Does the Alcubierre drive shorten distances?

[PAllen]

Yes, that's the case I was referring to.

Just to be clear, the time inside shell of matter is slower than outside far away, even though both are flat. That is the typical case. The warp spacetime is unusual in that the times are the same.

 

[PAllen]

Ok so this means that distance from a bubble center observer to a world line just to right of the whole bubble region, per fig. 3, starts at e.g 4 ly and becomes small at the end, when v=0. This is true both along foliation slices, or spacelike geodesics orthogonal to the bubble center. However, almost all this distance is outside the bubble. The only one of my notions that measures inside the bubble is the odometer. This measures the distance as 0 while in the bubble. If you add a little travel into the left side, and out of the right, you get a small distance traveled by odometer - but all of it is in the process of entering and leaving the bubble region.

 

[PeterDonis]

This is true both along foliation slices, or spacelike geodesics orthogonal to the bubble center.

These are the same.

almost all this distance is outside the bubble.

Yes, agreed. On any foliation slice (all of which are orthogonal to the bubble center), the bubble only occupies a region in space much, much smaller than 4.3 ly.

The only one of my notions that measures inside the bubble is the odometer. This measures the distance as 0 while in the bubble.

Yes. One way of writing this down explicitly is to transform to coordinates comoving with the bubble center, as is suggested in the Natario paper just before Definition 1.9. The paper doesn't explicitly write down the metric in these coordinates, but it's easy to do. I'll leave out two spatial dimensions for brevity.

We have a new spatial coordinate $\xi$ that replaces $x$ (or $z$ as it's called in the paper with the spacetime diagram in it, the "horizontal" dimension in that diagram), defined as $\xi = x - x_s$, where $x_s(t)$ is the position of the bubble center. This means $d \xi = dx - v_s dt$, where $v_s$ (called just $v$ in the paper with the diagram in it) is $dx_s / dt$. Inverting this gives $dx = d \xi + v_s dt$, and we can then rewrite the metric as

$$
ds^2 = - dt^2 + \left[ d \xi + v_s \left( 1 - f \right) dt \right]^2
$$

Inside the bubble, where $f = 1$, this reduces to just the ordinary flat line element

$$
ds^2 = - dt^2 + d \xi^2
$$

Far outside the bubble, where $f = 0$, however, it becomes

$$
ds^2 = - \left( 1 - v_s^2 \right) dt^2 + 2 v_s dt d \xi + d \xi^2
$$

Does that look familiar? It looks like the Painleve metric! The only difference is that $v_s$ here is a function of time, where in the case of the Schwarzschild metric in Painleve coordinates, the "speed" is a function of the areal radius $r$. And since in the case of interest, $v_s > 1$, this corresponds to the Painleve metric inside the horizon of a black hole, where the "rain" falls "faster than light", and curves "at rest" in the chart (meaning here $d\xi = 0$) are spacelike, not timelike.

So in these "comoving coordinates", where the center of the bubble is fixed at $\xi = 0$ and the bubble as a whole is at rest in a small region centered on that point, spacetime inside the bubble is just ordinary flat spacetime, but outside the bubble, it has flat spatial slices, but there is a "rain", so to speak, in the negative $\xi$ direction--a family of free-falling observers that are all moving in that direction at speed $v_s$. (These are the observers that are at rest far outside the bubble in the coordinates of the diagram in Figure 3 of the first paper.)

"Odometer distance" in these coordinates, I think, is just $\xi$.

 

[Jaime Rudas]

No we are in agreement that the peculiar velocity is c. The difference is the notion of distance traveled. You have not provided any part of their discussion that uses your notion of distance traveled. The section you quoted earlier make no mention distance. A feature of the comoving odometer definition is that even though peculiar velocity is approaching c, proper time along the journey is approaching 0, so amount of comoving road going by approaches 0.

As I understand it, the peculiar speed is relative to a comoving observer, that is, with respect to an observer whose peculiar speed is zero. Thus, the distance traveled by the photon would be c times the time interval measured by that observer.

 

[PeterDonis]

the distance traveled by the photon would be c times the time interval measured by that observer.

Once again: as far as we can tell, there is no ground in any reference you've given for your interpretation of this quantity as "the distance traveled by the photon". If you want to justify that interpretation, you will need to either find a reference to support it, or at least give an argument that isn't just "well, it's $c$ times the time". We all agree on the math. It's the physical interpretation of the math that's at issue.

Specifically, to justify your interpretation, one would expect to be able to do something like: find a spacelike curve along which to integrate the local "distance traveled by the photon" with reference to each comoving observer. "Arc length along a spacelike curve" is normally what the term "distance" means in GR. But there is no such spacelike curve in your example. The integral you wrote down is along the photon's worldline, which is null, not spacelike.

