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There seem to be honest scientific exploration in the field of the alcubierre warp drive if you believe some articles on the net. So to be a little bit scientific accurate in a story using it, I wonder about the following:

The warp drive deflates space in front and inflates it to the back of the ship, therefore shortening the distance to your destination and circumvent the light barrier problem.

But what now? Do I have to move in the direction by conventional thrusters and my (real) speed is multiplied by the compression factor (meaning I have to achieve either high real speed or high compression)? Do I have to "step over" the inflated space and then the drive will relax the space and inflate the next part? Do I have to do this with a very high frequency to obtain more relative speed? Or does some part of the alcubierre drive move the space bubble I'm in in the desired direction?

So, what (theoretically) would/could define the speed and direction of such a drive? What would the ship have to do to change speed and/or direction?

I know the exotic matter and "everything will be destroyed at your destination" debate. But for the sake of this question, please ignore it.

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    $\begingroup$ How about distance (measured via a stationary observer)/time (measured via a stationary observer). $\endgroup$ – PyRulez Aug 15 '16 at 11:58
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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.

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  • $\begingroup$ You made me curious. That means it's an accelerated flight? With no speed limit? Then the question is, what determines the acceleration? So there is no "real" movement (in space) involved and the ship has essentially speed 0 in space? Why one has to stop first to change direction? If I can accelerate and decelerate, am I not able to accelerate in another direction changing my vector? The ship has to stop before creating the bubble? To stop in which frame? What happens if we turn off the drive (the bubble?) before we stopped? Or is the bubble self preserving until stop? Again in which frame? $\endgroup$ – Hothie Aug 15 '16 at 14:18
  • $\begingroup$ The alcubierre drive is momentum-less, there is no acceleration or gravity to be heard of (beyond undesirable side-effects). The drive works by moving the space surrounding the ship - "the ship and its inhabitants would experience no proper acceleration." Meaning that your penultimate paragraph, and much of the prior paragraph, are exactly the opposite explanation of what is really happening. The ship would return to its original subluminal speed the very instant that the drive is turned off. $\endgroup$ – Jonathan Dickinson Aug 15 '16 at 14:49
  • $\begingroup$ @JonathanDickinson When I say 'gravity', I mean that it's moving in a way that's explained by general relativity (hence the preamble). There is co-ordinate acceleration although no proper acceleration. I'll edit the question to include this. Again by velocity, it's the co-ordinate velocity that I'm using - if $v_s$ is set to $0$, then we recover the Minkowski metric. If the bubble is suddenly turned off while $v_s \neq 0$, then the change in the metric isn't smooth. I'll edit the answer to include the other questions $\endgroup$ – Mithrandir24601 Aug 15 '16 at 15:04
  • $\begingroup$ @Mithrandir24601 it occurs to me that you might be talking about the frame outside of the bubble, rendering my comment moot. $\endgroup$ – Jonathan Dickinson Aug 15 '16 at 15:06
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    $\begingroup$ @JonathanDickinson Essentially, yep :) Most of this uses the co-ordinate time, $t$, and co-ordinate values which are the co0ordinatess of a distant observer i.e. an observer that's far enough away from the bubble so as to not experience any of the effects of the bubble. While the proper distance the ship has travelled is $0$, to a distant observer, the ship has moved. $\endgroup$ – Mithrandir24601 Aug 15 '16 at 15:09

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