Here's my latest: https://vixra.org/abs/2308.0045

Title: A Proof that the Separation Between Accelerating Rockets is Constant

Author: Michael Leon Fontenot

email: PhysicsFiddler@gmail.com

____________________________________________________________________________________

Abstract:

For rockets whose accelerometers show identical, constant readings,
their separation is constant. The proof of that fact makes use of the
limit of a sequence of accelerations "A", lasting for a time "delta_t",
such that the total change in rapidity "A*delta_t", and therefore the
total change in the velocity, don't vary for each iteration of the
sequence of accelerations. In the limit, as "A" goes to infinity, and
"delta_t" goes to zero, the velocity of the rockets changes
instantaneously, and their separation doesn't change. The result is
analogous to the CoMoving Inertial Frame (CMIF) simultaneity method of
Special Relativity, which says that, according to the traveling twin
(him), the home twin (she) instantaneously gets older during his
instantaneous turnaround. Likewise, the ages of the people on the
leading rocket instantaneously get older during their instantaneous
velocity change.
____________________________________________________________________________________

The above abstract really says all that needs to be said about the proof
that the separation between accelerating rockets is constant. The only
thing that would be useful to add, is to elaborate a bit about the
CoMoving Inertial Frame (CMIF) simultaneity method used to resolve the
twin paradox, and to give the "delta_CADO" equation that makes the CMIF
method especially easy and quick to use.

The CMIF simultaneity method says that the accelerating person (the
"AP") must agree with the inertial person ("IP") who is momentarily
stationary with respect to the accelerating person at any given instant.
In the case of the instantaneous turnaround, there is an IP1 immediately
before the turnaround, and an IP2 immediately after the turnaround. For
each of those IP's, their line of simultaneity ("LOS") can be plotted on
a Minkowski diagram. Where those two LOS's intersect the home twin's
world line then give her age, according to the AP, immediately before
and after the turnaround. I.e., that gives the amount by which she
instantaneously ages during his turnaround, according to him.

It's even easier to get that instantaneous age change by using the
"delta_CADO" equation:

delta_CADO = - L * delta(v),

where

delta(v) = v_after_turnaround - v_before_turnaround,

and "L" is their separation, according to HER. Velocities are positive
when directed away from her, and negative when directed toward her.

On Thursday, August 10, 2023 at 8:31:43 AM UTC-7, Mike Fontenot wrote:
> rubbish<
Not again

On 8/10/23 10:31 AM, Mike Fontenot wrote:
> [...] For rockets whose accelerometers show identical, constant
> readings, their separation is constant.

You MUST state in which coordinates or frame this holds. Without that,
you just completely confuse yourself. In the Bell's spaceship paradox,
this constant separation holds in the inertial frame from which they start.

> The proof of that fact makes use of [... bunch of nonsense]

Let their initial inertial frame be called S with coordinates (t,x). The
rockets have constant proper acceleration a (along x), and they start
from points (t=0,x=0) and (t=0,x=D) in S. Their positions as a function
of t (>=0), are [#]:
x1(t) = 0 + (1/a)(sqrt(1+(at)^2) - 1)
x2(t) = D + (1/a)(sqrt(1+(at)^2) - 1)
Manifestly their separation (x2(t)-x1(t)) = D for any t>=0, measured in S.

If the rockets cease accelerating at identical values of proper time,
they will come to rest in the same inertial frame, and measured in that
frame their separation will be larger than D.

(Hint: "length contraction" applies.)

It certainly is possible to consider this from the standpoint of the
successive instantaneously co-moving inertial frames of each rocket, but
that is rather complicated. It is QUITE CLEAR that you have not done
that correctly.

[#] This is derived in many places, including:
https://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

Tom Roberts

On Sunday, 13 August 2023 at 02:31:30 UTC+2, Tom Roberts wrote:
> On 8/10/23 10:31 AM, Mike Fontenot wrote:
> > [...] For rockets whose accelerometers show identical, constant
> > readings, their separation is constant.
> You MUST state in which coordinates or frame this holds.

You're FORCED!!! Cause it's THE BEST WAY!!!

