Does the Alcubierre drive shorten distances?

[javisot]

I want to highlight one point: in warp, the energy conditions are violated, and it is precisely the violation of the energy conditions that produces the effective displacement.
https://arxiv.org/abs/2103.05610

 

[PAllen]

So, the easy part is as follows: my traveler has reached the left edge of the right bubble wall at t (coord time) of 4.95, z of 16.295, $\tau=4.95\sqrt{.99}$, odometer reading of $.495\sqrt{.99}$.

The next part involves a slightly messy differential equation to track traveler through wall (rather than just integration for first part). It is first order and linear, though, given the simplified f i am using. So it can be solved explicitly.

 

[PeterDonis]

in wall Eulerian observer all get asymptotically trapped against the back of the forward wall until the bubble slows down.

Yes. And in the back wall, Eulerian observers asymptotically get "peeled off" of the wall and released into the region behind the bubble.

 

[PAllen]

So, the easy part is as follows: my traveler has reached the left edge of the right bubble wall at t (coord time) of 4.95, z of 16.295, $\tau=4.95\sqrt{.99}$, odometer reading of $.495\sqrt{.99}$.

The next part involves a slightly messy differential equation to track traveler through wall (rather than just integration for first part). It is first order and linear, though, given the simplified f i am using. So it can be solved explicitly.

Before the full solution, I can put an upper bound on the odometer reading. It is easy to compute, using the full metric, rigorously, that for my definition of T, at all time, $d\tau/dt=\sqrt{.99}$. Thus, even if the traveler is assumed trapped at the beginning of the wall until the bubble disappears, this only adds $.505\sqrt{.99}$ to the odometer reading (this is because the bubble wall is reached at t of 4.95, and the bubble is gone after t of 10, and the odometer integrand is always .1). Then the bubble wall is traversed normally. So the total odometer reading is less than 1.1 on escape from the bubble. This has taken just over 10 units of time. The external distance traveled is a little over 32. So this fully validates the notion that if measure distance traveled internal to the bubble, using the traveler’s measurements, you have only traveled subliminally. Meanwhile, with the parameters I chose, the external speed of the bubble is an average of 3.2c. Note, as I claimed earlier, the distance traveled internally is essentially just the bubble width.

But wait,the match against bubble width is a bit of a coincidence of parameters chosen. If you imagine that the steady 4c phase of the bubble continues for 100 units (instead of the 8 in the model I proposed earlier) and T trapped at the wall for most of this phase, the odometer would read about 10 for this phase. So the internal measurement remains always subliminal, but it is not just related to the bubble width. Instead, a major part of it is simply proportional to the trapping time.

 

[PAllen]

Just to take up a detail discussed earlier: whether There is a difference between computing distance using a geodesic of a Euclidean slice, vs a spacetime geodesic orthogonal to a bubble rider. The claim was made that these are the same. I do not believe this is true. In particular, I think the orthogonal geodesic from bubble center rider diverges from Euclidean slice during traversal through the wall. That is, in the wall region, geodesics of the Euclidean slice are not spacetime geodesics. Thus these distance definitions would disagree, but not by any large amount (I think).

 

[PeterDonis]

I think the orthogonal geodesic from bubble center rider diverges from Euclidean slice during traversal through the wall.

Any curve in a surface of constant time in the warp coordinates, and only such curves, are orthogonal to all Eulerian worldlines. That fact is obvious by inspection of the line element and the definition of Eulerian worldlines.

However, that in itself does not guarantee that geodesics of the Euclidean slices are always spacetime geodesics. There are two geodesic equations to check, and it's only obvious by inspection that geodesics of the Euclidean slices satisfy one (the second one given in the Natario paper, between Propositions 1.3 and 1.4). If the other one is not satisfied within the bubble wall, there might be no orthogonal geodesics to Eulerian worldlines within the wall (because the only available curves that are orthogonal to those worldlines within the wall would not be geodesics).

 

[PAllen]

Any curve in a surface of constant time in the warp coordinates, and only such curves, are orthogonal to all Eulerian worldlines. That fact is obvious by inspection of the line element and the definition of Eulerian worldlines.

