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Gauge transformations in GR

  1. May 17, 2007 #1
    I have been told that using a metric

    [tex]g_{00} = -a^2(\eta)(1+2\psi)[/tex]
    [tex]g_{oi} = g_{i0} = a^2(\eta)\omega_i[/tex]
    [tex]g_{ij} = a^2(\eta) \left[(1+2\phi)\gamma_{ij} + 2\chi_{ij} \right][/tex]

    and a gauge transformation

    [tex]x^{\bar{\mu}} = x^{\mu} + \xi^{\mu}[/tex]

    with

    [tex]\xi^0 = \alpha[/tex]
    [tex]\xi^i = \beta^j[/tex]

    gives the changes in the amplitude as

    [tex]\delta \psi = \alpha' + \frac{a'}{a} \alpha[/tex]

    and so on.

    But how do I calculate these changes? How do I start?
     
  2. jcsd
  3. May 17, 2007 #2

    Dick

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    The 'gauge' transformation is a coordinate transformation. You know how the metric changes in a coordinate transformation. That should certainly be enough info to start.
     
  4. May 17, 2007 #3
    Yes like

    [tex]g_{\bar{\mu} \bar{\nu}} = \frac{\partial x^{\mu}}{\partial x^{\bar{\mu}}} \frac{\partial x^{\nu}}{\partial x^{\bar{\nu}}} g_{\mu \nu}[/tex]

    but I can't get it right. I get that the perturbations change like

    [tex]\delta g = -\partial_{\bar{\nu}} \xi^{\nu} -\partial_{\bar{\mu}} \xi^{\mu}[/tex]

    but then what?
     
  5. May 17, 2007 #4
    If

    [tex]g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}[/tex]

    and

    [tex]g_{00} = -(1+2\psi)[/tex]

    then

    [tex]h_{\bar{\mu} \bar{\nu}} = h_{\mu \nu} -\partial_{\mu} \xi_{\nu} -\partial_{\nu} \xi_{\mu}[/tex]

    and

    [tex]h_{00} = -2\psi[/tex].

    But how do I get that

    [tex]\psi \rightarrow \psi + \alpha' + \frac{a'}{a}\alpha[/tex]??

    If I use the above equation for [itex]h_{\bar{\mu} \bar{\nu}}[/itex] I just get that

    [tex]\psi \rightarrow \psi + \alpha'[/tex].

    Please help someone!
     
  6. May 17, 2007 #5

    Dick

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    As usual, I'm pounding my head over the tex. The extra term comes from the effect of the transformation on the overall scale factor a. a(eta) -> a(eta+alpha) -> a(eta)+alpha*a'(eta) -> a(eta)(1+(a'/a)*alpha). (In case I never get the tex straightened out).
     
    Last edited: May 17, 2007
  7. May 17, 2007 #6
    I don't get it. Where does this come from?
     
  8. May 17, 2007 #7

    Dick

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    The scale factor 'a' changes under the transformation.
     
  9. May 17, 2007 #8
    That I got, but how do I get your transformation?
     
  10. May 17, 2007 #9

    Dick

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    a(eta) goes to a(eta+alpha). I just took the first term of the taylor series expansion of a(eta).
     
  11. May 17, 2007 #10
    I don't follow. How does this couple with my metric transformation?
     
  12. May 17, 2007 #11

    Dick

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    I have to confess, I've only had to deal with this metric perturbation formalism once. And I found it pretty confusing myself. So I'm not sure I can clearly answer your question. But I do know that that is where your extra term comes from. It seems to me there is a review paper around by Brandenberger and Muhkanov that was pretty handy. But I don't have access to it right now.
     
  13. May 17, 2007 #12
    Ok. I think have all papers ever written about this here, but all they say is that "one can easily see that..."
     
  14. May 17, 2007 #13

    Dick

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    Annoying, isn't it?
     
  15. May 17, 2007 #14
    Yes. Very.
     
  16. May 17, 2007 #15

    nrqed

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    I guess the point is what is [itex] \eta [/itex] ??
     
  17. May 17, 2007 #16

    Dick

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    In problems like this eta is usually the conformal time. Just a specific parametrization of the time coordinate.
     
  18. May 17, 2007 #17
    Correct...
     
  19. May 17, 2007 #18

    nrqed

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    Ok. Thanks. What is the definition? What is the relation with [itex] x_0 [/itex]?
     
  20. May 17, 2007 #19
    Conformal time is defined as

    [tex]\eta = \int_0^{x_0} \frac{dx_0'}{a(x_0')}[/tex]
     
  21. May 17, 2007 #20

    nrqed

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    Ok. Then why not simply do the change of coordinates in that expression?
     
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