- #36
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Yes, my first transformation is valid, although there is a typo, it should be ##\vec X = t \vec x## in order to get the scale factor ##t^2## in front of ##d\vec X^2## in the metric, otherwise it would be scaling in the wrong direction.
This results in the following:
$$
d\vec x^2 = d(\vec X/t) = (d\vec X)/t - (\vec X/t^2) dt.
$$
The line element is therefore given by
$$
ds^2 = dt^2 - t^4 [(d\vec X)/t - (\vec X/t^2) dt]^2 = \left(1 - \vec X^2\right) dt^2 + 2t (\vec X \cdot d\vec X)\, dt - t^2 d\vec X^2,
$$
which has the spatial part scaling linearly with ##t## but definitely is not a metric on the FRW form due to the cross terms and the altered time-time component and therefore also represents a universe distinct from that of a linearly expanding one as I stated earlier:
There simply is no transformation that will make your FRW universe with an arbitrary scale factor equivalent to one where the scale factor is linear with the cosmological time t.
This results in the following:
$$
d\vec x^2 = d(\vec X/t) = (d\vec X)/t - (\vec X/t^2) dt.
$$
The line element is therefore given by
$$
ds^2 = dt^2 - t^4 [(d\vec X)/t - (\vec X/t^2) dt]^2 = \left(1 - \vec X^2\right) dt^2 + 2t (\vec X \cdot d\vec X)\, dt - t^2 d\vec X^2,
$$
which has the spatial part scaling linearly with ##t## but definitely is not a metric on the FRW form due to the cross terms and the altered time-time component and therefore also represents a universe distinct from that of a linearly expanding one as I stated earlier:
Orodruin said:Rescaling to make a(t) linear in t for the spatial part is going to introduce cross terms between dt and the spatial differentials into the metric and so is distinctly different from if you had just put a(t) linear in t from the beginning.
There simply is no transformation that will make your FRW universe with an arbitrary scale factor equivalent to one where the scale factor is linear with the cosmological time t.