Two perpetually inertial observers (IO1 and IO2), perpetually mutually
stationary with one another, are initially co-located with two separated
observers (AO1 and AO2), with separation "L". AO1 and AO2 are about to
begin a constant (according to them) acceleration "A" (with the
separation in the direction of their acceleration). AO1 and AO2 KNOW
that their acceleration is "A", because they each are carrying an
accelerometer that confirms it. IO1 and IO2 will conclude that AO1
and AO2 maintain the separation "L" during the accelerations. And AO1
and AO2 will agree with that: AO1 and AO2 conclude that their separation
remains constant at "L" during the acceleration. But two other inertial
observers, IO3 and IO4, who are momentarily co-located with AO1 and AO2
at any time later in the trip, will NOT agree that the separation "L" is
constant: they will say that it has increased since the start of the trip.

Each accelerometer directs its attached rocket to accelerate at exactly
"A" lightyears per year per year. NEWTONIAN physics would say that the
velocity of AO1 and AO2 would increase linearly with time, forever:

v = A * t. (incorrect)

But that means that "v" would go to infinity as "t" goes to infinity,
which we know can't be true in special relativity. So the above
equation is clearly wrong. Special relativity says the quantity "(A *
t)", which it calls the " rapidity" (denoted by the variable "theta"),
is related to the velocity "v" by

velocity = v = tanh(rapidity) = tanh(theta) = tanh(A * t).
(correct)

That says that with a constant acceleration "A", "v" approaches (but
never equals) the speed of light, "c", as "t" goes to infinity.

The distance "D" each rocket moves, according to AO1 and AO2, is

D = integral {from 0 to tau} [ v dt ] .

= integral {from 0 to tau} [ tanh(A*t) dt ] .

The integral of tanh(x) is equal to log[cosh(x)], so

D = log[cosh(tau)] - log[cosh(0)] .

D = log[cosh(tau)] .

So "D" grows forever, but it's RATE of growth decreases as tau increases.

The distance "D" each rocket moves during the acceleration is EXACTLY
the same, so the separation "L" between AO1 and AO2, according to THEM,
can't change during the acceleration.

The INERTIAL observers (IO1 and IO2) also conclude that the separation
"L" between AO1 and AO2 stays constant during the acceleration. But two
inertial observers (IO3 and IO4) who are momentarily co-located with AO1
and AO2 at some later instant in the trip will conclude that the
separation between AO1 and AO2 is LARGER than it was when the
acceleration started. And that larger separation continues to increase
as the trip progresses, according to the INERTIAL observers momentarily
co-located with AO1 and AO2 later in the trip.

So the accelerating observers (AO1 and AO2) say that their separation is
CONSTANT during their trip. The inertial observers (IO3 and IO4) say
that the separation of AO1 and AO2 INCREASES during the trip. Those two
groups of observers DISAGREE. That's just the way special relativity IS.

But it's normal in special relativity for an accelerating observer to
agree with the inertial observer who is momentarily co-located with him
at at some instant ... that's what the CMIF simultaneity method IS. The
inertial observer IO3 is momentarily co-located with AO1, and IO3 tells
AO1 that the separation between AO1 and AO2 is larger than it was when
the separation began. Does that contradict my above argument? No, it
doesn't, because the scenarios are themselves different: the actual
accelerations are slightly different. How do the people producing the
scenario with all the inertial observers achieve the acceleration "A"?
It's based on the CALCULATIONS by the inertial people: they measure
positions of AO1 and AO2 versus the time on their own watches, and
COMPUTE the acceleration. It is NOT based on any accelerometer, and it
differs from what AO1 and AO2 read on their accelerometers.

On Monday, March 6, 2023 at 9:25:30 AM UTC-8, Mike Fontenot wrote:
> How do the people producing the
> scenario with all the inertial observers achieve the acceleration "A"?
> It's based on the CALCULATIONS by the inertial people: they measure
> positions of AO1 and AO2 versus the time on their own watches, and
> COMPUTE the acceleration. It is NOT based on any accelerometer, and it
> differs from what AO1 and AO2 read on their accelerometers.

The above proves once more (as if it were necessary) that you are an ignorant crank who doesn't know the difference between coordinate acceleration and proper acceleration.

On Monday, March 6, 2023 at 6:25:30 PM UTC+1, Mike Fontenot wrote:
> Two perpetually inertial observers (IO1 and IO2), perpetually mutually
> stationary with one another, are initially co-located with two separated
> observers (AO1 and AO2), with separation "L". AO1 and AO2 are about to
> begin a constant (according to them) acceleration "A" (with the
> separation in the direction of their acceleration). AO1 and AO2 KNOW
> that their acceleration is "A", because they each are carrying an
> accelerometer that confirms it. IO1 and IO2 will conclude that AO1
> and AO2 maintain the separation "L" during the accelerations. And AO1
> and AO2 will agree with that: AO1 and AO2 conclude that their separation
> remains constant at "L" during the acceleration.

No. They will discover, instead, that if they started their journey with a
rope pulled taut between them, that rope will start stretching (developing
an internal strain) and if AO1 and AO2 want to prevent the rope from
breaking, they will have to add more and more length or that rope
(keeping it taut throughout). Assuming one end of the rope is on
a spool, that spool will start unwinding until all the rope is gone from it..

