What would define the "speed" (and direction?) of an alcubierre like warp drive?

Preamble

Lets start from the beginning here.

The Alcubierre drive is a result of general relativity. This has what's known as a metric. Minkowski spacetime (lets call this 'flat') has the metric $dS^2 = -dt^2 + dx^2 + dy^2 + dz^2$ (where the speed of light has been taken to be $1$). This leads to special relativity - i.e. a spacetime with no mass.

Things are easiest to explain from this - take a 'test particle with mass $m$' (something that has a mass that somehow doesn't have an effect on spacetime for mathematical convenience. This will give you the general idea of how things work to a certain extent). Now, this 'test particle' follows what is known as a geodesic. Lets say this particle is moving in the $x$-direction: the metric becomes $dS^2 = -dt^2 + dx^2$. For a massless particle, this becomes $dt^2 = dx^2$ and so $\frac{dx}{dt} = \pm 1$. This is the velocity of a photon.

For a massive particle, this gets a little more complicated - use Lagrangian mechanics $\left(\dot{x} = \frac{dx}{d\tau}\right)$ where $\tau$ is the 'proper time' of the test particle i.e. $t$ is the time of the observer, $\tau$ is the time of the test particle: $$\mathcal{L} = -1 = -\dot{t^2} + \dot{x}^2$$ which gives $$\frac{\partial\mathcal{L}}{\partial t} = \frac{\partial\mathcal{L}}{\partial x} = 0$$ and $$\frac{\partial\mathcal{L}}{\partial \dot{t}} = -2\dot{t} = constant, \frac{\partial\mathcal{L}}{\partial \dot{x}} = 2\dot{x} = constant$$ giving $\frac{dx}{dt} = constant$. Again, the velocity of something is constant.

Sticking something massive in there (that's not a test particle) and saying that it is stationary and static gives the Schwarzchild Metric $$dS^2 = -Fdt^2 + F^{-1}dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\varphi^2$$ where $F = F\left(r\right) = 1 - \frac{r_s}{r}$ is a function of the mass of the (non-test) object. If this object has a charge, then $F$ is also a function of the charge - this is known as the Reissner"“Nordström metric. However, as there is now an $r$-dependence, the equations of motion have an additional term. This leads to what we call Gravity - due to the curvature of spacetime, things no longer travel in what you would normally consider to be a straight line.

Answer

$t$ is the time of an observer at infinity, $\tau$ is the time of the test particle (ship), so $\frac{dr}{dt}$ is the velocity according to an observer at infinity

Now, the Alcubierre metric is formulated to have a spaceship at $x_s\left(t\right)$ and can be written as $$dS^2 = -dt^2 + \left(dx - v_s\left(t\right)f\left(r_s\right)dt\right)^2 + dy^2 + dz^2$$ for a ship travelling in the $x$-direction. $f\left(r_s\right)$ is the 'warp bubble' function. $v_s\left(t\right) = \frac{dx_s}{dt}$ is the velocity of the ship and is intrinsic to the metric being used. In other words, the ship travels under gravity, albeit gravity created by exotic matter. The ship stops a distance from wherever you're travelling from and creates a warp bubble using exotic matter (matter with a negative energy density), or perhaps creates exotic matter as a result of creating the bubble. This bubble propels the ship along, just like gravity, only the ship takes a much shorter time to reach the destination than light (and you don't experience g-forces as you're in free-fall). Half-way along the trip, you reverse the direction of the warp bubble so that a certain distance away from the destination, you're now stationary and can switch the bubble off. During the entire journey, you haven't violated any laws of relativity, except for the energy conditions, which can also be violated in QFT.

It should also be noted that the speed of the ship is not constant, but the acceleration $\left(a\right)$ is. Half-way, the acceleration changes from $a$ to $-a$. The ship always remains within the bubble - either multiple bubbles are pre-set in space which are triggered by the ship, or the ship creates its own bubble which travels with the ship. A change in 'proper time' (of the ship) = A change in 'co-ordinate time' (of a distant observer) i.e. There is no time dilation or anything because the warp bubble warps the space in front of the ship giving $dt = d\tau$. This is essentially what allows the ship to appear to go FTL.

If you want to change the direction, then the best idea would probably be to reverse the direction of the bubble until you stop, then create a new bubble in the new direction you want to go in.

Edit to answer questions from the comments:

There are two types of co-ordinates (as far as this question is concerned anyway): That of the ship and that of a distant observer (someone so far from the bubble that they are completely unaffected by its existence or lack thereof). In the ship's frame, $dx = vdt$ and $f\left(r_s\right) \simeq 1$ in a small area around the ship and so $d\tau = dt$, directly giving no time dilation. As it moves on a geodesic by design, the proper acceleration is $0$, just like when under free-fall in a 'gravitational field' (such as on Earth) even though the bubble (and so, ship) is accelerating from the perspective of a distant observer. In other words, you feel no acceleration (literally, you're still in 0-gravity), yet a distant observer will see you accelerating. As you're not actually accelerating, your proper velocity is also $0$, despite the fact that you're moving arbitrarily fast to a distant observer! A neat comparison to this is that distant galaxies appear to be travelling away from us at relativistic velocities, only (on average) they're not - it's the expansion of space causing this effect. That's just what's happening here - the space in front of the ship is contracting and the space behind is expanding again, so it's very like the cosmological constant.

You might be able to change direction without stopping, but the change to the metric is no longer smooth. Having said that, adding another bubble in an orthogonal direction should be OK. So, if you're travelling in the $x$-direction and want to travel in the $xy$-plane at 45 degrees to that, add an equally strong bubble in the $y$-direction, as opposed to rotating the bubble. Maybe you can rotate the bubble - it might be OK, but the changes may not be smooth, so I couldn't guarantee it. I suppose, considering that you're ignoring end effects of creating/getting rid of the bubble, it could be valid to arbitrarily rotate the bubble. Again, the whole 'ship stopping' thing is just to keep changes to the metric smooth, although, really, it should be fine to create the bubble while moving (if not, then indeed the question is what are you stationary relative to?). I don't really know what accelerating (using e.g. rockets to give a proper acceleration) would do though. It's easier to refer to the velocity it created the bubble at as 'stopped' though as creating the bubble at a different velocity to the one where it's destroyed could cause problems with the metric, so again, how exactly this would work isn't really known. I'd guess either the bubble is self-preserving, it doesn't matter or it's likely that you'd end up with a black hole or something? Certainly, the tidal forces at the edge of the bubble are massive, so you don't want to get near to the edge of the bubble.

posted about 8 years ago

Mithrandir24601‭ staff

104 reputation 3 6 17 1