Bell's Spaceship Paradox, Born Rigidity, and the 1-Way Speed of Light

[Freixas]

I recently revisited Bell's Spaceship Paradox, and I have a few questions about it that I may ask later, but there's one that I can ask now

In Bell's Spaceship Paradox, two ships, separated by space, must begin accelerating simultaneously. After learning about simultaneity conventions and the 1-way speed of light, anything that requires simultaneous actions by observers not co-located gives me pause.

If the front ship has a specific lower acceleration that the rear ship, then the string won't break. I worked out the formula for calculating the acceleration difference and then discovered that this formula is well known as a condition for Born rigidity. My interpretation is that, if the two ships in Bell's Spaceship Paradox accelerate at proper accelerations that satisfy the Born rigidity constraint, the string won't break.

Once under way, the situation seems independent of any simultaneity convention. But the paradox requires that two ships, at rest relative to each other and some other observer, begin accelerating at the same time. If the two ships do not begin accelerating simultaneously, I would think the string would break. A string breaking is an invariant, independent of any simultaneity convention. Could we use this to determine the 1-way speed of light? We could assume various 1-way speeds, synchronize the engine firings using each convention, and then check to see if the string breaks.

Usually, most experiments that try to measure the 1-way speed of light require knowing the 1-way speed in order to measure it. This requirement is often buried somewhere in the setup. This must be the case here, but I'm having trouble seeing where it's hiding.

 

[PeterDonis]

In Bell's Spaceship Paradox, two ships, separated by space, must begin accelerating simultaneously.

Simultaneously in the inertial frame in which they are initially both at rest.

If the front ship has a specific lower acceleration that the rear ship, then the string won't break. I worked out the formula for calculating the acceleration difference and then discovered that this formula is well known as a condition for Born rigidity.

That's correct.

My interpretation is that, if the two ships in Bell's Spaceship Paradox accelerate at proper accelerations that satisfy the Born rigidity constraint, the string won't break.

That's wrong. Bell's Spaceship Paradox is defined as the case where both spaceships have the same proper acceleration. That is inconsistent with the Born rigidity constraint.

Could we use this to determine the 1-way speed of light?

No. You have simply missed some key aspects of the definition of the scenario. See above.

 

[Nugatory]

My interpretation is that, if the two ships in Bell's Spaceship Paradox accelerate at proper accelerations that satisfy the Born rigidity constraint, the string won't break.

I think that you are saying that if we set up two ships connected by a string, as in Bell's thought experiment, but then choose different accelerations so are considering a different thought experiment not Bell's, we can arrange for the string to not break? That is, you're not describing Bell's paradox, but rather a different situation inspired by thinking about Bell's paradox?

 

[Freixas]

I think that you are saying that if we set up two ships connected by a string, as in Bell's thought experiment, but then choose different accelerations so are considering a different thought experiment not Bell's, we can arrange for the string to not break? That is, you're not describing Bell's paradox, but rather a different situation inspired by thinking about Bell's paradox?

Yes.

 

[PeterDonis]

Yes.

So what scenario, exactly, are you asking about?

 

[PeterDonis]

A string breaking is an invariant, independent of any simultaneity convention.

That is correct. But whether or not the string breaks does not depend on the one-way speed of light. It depends on the expansion of the congruence of worldlines that describes the string. That is an invariant and does not depend on the one-way speed of light.

 

[Nugatory]

Could we use this to determine the 1-way speed of light? We could assume various 1-way speeds, synchronize the engine firings using each convention, and then check to see if the string breaks.

We'd still be making an arbitrary choice of simultaneity convention.
Both accelerating ships are tracing out their own hyperbolic paths on a standard Minkowski diagram. These paths are determined by the acceleration profiles of each ship as a function of its proper time, and we've chosen these profiles such that the string won't break.

That's fine and easily stated in frame-independent invariant language. But to get a clock synchronization out of this we have to take another step: Somehow associate coordinate time of an event on one hyperbola with the coordinate time on the other. That's where our arbitrary choice of simultaneity convention reappears.