On 8/12/23 6:31 PM, Tom Roberts wrote:
> On 8/10/23 10:31 AM, Mike Fontenot wrote:
>> [...] For rockets whose accelerometers show identical, constant
>> readings, their separation is constant.
>
> You MUST state in which coordinates or frame this holds. Without that,
> you just completely confuse yourself. In the Bell's spaceship paradox,
> this constant separation holds in the inertial frame from which they start.
>

The frame is according to the people on the trailing rocket.

On Sunday, August 13, 2023 at 9:46:54 AM UTC-7, Mike Fontenot wrote:
> On 8/12/23 6:31 PM, Tom Roberts wrote:
> > On 8/10/23 10:31 AM, Mike Fontenot wrote:
> >> [...] For rockets whose accelerometers show identical, constant
> >> readings, their separation is constant.
> >
> > You MUST state in which coordinates or frame this holds. Without that,
> > you just completely confuse yourself. In the Bell's spaceship paradox,
> > this constant separation holds in the inertial frame from which they start.
> >
> The frame is according to the people on the trailing rocket.

According to that frame the distance increases. This is elementary
calculus, very much like it's done in Newtonian mechanics.

--
Jan

On 8/13/23 12:51 PM, JanPB wrote:
> On Sunday, August 13, 2023 at 9:46:54 AM UTC-7, Mike Fontenot wrote:

>> The frame is according to the people on the trailing rocket.
> (And Jan wrote):
> According to that frame the distance increases. This is elementary
> calculus, very much like it's done in Newtonian mechanics.
>
> --
> Jan

I'd like to see you do that. You have the floor.

On 8/13/23 11:46 AM, Mike Fontenot wrote:
> On 8/12/23 6:31 PM, Tom Roberts wrote:
>> On 8/10/23 10:31 AM, Mike Fontenot wrote:
>>> [...] For rockets whose accelerometers show identical, constant
>>> readings, their separation is constant.
>>
>> You MUST state in which coordinates or frame this holds. Without
>> that, you just completely confuse yourself. In the Bell's spaceship
>> paradox, this constant separation holds in the inertial frame from
>> which they start.
>>
>
> The frame is according to the people on the trailing rocket.

There is no "frame" in which the trailing rocket is at rest [#].

There is a series of instantaneously co-moving inertial frames in which
the trailing rocket is momentarily at rest. In successive such frames
the separation gets larger and larger:
a) they are moving faster and faster relative to the initial
inertial frame
b) the separation is constant in that initial inertial frame
c) "length contraction" applies

[#] "Frame" implies orthogonal coordinate axes. For
coordinates with nonzero acceleration along the x axis,
the g_xt metric component is necessarily nonzero, so
the axes are not orthogonal.

As I said before, it is rather complicated to describe this in terms of
accelerated coordinates. It's quite clear that you made a mistake. For
instance, if you attempt to apply the equivalence principle (as you did
in posts around here), that's a mistake.

Tom Roberts

On Sunday, August 13, 2023 at 1:29:33 PM UTC-7, Mike Fontenot wrote:
> On 8/13/23 12:51 PM, JanPB wrote:
> > On Sunday, August 13, 2023 at 9:46:54 AM UTC-7, Mike Fontenot wrote:
>
> >> The frame is according to the people on the trailing rocket.
> > (And Jan wrote):
> > According to that frame the distance increases. This is elementary
> > calculus, very much like it's done in Newtonian mechanics.
> >
> > --
> > Jan
> I'd like to see you do that. You have the floor.

This is easy to see without much calculation. Consider the two
accelerated worldlines. According to the observer at rest (situated
at the launch pad) the two worldlines are two hyperbolas offset by
some distance D. Assume the launch occurs at his time t = 0.
Now imagine a momentary simultaneity space according to the trailing
rocket at some time t0 > 0.

Let's say the x-axis is horizontal and the t-axis is vertical.

If you draw a simultaneity space according to the launchpad observer,
it will look line the line t = t0.

The length of the segment of that line connecting the two worldlines
is D.

Now the claim is that if you draw the line representing the momentary
simultaneity space of the trailing observer at t = t0, the length of the
analogous connecting segment will be greater than D.

