# Alcubierre Drive - Clarification on relativistic effects

On the Wikipedia article on the Alcubierre drive, it says:

Since the ship is not moving within this bubble, but carried along as the region itself moves, conventional relativistic effects such as time dilation do not apply in the way they would in the case of a ship moving at high velocity through flat spacetime relative to other objects.

And...

Also, this method of travel does not actually involve moving faster than light in a local sense, since a light beam within the bubble would still always move faster than the ship; it is only "faster than light" in the sense that, thanks to the contraction of the space in front of it, the ship could reach its destination faster than a light beam restricted to travelling outside the warp bubble.

I'm confused about the statement "conventional relativistic effects such as time dilation do not apply".

Say Bob lives on Earth, and Jill lives on a planet in Andromeda, and we'll say for the sake of argument that they're stationary. If I were to travel from Bob to Jill using an Alcubierre drive such that the journey would take me, say, 1 week from my reference frame... how long would Jill have to wait from her reference frame? Do the time dilation effects cancel out altogether? Would she only wait 1 week?

Spacetime can dynamically evolve in a way which apparently violates special relativity. A good example is how galaxies move out with a velocity v = Hd, the Hubble rule, where v = c = Hr_h at the de Sitter horizon (approximately) and the red shift is z = 1. For z > 1 galaxies are frame dragged outwards at a speed greater than light. Similarly an observer entering a black hole passes through the horizon and proceeds inwards at v > c by the frame dragging by radial Killing vectors.

The Alcubierre warp drive is a little spacetime gadget which compresses distances between points of space in a region ahead of the direction of motion and correspondingly expands the distance between points in a leeward region. If distances between points in a forwards region are compressed by a factor of 10 this serves as a “warp factor” which as I remember is $w~=~1~+~ln(c)$, so a compression of 10 is a warp factor 3.3. The effect of this compression is to reduce the effective distance traveled on a frame which is commoved with the so called warp bubble. This compression of space is given by $g_{tt}$ $=~1~-~vf(r)$.

Of course as it turns out this requires exotic matter with $T^{00}~<~0$, which makes it problematic. Universe is also a sort of warp drive, but this is not due to a violation of the weak energy condition $T^{00}~\ge~0$. Inflationary pressure is due to positive energy. The gravity field is due to the quantum vacuum, and this defines an effective stress-energy tensor $T^{ab}$ with components $T^{00}~=~const*\rho$, for $\rho$ energy density, and $T^{ij}~=~const*pu^iu^j$, for $i$ and $j$ running over spatial coordinates $u^i$ velocity and $p$ pressure density. For the de Sitter spacetime the energy density and pressure satisfies a state $p~=~w*\rho$ where $w~=~-1$. So the pressure in effect is what is stretching out space and frame dragging galaxies with it. There is no need for a negative energy density or exotic matter.

Negative energy density or negative mass fields have serious pathologies. Principally since they are due to quantum mechanics the negative eigen-energy states have no lower bound. This then means the vacuum for these fields is unstable and would descend to ever lower energy levels and produce a vast amount of quanta or radiation. I don’t believe this happens. The Alcubierre warp drive then has a serious departure between local laws of physics and global ones, which is not apparent in the universe or de Sitter spacetime. The Alcubierre warp drive is then important as a gadget, along with wormholes as related things, to understand how nature prevents closed timelike curves and related processes.

The question was asked about the redshift factor and the cosmological horizon. This requires a bit more than a comment post. On a stationary coordinate region of the de Sitter spacetime $g_{tt}~=~1~-~\Lambda r^2/3$. This metric term is zero for $r~=~\sqrt{3/\Lambda}$, which is the distance to the cosmological horizon.

