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Transpose Inverse Property (Dual Vectors)

  1. Sep 19, 2014 #1
    Hello,

    While studying dual vectors in general relativity, it was written as we all know that dual vectors (under Lorentz Transformation) transform as follows:

    [itex]\tilde{u}[/itex][itex]_{a}[/itex] = [itex]\Lambda[/itex][itex]^{b}_{a}[/itex]μ[itex]_{b}[/itex]

    where [itex]\Lambda[/itex][itex]^{b}_{a}[/itex]= η[itex]_{ac}[/itex]L[itex]^{c}[/itex][itex]_{d}[/itex]η[itex]^{db}[/itex]

    I was wondering if one can prove the latter or we take it as is.

    This to a certain extend can be related to [itex]\Lambda[/itex] = ηLη[itex]^{-1}[/itex], so is it that they took this relation and placed indices in a way if they are summed over we get [itex]\Lambda[/itex][itex]^{b}_{a}[/itex]? Or is there any clearer procedure?

    Thanks!
     
  2. jcsd
  3. Sep 20, 2014 #2

    Fredrik

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    I know I have proved this several times in this forum, but now I can't find it. Oh well, here we go again... (I should probably turn this into a FAQ post).

    I will use the notation ##M^\mu{}_\nu## for the component on row ##\mu##, column ##\nu## of an arbitrary 4×4 matrix M. To understand this post, you will need to understand the relationship between linear operators and matrices explained in https://www.physicsforums.com/threads/matrix-representations-of-linear-transformations.694922/ [Broken]. You also need to understand dual spaces and dual bases. In our notation, the definition of matrix multiplication is ##(AB)^\mu{}_\nu =A^\mu{}_\rho B^\rho{}_\nu##. The vectors in this post are elements of a 4-dimensional vector space over ##\mathbb R##. I will denote the vector space by V, and its dual space by V*. V is either ##\mathbb R^4## or the tangent space of spacetime at some event, depending on whether you prefer to define Minkowski spacetime as a vector space or as a manifold. The statements I make that involve an index that isn't summed over are "for all" statements, even if I don't say so explicitly. For example, when I say that ##x'^\mu=\Lambda^\mu{}_\nu x^\nu##, I mean that this equality holds for all ##\mu\in\{0,1,2,3\}##.

    Let ##\Lambda## be a Lorentz transformation. Let ##(e_\mu)_{\mu=0}^3## and ##(e'_\mu)_{\mu=0}^3## be ordered bases. Let v be an arbitrary vector. Let x be the matrix of components of v with respect to ##(e_\mu)_{\mu=0}^3##. Let x' be the matrix of components of v with respect to ##(e'_\mu)_{\mu=0}^3##. We have ##v=x^\mu e_\mu =x'^\mu e'_\mu##.

    Suppose that these ordered bases are such that the relationship between x' and x is given by ##x'=\Lambda x##, where ##\Lambda## is a Lorentz transformation. The component form of this is ##x'^\mu=\Lambda^\mu{}_\nu x^\nu##. We will determine the relationship between the two ordered bases. Let T be the unique linear operator such that ##e'_\mu=Te_\mu##. We will express the right-hand side as a linear combination of the ##e_\mu##, and then use the formula for the components of a linear operator with respect to an ordered basis.
    $$e'_\mu=Te_\mu=(Te_\mu)^\nu e_\nu =T^\nu{}_\mu e_\nu.$$ This implies that
    $$v=x^\mu e_\mu =x'^\mu e'_\mu =\Lambda^\mu{}_\rho x^\rho T^\sigma{}_\mu e_\sigma.$$ Since a basis is linearly independent, this implies that
    $$x^\rho =\Lambda^\mu{}_\rho x^\rho T^\sigma{}_\mu = T^\sigma{}_\mu \Lambda^\mu{}_\rho x^\rho =(T\Lambda)^\sigma{}_\rho x^\rho.$$ Since v (and therefore x) is arbitrary, this implies that ##(T\Lambda)^\sigma{}_\rho=\delta^\sigma_\rho##, i.e. that ##T\Lambda=I##. This implies that ##T=\Lambda^{-1}##. So we have ##e'_\mu=(\Lambda^{-1})^\nu{}_\mu e_\nu =\Lambda_\mu{}^\nu e_\nu##. The notation ##\Lambda_\mu{}^\nu## is explained in this post.

