Why is this motion possible

[sep...@yahoo.com]

Two identical spaceships are at rest on the x-axis 100 meters apart. When either spaceship accelerates it accelerates at a fixed proper rate of 0.01g in the positive x direction. Now if one spaceship starts accelerating in the positive x direction toward the other spaceship before the other spaceship starts accelerating in the positive x direction, the two spaceships will get closer and closer to each other as the accelerations continue (as measured by the crews on board each ship). But during these constant accelerations of the two spaceships instead of continually getting closer and closer to each other they will suddenly start moving further and further apart even though neither spaceship experienced any change in their acceleration nor any other force affecting the motion of either spaceship. Is motion like that possible? If so, what caused that change in motion between the two spaceships?
Thanks,
David Seppala
Bastrop TX

[Richard Hachel]

Problème intéressant.

Mais il est clair intuitivement que le second vaisseau rattrapera
toujours le premier puisque sa vitesse sera toujours plus grande.

R.H.

[sep...@yahoo.com]

According to Einstein's concepts the first spaceship will not catch up under certain conditions. It will start getting closer and closer but then suddenly the two spaceships will start moving further and further apart with no change in the accelerations or any other force acting on the motion of the two spaceships.
David Seppala
Bastrop TX

[Dickie Spiker]

sep...@yahoo.com wrote:

> According to Einstein's concepts the first spaceship will not catch up
> under certain conditions. It will start getting closer and closer but
> then suddenly the two spaceships will start moving further and further
> apart with no change in the accelerations or any other force acting on
> the motion of the two spaceships. David Seppala Bastrop TX

not true. Proofs:

Kiev demands $55 billion worth of weapons from West
https://%72t.com/%72%75%73sia/577421-kiev-demand-billion-weapons-west/
Ukraine needs *_at_least_50_Patriot_air_defense_system_batteries_* to
protect its cities and troops, actor "President" Vladimir Zelensky says

[Trevor Lange]

On Saturday, June 3, 2023 at 11:11:41 AM UTC-7, sep...@yahoo.com wrote:
> Two

This is exactly the same question you asked previously, and the full explanation has been carefully and patiently provided to you, over a dozen times. Each time, you just run away... only to return some months later to ask the same question again, and again, and again...

The only way you will ever understand this is if you stop running away, and actually read the explanation, and if you have any questions *about the explanation*, go ahead and ask. Note: This will require you to actually read the explanation, and refer to it in your follow-up questions. On the other hand, if your intent is to forever ignore the answer to your question, and run away from it, then why even ask?

[whodat]

There is apparently quite a lot of attention seeking behavior
in the sci.physics newsgroups. I'm a little surprised that
so much of it sounds like tavern talk (not my forte.)

[Richard Hachel]

There are some great human problems today. The first is arrogance, and the
second is laziness.

It is the explanation of the sacred text concerning the genesis of
humanity. Adam is arrogant, and wants to be God instead of God, and then
everything sinks into disgust. Cain is lazy, does lazy works, is
reproached for being lazy, and becomes his brother's murderer.

It's a bit what I blame physicists for. They start loving Albert Einstein
out of arrogance and laziness. It frees them from racking their brains
elsewhere. They then sometimes experiment with thoughts, but none have the
courage to point out that they most often lead to absurdity and lies.

R.H.

[Python]

Le 04/06/2023 à 12:15, M.D. Richard "Hachel" Lengrand wrote:
...

> There are some great human problems today. The first is arrogance, and
> the second is laziness.

You are arrogant. You are lazy.

> [snip idiotic rang] ... but none

> have the courage to point out that they most often lead to absurdity and
> lies.

Your claims are full of absurdities. Your posts are full of lies.

[Richard Hachel]

Le 04/06/2023 à 15:52, Python a écrit :
> Le 04/06/2023 à 12:15, M.D. Richard "Hachel" Lengrand wrote:
> ...
>> There are some great human problems today. The first is arrogance, and
>> the second is laziness.
>
> You are arrogant. You are lazy.

Ah merde...

>
>> [snip idiotic rang] ... but none
>> have the courage to point out that they most often lead to absurdity and
>> lies.
>
> Your claims are full of absurdities. Your posts are full of lies.

Oh... no...

R.H.

[sep...@yahoo.com]

Not true?
There are three inertial reference frames, F0, F1 and F2. F1 is moving in the negative x direction with velocity V=c*sqrt(3)/2 wrt to F0 and F2 is moving in the positive x direction with velocity V=c*sqrt(3)/2 wrt to F0. In F1 there are two spaceships, at rest, 100 meters apart along the x-axis. Each spaceship can accelerate at a proper rate of 0.01g along the positive x-axis. Observers in F0 start the accelerations of the two spaceships simultaneously. In F1 one spaceship started accelerating in the positive x direction before the other spaceship started accelerating in the same direction.
On board each spaceship is a rod 100 meters in length that is generally kept perpendicular to the x-axis. Before the start of the accelerations the rod is aligned with the x-axis. It extends from the tip of one spaceship to the tip of the other spaceship. The rod is then returned to its perpendicular position relative to the x-axis. Every now and then similar measurements are made by the crews as the spaceships accelerate. The separation between the two spaceships becomes shorter and shorter compared to the length of the rod. Remember after each measurement, the rod is returned to its perpendicular orientation relative to the x-axis. When the two spaceships have zero velocity wrt to F0, their separation is half the length of the rod. As the acceleration continues, the separation of the two spaceships starts to increase relative to the length of the measuring rod that the crews are using. When the two spaceships have zero velocity wrt to F2, their separation is the same length as the measuring rod. Note that optical devices can be used by the crews to make these same measurements.
The crews never experience any change in their acceleration or any other force during this journey.
Regards,
David Seppala
Bastrop TX

[Trevor Lange]

On Sunday, June 4, 2023 at 9:24:02 AM UTC-7, sep...@yahoo.com wrote:
> There...

Again, all your questions about this have been thoroughly answered multiple times in previous threads. (Your brain does remember those discussions, and the explanation that's been provided to you.... right?) If this has cleared things up for you, then, you're welcome. If, on the other hand, you're still unclear about any aspect of the explanation, then go ahead and ask.

[sep...@yahoo.com]

Show me the link where anyone explains how the crews measure that decreasing separation and then increasing separation without any change in their accelerations. Or better yet, why don't you just tell me.
David Seppala
Bastrop TX

[Trevor Lange]

On Sunday, June 4, 2023 at 9:48:56 AM UTC-7, sep...@yahoo.com wrote:
> > Again, all your questions about this have been thoroughly answered multiple times in previous threads. (Your brain does remember those discussions, and the explanation that's been provided to you.... right?) If this has cleared things up for you, then, you're welcome. If, on the other hand, you're still unclear about any aspect of the explanation, then go ahead and ask.
>
> Show me the link where anyone explains how the crews measure that decreasing
> separation and then increasing separation without any change in their accelerations.

The most recent (in a long series of threads) is here:

https://groups.google.com/g/sci.physics.relativity/c/poYb_M1lqEM/m/sWWMSsQ_BQAJ

> why don't you just tell me.

Already done, in a dozen previous threads, including the one cited above. In each thread, all your questions were thoroughly answered multiple times, in complete detail, pointing out every one of your misconceptions and misunderstandings. Remember?

[Maciej Wozniak]

So are yours, and the mumble of your idiot
guru wasn't even consistent.

[sep...@yahoo.com]

In that thread your explanation of the measuring rod used by the crew wasn't clear to me. Answer the following. Let two spaceships be at rest in F1 along the x-axis and separated by 100 meters (tip to tip) as measured in frame F1 but observers in frame F0 measure the separation as 50 meters due to the relative velocity between F0 and F1 as V = c*sqrt(3)/2. Then F0 observers start the accelerations of the two spaceships simultaneously. The two spaceships accelerate identically at a proper rate of 0.01g. When they achieve zero velocity with respect to F0, they are 50 meters apart. On board the spaceship is a rod 100 meters in length aligned perpendicular to the direction of acceleration. Before the accelerations started, when the rod was aligned along the x-axis its length was identical as the tip to tip distance of the two spaceships. During the acceleration the rod is kept perpendicular to the x-axis (the direction of the acceleration). When the two spaceships achieve zero velocity with respect to F0, they are 50 meters apart (from tip to tip), so if that measuring rod is now aligned with the x-axis, that measuring rod is twice that 50 meters from tip to tip, since it is 100 meters long.
Do you agree with that?

[Trevor Lange]

On Sunday, June 4, 2023 at 1:43:20 PM UTC-7, sep...@yahoo.com wrote:
> Answer the following... [Insert insane repetition of the same question asked and answered dozens of times before.]

> During the acceleration the rod is kept perpendicular to the x-axis...

Again, that is utterly brain-dead, since it makes absolutely no difference, as explained in detail in the referenced thread. What on earth is wrong with you? Have you suffered a severe head injury?

> When the two spaceships achieve zero velocity... Do you agree with that?

Of course, this is precisely what has been explained to you dozens of times, including in the very thread that you supposedly just re-read, in complete detail, along with explaining each of your misunderstanding and misconceptions. Is anything about the explanation still unclear to you?

[sep...@yahoo.com]

So you say that "it makes absolutely no difference" whether the rod is aligned with the x-axis during the acceleration or whether it is perpendicular to the acceleration when going from F1 to F0 and then aligned with the x-axis when the spaceships have zero velocity with respect to F0. Please don't garble your response. Just say, yes it doesn't make any difference or no it makes a big difference.

[Trevor Lange]

On Sunday, June 4, 2023 at 2:51:48 PM UTC-7, sep...@yahoo.com wrote:
> So you say [for the 25th time] that "it makes absolutely no difference" whether
> the rod is aligned with the x-axis during the acceleration... Just say, yes it doesn't

> make any difference or no it makes a big difference.

For the 26th time: It. Makes. No. Difference.

Here are quotes from the other thread, from months ago, that you claim to have read:

************

On Thursday, April 6, 2023 at 3:29:26 PM UTC-7, sep...@yahoo.com wrote:
> You stated, " First, periodically re-orienting the rod is totally brain-dead, since it doesn't give any different readings than if you always leave it aligned with the objects."

Right.

> So you are saying that if the 10 meter rod is perpendicular to the direction
> of a low level acceleration it will remain at a length of 10 meters throughout
> the journey, hence no "catching up" will happen but if it is aligned in the
> direction of the acceleration then "catching up" must occur...

No, that's complete gibberish. Look, suppose there are two rods, one of which is always aligned with the objects, with it's left end held at the position of the left object. The other (identical) rod is initially side by side with the first, but ocassionally you re-orient this second rod, and then orient it back in alignment with the first rod. Whenever you have the two rods lined up side by side, their ends coincide. So re-orienting one of them, between measurements, makes absolutely no difference to the measurements you make when you align it with the objects. With such low acceleration, the rod is always essentially in equilibrium, so it is undergoing Born rigid motion.. regardless of whether you re-orient it ocassionally. Do you understand now why the re-orientation is totally brain-dead?

> Explain what "catching up" means...

It means that the trajectory traced out by the right end of the rod (when aligned with the objects, with the left end held at the left object) is undergoing hyperbolic motion with a lesser leftward initial velocity and a lesser proper acceleration than the right hand object. Hence if they are coinciding initially, they will begin to diverge until reaching a maximum (when the right object is at the midpoint of the rod) and then they will converge as the right object catches up with the right end of the rod.

Think of two particles, one starts faster buit the other has greater acceleration, so the first pulls out ahead but eventually the second catches up. This is elementary kinematics. Why are you unable to understand this? Were you kicked in the head by a mule?
*********

Now do you understand? And do you see why you are called a liar, when you claim that these things have never been explained to you before?

