Does the Alcubierre drive shorten distances?

[PeterDonis]

the odometer relative to the Eulerian road only measures distance during the short trip into the bubble and out of the bubble.

I'm not sure what you mean by this either. If we want to find a $\hat{v}$ with nonzero odometer distance by your definition, we just need to construct it so that it has $\gamma > 1$ by your definition. That's straightforward. I'll work the construction for $\gamma = 2$, which gives for your odometer distance

$$
d = \frac{\sqrt{3}}{2} t
$$

where $t$ is the elapsed coordinate time in the warp coordinates given in the first paper (the one with the diagram we've referred to).

We have $\hat{u} = \left( 1, v_s f \right)$, where I'll give vector components in $(t, x)$ form. We want to find $\hat{v}$ such that $\hat{u} \cdot \hat{v} = 2$ and $| \hat{v} |^2 = 1$ (i.e., $\hat{v}$ is a timelike unit vector--I'm using the timelike signature convention in this post, which is a sign switch from the spacelike convention used in the paper, but makes this computation easier).

Let $\hat{v} = (v^t, v^x)$. The line element is

$$
ds^2 = dt^2 - \left( dx - v_s f dt \right)^2
$$

Then we have

$$
\hat{u} \cdot \hat{v} = v^t
$$

(because the last term on the RHS of the line element conveniently vanishes when we take any dot product with $\hat{u}$), and

$$
| \hat{v} |^2 = \left( v^t \right)^2 - \left( v^x - v_s f v^t \right)^2
$$

Since we want $\gamma = 2$, that gives $v^t = 2$, and plugging that into the equation for the squared norm of $\hat{v}$, and choosing $v^x$ so that it's moving to the left outside the bubble, we get $\hat{v} = \left( 2, 2 v_s f - \sqrt{3} \right)$. This describes a worldline that, far outside the bubble, moves to the left (the minus $x$ direction) at 87% of the speed of light, but inside the bubble, gets "dragged" to the right, but is still moving to the right (in the Euclidean space of the warp coordinates in the paper) slower than the Eulerian observers inside the bubble, so it's still accumulating leftward odometer distance relative to them.

 

[PAllen]

I'm not sure what you mean by this. There are Eulerian worldlines outside the bubble as well, and using them as the "road" doesn't restrict the "road" to just the bubble and its boundaries.

I mean that with the other distance definitions we used, the existence of the bubble was mostly irrelevant. The experience inside the bubble made minimal contribution. This is fine from the point of view of non bubble observers, but it doesn’t capture the experience of the observer mostly inside the bubble.

 

[PAllen]

I'm not sure what you mean by this either. If we want to find a $\hat{v}$ with nonzero odometer distance by your definition, we just need to construct it so that it has $\gamma > 1$ by your definition. That's straightforward. I'll work the construction for $\gamma = 2$, which gives for your odometer distance

$$
d = \frac{\sqrt{3}}{2} t
$$

where $t$ is the elapsed coordinate time in the warp coordinates given in the first paper (the one with the diagram we've referred to).

We have $\hat{u} = \left( 1, v_s f \right)$, where I'll give vector components in $(t, x)$ form. We want to find $\hat{v}$ such that $\hat{u} \cdot \hat{v} = 2$ and $| \hat{v} |^2 = 1$ (i.e., $\hat{v}$ is a timelike unit vector--I'm using the timelike signature convention in this post, which is a sign switch from the spacelike convention used in the paper, but makes this computation easier).