The above is based on calculations done in the accelerated (curvilinear)
coordinates (not inertial). We may have to go through the computational
details for this, we'll see.

> But two other inertial
> observers, IO3 and IO4, who are momentarily co-located with AO1 and AO2
> at any time later in the trip, will NOT agree that the separation "L" is
> constant: they will say that it has increased since the start of the trip..

AO1, AO2 and IO3, IO4 agree the rope will behave as above. The relevant
momentarily co-moving inertial coordinates produce the same
curves representing the rope.

IO1, IO2 will OTOH see the L separation constant with the rope
starting to unwind from the spool due to its Lorentz contraction.

--
Jan

On 2023-03-06 17:25:27 +0000, Mike Fontenot said:

> The distance "D" each rocket moves during the acceleration is EXACTLY
> the same, so the separation "L" between AO1 and AO2, according to THEM,
> can't change during the acceleration.

The two observers AO1 and AO2 can measure the distance between them
and the change of the distance with a simple radar: one of them sends
a radar pulse and measures the duration to the arrival of the refloection
from the other observer. The other observer can do the same. Both will
observe that the distance increases after the start of their motion.

Mikko

Here's an update to by above post:

Two perpetually inertial observers (IO1 and IO2), perpetually mutually
stationary with one another, are initially co-located with two separated
observers (AO1 and AO2), with separation "L". AO1 and AO2 are about to
begin a constant (according to them) acceleration "A" (with the
separation in the direction of their acceleration). AO1 and AO2 KNOW
that their acceleration is "A", because they each are carrying an
accelerometer that confirms it. IO1 and IO2 will conclude that AO1 and
AO2 maintain the separation "L" during the accelerations. And AO1 and
AO2 will agree with that: AO1 and AO2 conclude that their separation
remains constant at "L" during the acceleration. But two other inertial
observers, IO3 and IO4, who are momentarily co-located with AO1 and AO2
at any time later in the trip, will NOT agree that the separation "L" is
constant: they will say that it has increased since the start of the trip.

NOTE: My use of the phrase "Proper Separation" in the title of this
submission means that it is the separation of the two people undergoing
the acceleration, ACCORDING TO THOSE TWO PEOPLE THEMSELVES. And the
REFERENCE FRAME of an accelerating observer, say AO1's reference frame,
is constructed in the same manner as an inertial observer constructs his
reference frame (as Einstein explained to us). AO1's helpers just lay
out yardsticks, end-to-end, in the direction of the acceleration. To
keep the yardsticks in place, they each are attached to an
accelerometer-controlled rocket, so they each are accelerating at "A"
lightyears/year/year. And between every pair of yardsticks, there is a
clock. The clocks were initially synchronized before the the constant
acceleration started. Once the acceleration starts, the clocks don't
remain synchronized ... clocks located farther in the direction of the
acceleration tic faster, by an amount originally given incorrectly by
Einstein (his exponential equation), but which has now been corrected.
(END NOTE)

Each accelerometer directs its attached rocket to accelerate at exactly
"A" lightyears per year per year. NEWTONIAN physics would say that the
velocity of AO1 and AO2 would increase linearly with time, forever:

v = A * t. (incorrect)

But that means that "v" would go to infinity as "t" goes to infinity,
which we know can't be true in special relativity. So the above
equation is clearly wrong. Special relativity says the quantity "(A *
t)", which it calls the " rapidity" (denoted by the variable "theta"),
is related to the velocity "v" by

velocity = v = tanh(rapidity) = tanh(theta) = tanh(A * t).
(correct)

That says that with a constant acceleration "A", "v" approaches (but
never equals) the speed of light, "c", as "t" goes to infinity.

The distance "D" each rocket moves, according to AO1 and AO2, is

D = integral {from 0 to tau} [ v dt ] .

= integral {from 0 to tau} [ tanh(A*t) dt ] .

The integral of tanh(x) is equal to log[cosh(x)], so

D = log[cosh(tau)] - log[cosh(0)] .

D = log[cosh(tau)] .

So "D" grows forever, but it's RATE of growth decreases as tau increases.

The distance "D" each rocket moves during the acceleration is EXACTLY
the same, so the separation "L" between AO1 and AO2, according to THEM,
can't change during the acceleration.

The INERTIAL observers (IO1 and IO2) also conclude that the separation
"L" between AO1 and AO2 stays constant during the acceleration. But two
inertial observers (IO3 and IO4) who are momentarily co-located with AO1
and AO2 at some later instant in the trip will conclude that the
separation between AO1 and AO2 is LARGER than it was when the
acceleration started. And that larger separation continues to increase
as the trip progresses, according to the INERTIAL observers momentarily
co-located with AO1 and AO2 later in the trip.

So the accelerating observers (AO1 and AO2) say that their separation is
CONSTANT during their trip. The inertial observers (IO3 and IO4) say
that the separation of AO1 and AO2 INCREASES during the trip. Those two
groups of observers DISAGREE. That's just the way special relativity IS.