However, that in itself does not guarantee that geodesics of the Euclidean slices are always spacetime geodesics. There are two geodesic equations to check, and it's only obvious by inspection that geodesics of the Euclidean slices satisfy one (the second one given in the Natario paper, between Propositions 1.3 and 1.4). If the other one is not satisfied within the bubble wall, there might be no orthogonal geodesics to Eulerian worldlines within the wall (because the only available curves that are orthogonal to those worldlines within the wall would not be geodesics).

There is always a geodesic orthogonal to any world line at any point. What happens is the spacetime geodesic and the slice geodesic are tangent at that point, but diverge elsewhere. This is easily possible. Consider a diameter line in a ball. At the 2 sphere surface, a great circle and a tangent line of the space are tangent. One is a geodesic of the 2-sphere, the other of the overall space. Both are orthogonal to the diameter. Same thing happens here in the wall.

 

[PeterDonis]

There is always a geodesic orthogonal to any world line at any point.

Yes, you're right, I was too pessimistic there.

However, any such geodesic orthogonal to an Eulerian worldline would have to lie within a surface of constant time in the warp coordinates. Again, that's obvious by inspection of the line element and the definition of an Eulerian worldline. If a geodesic of the Euclidean surfaces isn't a spacetime geodesic within the wall, then a spacetime geodesic within the wall that's orthogonal to an Eulerian worldline there would have to be some non-geodesic curve when viewed from within the Euclidean surface--but still confined to that surface.

 

[PAllen]

Yes, you're right, I was too pessimistic there.

However, any such geodesic orthogonal to an Eulerian worldline would have to lie within a surface of constant time in the warp coordinates.

This I don’t believe. One thing that is confusing about the notation in some of these papers is that f is a function of all 4 spacetime coordinates. That leads to different Euler Lagrange equations than if you don’t notice this.

 

[PeterDonis]

This I don’t believe.

Any curve whatsoever that is orthogonal to an Eulerian worldline must have a zero $t$ component in the standard warp coordinates, which means it must lie in a Euclidean surface of constant time. I've said several times that that's obvious by inspection of the line element and the definition of an Eulerian worldline, but now I'll go ahead and show it explicitly.

The line element is (for a bubble moving in the $z$ direction, as in the first paper)

$$
ds^2 = - dt^2 + dx^2 + dy^2 + \left( dz - v f dt \right)^2
$$

The 4-velocity of an Eulerian worldline is $u = (1, 0, 0, vf)$. So the inner product of a Eulerian worldline's 4-velocity with an arbitrary vector $e = (e^t, e^x, e^y, e^z)$ is

$$
u \cdot e = 1 \cdot e^t + 0 \cdot e^x + 0 \cdot e^y + \left( vf - vf \cdot 1 \right) \cdot \left( e^z - vf e^t \right) = e^t
$$

Therefore, $e^t = 0$ is a necessary and sufficient condition for an arbitrary vector to be orthogonal to the 4-velocity of an Eulerian worldline.

None of the above depends on what $f$ is a function of. In the Natario paper they appear to allow for $X^i$ (which is their general notation for the spatial part of the Eulerian observer's 4-velocity, and which is $v f$ in the $z$ direction for the Alcubierre case) to be a function of all four coordinates, so I don't see an issue there. In any case, what curves solve the geodesic equation is a separate question from what curves are orthogonal to Eulerian worldlines.

 

[PAllen]

Any curve whatsoever that is orthogonal to an Eulerian worldline must have a zero $t$ component.

A curve orthogonal to all Eulerian world lines it touches must have zero t component. But one that is orthogonal to just one need not; its tangent at that point has no t component. Note, besides my own work, which may have errors. I have seen no claim in any of the papers that the geodesics of the Euclidean slices are spacetime geodesics. My claim is that within the wall, what happens is exactly like the sphere example I gave - one curve orthogonal to a Eulerian world line that remains orthogonal others (and is a geodesic of the slice), and another curve, tangent at that point, orthogonal to that Eulerian world line at that point, but to no other, that is a geodesic of the spacetime and does not remain in the slice.