 

[Ibix]

Essentially the "string never breaks nor goes slack" condition is Born rigidity. The worldlines of the ships in this case are concentric hyperbolae and starting simultaneously in their initial rest frame is one way of specifying the concentric requirement.If you want the curved portions of the worldlines to be concentric then the acceleration must start on a straight line orthogonal to the inertial sections of the ships' motion.

Einstein simultaneity also uses lines orthogonal to a chosen family of inertial clocks. But there's still nothing stopping you using a different criterion.

 

[Freixas]

So what scenario, exactly, are you asking about?

Consider that we have two ships lined up as in Bell's Spaceship Paradox and with a taut string attached between them. One ship is to my left, one to the right and both are equidistant to me. Both ships will begin accelerating at some instant in time, headed to my right. When they accelerate, the rear ship will have more proper acceleration than the ship in front. The difference in acceleration will satisfy the Born rigidity constraint.

To match Bell's paradox, both ships must begin accelerating simultaneously. Arranging this requires that everyone agrees on a simultaneity convention. The paradox, as I've seen it, seems to assume that speed of light is isotropic.

To clarify my thinking, I can simultaneously launch a photon both left and right. When the ships receive the photon, they could begin accelerating. They will consider their actions simultaneous, but it is only simultaneous due to an arbitrary choice of an isotropic simultaneity convention. If, for example, we choose a convention in which light travels faster to my right, then the right-most ship should not begin accelerating on receiving its photon since the left-most ship would not yet have received its.

As I said, I don't think there is any issue if we have two ships that have been accelerating while maintaining Born rigidity since the beginning of time. The issue is with the transition between constant and accelerating velocity. If the two ships begin accelerating simultaneously, one would think the string wouldn't break. Conversely, if the acceleration does not begin simultaneously, it would seem that the string would quickly break, even when the accelerations, when they occur, are consistent with Born rigidity.

This would appear to give us a tool to determine which simultaneity convention is the "correct" one. I'm asking why it would not give us such a tool.

 

[Ibix]

The paradox, as I've seen it, seems to assume that speed of light is isotropic.

The usual setup is stated in those terms, yes, because it was supposed to point out a flaw in people's "intuitive" thinking so deliberately used standard conventions. But the same scenario can easily be stated using any simultaneity convention you want.

This would appear to give us a tool to determine which simultaneity convention is the "correct" one. I'm asking why it would not give us such a tool.

No, they just tell you that Einstein simultaneity is a plane orthogonal to the worldlines. Starting acceleration on such a plane is part of the condition you require for two straight parallel worldlines starting to curve if their curved sections are to be concentric.

 

[PeterDonis]

To match Bell's paradox, both ships must begin accelerating simultaneously. Arranging this requires that everyone agrees on a simultaneity convention.

Yes, and that convention is already specified in the definition of the Bell Spaceship Paradox scenario, as has already been posted in this thread.

The paradox, as I've seen it, seems to assume that speed of light is isotropic.

The simultaneity convention that is specified in the definition does have that implication, yes, since it's based on Einstein clock synchronization.

However, as @Ibix has posted, the specification of when each ship starts accelerating can actually be done using invariants, i.e., with any simultaneity convention you want. The intent is to specify a particular congruence of worldlines, and that is an invariant.

I'm asking why it would not give us such a tool.

Because, as I've already posted, what actually determines whether or not the string breaks has nothing to do with any simultaneity convention. It is the expansion of the congruence of worldlines that describes the string, which is an invariant and can be calculated in any coordinate chart you want, with any simultaneity convention you want. It tells you nothing about simultaneity conventions at all.

 

[PeterDonis]

I can simultaneously launch a photon both left and right. When the ships receive the photon, they could begin accelerating.

Yes, and note that this is an invariant specification--it does not depend on any simultaneity convention, because when you use the word "simultaneously" in this particular case, you do not mean what "simultaneous" usually means in relativity discussions. You are not referring to two spacelike separated events, but to one event--the event at which you launch both photons. Two photons both being launched from a single event is an invariant.