But it's normal in special relativity for an accelerating observer to
agree with the inertial observer who is momentarily co-located with him
at at some instant ... that's what the CMIF simultaneity method IS. The
inertial observer IO3 is momentarily co-located with AO1, and IO3 tells
AO1 that the separation between AO1 and AO2 is larger than it was when
the separation began. Does that contradict my above argument? No, it
doesn't, because the scenarios are themselves different: the actual
accelerations are slightly different. How do the people producing the
scenario with all the inertial observers achieve the acceleration "A"?
It's based on the CALCULATIONS by the inertial people: they measure
positions of AO1 and AO2 versus the time on their own watches, and
COMPUTE the acceleration. It is NOT based on any accelerometer, and it
differs from what AO1 and AO2 read on their accelerometers.

Note that the above paragraph is relevant to the well-known (and much
misunderstood) Bell's Spaceship Paradox:

 https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox

(read the WHOLE thing). Does the string break or not? The answer
depends on how the acceleration is measured. If the acceleration is
determined by calculations and measurements made by various inertial
observers, the string WILL break. But if the acceleration is what
accelerometers attached to the spaceships display, the string will NOT
break.

On Tuesday, March 7, 2023 at 12:13:27 AM UTC+1, Mike Fontenot wrote:
>
> Note that the above paragraph is relevant to the well-known (and much
> misunderstood) Bell's Spaceship Paradox:
>
> https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox
>
> (read the WHOLE thing). Does the string break or not? The answer
> depends on how the acceleration is measured. If the acceleration is
> determined by calculations and measurements made by various inertial
> observers, the string WILL break. But if the acceleration is what
> accelerometers attached to the spaceships display, the string will NOT
> break.

Something is wrong here. The string breaking or not cannot
depend on what measurements are being made. It either breaks or it
doesn't, according to all observers, accelerated or not.

Any standard relativistic analysis shows that it will break(*), although
different observers will assign the cause differently, due to the
kinematic effects. It's a bit like different observers assigning
the E and B fields differently. But in all cases the actual physical
results are always predicted consistently in the end.

(*)Wikipedia quotes some people who should have known better.

--
Jan

Den 06.03.2023 18:25, skrev Mike Fontenot:
>
> Two perpetually inertial observers (IO1 and IO2), perpetually mutually
> stationary with one another, are initially co-located with two separated
> observers (AO1 and AO2), with separation "L".  AO1 and AO2  are about to
> begin a constant (according to them) acceleration "A" (with the
> separation in the direction of their acceleration).  AO1 and AO2 KNOW
> that their acceleration is "A", because they each are carrying an
> accelerometer that confirms it.    IO1 and IO2 will conclude that AO1
> and AO2 maintain the separation "L" during the accelerations.  And AO1
> and AO2 will agree with that: AO1 and AO2 conclude that their separation
> remains constant at "L" during the acceleration.  But two other inertial
> observers, IO3 and IO4, who are momentarily co-located with AO1 and AO2
> at any time later in the trip, will NOT agree that the separation "L" is
> constant: they will say that it has increased since the start of the trip.
>
> Each accelerometer directs its attached rocket to accelerate at exactly
> "A" lightyears per year per year.  NEWTONIAN physics would say that the
> velocity of AO1 and AO2 would increase linearly with time, forever:
>
>   v  =  A * t.    (incorrect)
>
> But that means that "v" would go to infinity as "t" goes to infinity,
> which we know can't be true in special relativity.  So the above
> equation is clearly wrong.  Special relativity says the quantity "(A *
> t)", which it calls the " rapidity" (denoted by the variable "theta"),
> is related to the velocity "v" by
>
>   velocity  =  v  =  tanh(rapidity)  =  tanh(theta)  =  tanh(A * t).
>  (correct)
>
> That says that with a constant acceleration "A", "v" approaches (but
> never equals) the speed of light, "c", as "t" goes to infinity.
>
> The distance "D" each rocket moves, according to AO1 and AO2, is
>
>   D  =  integral {from 0 to tau} [ v dt ] .
>
>        =  integral {from 0 to tau} [ tanh(A*t) dt ] .
>
> The integral of tanh(x) is equal to log[cosh(x)], so
>
>   D   =  log[cosh(tau)]  -  log[cosh(0)] .
>
>   D   =  log[cosh(tau)] .
>
> So "D" grows forever, but it's RATE of growth decreases as tau increases.
>
> The distance "D" each rocket moves during the acceleration is EXACTLY
> the same, so the separation "L" between AO1 and AO2, according to THEM,
> can't change during the acceleration.
>
> The INERTIAL observers (IO1 and IO2) also conclude that the separation
> "L" between AO1 and AO2 stays constant during the acceleration.  But two
> inertial observers (IO3 and IO4) who are momentarily co-located with AO1
> and AO2 at some later instant in the trip will conclude that the
> separation between AO1 and AO2 is LARGER than it was when the
> acceleration started.  And that larger separation continues to increase
> as the trip progresses, according to the INERTIAL observers momentarily
> co-located with AO1 and AO2 later in the trip.
>
> So the accelerating observers (AO1 and AO2) say that their separation is
> CONSTANT during their trip.  The inertial observers (IO3 and IO4) say
> that the separation of AO1 and AO2 INCREASES during the trip.  Those two
> groups of observers DISAGREE.  That's just the way special relativity IS.
>
> But it's normal in special relativity for an accelerating observer to
> agree with the inertial observer who is momentarily co-located with him
> at at some instant ... that's what the CMIF simultaneity method IS.  The
> inertial observer IO3 is momentarily co-located with AO1, and IO3 tells
> AO1 that the separation between AO1 and AO2 is larger than it was when
> the separation began.  Does that contradict my above argument?  No, it
> doesn't, because the scenarios are themselves different: the actual
> accelerations are slightly different.  How do the people producing the
> scenario with all the inertial observers achieve the acceleration "A"?
> It's based on the CALCULATIONS by the inertial people: they measure
> positions of AO1 and AO2 versus the time on their own watches, and
> COMPUTE the acceleration.  It is NOT based on any accelerometer, and it
> differs from what AO1 and AO2 read on their accelerometers.
>

This is quite simple.

