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

## Question

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:

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

Thus,

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

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

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.