 

[PeterDonis]

one that is orthogonal to just one need not

Ah, I see. So this is not the "orthogonal foliation" notion of distance, or the "odometer" notion of distance. It's the other one from your previous post (a while back now).

one curve orthogonal to a Eulerian world line that remains orthogonal others (and is a geodesic of the slice), and another curve, tangent at that point, orthogonal to that Eulerian world line at that point, but to no other, that is a geodesic of the spacetime and does not remain in the slice.

From looking at the geodesic equations in the Natario paper, yes, you would have to have $\ddot{t} \neq 0$ at the point of tangency for the curve to be a spacetime geodesic inside the wall. At that point, $\dot{t} = 0$, so most of both geodesic equations vanish, and what's left is $\ddot{x} = 0$ and $\ddot{t} = - \dot{X}^i \dot{x}^i$, and $\dot{X}^i \neq 0$ inside the wall (which in the notation of the first paper corresponds to $f$ changing inside the wall).

 

[PAllen]

I would like to summarize conclusions scattered across this thread.

The following are key references:

The main document that generalized Alcubierres original to a bubble that starts and stops, and derives many featrues:
https://scipost.org/SciPostPhysLectNotes.10/pdf
A paper @PeterDonis has used (I haven't looked at this one):
https://arxiv.org/pdf/gr-qc/0110086
My reference post formalizing some definitions of distance in GR:

P

Undergrad Post in thread 'Does the Alcubierre drive shorten distances?'

Apr 30, 2026

Getting back to the notion of distance, and how to make sense of it in relation to the warp drive metric, a few common conventions in GR as:

1) If you have a timelike congruence, and an orthogonal foliation for it, then defining distance in terms ot the geodesics of the induced 3-metric of the foliation makes a lot of sense. In particular, this is true FLRW spacetime and Minkowski spacetime. It is also possible for the stationary congruence outide a BH. It is rare, in the general case, and is not possible in the warp spacetime. Note that of these 3 examples, only in Minkowski...

• PAllen

• 

My post formalizing a computable complete model of a warp bubble starting and stopping (heavily based on the details in the first reference above):

P

Undergrad Post in thread 'Does the Alcubierre drive shorten distances?'

May 1, 2026

I’m working through a concrete example. I’ll post the details of what I am assuming in case anyone else wants to work the same case. This example has the feature that the traveler arrives at the origin just as bubble center starts increasing from zero - which is actually formation of the bubble. Thus the traveler is at the center of the bubble as soon as the bubble starts to exist, effectively. I’ve traced it part way through the bubble, but not up to the wall yet.

For f I am using the piecewise linear function proposed in(211) of the paper we’ve been using. I suppress x and y, leaving...

• PAllen

The starting question of the thread was "Does the Alcubierre drive shorten distances"? The main answer, is that using the only definition in the reference post above that does measurements inside the bubble and also captures length/distance contraction exactly as it occurs in SR, finds that distance is highly contracted compared to distance per a global, hypersurface orthogonal foliation. This is exactly as expected, but the details are quite subtle compared to SR. Note that Alcubierre's answer simply posits that a bubble rider chooses to use the global distance definition, which is not the relevant choice. If a rocket traveler in SR, rather than measuring according to their changing speed per the global fame, uses the global foliation, they, of course don't find any distance contraction either. This is the high level summary.

Nothing much will be said about the global foliation distance. This simply used the hypersurface orthogonal Euclidean slices presented in all the original references. The only comment is that in the concrete model in the second reference post above, I have set it up so that the global distance traveled by the bubble center is 32 units (say light years), and the main coasting phase occurs at "warp 4": 4 times light speed, and covers 24 of the 32 light years total (4 ly covered in startup acceleration phase, 4 ly covered in bubble deceleration phase).