[sep...@yahoo.com]

If I travel with a rod 10 meters long perpendicular to the x-axis and I am accelerating at 0.01g and there is a 5 meter rod aligned with the x-axis when I have zero velocity with respect to that rod, if I rotate my 10 meter rod so that it is aligned parallel to the 5 meter rod that is at rest in F0, my rod will be 10 meters in length, twice the span of the 5 meter rod at rest in F0. You can try that experiment and see for yourself. When something is 10 meters long in the y direction, every frame that only has relative velocities to each other in the x direction measures that rod that in the y direction to be 10 meters in length.
David Seppala
Bastrop TX

[Trevor Lange]

On Sunday, June 4, 2023 at 3:46:34 PM UTC-7, sep...@yahoo.com wrote:
> > Suppose there are two rods, one of which is always aligned with the objects, with it's left end held at the position of the left object. The other (identical) rod is initially side by side with the first, but ocassionally you re-orient this second rod, and then orient it back in alignment with the first rod. Whenever you have the two rods lined up side by side, their ends coincide. So re-orienting one of them, between measurements, makes absolutely no difference to the measurements you make when you align it with the objects. With such low acceleration, the rod is always essentially in equilibrium, so it is undergoing Born rigid motion.. regardless of whether you re-orient it ocassionally.
>
> If I travel with a rod 10 meters long perpendicular to the x-axis ... and there is a 5 meter rod aligned with the x-axis...

No, the situation at any particular moment is that there are two co-moving 10 meter rods (rest lengths), one aligned with the x axis and the other with some other orientation, and then we align the two 10 meter rods and find... sure enough... they match their lengths. The rest lengths of the rods are always 10 meters. The acceleration is low enough that they are never appreciably distended from their equilibrium lengths. Why can you not understand this? Did you get kicked in the head by a mule?

Here's another quote from the archives... from nearly *two years* ago:

*********
On Tuesday, September 28, 2021 at 9:19:11 AM UTC-7, sep...@yahoo.com wrote:
> That rod is generally perpendicular to the x-axis...

No, you've been told countless times that re-orienting the rod is utterly pointless and superfluous. The acceleration rate is extremely low, so the rod is always essentially in mechanical equilibrium, i.e., Born rigid motion with constant proper length, so re-orienting it has absolutely no effect at all. Until you understand this, you are just spinning your wheels.
*********

Notice that it says already two years ago that you've been told countless times. And here you are claiming no one ever told you this before. Sheesh.

[Tom Roberts]

On 6/3/23 1:11 PM, sep...@yahoo.com wrote:
> [...]

You just keep repeating the same mistake: not specifying the inertial
frame relative to which words like these refer:
moving
before, after
at the same time
get closer and closer

start moving further and further apart

You repeatedly confuse both yourself and your readers by not being
precise enough, and by using wishy-washy, ambiguous words that have no
specific meaning (such as the above words without specifying the frame
being used). For just about everything you post, specifying the inertial
frame for such words would resolve your confusion.

Note that for the basic Bell's spaceship scenario, one cannot really
claim the spaceships are "moving further apart", because "moving" must
reference a SINGLE coordinate system, and that simply does not apply: as
the spaceships accelerate and change inertial rest frames, in successive
frames they are further apart -- using "motion" for this process is a
PUN and only serves to confuse both you and your readers. Note also that
relative to their starting inertial frame they maintain a constant
separation and are NOT moving further apart.

[Read the above paragraphs carefully and repeatedly until
you understand why I can use the word "moving" in its last
sentence.]

Tom Roberts

[sep...@yahoo.com]

Okay, so you agree that the crews on each spaceship using a measuring rod on board their spaceships can determine that the two spaceships are moving closer and closer to each other during the leg of the journey from F1 to F0 and can determine that the two spaceships are moving further and further apart during the leg of the journey from F0 to F2. Correct?
David Seppala
Bastrop TX

[Trevor Lange]

On Monday, June 5, 2023 at 5:45:00 PM UTC-7, sep...@yahoo.com wrote:
> The crews on each spaceship using a measuring rod on board their

> spaceships can determine that the two spaceships are moving closer

> and closer to each other during the leg of the journey from F1 to F0 ...

for the 20th time, holding one end of the rod (zero marking) at the left hand object, aligned with the x axis, the right hand object is initially aligned with 10 meter mark on the rod, and then it is aligned with the 5 meter mark on the rod, and then it is aligned with the 10 meter mark on the rod. This is because the rod is essentially always in mechanical equilibrium in its current state of motion, so the locus of positions of the right end of the rod trace a hyperbolic path with a slightly lesser proper acceleration than the right hand object (Born rigid motion), but a lesser leftward initial speed (in terms of F0), and hence that locus and the trajectory of the right object initially diverge, then come to mutual rest, and then re-converge... simple kinematics. Now do you finally understand?

[sep...@yahoo.com]

I understand that, but why do the crews say that happens when neither spaceship changes its acceleration. How do they explain those measurements. That's what I'm trying to understand.
David Seppala
Bastrop TX

[Trevor Lange]

On Tuesday, June 6, 2023 at 4:02:56 AM UTC-7, sep...@yahoo.com wrote:
> > For the 20th time, holding one end of the rod (zero marking) at the left hand object, aligned with the x axis, the right hand object is initially aligned with 10 meter mark on the rod, and then it is aligned with the 5 meter mark on the rod, and then it is aligned with the 10 meter mark on the rod. This is because the rod is essentially always in mechanical equilibrium in its current state of motion, so the locus of positions of the right end of the rod trace a hyperbolic path with a slightly lesser proper acceleration than the right hand object (Born rigid motion), but a lesser leftward initial speed (in terms of F0), and hence that locus and the trajectory of the right object initially diverge, then come to mutual rest, and then re-converge... simple kinematics. Now do you finally understand?
>
> I understand that...

That would be great... but the rest of your sentence falsifies that claim.

> ...but why do the crews say that happens when neither spaceship changes
> its acceleration.

For the 21st time: (1) When you say "neither ship changes its acceleration you are failing to distinguish between coordinate acceleration and proper acceleration, and that the relations between the entities (ships and ends of rod) are precisely in accord with the proper and the coordinate accelerations of every particle of every object. (2) You fail to recognize that the free end of the rod is undergoing different accelerations (from a different initial velocity) than the right hand ship, which fully accounts for the facts described above.

Moreover, as explained to you a dozen times before, relativity is not a subjectivist theory, there is a single objective set of facts, and the facts are the same, regardless of anyone's state of motion, as just explained in (1) and (2). Talking about "what the crews say" is just your stupid way of asking about the description of the unique and unambiguous events (that manifestly follow all the normal laws of physics) in terms of an accelerating coordinate system, in terms of which the equations of physics include extra terms (precisely analogous to coriolis and centrifugal accelerations terms of rotating coordinates).

> That's what I'm trying to understand.

That doesn't appear to be true. You seem to be trying to NOT understand it. This has all been explained to you in detail many times. Do you finally understand? If not, what remains unclear to you?

[Mikko]

On 2023-06-03 18:11:39 +0000, sep...@yahoo.com said:

> Two identical spaceships are at rest on the x-axis 100 meters apart.
> When either spaceship accelerates it accelerates at a fixed proper rate
> of 0.01g in the positive x direction. Now if one spaceship starts
> accelerating in the positive x direction toward the other spaceship
> before the other spaceship starts accelerating in the positive x
> direction, the two spaceships will get closer and closer to each other
> as the accelerations continue (as measured by the crews on board each
> ship). But during these constant accelerations of the two spaceships
> instead of continually getting closer and closer to each other they
> will suddenly start moving further and further apart even though
> neither spaceship experienced any change in their acceleration nor any
> other force affecting the motion of either spaceship. Is motion like
> that possible? If so, what caused that change in motion between the
> two spaceships?

The spaceship that starts first is always faster. It first approaches the
one in front, then catches up and flies past. After that the faster ship
is ahead and the separation between ships keeps increasing.

Mikko

[Mike Fontenot]

On 6/5/23 10:18 AM, Tom Roberts wrote:

>
> Note that for the basic Bell's spaceship scenario, one cannot really
> claim the spaceships are "moving further apart", because "moving" must
> reference a SINGLE coordinate system, and that simply does not apply: as
> the spaceships accelerate and change inertial rest frames, in successive

> frames they are further apart ...

That's an almost universal contention, but it's not true. In the frame
of the trailing spaceship (and there IS such a frame), the separation of
the spaceships is constant. The string doesn't break.

>Note also that
> relative to their starting inertial frame they maintain a constant
> separation and are NOT moving further apart.
>

Relative to that initial inertial frame, the two rockets get closer
together. The string doesn't break.

Details are in:

https://vixra.org/abs/2304.0223


Michael Leon Fontenot

[Dono.]

Vixra is the repository of idiocies written by cranks. Like you, Mike.

[Mikko]

On the page

https://uploads.disquscdn.com/images/7d6b7c8ae54383686bb1c5935af03c79dddfb88d1af765e126d132fca2eaed5f.png


there is a diagram that illustrates the situation.

Mikko

[Trevor Lange]

That isn't the scenario Dave is talking about. He's essentially talking about two fully hyperbolic trajectories, giving two copies of Bell's spaceship scenario, one forward and one backwards. In terms of the inertial coordinate system he calls F0, the two ships are both stationary at a given time at the middle of the scenario, at a distance of 5 meters apart. Both ships are and have been undergoing equal constant proper acceleration, both before and after this mutual resting condition, so their mutual distance (in terms of F0) is constant. The ships never pass each other. The acceleration is extremely slow. Originally, when both were moving to the left, there was a condition when a measuring rod at rest with one of them reads 10 meters separation, and then later that rod shows 5 meter separation, and then later still it is back up to 10 meters. Again, they never pass each other. This has been fully explained to him many many times.

Please don't respond to him until/unless you actually understand what he is talking about. That just encourages him, by giving him the opportunity to correct you, which makes him feel smart. Don't do that.

[Tom Roberts]

On 6/6/23 11:14 AM, Mike Fontenot wrote:
> On 6/5/23 10:18 AM, Tom Roberts wrote:
>> Note that for the basic Bell's spaceship scenario, one cannot
>> really claim the spaceships are "moving further apart", because
>> "moving" must reference a SINGLE coordinate system, and that simply
>> does not apply: as the spaceships accelerate and change inertial
>> rest frames, in successive frames they are further apart ...
>
> That's an almost universal contention, but it's not true.

Yes it is.

> In the frame of the trailing spaceship (and there IS such a frame),

While it is accelerating, there is no frame. Because "frame" implies
orthogonal axes (i.e. all off-diagonal components of the metric are
zero); coordinates accelerated along x necessarily have nonzero g_xt and
g_tx components of the metric.

That's why the Bell scenario has the spaceships stop accelerating after
a given elapsed proper time -- that puts them at rest in a single
inertial frame, and in that frame they are further apart than they were
in the initial frame when they started. (One must do the calculation to
show that; it is available in many places on the net.)

>> Note also that relative to their starting inertial frame they
>> maintain a constant separation and are NOT moving further apart.
>
> Relative to that initial inertial frame, the two rockets get closer
> together. The string doesn't break.

Both of these claims are wrong.

It is easy to show that the two spaceships have constant separation as
measured in their initial inertial frame: they start accelerating
simultaneously in that frame, and they have identical accelerations
relative to that frame; integrate their acceleration twice to get
position relative to this frame, and since the accelerations are
identical, the integrals are identical, so their separation is constant
(the same as when they started). So at any time in the initial frame
their separation is the same as when they started (of course they are
observed simultaneously in that frame). Stopping acceleration at a given
value of elapsed proper time is simultaneous in their initial frame.

When they stop accelerating they come to rest in some other inertial
frame, and are further apart in that frame than they were in the initial
frame when they started. The string also comes to rest in that other
frame, and since the spaceships are further apart it must break.

> Details are in: https://vixra.org/abs/2304.0223

I have no interest in spending time to find your mistake. But if you
attempt to make a calculation while the spaceships are accelerating it
is VERY difficult to avoid mistakes. Even if you make no overt mistake,
the results will be ambiguous....

Tom Roberts

[Mike Fontenot]

On 6/6/23 1:29 PM, Tom Roberts wrote:
> On 6/6/23 11:14 AM, Mike Fontenot wrote:
>>
>> In the frame of the trailing spaceship (and there IS such a frame),

> (and Tom Roberts continues):

>While it is accelerating, there is no frame.

That is an absurd statement. Tell that to the commander of the trailing
spaceship ... he certainly knows that he has his own frame, and you'll
never convince him otherwise.

