# 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?

• You would need to explain what $v_s$ and $f(r_s)$ are. Jan 17, 2020 at 14:20
• Do you understand how to put the $c$ into $ds^2=-dt^2+dx^2+dy^2+dz^2$? Jan 17, 2020 at 14:24
• @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. Jan 17, 2020 at 14:37

## 1 Answer

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.)