Bell’s Spaceship Paradox. Does the string break?

[DaveC426913]

Most of the recent disagreement between me and others concerns the question of whether two separated rockets with equal accelerometer readings will maintain the same separation as the acceleration proceeds. This is the "Bell's Paradox". Most people (these days) say the separation increases. I say it doesn't, and I've given a proof of that, which has provoked the most recent controversy on this and other forums.

Of this I am aware. I check in every few days. I guess you can't really go back to PF and try again.

 

[Mike_Fontenot]

Previously, I (Mike Fontenot) had said:

"Most of the recent disagreement between me and others concerns the question of whether two separated rockets with equal accelerometer readings will maintain the same separation as the acceleration proceeds. This is the "Bell's Paradox". Most people (these days) say the separation increases. I say it doesn't, and I've given a proof of that, which has provoked the most recent controversy on this and other forums."

Of this I am aware. I check in every few days. I guess you can't really go back to PF and try again.

I DO send the Physics Forum emails from time to time, giving them some recent results, but I never hear back from them.

AS far as the Bell's Paradox is concerned, it's important to understand that it is a DIFFERENT scenario from the one we've been discussing. Neddy was the one who pointed that out to me. In that scenario, it is FORCED that the initial inertial observers, by doing whatever is necessary, keep their view of the rocket separation constant during the acceleration. That results (via the length contraction equation) in the conclusion that, according to the people on the trailing rocket, the separation between rockets INCREASES during the acceleration. Accelerometers on the two rockets will NOT show the same readings, in the Bell scenario.

The scenario we've been discussing is different. The accelerometers on our two rockets DO show the same acceleration during the acceleration. In our scenario, we utilize an instantaneous velocity change, with a Dirac delta function for the acceleration (an infinite acceleration lasting for an infinitesimal duration). But that is just a limit of a sequence of finite accelerations, each of which lasting a finite time, such that the speed change is the same for each iteration. In our different scenario, the separation between the two rockets is constant, and so the home twin sees no instantaneous change in the LOCATION of the leading rocket. In contrast, if one incorrectly assumes that the separation of the rockets increases, the home twin WOULD see the leading rocket instantaneously move a finite distance away from her, which is absurd.

 

[phyti]

This is my corrected Minkowski diagram:

For this one, the outgoing frame moves left to right and has the positive axis on the right side of the origin. The incoming frame moves right to left and also has the positive axis on the right side of the origin.

The one I posted earlier had the positive axis on the left for the incoming frame, so some signs were backwards. The earlier one also had totally wrong coordinates for the event at the upper left corner of the diagram. This one is all correct, as far as I can tell.

You were also including his complication of twin rockets, which I did not.

 

[Mike_Fontenot]

[...]
2) At a given time in S, if two objects with identical velocity relative to S experience identical proper acceleration, they will undergo identical coordinate acceleration relative to S. (the red color was added by me (Mike Fontenot), to emphasize the fact that you are saying that the accelerometers show the same readings).
[...]

I disagree with your above statement. If the two rockets have the same velocity in S, their proper acceleration (i.e., the readings on their two accelerometers) will NOT be equal. People on the trailing rocket will say that the leading rocket has a greater acceleration than the trailing rocket, and that their separation is increasing. The length contraction equation (LCE) REQUIRES that result.

That is why the Bell scenario is fundamentally different from my scenario. In my scenario, the accelerometers all read the same, the separation between the rockets is constant, and the home twin (she) DOESN'T see the leading rocket instantaneously move from c0located with her to a finite distance away from her (in the direction opposite from the trailing rocket). But if it is instead (incorrectly) assumed that the separation increases between the two rockets, then she WOULD see the leading rocket instantaneously move a finite distance from her (in the direction away from the trailing rocket), and THAT is an ABSURD situation. So the assumption that the separation increases between the two rockets is INCORRECT. THAT is what my proof shows.

 

[Mike_Fontenot]

[...]
Most of the recent disagreement between me and others concerns the question of whether two separated rockets with equal accelerometer readings will maintain the same separation as the acceleration proceeds. This is the "Bell's Paradox". Most people (these days) say the separation increases. I say it doesn't, and I've given a proof of that, which has provoked the most recent controversy on this and other forums.

That paragraph quoted above could have been worded better, like I did in my post #164. I'll repeat those earlier two paragraphs here:

"If the two rockets have the same velocity in S, their proper acceleration (i.e., the readings on their two accelerometers) will NOT be equal. People on the trailing rocket will say that the leading rocket has a greater acceleration than the trailing rocket, and that their separation is increasing. The length contraction equation (LCE) REQUIRES that result."

