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Math Newb Wants to know what a Tensor is

  1. Jul 20, 2004 #1
    Math "Newb" Wants to know what a Tensor is

    Hey, I'll be entering my senior year of High School next year and this summer I'm taking Multivariable Calculus at UCLA. In September I'll start AP Stats.

    Anyway, what is a Tensor? I've always wondered this question. From what I've gathered, It's a Vector with two bits of information.

    What are they useful for. How do you "play" with them?

    Thanks, guys!
     
  2. jcsd
  3. Jul 20, 2004 #2

    HallsofIvy

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    A "tensor" is a generalization of a "vector". The crucial point about tensors (as well as vectors) is that they change "homogeneously" under change of coordinates. Specifically, if a tensor is all zeroes in one coordinate system then it is zero in all possible coordinate systems. That means that if the equation A= B (where A and B are coordinates) is true in one coordinate system (A= B is the same as A-B=0) then it is true in any coordinate system. Since (obviously) a physical law does not depend upon an (arbitrary) choice of coordinates system, it follows that physical laws must be expressed in terms of tensors.
     
  4. Jul 20, 2004 #3

    Janitor

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    HallsofIvy nicely captured the main idea. As he could also tell you, a vector is a special case of tensor, namely a vector is a tensor of rank 1. A scalar can be viewed as a tensor of rank zero.
     
  5. Jul 21, 2004 #4

    jcsd

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    For a mathematical object to be a tensor it must obey certain relationships when chgnaging between coordinate systems (as Halls of Ivy says tensor analysis is essientially revolves aropuind the idea that physical laws should be indepedndet of coordiante system). The exact relationship it obeys depends on it's covariant and contravariant orders.

    As Janitor says scalars can be considered tensors of rank 0 and vectors are tensors of rank 1. As you know a scalar is just a single number where as an n-dimensional vector vector can be fully described with n numbers, for example:

    A = 2i - 3j + 5k

    Fully desccribes the three vector A (in terms of it's rectangular component vectors).

    The number of components of a tensor of N diemsions and rank p (i.e the numbers needed to describe a tensor) is equal to Np (thpough of course there's nothing to stop some or all of these components from being zero)
     
    Last edited: Jul 21, 2004
  6. Jul 22, 2004 #5
    I spoke with my aunt's husband who has a degree in engineering as well as read up on this in a physics book. My understanding is that a tensor is sort of like a vector, where the magnitude is a function of the direction.

    The example he gave me was the force acting on a surface with a force that makes some angle theta with the surface. More specifically, the component of force that acts against the surface (the perpendicular component of the force) is a vector but its magnitude depends on direction (angle), more specifically, for example the force of gravity on an incline that works against the ground is the weight of the object times the cosine of the angle.
     
  7. Jul 22, 2004 #6

    chroot

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    Even more generally, a tensor is a sort of mathematical machine that takes one or more vectors and produces another vector or number.

    A tensor of rank (0,2), often just called rank 2, is a machine that chews up two vectors and spits out a number.

    A tensor of rank (0,3) takes three vectors and produces a number.

    A tensor of rank (1,2) takes two vectors and produces another vector.

    Hopefully you see the pattern.

    You actually already know what a (1,1) tensor is -- it's nothing more than a good ol' matrix. It accepts one vector and produces another vector.

    If you're working in three dimensions, a (1,1) tensor can be represented by its nine components. Here's a simple (1,1) tensor.

    [tex]
    T = \left(
    \begin{array}{ccc}
    1 & 0 & 0\\
    0 & 1 & 0\\
    0 & 0 & 1
    \end{array}
    \right)
    [/tex]

    You already know what this guy does -- it takes a vector and gives you back the exact same vector. It's the identity matrix, of course. You would use it as such:

    [tex]\vec v = T \vec v[/tex]

    If the world were full of nothing but (1,1) tensors, it'd be pretty easy to remember what T means. However, there are many different kinds of tensors, so we need a notation that will help us remember what kind of tensor T is. We normally use something "abstract index notation," which sounds more difficult than it is. Here's our (1,1) tensor, our identity matrix, laid out in all its regalia:

    [tex]T^a_b[/tex]

    The a and b are referred to as indices. The one on the bottom indicates the tensor takes one vector as "input." The one of the top indicates it produces one vector as "output."

