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A vector whose components are vectors?

  1. Nov 26, 2014 #1

    BiGyElLoWhAt

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    This question just recently came up in a physics problem, but it's actually more of a math problem, so I'm going to ask it here.

    Let me provide some context.

    The HW was to show mathematically that the divergence of the curl was zero for the first part, and to explain it physically/conceptually.

    The assignments done, I'm not asking for help with that, but...

    I was thinking, the determinant grants an nth dimentional volume of the parallelipiped created by the vectors containing it, and this gets really interesting when you consider how we calculate curl (and cross products) with a determinant.

    Let's have a vector field in R^3 ##F:=<F_x,F_y,F_z>## or ## F_x \hat{i} +F_y \hat{j} + F_z \hat{k}##
    and we want to calculate the curl.
    ##\left | \begin{array} c
    \hat{i} &\hat{j} & \hat{k} \\
    \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
    F_x & F_y & F_z \\
    \end{array} \right |##

    Now the bottom line is just our vector field, expressed as a row vector, but each entry is the magnitude of each component. In other words, each entry (by itself) is a scalar.
    The middle row is our vector operator del, expressed as a row vector, but, again, each entry is the magnitude of each component. Each entry is also a scalar. The magnitude of each component of our bottomm two row vectors is a scalar, as expected, and also (to me) as makes sense.

    Now we get to that pesky top row. i j k hat, seems harmless enough, but think about it. What that row is saying, is that in our vector-space R^3, there exists a vector such that the magnitude of the i hat component is i hat (not 1, i hat), the magnitude of the j hat component is j hat (again, j hat, not 1), and the magnitude of the z component is k hat!
    This to me seems bothersome. There is a 3x3 vector array within that top row vector, in otherwords, we have a vector such that:
    ##v :=
    \left [ \begin{array} c
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    \end{array} \right ] ##
    maybe I didn't express that last vector array correctly, but the first column (or row for that matter, it's all the same in this case) is the x component of our vector, the second column is our y component, and the third is the z. The components of our vector are in fact vectors.

    So if I am to interpret this physically, it starts to get really abstract. First off, if we have 3 vectors, just normal vectors in R^3, and we calculate the determinant of their matrix, we get a volum, but this volume doesn't exist in R^3, this is a volume of the vector space that exists in R^3 (the vector corresponding to a field, for example, doesn't actually extend into physical space, that vector only represents the field at a point, but we're using it's magnitude and direction to gather the volume generated within this vector space).

    So that's a little bit abstract, but not too bad. Now when we add in our vector array, the only way I can see to interpret this, is that this vector, itself, contains within iteself a vector space. This is not a direct subspace however, this is a dimension within a dimension. On top of that, it's a spatial dimention within a spacial dimension! If that wasn't enough, it's actually 3 spacial dimensions within 1 spacial dimension. So in calculating the curl of a field, are we actually calculating the volume of some messued up vector space which has "tiny dimensions" within itself? If it is, is this what's meant by the tiny dimensions that you hear in reference to string theory? (That's a completely off topic and random question that popped up when I wrote that line, I'd like to hear an answer, but not to get bogged down on it).

    So, the curl(/cross product): the volume between a 2 vectors in R^3 and a vector space that exists within the dimensions of R^3? Seems a little messed up to me, but hopefully I can be enlightened.

    Ok also, I used to start my arrays with \left [ \begin{array} ccc for a 3x3, cc for a 2x2... etc, but that's giving me some random c's in my array now? What is that c actually for? I figured it was the number of columns, but I guess not?
     
  2. jcsd
  3. Nov 26, 2014 #2

    ShayanJ

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    Cool down man!
    You're taking that formula too far. That's just a way of remembering the formula for curl easier. Nothing more!!! That's not a rigorous thing so there is no problem with it and there is no need for further interpretations.
     
  4. Nov 26, 2014 #3

    BiGyElLoWhAt

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    =[
    Aww, I thought I was on to something. I still think that it's interesting.
    Is it completely unsensable to try to put a physical interpretation to this, though?
     
  5. Nov 26, 2014 #4

    Stephen Tashi

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    Perhaps, it would be simpler to pose the problem as mathematics and deal with the context later!

    The title of your thread appears to ask the mathematical question of whether one may define a type of vector whose components are vectors- and if it is defined, would this idea have any mathematical use? If we have a finite dimensional vector space and a given basis for it, then any vector in the space is a linear combination of the basis vectors. The terms in this linear combination are the vector's components, so each vector has components that are vectors - that's just the definition of "component".

    I think your question actually concerns the "coordinates" of a vector. For example, in 2-D, we understand that calling (2,3) a "vector" represents the vector that is the sum of 2 times the unit vector in the x-direction and 3 times the unit-vector in the y-direction. The (2,3) is a coordinate representation of the vector. The "coordinates" of the vector are real numbers. The "components" of the vector are vectors.