"That is why the Bell scenario is fundamentally different from my scenario. In my scenario, the accelerometers all read the same, the separation between the rockets is constant, and the home twin (she) DOESN'T see the leading rocket instantaneously move from c0located with her to a finite distance away from her (in the direction opposite from the trailing rocket). But if it is instead (incorrectly) assumed that the separation increases between the two rockets, then she WOULD see the leading rocket instantaneously move a finite distance from her (in the direction away from the trailing rocket), and THAT is an ABSURD situation. So the assumption that the separation increases between the two rockets is INCORRECT. THAT is what my proof shows."

 

[Halc]

And I THINK Neddy always agreed with me that the resolution of the twin paradox is that the traveling twin (he) says that the home twin (she) instantaneously gets much older when he instantaneously reverses course. I don't think that Halc believes that, or at least he doesn't like to emphasize that, preferring instead to just use the spacetime interval.

Just want to clarify this. Your explanation (were it to properly include frame references, which it doesn't) isn't wrong. My explanation using intervals is also not 'the' correct one, it just has the property of generalizing for any situation, which is an unnecessary requirement for an explanation intended for a layman.

Probably the simplest explanation (more intuitive than the interval explanation) for the twins thing is to pick an inertial frame (any one will do) and then just compute how fast each clock moves relative to that frame and for how long. Regardless of frame chosen, the outcome will be the same with the one twin (acting merely as a living clock, never needing to observe, conclude, or believe anything) having the same age differential at the reunion.


Below is from your reply to Neddy's updated spacetime diagram.

That's an improvement, but you've still got the initial conditions wrong: my proof starts with the traveling twin stationary with respect to the home twin.

I will try to draw one with that included, but nowhere near as neat as the machine-generated picture Neddy provided. I don't have very good tools.

At the beginning of the scenario, they are separated by 34.64 ly. She is 40, and he is 20. They stay that way until she is 50 and he is 30.

OK, let me draw that. S (she) is the inertial frame Neddy depicts. She stays put. Both rockets go off to the left in S, with him riding his rocket.


This depicts frame S, the frame in which Alice is stationary. Purple worldline is leading rocket. It's the same as Neddy's pic except it adds the 10 year wait as you specify, and it doesn't include Bob getting out there (the S' outbound inertial frame). Alice follows the black vertical time line labeled with her age. The brown line is the trailing rocket with Bob on it. Frame S" is the 'return' inertial frame in which Bob is stationary after the acceleration. I chose the origin so that Bob's age was 30 and his x" coordinate was identical to his x coordinate at his acceleration event.
It shows the two of them meeting at the event where Alice is 90 and Bob is 50. Perhaps Bob gets off there, but I let the rocket continue left off the page. My apologies that the event where the S" line of simultaneity (green) meets the lead rocket worldline is off the page.
Alice does not witness the ship teleporting away. It just moves away at 0.866c as soon as she turns 50, just like you say.

At that instant, he instantaneously changes his speed wrt her from zero to 0.866 (directed toward her). And that is also the instant when the two separated rockets (colocated with him and her) instantaneously change speed from zero to 0.866 ly/y (by prior agreement), both directed along the line from him and through her position (but not quite hitting her!). All of the action of interest in this proof, though, happens when she is exactly 50. If you CORRECTLY assume that the leading rocket and the trailing rocket maintain the same separation during their instantaneous speed changes, then she sees nothing absurd (the rocket colocated with her doesn't spontaneously disappear from her position and re-appear a finite distance from her [in the direction away from the traveling twin] ... it just instantaneously starts moving at a fixed speed away from her).

It was brief, but the green bit is a frame reference. So that means I actually agree with all this. It matches what SR says. The separation of the rockets in S is constant at 34.64, both before and after the simultaneous speed change.

Is the drawing correct? It can be altered if I got the scenario wrong, or you could draw your own if I mucked up what you're suggesting.

 

[Neddy Bate]

Halc,

Of course the rockets maintain the same separation distance in frame S, that is given at the very start of Bell's scenario. What Mike is claiming pertains to the trailing rocket's frame (or traveling twin's frame since they are identical after the turnaround). He claims the separation remains constant in those two frames both before and after the acceleration of the trailing rocket.

In my diagram, that would require the leading rocket to be co-located with the home twin at both t=40 and t=70. My diagram is the same scenario except without the unnecessary 10 year pause at turn-around, so for Mike that would be t=50 and t=80. Surely you must agree that the leading rocket is not co-located with her at both times.

It is given in Mike's scenario that the leading rocket departs from her at her own time t=40 (mine) or t=50 (Mike's) so the leading rocket indisputably cannot be co-located with her at either t=70 (mine) or t=80 (Mike's). Look at my simultaneity line shift from horizontal to slanted. Those simultaneity lines are what Minkowski diagrams are useful for.

 

[Halc]

Of course the rockets maintain the same separation distance in frame S, that is given at the very start of Bell's scenario.

