What is the correct gamma factor in FLRW metric in curved spacetime?

So I'm quite perplexed my this paper. It seems to be using the Lorentzian gamma factor (equation $3.11$) but for FLRW metric.

$$ \Gamma = (1- \vec U \cdot \vec U/c^2 )^{-1/2}$$

where $\vec U$ is the spatial components of the $4$ velocity and $c$ is the speed of light. But I get a different $\Gamma$ factor.

My Attempt

Starting with the FLRW metric in flat spacetime:

$$ ds^2 = -c^2 dt^2 + a(t)^2 d\vec r \cdot d \vec r$$

Let us look at the proper time $\tau$:

$$ - c^2 d \tau^2 = -c^2 dt^2 + a(t)^2 d\vec r \cdot d \vec r$$

Dividing by proper time:

$$ c^2 = (c^2 (\frac{dt}{d \tau})^2 - a(t)^2 \vec U \cdot \vec U) $$


$$ \Gamma = \frac{d \tau}{dt} = (1 - a(t)^2 \frac{ \vec U \cdot \vec U}{c^2})^{-1/2} $$

  • $\begingroup$ In the attached paper, $\vec{U}$ is defined as $\frac{d\vec{x}}{d\tau}$ but here you have defined it as $\frac{d\vec{x}}{dt}$ $\endgroup$
    – paul230_x
    Jun 7, 2022 at 2:29
  • $\begingroup$ @KP99 fixed now the definitions of $\vec U$ are consistent . $\endgroup$ Jun 7, 2022 at 4:49

1 Answer 1


In (3.11), $\vec U$ is a 4-vector, and $\vec U\cdot\vec U$ means $\displaystyle \sum_{μ,ν=0}^3 g_{μν}U^μU^ν$, where, in your case, $g = \mathrm{diag}(-1,a^2,a^2,a^2)$. In the FLRW metric as you've written it, $d\vec r$ is a vector in $\mathbb R^3$ with the standard $\mathbb R^3$ inner product, so $d\vec r\cdot d\vec r$ means $\displaystyle \sum_{i=1}^3 (dr^i)^2$.

If the $\cdot$ in the metric was the true inner product (the one you're in the process of defining), then the metric would look like $ds^2=-c^2dt^2+d\vec r\cdot d\vec r$, but that isn't very useful, as it just says that the spatial part is equal to itself.


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