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Electric and magnetic constants are tensors

  1. Jan 9, 2006 #1
    What a tensor is .? I have found a text in my book that says that the electric and magnetic constants are tensors.. Do u have something in mind?
    Thx a lot
  2. jcsd
  3. Jan 9, 2006 #2
    A tensor is simply a multilinear map (a map thats linear in each variable) from a vector space and the dual of the vector space to the Reals.
    A very simple example is the dot product. It takes in two 2 vectors and gives a Real number.It is linear in both varibales. Thus the dot product is a (0 2) tensor.
  4. Jan 9, 2006 #3
    I am not sure i have completely understabd
  5. Jan 10, 2006 #4
    No, that is not right. The dot product is not a tensor, nor is the result of a dot product a (0,2) tensor-it is a (0,0) tensor a.k.a. scalar.

    You know that we can take several numbers and form a vector. Simmiliarly we can take N vectors of length N and produce an N by N matrix. One way we could do this is like this:
    We have a vector, [itex]\vec{A}[/itex] whose components in some coordinate system are [itex]A_1, A_2, A_3...A_N[/itex]. When I write Ai, I mean some particular component of [itex]\vec{A}[/itex]. Also we have the vector [itex]\vec{B}[/itex] with components [itex]B_1, B_2, B_3...B_N[/itex]. Now we can form the matrix D by saying that the element in the i'th row and the jth collumn is:
    Similarly, I could create an object with three indices:
    And so on. Are these tensors? Not necessarily. Notice that the components of the vectors were defined with respect to some coordinate system. We have said nothing about these components would change if we were to change the coordinate system. What makes a tensor a tensor is the way its components change when you change coordinate systems. The rank of a tensor is the number of indices required to specify its compnents.

    If we are talking about rectangular coordinates then the components transform in the obvious way: First the components are written as the projections of the vector on each of the basis vectors. When you change coordinate systems you are changing basis vectors, so now you just find the new components along the new basis vectors. This transormation can be represented as a matrix. If the components of this matrix are [itex]a_{ij}[/itex] and the components of [itex]\vec{A}[/itex] in the new coordinate system are written as [itex]\hat{A_1}, \hat{A_2}, \hat{A_3}...\hat{A_N}[/itex], then the new coordinates are related to the old by:
    [tex]\hat{A_i}=\sum_{j=1}^n a_{ij}A_j[/tex]
    For a tensor of rank greater than one we have:
    [tex]\hat{D_{ik}}=\sum_{j=1}^n \sum_{m=1}^n a_{ij}a_{km}D_{jm}[/tex]
    This defines the transformation laws tensors must satisfy in rectangular geometry. The law is generalized to tensors of higher rank in the obvious way.

    Now when we move to curvilinear coordinates the situation becomes more complicated. There end up being two transformation laws that are useful at various times. One is the contravariant transformation law, and the other is the covariant transformation law. Tensors can be 'mixed' in the sense that they have some components that follow one transformation and some that follow the other. The situation is complicated by the fact that the basis vectors vary from point to point in curvilinear coordinates. If you want to know about how these concepts are generalized, I will explain, but it will take a while.
  6. Jan 10, 2006 #5
    I think you are getting confused between what a tensor is, and what the COMPONENTS of a tensor are. Things like [tex]E_{ijk}[/tex] notationaly refer to the components of a tensor (0 3) tensor E, and not the tensor itself. The dot product is a linear FUNCTION on two vectors and hence IS a (0 2) tensor. Tensors are multilinear FUNCTIONS, that's it.
  7. Jan 10, 2006 #6


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    Let's start with a linear algebra review.

    You know about vectors. Hopefully you're comfortable with abstract vectors spaces, but I'm just going to work with n-tuples for now.

    For this entire post, I'll assume we're working up from an n-dimensional vector space.

    So, a vector is simply an nx1 matrix. It has n rows and 1 column.

    Then, you have covectors (a.k.a. dual vectors). In this setting, a covector is simply a 1xn matrix.