Let $\hat{v} = (v^t, v^x)$. The line element is

$$
ds^2 = dt^2 - \left( dx - v_s f dt \right)^2
$$

Then we have

$$
\hat{u} \cdot \hat{v} = v^t
$$

(because the last term on the RHS of the line element conveniently vanishes when we take any dot product with $\hat{u}$), and

$$
| \hat{v} |^2 = \left( v^t \right)^2 - \left( v^x - v_s f v^t \right)^2
$$

Since we want $\gamma = 2$, that gives $v^t = 2$, and plugging that into the equation for the squared norm of $\hat{v}$, and choosing $v^x$ so that it's moving to the left outside the bubble, we get $\hat{v} = \left( 2, 2 v_s f - \sqrt{3} \right)$. This describes a worldline that, far outside the bubble, moves to the left (the minus $x$ direction) at 87% of the speed of light, but inside the bubble, gets "dragged" to the right, but is still moving to the right (in the Euclidean space of the warp coordinates in the paper) slower than the Eulerian observers inside the bubble, so it's still accumulating leftward odometer distance relative to them.

Not remotely what I meant. Considering the famous fig. 3, I assume the world line of earth is vertical and just to the left of the bubble region, Alpha Centauri vertical and just to the right of it. I want travel from one to the other. Starting from point on earth world line, i move at modest speed (slanted to the right in the diagram) arriving at the bubble center just as it begins to accelerate. Then stay put until the bubble center has decelerated near appliance centauri. Then move to the right a small amount at modest speed. My odometer distance traveled will be completely determined by the short journies into and out of the bubble.

 

[PeterDonis]

I mean that with the other distance definitions we used, the existence of the bubble was mostly irrelevant. The experience inside the bubble made minimal contribution.

Ah, I see. Well, to the extent that your odometer distance inside the bubble is a lot less than, say, 4.3 light years, that's still true for your odometer definition. But I would agree that with your definition, it's a lot easier to understand the relative contributions, since they're driven basically by how much time a given worldline spends inside vs. outside the bubble, which makes sense.

 

[PAllen]

Ah, I see. Well, to the extent that your odometer distance inside the bubble is a lot less than, say, 4.3 light years, that's still true for your odometer definition. But I would agree that with your definition, it's a lot easier to understand the relative contributions, since they're driven basically by how much time a given worldline spends inside vs. outside the bubble, which makes sense.

Right, and my definition justifies the intuition that travel distance using the bubble to get from earth to Alpha Centauri is small.

 

[PeterDonis]

Not remotely what I meant.

Then I'm really confused about what you meant, since I just took your odometer distance definition and constructed a worldline with a particular $\gamma$ by your definition. How is that not what you meant?

Considering the famous fig. 3, I assume the world line of earth is vertical and just to the left of the bubble region, Alpha Centauri vertical and just to the right of it. I want travel from one to the other.

Sure, just use the worldline I gave, starting from the right edge of the diagram and going to the left edge. In that diagram there will be a segment of the worldline that gets dragged to the right by the bubble, but it will still make it to the left edge eventually. Just compute how much coordinate time that takes and plug into the formula I gave.

If you want to go from Earth to Alpha Centauri instead of Alpha Centauri to Earth, then we can just construct a worldline with the same $\gamma$ as I gave, but moving to the right instead of to the left. Since the bubble itself moves to the right, this worldline will take less time to go across the diagram, so its odometer distance will be less--which means that the odometer distance from Earth to Alpha Centauri by your definition is not isotropic, but that's to be expected since you're using the Eulerian worldlines as the "road" and they're not isotropic.

 

[PeterDonis]

my definition justifies the intuition that travel distance using the bubble to get from earth to Alpha Centauri is small

It would if, for example, a worldline with the $\gamma$ I gave, or at least some reasonable $\gamma$, but moving to the right instead of to the left, spends most of its time inside the bubble, not outside. So the time it takes such a worldline to accumulate enough odometer distance (to the right this time) to cross the bubble would have to be about the same as the time it takes the bubble to make the trip. I'll have to run numbers to see what that looks like when I get a chance.

 

[PeterDonis]

Then stay put

But if you stay put, obviously you're accumulating zero odometer distance--you're just letting the bubble carry you. That seems to me like trying to get out on a technicality, so to speak.

The question, to me, would be whether a worldline that did not stay put--that still tried to move to the right, even inside the bubble--would still accumulate a very small amount of odometer distance (roughly the spatial extent of the bubble, plus a small amount for moving in and out of it). My previous post just now described how I would approach that question.