This is easily seen geometrically (messy to calculate): start with the
previous line of simultaneity (the "launchpad" one). Now draw the locus
of all points separated from the trailing rocket by D. It is a hyperbola
whose slope starts going straight up (vertical, or parallel to the t-axis),
where by "starts" I mean again starting with the "launchpad" simultaneity
line.

But the worldline of the leading rocket has non-vertical slope for t0 > 0.

Hence the segment of the trailing rocket's momentary simultaneity line
connecting the two worldlines must hit the aforementioned locus of
equidistant points FIRST (when the distance is equal to D).

Therefore, the distance between the two rockets according to the
trailing rocket's momentary simultaneity line is larger than D.

--
Jan

On Sunday, August 13, 2023 at 5:02:27 PM UTC-5, Tom Roberts wrote:
> On 8/13/23 11:46 AM, Mike Fontenot wrote:
> > On 8/12/23 6:31 PM, Tom Roberts wrote:
> >> On 8/10/23 10:31 AM, Mike Fontenot wrote:
> >>> [...] For rockets whose accelerometers show identical, constant
> >>> readings, their separation is constant.
> >>
> >> You MUST state in which coordinates or frame this holds. Without
> >> that, you just completely confuse yourself. In the Bell's spaceship
> >> paradox, this constant separation holds in the inertial frame from
> >> which they start.
> >>
> >
> > The frame is according to the people on the trailing rocket.
> There is no "frame" in which the trailing rocket is at rest [#].
>
> There is a series of instantaneously co-moving inertial frames in which
> the trailing rocket is momentarily at rest. In successive such frames
> the separation gets larger and larger:
> a) they are moving faster and faster relative to the initial
> inertial frame
> b) the separation is constant in that initial inertial frame
> c) "length contraction" applies
>
> [#] "Frame" implies orthogonal coordinate axes. For
> coordinates with nonzero acceleration along the x axis,
> the g_xt metric component is necessarily nonzero, so
> the axes are not orthogonal.
>
> As I said before, it is rather complicated to describe this in terms of
> accelerated coordinates.

As you state, the variant of Bell's spaceship paradox where both
spaceships accelerate "forever" is complicated to describe. That
is why I present a variant where the acceleration of the two ships
is turned off after each been accelerating for an equal proper time,
so that they are comoving and inertial after the constant
acceleration phase.

With this simplification, the scenario is almost trivial to analyze.
https://en.wikipedia.org/wiki/Spacetime#Dewan%E2%80%93Beran%E2%80%93Bell_spaceship_paradox

> It's quite clear that you made a mistake. For
> instance, if you attempt to apply the equivalence principle (as you did
> in posts around here), that's a mistake.
>
> Tom Roberts

On 8/13/23 4:02 PM, Tom Roberts wrote:
> On 8/13/23 11:46 AM, Mike Fontenot wrote:

>> The frame is according to the people on the trailing rocket.

> (and Tom responded):

> There is no "frame" in which the trailing rocket is at rest.

I disagree. Someone, she, riding on the trailing rocket WILL have a
conclusion about the distance to the leading rocket, as a function of
her age (or the reading on her watch). THAT constitutes HER frame.

She certainly might not describe her state as "being at rest", because
she can feel the "g" forces produced by the acceleration (just like you
feel them when accelerating in a fast sports car). But she WILL regard
it as "her frame". She might reasonably refer to it as "my accelerated
frame of reference". And part of her tools for exploring her frame
would be a long rod that connects her rocket to the leading rocket, with
length marks etched along the entire rod. (And, that rod can be such
that it can slide back and forth where it reaches the leading rocket ...
the two rockets are not CONSTRAINED by the rod to be a fixed distance
apart). She can enlist a helper to crawl along that rod between the two
rockets, and report back to her what the length of that rod is. I
maintain that the answer she'll get is that the distance is the same as
it was before the two rockets started their (equal) accelerations. And,
contrarily, the initial inertial observers will say that the two
rockets are getting closer together as the acceleration progresses.

On 8/13/23 6:28 PM, Mike Fontenot wrote:
> On 8/13/23 4:02 PM, Tom Roberts wrote:
>> On 8/13/23 11:46 AM, Mike Fontenot wrote:
>>> The frame is according to the people on the trailing rocket.
>> (and Tom responded): There is no "frame" in which the trailing
>> rocket is at rest.
>
> I disagree. Someone, she, riding on the trailing rocket WILL have a
> conclusion about the distance to the leading rocket, as a function of
> her age (or the reading on her watch). THAT constitutes HER frame.