The red shift factor can be considered as the expansion of a local volume of space, where photons that enter and leave this “box” can be thought of as a standing wave of photons. The expansion factor is then given by the scale factor for the expansion of the box $$z~=~\frac{a(t_0)}{a(t)}~-~1$$ The dynamics for the scale factor is given by the FLRW metric $$\Big(\frac{\dot a}{a}\Big)^2~=~\frac{8\pi G\rho}{3}$$ for $k~=~0$. The left hand side is the Hubble factor, which is constant in space but not time. Writing the $\Lambda g_{ab}~=~8\pi GT_{ab}$ as a vacuum energy and $\rho~=~T_{00}$ we get $$\Big(\frac{\dot a}{a}\Big)^2~=~H^2~=~\frac{\Lambda}{3}$$ the evolution of the scale factor with time is then $$a(t)~=~\sqrt{3/\Lambda}e^{\sqrt{\Lambda /3}t}.$$ Hence the ratio is $a(t)/a(t_0)~=~ e^{\sqrt{\Lambda /3}(t-t_0)}$.The expansion is this exponential function, which is Taylor expanded to give to first order the ratio above $$a(t)/a(t_0)~\simeq~1~+~H(t_0)(t_0-t)~=~1~+~H(t_0)(d-d_0)/c$$ which gives the Hubble rule. $z~=~a(t)/a(t_0)~-~1$. It is clear that from the general expression that $a(t)$ can grow to an arbitrarily large value, and so can $z$. On the cosmological horizon for $d~-~d_0~=~r_h~=~\sqrt{3/\Lambda}$ we have $z~=~1$.

Looking beyond the cosmological horizon $r_h~\simeq~10^{10}$ly is similar to an observer in a black hole looking outside to the exterior world outside the black hole horizon. People get confused into thinking the cosmological horizon is a black membrane similar to that on a black hole. Anything which we do observe beyond the horizon we can never send a signal to, just as a person in a black hole can see the exterior world and can never send a message out.

• Isn't $z=\infty$ the redshift of the horizon? $1+z = \frac{\lambda_{now}}{\lambda_{then}}$ Apr 19, 2011 at 13:18
• May I ask what you do for a living Lawrence B. Crowell? Jun 16, 2014 at 14:21
• So, with all this mathematically gobbilty-gook you've written (that I barely comprehend) are you saying in terms of outer space, gravity is such a weak force that it wouldn't take "much" energy for the warp drive to expand/contract space? Similar to the way cleaning vacuums create very high suction by only generating a very small amount of negative pressure? Jun 16, 2014 at 14:26
• While you exposé of the physics behind the Alcubierre Drive is interesting, I am still curious about the answer to the actual question: How much time will the person travelling from Bob to Jill experience compared to the amount of time experienced by Jill (assuming the traveller is moving at non-relativistic speeds within the bubble)? Jul 12, 2017 at 20:02

Relativistic effects happen when you travel near light speed through space-time. With the Alcubierre drive, you don't travel through space-time, but remain stationary and the space-time around you is warped in a way that brings you closer to your destination. So if it only took a week for you, it would only take a week for your observers.

It'd be like if you put two objects on a sheet, and instead of moving them across the sheet, you scrunched up the part of the sheet between them--moving them closer together even though they both remained stationary relative to the sheet.

• 2 things about your 2nd paragraph: 2nd, if you move objects around on a sheet, there's no reason time dilation would enter into the situation unless we're talking about a very large sheet, and very large speeds. 2nd, I'm not really able to visualise the "scrunching". Assuming the sheet is elastic, you could move a region of the sheet closer to a point. In that case, there would be compressed space in between, and expanded behind, but the only way the region can stay there is if the distortions stay. Only solution I can visualise would be to cut the region, and place it at the destination? Apr 19, 2011 at 22:51
• its not a particularily enlightening analogy in this case.. usually sci-fi movies about wormholes try to use the "folded paper with a pen through it"-trick when they explain how to go from A to B in space by topology though :) Apr 20, 2011 at 0:01
• I was just excited to see a question on this site that I could at least partially answer :) Apr 21, 2011 at 2:31

I'm confused about the statement "conventional relativistic effects such as time dilation do not apply".

The Wikipedia section: "Alcubierre Metric" has been updated since your question was asked, but I suspect it's no clearer to you than what you quoted:

The Alcubierre metric defines the warp-drive spacetime. It is a Lorentzian
manifold that, if interpreted in the context of general relativity, allows
a warp bubble to appear in previously flat spacetime and move away effectively
faster than lightspeed. The interior of the bubble is an inertial reference
frame and inhabitants suffer no proper acceleration. This method of transport
does not involve objects in motion at speeds faster than light with respect
to the contents of the warp bubble; that is, a light beam within the warp
bubble would still always move faster than the ship. Because objects within
the bubble are not moving (locally) faster than light, the mathematical
formulation of the Alcubierre metric is consistent with the conventional
claims of the laws of relativity (namely, that an object with mass cannot
attain or exceed the speed of light) and conventional relativistic effects
such as time dilation would not apply as they would with conventional motion
at near-light speeds.