    The important feature of a covector is

    (covector) * (vector) = (scalar)

    Then, we have nxn matrices. The important feature of nxn matrices is that

    (matrix) * (vector) = (vector)

    We also have some side benefits, though:

    (covector) * (matrix) = (covector)


    (covector) * (matrix) * (vector) = (scalar)

    We also have another interesting feature:

    (vector) * (covector) = (matrix)

    This is your first nontrivial example of a tensor product.

    You might wonder about other sorts of combinations such as:

    (????) * (vector) = (covector)

    Or even products with more than one term, like:

    (????) * (vector , vector) = (scalar)

    You've actually seen an example of this last thing: the dot product is a good example. Traditionally write the dot between the other two arguments:

    (vector) (dot) (vector) = (scalar)

    though someone fond of indices would actually write it like this:

    [tex]g_{ij} v^i w^j = s[/tex]

    Where g is the dot, v and w are the vectors, and s is the scalar. (In this notation, you could actually write the three terms in any order you choose -- the indices specify how they're "glued together" for the product)

    Anyways, in general, these more complicated things exist, and we call them tensors. A rank (p, q) tensor is something that takes p covectors and q vectors, and gives you a scalar.

    For example, an nxn matrix is a rank (1,1) tensor, since we could do:

    (covector) * (matrix) * (vector) = (scalar)

    to produce a scalar.

    A vector is a rank (1,0) tensor, since we can do:

    (covector) * (vector) = (scalar)

    And a covector is a rank (0,1) tensor for the same reason.

    Of course, just like matrices, we can put things together in all sorts of interesting ways. A rank (1,1) tensor can operate on a rank (1,0) tensor and produce a rank (1,0) tensor, and all sorts of other stuff.

    We can build higher-rank tensors out of lower-rank tensors. For example, remember the tensor product I mentioned before:

    (vector) * (covector) = (matrix)

    We've taken a (1,0) tensor and a (0,1) tensor and produced a (1,1) tensor!

    In general, given a rank (p,q) tensor and a rank (r,s) tensor, we can take their tensor product which is a rank (p+r,q+s) tensor.

    These things aren't all that fun to write out in full, but I'll give a simple example of a tensor product ([itex]\otimes[/itex] is the symbol for tensor product):

    \begin{array}{ccc} 2 & 3 & 5 \end{array}
    \begin{array}{ccc} 7 & 11 & 13 \end{array}
    14 & 22 & 26 & 21 & 33 & 39 & 35 & 55 & 65

    That last thing is supposed to be read as a "partitioned" matrix -- it a 1x3 matrix, whose entries are 1x3 matrices.

    (actually, I may have the terms of the product backwards -- I always forget what convention people like to use when writing these things)

    Michael_McGovern talked a lot about transformation laws. Contrary to what he says, such things are not of fundamental importance to the notion of a tensor -- they are merely a consistency check for a particular method of using tensors.

    As you might recall from talk about abstract vector spaces, you cannot talk about the components of a vector until you've selected a basis for that vector space. Then, you talked about comparing two different bases, and worked out a change of basis transformation.

    Most of what Michael_McGovern talked about is simply the change of basis transformations for tensors.

    To be fair, these things are important, because often times, physicists will construct the components of a tensor in some apparently basis-dependent manner -- but bases aren't supposed to matter to physics! So they have to carefully prove that their construction properly respects the change of basis transformations before they can use their tensors.

    Also, in physics, one is interested in tensor fields which is yet another layer of complexity! You've probably heard about vector fields (such as the [itex]\vec{E}[/itex] field) -- to each point of space you associate a vector. Well, you can do the same sort of thing with tensors, and have to worry about that.

    Since Michael_McGovern likes to do things in index notation (i.e. using coordinates), he has to select a basis. However, for technical reasons, just like a vector field assigns a vector to each point in space, you must also specify a basis for every point in space!

    Of course, if you are comfortable doing your linear algebra "abstractly" (i.e. you're happy manipulating the vector [itex]\vec{v}[/itex] as opposed to insisting on picking a basis and manipulating the n-tuple [itex](v_1, \ldots, v_n)[/itex]), you avoid just about all of the issues Michael_McGovern discussed in his post.