The second distance I described in the reference post above was distance along a spacelike geodesic orthogonal to an observer's world line. This does the 'best' generalization of a local inertial frame to large scope in GR. It often provides insight in GR contexts (for example, it shows that cosmological redshift up to 'medium' distances can be treated to first order as ordinary redshift due to motion). Unfortunately, in the warp spacetime, there is a serious complication. The path of this geodesic originating from a central bubble rider is primarily determined by details of the wall metric function, that one generally wants to treat as irrelevant. While inside the bubble, this geodesic would follow the Euclidean slice. In the wall it would get deflected in a way determent by the details of the wall metric. On leaving the wall, it would again be following a spacelike geodesic of Minkowski space, but with a deflected time component compared to the global foliation (because the bubble is taken to be spherically symmetric, the only possible deflection is in the 'time' direction). What this means is that it would reflect simultaneity of a boosted frame. Potentially, you could get any distance you want (from near 0 to 32) for the measure of this geodesic depending on how your tailored wall metric deflects this geodesic. IMO, this is not very meaningful.

One other common, less geometric approach to distance is to interpret light signal bounces. However this also runs into a complication with the warp spacetime. A light signal emitted by the bubble rider once the bubble is superluminal simply gets trapped in the wall until the bubble becomes subluminal. Then, it can escape and return as the bubble rider is approaching the target (once the bubble has slowed down). In the example in the last post reference above, light emitted when the bubble was just up to speed would take about .5 year to reach the wall, then be trapped for 9 years (that's how long before the bubble just drops into subluminal speed), then take about a 1 year round trip to the traveler heading at modest speed out of the now stopped bubble (assuming the target is just outside the bubble radius). If you insisted on interpreting this 'normally' you would conclude the that the target was 5.5 ly away at the the middle of your journey. Again, I don't find this very meaningful. The warp space time, as it were, is throwing up obstacles to the most common ways of talking about distance in relativity (other than the global foliation).

So we are left with relativistic odometer I described. It is worth mentioning why so called Eulerian observers are the most reasonable choice as a reference for measuring travel. Physically, these are the free fall observers that are always stationary per the global Euclidean foliation. This is captured in the statement that their world lines are everwhere orthogonal to the Euclidean slices of constant t coordinate. As a result, they are also coordinate stationary in the Gaussian-Normal coordinates produced by this foliation. Using this idealized device, you get some behavior similar to an SR rocket travler, and some behavior radically different due to the features of the warp spacetime. The part similar to SR, is that if you maintain always a high speed relative (above a threshold discussed below) to Eulerian observers, then the higher such speed, the smaller the odometer distance, just as for a rocket in SR. This distance will approach zero for near lightspeed relative to Eulerian observers. However, for smaller relative speeds, it is time for something completely different (doing justice to Monty Python). You can get many different results depending on the exact pattern of your motion relative to Eulerian observers.

First, there is another way to get near zero distance. Simply go at near light speed from the start (just outside the bubble) to the center of where the bubble starts forming. Then stop dead relative to this Eulerian observer. Wait until the bubble has stopped and disappeared. Then go at near light speed to the destination assumed just outside the other side of where the bubble was. This will also get near zero distance. However, your clock time will be completely different than the prior case. In the case of always high relative speed, you will spend most of your time frozen in the bubble wall, with your clock reading very slow compared to external time (note, that even though you are frozen in the wall, at all times your local speed relative to Eulerian observers is very high). [remember, it is impossible to leave the bubble while it is superluminal; and if you travel at some speed relative Eulerian observers, you can make no progress through the wall until the bubble is globally slower than that relative speed]. So your clock will read very little time. However, for the second way of measuring near zero distance, you clock will read about 10 years, as you simply ride the bubble center at rest with respect to the central Eulerian observer. So, though both these strategies measure near zero distance, they end up with completely different watch times at the end.