 

[PAllen]

Then I'm really confused about what you meant, since I just took your odometer distance definition and constructed a worldline with a particular $\gamma$ by your definition. How is that not what you meant?


Sure, just use the worldline I gave, starting from the right edge of the diagram and going to the left edge. In that diagram there will be a segment of the worldline that gets dragged to the right by the bubble, but it will still make it to the left edge eventually. Just compute how much coordinate time that takes and plug into the formula I gave.

If you want to go from Earth to Alpha Centauri instead of Alpha Centauri to Earth, then we can just construct a worldline with the same $\gamma$ as I gave, but moving to the right instead of to the left. Since the bubble itself moves to the right, this worldline will take less time to go across the diagram, so its odometer distance will be less--which means that the odometer distance from Earth to Alpha Centauri by your definition is not isotropic, but that's to be expected since you're using the Eulerian worldlines as the "road" and they're not isotropic.

There is no reason for gamma to be constant. My definition was, in fact, partly motivated by a bug scampering over the surface of a spinning record, both with arbitrary speeds. Problem: describe what an ideal odometer would measure.
So the world line i describe is what I would take to be the natural definition of travel from earth to Alpha Centauri using the bubble. You could use yours, but mine captures what I would think the average person would want for a travel experience.

 

[PAllen]

But if you stay put, obviously you're accumulating zero odometer distance--you're just letting the bubble carry you. That seems to me like trying to get out on a technicality, so to speak.

The question, to me, would be whether a worldline that did not stay put--that still tried to move to the right, even inside the bubble--would still accumulate a very small amount of odometer distance (roughly the spatial extent of the bubble, plus a small amount for moving in and out of it). My previous post just now described how I would approach that question.

Not a technicality. It is the natural result of using the bubble to efficiently get to your intended destination. Note, this would be visually clear in the chart where all Eulerian world lines are vertical, and the metric is diagonal. Note that I think light cones inside the bubble in this chart would be extremely skinny, so you would have no choice but to “move” near vertical inside the bubble.

 

[PeterDonis]

There is no reason for gamma to be constant.

Of course worldlines exist where it isn't constant. But if you're trying to capture the idea of "moving at a certain speed along the road" with your definition, $\gamma$ captures what that speed is. And that's the simplest case of trying to understand what your odometer distance is telling you.

If you let $\gamma$ vary, you're letting your speed relative to the road vary--but that should also make the time it takes you to travel vary. A lower speed takes a longer time to cover the same odometer distance. Again, you can certainly analyze such a case, but it's more complicated. I wanted to keep it simple.

 

[PeterDonis]

It is the natural result of using the bubble to efficiently get to your intended destination.

The "push" to the right that the bubble gives any worldline, regardless of $\gamma$, of course helps you to get to the destination faster. That's not the technicality.

The technicality is having your observer who's trying to get to the destination just stop, relative to the road, and then claiming that, well, that makes the odometer distance smaller. Of course it does. But that's the technicality. To really see the effect of the bubble, you want to have an observer who doesn't stop--who keeps moving along the road--but also gets the extra push because the bubble is moving (or, more pertinently, the road itself is moving) in the same direction the observer is trying to go, and therefore still covers a lot less odometer distance.

The bubble, in other words, is kind of like a moving walkway--yes, you can just stop on it and still get to your destination, and cover zero "odometer distance" in doing so. But again, that's what seems to me to be a technicality. It seems to me that a better question would be whether you can keep walking along the moving walkway, but still only cover an odometer distance about the same as the spatial extent of the bubble.

 

[PAllen]

Of course worldlines exist where it isn't constant. But if you're trying to capture the idea of "moving at a certain speed along the road" with your definition, $\gamma$ captures what that speed is. And that's the simplest case of trying to understand what your odometer distance is telling you.

If you let $\gamma$ vary, you're letting your speed relative to the road vary--but that should also make the time it takes you to travel vary. A lower speed takes a longer time to cover the same odometer distance. Again, you can certainly analyze such a case, but it's more complicated. I wanted to keep it simple.