    There can be different coordinate systems for representing the same thing. The usual cartesian coordinate system for vectors uses coordinates that are real numbers. So I think the mathematical interpretation of your question is whether there is a different coordinate system for vectors that uses vectors as coordinates instead of real numbers for coordinates. (Of course, I haven't read all your physics. Maybe your question is something else.)
     
  6. Nov 26, 2014 #5

    BiGyElLoWhAt

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    In a sense, that is my question, but not about the mathematical usefulness, rather the physical implications. @Stephen Tashi
    What my concern is is this: can a vector component, so i hat, be comprised of a vector itself. I don't mean a linear combination. something like this:
    ## \vec{F} = <a*e_1,b*e_2,c*e_3,>\hat{i} + <d*e_1,e*e_2,f*e_3,>\hat{j} + <g*e_1,h*e_2,i*e_3,>\hat{k} ##
     
  7. Nov 26, 2014 #6

    ShayanJ

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    That's not what a change of coordinate can do Tashi! Actually here you're talking about the space, [itex] (\mathbb R^n)^m [/itex] which is isomorphic to [itex] \mathbb R^{nm} [/itex] and so nothing new is gained by examining it. Also a vector in [itex] (\mathbb R^n)^m [/itex] is of the form [itex] ((v_1^1,v_2^1,...,v_n^1),(v_1^2,v_2^2,...,v_n^2),(v_1^3,v_2^3,...,v_n^3),...,(v_1^m,v_2^m,...,v_n^m)) [/itex]. But this means the basis vectors can't be the same as the ones for [itex] \mathbb R^n [/itex] and so what the OP wants can't be realized mathematically.
     
    Last edited: Nov 26, 2014
  8. Nov 26, 2014 #7

    Fredrik

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    You can think of the determinant formula as talking about an actual determinant if we extend the definition of that term by using the standard formula ##\det A=\sum_P (\operatorname{sgn}P)A_{1,P1}A_{2,P2}A_{3,P3}## (where the sum is over all permutations of ##\{1,2,3\}##), and allowing the ##A_{k,Pk}## factors to belong to different vector spaces.
     
    Last edited: Nov 27, 2014
  9. Dec 9, 2014 #8
    I have thought about the same problem 20 years ago, so I am happy to tell you my conclusions.
    You are right when you say that this strange matrix is not only a "way" to remember a formula, as someone said above : indeed, it does explain some properties of the curl. In fact, there is a similar strange matrix to define the cross product of two vectors a and b:
    it is simply the determinant of the matrix whose two first lines are the components of the vectors a and b, and the third one is formed by the vectors i, j, k. The
    usual properties of the determinant give you immediately that the cross product of a and b is the opposite of the cross product of b and a, that the cross product of a with a is 0, and more generally that the cross product is bilinear and alternated.
    You could similarly define a n-linear alternated form by writing the n-1 lines of a n x n matrix using the components of n-1 vectors of length n, letting the last row of the matrix be the n vectors of a basis of Rn, and taking the determinant.
    There is also a strange matrix determinant trick that define the triple product (=scalar product of c with a x b), that explains the properties of the triple product, but I don't remember it.
    Why all this works? because matricial calculus does not reduce to linear algebra, as is generally believed: it can handle heterogeneous objects, AS LONG AS THE OPERATIONS ARE DEFINED AND BEHAVE AS EXPECTED. But since such a condition seems to be difficult to define rigorously, there is no general theory (to my knowledge) that handles this kind of problems; rather, they let you "feel" by yourself that everything works (well, this is most used in physics, isn't it :-)).

    Good luck.
     
  10. Dec 9, 2014 #9

    Stephen Tashi

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    Try "geometric algebra".
     
  11. Dec 9, 2014 #10
    Isn't a vector whose components are vectors, a pretty natural way to think of a 2nd order tensor?

     
  12. Dec 10, 2014 #11
    Well, it is true that geometric algebra has something to do with this, but I don't think that matricial calculus can be reduced to geometric algebra: matricial calculus remains valid even if you replace the entries of a matrix with heterogeneous objects (like, numbers, vectors, differential operators), as long as the operations make senses and behave as they have to behave. Geometric algebra deals with a defined category of objects.
     
  13. Dec 10, 2014 #12
    You are right, but the problem here is that the matrix contains heterogeneous objects, not only vectors whose components are vectors.
     
  14. Dec 10, 2014 #13

    BiGyElLoWhAt

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    @1MileCrash Interesting video. He spent a lot of time on vectors for a video titled what's a tensor, though.
    @coquelicot By heterogeneous objects, are you referring to the fact that 1 row is has vector components, while the other 2 contain scalar components? If that's the case I definitely agree. Thanks for bringing this back up by the way. I thought this was a dead thread and wasn't completely satisfied with conclusion XD
     
  15. Dec 10, 2014 #14
    Yes, that's what I mean. And this can be even other objects like operators, matrices etc.
     
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