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Dummy's definition of a tensor?

  1. Jul 19, 2005 #1
    Can someone please give me a dummy's definition of a tensor? We always use the stress/strain tensor in my structures class and the inertia tensor in my dynamics class. The only thing I really see between the both of these is that they are symmetric 3x3 matrices, so whats so special about them that makes them tensors?

    Here's what I know

    A scalar is tensor of order 0, a vector has order 1, and a matrix has order 2

    Also I think I remember seeing something about how tensors can always be transformed to a different frame but retain some property? This is consistent with all the transformations we do on these matrices. But, is the fact that they are symmetric matrices that can be transformed to different basis the reason why they are tensors?
  2. jcsd
  3. Jul 19, 2005 #2


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    this has been answered 200987 times on here by now. so you can probably find the answer quicker by searching than you can find someone wanting to answer it again. i myself have written zillions of pages on it from every possible viewpoint and level of spohistication. so you could probably just peruse my posts.
  4. Jul 19, 2005 #3


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    here is my original answer, posted almost one year ago.

    "........ In algebra, a tensor is just a way of multiplying vectors. So the dot product is one example of a tensor. hence you have already seen a tensor.

    Moreover note that a taylor series approximates a function by polynomials of various orders, of which the first order term is linear, i.e. a vector. The second and higher order terms are bilinear, and trilinear, and so on, since they involve multiplying 2 or 3 or more vectors.

    Recall that the curvature of a curve involves the second derivative. Thus in higher dimensions, curvature is a tensor. E.g. the curvature of a surface is a sort of product defined on tangent vectors to the surface."
    Last edited: Jul 19, 2005
  5. Jul 19, 2005 #4


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    here is another attempt, only 4 months ago:

    vectors are to tensors, as linear polynomials are to higher degree polynomials, except with tensors multiplication is also allowed to be non commnutative.

    for example the dot product, a second degree polynomial in the entries of the two vector arguments, is a symmetric (i.e. commutative) tensor.

    "short course on tensors:

    let elements of R^2 be called vectors, and write them as length 2 column vectors.

    then we can define "covectors" as row vectors of length 2. then row vectors may be considered as linear functions on vectors, i.e. as linear functions f:R^2-->R.

    some people call vectors, tensors of rank 1 and type (1,0), and call covectors, tensors of rank one and type (0,1).

    now consider two covectors f and g. they yield a bilinear function f(tensor)g of two vector variables, by multiplication, i.e. f(tensor)g (v,w) = f(v)g(w), a number.

    this product is not commutative since g(tensor)f (v,w) = g(v)f(w).

    for the same reason it gives a different answer when applied to (w,v), as compared to when applied to (v,w).

    some people call f(tensor)g a rank 2 tensor of type (0,2).

    if we add up several such products, e.g. f(tensor)g + h(tensor)k, we still have a bilinear function of two vector variables, hence another rank 2 tensor of type (0,2).

    now we could consider also a product v(tensor)f, of a vector and a covector. if we apply this to a vector w we get a vector: namely v times the scalar f(w).

    again a sum of such things is another: v(tensor)f + u (tensor)g, applied to w is

    f(w) times v + g(w) times u.

    some people call such a thing a rank 2 tensor of type (1,1).

    since as a function on the vector w, this object is linear, it could be represented as a 2 by 2 matrix, whose columns were the vector values taken by this function at the standard basis vectors (1,0) and (0,1).

    the ordinary dot product is a tensor of rank 2 and type (0,2), since it takes two vectors and gives out a number, and is bilinear.

    i.e. if f is the linear function taking the vector (x,y) to x, and g is the linear function taking the vector (x,y) to y, then the dot product equals f(tensor)f + g(tensor)g.