That's fine, but then, as you noted, the relevant travel direction is to the right in terms of that fig.3. It is now a case of "I like my world line, you like yours, we don't have to agree". Generally, for any "road definition" distance traveled will be essentially independent of relative speed until the relative gamma becomes large, matching simple SR intuition. As relative gamma goes to infinity, odometer distance goes to zero because the integrand is bounded above, while the bounds of integration go to zero.

Note, this means that my "wait in the bubble" is likely to be the upper bound on odometer distance measured for the trip using Eulerian road.

 

[PeterDonis]

as you noted, the relevant travel direction is to the right in terms of that fig.3.

Yes, and that raises an issue that the left-moving case doesn't raise: when the right-moving observer gets to the right edge of the bubble, they can't really get out while the bubble is moving with $v_s > 1$. Basically, Eulerian observers get crunched up against the right edge of the bubble. In terms of the math I posted before, $f$ is a function of both $t$ and $x$, and it's not really possible for an observer with $\beta < 1$ to move into a region where $f \to 0$ while $v_s > 1$--no matter how they try, they'll keep getting pushed along by the bubble so they stay in the region where $f \approx 1$. Only when the bubble comes to a stop at Alpha Centauri can they then step out of it.

for any "road definition" distance traveled will be essentially independent of relative speed until the relative gamma becomes large, matching simple SR intuition.

I'm not sure I understand the "simple SR intuition" you're relying on here. If $\beta = 0$, odometer distance is zero. For small $\beta$, odometer distance goes up linearly with $\beta$; it's not independent of $\beta$. This is just the case where the time it would take to cross the bubble at speed $\beta$ is longer than the time it takes the bubble to make the trip, so the observer is just moving in the flat space inside the bubble while the bubble moves, and their odometer distance is just how far they manage to move within the bubble by the time the trip ends. The trip time itself is constant, since it's determined by the bubble's speed $v_s$ and the distance from Earth to Alpha Centauri in the underlying Euclidean space; so the trip time is independent of $\beta$, but the odometer distance is not--it's just $\beta$ times the constant trip time, which is linear in $\beta$.

As relative gamma goes to infinity, odometer distance goes to zero because the integrand is bounded above, while the bounds of integration go to zero.

I don't see why the bounds of integration would go to zero. The bubble still takes a finite time to make the trip, and that time is the bounds of integration. It's not really possible for any right-moving observer with $\beta < 1$ to make the trip quicker than the bubble does--that's one of the key implications of my comments at the start of this post.

For the case of $\beta \to 1$, it's true that the time to cross the bubble would be much shorter than the time the bubble takes to make the trip--but the bubble is still moving with $v_s > 1$ in the Euclidean space of the warp coordinates, and, per my comments at the start of this post, even the $\beta \to 1$ observer will end up getting crunched up at the right wall of the bubble. But because the Eulerian observers are also crunched up there, the $\beta \to 1$ observer is still accumulating odometer distance relative to them as they all get pushed along by the bubble.

this means that my "wait in the bubble" is likely to be the upper bound on odometer distance measured for the trip using Eulerian road.

I'm not sure I agree. See above.

 

[PAllen]

Yes, and that raises an issue that the left-moving case doesn't raise: when the right-moving observer gets to the right edge of the bubble, they can't really get out while the bubble is moving with $v_s > 1$. Basically, Eulerian observers get crunched up against the right edge of the bubble. In terms of the math I posted before, $f$ is a function of both $t$ and $x$, and it's not really possible for an observer with $\beta < 1$ to move into a region where $f \to 0$ while $v_s > 1$--no matter how they try, they'll keep getting pushed along by the bubble so they stay in the region where $f \approx 1$. Only when the bubble comes to a stop at Alpha Centauri can they then step out of it.

Right, and that is equivalent to my statement that in orthogonal (Gaussian Normal) ccordinates, the light cones well within the bubble would be very skinny and vertical.