If, instead, you move slowly (e.g. less than .1 c) (I will no longer bother saying relative to Eulerian observers) to the buble center from the start, then wait, then move slowly to destination once bubble has stopped, you will measure 1+ ly - the diameter of the bubble plus a tiny bit on each end. If instead, we go .1c relative speed the whole time, we get a similar distance (this was the case analyzed a few posts ago), with similar clock time, but this is mostly a coincidence of the model parameters I chose. If instead, you go .2c the whole time, you get a bit less than 3 ly as your total distance traveled. The distance measured by the odometer starts to be dominated by proper time spent trapped in the wall times relative speed of Eulerian observers going by (you still see this speed happening locally, despite extreme compression from the external point of view). To see this, note that the vector $\gamma(u)(1,vf+u)$, for all $u<c$ is a timelike unit vector with relative speed u compared to the Eulerian observer $(1,vf)$. But this behavior (increasing measured distance with increasing speed) doesn't continue. The wall trapped odometer reading is a competition between increasing speed of passage of Eulerian observers with decreasing proper time. The maximum is reached for speed $c/\sqrt{2}$. At this speed, the wall trapped period produces around 5 ly of measured distance. Higher speed than than this steadily decreases measured distance toward zero, as the decrease in proper time dominates the increasing relative speed of the Eulerian observers.

For me, thus sums up the major findings of this thread. Major thanks to @PeterDonis for clarifying discussions, and the push to look at the case of constant speed relative Eulerian observers over the whole trip (rather than only the start/stop cases I was initially considering).

 

[PeterDonis]

Physically, these are the free fall observers that are always stationary per the global Euclidean foliation.

Not on the obvious layman's meaning of "stationary": not all Eulerian observers have constant spatial coordinates with respect to this foliation. Note also that this spacetime is not stationary by the usual meaning of that term in GR--there is no timelike Killing vector field.

This is captured in the statement that their world lines are everwhere orthogonal to the Euclidean slices of constant t coordinate.

"Orthogonal foliation" is a separate concept from "stationary".

For comparison (apt considering what the metric looks like in the alternate coordinate chart I described earlier in the thread, using $\xi$ instead of $x$ as the spatial coordinate in the direction of the bubble's motion, so that Eulerian observers inside the bubble have constant spatial coordinates, while those outside the bubble don't), consider Painleve coordinates in Schwarzschild spacetime. Painleve observers, who free-fall radially inward from rest at infinity, have worldlines that are everywhere orthogonal to the surfaces of constant Painleve coordinate time; but those Painleve observers are not at rest in those coordinates.

 

[PAllen]

Not on the obvious meaning of "stationary": not all Eulerian observers have constant spatial coordinates with respect to this foliation. Note also that this spacetime is not stationary by the usual meaning of that term in GR--there is no timelike Killing vector field.


"Orthogonal foliation" is a separate concept from "stationary".

For comparison (apt considering what the metric looks like in the alternate coordinate chart I described earlier in the thread, using $\xi$ instead of $x$ as the spatial coordinate in the direction of the bubble's motion, so that Eulerian observers inside the bubble have constant spatial coordinates, while those outside the bubble don't), consider Painleve coordinates in Schwarzschild spacetime. Painleve observers, who free-fall radially inward from rest at infinity, have worldlines that are everywhere orthogonal to the surfaces of constant Painleve coordinate time; but those Painleve observers are not at rest in those coordinates.

good catch, I'll fix.

 

[PAllen]

Not on the obvious layman's meaning of "stationary": not all Eulerian observers have constant spatial coordinates with respect to this foliation. Note also that this spacetime is not stationary by the usual meaning of that term in GR--there is no timelike Killing vector field.


"Orthogonal foliation" is a separate concept from "stationary".

For comparison (apt considering what the metric looks like in the alternate coordinate chart I described earlier in the thread, using $\xi$ instead of $x$ as the spatial coordinate in the direction of the bubble's motion, so that Eulerian observers inside the bubble have constant spatial coordinates, while those outside the bubble don't), consider Painleve coordinates in Schwarzschild spacetime. Painleve observers, who free-fall radially inward from rest at infinity, have worldlines that are everywhere orthogonal to the surfaces of constant Painleve coordinate time; but those Painleve observers are not at rest in those coordinates.

Not on the obvious layman's meaning of "stationary": not all Eulerian observers have constant spatial coordinates with respect to this foliation. Note also that this spacetime is not stationary by the usual meaning of that term in GR--there is no timelike Killing vector field.


"Orthogonal foliation" is a separate concept from "stationary".