[Prokaryotic Capase Homolog]

On Tuesday, June 6, 2023 at 11:14:41 AM UTC-5, Mike Fontenot wrote:

> Relative to that initial inertial frame, the two rockets get closer
> together. The string doesn't break.
>
> Details are in:
>
> https://vixra.org/abs/2304.0223

It is immediately evident from your axes flipped "timespace"
diagram in your vixra article what you are doing wrong and
how you have misinterpreted the setup. Compare with my
spacetime diagram and discussion here:
https://en.wikipedia.org/wiki/Spacetime#Dewan%E2%80%93Beran%E2%80%93Bell_spaceship_paradox

[Tom Roberts]

On 6/6/23 2:45 PM, Mike Fontenot wrote:
> On 6/6/23 1:29 PM, Tom Roberts wrote:
>> On 6/6/23 11:14 AM, Mike Fontenot wrote:
>>> In the frame of the trailing spaceship (and there IS such a
>>> frame),

>> While it is accelerating, there is no frame.
>
> That is an absurd statement.

No, it is a correct statement, as I described.

> Tell that to the commander of the trailing spaceship ... he certainly
> knows that he has his own frame, and you'll never convince him
> otherwise.

Only if he is ignorant of very basic GR. (This isn't about "convincing"
anybody, this is about the basic math of GR.)

He has a APPROXIMATE frame, that may well be good enough to use as if it
were a true frame inside his spaceship [#] (of course he would have to
deal separately with the acceleration). That is certainly not good
enough to extend to the other spaceship.

[#] Similar to the way the LHC experiments use the
approximate frame of their apparatus and analyze each
event using SR. That works because each event lasts no
longer than ~100 ns and is confined to their cavern --
the approximations involved are very much smaller than
their measurement resolution.

Note that having three orthogonal rulers is NOT enough, the time
coordinate must also be orthogonal to each of them, and for accelerating
coordinates that is not possible.

Tom Roberts

[RichD]

On June 6, Mike Fontenot wrote:
> In the frame of the trailing spaceship (and there IS such a frame),
> the separation of the spaceships is constant. The string doesn't break.

> Relative to that initial inertial frame, the two rockets get closer
> together. The string doesn't break.

> https://vixra.org/abs/2304.0223

This is a vacuous debate. You have distorted Bell's paradox
into something else.

The problem, as CORRECTLY stated, stipulates that, in the
external inertial frame, the ships maintain a constant
separation, yet the string breaks.

These facts are stipulated, a GIVEN. The ships DESIGN their
velocity profiles to achieve this condition. The challenge for
the student is to explain the break.

Not "does the string break, or does it not?"

--
Rich

[Mike Fontenot]

On 6/6/23 2:51 PM, Tom Roberts wrote:

> While it is accelerating, there is no frame.

>> Mike Fontenot wrote:
>> That is an absurd statement.
>
> No, it is a correct statement, as I described.
>
>> Tell that to the commander of the trailing spaceship ... he certainly
>> knows that he has his own frame, and you'll never convince him
>> otherwise.
>
> Only if he is ignorant of very basic GR.

This ISN'T a GR scenario! It is purely an SR scenario. There are
accelerations, but there are no gravitational fields. Einstein (in his
1907 paper) was very clear: Two observers accelerating with the same
constant acceleration (as confirmed by accelerometers attached to the
two rockets) will remain a fixed distance apart (according to the
observer in the trailing rocket).

[Mike Fontenot]

In the scenario you describe, accelerometers attached to the two rockets
would show different accelerations. If you want to define the scenario
that way, go ahead, but it is of no interest to me. It is a different
scenario than the scenario I am interested in. And I think it is a
different scenario than most participants in the Bell Spaceship debate
are assuming.

[Tom Roberts]

On 6/6/23 4:58 PM, Mike Fontenot wrote:
> On 6/6/23 3:20 PM, RichD wrote:
>> The problem, as CORRECTLY stated, stipulates that, in the external
>> inertial frame, the ships maintain a constant separation, yet the
>> string breaks.
>>
>> These facts are stipulated, a GIVEN. The ships DESIGN their
>> velocity profiles to achieve this condition. The challenge for
>> the student is to explain the break.
>
> In the scenario you describe, accelerometers attached to the two
> rockets would show different accelerations.

This is NOT true.

To maintain a constant separation in their initial inertial frame, the
two rockets must have identical accelerations relative to that frame
(when measured simultaneously in that frame). Since they start
simultaneously in that frame, they have identical velocities relative to
that frame (when measured simultaneously in that frame). That implies
they have identical proper accelerations as a function of their proper
times, which is what accelerometers attached to the rockets would show.

Note that if they have identical and constant proper accelerations, then
they have identical accelerations relative to their initial frame (when
measured simultaneously in that frame), but these are not constant --
they decrease asymptotically to zero as the rockets asymptotically
approach c relative to that frame.

Note also that they cannot have constant accelerations relative to their
initial frame -- as they approach c relative to that frame, the constant
acceleration simply cannot be maintained as that would require more than
infinite proper acceleration (thrust).

> If you want to define the scenario that way, go ahead, but it is of
> no interest to me.

The standard Bell's spaceship paradox states the rockets have identical
proper acceleration profiles as a function of their proper time. Since
they start simultaneously in their initial inertial frame, they maintain
a constant separation in that frame (when measured simultaneously in
that frame). RichD describes the same scenario, differently. Of course
the string breaks.

This is all well known. For someone who has been obsessing with
accelerated motion for so many years, it is sad that you still don't
understand it.

Tom Roberts

[Tom Roberts]

On 6/6/23 4:45 PM, Mike Fontenot wrote:
> On 6/6/23 2:51 PM, Tom Roberts wrote:
>
>> While it is accelerating, there is no frame.
>
>>> Mike Fontenot wrote: That is an absurd statement.
>>
>> No, it is a correct statement, as I described.
>>
>>> Tell that to the commander of the trailing spaceship ... he
>>> certainly knows that he has his own frame, and you'll never
>>> convince him otherwise.
>>
>> Only if he is ignorant of very basic GR.
>
> This ISN'T a GR scenario! It is purely an SR scenario.

But the same limitation on LOCALLY inertial frames applies.

> Einstein (in his 1907 paper) was very clear: Two observers
> accelerating with the same constant acceleration (as confirmed by
> accelerometers attached to the two rockets) will remain a fixed
> distance apart (according to the observer in the trailing rocket).

I don't know if Einstein really said that, but I'm skeptical because it
is poorly stated and he was always very careful and precise. Note that
claim is WRONG if that trailing observer uses their [#] instantaneously
co-moving inertial frame to measure the distance between rockets. (There
is no locally inertial frame that includes both rockets at rest and is
accurate enough to measure the change in separation.)

[#] gender-neutral pronoun, not plural.

Tom Roberts

[Trevor Lange]

On Tuesday, June 6, 2023 at 4:02:56 AM UTC-7, sep...@yahoo.com wrote:

> > For the 21st time, holding one end of the rod (zero marking) at the left hand object, aligned with the x axis, the right hand object is initially aligned with 10 meter mark on the rod, and then it is aligned with the 5 meter mark on the rod, and then it is aligned with the 10 meter mark on the rod. This is because the rod is essentially always in mechanical equilibrium in its current state of motion, so the locus of positions of the right end of the rod trace a hyperbolic path with a slightly lesser proper acceleration than the right hand object (Born rigid motion), but a lesser leftward initial speed (in terms of F0), and hence that locus and the trajectory of the right object initially diverge, then come to mutual rest, and then re-converge... simple kinematics. Now do you finally understand?
>
> I understand that...

That would be good... but the rest of your sentence falsifies that claim.

> ...but why do the crews say that happens when neither spaceship changes
> its acceleration.

For the 22nd time: (1) When you say "neither ship changes its acceleration you are failing to distinguish between coordinate acceleration and proper acceleration, and that the relations between the entities (ships and ends of rod) are precisely in accord with the proper and the coordinate accelerations of every particle of every object. (2) You fail to recognize that the free end of the rod is undergoing different accelerations (from a different initial velocity) than the right hand ship, which fully accounts for the facts described above.

Moreover, as explained to you many times before, relativity is not a subjectivist theory, there is a single objective set of facts, and the facts are the same, regardless of anyone's state of motion, as just explained in (1) and (2). Talking about "what the crews say" is just your stupid way of asking about the description of the unique and unambiguous events (that manifestly follow all the normal laws of physics) in terms of an accelerating coordinate system, in terms of which the equations of physics include extra terms (precisely analogous to coriolis and centrifugal accelerations terms of rotating coordinates).

> That's what I'm trying to understand.

That doesn't appear to be true. You seem to be trying to NOT understand. This has all been explained to you in detail many times before. Do you finally understand? If not, what remains unclear to you? As always, instead of seeking to finally understand, you just run away again. Very strange.

[Maciej Wozniak]

On Tuesday, 6 June 2023 at 22:51:42 UTC+2, Tom Roberts wrote:
> On 6/6/23 2:45 PM, Mike Fontenot wrote:
> > On 6/6/23 1:29 PM, Tom Roberts wrote:
> >> On 6/6/23 11:14 AM, Mike Fontenot wrote:
> >>> In the frame of the trailing spaceship (and there IS such a
> >>> frame),
> >> While it is accelerating, there is no frame.
> >
> > That is an absurd statement.
> No, it is a correct statement, as I described.
> > Tell that to the commander of the trailing spaceship ... he certainly
> > knows that he has his own frame, and you'll never convince him
> > otherwise.
> Only if he is ignorant of very basic GR. (This isn't about "convincing"
> anybody, this is about the basic math of GR.)

Speaking of basic math, it's always good to remind
that your GR shit had to annouince it false, as it didn't
want to fit the insane postulates of your idiot guru.

>
> He has a APPROXIMATE frame,

And he is FORCED!!! To THE BEST WAY!!!!

[Mikko]

On 2023-06-06 18:54:46 +0000, Trevor Lange said:

> On Tuesday, June 6, 2023 at 7:17:48 AM UTC-7, Mikko wrote:
>> On 2023-06-03 18:11:39 +0000, sep...@yahoo.com said:>> > Two identical
>> spaceships are at rest on the x-axis 100 meters apart.> > When either
>> spaceship accelerates it accelerates at a fixed proper rate> > of 0.01g
>> in the positive x direction. Now if one spaceship starts> >
>> accelerating in the positive x direction toward the other spaceship> >
>> before the other spaceship starts accelerating in the positive x> >
>> direction, the two spaceships will get closer and closer to each other>
>> > as the accelerations continue (as measured by the crews on board
>> each> > ship). But during these constant accelerations of the two
>> spaceships> > instead of continually getting closer and closer to each
>> other they> > will suddenly start moving further and further apart even
>> though> > neither spaceship experienced any change in their
>> acceleration nor any> > other force affecting the motion of either
>> spaceship. Is motion like> > that possible? If so, what caused that
>> change in motion between the> > two spaceships?
>>
>> The spaceship that starts first is always faster. It first approaches
>> the> one in front, then catches up and flies past. After that the
>> faster ship> is ahead and the separation between ships keeps increasing.

> The ships never pass each other.

Whether one ship catches up and passes the other depends on how much
earlier it starts. That was not specifed in the question.

Mikko

[Trevor Lange]

It didn't need to be, because his trolling question was "Is motion like that possible?", to which the answer is yes, motion "like that" obviously occurs if (and only if) the ships begin their accelerations at spacelike-separated events. You see, he wants to find people who don't realize that this is possible, and who think (for example) that the only way a sequence of events such as he described could occur is if the ships pass each other (which he arguably failed to preclude with his wording), and then he can "spring his trap". Had he not run away, his first reply to you would have been to stipulate that the ships never pass each other, and then ask again if "that kind of motion" is possible. He has posed this same question in over a dozen previous threads.

[Mike Fontenot]

On 6/6/23 8:23 PM, Tom Roberts wrote:

>
> To maintain a constant separation in their initial inertial frame, the
> two rockets must have identical accelerations relative to that frame
> (when measured simultaneously in that frame).

If the rockets' accelerometers show the same constant acceleration
(which is the case in the scenario under consideration), the initial
inertial frame MUST conclude that the two rockets get closer together as
the acceleration progresses. This follows from the well-known length
contraction equation, which says an inertial observer will always
conclude that a moving yardstick is shorter than his own yardstick, by
the gamma factor. (The "moving yardsticks" in the above statement
consist of the tape measure that permanently extends from the trailing
rocket to the leading rocket). Details are in

https://vixra.org/abs/2304.0223

[Mike Fontenot]

On 6/6/23 8:35 PM, Tom Roberts wrote:
> On 6/6/23 4:45 PM, Mike Fontenot wrote:

>> Einstein (in his 1907 paper) was very clear:  Two observers
>> accelerating with the same constant acceleration (as confirmed by
>> accelerometers attached to the two rockets) will remain a fixed
>> distance apart (according to the observer in the trailing rocket).
>
>

> Note that
> claim is WRONG if that trailing observer uses his instantaneously

> co-moving inertial frame to measure the distance between rockets.