They will consider their actions simultaneous, but it is only simultaneous due to an arbitrary choice of an isotropic simultaneity convention.

Whether or not they consider their actions simultaneous is irrelevant. The fact that they both start accelerating when they receive your photons is an invariant.

Also, the fact that the simultaneity convention that results from the two events in question--the reception of the photons by the two spaceships--is isotropic is not arbitrary: it's because you chose the event where you launched the photons in order to make it that way. In other words, you chose to launch the photons from a point equidistant between the two spaceships which were at rest relative to each other. You didn't specify that, but it's necessary in order for the resulting simultaneity convention to be isotropic. Try considering a scenario where the photon launch point is not equidistant and see what it implies.

 

[Ibix]

Also, the fact that the simultaneity convention that results from the two events in question--the reception of the photons by the two spaceships--is isotropic is not arbitrary: it's because you chose the event where you launched the photons in order to make it that way. In other words, you chose to launch the photons from a point equidistant between the two spaceships which were at rest relative to each other. You didn't specify that, but it's necessary in order for the resulting simultaneity convention to be isotropic. Try considering a scenario where the photon launch point is not equidistant and see what it implies.

Another way to think of this, @Freixas, is to think of how you would interpret the acceleration starts if you remain at the center but think about it with different anisotropies. Nothing changes physically, but as you change your chosen anisotropy you change your interpretation of whether the ships start simultaneously.

The invariant fact is that the inner product of the spacelike line joining the acceleration events is orthogonal to the inertial portions of the ships' worldlines. That's the actual requirement for what you want to do. Whether you choose that spacelike line to lie in your planes of simultaneity or not is uo to you.

 

[Freixas]

Whether or not they consider their actions simultaneous is irrelevant. The fact that they both start accelerating when they receive your photons is an invariant.

Ok, I think I've got it. To translate everything everyone said to something I could follow, I diagrammed the problem using the normal isotropic convention, then converted this to use a different convention by rotating the X axis by 15 degrees and then skewing the result -15 degrees.

I'm showing the original X axis (now skewed) just to show the X units. The real X axis is now horizontal.

And there we are: the ships, using this convention, do not begin accelerating simultaneously, but the string doesn't break, nor can we use this to determine anything about the 1-way speed of light.

A follow-up question: using the convention above, is there any way to begin the acceleration so that it is considered simultaneous with respect to this convention—and not break the string?

 

[PeterDonis]

the ships, using this convention, do not begin accelerating simultaneously

Simultaneously according to what convention?

Since simultaneity conventions keep tripping you up, you really should try to specify your scenario in terms of invariants. For example, how would you launch light signals to get the spaceships to start accelerating as your diagram shows? Your diagram has what I suppose are intended to be light signal worldlines on it, but they don't look like they are traveling on lightlike paths.

the ships, using this convention, do not begin accelerating simultaneously, but the string doesn't break

How do you know?

nor can we use this to determine anything about the 1-way speed of light.

You would not be able to use this to determine the one-way speed of light no matter how you have the ships accelerate and whether or not the string breaks. None of those things tell you anything about the one-way speed of light, as has already been posted. There are an infinite number of possible scenarios and picking just one of them, or indeed any finite number of them, and analyzing those can't possibly give you a general conclusion. For a general conclusion you need to grasp the general fact that the one-way speed of light is a convention, not a physical thing.

 

[Freixas]

Simultaneously according to what convention?

A simple way to view a standard Minkowski diagram with respect to some other convention is to skew the X axis. The diagram I posted rotated the X axis by 15 degrees. I then skewed the rotated X axis back to its usual horizontal position.

You can approximate the 1-way speed of light in each direction by looking at the grid or you could do the math to calculate it.

Your diagram has what I suppose are intended to be light signal worldlines on it, but they don't look like they are traveling on lightlike paths.

When you rotate the X axis and the skew it back horizontal, light paths will look distorted.

How do you know?

