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Dumb question on FRW metric

  1. Dec 4, 2014 #1
    Hello everyone, I already know that the solution to this question is obvious but I can't find it.
    Consider an expanding universe following the FRW metric [itex] ds^2=-dt^2-a^2(t)dx^2[/itex] (1 space dimension for simplicity). We know that the physical spatial distance [itex]x_p[/itex] is related to the comoving spatial distance [itex]x_c[/itex] by [itex]x_p=ax_c[/itex]. First of all, is this correct?

    Now we also know that the metric in the physical coordinates is related to the metric in the comoving coordinates by:
    $$
    g_{\mu\nu}=\frac{\partial x_c^\alpha}{\partial x_p^\mu}\frac{\partial x_c^\beta}{\partial x_p^\nu}\eta_{\alpha\beta},
    $$
    where [itex]\eta_{\alpha\beta}[/itex] is the flat metric. Now, computing explicitely the transformation matrix I got:
    $$
    \frac{\partial x_c^\alpha}{\partial x_p^\mu}=\left(\begin{array}{cc}
    1 & 0 \\ 0 & 1/a \end{array}\right).
    $$
    Is this correct?
    But if this is true then it seems to me that applying it to the equation for [itex]g_{\mu\nu}[/itex] one gets:
    $$
    g_{\mu\nu}=\left(\begin{array}{cc} -1 & 0 \\ 0 & 1/a^2 \end{array}\right).
    $$
    How is this possible? What am I doing wrong?
     
  2. jcsd
  3. Dec 4, 2014 #2

    Matterwave

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    First, you have a sign error in the metric, you can't have both the spatial components and the time component being negative.

    In addition, if you are making the transformation ##t_p=t_c,~~x_p=ax_c## you need to consider also terms like ##\partial x_c/\partial t_p## which will get you terms with ##a'(t_p)## in them. Also, why are you sticking in the Minkowski metric for your transformation, instead of the FRW metric?

    Remember that with your coordinate transformation as you defined them, the transformation is a function of ##t## since ##a## is a function of t. So you have ##x_p=x_p(t,x_c)## not just ##x_p=x_p(x_c)## as it seems you are doing judging by the lack of off-diagonal terms in your transformation matrices.
     
  4. Dec 5, 2014 #3
    Oh I see, you are right. The first sign is just a typo but now I understand (even if I still don't know if I can get the FRW). However, I'm no sticking in the Minkowski metric, the point is that in the comoving system the metric is Minkowski while in the physcal system is FRW. I wanted to go from one to another with usual tensor trasformation but I'm haveing problem finding the matrices [itex] \partial x_c^\alpha/\partial x_p^\mu[/itex]. Thanks for you help!
     
  5. Dec 5, 2014 #4
    How can a coordinate transformation take you from FRW metric to Minkowski metric? The FRW doesn't represent flat space time so it can't be equivalent to Minkowski.
     
  6. Dec 5, 2014 #5
    Now I see that this is in 2D spacetime. Are they equivalent in 2D?
     
  7. Dec 5, 2014 #6
    Maybe I'm saying something really stupid but my understanding is that FRW is the metric seen in the physical coordinates, but if you go to a comoving system (i.e. perform a change of variables [itex]t_p=t_c[/itex] and [itex]x_p=a x_c[/itex]) the metric in this comoving system is locally Minkovski so I should be able to go from one to another. Being in 2D is not relevant, it was just to simplify.
     
  8. Dec 5, 2014 #7
    So you mean locally Minkowski. That is a different matter.
     
  9. Dec 5, 2014 #8

    Matterwave

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    Actually you have it the wrong way. FRW is seen in co-moving coordinates. The coordinates explicitly follow the family of observers who are free-floating. That is why the ##dx^2## term has a ##a(t)^2## in front of it to signal that these free-floating observers are moving apart isotropically.