Not true. See

https://vixra.org/abs/2304.0223

[Mikko]

On 2023-06-07 16:29:35 +0000, Mike Fontenot said:

> On 6/6/23 8:23 PM, Tom Roberts wrote:
>
>> To maintain a constant separation in their initial inertial frame, the
>> two rockets must have identical accelerations relative to that frame
>> (when measured simultaneously in that frame).
>
> If the rockets' accelerometers show the same constant acceleration
> (which is the case in the scenario under consideration), the initial
> inertial frame MUST conclude that the two rockets get closer together
> as the acceleration progresses.

Write the position of each rocket as a function of time,
then compute the speed and acclereation and the readint of the
accelerometers of each rocket as a function of time,
then show that the accelerations are equal and the distance
between the rockets shrinking.
Until you have done all this everyone is free to think that you are wrong.

Mikko

[Tom Roberts]

On 6/7/23 11:29 AM, Mike Fontenot wrote:
> On 6/6/23 8:23 PM, Tom Roberts wrote:
>> To maintain a constant separation in their initial inertial frame,
>> the two rockets must have identical accelerations relative to that
>> frame (when measured simultaneously in that frame).
>
> If the rockets' accelerometers show the same constant acceleration
> (which is the case in the scenario under consideration), the initial
> inertial frame MUST conclude that the two rockets get closer together
> as the acceleration progresses.

This is just plain wrong. This is just Bell's spaceship paradox, and the
analysis is well known:

Since the rockets' proper accelerations are equal, and since they began
accelerating simultaneously in their initial inertial frame, their speed
relative to that frame is always equal (when measured simultaneously in
that frame), but increasing. Call their acceleration and velocity
relative to their initial inertial frame a(t) and v(t), with t that
frame's time coordinate:
v(t) = \integral_0^t dt' a(t') (same for both)

One can compute their positions in that frame as a function of t,
presuming the rear rocket starts at x=0, the front rocket starts at x=L,
and both start accelerating with value a(t) in the +x direction at t=0:
x_rear(t) = \integral_0^t dt' v(t') + 0
x_front(t) = \integral_0^t dt' v(t') + L

CLEARLY their separation in this frame is L, independent of t, and
independent of the details of a(t).

[They have constant proper acceleration, which implies
a(t) varies. Clearly a(0) = their proper acceleration,
and a(t) decreases as t increases, asymptotically
approaching 0 as v(t) asymptotically approaches c.
Clearly the variations with t of a(t) and v(t) do not
affect their separation in this frame.]

Note I did not use SR at all here, just basic Euclidean geometry in
their initial frame. SR is needed to calculate a(t) from their proper
acceleration, but as the conclusion is independent of the details of
a(t), I did not need to do that.

> This follows from the well-known length contraction equation, [...]

You got that wrong.

If, at a given value of their elapsed proper time, both rockets cease
accelerating, that implies they stopped accelerating at the same value
of t in the above equations; call it T. Clearly they are now both at
rest in the inertial frame moving with speed v(T) relative to their
initial frame. Since their separation in their initial frame is L, their
separation in their new rest frame must be LARGER than L, because
"length contraction" applies. That means the string between them must break.

Tom Roberts

[Maciej Wozniak]

On Wednesday, 7 June 2023 at 20:33:24 UTC+2, Tom Roberts wrote:
> On 6/7/23 11:29 AM, Mike Fontenot wrote:
> > On 6/6/23 8:23 PM, Tom Roberts wrote:
> >> To maintain a constant separation in their initial inertial frame,
> >> the two rockets must have identical accelerations relative to that
> >> frame (when measured simultaneously in that frame).
> >
> > If the rockets' accelerometers show the same constant acceleration
> > (which is the case in the scenario under consideration), the initial
> > inertial frame MUST conclude that the two rockets get closer together
> > as the acceleration progresses.
> This is just plain wrong. This is just Bell's spaceship paradox, and the
> analysis is well known:
>
> Since the rockets' proper accelerations are equal,

Since they're equal - we're FORCED!!! To
THE BEST WAY!!!!

[Mike Fontenot]

On 6/7/23 12:33 PM, Tom Roberts wrote:
> On 6/7/23 11:29 AM, Mike Fontenot wrote:

>> If the rockets' accelerometers show the same constant acceleration
>> (which is the case in the scenario under consideration), the initial
>> inertial frame MUST conclude that the two rockets get closer together
>> as the acceleration progresses.

>
> This is just plain wrong.
>

OK, I'll walk you through it.

First, open another window so you can switch back and forth between what
I say here, and what I'll walk you through in my viXra paper's diagram.
In that new window, bring up

https://vixra.org/abs/2304.0223

Look at the diagram on the second page of that viXra paper. (I computed
and printed that diagram about 20 or 30 years ago, and I believed it was
correct until very recently. The two curved lines supposedly show the
position of each rocket, according to the initial inertial observers who
are stationary wrt the rockets immediately before the rockets are
ignited. The horizontal axis gives the current age of those initial
inertial observers. The vertical axis gives the distance from the
initial position of the trailing rocket.

A question for YOU (Tom): Is THAT the diagram you have been referring
to? I THINK it is consistent with what you have been arguing, i.e.,
that the initial inertial observers say that the distance between the
two rockets is constant, as time increases during the accelerations.

If your answer is yes, then you and I agreed until just recently. But I
recently realized that that diagram isn't correct. That diagram says
that, according to those initial inertial observers, the distance
between the two rockets is constant during the accelerations. But the
length contraction equation (LCE) of special relativity says that ANY
inertial observer will conclude that a yardstick moving wrt himself is
always SHORTER than his own yardsticks, by the gamma factor. For
example, at a speed of 0.866 ly/y, the gamma factor is equal to 2.0, so
in that case, the moving yardsticks (linking the spaceships) are only
half as long at the initial inertial observers' own yardsticks.
Therefore the vertical distance between those two curved lines CAN'T be
constant as the acceleration progresses. So the existing curves are
incorrect, and must somehow be modified so that they get closer together
as the acceleration progresses. The bottom curve can't be changed so
that it curves toward the upper curve, because the lower curve already
gets arbitrarily close to the speed of light (1.0 ly/y). The only way
we can modify that diagram so that it is consistent with the length
contraction equation is to lower the upper curve, so that its vertical
distance above the lower curve (at each instant "t") is equal to the
original distance between the curves, divided by the gamma factor at
that instant (the gamma factor is always greater than or equal to 1.0).
As "t" goes to infinity, and the velocity "v" gets arbitrarily close to
1.0 ly/y, gamma gets arbitrarily close to infinity, and so the upper
curve will get arbitrarily close to (but always above) the lower curve.
I.e., the two rockets will get arbitrarily close together as "t" goes to
infinity.

[RichD]

On June 7, Mike Fontenot wrote:
>>> If the rockets' accelerometers show the same constant acceleration,

>>> the initial inertial frame MUST conclude that the two rockets get closer together
>>> as the acceleration progresses.

> But the length contraction equation of special relativity says that ANY

> inertial observer will conclude that a yardstick moving wrt himself is
> always SHORTER than his own yardsticks, by the gamma factor.

Well done, Mike. You have validated John Bell's invention.
Failure to solve a puzzle is a tribute to the puzzle maker.

The lead craft RUNS AWAY from the trailer, because he starts
his engine first, in their frame. Thus the string stretches,
duh! It's as simple as that. Later, the trailer starts, but the
leader maintains a constant velocity, relative to the trailer.
The spacing between the ships isn't a yardstick.

Now do you see where length contraction enters, crucially?

You can twist this any way you want, but then it's no
longer Bell's original problem.

--
Rich

[Mike Fontenot]

On 6/9/23 11:39 AM, RichD wrote:
>
> The lead craft RUNS AWAY from the trailer, because he starts
> his engine first, in their frame.

Not true. The two rocket commanders had synchronized their watches
during the period before they started their trip, and had agreed on
exactly what instant on their watches they would each start their
rockets. They both agree that they started their rockets at the same
instant.

[Mike Fontenot]

On 6/9/23 1:16 PM, Mike Fontenot wrote:
>
> Not true.  The two rocket commanders had synchronized their watches
> during the period before they started their trip, and had agreed on
> exactly what instant on their watches they would each start their
> rockets.  They both agree that they started their rockets at the same
> instant.
>
>

(In the above, I am referring to the scenario in Bell's Spaceship
Paradox, NOT to the scenario specified by the original poster, which is
entirely different.)

[RichD]

In that case, the string would appear to shrink, in Stationary Sam's
frame. Contrary to the stipulated condition.

Well done. Again.

--
Rich

[sep...@yahoo.com]

On Sunday, June 4, 2023 at 4:02:38 PM UTC-5, Trevor Lange wrote:
> On Sunday, June 4, 2023 at 1:43:20 PM UTC-7, sep...@yahoo.com wrote:
> > Answer the following... [Insert insane repetition of the same question asked and answered dozens of times before.]
>
> > During the acceleration the rod is kept perpendicular to the x-axis...
>
> Again, that is utterly brain-dead, since it makes absolutely no difference, as explained in detail in the referenced thread. What on earth is wrong with you? Have you suffered a severe head injury?
>
> > When the two spaceships achieve zero velocity... Do you agree with that?
>
> Of course, this is precisely what has been explained to you dozens of times, including in the very thread that you supposedly just re-read, in complete detail, along with explaining each of your misunderstanding and misconceptions. Is anything about the explanation still unclear to you?

Trevor,
If each identical spaceship is 100 meters in length as measured in F1, and both are aligned parallel to the x-axis, separated by 100 meters along the x-axis as measured in F1, and F1 is moving in the negative x direction with velocity of magnitude V = c*sqrt(3)/2 and they both start accelerating simultaneously as measured in F0, then when the two spaceships reach F0 the spaceship that starts accelerating first in the positive x direction will extend halfway past the base of the second spaceship. That means that the tip of the spaceship accelerates faster than the base of the spaceship where the thrusters are applying the acceleration force.
Please explain why the tip of the spaceship accelerates faster than the base of the spaceship where the acceleration force is applied.
Thanks,
David Seppala
Bastrop TX

[Trevor Lange]

On Tuesday, June 13, 2023 at 4:15:54 PM UTC-7, sep...@yahoo.com wrote:
> If each identical spaceship is 100 meters in length...

All you are doing is replacing the measuring rod with the spaceship. This changes nothing. The same reasoning that has been explained to you a dozen times before still applies.

> The tip of the spaceship accelerates faster than the base of the spaceship...

As always, you willfully refuse to specify whether you are talking about proper acceleration or coordinate systems in terms of some specified system of coordinates. Again, the proper acceleration of the front of each ship is less than (not greater than) the proper acceleration of the back of the ship. This is because the ship, like the rod, is accelerating so slowly that it is always essentially in equilibrium, so it is in Born rigid motion. Of course, the initial velocities are also different, and the combination of the initial velocities and the different accelerations results in the positioning of each end of each rocket.

If this clears things up for you, great. If anything is still unclear to you, go ahead and ask.

[sep...@yahoo.com]

You replied,

"Again, the proper acceleration of the front of each ship is less than (not greater than) the proper acceleration of the back of the ship."

If that is correct, if both spaceships have identical accelerations and started simultaneously in F0, then F0 observers measure that the distance between the thrusters in each spaceship remain constant. But the tip of spaceship that accelerated first traveled 50 meters further than the thruster of the second spaceship. Since both spaceships accelerated identically, starting at the same time, if the tip of the first spaceship traveled 50 meters further then the thruster of the second spaceship, doesn't that mean that the tip of the first spaceship had to accelerate faster than the thruster of the first spaceship?
Please clarify how the thrusters of the two spaceships start identical accelerations simultaneous, travel equal distance, but the tip of the first spaceship travels 50 meters further than the thruster of the first spaceship unless it traveled had a greater acceleration.
David Seppala
Bastrop TX

[Trevor Lange]

On Tuesday, June 13, 2023 at 8:19:05 PM UTC-7, sep...@yahoo.com wrote:
> > The proper acceleration of the front of each ship is less than (not greater than)

> > the proper acceleration of the back of the ship.
>

> If both spaceships have identical accelerations and started simultaneously in F0,

> then F0 observers measure that the distance between the thrusters in each
> spaceship remain constant.

