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Conformally Flat Space Time

  1. Jan 17, 2008 #1
    1. The problem statement, all variables and given/known data

    A space-time is said to be conformally flat if there is a frame in which the metric is [itex]g_{ab} = \Omega^2 \eta _{ab} \text{ with } \eta_{ab}[/itex] the metric in Minkowski.

    Is the general Friedmann-Robertson-Walkers space-time with line element
    [tex] ds^2 = -dt^2 +a^2(t)(dx_1^2 + \ldots + dx_n^2)[/tex]
    conformally flat?


    2. Relevant equations

    [tex] ds^2 = g_{ij} dx_i dx_j [/tex]


    3. The attempt at a solution

    We can easily use the symmetry of the metric and the line element to find that the metric in FRW spacetime is

    [tex] \begin{bmatrix}
    -1 & 0 & \ldots & 0\\
    0 & a^2(t) & \ldots & 0\\
    \vdots & \vdots &\ddots & \vdots\\
    0 & 0 & \ldots & a^2(t) \end{bmatrix} [/tex]

    Now we weren't told what [itex] \Omega^2 [/itex] was in the question, but when I asked my TA he said that it was a real-valued function from spacetime coordinates. He also told me that I should make a change of coordinates. I've tried using the definition of the line element, and taking derivatives with respect to time, but to no avail. I attempted using

    [tex] dt^2 = \Omega^2 d\tau^2[/tex]
    [tex] a^2(t)dx_i^2 = \Omega^2 d\xi_i^2 \quad \forall i = 1\ldots n [/tex]

    to try and solve for a coordinate transformation, but again I just ended up getting integrals of [itex]\Omega[/itex] which were at best, complex valued. A few of my class-mates and I have been pondering this question for a bit, and we just can't seem to find the trick. Any help would be appreciated.
     
    Last edited: Jan 17, 2008
  2. jcsd
  3. Jan 17, 2008 #2
    Try a change of variable t, i.e. [tex]t=f(\tau)[/tex]
     
  4. Jan 17, 2008 #3

    Dick

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    You have a problem with your metric expressions. You want e.g. [tex] dt^2 = \Omega(t)^2 d\tau^2[/tex]. Now just separate the variables to get a definition of tau. And just take [tex]\Omega(t)=a(t)[/tex].
     
  5. Jan 17, 2008 #4
    Yes yes, sorry about that typo. I'll go and fix that now. Anyway, like I've stipulated, I've already done that, but it didn't seem to offer me a reasonable solution; I'll have a look at it again.
     
  6. Jan 17, 2008 #5
    Why can we just assume that [itex]\Omega[/itex] is just a function of a single variable rather than over the whole space-time? Furthermore, wouldn't [itex]\Omega[/itex] need to be a function of [itex]\tau[/itex] rather than t? Since otherwise we would have that t is a function of itself - which is certainly possible, but makes it somewhat impossible to solve for t in order to properly evaluate the change of coordinates.
     
    Last edited: Jan 17, 2008
  7. Jan 17, 2008 #6
    If we do assume that [itex]\Omega[/itex] is a function of [itex]\tau[/itex] alone, then the associated metric has cross-terms that lie outside of the diagonal - making it very difficult to associate to the Minkowski metric. Granted, I might be doing all of this wrongly...
     
  8. Jan 17, 2008 #7
    I'll work something out and maybe you can tell me where I'm going wrong. Let's limit ourselves to two space.

    Using the previous equations, we get that
    [tex] \frac{d\tau}{dt} = \frac{1}{a(t)} [/tex]

    [tex]\frac{d\xi}{dx} = 1 [/tex]

    Thus
    [tex] \tau = \int \frac{dt}{a(t)} [/tex]

    [tex]\xi=x [/tex]

    Then if we define the vector [itex] F = \begin{bmatrix} \displaystyle\int \frac{dt}{a(t)} \\ x \end{bmatrix}[/itex], the metric is

    [tex] g^{ij} = \begin{bmatrix}
    \frac{dF}{dt}\frac{dF}{dt} & \frac{dF}{dx}\frac{dF}{dt} \\
    \frac{dF}{dt}\frac{dF}{dx} & \frac{dF}{dx}\frac{dF}{dx} \end{bmatrix}
    = \begin{bmatrix}
    \frac{1}{a^2(t)} & 0 \\
    0 & 1 \end{bmatrix} [/tex]
     
  9. Jan 17, 2008 #8

    Dick

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    What do the spatial coordinates have to do with it? Your metric is already conformally flat in them. You just have to redefine the time coordinate, which you have already done once you get rid of that confusing xi. tau is called 'conformal time'. And not without reason. If it's any help you can find any omega you want. It doesn't HAVE to be a function of the spatial coordinates. And for FRW a function only of time is fine.
     
    Last edited: Jan 17, 2008
  10. Jan 18, 2008 #9
    I'll take your word for it, but it just seems to me like it should've been more complicated. I understand what you're saying just fine and it makes sense. Perhaps I'm just over analyzing the problem.

    Edit: Yeah, now that I think about it I don't know why I assumed that the transformation needed to affect the spatial coordinates. Thanks for clearing that up.
     
    Last edited: Jan 18, 2008
  11. Jan 18, 2008 #10

    Dick

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    Well, you've defined tau such that [tex]dt = a(t) d\tau[/tex]. Just put that into the metric. You could also write down the change in the metric due to that coordinate change by doing the partial derivatives and the formal tensor change of variables thing. But you'll get the same answer. It really isn't as hard as you were expecting.
     
  12. Jan 19, 2008 #11
    Apparently it was indeed much simpler. Thanks.
     
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