Two 'observers' O1 and O2 are stationary in an inertial frame K(t,x).
The distance between them is L.
A the time t = 0 O1 and O2 simultaneously start their
rockets which give them a proper acceleration A.
Measured in K, their acceleration will obviously diminish
as their speeds in K increases, but their accelerations
and thus their speeds will always be equal measured in K.
That means that the distance between them measured in K will remain L.

At the time t their speeds in K are:
v₁ = v₂ = At/√(1+(At/c)²) (1)

Se equation (38) in:
https://paulba.no/pdf/TwinsByMetric.pdf

This means that at the time t in K, O1 and O2 will
both be instantly at rest in an inertial frame K'
which is moving with the speed v₁ = v₂ relative to K.

Note that the distance between O1 and O2 always must be
measured in their momentarily co-moving inertial frame.

It is then a trivial task to transform L from K to K'.
The distance between them will be L' = γ⋅L.

Since γ(t) = √(1+(At/c)²) we can set L'(t) = √(1+(At/c)²)⋅L

--
Paul

https://paulba.no/

On Tuesday, 7 March 2023 at 13:45:52 UTC+1, Paul B. Andersen wrote:
> Den 06.03.2023 18:25, skrev Mike Fontenot:
> >
> > Two perpetually inertial observers (IO1 and IO2), perpetually mutually
> > stationary with one another, are initially co-located with two separated
> > observers (AO1 and AO2), with separation "L". AO1 and AO2 are about to
> > begin a constant (according to them) acceleration "A" (with the
> > separation in the direction of their acceleration). AO1 and AO2 KNOW
> > that their acceleration is "A", because they each are carrying an
> > accelerometer that confirms it. IO1 and IO2 will conclude that AO1
> > and AO2 maintain the separation "L" during the accelerations. And AO1
> > and AO2 will agree with that: AO1 and AO2 conclude that their separation
> > remains constant at "L" during the acceleration. But two other inertial
> > observers, IO3 and IO4, who are momentarily co-located with AO1 and AO2
> > at any time later in the trip, will NOT agree that the separation "L" is
> > constant: they will say that it has increased since the start of the trip.
> >
> > Each accelerometer directs its attached rocket to accelerate at exactly
> > "A" lightyears per year per year. NEWTONIAN physics would say that the
> > velocity of AO1 and AO2 would increase linearly with time, forever:
> >
> > v = A * t. (incorrect)
> >
> > But that means that "v" would go to infinity as "t" goes to infinity,
> > which we know can't be true in special relativity. So the above
> > equation is clearly wrong. Special relativity says the quantity "(A *
> > t)", which it calls the " rapidity" (denoted by the variable "theta"),
> > is related to the velocity "v" by
> >
> > velocity = v = tanh(rapidity) = tanh(theta) = tanh(A * t).
> > (correct)
> >
> > That says that with a constant acceleration "A", "v" approaches (but
> > never equals) the speed of light, "c", as "t" goes to infinity.
> >
> > The distance "D" each rocket moves, according to AO1 and AO2, is
> >
> > D = integral {from 0 to tau} [ v dt ] .
> >
> > = integral {from 0 to tau} [ tanh(A*t) dt ] .
> >
> > The integral of tanh(x) is equal to log[cosh(x)], so
> >
> > D = log[cosh(tau)] - log[cosh(0)] .
> >
> > D = log[cosh(tau)] .
> >
> > So "D" grows forever, but it's RATE of growth decreases as tau increases.
> >
> > The distance "D" each rocket moves during the acceleration is EXACTLY
> > the same, so the separation "L" between AO1 and AO2, according to THEM,
> > can't change during the acceleration.
> >
> > The INERTIAL observers (IO1 and IO2) also conclude that the separation
> > "L" between AO1 and AO2 stays constant during the acceleration. But two
> > inertial observers (IO3 and IO4) who are momentarily co-located with AO1
> > and AO2 at some later instant in the trip will conclude that the
> > separation between AO1 and AO2 is LARGER than it was when the
> > acceleration started. And that larger separation continues to increase
> > as the trip progresses, according to the INERTIAL observers momentarily
> > co-located with AO1 and AO2 later in the trip.
> >
> > So the accelerating observers (AO1 and AO2) say that their separation is
> > CONSTANT during their trip. The inertial observers (IO3 and IO4) say
> > that the separation of AO1 and AO2 INCREASES during the trip. Those two
> > groups of observers DISAGREE. That's just the way special relativity IS.
> >
> > But it's normal in special relativity for an accelerating observer to
> > agree with the inertial observer who is momentarily co-located with him
> > at at some instant ... that's what the CMIF simultaneity method IS. The
> > inertial observer IO3 is momentarily co-located with AO1, and IO3 tells
> > AO1 that the separation between AO1 and AO2 is larger than it was when
> > the separation began. Does that contradict my above argument? No, it
> > doesn't, because the scenarios are themselves different: the actual
> > accelerations are slightly different. How do the people producing the
> > scenario with all the inertial observers achieve the acceleration "A"?
> > It's based on the CALCULATIONS by the inertial people: they measure
> > positions of AO1 and AO2 versus the time on their own watches, and
> > COMPUTE the acceleration. It is NOT based on any accelerometer, and it
> > differs from what AO1 and AO2 read on their accelerometers.
> >
> This is quite simple.
>
> Two 'observers' O1 and O2 are stationary in an inertial frame K(t,x).
> The distance between them is L.