Correction: Since you have stipulated that, in terms of S0, both objects (which you are now calling thrusters) were initially at edqual speeds and then simultaneoutaneously began equal constant proper acceleration, they maintain constant spatial distance from each other. This is self-evident.

> The tip of spaceship that accelerated first traveled 50 meters further than

> the thruster of the second spaceship.

Nope, in terms of S0 the right hand end of the measuring rod (which you are now calling the tip of a spaceship) travels 50 meters LESS far than the left end of the rod and the left object (which you are now calling a thruster). Of course, in terms of an acceleratring coordinate system in which the tip of the ship is at rest, it has not moved at all, and in terms of a coordinate system in which it has moved 10000 meters, it has moved 10000 meters, and so on. Do you understand this?

> Since both spaceships accelerated identically...

Correction: Both objects undergo equal constant proper acceleration, once they respectively begin to accelerate, but they obviously do not undergo equal (or even time-shifted similar) acceleration in terms of arbitrary systems of accelerating coordinates, such as the one that you are trying maniacally to smuggle into the discussion. Remember, in terms of the accelerating coordinate system in which the left object is continuously at rest, the right object is accelerating leftward, then coming to a stop, and then accelerating rightward. Do you understand this? It has been explained to you, in complete detail, over a dozen times.

> if the tip of the first spaceship traveled 50 meters further then the thruster
> of the second spaceship

It didn't. In terms of S0 it traveled 50 meters LESS far than the left object. Of course, in terms of the accelerating coordinates in which it is stationary, it didn't move at all. And so on. Do you understand this?

> Please clarify how the thrusters of the two spaceships start identical PROPER
> accelerations simultaneously [hence identical coordinate accelerations IN TERMS
> OF S0], travel equal distance IN TERMS OF S0, but the tip of the first spaceship
> travels 50 meters further [in terms of S0] than the thruster of the first spaceship...

Again, in terms of S0, it traveled 50 meters LESS, not more. What is wrong with you?

[sep...@yahoo.com]

You seem to be describing the scenario when the spaceships accelerate from F0 to F2 where F2 is a velocity in the positive x-direction relative to F0. Whereas in the scenario in this thread, the two spaceships are at rest in F1 where the relative velocity between F1 and F0 is in the negative x direction relative to F0.

If not, then
Two spaceships at rest in F1 have a proper length from their base to their tip of 100 meters. The two spaceships are parallel to the x-axis. As measured in F1 they are separated by 100 meters along the x-axis. FO observers measure the separation between the bases of each of the two spaceships while they are at rest in F1 as 50 meters. Observers in F0 start the identical accelerations of the two spaceships simultaneously. When the two spaceships reach zero velocity with respect to F0 do you agree that the F0 observers measure the separations between the bases of the two spaceships to be 50 meters?

If so, what do the F0 observers measure the proper length of each of the two spaceships when they now have zero velocity with respect to F0? If it is 100 meters, then draw a simple diagram showing me how the bases of the two spaceships can be 50 meters apart and the proper length of each of the two spaceships 100 meters such that the tip of the first spaceship (the one on the less) traveled 50 meters less then the base of the second spaceship .

[Trevor Lange]

On Wednesday, June 14, 2023 at 7:14:53 AM UTC-7, sep...@yahoo.com wrote:
> > > if the tip of the first spaceship traveled 50 meters further then the thruster
> > > of the second spaceship
> > It didn't. In terms of S0 it traveled 50 meters LESS far than the left object. Of course, in terms of the accelerating coordinates in which it is stationary, it didn't move at all. And so on. Do you understand this?
> >
> > > Please clarify how the thrusters of the two spaceships start identical PROPER
> > > accelerations simultaneously [hence identical coordinate accelerations IN TERMS
> > > OF S0], travel equal distance IN TERMS OF S0, but the tip of the first spaceship
> > > travels 50 meters further [in terms of S0] than the thruster of the first spaceship...
> >
> > Again, in terms of S0, it traveled 50 meters LESS, not more. What is wrong with you?
>

> You seem to be describing the scenario when the spaceships accelerate from F0 to F2 ...

No, I'm referring to the ships beginning at rest in S1 (which is moving to the left in terms of S0) and then accelerating to rest in S0. During that process, in terms of S0, the right end of the rod travels LESS far than the right hand object. (In terms of your pointless new terminology, the tip of the left ship travels less far than the right thruster.) That's why the left object is at the middle of the rod when they all come to rest in terms of S0: The right object travels further to the left than the right end of the rod does. Now do you understand?

[sep...@yahoo.com]

Trevor,
The spaceships when they are at rest in F1 are 100 meters apart along the x-axis and 100 meters in length as measured in F1.
When observers in F0 start the two identical spaceships with identical accelerations simultaneously, when the spaceships reach F0 and have zero relative velocity wrt F0 and the base of the spaceship that started first (leftmost) is at x=0 in the F0 coordinate system and the tip of that spaceship is at x=100 meters as measured in F0, then tell me in terms of the F0 coordinate system the x coordinate of of the base of the second spaceship and the x coordinate of the tip that second spaceship.

[Trevor Lange]

On Wednesday, June 14, 2023 at 4:05:25 PM UTC-7, sep...@yahoo.com wrote:
> The spaceships when they are at rest in F1 are 100 meters apart along the x-axis
> and 100 meters in length as measured in F1.

You failed to specify how the "apartness" is defined, e.g., between the tip of one and the thruster of the other, or between the tips, etc. How your brain could fail to recognize this ambiguity is a mystery. And why you have idiotically re-cast everything from two objects and a measuring rod to two thrusters and a measuring ship is also a mystery. And why you continue to talk in terms of frames (F) instead of coordinate systems (S), as required for any strictly meaningful specification, is yet another mystery.

Since your brain is simply incapable of even formulating your question coherently, I will edit it for you:

> The two spaceships, initially at rest in terms of S1, and their thrusters begin equal constant proper acceleration simultaneously in terms of S0. When the spaceships reach rest in terms of S0 the thruster of the lefthand ship is at x=0 in terms of S0, and the tip of that spaceship is at x=100 meters (i.e., x=1 spaceship length) in terms of S0.

> Please tell me (for the 25th time) in terms of S0 the x coordinate of the

> base of the second spaceship and the x coordinate of the tip that second
> spaceship.

For the 25th time, the x coordinates of the right hand spaceship thruster is x=50 meters when it is at rest in terms of S0, and the tip of that ship is at 150 meters. Hence in terms of S0 the tips have moved 50 meters LESS far than the thrusters have moved. Do you understand this?

[sep...@yahoo.com]

Trevor,
Now you can answer the question that I've been trying to get you to answer. When the spaceships have zero velocity with respect to F0, the tip of the first spaceship is at x=100 meters. The base of the second spaceship is at 50 meters in terms of F0. The tip of the first spaceship and the base of the second spaceship started at the same x coordinate. So you are saying that the tip of the first spaceship accelerated faster than the base of the second spaceship, while the base of each spaceship accelerated at exactly the same rate. Correct?

[Trevor Lange]

On Wednesday, June 14, 2023 at 7:11:00 PM UTC-7, sep...@yahoo.com wrote:
> > > Please tell me in terms of S0 the x coordinate of the

> > > base of the second spaceship and the x coordinate of the tip that second
> > > spaceship.
> >

> >The x coordinates of the right hand spaceship thruster is x=50 meters when it is at rest in terms of S0, and the tip of that ship is at 150 meters. Hence in terms of S0 the tips have moved 50 meters LESS far than the thrusters have moved.
>
> [Thank you for patiently answering all my questions.]

You're welcome.

> [Allow me to summarize what you have just taught me:] When the spaceships have
> zero velocity in terms of S0, the tip of the first spaceship is at x=100 meters. The
> base of the second spaceship is at 50 meters in terms of S0.

Right! You have successfully repeated what I just told you. Bravo.

> The tip of the first spaceship and the base of the second spaceship
> started at the same x coordinate.

Right, this was the part of the scenario specification that you omitted, and I supplied. Again, you're welcome.

> So you are saying that the tip of the first spaceship accelerated faster

> than the base of the second spaceship...

The tips of the spaceships undergo LESS proper acceleration, and also less coordinate acceleration in terms of S0, because the ships are in essentially Born rigid motion. How many dozen times has this been explained to you? And how many more times will it need to be explained to you until you grasp it?

> while the base of each spaceship accelerated at exactly the same rate.

The two bases, by stipulation, are undergoing equal constant proper acceleration, and the two tips are undergoing slightly LESS equal constant proper acceleration (and initially are imparted with slightly less leftward velocity in terms of S0).

This has all been explained to you a dozen times before. Why are you unable to grasp this? Were you kicked in the head by a mule?

[sep...@yahoo.com]

Trevor
If the " The two bases, by stipulation, are undergoing equal constant proper acceleration, and the two tips are undergoing slightly LESS equal constant proper acceleration.. " as you say, please explain why the base of the second spaceship travels a shorter distance than the tip of the first spaceship if they start at the same x coordinate and the tip is undergoing slightly LESS acceleration.

You add that the tips, "initially are imparted with slightly less leftward velocity in terms of S0." How are they imparted with slightly less leftward velocity if they start accelerating to the right after the base of each spaceship (where the thrusters are) starts accelerating to the right first?

David Seppala
Bastrop TX

[Trevor Lange]

On Wednesday, June 14, 2023 at 8:03:28 PM UTC-7, sep...@yahoo.com wrote:
> Please explain why the base of the second spaceship travels a shorter distance

> than the tip of the first spaceship if they start at the same x coordinate and the
> tip is undergoing slightly LESS acceleration.

As already explained to you a dozen times before, the tips are initially are imparted with slightly less leftward velocity in terms of S0.

> How are they imparted with slightly less leftward velocity if they start

> accelerating to the right after the base of each spaceship starts

> accelerating to the right first?

The acceleration you've specified is extremely slow, so the spaceships (previously measuring rods) are always in mechanical equilibrium over the time scale of relevance, and hence are effectively in Born rigid motion. The deviations from perfect Born rigid motion for the parameters you specified are totally imperceptible, and hence the tips are continuously forced by mechanical equilibrium to maintain a constant proper length, which entails that the tips are forced to have a slightly less leftward initial velocity and a lower proper acceleration, all in accord with Born rigid motion.

This has all been explained to you multiple times before. Are you beginning to understand?

[sep...@yahoo.com]

Trevor,
If a spaceship (or a rod) has a constant velocity in the negative x direction relative to F0, and you start to accelerate the left end (most negative x coordinate) of the spaceship (or rod), the right end of the rod will not change its velocity instantaneously. Hence it will have a greater velocity toward the left (negative direction) then the left end of the spaceship (or rod) that just started accelerating in the positive x direction. Doesn't that make any sense to you?
David Seppala
Bastrop TX

[Trevor Lange]

On Wednesday, June 14, 2023 at 9:40:16 PM UTC-7, Trevor Lange wrote:

On Wednesday, June 14, 2023 at 8:53:07 PM UTC-7, sep...@yahoo.com wrote:
> If a spaceship (or a rod) has a constant velocity in the negative x direction

> in terms of S0, and you start to accelerate the left end (most negative x

> coordinate) of the spaceship (or rod), the right end of the rod will not change
> its velocity instantaneously.

Right, there will be a speed of sound delay, but that is negligible because the acceleration you've specified is so slow that the rod or spaceship is essentially in mechanical equilibrium, meaning it is not perceptibly stressed at all, and this state of mechanical equilibrium is maintained through the very long period of very slow acceleration. Hence it is always essentially in Born rigid motion.

> Hence it will have a greater velocity toward the left (negative direction)

> than the left end of the spaceship (or rod) that just started accelerating

> in the positive x direction.

Again, the acceleration rate is so slow that the ship or rod is always essentially in mechanical equilibrium, and any strains (transient or steady-state) due to acceleration are imperceptibly small. At any point along the journey you could have an inertial ship side-by-side with this slowly accelerating ship with matched speed at that moment, and you would not be able to measure the difference between them, because they are both essentially unstressed. The force you are applying to the end of the steel ship to accelerate it at 0.01g is not even *remotely* large enough to produce any appreciable mechanical compression. Repeat after me: The ship is never perceptibly far from mechanical equilibrium at any time.