I'm not sure I understand the "simple SR intuition" you're relying on here. If $\beta = 0$, odometer distance is zero. For small $\beta$, odometer distance goes up linearly with $\beta$; it's not independent of $\beta$. This is just the case where the time it would take to cross the bubble at speed $\beta$ is longer than the time it takes the bubble to make the trip, so the observer is just moving in the flat space inside the bubble while the bubble moves, and their odometer distance is just how far they manage to move within the bubble by the time the trip ends. The trip time itself is constant, since it's determined by the bubble's speed $v_s$ and the distance from Earth to Alpha Centauri in the underlying Euclidean space; so the trip time is independent of $\beta$, but the odometer distance is not--it's just $\beta$ times the constant trip time, which is linear in $\beta$.

No the trip time is not constant. The faster you move across the bubble, the sooner you get out of it. The amount may be less than expected due to the feature discussed just above, but it still true that trip time decreases for increasing relative gamma. Also, remember we are talking about proper time along the traveler world line - which can be much less than coordinate time.

I don't see why the bounds of integration would go to zero. The bubble still takes a finite time to make the trip, and that time is the bounds of integration. It's not really possible for any right-moving observer with $\beta < 1$ to make the trip quicker than the bubble does--that's one of the key implications of my comments at the start of this post.

Imagine a locally SR frame around an event on e.g. a Eulerian world line. Imagine that the Eulerian world line is the time axis of this local frame. Then as relative gamma goes to infinity, the proper time for the traveler to traverse this local frame goes to zero. Repeat for every such local frame. Note that the limit is efffectively a null path which has no proper time.

 

[PAllen]

I have an idea to make this explicit. I think it will be fun. In the article containing the famous fig.3, they give, later on, a piecewise linear f(r). This should make the math easier, without loosing anything essential. Similarly, lets use a piecewise v(t) e.g. starting with v=0, then $\dot v$ constant, then v=4, repeating on the other side. Then we can make this whole example explicit with tractable math (I hope). We can race to see who finishes first.

 

[PeterDonis]

the trip time is not constant. The faster you move across the bubble, the sooner you get out of it.

But you don't get out of it--that's the point. The bubble is moving at $v_s > 1$ rightward. If you are outside the bubble to the right of it, you're moving at $\beta < 1$ rightward--meaning the bubble catches you. That happens as soon as you try to move out of the bubble to the right. So you can't. That's why you can't get there faster than the bubble does.

Also, remember we are talking about proper time along the traveler world line - which can be much less than coordinate time.

I know that's how you wrote the integral, and I see the rationale for it, but I'm still trying to reconcile it with how things look in the underlying Euclidean space. I'll probably have to work out what the time dilation factor looks like in those coordinates for $\beta \to 1$.

I do agree that if the limits of integration are in terms of proper time, the trip time is not necessarily bounded below by the bubble trip time (which is determined on the basis of zero time dilation for Eulerian observers). I didn't take that into account in my previous post.

 

[PAllen]

But you don't get out of it--that's the point. The bubble is moving at $v_s > 1$ rightward. If you are outside the bubble to the right of it, you're moving at $\beta < 1$ rightward--meaning the bubble catches you. That happens as soon as you try to move out of the bubble to the right. So you can't. That's why you can't get there faster than the bubble does.

You get out of it sooner, just not as much as expected. E.g. when relative gamma corresponds to a speed greater than v for that slice, you make progress. Once you start making progress, you do better and better, because the 'flow' is vf, and f starts decreasing, so you readily get out of the bubble.

 

[PeterDonis]

I'll probably have to work out what the time dilation factor looks like in those coordinates

Actually, I realized I don't have to work this out--it is $\gamma$, because that's the time dilation factor relative to Eulerian observers, and Eulerian observers' clocks tick at the same rate as coordinate time in those coordinates. That helps.