This is just a remapping of a diagram in which Born rigidity is maintained. If maintaining Born rigidity keeps the string from breaking, then remapping the coordinates is not going to change anything.

Sorry, I thought you would understand the diagram. If it helps, here is the same thing, with the rotated X axis but unskewed:

For a general conclusion you need to grasp the general fact that the one-way speed of light is a convention, not a physical thing.

Yes, I get that, which is why I asked the people here to help point out the problem in my analysis.

 

[Dale]

Could we use this to determine the 1-way speed of light? We could assume various 1-way speeds, synchronize the engine firings using each convention, and then check to see if the string breaks.

With every one way convention there exists a time offset, $\Delta t$, that they can start the engines to avoid breaking the string. For the isotropic one way convention that time is $\Delta t=0$. For an anisotropic convention $\Delta t$ is a specific non zero value.

If a given pair of ships meets that zero criterion under the isotropic convention then that same pair will also meet the specific non-zero criterion for every anisotropic convention. All conventions will agree on the outcome and so all outcomes equally support each convention.

 

[Freixas]

With every one way convention there exists a time offset, $\Delta t$, that they can start the engines to avoid breaking the string. For the isotropic one way convention that time is $\Delta t=0$. For an anisotropic convention $\Delta t$ is a specific non zero value.

If a given pair of ships meets that zero criterion under the isotropic convention then that same pair will also meet the specific non-zero criterion for every anisotropic convention. All conventions will agree on the outcome and so all outcomes equally support each convention.

To repeat back: the nature of this problem is such that only an isotropic convention allows the firing delta to be 0 (i.e. simultaneous with respect to the convention).

Still, it feels a little disturbing that there is no way to fire the engines simultaneous with a non-isotropic convention and keep the string from breaking. It seems to make the isotropic convention special: the problem links simultaneity to an invariant (the string breaking or not breaking).

It would feel less disturbing if I could create a version of the problem using a non-isotropic convention where the engine firings were simultaneous (with respect to the chosen convention) and the string didn't break. But that doesn't seem possible. Another option would be to find a different problem where simultaneous events using a non-isotropic convention would lead to some unique invariant result only for that convention.

 

[Ibix]

It seems to make the isotropic convention special:

The isotropic convention is nice because in that case alone the simultaneity planes are orthogonal to the worldlines of clocks at rest in that coordinate system. That makes all the cross-terms drop out of the metric and everything is mathematically simple.

This particular problem requires the spacelike line between the acceleration starts to be orthogonal to the hyperbolae so that the hyperbolae can share radii (in the same sense that concentric circles share radii). That means that this coordinate system is well-adapted to this particular problem, but it doesn't make it special.

But that doesn't seem possible.

It's sort of possible. Choose an anisotropy. Pick one spaceship and call it "at rest". Let the other spaceship be moving with some velocity $v$ such that its worldline is orthogonal to the (anisotropic) simultaneity planes of the first ship. Both ships start accelerating at their anisotropic time zero. They will maintain separation in their own rest frame, but because they were initially moving they can't be tied with a string (unless you do something like have the rear spaceship grab the string at the moment it accelerates).

Taking your Minkowski diagram, just delete one inertial worldline and a part of the bottom of the corresponding hyperbola. Then extend a straight worldline tangent to the hyperbola at the event you truncated to.

 

[pervect]

One comment. In order to make the low-velocity limit of special relativity obey Newton's laws, one needs to use the Einstein synchronization conventions. While one can use different conventions, one needs to sacrifice Newton's laws in the low speed limit.

Consider for example two equal masses, each a distance d away from a center point at x=0. At t=0 in this conventional system, both masses start moving, one with a velocity of +v and one with a velocity of -v, and they both arrive at the center point at the same time, t = d/v.

The total momentum is p=mv + m(-v) = 0, before and after the inelastic collision, so when the masses collide, they come to a stop.

