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How to get the universe’s volume from the universe’s scale factor? Taking into account that the scale factor is dimensionless and the volume has dimensions of $cm^3$ or $m^3$ ?

In this paper for instance: Fine Tuning Problem of the Cosmological Constant in a Generalized Randall-Sundrum Model

They got the scale factor from equations (3.4) and (3.5). And then got the volume in equation (3.8). Any help to understand the paragraph after equation (3.8)?

The brane just formed is of order $10^{35}$ in planck unit ($\sim 1$ m), in order to form the presently observed 3D scale of order $10^{61}$, we obtain the scale of extra dimensions is of order $10^9$ with $n_2= 3$.

  • First: how they gave the scale factor a dimension whereas it’s dimensionless. Or If a(t) $\propto t$ , then the dimensions of the scale factor will be the dimensions of the time, seconds or minutes. How to get the dimensions of the scale factor as distance’s dimensions like meters or cm? So that $a^3$ represents the volume?

  • Second: from this page Scale factor the current value of the universe’s scale factor is normalized to $a(t_0)=1$, so what did they mean by (presently observed 3D scale of order $10^{61}$)

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  • $\begingroup$ Scale factor can have dimension. It’s just depends on how to define/take the metric and distances. $\endgroup$
    – seVenVo1d
    Sep 25, 2022 at 16:29
  • $\begingroup$ If a(t) $\propto t$ , then the dimensions of the scale factor will be the dimensions of the time, seconds or minutes. How to get the dimensions of the scale factor as distance’s dimensions like meters or cm? So that $a^3$ represents the volume? $\endgroup$
    – Dr. phy
    Sep 25, 2022 at 16:34
  • $\begingroup$ See here. $\endgroup$
    – J.G.
    Sep 25, 2022 at 16:45

1 Answer 1

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We can write the FRW metric as

$$ds^2 = dt^2 - R^2(t)\Big[\frac{d\tilde{r}^2}{1-k\tilde{r}^2} + \tilde{r}^2d\Omega^2\Big]~~(1)$$.

Where $R(t)$ is the scale factor, $\tilde{r}$ is the comoving coordinate and $k$ is the curvature parameter and can take only three values $k \in \{-1,0,1\}$. Notice that we can also re-scale these three parameters via;

$$a(t) = R(t)/R_0,~r=R(t)\tilde{r}~~\text{and}~~\kappa = k/R_0$$.

Here $a(t)$ is the again scale factor, $r$ is the comoving coordinate, and $\kappa$ is the gaussian curvature (sometimes referred to as curvature parameter. The notation in cosmology textbooks is wildly different from one book to another.)

In this case, the FRW metric becomes;

$$ds^2 = dt^2 - a^2(t)\Big[\frac{dr^2}{1-\kappa r^2} +r^2d\Omega^2\Big]~~(2)$$

So in summary, in equation $(1)$, the scale factor $R(t)$ and has dimension of length and $\tilde{r}$ and $k$ are dimensionless with $k$ being only $(-1,0,1)$. This is coming from the derivation of the metric.

Meanwhile, in equation $(2)$, the scale factor $a(t)$ is dimensionless such that it's set to unity, i.e., $a_0 \equiv a(t_0) = 1$. In this case, $r$ and $\kappa^{-1/2}$ have units (length), and $\kappa$ can take any real values.

If $a(t) \propto t$, then the dimensions of the scale factor will be the dimensions of the time, seconds, or minutes.

That is just the proportionality. In general, there's always a constant value in the scale factor that makes it dimensionless. For instance, in matter dominated universe it's usually said $a(t) \propto t^{2/3}$ (and that is true) but the actual equation is $a(t) = (t/t_0)^{2/3}$, which has no dimension.

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  • $\begingroup$ Hi, if I’m solving Friedmann equation according to the formula in this thread answer: physics.stackexchange.com/questions/92805/… . Then what will be the dimensions of a(t)? Can it be considered has the dimensions of length? $\endgroup$
    – Dr. phy
    Sep 25, 2022 at 20:32
  • $\begingroup$ @Dr.phy In general if you are normalizing the scale factor to unity, then it should be automatically unitless. If that is the case in that question then it cannot have dimension of length. In cosmological calculations, we usually set $a(t_0)=1$ for simplicity. $\endgroup$
    – seVenVo1d
    Sep 26, 2022 at 6:09

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