In flat spacetime, or indeed in any stationary spacetime, one could finesse that issue, because one could foliate the spacetime by spacelike surfaces of constant time, such that every such spacelike surface is identical. Then one could just time translate each small piece of spacelike curve in the local inertial frame around each "observer at rest" that the photon passes along its worldline, into the same surface of the foliation, and assemble a single spacelike curve whose total arc length is the same as the result of the integral you take along the photon's worldline.

But the only reason you can get away with this is that the spacetime is stationary--that each such spacelike surface is identical. In the case of a photon traveling through an expanding universe, they're not. There is no constant "space" you can use to interpret the photon as traveling through; "space" is expanding as the photon travels, and that prevents you from interpreting the sum of all the little spacelike pieces along the photon's worldline as a "distance", or at least as a distance with any actual physical meaning.

To put this another way, your interpretation "c times travel time equals distance" assumes, without you realizing it, that there is a fixed "space" that the photon moves through. In the case of flat Minkowski spacetime, or any stationary curved spacetime, you can indeed find such a thing. But in an expanding universe, you can't. That's the fundamental issue underlying the objections you have gotten in this thread to your claim about the photon in an expanding universe.

 

[PAllen]

These are the same.


Yes, agreed. On any foliation slice (all of which are orthogonal to the bubble center), the bubble only occupies a region in space much, much smaller than 4.3 ly.


Yes. One way of writing this down explicitly is to transform to coordinates comoving with the bubble center, as is suggested in the Natario paper just before Definition 1.9. The paper doesn't explicitly write down the metric in these coordinates, but it's easy to do. I'll leave out two spatial dimensions for brevity.

We have a new spatial coordinate $\xi$ that replaces $x$ (or $z$ as it's called in the paper with the spacetime diagram in it, the "horizontal" dimension in that diagram), defined as $\xi = x - x_s$, where $x_s(t)$ is the position of the bubble center. This means $d \xi = dx - v_s dt$, where $v_s$ (called just $v$ in the paper with the diagram in it) is $dx_s / dt$. Inverting this gives $dx = d \xi + v_s dt$, and we can then rewrite the metric as

$$
ds^2 = - dt^2 + \left[ d \xi + v_s \left( 1 - f \right) dt \right]^2
$$

Inside the bubble, where $f = 1$, this reduces to just the ordinary flat line element

$$
ds^2 = - dt^2 + d \xi^2
$$

Far outside the bubble, where $f = 0$, however, it becomes

$$
ds^2 = - \left( 1 - v_s^2 \right) dt^2 + 2 v_s dt d \xi + d \xi^2
$$

Does that look familiar? It looks like the Painleve metric! The only difference is that $v_s$ here is a function of time, where in the case of the Schwarzschild metric in Painleve coordinates, the "speed" is a function of the areal radius $r$. And since in the case of interest, $v_s > 1$, this corresponds to the Painleve metric inside the horizon of a black hole, where the "rain" falls "faster than light", and curves "at rest" in the chart (meaning here $d\xi = 0$) are spacelike, not timelike.

So in these "comoving coordinates", where the center of the bubble is fixed at $\xi = 0$ and the bubble as a whole is at rest in a small region centered on that point, spacetime inside the bubble is just ordinary flat spacetime, but outside the bubble, it has flat spatial slices, but there is a "rain", so to speak, in the negative $\xi$ direction--a family of free-falling observers that are all moving in that direction at speed $v_s$. (These are the observers that are at rest far outside the bubble in the coordinates of the diagram in Figure 3 of the first paper.)

"Odometer distance" in these coordinates, I think, is just $\xi$.

Not what I meant. An odometer measures passing road. The road is defined by a chosen timelike congruence. If I close the eulerian congruence to be the road, then a bubble center observer sees no road go by, since it is, itself in the congruence. Only during a short span entering and leaving the bubble is any road measured as passing.

 

[Jaime Rudas]

Once again: as far as we can tell, there is no ground in any reference you've given for your interpretation of this quantity as "the distance traveled by the photon".

Once again: I made a mistake when I said I was referring to the distance traveled by the photon.

 

[PAllen]

As I understand it, the peculiar speed is relative to a comoving observer, that is, with respect to an observer whose peculiar speed is zero. Thus, the distance traveled by the photon would be c times the time interval measured by that observer.

I believe I have clearly expressed the problem I see with this multiple times. Again, you want sum a bunch of measurements made by different instruments at different cosmological times and places, and call it a distance traveled. Noting that the spacetime is dynamic, and the distance between these devices is not static, I do not find this definition meaningful. I will not repeat myself again.