    Tensors don't have to accept just vectors or produce just vectors -- vectors are themselves just a type of tensor. Vectors are tensors of type (0,1). In full generality, tensors can accept other tensors, and produce new tensors. Here are some complicated tensors:

    [tex]
    R^a{}_{bcd}\ \ \ \ G_{ab}
    [/tex]

    The second one, [itex]G_{ab}[/itex] is a neat one to understand. You should already understand from its indices that it is a type (0,2) tensor, which means it accepts two vectors as input and produces a number as output. It's called the metric tensor, and represents an operation you already know very well -- the dot product of two vectors!

    In normal Euclidean 3-space, [itex]G_{ab}[/itex] is just the identity matrix. You can easily demonstrate the following statement is true by doing the matrix multiplication by hand:

    [tex]\vec u \cdot \vec v = G_{ab} \vec u \vec v[/tex]

    The metric tensor is more complicated in different spaces. For example, in curved space, it's certainly not the identity matrix anymore -- which means the vector dot product is no longer what you're used to either when you're near a black hole. Tensors are used extensively in a subject called differential geometry, which deals with, among other topics, curved spaces. General relativity, Einstein's very successful theory which explains gravity as the curvature of space, is cast in the language of differential geometry.

    So there you have it: tensors are the generalization of vectors and matrices and even scalars. (Scalars, by the way, are considered to be type (0,0) tensors.)

    I should mention that there not all mathematical objects with indices are tensors -- a tensor is a specific sort of object that has the transformation properties described by others in this thread. To be called a tensor, an object much transform like a tensor. Don't worry though, you're not going to run into such objects very often.

    - Warren
     
  8. Jul 23, 2004 #7

    mathwonk

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    wow! this is the simplest clearest explanation of tensors i have seen. thank you.
     
  9. Jul 24, 2004 #8
  10. Jul 24, 2004 #9
    That's the definition of a general tensor. You neglected to add that the machine must be linear. There are also general tensors which accepts one-forms as an arguement.

    An (M,N) tensor is a linear function of M one-forms and N-vectors into the real numbers.

    An affine tensor is another kind of tensor. Examples of affine tensors are Cartesian tensors and Lorentz tensors. These are defined according tothow their components transform under a restricted transformation in a given space. E.g. the transfrormation relavent to a Cartesian tensor is the orthogonal transformation. See -- http://www.geocities.com/physics_world/ma/orthog_trans.htm

    The transformation relavent to a Lorentz tensor is the Lorentz transformation.

    For details see

    Tensors, Differential Forms, and Variational Principles, Lovelock & Rund, Dover Pub., (1989)

    Pete
     
  11. Jul 24, 2004 #10

    chroot

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    Of course I left out a bit pmb, I wasn't trying to get into the concepts of one-forms versus vector fields just yet. :smile: Thanks for the additions, though.

    - Warren
     
  12. Jul 24, 2004 #11
    You're welcome chroot. I had a feeling you were being sparse intentionaly but wanted to give the StonedPanda more of the details just in case he was overly stoned and needed that much more. :biggrin:

    The part about affine tensors I find to be an important addition because not knowing the use of the different kinds of tensors can be confusing. For instance: In D'Inverno's text Introducing Einstein's Relativity even the author goofed up on this subtle point. I.e. D'Inverno defines tensors in a manner which is equivalent to how you have defined them, i.e. as general tensors (by "equivalent" I refer to the definition of tensor as that whose components transform in a particular way). D'Inverno then goes on to define the angular momentum tensor (an important tensor in SR) on page 118

    [tex]L^{\mu\nu} = x^{\mu}p^{\nu} - x^{\nu}p^{\mu} [/tex]

    This is certainly not a general tensor since xu isn't a general tensor. However it is a Lorentz tensor and therefore Luv is a Lorentz tensor and not a general tensor.

    Pete
     
    Last edited: Jul 24, 2004
  13. Jul 26, 2004 #12

    chroot

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  14. Aug 13, 2004 #13
    I definitely picked the wrong name for this forum!

    But thanks!

    At what level or mathematics does one encounter tensors?
     
  15. Aug 19, 2004 #14
    Is there any chance you could enlighten me here - I follow what was said earlier regarding sub/superscript representing the vectors that are input and output. But, in the equation quoted there appears to be no input vector, but an output - what is going on there?

    I'm basically asking for a clarification of the meaning of the notation when you just have superscripts, not for any particularly detailed mathematical explanation, as some of these maths threads are way over my head.

    The other thing is I'm a bit unclear on the exact meanings of covariant and contravariant. Does anyone know a website, or can give an explanation where these terms are explained geometrically rather than in mathspeak. (To put it in context here - I love physics, but I'm not particularly good at maths when it is explained to me in terms of maths.... I guess I just need to see what is happening physically when these terms are defined.)
     