Well, that isn't given, it is derived. I quoted the D&B wording in post 145, and what is given there is that two identically constructed rockets are simultaneously (in S) fired, resulting in identical acceleration profiles and thus identical velocity at any time in S. The rest (velocity included) is derived from that. I guess that means that the picture I drew of Mike's scenario as it was described qualifies as meeting this condition. The 'rocket' is actually just an artillery piece that fires a projectile (one with Bob in it, the other with a dummy of Alice to keep the weight identical) at 0.866c, which is inertial thereafter. Bob gets to Earth and jumps off into a ball pit allowing him to be with Alice and letting his rocket continue in that line we both drew.

What Mike is claiming pertains to the trailing rocket's frame (or traveling twin's frame since they are identical after the turnaround). He claims the separation remains constant in those two frames both before and after the acceleration of the trailing rocket.

I know that. But he won't draw a picture of that, or provide numbers. One step at a time. He described not a turnaround, but both separated and relatively stationary for 10 years, so I drew that, and am waiting to see if Mike agrees with the drawing or if not, if he could possibly provide one himself. He says that in what I call the S" frame (his return inertial frame), Alice is 80 years old at the time of his rocket taking off for the return trip. The picture shows that, but with the exception of the green LoS line and other S" numbers, it's all from Alice's perspective.

My diagram is the same scenario except without the unnecessary 10 year pause at turn-around

But Mike wanted it, so I put in in there since he won't engage if there's a nit. A pause before the return gives it the same initial condition as Bell's, so I think that has a lot to do with why he put it there, and 10 years to keep all the ages to nice round numbers.

 

[Mike_Fontenot]

I will draw the Minkowski diagram that I have been using, and post it ASAP.

 

[Mike_Fontenot]

 

[Mike_Fontenot]

In the above chart, I forgot to write that the slope of the dashed line leaving the origin is 0.866 ly/y, which is the initial speed of the traveling twin (he) when he leaves the home twin (her). That's why their separation (according to her) immediately before he changes his speed wrt her to zero, is 34.64 ly/y.

Also, I should have said that the "LOS" lines that I show several places in the chart are his "Lines Of Simultaneity", where they cross his worldline.

 

[Mike_Fontenot]

One other addition:

Although it doesn't matter for the proof, if you're curious about what the track of the leading rocket is, immediately after it instantaneously changes it's speed from zero to 0.866 ly/y, it starts from the point T = 50 and X = zero, and follows a line headed toward the lower right, parallel to the dashed line in the upper right labeled "His Return Trip". As that happens, her age slowly increases along the horizontal axis, starting from the point T = 50, but her position is always constant at X = zero, as she sees the leading rocket moving away from her (in the direction opposite from the traveling twin) at the constant speed 0.866 ly/y.

 

[Neddy Bate]

Mike does not even want to look at the traveling twin's LOS that points to the home twin being 80 years old. Maybe that is because the leading rocket is long gone away from her when her age is 80. Sure there is also a vertical LOS pointing to her being 50 years old with the leading rocket co-located with her, but that is no longer his LOS once he is 30+. Deceptive to say the least.

 

[Mike_Fontenot]

My (Mike Fontenot's) response is in red:

Neddy said:

"Mike does not even want to look at the traveling twin's LOS that points to the home twin being 80 years old.
On my chart, the lower dashed line starting from the traveling twin"s (his) location when he is 30+ is precisely his LOS you're describing above. I ran out of room on that plot, but you should be able to tell that that LOS intersects the "T" axis at T = 80.

Maybe that is because the leading rocket is long gone away from her when her age is 80.

There's no problem with that.

 

[Neddy Bate]

Here is my Minkowski diagram with Mike's 10 year resting period included.

I mirrored and rotated so it would be in Mike's preferred orientation.

Anyone who can read Minkowski diagrams would interpret this to mean that when the traveling twin is 30 years old and at rest with respect to the home twin, he would say the home twin is 50 years old. Then the instant he accelerates toward her, he would say she is 80 years old.

What kills Mike's "proof" is that although the leading rocket is co-located with her when she is 50, it is long gone and far away from her when she is 80, demonstrating that the traveling twin will say it moved far away from her in the instant he accelerated. The very thing he wanted to prove does not happen, is shown to happen.

 

[Halc]

Thank you Mike so much for that.

Your picture is pretty much the same as mine, except with the axes reversed, you putting time horizontal. That's a bit unconventional, but not wrong in any way.
My picture omits his worldline before his age 20 (his outbound trip, stationary in frame S') since your description in post 157 didn't dwell on that, but instead concentrated on what happens after that. I also didn't put in any LoS lines for frame S since they're all horizontal. Neddy put them in his though, and he obligingly swapped the axes in his latest one to look more like yours, annoyingly making the text hard to read since the mirror image was done after much of the picture was already labeled.
None of the worldlines (Alice, Bob, and the two rockets) have discontinuities and nothing moves faster than c, so there is no 'instant teleporting' of any object witnessed by anybody, which would indeed be absurd.