But that is NOT a frame. Calling it one doesn't make it one.

I repeat: a frame has orthogonal coordinate axes, and in SR that
can happen ONLY for inertial frames.

> [... repetition of previous mistakes]

Claiming a measuring ruler laid between the rockets would indicate a
constant distance between them does not make it so.

[That is a VERY BAD approach, as you don't know
how the material of that ruler responds to the
acceleration.]

(This is getting overly repetitious. Don't expect me to continue until
you learn how to make a correct calculation. Hint: do it in the initial
inertial frame; forget doing it in coordinates of either rocket, as that
is fraught with difficulties which you have CLEARLY not understood.
Indeed I already did it earlier in this thread.)

Tom Roberts

On Monday, 14 August 2023 at 05:43:26 UTC+2, Tom Roberts wrote:
> On 8/13/23 6:28 PM, Mike Fontenot wrote:
> > On 8/13/23 4:02 PM, Tom Roberts wrote:
> >> On 8/13/23 11:46 AM, Mike Fontenot wrote:
> >>> The frame is according to the people on the trailing rocket.
> >> (and Tom responded): There is no "frame" in which the trailing
> >> rocket is at rest.
> >
> > I disagree. Someone, she, riding on the trailing rocket WILL have a
> > conclusion about the distance to the leading rocket, as a function of
> > her age (or the reading on her watch). THAT constitutes HER frame.
> But that is NOT a frame. Calling it one doesn't make it one.

Too bad nobody has informed your idiot guru
about it.

On 8/13/23 5:20 PM, JanPB wrote:
>
> This is easily seen geometrically (messy to calculate):

I wasn't able to follow that description.

According to the initial inertial observers (the inertial observers who
are stationary wrt the rockets before they are ignited), the velocity of
the trailing rocket is

v(t) = tanh(A*t).

(ref: Taylor and Wheeler, "Spacetime Physics", p. 98.)

The distance traveled by the trailing rocket is just the integral of
v(t), which is

d(t) = (1/A) * ln( cosh[A*t] ).

The equation d(t) gives the lower curve (the curve for the trailing
rocket) on the diagram I have referred to recently, and which is given
in https://vixra.org/abs/2307.0151 . The upper curves on that diagram
can be considered to be either various choices for the leading rocket,
or an array of rockets ahead of the trailing rocket.

As required by the famous length contraction equation (the LCE), the
initial inertial observers say that the rockets get closer together, by
the factor

gamma = 1 / sqrt(1 - v*v),

and so a thread initially stretched between the rockets won't break
during the acceleration, because the thread also shrinks by the same
factor gamma.

The accelerating observers riding in the trailing rocket say the
distance between the rockets is constant during the acceleration, and
that the thread has a constant length, so the thread doesn't break.

On Monday, August 14, 2023 at 9:14:59 AM UTC-7, Mike Fontenot wrote:
> On 8/13/23 5:20 PM, JanPB wrote:
> >
> > This is easily seen geometrically (messy to calculate):
> I wasn't able to follow that description.
>
> According to the initial inertial observers (the inertial observers who
> are stationary wrt the rockets before they are ignited), the velocity of
> the trailing rocket is
>
> v(t) = tanh(A*t).
>
> (ref: Taylor and Wheeler, "Spacetime Physics", p. 98.)
>
> The distance traveled by the trailing rocket is just the integral of
> v(t), which is
>
> d(t) = (1/A) * ln( cosh[A*t] ).
>
> The equation d(t) gives the lower curve (the curve for the trailing
> rocket) on the diagram I have referred to recently, and which is given
> in https://vixra.org/abs/2307.0151 . The upper curves on that diagram
> can be considered to be either various choices for the leading rocket,
> or an array of rockets ahead of the trailing rocket.
>
> As required by the famous length contraction equation (the LCE), the
> initial inertial observers say that the rockets get closer together, by
> the factor
>
> gamma = 1 / sqrt(1 - v*v),

No. This is the same mistake people
made in the early days of relativity when they thought the circumference of
the rotating disc was shortened by the gamma factor. But it stays
unchanged (while the rigid sticks laid along the circumference *do*
contract and thus develop gaps separating them, so *more* then 2 pi R
of them will fit.

