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Is a contravariant vector

  1. Apr 29, 2012 #1
    1. The problem statement, all variables and given/known data
    B is a third order tensor. Show that [itex]B^{ij}_{i}[/itex] is a contravariant vector.

    3. The attempt at a solution
    Well... I just thought about a simple solution but I don't think I'm right. But anyways.
    Considering [itex]B^{ij}_{i}[/itex]. If I raise the index i: [itex]g^{ij}B^{ij}_{i} = B^{ijj}[/itex]
    And so I can say that B is a first order tensor = vector. And this is a contravariant vector.
    Is this right?
     
  2. jcsd
  3. Apr 29, 2012 #2

    HallsofIvy

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    Re: Tensors

    You can't have the same index three times! Also, using the "summation convention" you sum over one "upper" and one "lower" index so even if you wrote [itex]g^{kj}B^{ij}_k[/itex], you would not be summing. Perhaps what you mean is [itex]g^k_jB^{ij}_k= B^i[/itex]

    Well, what is the definition of a contravariant vector? Can you show that this satisfies that definition? Remember that it is NOT enough just to show that something can be written with a single index- you can write an array, in a given coordinate system, indexed with a single index- that does not make it a "vector". A vector must change coordinates correctly as the coordinate system changes. If I remember correctly (it's been a while!) one test for a contravariant vector, written as [itex]v^i[/itex], is that the combination [itex]g_{ij}v^iv^j[/itex] (essentially the vector length) is a scalar. Which, again, does not just mean "a number" in a given coordinate system. A scalar (0 order tensor) must not change when you change coordinate systems.
     
  4. Apr 30, 2012 #3
    Re: Tensors

    Equally well you can show that [itex] B^{ij}_i [/itex] transforms as a vector, ie. if I do a coordinate transform [itex] x^{i} \rightarrow x^{i'} [/itex], B should transform as
    [tex] B^{i' j'}_{i'} = \frac{\partial x^{j'}}{\partial x^{j}} B^{ij}_i [/tex]
     
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