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Homework Help: Change of basis- contravariant, covariant components of a vector.

  1. Aug 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Let e[tex]_{i}[/tex] with i=1,2 be an orthonormal basis in two-dimensional Euclidean space ie. the metric is g[tex]_{ij}[/tex] = [tex]\delta _{ij}[/tex]. In the this basis the vector v has contravariant components [tex]v^{i}[/tex] = (1,2). Consider the new basis
    [tex]e_{1}^{'} = 5e_{1} - 2e_{2}[/tex]
    [tex]e_{2}^{'} = 3e_{1} - e_{2}[/tex]

    a)Calculate the contravariant components of v in the new basis

    a)Calculate the components of the metric in the new basis

    a)Calculate the covariant components of v in the new basis

    3. The attempt at a solution
    Just a quick intro- I have no textbooks on linear algebra just the accompanying textbook to the module this question is from, unfortunately it doesn't get into much detail at all on the linear algebra side of things (this is a special relativity course).

    Nevertheless, what I'm thinking is this:
    I believe for part a) you just multiply the new basis transformations things by the old vector so that you end up with a new vector with components (1,1)? I've no idea what the contravariance has to do with this... I've read the definition of contravariant and covariant components and it seems to be to do with the derivatives between the two bases...

    As for the metric...hmm I'm less sure about this. I know a metric defined the distance between two points in a particular space, such as in Euclidean its the kronecker delta so that it results from pythag: [tex]ds^{2}=dx^{2}+dy^{2}...[/tex]. Fairly irrelevent anyway, my thinking is you express the metric as a 2x2 matrix (as its 2-d space) and then transform it using the given change of bases?
    This would result in(I think):
    ( 5 3)
    (-2 -1)

    This is probably all wrong :p

    And as for determining the covariantcomponents I haven't even got a clue ^_^

    This is pretty much all wrong but I haven't really got any idea, all I can find in the book I have on this module is the definition of metric, contravariant and covariant components with respect to tensors.

    Thanks :biggrin:
  2. jcsd
  3. Aug 28, 2008 #2


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    Science Advisor

    First, you don't need to worry about the distinction between "contravariant" and "covariant" components. In Euclidean space, or any space in which the coordinate axes are always orthogonal, they are the same. So just think "components".

    You are given that [itex]v^{i}[/itex] = (1,2) and[itex]e_{1}^{'} = 5e_{1} - 2e_{2}[/itex] and [itex]e_{2}^{'} = 3e_{1} - e_{2}[/itex]. You need to solve [itex]a a_1'+ be_2'= e_1+ 2e_2[/itex]. that is, [itex]a(5e_1- 2e_2)+ b(3e_1- e_2)= e_1+ 2e_2[/itex] Multiplying that out and setting coefficients of [itex]e_1[/itex] and [itex]e_2[/itex] equal gives you two equations for a and b.

    As for the metric, you can think of it as given by [itex]ds^2= dx^2+ dy^2= de_1^2+ de_2^2[/itex]. Solving the given equations for [itex]e_1[/itex] and [itex]e_2[/itex] as functions of [itex]e_1'[/itex] and [itex]e_2'[/itex], you get [itex]e_1= -e_1'+ 2e_2'[/itex] and [itex]e_2= -3e_1'+ 5e_2'[/itex]. Then [itex]de_1= -de_1'+ 2de_2'[/itex] so [itex]de_1^2= de_1'^2- 4de_1'de_2'+ 4de_2'^2[/itex] and [itex]d_2= -3de_1'+ 5de_2'[/itex] so [itex]de_2^2= 9de_1'^2- 30de_1'de_2'+ 25de_2'^2[/itex]. Adding those, [itex]ds^2= 10de_1'^2-34de_1'de_2'+ 29de_2'^2[/itex]. Since the metric tensor is always symmetric, it is
    [tex]\left[\begin{array}{cc}10 & -17 \\ -17 & 29\end{array}\right][/tex].

    Of course, you are close with looking at the transformation matrix. The transformation matrix, from [itex]e_i[/itex] to [itex]e_i'[/itex] is
    [tex]\left[\begin{array}{cc}5 & -2 \\ 3 & -1\end{array}\right][/tex]
    so the transformation matrix the other way is its inverse
    [tex]\left[\begin{array}{cc}-1 & 2 \\ -3 & 5\end{array}\right][/tex]

    Multiplying that by its adjoint gives
    [tex]\left[\begin{array}{cc}-1 & -3 \\ 2 & 5\end{array}\right]\left[\begin{array}{cc}-1 & 2 \\ -3 & 5\end{array}\right]= \left[\begin{array}{cc}10 & -17 \\ -17 & 29\end{array}\right]
  4. Sep 14, 2008 #3
    hello HallsofIvy,

    a question for you. What is this covariance and contravariance?
    Is it about the components of the vector (tensor) or the basis vectors of the chosen coord system?
    Can you give a layman explanation?
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