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I Conformal transformation of the line element

  1. Jul 22, 2016 #1
    Let us see how the line element transforms under conformal transformations. Consider the Minkovski metric gij, a line element ds2=dxigijdxj, and a conformal transformation

    δk(x)=ak + λ xk + Λklxl + x2sk - 2xkx⋅s

    We have δ(dxk)=dδ(x)k=λ dxk + Λkldxl + 2 x⋅dx sk - 2dxkx⋅s - 2xkdx⋅s

    And so the line element transforms by δds2=δ(dxi)gijδ(dxj)=
    (λ dxi + Λildxl + 2 x⋅dx si - 2dxix⋅s - 2xidx⋅s) gij (λ dxj + Λjrdxr + 2 x⋅dx sj - 2dxjx⋅s - 2xjdx⋅s)

    How can we see that δds2=(2λ-2x⋅s)ds2
     
  2. jcsd
  3. Jul 22, 2016 #2

    samalkhaiat

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    Science Advisor

    The variation symbol [itex]\delta[/itex] is a derivation. So, you should consider

    [tex]\delta \left(ds^{2}\right) = \eta_{\mu\nu} \ \delta \left(dx^{\mu}\right) \ dx^{\nu} + \eta_{\mu\nu} \ dx^{\mu} \ \delta \left(dx^{\nu}\right) . \ \ \ (1)[/tex]

    Now, for the infinitesimal conformal transformation

    [tex]\delta x^{\mu} = a^{\mu} + \lambda x^{\mu} + \omega^{\mu}{}_{\nu}x^{\nu} + c^{\mu}x^{2} - 2 (c \cdot x ) x^{\mu} ,[/tex]

    if we take the partial derivative with respect to [itex]x^{\sigma}[/itex], we get

    [tex]\partial_{\sigma} (\delta x^{\mu}) = \delta^{\mu}_{\sigma} \left( \lambda - 2 c \cdot x \right) + \eta_{\sigma \tau} \left( \omega^{\mu \tau} + 2 ( c^{\mu}x^{\tau} - c^{\tau}x^{\mu}) \right) . \ \ (2)[/tex]

    In terms of the following local parameters

    [tex]\Lambda (x) = \lambda - 2 c \cdot x ,[/tex] [tex]\Omega^{\mu \tau}(x) = - \Omega^{\tau \mu}(x) = \omega^{\mu \tau} + 2 (c^{\mu}x^{\tau} - c^{\tau}x^{\mu}) ,[/tex] equation (2) becomes

    [tex]\partial_{\sigma} (\delta x^{\mu}) = \delta^{\mu}_{\sigma} \ \Lambda (x) + \eta_{\sigma \tau} \ \Omega^{\mu \tau}(x) .[/tex]

    From this, you get

    [tex]d (\delta x^{\mu}) = \Lambda (x) \ dx^{\mu} + \eta_{\sigma \tau} \ \Omega^{\mu \tau} \ dx^{\sigma} . \ \ \ \ \ (3)[/tex]

    Substituting (3) in (1), we find

    [tex]\delta \left(ds^{2}\right) = 2 \Lambda (x) \ \eta_{\mu\nu} \ dx^{\mu} dx^{\nu} + \Omega_{\mu \nu}(x) \ dx^{\mu} dx^{\nu} + \Omega_{\nu \mu} (x) \ dx^{\mu}dx^{\nu} .[/tex]

    The last two terms vanish because [itex]\Omega_{\mu\nu} = - \Omega_{\nu\mu}[/itex]. So you are left with

    [tex]\delta \left(ds^{2}\right) = 2 \Lambda (x) \ ds^{2} .[/tex]
     
  4. Jul 25, 2016 #3
    Right, variations satisfy the Leibnitz rule. Thank you very much, sir.
     
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