And in the meantime in the real world, forbidden
by your bunch of idiots improper clocks keep
measuring forbidden by your bunch of idiots
improper t'=t in forbidden by your bunch of idiots
old seconds.

Le 06/03/2023 à 19:33, Mikko a écrit :
>
>> The distance "D" each rocket moves during the acceleration is EXACTLY
>> the same, so the separation "L" between AO1 and AO2, according to THEM,
>> can't change during the acceleration.
>
> The two observers AO1 and AO2 can measure the distance between them
> and the change of the distance with a simple radar: one of them sends
> a radar pulse and measures the duration to the arrival of the refloection
> from the other observer. The other observer can do the same. Both will
> observe that the distance increases after the start of their motion.

To fully understand the theory of relativity, you have to go slowly,
because it is full of little traps.

It is necessary to use the method advocated by Sir Paul B.Andersen which
is "the method of small baby steps".

If we place ourselves in the initial reference frame and we observe A and
B accelerating, the distance L is invariant for a tele-transverse
observer.

This is a certainty.

Then you have to go very slowly.

In which inertial reference frame am I? Where am I in this?

R.H

--
"Mais ne nous trompons pas :
il n'y a pas de violence qu'avec des armes : il y a des situations de
violence."
Abbé Pierre
Δ₀₁₂"<http://news2.nemoweb.net/?DataID=Bxl8OcBglfzrNDsGTRi__7RxsGY@jntp>

On Tuesday, March 7, 2023 at 12:13:27 AM UTC+1, Mike Fontenot wrote:
> Here's an update to by above post:
> Two perpetually inertial observers (IO1 and IO2), perpetually mutually
> stationary with one another, are initially co-located with two separated
> observers (AO1 and AO2), with separation "L". AO1 and AO2 are about to
> begin a constant (according to them) acceleration "A" (with the
> separation in the direction of their acceleration). AO1 and AO2 KNOW
> that their acceleration is "A", because they each are carrying an
> accelerometer that confirms it. IO1 and IO2 will conclude that AO1 and
> AO2 maintain the separation "L" during the accelerations.

Yes.

> And AO1 and
> AO2 will agree with that:

No. See below.

> AO1 and AO2 conclude that their separation
> remains constant at "L" during the acceleration.

No. I suggested one simple way to demonstrate this experimentally:
have a rope pulled taut between the spaceships. Or better yet: a
rubber band. And see what happens. And whatever happens to the
rubber band, all observers must necessarily agree will happen to it.
Namely: *all* observers will see the rubber band beginning to
experienced strain (this can be ascertained by strain sensors embedded
in it, the readings of which will necessarily be agreed upon by all
the observers).

I think I can prove it without referring to any coordinate simultaneity spaces
(which I know you don't trust in this context).

This method involves calculating the metric of what's sometimes
called "relative space" which is a 3D *spatial metric* (no time)
representing the behaviour or real measuring implements like
sticks, ropes, rubber bands, etc. One classic example of this
is the metric on a rotating disc which happens to be hyperbolic.
The "relative space" metric is obtained by stitching together
infinitesimal pieces of 3D "space elements" that are at each
event Lorentz-orthogonal to the worldlines of the material
particles representing the aforementioned measuring implements.
The resulting 3D surface is NOT necessarily embeddable in the
spacetime under consideration (as in the rotating disc case:
a hyperbolic 3D manifold cannot be embedded in a 4D flat
Minkowski space). (I keep saying "3D" for generality, of course
in the rotating disc case 2D is sufficient (suppressing the
z-coordinate), in the case you are inquiring about only
1D metric is needed (suppressing the y- and z-coordinates)).

The fact that "relative space" is not necessarily an isometric
subset of the given spacetime is the source of endless
confusion BTW, viz. the endless discussions like this
regarding accelerated observers, Sagnac, etc.
It is, instead, a topological 3D quotient of the given
4D spacetime by the relevant material particles worldlines
and then the metric is quilted together as described above.

I need a bit of time to write it down but the result is
that the relevant 3D relative space for your scenario
has the following metric (the rubber band is along
the x-direction, the y- and z- are irrelevant):

ds^2 = (1+t^2/C) dx^2 + dy^2 + dz^2

....where C is a constant depending on the initial
conditions. The t here is to be understood not as
a coordinate but a time-dependent parameter. This
means that the rubber band of length L initially
has the following length at t:

L' = L * sqrt(1+t^2/C)

....so it increases. Note that no mention is made here of
any particular observer, it's all based on assuming the
material points of the rubber band follow both rockets
proportionally.

