Alcubierre metric without geometrized units In the beginning of Alcubierre's paper, he defines his metric with the use of geometrized units ($G = c = 1$). While it makes the math simpler, it seems to hide the physical numbers away, leaving me pretty confused.
As I understand things, Alcubierre starts with a general spacetime metric:
$$
ds^2 = -(\alpha^2-\beta_i\beta^i) dt^2 + 2\beta_idx^idt + \gamma_{ij}dx^idx^j
$$
And then defines several parameters with the use of geometrized units, arriving at his specific form of the metric:
$$
ds^2=-dt^2+\big(dx-v_sf(r_s)dt\big)^2+dy^2+dz^2
$$
Where:
$$
v_s(t)=\frac{dx_s(t)}{dt}$$$$r_s(t)=\Big[\big(x-x_s(t)\big)^2+y^2+z^2\Big]^{1/2}
$$
and where $f$ is the function:
$$
f(r_s)=\frac{\tanh\big(\sigma(r_s+R)\big)-\tanh\big(\sigma(r_s-R)\big)}{2\tanh(\sigma R)}
$$
With arbitrary arguments $\sigma$ and $R$.
My question is, what would this metric look like with any geometrized units written explicitly, and where do these units come from in the derivation of the metric?
 A: Starting with 
$$
ds^2=-dt^2+\big(dx-v_sf(r_s)dt\big)^2+dy^2+dz^2
$$
you need to get each term on the right-hand-side to have units of length-squared. 
The first term currently has units of time-squared, so you'll need a factor of $c^2$ there.
The second term is ok provided that $f$ is unitless and $v_s$ has units of velocity. The latter is true upon inspection of the expression that you gave without further adjustment.  The former is also true since it is a ratio of hyperbolic tangents, which are unitless, although we now need to check the units of the free parameters $\sigma$ and $R$ to make sure that the arguments to those hyperbolic tangents are also unitless.
From the expressions like $r_s \pm R$, we can see that $R$ is a length, and then by inspection we can see that $\sigma$ has units of inverse ("one-over") length so that the products $\sim \sigma R$ are unitless.
So the only modification that you needed was a factor of $c^2$ in the first term of the metric
$$
ds^2=-c^2 dt^2+\big(dx-v_sf(r_s)dt\big)^2+dy^2+dz^2
$$
plus the understanding that $R$ is a length and $\sigma$ is one-over-length.
(If you went back to the first expression that was more generally expressed in terms of lapse $\alpha$ and shift $\beta^i$ you could follow similar arguments to find that you need an overall factor of $c^2$ on the $dt^2$ term and a factor of $c$ on the $dx\,dt$ term.)
