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Homework Help: How to show something is a covector?

  1. Mar 12, 2006 #1
    Hi, think this question will be relatively simple for those familiar with tensor analysis

    Need to show that (-x, -y, -z, t) is a covector. We're given a hint that we should apply a lorentz boost in the x direction but i don't see the point of this. If that is a covector then surely you can say

    x
    y
    z
    t

    is a vector and just take the scalar product to find the interval?

    There is also a samilar question where by I have to show that

    (df/dx, df/dy,df/dz,df/dt) (f is a scalar field) is a covector

    none of these questions are for credit or anything it's just to help with our understanding...but I don't :(

    Thanks
     
  2. jcsd
  3. Mar 12, 2006 #2

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    Well, there are two methods.

    (1) You could, as you said, apply a coordiante transformation, and make sure the components transform appropriately.

    (2) You could show that they are (coordiante-independent!) linear functionals on the space of vectors.


    Your hint was to apply method (1). It has the advantage of being a mindless, straightforward algorithm.


    But method (2) is often nicer, as you suggest. You know that [itex]\mathbf{x} := [x, y, z, t]^T[/itex] is a vector, and left multiplying by [-x, -y, -z, t] is, as you say, the (coordinate independent!) function:

    [tex]
    f(\mathbf{y}) = \langle \mathbf{x}, \mathbf{y} \rangle
    [/tex]

    At least I think that's what you said -- I hope you didn't mean to simply take the product1 of [-x, -y, -z, t] and [itex][x, y, z, t]^T[/itex]: sure, the result is an invariant, but you need to show it's invariant no matter what vector you use on the right.


    For the second problem, you know a suitable geometric interpretation of the gradient, and what you can do with the gradient and a vector.



    1: This is not a scalar product! The scalar product, at least in mathspeak, is a function of two vectors. When you have a covector and a vector, you're just plugging the vector into the covector. (Since a covector is a linear functional) In the coordinate representation, it's multiplying a 1xN matrix by an Nx1 matrix.
     
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