"There, it was noticed that to an observer within the warp-drive bubble, the backward and forward walls (along the direction of motion) look, respectively, like the horizon of a black hole and of a white hole.”

PHYSICAL REVIEW D 79, 124017 (2009

Semiclassical instability of dynamical warp drives

Stefano Finazzi,1,* Stefano Liberati,1,† and Carlos Barcelo´ 2,‡

**It looks like Alcubierre and everyone after him made a really stupid math mistake that completely changes the superluminal warp drive horizon picture when done correctly.**

Introducing Physical Warp Drives

Alexey Bobrick, Gianni Martire

Advanced Propulsion

They use Galilean transformation

dx’ = dx - vdt

In transforming from the warp bubble rest frame to the observer frame outside the warp bubble and then take v > c.

This is clearly inconsistent.

In the simple 1 + 1 ST case

It’s only because they use the Galilean transformation that they get superluminal horizons

g00 = 0

From

g00 = (c + v)(c - v)

**The metric inside the Tic Tac warp bubble Oin = Oout,co**

**is simply the flat**

ds^2 = -c^2dt’2 + dx’^2

Because the warp shape function f(r) = 1 and v = 0 in the rest frame of the Tic Tac.

Therefore,

**what Cmndr Fravor Oout sees outside the warp bubble must use the Lorentz transformations**dt' = (1 - (v/c)^2)^-1/2 [dt - (v/c^2)dx^2]

dx' = (1 - (v/c)^2)^-1/2 [dx - vdt]

ds^2 = - c^2 {(1 - (v/c)^2)^-1/2 [dt - (v/c^2)dx^2]}^2 + {(1 - (v/c)^2)^-1/2 [dx - vdt]}^2

= - c^2 (1 - (v/c)^2)^-1 [dt - (v/c^2)dx^2]^2 + (1 - (v/c)^2)^-1 [dx - vdt]^2

g00 = - (1 - (v/c)^2)^-1{- c^2 + v^2}

= - c^2

That is the time dilation and Lorentz contraction in the frame transformation from the interior to the exterior cancel out any horizons from the warp bubble. The apparent superluminal horizons are simply a false artifact from using the incorrect Galilean relativity transformation rather than the special relativity Lorentz transformation.

**NO SUPERLUMINAL HORIZONS!**Check my algebra - did I make a mistake?