On my chart, the lower dashed line starting from the traveling twin"s (his) location when he is 30+ is precisely his LOS you're describing above. I ran out of room on that plot, but you should be able to tell that that LOS intersects the "T" axis at T = 80.

Agree. We can see that it is on the way to intersecting her worldline at t=80. That event is in my chart, but the event where the same LoS line intersects the leading ship is a bit off my chart, but I give the coordinates of it anyway, with an arrow pointing to the upper left. Neddy's drawing is panned back far enough to show this event.

So I notice that if you didn't run out of space at the edge of your chart, your LoS line intersects the leading ship worldlineat a separation distance of 69.28 in the S" frame in which both ships are stationary after the acceleration events, which is twice the separation distance (34.64) between the ships in frame S in which both ships were stationary before the acceleration events. So your drawing shows any string between them needing to be twice as long in order to not break.

Maybe that is because the leading rocket is long gone away from her when her age is 80.

There's no problem with that.

Agree. We all seem to be more in agreement lately. Progress!

 

[Mike_Fontenot]

I (Mike Fontenot) am responding to Neddy in red:

I mirrored and rotated so it would be in Mike's preferred orientation. Thanks.

Anyone who can read Minkowski diagrams would interpret this to mean that when the traveling twin (he) is 30 years old and at rest with respect to the home twin (her), he would say the home twin (she) is 50 years old. Yes ... that's what I call "30-" for his time. Then the instant he accelerates toward her (that's when he is 30+) , he would say she is 80 years old (that's correct).

What kills Mike's "proof" is that although the leading rocket is co-located with her when she is 50 (true), it is long gone and far away from her when she is 80 (true), demonstrating that the traveling twin (he) will say it (the leading rocket) moved far away from her in the instant he accelerated (that's true ... it moved in the direction opposite to the traveling twin [him]). The very thing he wanted to prove does not happen, is shown to happen. No. What I wanted to prove, and DID prove, is that the leading rocket does NOT instantaneously move a finite distance from her (according to her), under the (correct) assumption that the separation between the two rockets is constant. At 50+ of her time, the two rockets instantaneously increase their speed from zero to 0.866 ly/y, but the two rockets haven't moved yet ... the leading rocket is still colocated with her. Which is perfectly fine. THAT is the outcome when the original assumption is that the separation between the two rockets is constant, which is true. But when you use the incorrect assumption that the separation between the leading rocket has increased the separation between the two rockets, then the leading rocket will INSTANTANEOUSLY move a FINITE distance from her (according to her), which is IMPOSSIBLE, thus proving that the leading rocket does NOT instantaneously get farther from the trailing rocket, which was the whole point of this exercise.

 

[DaveC426913]

...thus proving that the leading rocket does NOT instantaneously get farther from the trailing rocket, which was the whole point of this exercise.

But wasn't this your surmise in the first place?

IOW, you interpreted SR to mean that he would see an instantaneous aging, despite everyone telling you that wasn't true.
Now you've proven that your surmise was wrong, and that you agree with everyone.
(Have I got that right? It's been a long thread.)

 

[Mike_Fontenot]

My (Mike Fontenot) responses are in red.

Your picture is pretty much the same as mine, except with the axes reversed, you putting time horizontal. That's a bit unconventional, but not wrong in any way.
My picture omits his worldline before his age 20 (his outbound trip, stationary in frame S') since your description in post 157 didn't dwell on that, but instead concentrated on what happens after that. I also didn't put in any LoS lines for frame S since they're all horizontal. Neddy put them in his though, and he obligingly swapped the axes in his latest one to look more like yours, annoyingly making the text hard to read since the mirror image was done after much of the picture was already labeled.
None of the worldlines (Alice, Bob, and the two rockets) have discontinuities and nothing moves faster than c, so there is no 'instant teleporting' of any object witnessed by anybody, which would indeed be absurd.

I agreed with everything you said in here.

But I DID disagree with you here:

So I notice that if you didn't run out of space at the edge of your chart, your LoS line intersects the leading ship worldlineat a separation distance of 69.28 in the S" frame in which both ships are stationary after the acceleration events, which is twice the separation distance (34.64) between the ships in frame S in which both ships were stationary before the acceleration events. So your drawing shows any string between them needing to be twice as long in order to not break.

I've run out of time for today. I'll address your above paragraph tomorrow.

 

[Neddy Bate]

We all drew the same spacetime diagram, so the geometry of the events is not in dispute.