 

[PAllen]

So, at the center of the bubble, letting u be coordinate speed of a traveler world line at that point (passing through bubble center), v be the bubble speed at that point, I get:
$$d\tau / d t = \sqrt{1-(u-v)^2}$$
which makes perfect sense. u=v is 'fastest' rate of proper time, other choices are slower. This immediately implies you can have significantly lower journey proper time, leading, effectively to 'more length contraction' than just riding the bubble. This also means you can't go net leftwards in the center of the bubble, but you can go rightwards much faster than the bubble center. Actually, supposing v=2, then the range of valid u is (1,3), which makes sense to me.

 

[PeterDonis]

you can go rightwards much faster than the bubble center.

While you're inside the bubble, yes. The question is what happens as you move through the bubble wall.

The respective 4-velocities are $\left( 1, v_s f \right)$ for an Eulerian observer (at the bubble center $f = 1$ and this observer moves rightward at $v_s$) and $\left( \gamma, \gamma \left[ v_s f + \beta \right] \right)$ for the observer trying to cross the bubble to the right.

The comparative rightward speeds are simply $v_s f$ vs. $v_s f + \beta$. So the "right-moving" observer is always moving rightward faster than the Eulerian observer he's next to, yes (since $f$ is the same for both of them). But the observer at the bubble center always has $f = 1$, so they're always moving rightwards at $v_s$. So for $v_s = 2$, for example, and $\beta \to 1$, when $f < 0.5$, this observer's net rightward speed is slower than that of the bubble center, and it continues to get slower as $f$ decreases. (Of course the net rightward speeds of Eulerian observers in this regime are getting slower even faster, so to speak, since they don't have the extra $\beta$ added to their speed to the right.)

The question is what all this means, physically. It doesn't seem possible that an observer whose net rightward movement is less than that of the bubble center can stay outside the bubble on the right. I might have to try to visualize it in the $\xi$ coordinates I used earlier to understand what's going on.

 

[PeterDonis]

the 'flow' is vf, and f starts decreasing, so you readily get out of the bubble.

This assumes that $f$ is constant along all Eulerian worldlines, so that once you've moved past a Eulerian worldline with a given $f$, you'll never see an $f$ that high again.

But $f$ is not constant along all Eulerian worldlines. It's constant along some set of them inside the bubble; but it's obviously not constant along the ones that start outside the bubble to the right, since those start out with $f = 0$ but sooner or later get swept up by the bubble and have their $f$ increase. So there has to be some transition point, somewhere in the bubble wall, where you can no longer depend on $f$ continuing to decrease as you move past Eulerian worldlines.

 

[PeterDonis]

there has to be some transition point, somewhere in the bubble wall, where you can no longer depend on $f$ continuing to decrease as you move past Eulerian worldlines.

I think I might be getting at something similar here to what the paper discusses in section 5.5 about an event horizon. That section uses right-moving light as an example, and shows that even that light can't escape the bubble to the right--it gets stuck in the bubble wall. Their stopping criterion is that the light can't move any further through the bubble wall when it reaches $f = 1 - 1 / v_s$. This isn't quite the same as the reasoning I came up with, but it shows that we can't just assume that anything moving to the right relative to Eulerian observers can escape the bubble to the right.

 

[PeterDonis]

Does that look familiar? It looks like the Painleve metric!

And in that previous post I basically rediscovered what's presented in section 5.5 of the paper that talks about the event horizon. What I didn't pick up on when I made that post is the obvious next step that the paper takes: transform to the equivalent of Schwarzschild coordinates, in which the metric is diagonal and in fact is static. The interesting difference between this and Schwarzschild spacetime is that here, the "normal" region where curves of constant $r$ are timelike is inside the horizon instead of outside, i.e., inside the bubble. The horizon, where curves of constant $r$ are null, is somewhere within the bubble wall, and the region where curves of constant $r$ are spacelike is outside that.

Another difference from Schwarzschild spacetime, of course, is that this spacetime is not spherically symmetric; the right side of the wall is different from the left side. The paper doesn't really go into that.

