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    You can think of tensors as being objects (a lot like the objects in object-oriented programming) with the following properties.

    1) They are defined at one point (and its nearby neighborhood).

    2) They have methods which allow them to transform themselves into different coordinates. This makes tensors independent of coordinates. The new coordinates may in general be moving or even accelerating with respect to the old coordinates.

    An example of things which are and are not tensors may helpl.

    The electric field is not a tensor. It is incomplete - if you know the electric field at a point, you do not know and cannot know the electric field at that point from the point of view of a moving observer.

    The magnetic field is also not a tensor.

    The electromagnetic field CAN be represented by a tensor, the Faraday tensor, which combines information on both the electric and magnetic fields into one "object". When you know both the electric and magnetic fields at a point, you can compute both of them at that same point from the POV of any moving observer.

    Note that a moving electric field generates a mangetic field, and a moving magnetic field generates an electric field. This is why neither one of them is a "complete" object alone.

    These are probably the most important quantities of tensors. Another important quality is more difficult to define unless one uses mathematics - they are linear. This was important to the electromagnetic example above, but I won't go into that in detail as this is just a very rough overview.

    Because they are linear, they are a bit like the matrices of linear algebra. In fact, matrices are rank-2 tensors (and vectors are rank 1 tensors). Tensors are more general than matrices and vectors, though, because tensors of up to rank 4 are needed in general relativity.

    This is probably a little over-simplified, but it should give you some general idea of what tensors are about. Mathematically, as the books you've read no doubt indicate, they are defined by their transformation properties.

    To understand tensors at a mathematical level, I would suggest reviewing linear algebra, with a close attention to the duals of vectors. Duality is a key concept.

    You'll definitely need calculus - if by some chance you don't have calculus yet, start learning it first (before tensors).
  6. Nov 13, 2006 #5
    Googled up what looks like a pretty good discussion:


    A good example to study is the stress tensor from continuum mechanics. This is a linear function that, given a normal to a element of surface area, gives the net force per unit area across that element. It's a linear function from vectors to vectors, or on pairs of vectors to scalars.
  7. Nov 13, 2006 #6


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    I think some of these statements need some clarification to be taken literally.

    The electric and magnetic fields are tensors in [an observer's] three-dimensional Euclidean space... as one studies in any intro physics textbook. We usually refer to them as Electric Field Vectors and Magnetic Field Vectors... these vectors are particular types of tensors in 3D Euclidean space.

    There is no observer-independent electric field tensor in four-dimensional spacetime, and similarly for a magnetic field tensor.

    Given THE electromagnetic field tensor [tex]F_{ab}[/tex] and an observer [tex]u^a[/tex], that "electric field according to that observer" is [tex]E_b=u^aF_{ab}[/tex] is a tensor-field in spacetime (which is everywhere orthogonal to [tex]u^a [/tex]).... however, it's observer-dependent (since it depends on [tex]u^a[/tex]).
  8. Nov 14, 2006 #7


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    Would it help if I qualfiied my statements with "in special relativity, the electric field alone is not a tensor"?

    I do want to make my illustrative example as non-controversial as possible. At low velocities, one can think of the electric field as being independent of the state of motion of the observer, but at high (relativistic) velocities this just doesn't work - for example, someone sitting stationary next to a charge will measure one value of the electric field, but someone moving at a relativistic velocity in a direction transverse from the charge will measure a different electric field (the transverse field will be boosted by a factor of gamma).

    Thus I would argue that the electric field in isolation is not actually a tensor, though it's treated as one in some low-velocity approximations.
  9. Nov 14, 2006 #8


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    One key point of my comment is that there is an underlying vector space associated with a tensor. So, that needs to be defined...at least implicitly. That's why I was trying to distinguish between, say, 3-dimensional Euclidean space and (3+1)-Minkowski spacetime. Once established, we can then define structures that model physical quantities. Because of the ambiguities in language, especially between purely Newtonian-like spatial-thinking and relativistic spacetime-thinking, some unqualified statements may be misinterpreted.
  10. Nov 14, 2006 #9

    I've been reading the various responces to my question. Thank you for the various links - I'd come across one of them before but had lost it somewhere along the way.

    The thing that has become clear to me is that there does not seem to be a single simple answer to the question. That's Ok because I had been wondering whether I was just missing some obvious point.

    Well I have got a fairly good background in calculus and with the various books I've already got, together with these new PDF files I've downloaded I reckon that I've got enough to be going on with for the moment. So thanks again.


  11. Nov 14, 2006 #10
    The single and right answer the question is that tensors are multilinear mappings (formed by tensor products of vectors and dual vectors) from a vector space to real numbers.

    Important here is to notice, that vectors are linear mappings from the dual vector space to real numbers and vv. Tensor product of two dual vectors map the vectors of the cartesian product of twovector spaces to real numbers i.e.
    if v_1 \in V_1 and v_2 \in V_2 and y^1 \in V*_1, y^2 \in V*_2 we have

    y_1 x y_2 (v_1,v_2) = y^1(v_1)y^2(v_2), where x is the tensor product and
    * means the dual of the space.

    Now it is easy to think about the metric tensor as the symmetric bilinear form of the vector (tangent) space (bilinear means two components and form means it is made of dual vectors).
    First thing is to make the tensor symmetric and for this we use the notation y^i y^j =y^i x y^j + y^j x y^i (<- the ultimate and unbeatable symmetrization procedure).
    As we all know form somewhere, y^i(v_i) = 1. To add some structure, we can stay inside the definition by adding some linear rule for the value of these products, for example:
    g(t,t) = g_ij y^i(t) y^j(t), where g_ij is a matrix telling the values for the component mappings. If we choose g_ij= id_3 i.e. the euclidean case, we have:
    g(t,t) = g_ij y^i(t) y^j(t) = g_ij y^i(t^k v_k)y^j(t^l v_l) = g_11v_1v_1+g_22v_2v_2+g_33v_3v_3 = the usual well known inner product (y^i(t^k(v_k)) = t_i).

    What? Soo, g(t,t) is the tensor formed by the tensor products of the dual vectors and is a map from V x V (x is the cartesian product) to real numbers. (btw. some weird people call the cartesian product a tensor product).

    And so on. And be careful, some bs even in graduate level gravity books.
    Last edited: Nov 14, 2006
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