 

[PAllen]

Once again: I made a mistake when I said I was referring to the distance traveled by the photon.

But you said exactly this in your last post, just above!

 

[PeterDonis]

If I close the eulerian congruence to be the road, then a bubble center observer sees no road go by

Yes, and such an observer is at constant $\xi$, so this in itself does not argue against $\xi$ being the "odometer distance".

However, outside the bubble, worldlines in the Eulerian congruence are not at constant $\xi$, so $\xi$ as "odometer distance" would not work for them by your definition.

 

[PeterDonis]

I made a mistake when I said I was referring to the distance traveled by the photon.

Okay. Then what do you think "$c$ times the time the photon has traveled" means, physically? You must think it means something relevant to this discussion, since you keep bringing it up.

 

[PAllen]

Yes, and such an observer is at constant $\xi$, so this in itself does not argue against $\xi$ being the "odometer distance".

However, outside the bubble, worldlines in the Eulerian congruence are not at constant $\xi$, so $\xi$ as "odometer distance" would not work for them by your definition.

Right. Entering and leaving the bubble you would measure the rate of eulerian observers going past. Note that my definition is coordinate independent. It could easily be computed in the regular warp coordinates.

 

[Dale]

Thus, the distance traveled by the photon would be c times the time interval measured by that observer

I don’t think you can claim this as “the distance”. It is a quantity that you have defined which has units of distance. But I have never seen anyone else anywhere use that definition.

 

[PeterDonis]

my definition is coordinate independent. It could easily be computed in the regular warp coordinates.

We've already computed it in the regular warp coordinates: it's just Euclidean distance in the surfaces of constant coordinate time, since those surfaces are everywhere orthogonal to your chosen timelike congruence.

 

[PAllen]

We've already computed it in the regular warp coordinates: it's just Euclidean distance in the surfaces of constant coordinate time, since those surfaces are everywhere orthogonal to your chosen timelike congruence.

Correct, as long as the entry and exit speed is small. My formula builds in length contraction, so if entry and exit are done very fast relative to the eulerian observers, the odometer distance would be less.

So, entering at not too fast speed ( relative to eulerian observers) into the bubble in a region where v(t) is 0, would just measure x coordinate difference from earth into bubble. Similarly, at the destination. Once centered in the bubble, following the eulerian world line of the center, no distance at all would be measured.

 

[PeterDonis]

Correct

But by that definition of "distance", the observer at the center of the bubble covers a nonzero distance: $v_s t$, don't they?

as long as the entry and exit speed is small. My formula builds in length contraction, so if entry and exit are done very fast relative to the eulerian observers, the odometer distance would be less.

I'm not sure I understand this part.

 

[Jaime Rudas]

Okay. Then what do you think "$c$ times the time the photon has traveled" means, physically? You must think it means something relevant to this discussion, since you keep bringing it up.

Once again: I made a mistake

 

[PeterDonis]

Once again: I made a mistake

In other words, you now agree that "$c$ times the time the photon has traveled" has no physical meaning in an expanding universe?

 

[PAllen]

But by that definition of "distance", the observer at the center of the bubble covers a nonzero distance: $v_s t$, don't they?


I'm not sure I understand this part.

Go back and use my formula to compute odometer distance between traveling inertially between two constant position world lines in an inertial frame in SR. That is, in some frame, constant x lines form the road congruence. Then measure distance from position 0 to position L, at different t inertial speeds per my formula. You find that the odometer distance is $L/\gamma$. you can then try it in the traveler frame, getting the same result. Everything is determined by road congruence and odometer world line.

The way I wrote it in the earlier post is manifestly invariant.

 

[PeterDonis]

Go back and use my formula

Ah, I see; I had gotten mixed up between your "orthogonal to foliation" notion of distance and your "odometer" notion of distance. Of course in your SR example, your "odometer distance" for an object that's moving relative to your chosen congruence will correspond to spatial distance in the surface of a foliation--just not the one that's orthogonal to the timelike congruence that defines the "road".

The way I wrote it in the earlier post is manifestly invariant.

Yes, agreed, you're just taking the inner product of two vectors, which is of course invariant, and plugging it into a formula with everything else being known constants.

 

[PAllen]

But by that definition of "distance", the observer at the center of the bubble covers a nonzero distance: $v_s t$, don't they?

No, because if you are following a eulerian world line, you pass no other eulerian world lines, so zero road passes by.

 

[PAllen]

Ah, I see; I had gotten mixed up between your "orthogonal to foliation" notion of distance and your "odometer" notion of distance. Of course in your SR example, your "odometer distance" for an object that's moving relative to your chosen congruence will correspond to spatial distance in the surface of a foliation--just not the one that's orthogonal to the timelike congruence that defines the "road".