For comparison (apt considering what the metric looks like in the alternate coordinate chart I described earlier in the thread, using $\xi$ instead of $x$ as the spatial coordinate in the direction of the bubble's motion, so that Eulerian observers inside the bubble have constant spatial coordinates, while those outside the bubble don't), consider Painleve coordinates in Schwarzschild spacetime. Painleve observers, who free-fall radially inward from rest at infinity, have worldlines that are everywhere orthogonal to the surfaces of constant Painleve coordinate time; but those Painleve observers are not at rest in those coordinates.

True, but if you switch to coordinates that manifest the orthogonality, then the Eulerian world lines do have constant spatial coordinates. Similar to going from Gullestrand-Panlieve to Lemaitre. So now I think there is something I want to capture here, but need better wording that what I said first. The key point is that if you make a local frame with a Eulerian world line and the constant t Euclidean slice, the Eulerian world line is stationary in this local frame (precisely because it is orthogonal).

Another way of saying this: consider the congruence of Eulerian world lines, and the orthgonal Euclidean slices. The natural orthogonal coordinates this produces would have each Eulerian world line keeping constant spatial coordinates. This would be the Gaussian-Normal form of the metric.

 

[PeterDonis]

if you switch to coordinates that manifest the orthogonality, then the Eulerian world lines do have constant spatial coordinates.

Yes, I've had a couple of tries at constructing such a chart, but haven't had enough time to get there. It would be interesting to see what the line element looks like. One obvious complication is that the "stretching" or "compressing" of the Eulerian worldlines so that they all have constant spatial coordinates is time-dependent.

The key point is that if you make a local frame with a Eulerian world line and the constant t Euclidean slice, the Eulerian world line is stationary in this local frame (precisely because it is orthogonal).

Yes, whichever Eulerian worldline you "center" your frame on will be at rest in a chart constructed this way.

This seems to me to have some similarities with comoving observers in FLRW spacetime in standard FLRW coordinates.

 

[PAllen]

My current view is back to the idea that you can have synchronous, but not diagonal. However, the trick used by Hiscock for diagonalization does not work in the full 3+1 context, so diagonalization really is not possible, even in a finite patch (in the wall, that is). But Gaussian normal form should be achievable globally. It will have irremovable space-space terms.

 

[PeterDonis]

My current view is back to the idea that you can have synchronous, but not diagonal.

It seems to me that there's an obvious diagonalization: just define a coordinate $\zeta$ such that $d \zeta = dx - v f dt$. Then $d \zeta = 0$ along every Eulerian worldline, and the metric is manifestly diagonal since its fourth term just becomes $d \zeta^2$.

The expression for $\zeta$ itself would look messy, though.

 

[PAllen]

It seems to me that there's an obvious diagonalization: just define a coordinate $\zeta$ such that $d \zeta = dx - v f dt$. Then $d \zeta = 0$ along every Eulerian worldline, and the metric is manifestly diagonal since its fourth term just becomes $d \zeta^2$.

The expression for $\zeta$ itself would look messy, though.

I don’t think that works. I tried that. Remember f is a function (hidden within r) of all 4 coordinates. That means zeta must be as well. Then its total derivative could not look like that.

One way to check a method is to try it on the Kerr metric. If it seems to be able to work on the Kerr metric, then it is not valid, because the Kerr metric is known to be non diagonalizable.

One obvious point is that as a general rule for transforming a metric (covariant) is that you want old in terms of new; then when you take the differential, you get the right partials for the transform.

 

[Jaime Rudas]

Davis & Lineweaver present their equations 19 and 20 as follows:
$$\dot D= \dot R\chi+R \dot \chi$$
$$v_{tot}=v_{rec}+v_{pec}$$

Regarding the D&L paper, I think there might be an error in footnote 11 on page 19.

¹¹The recession velocity at the time of emission is $v_{rec}(t_{em}) = R(t_{em})\chi(z)$ where $R(t_{em}) = R(t)$ as defined in Eq. 23

Shouldn't it be $v_{rec}(t_{em}) =\dot R(t_{em})\chi(z)$?