  16. Aug 19, 2004 #15

    chroot

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    A tensor with just a single superscript is nothing more than an ordinary vector.

    [tex]x^\mu = \left(
    \begin{array}{c}
    x^0\\
    x^1\\
    x^2\\
    x^3
    \end{array}
    \right)[/tex]

    where [itex]x^0[/itex] and so on are just numbers (scalars).

    The contravariant vector is the normal vector you're used to working with. Covariant vectors are dual to contravariant vectors. Contravariant vectors are column vectors, while variant vectors are row vectors. If you multiply a covariant vector by a contravariant vector, you get a number out:

    [tex]x^\mu x_\mu = a[/tex]

    - Warren
     
    Last edited: Aug 20, 2004
  17. Aug 20, 2004 #16
    Note: Scalars are defined as tensors of rank one. All scalars are numbers, but not all numbers are scalars.

    Pete
     
    Last edited by a moderator: Aug 20, 2004
  18. Aug 20, 2004 #17

    chroot

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    I'm actually interested in MoonUnit's last paragraph:
    I'm not too clear on the geometrical meaning myself, though I can go through the motions and manipulate tensor expressions just fine.

    A contravariant vector exists in the tangent space of a specific point in the manifold being considered. In other words, if you have a basketball (the surface of which is a 2D manifold), and you glue little toothpicks tangent to it, each toothpick is a contravariant vector defined at the point it is glued to the basketball. This much makes intuitive sense to me -- I can just as easily think "tangent" whenever someone says "contravariant." If you take any point on the basketball, the set of all tangent vectors you could glue to it there forms a (2-dimensional) tangent space at that point. Contravariant vectors at that point belong to that tangent space.

    Now, I know covariant vectors live in cotangent spaces, but I'm not really clear on how to visualize a cotangent space. I have read most of John Baez' book "Gauge Fields, Knots, and Quantum Gravity," in which he makes an earnest attempt to help the reader visualize a covariant vector -- but it falls flat on me. I just can't understand what he's trying to say.

    Does anyone have a clear explanation of how to "visualize" a covariant vector? Is it really even possible to visualize it?

    - Warren
     
  19. Aug 20, 2004 #18

    robphy

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    A covector can be visualized as a stack of parallel-planes. (Generally, a pair of parallel planes is sufficient. Visualize "twin blades".)
    The spacing between planes is inversely-proportional to the size of the covector.
    Think: local approximations of "equipotential surfaces".

    The contraction of a vector and a covector can be interpreted in terms of the number of piercings (MTW's "bongs of a bell") of the stack by the vector.
     
    Last edited: Aug 20, 2004
  20. Aug 20, 2004 #19

    chroot

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    robphy,

    That's basically how Baez tried to explain it, and I basically still don't get it. Where are these parallel (hyper)planes supposed to be? Equipotential of what?

    Are the hyperplanes of the covariant vector perpendicular to the contravariant vector?

    - Warren
     
  21. Aug 20, 2004 #20

    robphy

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    Take the electrostatic field as a physical application.

    The electric potential [itex]\phi [/itex] is a scalar field.
    The object [itex]-\nabla_a \phi [/itex] is covector field (a field of one-forms).
    Visually, the covectors are these twin-blades tangent to the equipotentials of [itex]\phi [/itex].

    Consider a displacement vector [itex]d^a[/itex].
    Suppose that [itex]d^a[/itex] is parallel to an equipotential. The contraction [itex]-\nabla_a \phi d^a[/itex] (that is [itex]\vec E\cdot \vec d[/itex]) is zero. Thus, the potential difference from tail to tip is zero.

    Suppose that [itex]d^a[/itex] is perpendicular to an equipotential (i.e. parallel to the gradient vector). The contraction [itex]-\nabla_a \phi d^a[/itex] (that is [itex]\vec E\cdot \vec d[/itex]) is nonzero because [itex]d^a[/itex] pierces some planes in the stack. Thus, the potential difference from tail to tip is nonzero.

    With [itex]d^a[/itex] fixed, suppose that electric potential [itex]\phi[/itex] is doubled in strength. We expect the potential difference from tail to tip to double. To obtain twice the number of piercings for the same [itex]d^a[/itex], the spacing between the twin-blades must be halved.

    Here is a nice article by Jancewicz:
    http://arxiv.org/abs/gr-qc/9807044

    Other tensors (e.g., differential forms) can be visualized similarly.
     
    Last edited: Aug 20, 2004
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