Yes, agreed, you're just taking the inner product of two vectors, which is of course invariant, and plugging it into a formula with everything else being known constants.

Note that the formula works fine to define the odometer measurement by a relativistic drunk bug scampering over a relativistic spinning record. The record surface would be the Langevin congruence (for example) and the bug world line could be anything at all (that stays on the surface, for the purposes of this example).

 

[Jaime Rudas]

In other words, you now agree that "$c$ times the time the photon has traveled" has no physical meaning in an expanding universe?

I don't know how to determine whether something has physical meaning or not. For example, I can't understand why the speed at which an object moves can have physical meaning, while the speed at which an object's length expands cannot.

 

[Dale]

I don't know how to determine whether something has physical meaning or not.

The best rule of thumb is to think of an experiment whose actual measured outcome depends on the thing in question.

 

[Jaime Rudas]

The best rule of thumb is to think of an experiment whose actual measured outcome depends on the thing in question.

For example, if I have an elastic band that is one centimeter long and I stretch it for one hundredth of a second, its final length will depend on the speed at which its length expands. Does that mean that the speed of expansion of length has physical meaning?

 

[PAllen]

The following may be said about correcting the light cones in fig.3 of the paper @PeterDonis linked: the slopes should be $v\pm1$ well within the bubble, and the angle between the light cone bounds is given by $\cos\theta=v^2/\sqrt{v^4+4}$. This gives 90 degrees when v=0, and approaches 0 for v very large.

 

[Dale]

For example, if I have an elastic band that is one centimeter long and I stretch it for one hundredth of a second, its final length will depend on the speed at which its length expands. Does that mean that the speed of expansion of length has physical meaning?

The speed of expansion of the length of an elastic band certainly does. You could measure the stress or the strain in the band.

 

[PeterDonis]

if you are following a eulerian world line, you pass no other eulerian world lines, so zero road passes by.

Yes, I get that, but it seems to me that there is a fundamental tension between using that as the definition of "the road" and trying to interpret what's going on as motion, or lack thereof, in "Euclidean space", because in the "Euclidean space" of the usual warp coordinates, at least some Eulerian worldlines are moving--they do not have constant spatial coordinates in that chart. That's also true in the modified chart using $\xi$ as the spatial coordinate instead of $x$.

I suppose one could try to just construct a chart using the Eulerian worldlines themselves as the timelike "grid lines" and see what "space" (surfaces of constant time) looks like in that chart. It might be that even in such a chart (which would be neither of the ones we've discussed so far, but sort of a mismash of the two), "space" would still be Euclidean, since those horizontal surfaces are indeed orthogonal to all the Eulerian worldlines. If that's true, this whole business of transforming between these different coordinate charts would be something like a Galilean transformation in Newtonian physics--but with the "relative speed" of the transformation varying in space.

 

[PAllen]

Yes, I get that, but it seems to me that there is a fundamental tension between using that as the definition of "the road" and trying to interpret what's going on as motion, or lack thereof, in "Euclidean space", because in the "Euclidean space" of the usual warp coordinates, at least some Eulerian worldlines are moving--they do not have constant spatial coordinates in that chart. That's also true in the modified chart using $\xi$ as the spatial coordinate instead of $x$.

If you want a measure of travel distance inside the bubble, the only simple choice is relative to Eulerian world lines. Other simple choices have you measuring almost all outside the bubble.

I suppose one could try to just construct a chart using the Eulerian worldlines themselves as the timelike "grid lines" and see what "space" (surfaces of constant time) looks like in that chart. It might be that even in such a chart (which would be neither of the ones we've discussed so far, but sort of a mismash of the two), "space" would still be Euclidean, since those horizontal surfaces are indeed orthogonal to all the Eulerian worldlines. If that's true, this whole business of transforming between these different coordinate charts would be something like a Galilean transformation in Newtonian physics--but with the "relative speed" of the transformation varying in space.

To me, this was a key insight some time ago - the Euclidean slices are orthogonal to the Eulerian world lines, everywhere. Thus this is the only orthogonal foliation. The metric would thus be diagonal in this chart. However, a benefit of my odometer definition is there is no need to transform to these coordinates. It would change nothing about the computation. The result would still be that the odometer relative to the Eulerian road only measures distance during the short trip into the bubble and out of the bubble.

 

[PeterDonis]

If you want a measure of travel distance inside the bubble, the only simple choice is relative to Eulerian world lines. Other simple choices have you measuring almost all outside the bubble.

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.