Now, change the clock synchronization. The masses start at different times now, but they still collide at the center point at the same time. And the distance they travel hasn't changed. Thus, their velocities are different using this new clock synchronization. But they still collide at the same time, and they still come to a stop when they collide.

The Newtonian formula p = mv no longer works to give a conserved momentum. One can meaningfully define a conserved momentum with the right techniques, for instance one might write a Lagrangian and transform it. But the result will be a different, anisotropic relationship between momentum and velocity.

Einstein clock synchronization is the only synchronization that will give an isotropic relationship between velocity and momentum. Other clock synchronizations will give a different, anisotropic relationship between velocity and momentum.

When we say that one "can use" different clock synchronizations, there is a caveat. One needs to revise many familiar formula from Newtonian physics to do so.

 

[PeterDonis]

The diagram I posted rotated the X axis by 15 degrees. I then skewed the rotated X axis back to its usual horizontal position.

What does the metric look like in your skewed coordinates?

 

[PeterDonis]

why I asked the people here to help point out the problem in my analysis.

The problem in your analysis is that you are ignoring the fact that changing simultaneity conventions is physically meaningless. If you want a physically different scenario, you need to change something that is physically meaningful. You are confusing yourself because you are thinking in terms of a physically meaningless thing instead of a physically meaningful thing.

 

[Freixas]

The isotropic convention is nice because in that case alone the simultaneity planes are orthogonal to the worldlines of clocks at rest in that coordinate system.

I'm clearly missing something. Isn't the orthogonality of simultaneity planes just a choice of a coordinate mapping?

Just to explain what I mean: look at my diagram back in #16. The red lines represent lines of constant time using an arbitrary anisotropic convention. The lines do not look orthogonal to the worldlines of clocks at rest because I have overlaid them on a normal Minkowski isotropic diagram, but I can map them (as in #14) so that they are now orthogonal.

However, I agree that the isotropic convention makes the math a lot easier!

 

[PeterDonis]

Isn't the orthogonality of simultaneity planes just a choice of a coordinate mapping?

Whether or not a particular plane is orthogonal to a particular curve in spacetime is invariant.

Whether or not you call a particular plane a "simultaneity plane" is a convention that depends on your choice of coordinates.

As I have already posted, if you continue to focus on "simultaneity" and other physically meaningless concepts, you will continue to confuse yourself. You need to focus on physically meaningful things. Orthogonality is physically meaningful, but "simultaneity" is not.

 

[Dale]

It seems to make the isotropic convention special: the problem links simultaneity to an invariant (the string breaking or not breaking).

You can make many such “invariants”. The most basic being that two lights that flash simultaneously arrive at the midpoint together, which is an invariant. This also singles out the isotropic convention.

Notice what these “invariants” have in common. They start with a statement of simultaneity. In other words, you have given your scenario an invariant ending condition, but a variant starting condition. Is it really an “invariant” then?

It any case, given a variant starting condition it is not hard to construct invariant ending conditions that only occur with any one specific simultaneity convention.

Edit: for example, consider a point 2/3 between two light sources. There would be only one simultaneity convention such that the light emitted simultaneously from the ends arrives together (an invariant) at the 2/3 point.

 

[PeterDonis]

It seems to make the isotropic convention special: the problem links simultaneity to an invariant

No, it doesn't. You have shown no such link. You have picked one particular starting condition that happens to obey your "isotropic convention", in which the string does not break--because of the way you picked the proper accelerations (the Born rigid way). But the standard Bell spaceship paradox has the same "isotropic convention" for its starting condition, but picks different proper accelerations (the same for both spaceships), and the string breaks.

One could similarly pick some "anisotropic convention" for when the ships start accelerating, and pick accelerations such that the string would break, or that it would not.

In short, as I said before, you are picking a special case rather than trying to find a general argument, and you are focusing on the wrong thing even in that special case.

 

[Ibix]

I'm clearly missing something. Isn't the orthogonality of simultaneity planes just a choice of a coordinate mapping?

Whether they're drawn orthogonal on your diagram is up to you. Whether they're orthogonal in spacetime is an invariant.

