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Is our locally measured time actually conformal time?

  1. Apr 13, 2015 #1
    The FRW metric at the origin ##r=0##, with ##c=1##, is given by:
    $$ds^2=-dt^2+a(t)^2dr^2$$
    Now one can change variables so that near the origin the FRW metric is approximated by the Minkowski metric describing flat spacetime:
    $$dS^2=-dT^2+dR^2$$
    where:
    $$dT=\frac{dt}{a(t)}$$
    $$dS=\frac{ds}{a(t)}$$
    $$dR=dr$$
    All the physics experiments that we perform locally are assumed to occur in flat spacetime as described above.

    Surely therefore our locally measured time is not the cosmological time ##t## but rather the conformal time ##T## ?
     
  2. jcsd
  3. Apr 13, 2015 #2

    Chalnoth

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    The local measured time, for any observer, is the proper time. The proper time is, in this case, almost identical to the cosmological time.
     
  4. Apr 13, 2015 #3

    PeterDonis

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    This isn't a correct transformation. You can't transform ##ds## to ##dS##; ##ds## is the actual, physical line element, and it has to be left invariant, otherwise you're comparing apples and oranges. The correct transformation gives:

    $$
    ds^2 = a^2(T) \left( - dT^2 + dR^2 \right)
    $$

    In order to make the metric Minkowski in our local vicinity, you have to define ##a(T_0) = 1##, where ##T_0## is our current value of ##T##. But that only works locally, and we can make observations that go beyond that local region of spacetime. And in any case, proper time for an observer at constant ##R## is given by ##a(T) dT##, not ##dT##; again, you can try to obscure this by defining ##a(T_0) = 1##, but we can make observations covering a larger region of spacetime than the local region covered by that definition.
     
  5. Apr 14, 2015 #4

    ChrisVer

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    You can make the following:

    [itex] ds^2= dt^2 - a^2(t) dr^2 = a^2(t) \big( \frac{dt^2}{a^2(t)} - dr^2 \big)[/itex]

    now you can write the first term in the parenthesis as a single variable by changing [itex] \frac{dt}{a(t)}= d( \ln a(t) ) \equiv dT[/itex].

    So:

    [itex]ds^2= a^2(t) \big[ dT^2 - dr^2 \big] [/itex] or to have everything in the same coordinates: [itex]ds^2= e^{2T} \big[ dT^2 - dr^2 \big] [/itex]
    since [itex]T(t)= \ln a(t) \Rightarrow a(t)= e^{T}[/itex]
     
  6. Apr 14, 2015 #5

    wabbit

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    But ## d(\ln a(t))=\frac{\dot a(t)}{a(t)}dt\neq\frac{dt}{a(t)} ##
    You need ## T=\int\frac{dt}{a(t)} ## instead
     
  7. Apr 14, 2015 #6

    ChrisVer

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    Oops I'm sorry!
     
  8. Apr 14, 2015 #7

    wabbit

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    No it's an interesting point, I have been using this ## T=\int\frac{dt}{a(t)} ## but I didn't know it was called conformal time - never really got to checking what that name was referring to, seemed a bit exotic... So now I know, and I can also see it's a useful thing - thanks.
     
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