    (Yes, I am a big fan of doing it "abstractly", why do you ask? :wink:)

    But the point I wanted to make in these closing remarks is all of these additional concerns are only concerns about how people use tensors (especially physicists) -- they are not concerns inherent to the tensor concept itself.
    Last edited: Jan 10, 2006
  8. Jan 11, 2006 #7
    Tensor is no more than a symbol with princepal we use to express what we want express with many useful
  9. Jan 12, 2006 #8
    That's intersesting Hurkyl. That is a much better definition of the tensor than the one I learned. I was trying to learn this stuff on my own and I got one book that was more physics-oriented and one that was more pure math. They both defined tensors in terms of the way their components transformed. I do see the advantage of your approach.
  10. Aug 1, 2006 #9
    What are Tensors?

    Hi, I have been hearing/reading the word "tensor" a lot lately, but I have no idea what it is or what is it used for. I also googled for it but I get bogged down by so much coplicated mathematics that I am unable to make any sense of it. All I know that tensors have something to do with matrices and special relativity, no more no less. Could someone please just give me a gist of what tensors are?
  11. Aug 1, 2006 #10


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    yes, I implied this question in my vector calc thread. I didn't even bother to look it up, because I'm afraid I'll draw misconceptions from laymen explainations (as I have done in the past with relativity and quantum mechanics).
  12. Aug 1, 2006 #11


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    i think if you will search on here over the last few years you will find thousands of words written on this question. maybe one thread was called what is a tensor?

    i myself have answered this question uncountably many times.
  13. Aug 1, 2006 #12


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    you might look in the tensor forum.
  14. Aug 1, 2006 #13
  15. Aug 1, 2006 #14
    Hey you guys, I am trying to learn about tensors on my own this summer and I would be really glad if someone would recommend a good book(s) on them. Preferably a book which gives you a physicist/engineer's perspective on tensors (not a mathematicaian's).

    Thanks in advance.
  16. Aug 1, 2006 #15


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    think of a taylor series expanded at each point of a space. the constant terms are the values of the function at each point. the linear etrms are the differentials of the function at each point. these are first order tensors. then the second order taylor polynomials are second order approximations to the functions ate ach point. these are second order symmetric tensors. etc...

    there are also anti synmmetric tensors, like the 1-forms that one integrates over parametrized curves, and the 2 forms that one integrates over parametrized surfaces.

    and there are more complicated ones. in general they are multilinear combinations of tangent vectors and cotangent vectors.
  17. Aug 1, 2006 #16


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    I can't even begin to express the difficulty imagining this.

    To me, Taylor series is a bunch of equations that 'zoom in' on a slope, but my only practical application with taylor series was using a runge-kutta technique to eliminate error in a computational physics class...

    Other than that, it was a completely abstract equation that came at a tough time in math for me, where blatent memorization was my relief from the 'fire hose'*

    *I have a physics professor who states that Physics 211/212 are like asking a student to take a drink from a fire hose. It fits with my experience. Maybe it's a local thing, but having had no physics background before 211/212, it was a crazy year; it flew through all kinds of different branches of physics while I was learning calc 2 and 3 as well. I totally lacked soak time. A lot of my mathematical concepts are severely underdeveloped.
  18. Aug 1, 2006 #17


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    well start with one point.

    i.e. think of a polynomial in two variables and collect etrms of the same total degree,like 3 + (x-4y) + (x^2 +6xy -y^2) + (x^3 +xy^2 -y^3).

    the constant term, 3, is the zeroth order approximation. the lienar terms (x-4y) are the first order approximation, i.e. approximatiion by a first iorder symmetric tensor. (all a symetric tensor is, is a homogeneous polynomial).

    the next etrms ((x^2 +6xy -y^2), the second order approximation, are by a second order tensor.

    this polynomial is expanded in powers of X,Y.