If a city lays roads out in an orthogonal grid, would you expect that fact to change if you drew a map and then sheared the map? That's exactly analogous to what you are doing here, which is picking a non-orthogonal grid and shearing your representation so it looks orthogonal on the diagram. The grid in reality is still what it is, independent of your diagram.

The scenario you are considering doesn't actually care about simultaneity at all, which is how your "start when they see a flash of light" version (a description entirely couched in invariants) can exist at all. It cares that the acceleration start events lie on a plane orthogonal to the inertial worldlines. We just find it mathematically convenient to use those orthogonal planes as simultaneity planes. Bell, of course, was trying to demonstrate that many physicists' mental short cuts led them to a paradox in his case, so he deliberately obscured the important geometric facts behind physicist-friendly language. You are still kind of falling into his trap.

 

[Freixas]

The isotropic convention is nice because in that case alone the simultaneity planes are orthogonal to the worldlines of clocks at rest in that coordinate system.

I'd like to focus on just this one statement. Perhaps these questions will help identify my confusion:

1. For each simultaneity convention, isn't there a unique set of simultaneity planes?
2. For each simultaneity convention, aren't these planes, by definition, orthogonal to the worldlines of clocks at rest in that coordinate system?

If both these statements are true, then the isotropic convention cannot be the only one in which simultaneity planes are orthogonal to the worldlines of clocks at rest.

To make my thinking clearer, here is a spacetime diagram. The simultaneity plane has been reduced to simultaneity line, shown as a red dashed line. I've also included the wordlines of various clocks at rest with respect to each other, and the diagram is drawn relative to an observer also at rest.

The line of simultaneity is orthogonal to the worldlines of the clocks. But until I draw something like the worldline of a photon, you have no idea what simultaneity convention the diagram is based on.

The grid in reality is still what it is, independent of your diagram.

I agree that a city is a real thing, independent of any diagram. However, simultaneity planes don't appear to be analogous to cities; simultaneity planes seem to be mathematical constructs only. The analogy doesn't really help.

Ironically, it was a similar statement that resolved one of my questions: the isotropic convention is the only convention where the loci of the concentric hyperbolas needed to meet the constraints of the problem reside on the same simultaneity plane.

 

[PeterDonis]

simultaneity planes don't appear to be analogous to cities; simultaneity planes seem to be mathematical constructs only.

Hyperplanes in spacetime are not just mathematical constructs. Nor is the geometry of spacetime as a whole. That is analogous to the city, and hyperplanes in spacetime are analogous to city streets or blocks--invariant geometric objects.

What is analogous to "simultaneity convention" is drawing map coordinates on the city. Calling a specific set of hyperplanes "simultaneity planes" is defining a simultaneity convention, not defining the hyperplanes themselves; they're there regardless of what you call them.

 

[PeterDonis]

I'd like to focus on just this one statement. Perhaps these questions will help identify my confusion:

1. For each simultaneity convention, isn't there a unique set of simultaneity planes?
2. For each simultaneity convention, aren't these planes, by definition, orthogonal to the worldlines of clocks at rest in that coordinate system?

If both these statements are true, then the isotropic convention cannot be the only one in which simultaneity planes are orthogonal to the worldlines of clocks at rest.

For standard inertial frames in flat spacetime, both statements are true. But that does not mean they both are true for all possible simultaneity conventions.

Statement 1 is arguably true by definition: defining a simultaneity convention is picking out a unique set of simultaneity surfaces in the spacetime.

Statement 2, however, is obviously false as a general statement covering all possible simultaneity conventions. You have already given a counterexample to it in this thread: your "skewed" convention, which doesn't change the worldlines of clocks at rest in the frame, it just tilts the simultaneity surfaces so they're no longer orthogonal to the worldlines of the clocks. That is the point @Ibix has been trying to get across to you.

 

[Ibix]

I'd like to focus on just this one statement. Perhaps these questions will help identify my confusion:

1. For each simultaneity convention, isn't there a unique set of simultaneity planes?
2. For each simultaneity convention, aren't these planes, by definition, orthogonal to the worldlines of clocks at rest in that coordinate system?