 

[PAllen]

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 only z and t. For R, I am using .5. For wall thickness, .01. Instead of a funny Greek letter, I define the position of the center of the bubble using p(t). For t less than 0 it is 0. For t between 0 and 2, $t^2$. For t between 2 and 8, 4t - 4. For t between 8 and 10, $28+4(t-8)-(t-8)^2$. For t greater than 10, p is 32. This determines v as well. r(z,t) is just abs(z-p(t)). Note that the metric is everywhere Minkowski for t less than 0, because all terms that differ from Minkowski include v as a multiplier. For the travelerT(t), I am using the defining equation $T’=vf+.1$, with T(0)=0.

Not that the metric for all z, t greater than 10, is again Minkowski, for the same reason as t less than 0: v is 0. This means for sure the traveler escapes the bubble, the question is when. As v drops below 1, things get less weird.

Next, the 4 velocity of T may be computed. While the Eulerian 4 velocity is simply (1,vf), for T it is hardly more complex: $(1,vf+.1)/\sqrt{.99}$, if I did my calculation right.

 

[PAllen]

And in that previous post I basically rediscovered what's presented in section 5.5 of the paper that talks about the event horizon. What I didn't pick up on when I made that post is the obvious next step that the paper takes: transform to the equivalent of Schwarzschild coordinates, in which the metric is diagonal and in fact is static. The interesting difference between this and Schwarzschild spacetime is that here, the "normal" region where curves of constant $r$ are timelike is inside the horizon instead of outside, i.e., inside the bubble. The horizon, where curves of constant $r$ are null, is somewhere within the bubble wall, and the region where curves of constant $r$ are spacelike is outside that.

Another difference from Schwarzschild spacetime, of course, is that this spacetime is not spherically symmetric; the right side of the wall is different from the left side. The paper doesn't really go into that.

Note that this section is using v constant and superluminal, which is not what we want. This basically says that if the bubble never slows down, you can never escape it.

 

[PeterDonis]

Note that this section is using v constant and superluminal

Yes, agreed; in the scenario in Figure 3 of the paper, where the bubble stops once it gets to Alpha Centauri, it can be escaped then. How much sooner than arrival at Alpha Centauri an observer could escape the bubble would depend on the details of how it decelerates.

 

[PAllen]

So it is quite easy to analyze that light gets trapped in the wall in the original coordinates. Using my proposed diff.eq. for traveler T, I can similarly show that it gets trapped in the wall until v falls below .1. Still, at all times it is progressing against Eulerian observers, that also get trapped. That is, in wall Eulerian observer all get asymptotically trapped against the back of the forward wall until the bubble slows down.

 

[PAllen]

So, my choice for traveler world line is very convenient. It turns out it produces a constant integrand for the odometer integral of .1 (everywhere!). So “all” that remains is determining the proper time bounds along the traveler worldline. That might take some time (proper time since I am my own world line).

 

[sbrothy]

With respect to the bubble, that is correct. The bubble itself is moving with respect to Earth and Alpha Centauri. Which means that, from the viewpoint of the ship inside the bubble, Earth and Alpha Centauri are moving. That doesn't change any of the logic I gave in my previous post.


My previous post should make it clear that that is the question I am answering. I even gave you the logic behind the answer.

Here's another way of getting the answer (which I described in the previous thread you linked to): suppose the ship emits a light beam towards Alpha Centauri. Since the ship's worldline is timelike, the light beam will arrive at Alpha Centauri before the ship itself does. Since the ship takes less than 4.3 years to make the trip, the light beam must take an even shorter time to make it by the ship's clock. That means the distance, according to the ship (which means as measured through the warp bubble) must be less than 4.3 light years.

I realize this is a thought experiment, but in "practice" would a light beam even be able to leave the bubble? What I'm driving at is that with exotic matter and the reverse solution of the EFEs, we're really talking non-phyics (or perhaps to be kind "scifi"), aren't we?

EDIT: I'm in way over my head again, I realize. It just struck me as complete nonsense.

But then again who am I to talk?

EDIT2: Oh, and there were a lot of hidden posts I had to activate..... Didn't see them initially.