> But two other inertial
> observers, IO3 and IO4, who are momentarily co-located with AO1 and AO2
> at any time later in the trip, will NOT agree that the separation "L" is
> constant: they will say that it has increased since the start of the trip..

Again, both IO3 and AO1 will agree, and so will IO4 and AO2.

--
Jan

Le 06/03/2023 à 18:29, "Dono." a écrit :

> the difference between coordinate acceleration and proper acceleration.

What is this?

R.H.

--
"Mais ne nous trompons pas :
il n'y a pas de violence qu'avec des armes : il y a des situations de
violence."
Abbé Pierre
Δ₀₁₂"<http://news2.nemoweb.net/?DataID=GJg9HfPhTLklwBCRKFRseR3ZUBM@jntp>

Le 06/03/2023 à 18:47, JanPB a écrit :
> No. They will discover, instead, that if they started their journey with a
> rope pulled taut between them, that rope will start stretching (developing
> an internal strain) and if AO1 and AO2 want to prevent the rope from
> breaking, they will have to add more and more length or that rope
> (keeping it taut throughout). Assuming one end of the rope is on
> a spool, that spool will start unwinding until all the rope is gone from it.

But no! why do you want the rope to stretch between A and B?

R.H.

--
"Mais ne nous trompons pas :
il n'y a pas de violence qu'avec des armes : il y a des situations de
violence."
Abbé Pierre
Δ₀₁₂"<http://news2.nemoweb.net/?DataID=CufYR1vmKESuVjXusY56si-UmOM@jntp>

Le 07/03/2023 à 00:13, Mike Fontenot a écrit :
> Two perpetually inertial observers (IO1 and IO2), perpetually mutually
> stationary with one another, are initially co-located with two separated
> observers (AO1 and AO2), with separation "L". AO1 and AO2 are about to
> begin a constant (according to them) acceleration "A" (with the
> separation in the direction of their acceleration). AO1 and AO2 KNOW
> that their acceleration is "A", because they each are carrying an
> accelerometer that confirms it. IO1 and IO2 will conclude that AO1 and
> AO2 maintain the separation "L" during the accelerations. And AO1 and
> AO2 will agree with that: AO1 and AO2 conclude that their separation
> remains constant at "L" during the acceleration. But two other inertial
> observers, IO3 and IO4, who are momentarily co-located with AO1 and AO2
> at any time later in the trip, will NOT agree that the separation "L" is
> constant: they will say that it has increased since the start of the trip.

Let's do it, I beg of you, I who am one of the best theorists in the world
concerning the kinetics of the theory of relativity of small baby steps.

Let's start with the beginning.

We place ourselves first, because it's simple, in the initial reference
frame of rest.

We then place ourselves very far and transversely, so that the starting
signal that I will send is simultaneous in A and B FOR ME.

I will notice, since they have strictly the same acceleration a,
that the distance AB, ie L, is an invariant throughout the duration of the
acceleration, and even afterwards.
A and B go from left to right and A will pass at a time when B was.

We all agree on this point.

Now let's go to A.

When I get the signal, the acceleration starts.

BUT, at this instant, FOR ME who am in A, the event-arrival of the signal
in B, does not exist.

It's not that it hasn't happened yet, it's that it doesn't exist.

It is wrong to think that he just existed there, but that I would only be
aware of it later.

It is this idea of ​​the isochronous world that I have been
criticizing for decades.

What will happen then? I will have to wait for a time T=AB/c (whatever the
value of the acceleration, it is independent), so that the start-event B
exists.

t(start B//A)=BA/c

at this moment finally B leaves for A (and whatever the acceleration a,
the point B, for A has not varied, it is always at the distance L.

Suddenly he starts to leave, but he does not leave with the speed of A
which is mine at the moment!

It starts with a low initial speed.

I will therefore observe a distance L which will very slowly shorten from
time t=AB/c
and which will shorten indefinitely and more and more according to the
acceleration.

I'll let you do the math.

Now let's take point B.

For A, B begins to move at time t=BA/c and ATTENTION: B sees A moving away
slowly, then faster and faster!

These effects are very difficult to conceptualize, and can seem very
confusing.

For O: L is constant.

For A: L becomes smaller and smaller.

For B: L becomes larger and larger.

This is why I have always said that it is not because my vision of RR is
the most beautiful, the most true, and the most obvious once understood
that everyone will understand it easily.

R.H.

--
"Mais ne nous trompons pas :
il n'y a pas de violence qu'avec des armes : il y a des situations de
violence."
Abbé Pierre
Δ₀₁₂"<http://news2.nemoweb.net/?DataID=IWCQvpp40CAwtsGKm7rmpt7D8rQ@jntp>

Le 06/03/2023 à 19:33, Mikko a écrit :
> On 2023-03-06 17:25:27 +0000, Mike Fontenot said:
>
>> The distance "D" each rocket moves during the acceleration is EXACTLY
>> the same, so the separation "L" between AO1 and AO2, according to THEM,
>> can't change during the acceleration.
>
> The two observers AO1 and AO2 can measure the distance between them
> and the change of the distance with a simple radar:

Yes.