The oversimplified answer to that point is: "The space moves, it's movement does not dilate your time, only your own movement dilates your time.".

Note the statement above: "... inhabitants suffer no proper acceleration".

To prove that the trajectory of the spaceship is indeed a timelike curve, regardless of the value of $v_s(t)$, we substitute $x = x_s(t)$ in the metric (8). It is then easy to see that for the spaceship's trajectory we will have: $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad d=dt .\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad (13)$$ This implies not only that the spaceship moves on a timelike curve, but also that its proper time is equal to coordinate time. Since coordinate time is also equal to the proper time of distant observers in the at region, we conclude that the spaceship suffers no time dilation as it moves. It is also straightforward to prove that the spaceship moves on a geodesic. This means that even though the coordinate acceleration can be an arbitrary function of time, the proper acceleration along the spaceship's path will always be zero.

So the Drive itself causes no time dilation for the occupants of the 'warp field bubble'.

Say Bob lives on Earth, and Jill lives on a planet in Andromeda, and we'll say for the sake of argument that they're stationary. If I were to travel from Bob to Jill using an Alcubierre drive such that the journey would take me, say, 1 week from my reference frame... how long would Jill have to wait from her reference frame? Do the time dilation effects cancel out altogether? Would she only wait 1 week?

For more info about the Alcubierre drive see Harold White's paper at NASA Warp Field Mechanics 101 (.PDF) or Miguel Alcubierre's paper at arXiv.org The warp drive: hyper-fast travel within general relativity.

From what I understand of the mathematical contrivance ...

From the Alcubierre paper:

• Page 2: "... one can actually make such a round trip in an arbitrarily short time as measured by an observer that remained at rest ...".

• Page 3: "... one can use an expansion of spacetime to move away from some object at an arbitrarily large speed. In the same way, one can use a contraction of spacetime to approach an object at any speed. This is the basis of the model for hyperfast space travel that I wish to present here: create a local distortion of spacetime that will produce an expansion behind the spaceship, and an opposite contraction ahead of it. In this way, the spaceship will be pushed away from the Earth and pulled towards a distant star by spacetime itself. One can then invert the process to come back to Earth, taking an arbitrarily small time to complete the round trip.".

• Page 7: Note 3 indicates, from a practical standpoint (if there's such a thing with an impractical device), that you'll want to move some distance from your starting point (Earth) and discontinue use of the drive some distance before arriving at your destination (the Andromeda Gallery, 2.54 ± 0.11 Mly away); failure to do so would disturb the spacetime of the endpoints.

The implication of that is that your motion will cause an extremely tiny time dilation that you might wish to disregard; but if I don't mention it someone may comment about the omission. The faster you travel to the origin's standoff location the sooner you can go to warp, which in turn increases the dilation; but still only by a relatively small amount. Upon arrival you'll want to approach Jill's location at a reasonable speed and landing or docking will take an additional unknown amount of time.

If you traveled 0.10 $c$ you would save (60 * 60) - (0.995 * 60 * 60) = 18 second per hour, and at 0.866 $c$ you'd save 30 minutes per hour, accelerating to that speed is going to take some time.

• Page 8: So it will only take: $$\tau \simeq T \simeq 2 \sqrt{\frac Da}$$

"It is now clear that $T$ can be made as small as we want by increasing the value of $a$. Since a round trip will only take twice as long, we find that we can be back at Earth after an arbitrarily small proper time, both from the point of view of the spaceship and from the point of view of the people on Earth (or at the destination). The spaceship will then be able to travel much faster than the speed of light. However, as we have seen, it will always remain on a timelike trajectory, that is, inside its local light-cone: light itself is also being pushed by the distortion of spacetime.".

So setup time is the only consideration, after obtaining exotic matter and building the ship and engines.

... how long would Jill have to wait from her reference frame? Do the time dilation effects cancel out altogether? Would she only wait 1 week?

If you get into position prior to contacting Jill you'll arrive in her vicinity well before she gets the message. If you drag debris from across the galaxy and impact your destination with them she may be ticked off when you arrive (if you survive the journey). It won't be a week and your unlikely to have an hour difference in your clocks. You'll need to determine standoff distance and how fast you can get there to be any more precise than that. 