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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?

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    $\begingroup$ You would need to explain what $v_s$ and $f(r_s)$ are. $\endgroup$
    – G. Smith
    Jan 17 '20 at 14:20
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    $\begingroup$ Do you understand how to put the $c$ into $ds^2=-dt^2+dx^2+dy^2+dz^2$? $\endgroup$
    – G. Smith
    Jan 17 '20 at 14:24
  • $\begingroup$ @G.Smith Added. I'm not entirely sure, but from what I've seen, you'd multiply $-dt^2$ with $c^2$ to get $-c^2dt^2$, right? And that would give you a component $g_{00}$ of $c^2$, I think. $\endgroup$ Jan 17 '20 at 14:37
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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.)

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