If both these statements are true, then the isotropic convention cannot be the only one in which simultaneity planes are orthogonal to the worldlines of clocks at rest.

I would say 1 is more or less true - more precisely, for a given simultaneity convention and choice of origin clock the planes are unique. Hence the Einstein simultaneity convention yields a different set of planes for two clocks in relative motion.

However, 2 is completely wrong. Sticking to 2d, two lines are orthogonal if vectors in those lines have zero inner product. Given a first vector, only vectors in one direction satisfy this, so if you choose a timelike axis then only one choice of simultaneity plane is orthogonal to it.

I suspect that you are assuming that the Euclidean $\vec v\cdot\vec u=v^xu^x+v^yu^y$ or Minkowski $\vec v\cdot\vec u=v^tu^t-v^xu^x$ hold in all coordinate systems. This is not correct - in non-orthogonal systems you get cross-terms as well. That's the root of why the maths is nastier.

The line of simultaneity is orthogonal to the worldlines of the clocks.

The line you drew is orthogonal in the Euclidean sense to the worldlines on the diagram, yes. But the line that represents in reality may or may not be orthogonal to the worldlines.

Assume for a moment that it is a normal Minkowski diagram using Einstein synchronisation. Add the axes of some other frame. They are orthogonal in reality. Are they orthogonal in the Euclidean sense on your diagram?

I agree that a city is a real thing, independent of any diagram. However, simultaneity planes don't appear to be analogous to cities; simultaneity planes seem to be mathematical constructs only.

I can physically realise a Euclidean line by building a street. I can physically realise a simultaneity plane by taing a flock of clocks, sychronising them by some process, and making them flash a light when they read some pre-agreed time. I wouldn't say either is more or less of a mathematical construct.

Ironically, it was a similar statement that resolved one of my questions: the isotropic convention is the only convention where the loci of the concentric hyperbolas needed to meet the constraints of the problem reside on the same simultaneity plane.

What do you mean by "loci" here? If you mean the locus of the transitions from inertial to accelerated motion then yes. The point is that the plane in which they lie must be orthogonal to the inertial worldlines. You may choose to use that plane as a simultaneity plane, but you are not required to do so and nothing changes physically if you do or do not do so.

 

[Freixas]

I suspect that you are assuming that the Euclidean v→⋅u→=vxux+vyuy or Minkowski v→⋅u→=vtut−vxux hold in all coordinate systems. This is not correct - in non-orthogonal systems you get cross-terms as well. That's the root of why the maths is nastier.

I think you've found the source of my confusion. Thanks!

What do you mean by "loci" here?

Oops, sorry, I realize I both misread this and then drifted down some nonsense path. Ignore my reference to loci. I understand what you wrote now.

This particular problem requires the spacelike line between the acceleration starts to be orthogonal to the hyperbolae so that the hyperbolae can share radii (in the same sense that concentric circles share radii).

 

[PeterDonis]

I think you've found the source of my confusion.

Whenever you make a coordinate change, the form of the metric changes as well. A change of simultaneity convention is a coordinate change.

 

[cianfa72]

I can physically realise a simultaneity plane by taking a flock of clocks, sychronising them by some process, and making them flash a light when they read some pre-agreed time. I wouldn't say either is more or less of a mathematical construct.

Using whatever physical process to synchronize a such flock of clocks filling the space, does the set of events when they flash a light at a pre-agreed time value always define a spacelike hypersurface in spacetime?

 

[Nugatory]

Using whatever physical process to synchronize a such flock of clocks filling the space, does the set of events when they flash a light at a pre-agreed time value always define a spacelike hypersurface in spacetime?

As long as all the flash events are spacelike-separated from all the others and they completely fill space, they form a spacelike hypersurface. Devise a coordinate system in which they all have the same $t$ coordinate and we can also say that that hypersurface is a simultaneity surface in that coordinate system.