    the hard
    part is when we try to expand it in powers of X-a, Y-b, for every point (a,b) in the plane. the resuklt is we get a wholelot of constants, a whole,lot of linear polynomials, a wholelot of quadratic polynomials, etc....

    people who do not know what tensors are, think the complicated notation that is used to express these latter families of objects are the tensors, and talk about families of coefficients with upper and lower indices as "being" tensors.

    from another point of view the simplest tensors are real vaued funtions, the next simplest are vector vakued functions, and the next simplest are matrix valued functions. etcc.....
  19. Aug 1, 2006 #18


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    lets get systematic. first take the local point of view. we discuss only covariant tensors in the classical sense, and symmetric ones at that. and let us stick to 2 variables.

    a covariant 0 tensor is a number. a field of these is a family of numbers, one at each point, i.e. a real valued function. ok?

    now a 1 cotensor is a linear polynomial like ax+by. so a field of 1 cotensors is a family of these, one at each point. this is harder to imagine, so just imagine ax+by, but where a and b are functions,

    i.e.a(p)x+b(p)y. so in a sense it is represented by the 2 coefficient functions, a(p) and b(p).

    now a (symmetric) 2 cotensor is a homogeneous quadratic polynomial like ax^2 + bxy + cy^2, and thus a field of them is the same thing where a,b,c are functions. so in a sense a symmetric 2 co tensor is represented by the three coefficient functions, a,b,c.

    now a 2 cotensor that may not be symmetric, will have also a yx term that cannot be combined with the xy term, so will look like ax^2 + bxy + cyx + dy^2, and a field of those is such a thing again where the coefficients are functions.

    thats about it. so the big deal I gave you at first was to imagine all these degrees at once. i.e. a symmetric tensor is just a (NOT NECESSARILY HOMOGENEOUS,oops sorry) polynomiaL.

    and a field of them is a polynomial same but with function coefficients. so a not necessarily symmetric cotensor is a sort of non commutative polynomial. and so on for a field of them, which then might be represented by a huge family of coefficient functions.

    thus for each n, a smooth function f defines a field of symmetric tensors with top degree n, namely the field of nth order taylor polynomials at each point. the associated family of functional coefficients is just the family of partials of f up to degree n.

    if we stick to a single homogeneous degree for our (still co-) tensors, we see there is one coefficient for each noncommutative monomial, i.e. in 2 variables there are the following eight degree 3 monomials xxx, xxy, xyy, yyy, yyx, yxx, xyx, yxy.

    now i will probably screw this up from my allergy to coordinates, but never fear many people love those best and will leap right in with help.

    but anyway, coordinate junkies will prefer to write something like (1,1,1) instead of xxx, and (1,1,2) instead of xxy, and so on, so they will describe an order three cotensor by giving only the coefficients of each term in the form {a(1,1,1), a(1,1,2),...etc...}, where since they always think of fields of cotensors, they will say the a(i,j,k) each represents a function.

    hence they will say that an order 3 cotensor is a family of functions of form {a(i,j,k)} and they will also give you rules for how to change coordinates. Indeed they must do this since they have not told you what the symbols mean, so you would have no way of figuring out for yourself how they transform.

    ok now i withdraw to a safe bunker to await comments from a classical tensor perspective:biggrin: .
    Last edited: Aug 1, 2006
  20. Aug 1, 2006 #19


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    oh really there are three stages in understanding tensors, tensors at a point, then families of tensors in R^n, then patching together such families to get families on a manifold. i have discussedonly the first two stages.

    the part many physicists leave out is stage one. they jump right into stages two and three, and hence do not understand what the objects are that they are globalizing, so are obliged to memorize the rulkes for changing coordinates instead of deriving them from conceptual definitions.

    this i not their fault of course since the books they read do not explain what is going on. my near hopeless lack of grasp of physics is also due to the books i read leaving that aspect out from the math.
  21. Aug 1, 2006 #20


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    Hey Hurkyl,

    conventionally, what is the result of

    [tex]\left[ \begin {array} {c} a \\ b \end {array} \right] \otimes \left[ \begin {array} {ccc} c & d \end {array} \right] [/tex]

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