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Negative scale factor RW metric with scalar field

  1. Nov 10, 2015 #1
    1. The problem statement, all variables and given/known data

    The aim is to find a solution for the scale factor in a Robertson Walker Metric with a scalar field and a Lagrange multiplier.

    2. Relevant equations

    I have this action

    [tex]S=-\frac{1}{2}\int d^{4}x\sqrt{-g}[R(g_{\mu\nu})+\lambda(g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi-1)+L_{m}][/tex]

    In conformal for Robertson Walker metric [tex]ds^{2}=a^{2}(\eta)(d\eta^{2}-\delta_{ij}dx^{i}dx^{j})[/tex]

    where [tex]\phi[/tex] is a scalar field (that depends only on conformal time ), [tex]\lambda[/tex] a Lagrange multiplier, [tex]L_m[/tex] lagrangian matter that I set to 0 and R the Ricci Scalar.


    So the Ricci Scalar is [tex]R=-6\frac{a''}{a^{3}}[/tex] ()' is derivative with respect to conformal time (I found the value from the value that one finds for R from universal proper time, that is [tex]R=-6(\frac{\ddot a}{a}+\frac{\dot a^2}{a^2})[/tex] )

    So the action is

    [tex]S=-\frac{1}{2}\int d^{4}xa^{4}[-6\frac{a''}{a^{3}}+\lambda(a^{-2}\eta^{00}\partial_{0}\phi\partial_{0}\phi-1)]=-\frac{1}{2}lim_{V\rightarrow\infty}V\int d\eta[-6\frac{a''}{a^{3}}+\lambda(a^{-2}\phi'^{2}-1)]a^{4}[/tex]

    3. The attempt at a solution

    I have the various [tex]L=L_{V}=-\frac{1}{2}V[-6\frac{a''}{a^{3}}+\lambda(a^{-2}\phi'^{2}-1)]a^{4}[/tex] so because there is only the multiplicative factor V they all lead to the same eq. of motion. So I can omit the limit and take V=1. So the action is now

    [tex]S=-\frac{1}{2}\int d\eta[-6\frac{a''}{a^{3}}+\lambda(a^{-2}\phi'^{2}-1)]a^{4}=-\frac{1}{2}\int d\eta[-6aa^{''}+\lambda a^{4}(a^{-2}\phi'^{2}-1)][/tex]

    And without considering surface elements the action become
    [tex]S=-\frac{1}{2}\int d\eta[6a'^{2}+\lambda a^{4}(a^{-2}\phi'^{2}-1)][/tex]

    Now I want to find eq. of motion:

    Variation w.r.t a gives:
    [tex]6a''-\frac{1}{2}[\frac{\partial(\lambda a^{4})}{\partial a}(a^{-2}\phi'^{2}-1)-a\lambda\phi'^{2}2]=0[/tex]

    W.r.t [tex]\lambda[/tex]:
    [tex]\frac{\phi^{'2}}{a^{2}}-1=0\rightarrow\phi'^{2}=a^{2}[/tex]

    W.r.t [tex]\phi[/tex]:
    [tex]\frac{d}{d\eta}(2\phi'\lambda a^{2})=0[/tex]

    [tex]\phi'\lambda a^{2}=\lambda a^{3}=\eta+K\rightarrow\lambda=\frac{\eta+K}{a^{3}}[/tex]
    K is a real constant


    And so in the end I have:

    [tex]6a^{''}+\eta+K=0[/tex]

    Integrating

    [tex]a'(\eta)=-\frac{\eta^{2}}{12}-\frac{K\eta}{6}+C[/tex]

    And again:

    [tex]a(\eta)=-\frac{\eta^{3}}{36}-\frac{K\eta^{2}}{12}+C\eta+B[/tex]

    B and C real constants.

    Now, if [tex]\eta[/tex] goes to [tex]+\infty[/tex] is a problem because the scale factor becomes negative. And so the only possibility for this case in a flat Robertson Walker metric is that I have(setting B=0) [tex]a(\eta)=0[/tex] and then after a finite interval again [tex]a(\eta)=0[/tex] becuase then it will be negative and I think there isn't a physical meaning for negative scale factor. So, where are the errors? Because I can't interpret well this result.
    If I imagine it as a 1-D motion of a particle I can study without any problems the system. But here I am studying the evolution of a particular case of the universe. How can I interpret negative scale factor(if all above is correct)?

    I hope that the question is written well. Thank you

     
  2. jcsd
  3. Nov 11, 2015 #2
    Can at least someone check if the variation of the action w.r.t a is correct?
     
  4. Nov 11, 2015 #3
    Ok, I think that in general what I did is right. But I made a big mistake!

    [tex]\frac{d}{d\eta}(2a^3\lambda)=0[/tex] gives [tex]a^3\lambda=A[/tex] that is obvoius! A is a constant

    And then I obtain [tex]a(\eta)=-\frac{A}{6}\eta^2+B\eta+C[/tex]. B and C are real constants.

    And how can I get the value of the constant A? If A is positive there is a good behaviour at [tex]\eta[/tex] that goes to +infinity. (and the universe expands for ever). But for A negative the scale factor is negative! So I can't give a physical interpretation for it!
     
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