    I.e. applied to (v,w), where v = (v1,v2) and w = (w1,w2) are vectors, it gives us v1w1 + v2w2. thus it is symmetric.

    the reason for distinguishing the vector variables from the covector variables, is that under a linear transformation T:R^2-->R^2, the vectors transform by v goes to Tv, and the covectors transform by f goes to fT, i.e. the multiplication occurs on the other side.

    or if you insist on writing a row vector, i.e. a covector as a column vector, then you must multiply it (from the left) by T* = transpose of T.

    multiplying more vectors and covectors together, and adding up, gives higher rank tensors.

    what i have described are tensor spaces "at a point". just as the tangent space to a sphere, say at one point, is a vector space, so also there are tensor spaces at every point of the sphere.

    then just as we can consider familes of tangent vectors, or tangent fileds, we can consider tensor fields, one tensor at each point.

    was that simple enough for you? thats about as simple as i can make it, and still be correct."

    I might add some people made it seem even simpler by omitting the meanings of the symbols and giving only the rules for their manipulation.
    Last edited: Jul 19, 2005
  6. Jul 19, 2005 #5


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    chroot manages to give both meanings and symbols in post 6 of the thread "Math "Newb" Wants to know what a Tensor is
    " july 20, 2004.
  7. Jul 20, 2005 #6


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    i'm sorry i was so impatient young person, did that help at all getting you started?
  8. Jul 21, 2005 #7
    aber...02 sounds lke an engineer and I find it's better to introduce engineers to tensors like this:

    Think of a body subject to a force vector. The force has a component 'scalar' in each direction in space. In 3-D space the vector has 3 'scalar' components.

    Now think of a body subject to a (3-D) stress. Slice throught the body with a cutting plane. The stress has a component *vector* across the cutting plane. A quantity that has a component vector in each direction in space is called a (2nd order) tensor. To define the stress state completely requires three component vectors - one associated with of three mutually orthogonal cutting planes. Each of the three component vectors has three component 'scalars', so the stress tensor can be resolved into nine component 'scalars'.

    So a 2nd order tensor is a kind of higher order vector and, as you mention, there is a hierarchy of different order tensors.

    Like a vector, a 2nd order tensor can be *represented* by a single letter (Gibb's notation) or by a matrix of component 'scalars'. Also, like a vector, it has properties that are independent of the coordinate system we chose to represent it. These properties are called invariants. The length of a vector is an invariant and doesn't change when we change our viewpoint. Tensors such as stress also have invariants which don't change when we change our viewpoint. This is a GOOD THING - it would be unfortunate if a bridge were to fall down simply because we were to change our viewpoint.

    The three principal stresses are the most familiar invariants of the stress tensor - these control fracture. They can however be combined to give other invariants: J2, for example, controls plasticity.

    Engineering stress and strain are almost always symmetric but not all tensors are. Vorticity, for example, is anti-symmetric.

    Now, if you want to put some maths on this, try:


    (sorry about the formatting though - I'm tidying this up for next year)
    Last edited: Jul 21, 2005
  9. Jul 21, 2005 #8


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    The basic concept of a tensor is that the numbers representing it change "homogeneously" with a change of coordinate system (the components in the new coordinate system is a sum of products of the components in the old coordinate system). In particular that means that if a tensor has all 0 components in some coordinate system then it has 0 coordinates in any coordinate system. As a result of that, if A= B in some coordinate system, then A-B= 0 in that coordinate system and so A-B= 0 => A= B in any coordinate system. As long as quantities are written in terms of tensors, any equation that is true in one coordinate system is true in any coordinate system.
  10. Jul 21, 2005 #9
    A tensor is a multilinear mapping from vectors and covectors to the reals

    [tex]T: V \times V \times V \times ... V^* \times V^* \times V^* \times ... \rightarrow \mathbb{R}[/tex]

    such that the tensor mapping is invariant with respect to the coordinate system chosen.