    The next step is to determine the relationship between the two dual ordered bases. This is very similar to the above. Let S be the unique linear operator such that ##e'^\mu=Se^\mu##. We have
    $$e'^\mu=Se^\mu =(Se^\mu)_\nu e^\nu = S^\nu{}_\mu e^\nu =(S^T)^\mu{}_\nu e^\nu,$$ and
    $$\delta^\mu_\nu =e^\mu(e_\nu)=e'^\mu(e'_\nu)=(S^T)^\mu{}_\rho e^\rho \big((\Lambda^{-1})^\sigma{}_\nu e_\sigma\big) =(S^T)^\mu{}_\rho(\Lambda^{-1})^\sigma{}_\nu \delta^\rho_\sigma =(S^T)^\mu{}_\rho(\Lambda^{-1})^\rho{}_\nu = (S^T\Lambda^{-1})^\mu{}_\nu.$$ This implies that ##S^T=\Lambda##. So we have ##e'^\mu =\Lambda^\mu{}_\nu e^\nu##.

    Now let ##\Omega\in V^*## be arbitrary. Let ##\omega## be the matrix of components of ##\Omega## with respect to the ordered basis ##(e^\mu)_{\mu=0}^3##. Let ##\omega'## be the (4×1) matrix of components of ##\Omega## with respect to the ordered basis ##(e'^\mu)_{\mu=0}^3##. We will determine the relationship between ##\omega## and ##\omega'##. We have
    $$\Omega=\omega_\mu e^\mu =\omega'_\mu e'^\mu =\omega'_\mu \Lambda^\mu{}_\nu e^\nu.$$ This implies that ##\omega_\nu=\Lambda^\mu{}_\nu \omega'_\mu##. This is the component form of ##\omega^T=(\omega')^T\Lambda##. This implies that ##(\omega')^T=\omega^T\Lambda^{-1}##. This implies that
    $$\omega'=(\omega^T\Lambda^{-1})^T =(\Lambda^{-1})^T\omega.$$ The component form is
    $$\omega'_\mu=((\Lambda^{-1})^T)^\mu{}_\nu \omega_\nu =(\Lambda^{-1})^\nu{}_\mu \omega_\nu =\Lambda_\mu{}^\nu \omega_\nu.$$
     
    Last edited by a moderator: May 6, 2017
  4. Sep 20, 2014 #3

    stevendaryl

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    A sort-of interesting relation that I came up with by fooling around with indices:

    To make the distinction between the primed and unprimed basis clear, let me use Latin characters for primed indices, and Greek characters for unprimed. Then the Lorentz transform taking unprimed vectors to primed vectors is:

    [itex]\Lambda^a_\mu v^\mu = v'^a[/itex]

    Now, suppose I want to know how a covector [itex]w_\mu[/itex] transforms under Lorentz transformations. What I can do is first convert it to a vector using the metric, then transform that vector, then transform back using the metric again:

    [itex]g^{\mu \nu} w_\mu = w^\nu[/itex]
    [itex]\Lambda^b_\nu g^{\mu \nu} w_\mu = \Lambda^b_\nu w^\nu = w'^b[/itex]
    [itex]g_{ab} \Lambda^b_\nu g^{\mu \nu} w_\mu = g_{ab}w'^b = w'_a[/itex]

    So the transform for covectors is

    [itex](\Lambda^{-1})^\mu_a = g_{ab} \Lambda^b_\nu g^{\mu \nu}[/itex]
     
  5. Sep 20, 2014 #4

    stevendaryl

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    Wow. I don't know whether it's the new PF software, or my home computer, or my browser, or what, but it takes a LONG time for my computer to render LaTex. Roughly 30 seconds. (which is an eternity in cyberspace)
     
  6. Sep 20, 2014 #5

    Fredrik

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    Greg is aware of that, but he's sleeping now.
     
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