Obviolouly.
> Mikko

R.H.

--
"Mais ne nous trompons pas :
il n'y a pas de violence qu'avec des armes : il y a des situations de
violence."
Abbé Pierre
Δ₀₁₂"<http://news2.nemoweb.net/?DataID=xao0f1TASlstFx_gjG1EaV0l5C8@jntp>

Le 07/03/2023 à 00:30, JanPB a écrit :
> On Tuesday, March 7, 2023 at 12:13:27 AM UTC+1, Mike Fontenot wrote:
>>
>> Note that the above paragraph is relevant to the well-known (and much
>> misunderstood) Bell's Spaceship Paradox:
>>
>> https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox
>>
>> (read the WHOLE thing). Does the string break or not? The answer
>> depends on how the acceleration is measured. If the acceleration is
>> determined by calculations and measurements made by various inertial
>> observers, the string WILL break. But if the acceleration is what
>> accelerometers attached to the spaceships display, the string will NOT
>> break.
>
> Something is wrong here. The string breaking or not cannot
> depend on what measurements are being made. It either breaks or it
> doesn't, according to all observers, accelerated or not.
>
> Any standard relativistic analysis shows that it will break(*), although
> different observers will assign the cause differently, due to the
> kinematic effects. It's a bit like different observers assigning
> the E and B fields differently. But in all cases the actual physical
> results are always predicted consistently in the end.
>
> (*)Wikipedia quotes some people who should have known better.
>
> --
> Jan

The rope will not break.

It will appear invariant for O (placed far and transversely), it will
appear smaller for A and larger for B (the effects only beginning to occur
after a time t=AB/c

These are relativistic effects similar to aberrations.

The rope will remain invariable in itself in the "neutral" frame of
reference if we look at things transversely. It does not break.

R.H.

--
"Mais ne nous trompons pas :
il n'y a pas de violence qu'avec des armes : il y a des situations de
violence."
Abbé Pierre
Δ₀₁₂"<http://news2.nemoweb.net/?DataID=zODxfjIm8MaM-SVR6Y5xdXQAO7o@jntp>

Clocks tick away at any distance.
It does not change time.

On Tuesday, March 7, 2023 at 10:41:58 PM UTC+1, Richard Hachel wrote:
> Le 06/03/2023 à 18:47, JanPB a écrit :
> > No. They will discover, instead, that if they started their journey with a
> > rope pulled taut between them, that rope will start stretching (developing
> > an internal strain) and if AO1 and AO2 want to prevent the rope from
> > breaking, they will have to add more and more length or that rope
> > (keeping it taut throughout). Assuming one end of the rope is on
> > a spool, that spool will start unwinding until all the rope is gone from it.
>
> But no! why do you want the rope to stretch between A and B?

Mike says the spaceship (accelerating) observers won't measure the
spaceship separation as increasing. So I proposed to simply get
a length of rope or runner band between them and see what happens.

--
Jan

Le 08/03/2023 à 00:13, JanPB a écrit :
>
> Mike says the spaceship (accelerating) observers won't measure the
> spaceship separation as increasing. So I proposed to simply get
> a length of rope or runner band between them and see what happens.

Exactly.

It is necessary to make experiences of thoughts of this type, and to see
for example, if absurdities do not appear.

I highly recommend these thought experiments, and I always find them very
interesting.

R.H.

--
"Mais ne nous trompons pas :
il n'y a pas de violence qu'avec des armes : il y a des situations de
violence."
Abbé Pierre
Δ₀₁₂"<http://news2.nemoweb.net/?DataID=7j-YZXcRhOuITSJ68H60s3f_A1w@jntp>

Le 08/03/2023 à 00:13, JanPB a écrit :
>
> Mike says the spaceship (accelerating) observers won't measure the
> spaceship separation as increasing. So I proposed to simply get
> a length of rope or runner band between them and see what happens.

N.B. Si à la place de la corde, on met une très longue fusée qui va de
A à B, la fusée ne cassera pas si on l'accélère.
C'est une expérience de pensée, mais on voit bien que si la grande
fusée unique ne casse pas, pourquoi la corde (rigide ou pas) casserait?

R.H.

--
"Mais ne nous trompons pas :
il n'y a pas de violence qu'avec des armes : il y a des situations de
violence."
Abbé Pierre
Δ₀₁₂"<http://news2.nemoweb.net/?DataID=nhVkwQgbFAol75d9y0Uhkm4FlM8@jntp>