This was all explained to you many times before. Hence each part of the ship or rod behaves exactly as I've explained to you a dozen times. Now do you finally understand?

[sep...@yahoo.com]

Trevor
If in F1 the 100 meter spaceships are initially at rest and aligned parallel to the x-axis with the tip of the leftmost spaceship sharing the same x coordinate as the thruster of the identical spaceship then in terms of F0 when observers in F0 simultaneously start the identical acceleration of both spaceships I get these results:
The base (thruster end ) of spaceship1 is at x=L and the tip of spaceship1 is at x=L+50 meters as observed in F0. The base (thruster end) of spaceship 2 is at x=L+50 meters and the tip of spaceship2 is at x=L+100 meters as observed in F0.
When observers in F0 start the identical acceleration of each spaceship simultaneously, the thruster of spaceship2 starts accelerating in the positive x direction before the tip of spaceship1 starts accelerating.
When the the two spaceships have zero velocity with respect to F0, they each travel a distance D as measured in F0. Therefore
the position of the base (thruster end) of spaceship1 is x=L+D, the tip of spaceship1 is at L+D+100. The position of the base (thruster end) of spaceship2 is at x=L+D+50 and the tip of spaceship2 is at x=L+D+150.
Therefore, as observed in F0 the tip of spaceship1 accelerated a further distance than the base of spaceship2 since they started at the same x coordinate but the tip of spaceship1 traveled 50 meters further than the base of spaceship2. (In one response that you posted here you said that wasn't true, but you are now aware that it is).
Therefore since the acceleration of the base of each spaceship is identical and started simultaneously, observers in F0 observe that the tip of spaceship1 accelerated faster than the acceleration of the base (thruster end) of spaceship1.
So during this leg of the scenario, the tip of spaceship1 is accelerating faster than the base (thruster end) of spaceship2. Please explain the "physics" effect as to why the base of spaceship2 suddenly starts accelerating faster than the tip of spaceship1 once they reached zero velocity with respect to F0 and continued their acceleration. What caused that to occur, other than that is a consequence of Einstein's theory.

[Trevor Lange]

On Thursday, June 15, 2023 at 7:29:45 PM UTC-7, sep...@yahoo.com wrote:
> If in terms of S1 the 100 meter spaceships are initially at rest and

> aligned parallel to the x-axis with the tip of the leftmost spaceship
> sharing the same x coordinate as the thruster of the identical spaceship

> then when the acclerations start simultaneously in terms of S0 I get

> these results:
> The base (thruster end ) of spaceship1 is at x=L and the tip of spaceship1

> is at x=L+50 meters in terms of S0.

Right as edited.

> The base (thruster end) of spaceship 2 is at x=L+50 meters and the tip of

> spaceship2 is at x=L+100 meters in terms of S0.

Right as edited.

> When the equal constant proper accelerations of the thrusters start simultaneously
> in terms of S0, the thruster of spaceship2 starts accelerating in the positive x direction

> before the tip of spaceship1 starts accelerating.

Not appreciably. Given the parameters you have specified, the strain and settling time for mechanical equilibrium to be reached is entirely negligible. Look, depending on the material properties, the proper length when accelerating is, say, 99.9992 meters, and this remains constant throughout the next years while it is slowly advancing through the scenario, and throughout all the years, with the irrelevant exception of the first fraction of a second as the acceleration is applied and is settling into mechanical equilibrium, the object is in perfect Born rigid motion. That's why the tips undergo lesser constant proper acceleration, and their trajectories have initially lower leftward speed, and they travel a lesser distance in terms of S0.

> When the the two spaceships have zero velocity in terms of S0, the thrusters have each
> traveled a distance D in terms of S0. Therefore the position in terms of S0 of the base

> (thruster end) of spaceship1 is x=L+D, the tip of spaceship1 is at L+D+100.

Well, they've been moving in the negative x direction, so D is obviously negative.

> Also, in terms of S0, the position of the base (thruster end) of spaceship2 is at

> x=L+D+50 and the tip of spaceship2 is at x=L+D+150.

Again, D is negative. Why are you reciting these trivialities (poorly)?

> Therefore, in terms of S0 the tip of spaceship1 accelerated a further distance
> than the base of spaceship2...

"Accelerated a distance"? Presumably your addled brain meant to say "traveled a distance", and again, you are wrong. In terms of S0, the tips of the ships traveled 50 meters LESS than the thrusters. Look, the thruster1 traveled from L to L+D, and the tip1 traveled from L+50 to L+D+100. So, if D equals -1000 (for example), the base traveled a distance of -1000 whereas the tip traveled only -950, which is 50 meters LESS. Can it be you overlooked that D is negative? Sheesh.

> since they started at the same x coordinate but the tip of spaceship1 traveled
> 50 meters further than the base of spaceship2. (In one response that you
> posted here you said that wasn't true, but you are now aware that it is).

LOL. See above. Now are you beginning to understand?

[sep...@yahoo.com]

Trevor,

You wrote:
"Well, they've been moving in the negative x direction, so D is obviously negative."

In the scenario F1 is moving in the negative x direction relative to F0, but when they start accelerating they accelerate in the positive x direction so D is obviously positive.
David Seppala
Bastrop TX

[Trevor Lange]

> You wrote: "Well, they've been moving in the negative x direction, so D is obviously negative."

> In the scenario S1 is moving in the negative x direction relative to S0,

> but when they start accelerating they accelerate in the positive x direction
> so D is obviously positive.

No, the ships begin at rest in S1, which means they are initially moving in the negative x direction at velocity -sqrt(3)/2, and then they accelerate in the positive direction until coming to rest in terms of S0. In other words, their velocity goes from the initial value -sqrt(3)/2 to 0. Throughout this time they are always have negative velocity, so they are moving in the negative x direction in terms of S0, at a slower and slower speed (i.e., less and less negative velocity). Thus they have moved a huge distance in the negative x direction in terms of S0.

Then, as they continue to accelerate in the positive x direction, they acquire positive velocity (after reaching rest in terms of S0), and their velocity increases from 0 to +sqrt(3/2), by which time they are back at the same x position where they started.

How can you possibly not understand this? Did you get kicked in the head by a mule?

[Trevor Lange]

On Wednesday, June 14, 2023 at 8:03:28 PM UTC-7, sep...@yahoo.com wrote:
> > Well, they've been moving in the negative x direction, so D is obviously negative.
>
> In the scenario S1 is moving in the negative x direction relative to S0,
> but when they start accelerating they accelerate in the positive x direction
> so D is obviously positive.

No, the ships begin at rest in S1, which means they are initially moving in the negative x direction at velocity -sqrt(3)/2, and then they accelerate in the positive direction until coming to rest in terms of S0. In other words, their velocity goes from the initial value -sqrt(3)/2 to 0. Throughout this time they are always have negative velocity, so they are moving in the negative x direction in terms of S0, at slower and slower speed (i.e., less and less negative velocity). Thus, by the time they come to rest in terms of S0, they have moved a huge distance in the negative x direction in terms of S0.

Then, as they continue to accelerate in the positive x direction, they acquire positive velocity (after reaching rest in terms of S0), and their velocity increases from 0 to +sqrt(3/2), by which time they are back at the same x position where they started. Now do you understand? Is this the point where you run away again? Sheesh.

[sep...@yahoo.com]

Trevor,
If the two identical spaceships are 100 meters in length and 100 meters apart (from base to base, with the tip of spaceship1 having the same x coordinate as the base of spaceship2) as measured in F1, if F0 observers simultaneously start the accelerations of the two spaceships, when they arrive with zero velocity with respect to F0, if in terms of F0 coordinates, the base (thruster end) of spaceship1 is at x=0 and the base (thruster end) of spaceship2 is at x= 50 meters, please post the coordinates of the tips of each of the two spaceships.
I get spaceship1 tip is at x=100 meters, and spaceship2 tip is at x=150 meters.
David Seppala
Bastrop TX

[Trevor Lange]

On Saturday, June 17, 2023 at 11:11:56 AM UTC-7, sep...@yahoo.com wrote:
> > > > Well, they've been moving in the negative x direction, so D is obviously negative.
> > >
> > > In the scenario S1 is moving in the negative x direction relative to S0,
> > > but when they start accelerating they accelerate in the positive x direction
> > > so D is obviously positive.
> > No, the ships begin at rest in S1, which means they are initially moving in the negative x direction at velocity -sqrt(3)/2, and then they accelerate in the positive direction until coming to rest in terms of S0. In other words, their velocity goes from the initial value -sqrt(3)/2 to 0. Throughout this time they are always have negative velocity, so they are moving in the negative x direction in terms of S0, at slower and slower speed (i.e., less and less negative velocity). Thus, by the time they come to rest in terms of S0, they have moved a huge distance in the negative x direction in terms of S0.
> >
> > Then, as they continue to accelerate in the positive x direction, they acquire positive velocity (after reaching rest in terms of S0), and their velocity increases from 0 to +sqrt(3/2), by which time they are back at the same x position where they started. Now do you understand?
>

> Yes I finally understand my stupid mistake. I don't know how I (or anyone!)
> could be so stupid. Thanks for being patient with me.

No problem.

> Okay, not that I understand and acknowledge that everything I thought and said
> up to now was idiotic drivel, and everything you've been telling me all along was
> correct, let me ask yet another nitwit question (as always, please edit my willfully
> erroneous text so that it at least parses meaningfully):
>
> When the ships arrive with zero velocity in terms of S0, if in terms of S0 coordinates

> the base (thruster end) of spaceship1 is at x=0 and the base (thruster end) of

> spaceship2 is at x= 50 meters, what are the coordinates of the tips of each of
> the two spaceships?

This has already been asked and answered several times up above. Again, in terms of S0, after starting out at x= 97 light years with velocity v=-sqrt(3)/2, the ships arrive at rest with base1 at x=0, base2 at x=50, tip1 at x=100, and tip2 at x=150. Therefore, the tips have traveled 50 meters LESS than the bases, in the negative x direction. This is all utterly trivial, and has been explained to you dozens of times before.

Do you now recognize and acknowledge that special relativity's account of the phenomena is perfectly logical, self-consistent, and coherent, and that all your wacky claims were due to nothing but misconceptions and misunderstandings?

[sep...@yahoo.com]

Trevor
If in F1 both identical spaceships are 100 meters in length and 100 meters apart with the tip of spaceship1 sharing the same x coordinate as the base (thruster end) of spaceship2, and they are started simultaneously by observers in F1, if spacship1 is at x=0 as measured in F0 when it achieves zero velocity with respect to F0, what are the coordinates in terms of F0 of that spaceship's tip and the base and the tip of spacecship2 when it reaches zero velocity with respect to F0?

[Trevor Lange]

On Saturday, June 17, 2023 at 2:55:36 PM UTC-7, sep...@yahoo.com wrote:
> > This has already been asked and answered several times up above. Again, in terms of S0, after starting out at x= 97 light years with velocity v=-sqrt(3)/2, the ships arrive at rest with base1 at x=0, base2 at x=50, tip1 at x=100, and tip2 at x=150. Therefore, the tips have traveled 50 meters LESS than the bases, in the negative x direction.
>

> While both ships are at rest in S1 they are both 100 meters in length and their bases are 100 meters apart, so tip1 shares the same x coordinate as base2. Then they are started simultaneously in terms of S1...

No, you specified that the ships or objects start accelerating simultaneously in terms of S0, not in terms of S1. If you want to talk about a different scenario now, in which they begin their accelerations simultaneously in terms of S1, then you are essentially just starting in the middle of the previous scenario, and re-labeling S0 and calling it S1.

Again, everything about your scenario (which is just a re-statement of Bell's spaceship scenario, in forward and reverse) has been answered and explained in complete detail, and all your elementary fallacies and misconceptions have been exposed and resolved. If you have any other questions, go ahead and ask.