    A coordinate system based on "physical distances" (between these co-moving observers) can only be done at "one point in time" because ##a(t)## is a function of ##t## and is not a constant. You can re-do this transformation at every point in time, but in this coordinate system, you will have off-diagonal terms in the metric, because in this weird-ish coordinate system, the flow is no longer isotropic. The metric looks quite weird in this new coordinate system, you can try it for yourself. Your transformation is not a local one, you are trying to explicitly transform the FRW metric globally into the Minkowski metric and that is futile. There is no global coordinate transformation that can change the FRW metric into the Minkowski metric. FRW is not a flat metric. The best that you can do is transform to a tetrad basis where the "metric" will be Minkowski, however, the tetrad basis will not be a coordinate basis.
     
  10. Dec 5, 2014 #9

    PAllen

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    To add a little to what Matterwave explained, the most you can achieve by coordinate transformation is:

    - along one chosen world line the metric is Minkowski and the connection components are zero (the connection being zero is possible because a comoving world line has no proper acceleration)

    However, any point away from this chosen origin world line will have non-Minkowski metric and non-vanishing connection components.
     
  11. Dec 5, 2014 #10
    I think I understand now. So, just to make things clear, the fact that I can write some kind of mathematical transformations that bring me from one set of coordinate to another does not mean necessarily that they also allow me to go from one metric to another right? This is true only if these transformations are diffeomorphism for which physics is invariant. Is that correct?
    Thanks a lot for you help!
     
  12. Dec 5, 2014 #11

    Matterwave

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    The metric tensor is a tensor, and just like every other tensor, it is independent of the coordinate system that you use. The fact that the FRW metric is the FRW metric and not the Minkowski metric means that the two tensors are fundamentally different. For example, the FRW metric has curvature, whereas the Minkowski metric does not. What you can do with changing around your coordinate systems (or, your basis vectors if you are using the tetrad formalism) is to change the components of the metric around. If you change to a tetrad basis, the components of your metric will be the Minkowski components in Cartesian coordinates (i.e. diag(-1,1,1,1)), but that does not mean the metric is the same as the Minkowski metric. You have not "changed the metric to the Minkowski metric" in a rigorous sense. But of course, in speaking, one may mean "changed the metric to the Minkowski metric" to mean "changed the components of the metric to those of the Minkowski metric in Cartesian coordinates". This is just sloppy language, but often used because to be explicit takes many words.

    As far as the question you started out with though, you should try your coordinate transformation, and include the off-diagonal terms, and you will find that actually the new metric will be non-diagonal as well. What happened is that when you change the coordinates in the way you did, the time-coordinate no longer is the time coordinate that rides along with your space coordinates, the time coordinate rides along with the comoving space coordinates not the "proper distance" space coordinates. Therefore, the basis vector in the time-direction will no longer be orthogonal to the basis vector in the space-direction, and you get a ##dxdt## term in the metric.
     
  13. Dec 5, 2014 #12

    PeterDonis

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    More correctly, the components will be exactly diag(-1, 1, 1, 1) at one event. They will not be exactly diag(-1, 1, 1, 1) at any other event.
     
  14. Dec 5, 2014 #13

    George Jones

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    Matterwave is correct. In relativity, "tetrad" usually means "set of four orthonormal vector fields", so the components of the metric with respect to a tetrad take on Minkowski values at all events. The difference between a general spacetime and Minkowski spacetime is that for Minkowski spacetime there are coordinate systems that are the integral curves of tetrads, and this isn't true for general spacetimes.
     
  15. Dec 5, 2014 #14

    Matterwave

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    You can use a tetrad basis throughout, and everywhere the metric will be diag(-1,1,1,1), it just won't happen to be a coordinate basis. The tetrad basis will twist around on itself to compensate (i.e. they will have non-zero Lie brackets with each other). And like George mentioned, this twisting will prohibit you from using integral curves of tetrad fields as coordinates. At least, that is my understanding.
     
  16. Dec 6, 2014 #15

    PeterDonis

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    Ah, ok. Yes, what I said only applies to a coordinate chart.
     
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