On Tuesday, March 7, 2023 at 11:27:48 PM UTC+1, Richard Hachel wrote:
> Le 07/03/2023 à 00:30, JanPB a écrit :
> > On Tuesday, March 7, 2023 at 12:13:27 AM UTC+1, Mike Fontenot wrote:
> >>
> >> Note that the above paragraph is relevant to the well-known (and much
> >> misunderstood) Bell's Spaceship Paradox:
> >>
> >> https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox
> >>
> >> (read the WHOLE thing). Does the string break or not? The answer
> >> depends on how the acceleration is measured. If the acceleration is
> >> determined by calculations and measurements made by various inertial
> >> observers, the string WILL break. But if the acceleration is what
> >> accelerometers attached to the spaceships display, the string will NOT
> >> break.
> >
> > Something is wrong here. The string breaking or not cannot
> > depend on what measurements are being made. It either breaks or it
> > doesn't, according to all observers, accelerated or not.
> >
> > Any standard relativistic analysis shows that it will break(*), although
> > different observers will assign the cause differently, due to the
> > kinematic effects. It's a bit like different observers assigning
> > the E and B fields differently. But in all cases the actual physical
> > results are always predicted consistently in the end.
> >
> > (*)Wikipedia quotes some people who should have known better.
> >
> > --
> > Jan
> The rope will not break.
>
> It will appear invariant for O (placed far and transversely), it will
> appear smaller for A and larger for B (the effects only beginning to occur
> after a time t=AB/c
>
> These are relativistic effects similar to aberrations.
>
> The rope will remain invariable in itself in the "neutral" frame of
> reference if we look at things transversely. It does not break.
> R.H.

No, according to relativity theory it's incorrect.

--
Jan

On Wednesday, March 8, 2023 at 12:21:30 AM UTC+1, Richard Hachel wrote:
> Le 08/03/2023 à 00:13, JanPB a écrit :
> >
> > Mike says the spaceship (accelerating) observers won't measure the
> > spaceship separation as increasing. So I proposed to simply get
> > a length of rope or runner band between them and see what happens.
> Exactly.
>
> It is necessary to make experiences of thoughts of this type, and to see
> for example, if absurdities do not appear.
>
> I highly recommend these thought experiments, and I always find them very
> interesting.

No absurdity appears this way and the conclusion is clear: the
rope will break / the rubber band will experience strain.

--
Jan

On Tuesday, 7 March 2023 at 23:21:02 UTC+1, Richard Hachel wrote:
> Le 07/03/2023 à 00:13, Mike Fontenot a écrit :
> > Two perpetually inertial observers (IO1 and IO2), perpetually mutually
> > stationary with one another, are initially co-located with two separated
> > observers (AO1 and AO2), with separation "L". AO1 and AO2 are about to
> > begin a constant (according to them) acceleration "A" (with the
> > separation in the direction of their acceleration). AO1 and AO2 KNOW
> > that their acceleration is "A", because they each are carrying an
> > accelerometer that confirms it. IO1 and IO2 will conclude that AO1 and
> > AO2 maintain the separation "L" during the accelerations. And AO1 and
> > AO2 will agree with that: AO1 and AO2 conclude that their separation
> > remains constant at "L" during the acceleration. But two other inertial
> > observers, IO3 and IO4, who are momentarily co-located with AO1 and AO2
> > at any time later in the trip, will NOT agree that the separation "L" is
> > constant: they will say that it has increased since the start of the trip.
> Let's do it, I beg of you, I who am one of the best theorists in the world
> concerning the kinetics of the theory of relativity of small baby steps.
>
> Let's start with the beginning.
>
> We place ourselves first, because it's simple, in the initial reference
> frame of rest.
>
> We then place ourselves very far and transversely, so that the starting
> signal that I will send is simultaneous in A and B FOR ME.
>
> I will notice, since they have strictly the same acceleration a,
> that the distance AB, ie L, is an invariant throughout the duration of the
> acceleration, and even afterwards.
> A and B go from left to right and A will pass at a time when B was.

And in the meantime in the real world, forbidden by
your bunch of idiots improper clocks keep measuring
t'=t in fotrbidden by your bunch of idiots improper
old seconds.

On 3/6/2023 6:30 PM, JanPB wrote:

> Any standard relativistic analysis shows that it will break(*), although
> different observers will assign the cause differently, due to the
> kinematic effects. It's a bit like different observers assigning
> the E and B fields differently. But in all cases the actual physical
> results are always predicted consistently in the end.

I always thought that's cool about relativity. All observers agree that
the same outcome occurs, but for different reasons. For example, both
someone on earth and an observer traveling with a cosmic muon both agree
it reaches the ground, but the ground observer sees a long lifetime but
the comoving observer sees the muon lasting its usual 2.2µS lifetime but
traveling only 600 meters through the atmosphere of a length contracted
earth. Same with the string breaking, electric/magnetic fields, the
bug/rivet "paradox" etc.

On 3/7/2023 5:21 PM, Richard Hachel wrote:

[snip]

> This is why I have always said that it is not because my vision of RR is
> the most beautiful, the most true, and the most obvious once understood
> that everyone will understand it easily.
>
"It doesn't matter how beautiful your theory is, it doesn't matter
how smart you are. If it doesn't agree with experiment, it's wrong."

-Richard P. Feynman

On Wednesday, 8 March 2023 at 07:56:13 UTC+1, Volney wrote:
> On 3/6/2023 6:30 PM, JanPB wrote:
>
> > Any standard relativistic analysis shows that it will break(*), although
> > different observers will assign the cause differently, due to the
> > kinematic effects. It's a bit like different observers assigning
> > the E and B fields differently. But in all cases the actual physical
> > results are always predicted consistently in the end.
> I always thought that's cool about relativity. All observers agree that
> the same outcome occurs, but for different reasons. For example, both
> someone on earth and an observer traveling with a cosmic muon both agree
> it reaches the ground, but the ground observer sees a long lifetime but

But when a relativistic moron takes a twins instead
a muon his mumble is changing significantly.