[sep...@yahoo.com]

Trevor,
In the original scenario, spaceship1's tip ends at the midpoint of spaceship2 when it reaches zero velocity with respect to F0. I'm just trying to compare things when spaceship1 and spaceship2 start their accelerations about 500 nanoseconds apart instead of simultaneously in terms of F0.
David Seppala
Bastrop TX

[Trevor Lange]

On Sunday, June 18, 2023 at 2:47:48 PM UTC-7, sep...@yahoo.com wrote:
> > > > This has already been asked and answered several times up above. Again, in terms of S0, after starting out at x= 97 light years with velocity v=-sqrt(3)/2, the ships arrive at rest with base1 at x=0, base2 at x=50, tip1 at x=100, and tip2 at x=150. Therefore, the tips have traveled 50 meters LESS than the bases, in the negative x direction.
> > >
> > > While both ships are at rest in S1 they are both 100 meters in length and their bases are 100 meters apart, so tip1 shares the same x coordinate as base2. Then they are started simultaneously in terms of S1...
> >
> > No, you specified that the ships or objects start accelerating simultaneously in terms of S0, not in terms of S1. If you want to talk about a different scenario now, in which they begin their accelerations simultaneously in terms of S1, then you are essentially just starting in the middle of the previous scenario, and re-labeling S0 and calling it S1.
> >
> > Again, everything about your scenario (which is just a re-statement of Bell's spaceship scenario, in forward and reverse) has been answered and explained in complete detail, and all your elementary fallacies and misconceptions have been exposed and resolved. If you have any other questions, go ahead and ask.
>

> In the original scenario, tip1 is at the midpoint of spaceship2 when it reaches
> zero velocity in terms of S0.

Right, this has been fully explained to you in detail, and all your misunderstandings and fallacies have been dispelled. You're welcome.

> I'm now trying to think about a different scenario, in which the bases begin their constant proper acceleration simultaneously in terms of S1 instead of S0.

Again, if you want to talk about a different scenario now, in which they begin their accelerations simultaneously in terms of S1, then you are essentially just starting in the middle of the previous scenario, and re-labeling S0 and calling it S1. Obviously your diseased brain is now causing you to "think" that a small difference in the timing of the initial accelerations shouldn't result in comparative differences on the order of 50 meters... but of course that's insane. You're talking about a difference of 50 meters out of 97 lightyears(!), accumulating over centuries. The effects of changes in initial conditions are perfectly proportionate.

So, now that you finally understand this, do you have any additional questions?

[Tom Roberts]

On 6/7/23 2:52 PM, Mike Fontenot wrote:
> First, open another window so you can switch back and forth between
> what I say here, and what I'll walk you through in my viXra paper's
> diagram. In that new window, bring up
> https://vixra.org/abs/2304.0223

I am following your text here, not that paper; I am only looking at the
diagram in that paper. These are point-like rockets, not the extended
rockets discussed by others in this thread.

I presume that the scenario stipulates that the two rockets experience
identical proper accelerations as a function of their elapsed proper
times, and that they start accelerating simultaneously in their initial
inertial rest frame. That is, their internal proper-time clocks are
synchronized in their initial inertial rest frame immediately before
they begin accelerating.

> Look at the diagram on the second page of that viXra paper. (I
> computed and printed that diagram about 20 or 30 years ago, and I
> believed it was correct until very recently. The two curved lines
> supposedly show the position of each rocket, according to the initial
> inertial observers who are stationary wrt the rockets immediately
> before the rockets are ignited. The horizontal axis gives the
> current age of those initial inertial observers. The vertical axis
> gives the distance from the initial position of the trailing rocket.
>
> A question for YOU (Tom): Is THAT the diagram you have been
> referring to? I THINK it is consistent with what you have been
> arguing, i.e., that the initial inertial observers say that the
> distance between the two rockets is constant, as time increases
> during the accelerations.

That diagram is at least correct in that it shows the salient feature:
at any horizontal position (time coordinate of the initial inertial
frame), the vertical separation between the two lines (separation
measured in the initial inertial frame) is constant. I cannot vouch for
the shapes of the curves nor for the labeling of axes (which appears to
be confused).

> [...] That diagram says that, according to those initial inertial
> observers, the distance between the two rockets is constant during
> the accelerations.

Yes. That is what SR predicts for this specific scenario. See my
description of the calculation below.

> But the length contraction equation (LCE) of special relativity says
> that ANY inertial observer will conclude that a yardstick moving wrt
> himself is always SHORTER than his own yardsticks, by the gamma
> factor.

That is VERY poorly stated, and seems to be the core of your confusion.

[It is so sad that you have been dabbling about this
for many years, yet STILL do not know how to specify
things precisely.]

In SR, "length contraction" implies that the MEASUREMENT in an inertial
frame will obtain a value smaller than the proper length of a yardstick
(or the proper distance between two rockets) that is moving relative to
the frame along its length.

[This, of course, presumes the standard method of
measuring the length of a moving object: mark both
ends simultaneously and then measure the distance
between marks.]

So in this case, since the separation of the rockets is constant in
their initial inertial frame, it necessarily follows that their proper
separation is INCREASING as their velocity relative to that frame
increases -- in this scenario that means that if both rockets cease
accelerating at the same value of their elapsed proper times, they will
then be at rest in a single inertial frame, and their separation
measured in that frame will be LARGER than their original separation
measured in their original inertial frame.

[That is, in Bell's original scenario the string breaks.]

> For example, at a speed of 0.866 ly/y, the gamma factor is equal to
> 2.0, so in that case, the moving yardsticks (linking the spaceships)
> are only half as long at the initial inertial observers' own
> yardsticks.

Again this is so poorly worded that it is tantamount to being wrong. You
have confused the yardsticks' total proper length with their total
length measured in an inertial frame relative to which they are moving.

[I repeat: it is so sad that you STILL do not know how to
specify things precisely. No wonder you are confused.]

> Therefore the vertical distance between those two curved lines CAN'T
> be constant as the acceleration progresses.

This is just plain wrong. In this scenario their separation measured in
their initial inertial rest frame MUST be constant. This is just basic
relativity and integral calculus. Let me use S to denote their initial
inertial rest frame, x,t as coordinates in S, and a as their proper
acceleration:

1. a is an arbitrary function of the rocket's elapsed
proper time, the same function for both. Both proper
times start at 0, simultaneously in S.
2. at any value of their elapsed proper time (>0) the two
rockets are at rest in the same instantaneously co-
moving inertial frame (ICIF). This must be true because
the only difference between them is their starting
position in S.
3. at any value of their elapsed proper time, the conversion
from a to d^2x/dt^2, and the conversion from proper time
to t, depend only on the velocity of their ICIF relative
to S. They are thus identical for the two rockets.
4. so d^2x/dt^2 must be the same function of t for both
rockets. Integrate it once (dt) to verify their velocity
relative to S is the same function of t. Integrate it
again to show that their separation in S is constant,
independent of t (these integrals must of course
incorporate the initial conditions).

> [... further nonsense based on the above error]

Tom Roberts

[sep...@yahoo.com]

Trevor,
When the base (thruster end) of spaceship1 reaches zero velocity with respect to F0, what is the velocity of the tip of spaceship1 with respect to F0?

[Mike Fontenot]

That the acceleration of the two (point) rockets is exactly the same and
constant, is guaranteed by the readings on the two accelerometers
mounted on the rockets.

You can visualize a measuring rod spanning the distance between the two
(point) rockets. (If you'd like, you can require that all along that
measuring rod, there are attached accelerometers and tiny rockets that
guarantee that each part of the measuring rod is undergoing the exact
same constant acceleration as the two rockets.)

So the very long measuring rod is undergoing a constant acceleration in
a single dimension. Therefore the famous length contraction equation
(LCE) of special relativity says that, according to the inertial person
who was co-located with the trailing end of the measuring rod
immediately before it started accelerating, the measuring rod
continually gets shorter as the acceleration continues.

Therefore the original diagram that shows a constant separation of the
two rockets, according to that inertial person, is incorrect.

Q.E.D.

[Tom Roberts]

On 6/20/23 4:36 PM, Mike Fontenot wrote:
> That the acceleration of the two (point) rockets is exactly the same
> and constant, is guaranteed by the readings on the two
> accelerometers mounted on the rockets.

Once again you do not specify things with sufficient precision.

I must assume you mean their PROPER accelerations are the same. (That is
sort-of implied by your wording.)

> You can visualize a measuring rod spanning the distance between the
> two (point) rockets. (If you'd like, you can require that all along
> that measuring rod, there are attached accelerometers and tiny
> rockets that guarantee that each part of the measuring rod is
> undergoing the exact same constant acceleration as the two rockets.)
>
> So the very long measuring rod is undergoing a constant acceleration
> in a single dimension.

OK to all that. But the "measuring rod" is stretching (contrary to the
usual expectations)....

IOW: this is not Born rigid motion (which is what we expect for
measuring rods).

> Therefore the famous length contraction equation (LCE) of special
> relativity says that, according to the inertial person who was
> co-located with the trailing end of the measuring rod immediately
> before it started accelerating, the measuring rod continually gets
> shorter as the acceleration continues.

No. Repeating your previous error does not make it correct. I have no
idea where you get this, as it is NOT part of any "length contraction"
discussion I have seen (except your incorrect ones). Note it does not
matter if that "person" is in front or behind, because the measurement
is made IN THE FRAME.

Since the rockets' proper accelerations are identical, the initial
inertial rest frame must measure their separation to remain constant
(they start accelerating simultaneously in that frame). See my previous
post for why.

In this new scenario the front rocket must continually pull the
measuring rod apart. We normally consider measuring rods to be rigid and
not changing length, so this scenario is internally INconsistent.
Discuss the rockets' separation, not a measuring rod supposedly set
between them (which is basically assuming what you are trying to
establish, and simply cannot hold for the usual rigid measuring rod).

> Therefore the original diagram that shows a constant separation of
> the two rockets, according to that inertial person, is incorrect.

Nope. Re-asserting a falsehood does not make it correct. No amount of
hand-waving on your part can change the fact that in this scenario the
initial inertial rest frame measures the separation between rockets to
be constant -- that is just basic relativity and integral calculus; see
my previous post. And "length contraction" implies that as the rockets'
velocity gets larger relative to that frame, their PROPER separation
must be increasing.

It is sad that after so many years on your part you STILL do not
understand this. I believe much of the blame is due to your imprecise
wordings....

Tom Roberts

[RichD]

On June 20, Mike Fontenot wrote:
> That the acceleration of the two (point) rockets is exactly the same and
> constant, is guaranteed by the readings on the two accelerometers
> mounted on the rockets.
>

> So the very long measuring rod is undergoing a constant acceleration in
> a single dimension. Therefore the famous length contraction equation
> (LCE) of special relativity says that, according to the inertial person
> who was co-located with the trailing end of the measuring rod
> immediately before it started accelerating, the measuring rod
> continually gets shorter as the acceleration continues.

https://www.istockphoto.com/photo/woman-pulling-gum-out-of-her-mouth-gm90309172-14244199

So, if the girl rides inside an accelerating vehicle, and stretches
the gum, the gum contracts, in the frame of her teeth? (she starts
pulling before the car starts to move)

Fontenot World is a strange place -


--
Rich

[Trevor Lange]

On Tuesday, June 20, 2023 at 1:04:57 PM UTC-7, sep...@yahoo.com wrote:
> When the base (thruster end) of spaceship1 reaches zero velocity with
> respect to F0, what is the velocity of the tip of spaceship1 with respect to F0?

Zero, neglecting any irrelevant non-idealized effects. Again, the ship is essentially in Born rigid motion, so the trajectories of the ends are two "concentric" hyperbolas. How could you not already know the answer to that question? Very strange.

[Mike Fontenot]

Here's another way to look at it:

Suppose that, at some given instant "t2" after the acceleration begins
(where "t2" corresponds to the age of the person on the trailing rocket,
who had just been born when the acceleration began), there is an
inertial person who is momentarily stationary with that accelerating
person on the trailing rocket ... call him inertial person #2, with
inertial person #1 being the inertial person who was stationary wrt the
trailing rocket immediately before the acceleration began.

Inertial person #2 agrees with the accelerating trailing person about
the separation of the rockets at that instant "t2". I.e., at that
instant "t2", he agrees about the length of the long measuring rod that
exactly fills that separation. How does he determine that distance? To
(eventually) determine that distance, inertial person #2 can enlist the
help of all the other inertial observers with whom he is perpetually
stationary, and who are collectively spread throughout that spatial
dimension. And inertial person #1 can do likewise.