    As mathwonk has pointed out, the familiar dot product is a tensor:

    [tex]T = dx^i \otimes dx^j[/tex]

    [tex]T(\vec{u}, \vec{v}) = dx^i (\vec{u}) dx^j (\vec{v}) = u^i v^j[/tex]

    it can be shown that if you change your coordinate system, your vector representation and your tensor representation will both change such that you get the same value for your dot product. this should be familiar to you, if you have taken linear algebra before.

    the motivation for all of this is that we desire our vectors, being geometric objects, to be the same no matter the coordinate system...the tensor is a generalization of this idea. take the area of a unit square: does the area depend upon the coordinate system? of course not, so why should we settle for less when it comes to vectors?

    it is for this reason that tensors have found wide use in physics, since it is important that the physical laws not depend upon coordinate system.
    Last edited: Jul 21, 2005
  11. Jul 21, 2005 #10


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    what puzzles me is that although i understod psot 4, (my own), it did not seem familiar to me, i.e. i do not remember writing it, nor of understanding tensors that clearly myself. perhaps this is an attribute, or deficit, of advancing age.

    ialos appreciate thew wonderful motivation added by others, of physical and geometric concepts that lend thjemselves to measurement by tensors, like area, stress, length, curvature,....
  12. Jul 21, 2005 #11


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    One aspect that hasn't been explicitly brought up (which might be helpful to the original poster) is that tensors may be used to describe anisotropy. For example, for spherically symmetric objects, the angular momentum vector [tex]\vec L[/tex] is proportional to the angular velocity vector [tex]\vec \omega[/tex] according to [tex]\vec L = I \vec\omega[/tex], where [tex]I[/tex] is the moment of inertia of the sphere (considered as a scalar in this simple case).

    To generalize this to more general objects, the moment of inertia must be a tensor, sometimes denoted [tex]\tilde I[/tex]. Now, a new situation can occur... [tex]\vec L=\tilde I\vec \omega[/tex] need not be proportional (in particular, parallel) to [tex]\vec \omega[/tex]... although (by linearity) doubling [tex]\vec \omega[/tex] will double [tex]\vec L[/tex].

    Just to continue the story a little more... it may turn out that, for some choices of [tex]\vec \omega[/tex], you find that [tex]\tilde I\vec \omega[/tex] is parallel to [tex]\vec \omega[/tex]. In this case, you've stumbled upon a "special" direction associated with [tex]\tilde I [/tex]. Then [tex]\vec \omega[/tex] is called an eigenvector (or, making contact with rdt2's post, "principal axis") of [tex]\tilde I [/tex]. Often calculations are simpler by choosing "coordinate axes" along these special directions (although you can still solve the problem in any coordinate system). These special directions are associated with symmetries in your problem.

    One can play a similar game with electrical conductivity in the microscopic version of Ohm's Law [tex]\vec \jmath=\tilde \sigma \vec E[/tex].
  13. Dec 25, 2011 #12
    Re: Tensors

    thnak you very much i really started to understand the meaning of tensors
  14. Dec 25, 2011 #13


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    Re: Tensors

    In physics a tensor is a quantity whose measurement is independent of the frame of reference. Observers in different frames will come up with the same answer. In General Relativity the basic tensor is intrinsic time which each observer calculates from different measurements but these measurements always combine to give the same quantity.

    The mathematics of tensors are as Mathwonk described.
  15. Dec 27, 2011 #14


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    Re: Tensors

    This helped me understand a bit about tensors : multilinear objects are difficult to work with, while linear objects are more easily understood. Tensor products ( in the case of vector spaces) create a new vector space Z:=V(x)W out of two vector spaces VxW , so that for every vector space Y, every bilinear map L:VxW-->Y becomes a linear map
    on V(X)W, into Y. In this sense, a tensor is the representative of a class of bilinear maps. This construction extends to product of n vector spaces.

    There are similar constructions for tensor products of other mathematical objects, all of them "universal constructions" , giving you a new space in which maps into a third space Y are simplified.

    @rdt2: Your link seems to be broken.
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