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What sort of vector is this?

  1. Jul 7, 2009 #1
    What sort of "vector" is this?

    A contravariant transforms as [tex]A^\rho \rightarrow {\Lambda^\rho}_\sigma A^\sigma[/tex].

    A covariant vector [tex]A_\mu[/tex] can be built from [tex]A^\rho[/tex] by [tex]A_\mu=\eta_{\mu\rho}A^\rho[/tex]. Then it transforms according to

    [tex]A_\mu\rightarrow\eta_{\mu\rho}{\Lambda^\rho}_\sigma A^\sigma=\Lambda_{\mu\sigma}A^\sigma={\Lambda_\mu}^\sigma A_\sigma[/tex].

    (Indeed, [tex]{\Lambda_\mu}^\sigma {\Lambda^\rho}_\sigma=\delta^\rho_\mu[/tex] can be considered the definition of a Lorentz transformation, that is, one which leaves [tex]A^\mu A_\mu[/tex] invariant.)

    But are there (covariant?) vectors whose components transform according to

    [tex]B_\sigma \rightarrow {\Lambda^\rho}_\sigma B_\rho[/tex]

    ?

    What sort of a "vector" is B? It is not dual to a contravariant vector in the ordinary sense. And (why I put this in relativity forum and not a math forum) what does this type of vector signify within relativity theory? Are there any physical quantities which transform like this?


    Edit:
    Ok. I think I get it. B is a covariant vector and the transformation being questioned represents transforming it by the inverse of the given Lorentz transform.

    But please correct if I am wrong. Thanks!
     
    Last edited: Jul 7, 2009
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  3. Jul 7, 2009 #2

    Hurkyl

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    Re: What sort of "vector" is this?

    Just in case you haven't noticed it:

    Those two expressions are the same expression, just with different variable names.
     
  4. Jul 7, 2009 #3

    HallsofIvy

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    Re: What sort of "vector" is this?

    Well, yes, it is. Just as you can construct a covariant vector that is dual to a contravariant vector (some texts would say "covariant" and "contravariant" components of the same vector), you can construct a covariant vector that is dual to a contravariant vector (that is basically what "dual" means- it works both ways). The dual to [tex]B_\sigma[/itex] is [tex]B^\gamma= \nu^{\gamma\sigma}B_\sigma[/tex].
    Assuming that your [itex]\nu_{\mu\rho}[/itex] is the metric tensor, [itex]\mu^{\gamma\sigma}[/itex] is the tensor such that [itex]\nu_{\mu\rho}\nu^{\mu\gamma}= \nu^{\mu\gamma}\nu_{\mu\rho}= \delta^\gamma_\rho[/itex]. Algebraically, [itex]\nu^{\gamma\sigma}[/itex] is represented by the matrix that is the inverse to [itex]\nu_{\gamma\sigma}[/itex].

    Yes, your edit is correct.
     
  5. Jul 7, 2009 #4
    Re: What sort of "vector" is this?

    Thanks!
     
  6. Jul 7, 2009 #5

    Fredrik

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    Re: What sort of "vector" is this?

    Actually, they're not, because we have

    [tex]{\Lambda_\mu}^\sigma=(\Lambda^{-1})^\sigma{}_\mu[/tex]
     
  7. Jul 7, 2009 #6

    Fredrik

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    Re: What sort of "vector" is this?

    Covariant transformation from (say) frame S to frame S':

    [tex]A'_\mu=\Lambda_\mu{}^\nu A_\nu[/tex]

    Multiply by [itex]\Lambda^\mu{}_\rho[/tex].

    [tex]\Lambda^\mu{}_\rho A'_\mu=\Lambda^\mu{}_\rho\Lambda_\mu{}^\nu A_\nu=\Lambda^\mu{}_\rho(\Lambda^{-1})^\nu{}_\mu A_\nu=\delta^\nu_\rho A_\nu=A_\rho[/tex]

    This is clearly just a covariant transformation from S' to S.
     
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