If inertial person #2 then communicates with inertial person #1,
they will find that the separation measured by inertial person #2 is
less than the separation measured by inertial person #1, by the factor
gamma > 1.0. For example, if the velocity of the trailing rocket is v2
= 0.866c at "t2", then the gamma factor is equal to 2.0, and thus the
separation between the two rockets at "t2" is half as much as it was
immediately before the acceleration started. And their conclusion is
exactly the same as that given by the well-known length contraction
equation (LCE).

Q.E.D.

[sep...@yahoo.com]

Trevor,
Spaceship1 is 100 meters long and spaceship2 is 100 meters long and the tip of spaceship1 shares the same x coordinate as the base of spaceship2 when at rest in F1. The two spaceships start their accelerations simultaneously as observed in F0. Let F0 observers put three clocks onboard each spaceship, one at the base, one at the midpoint and on at the tip and the F0 observers set the clocks to zero when they start the accelerations of the two spaceships simultaneously (call the clocks S1BC, S1MC, S1TC, S2BC, S2MC, S2TC). If the clock at the at the base of spaceship1 (S1BC) and the clock at the base of spaceship2 (S2BC) each read an elapsed time of T when they have zero velocity with respect to F0, please describe the time shown on each of the other four clocks. Is a clock reading equal to T, or is a clock slower than T or is a clock faster than T when the spaceships each have zero velocity with respect to F0.

[RichD]

On June 7, Tom Roberts wrote:
>>> To maintain a constant separation in their initial inertial frame,
>>> the two rockets must have identical accelerations relative to that
>>> frame (when measured simultaneously in that frame).
>This is just Bell's spaceship paradox, and the
> analysis is well known:
> Since the rockets' proper accelerations are equal, and since they began
> accelerating simultaneously in their initial inertial frame, their speed
> relative to that frame is always equal (when measured simultaneously in
> that frame), but increasing.
> CLEARLY their separation in this frame is L, independent of t, and
> independent of the details of a(t).
> [They have constant proper acceleration, which implies a(t) varies.

That isn't right. Constant proper accelerations cannot satisfy the
stipulated condition.

We consider the trailer's frame T, and the stationary frame S.
Denote the T frame with primed coordinates, and the stationary
frame unprimed.

v is the trailer's speed.
g is the Lorentz factor.
L: the string length (crafts' separation)
u: the leader's speed, relative to the trailer
a: the trailer's acceleration
You propose that a' is constant.

In T, the trailer sees the leader start, then after a brief delay,
he follows. Hence the leader flies ahead at constant speed u'.
S sees them start simultaneously.

In T, during an interval dt', the string stretches:
I) dL' = (dL'/dt') dt' = u' dt'

Therefore S sees:
II) dL = dL' / g

This must be nullified via an increase in v; call it the
augmented Lorentz contraction.
This change in length, during the interval dt <==> dt':
III) dL = (dL/dv) dv = (d(L' / g)/dv) dv
L' = ∫ v' dt' = ∫ a' t' dt' = a' (t')²
Hence
(d(L' / g)/dv) = a' (t')²/ d(1/g)/dv = H(t',v)
And v = v' / g = (a' t') / g
Hence H(t',v) is a messy non-linear function of t'.

Equating (II) and (III):
dL' / g = (dL/dv) dv = H(t',v) dv

This shows a' as an implicit function of t'.
It cannot be constant.

--
Rich

[Trevor Lange]

On Wednesday, June 21, 2023 at 9:46:17 AM UTC-7, sep...@yahoo.com wrote:
> Spaceship1 is 100 meters long and spaceship2 is 100 meters long and the tip of spaceship1 shares the same x coordinate as the base of spaceship2 when at rest in F1. The two spaceships start their accelerations simultaneously as observed in F0. Let F0 observers put three clocks onboard each spaceship, one at the base, one at the midpoint and on at the tip and the F0 observers set the clocks to zero when they start the accelerations of the two spaceships simultaneously (call the clocks S1BC, S1MC, S1TC, S2BC, S2MC, S2TC). If the clock at the at the base of spaceship1 (S1BC) and the clock at the base of spaceship2 (S2BC) each read an elapsed time of T when they have zero velocity with respect to F0, please describe the time shown on each of the other four clocks. Is a clock reading equal to T...?

Obviously the thruster clocks read 127.75 years, and the tip clocks read slightly more, because they are following trajectories of slightly less proper acceleration.

> or is a clock slower than T or is a clock faster than T when the spaceships each have zero velocity with respect to F0.

Tilt. First you asked if the tip clock reading is the same as that of the base clock, and then you add " or is a clock slower or faster than T..."? Remember, T is not a clock rate, it is a reading, so it makes no sense to talk about faster or slower than T. Duh. For each clock we have dtau/dt = sqrt(1-v^2) where v is the speed of the clock in terms of the system x,t. Thus when each clock has v=0 we have dtau/dt = 1.

This is all just a trivial homework exercise for high school students. You ought to be able to answer these questions yourself.

[sep...@yahoo.com]

Trevor,
Your right. I should have asked if the clocks at the midpoint and the clocks at the tip of each of the two spaceships show an equal elapsed time, or greater elapsed time, or lesser elapsed time than the clocks at the base of each the two spaceships show when they have zero velocity with respect to F0. I didn't mean to confuse people who respond.
David Seppala
Bastrop TX

[Mike Fontenot]

On 6/21/23 10:33 AM, (I, myself) Mike Fontenot wrote:
>
>   If inertial person #2 then communicates with inertial person #1, they
> will find that the separation measured by inertial person #2 is less

> than the separation measured by inertial person #1, by the factor gamma.


I just realized that my statement above can't be correct, because ANY
two inertial people who are moving at a constant speed relative to each
other MUST each conclude that the other's measuring rod is shorter than
their own measuring rod.

I haven't yet figured out how to revise my previous posting, though.

[Mike Fontenot]

Here's another way to look at it: (Revised Version)

(From before, with the only changes being gender terms):

Suppose that, at some given instant "t2" after the acceleration begins

(where "t2" corresponds to the age of the person (she) on the trailing

rocket, who had just been born when the acceleration began), there is an

inertial person (he) who is momentarily stationary with that
accelerating person (her) on the trailing rocket ... call him inertial
person #2, with inertial person #1 being the inertial person (she) who

was stationary wrt the trailing rocket immediately before the
acceleration began.

(New edition from this point on):

The reason for introducing this particular inertial person #2 (he) (who
agrees with the trailing accelerating person (she)) is because inertial
observers can easily determine the separation of the spaceships BY
MEASUREMENTS (at least conceptually).

Inertial person #2 (he) agrees with the accelerating trailing person
(she) about the separation of the rockets at that instant "t2". I.e., at
that instant "t2", he agrees with her about the length of the long

measuring rod that exactly fills that separation. How does he determine

that distance empirically? To (eventually) determine that distance,
inertial person #2 (he) can enlist the help of all the other inertial
persons (all guys) with whom he is perpetually stationary, and who are

collectively spread throughout that spatial dimension. And inertial

person #1 (she) can do likewise, with her set of mutually-stationary
inertial persons (all gals).

What does inertial person #2 determine from his (and his helpers')
measurements? He finds that the separation of the spaceships decreases,
by the factor gamma, during the acceleration. He knows what the
separation was immediately before the acceleration began, by
communicating with inertial person #1 (her) about inertial person #1's
(and her helpers') measurements. Inertial person #1 (she) was able to
(eventually) determine the initial separation by enlisting the help of
other inertial persons (all gals) with whom she is perpetually stationary.

For example, if the velocity of the trailing rocket is v2 = 0.866c at
"t2", then the gamma factor is equal to 2.0, and thus the separation
between the two rockets at "t2" is half as much as it was immediately

before the acceleration started. And that conclusion (which was
determined experimentally) is exactly the same as that given by the

[Trevor Lange]

On Thursday, June 22, 2023 at 7:33:58 AM UTC-7, sep...@yahoo.com wrote:
> > Do the clocks at the tip of each of the two spaceships show an equal,
> > greater, or lesser elapsed time than the clocks at the bases when they

> > have zero velocity with respect to F0.
>

> Obviously the thruster clocks read 127.75 years, and the tip clocks
> read slightly more, because they are following trajectories of slightly

> less proper acceleration. Of course, for each clock we have

> dtau/dt = sqrt(1-v^2) where v is the speed of the clock in terms of
> the system x,t. Thus when each clock has v=0 we have dtau/dt = 1.
>

> I didn't mean to confuse people who respond.

The person who responded to your message wasn't confused. The person who *posted* your message was confused, and the person who responded untangled the confusion (you're welcome) and provided the anwer to your question (you're welcome).

Now that all your questions have been answered, and all your confusions have been untangled, if there is anything still unclear to you, go ahead and ask. I'll be happy to answer any questions, if you still have any.

[Tom Roberts]

On 6/22/23 11:25 AM, Mike Fontenot wrote:
> ANY two inertial people who are moving at a constant speed relative
> to each other MUST each conclude that the other's measuring rod is
> shorter than their own measuring rod.

As stated this is self-contradictory and absurd. You REALLY need to
learn how to state things more precisely. That failure is the source of
most, if not all, of your confusions.

In SR, two inertial people who are moving at a constant speed relative
to each other will MEASURE the other's rod as shorter than their own,
when using the standard technique for measuring a moving object (mark
both ends simultaneously in their inertial frame, then measure the
distance between marks); this only applies along the direction of their
relative motion.

Tom Roberts

[Maciej Wozniak]

On Friday, 23 June 2023 at 18:03:21 UTC+2, Tom Roberts wrote:
> On 6/22/23 11:25 AM, Mike Fontenot wrote:
> > ANY two inertial people who are moving at a constant speed relative
> > to each other MUST each conclude that the other's measuring rod is
> > shorter than their own measuring rod.
> As stated this is self-contradictory and absurd. You REALLY need to
> learn

that we're FORCED!!! To THE BEST WAY!!!

[Mike Fontenot]

On 6/23/23 10:03 AM, Tom Roberts wrote:
> On 6/22/23 11:25 AM, Mike Fontenot wrote:
>> ANY two inertial people who are moving at a constant speed relative to
>> each other MUST each conclude that the other's measuring rod is
>> shorter than their own measuring rod.
>
> As stated this is self-contradictory and absurd.

It's not self-contradictory at all ... it is one of the two most
important results of special relativity. Those two results are the
famous time dilation equation (TDE) and the famous length contraction
equation (LCE).

But you DID get it right in what you said immediately below:

>
> In SR, two inertial people who are moving at a constant speed relative
> to each other will MEASURE the other's rod as shorter than their own,
> when using the standard technique for measuring a moving object (mark
> both ends simultaneously in their inertial frame, then measure the
> distance between marks); this only applies along the direction of their
> relative motion.
>

You contradicted yourself, in the same posting.

What do you have to say about my latest posting (my posting immediately
below your current posting)? It starts with "Here's another way to look
at it: (Revised Version)".

[Mike Fontenot]

On 6/23/23 10:03 AM, Tom Roberts wrote:

> On 6/22/23 11:25 AM, Mike Fontenot wrote:
>> ANY two inertial people who are moving at a constant speed relative to
>> each other MUST each conclude that the other's measuring rod is
>> shorter than their own measuring rod.
>
> As stated this is self-contradictory and absurd.

Can you please elaborate? Why exactly do you think my above statement
is self-contradictory and absurd? Why do you think my statement is
different in any way from the well-known length contraction equation (LCE)?

[Trevor Lange]

On Friday, June 23, 2023 at 2:41:57 PM UTC-7, Mike Fontenot wrote:
> >> ANY two inertial people who are moving at a constant speed relative to
> >> each other MUST each conclude that the other's measuring rod is
> >> shorter than their own measuring rod.
> >

> Why exactly do you think my above statement is self-contradictory and absurd?

Relativity is not about what people (or mice or donkeys) "conclude" about things, let alone what they MUST conclude about things. The correct statement of what you are trying to say is: For any two identically constructed rods with rest lengths L and moving at relative speed v (after sufficiently mild accelerations of one or both, not exceeding elastic deformation limits), each rod's spatial length is L*sqrt(1-v^2) in terms of the standard inertial coordinates in which the other rod is at rest.

This is an objective statement of fact. Notice there is no mention of people any anyone being forced to "conclude" different things, etc. Relativity is not a subjectivist theory. Anyone in any state of motion can determine all the objective facts about the events in terms of any coordinate system they like, regardless of whether they are at rest in terms of that coordinate system.