Lorentz contraction does not apply willy-nilly to all distances separating
all material points in sight regardless of their motions.

> and so a thread initially stretched between the rockets won't break
> during the acceleration, because the thread also shrinks by the same
> factor gamma.

The way Lorentz contraction works is that, in this scenario, the thread
contracts but the rocket separation does not. (Again, consider the rotating
disc and the sticks resting around its circumference.)

You can calculate the relevant distances in the corresponding Minkowski
diagram. It gets quite messy, that's why I described an easy geometric
argument in my previous post in which it's easy to see that extending a segment
of length D from the trailing rocket along its momentary space of simultaneity
will NOT reach the leading rocket: there will be a fit of length left over.

The reason is that the locus of all events separated by spacelike distance D from the
trailing rocket (at some instant) is a hyperbola which intersects the worldline
of the leading rocket (also a hyperbola) at an angle, so the string of length D
between the rockets whose other end by definition lies on that first hyperbola
will NOT reach the leading rocket (the second hyperbola, aka. the leading
rocket worldline).

That argument does not provide the exact numbers but it shows that the
separation distance between the rockets increases, so an inelastic string of
fixed length connecting them must break.

> The accelerating observers riding in the trailing rocket say the
> distance between the rockets is constant during the acceleration, and
> that the thread has a constant length, so the thread doesn't break.

That's violated by that little piece of Minkowski geometry I posted earlier..

--
Jan

On 8/14/23 12:05 PM, JanPB wrote:
>
> [...]
>

Let's cut to the chase. There are two reasons to believe that two
rockets accelerating with the same acceleration "A", as measured by
onboard accelerometers, separated in the direction of their
acceleration, will have a constant separation.

The first reason is that that is what Einstein said in his 1907 paper.
reference:
https://einsteinpapers.press.princeton.edu/vol2-trans/319 . He was
working a special relativity scenario (with constant and equal
acceleration of two rockets), and he indicated that the separation of
the two rockets was constant.

The second reason is what the equivalence principle says about
accelerated motion in special relativity versus a scenario with
motionless clocks in a gravitational field. Two accelerating rockets,
separated in the direction of their equal accelerations, will maintain a
constant separation, but clocks on the leading rocket will run faster
than clocks on the trailing rocket. Equivalently, for two stationary
clocks separated by a fixed distance in the direction of a constant and
uniform gravitational field, the clock farther from the source of the
field will run faster than the other clock, by the same factor as in the
special relativity scenario.

So why do so many people still insist that the rocket separation is NOT
constant in the Bell's Paradox scenario? The reason is that the
scenario I'm analyzing is DIFFERENT from the scenario in the Bell's
Paradox. In the WIKI article, it says

"The distance between the spaceships does not undergo Lorentz
contraction with respect to the distance at the start, because in S, it
is effectively defined to remain the same, due to the equal and
simultaneous acceleration of both spaceships in S."

"S" is the initial inertial frame.

So they didn't specify that the rockets had accelerometers that showed
equal constant readings. And, most important, they specified that the
separation was constant, ACCORDING TO THE INITIAL INERTIAL FRAME. That
rules out the separation being constant in the trailing rocket's frame
.... it REQUIRES that the accelerations are different, as measured by the
accelerometers The two scenarios are DIFFERENT, and it's not
surprising that their conclusions about the survival of the thread are
different.

On Monday, August 14, 2023 at 2:58:16 PM UTC-7, Mike Fontenot wrote:
> On 8/14/23 12:05 PM, JanPB wrote:
> >
> > [...]
> >
>
> Let's cut to the chase. There are two reasons to believe that two
> rockets accelerating with the same acceleration "A", as measured by
> onboard accelerometers, separated in the direction of their
> acceleration, will have a constant separation.

According to the stationary observer.

> So why do so many people still insist that the rocket separation is NOT
> constant in the Bell's Paradox scenario?

It's non-constant according to the rocket observer(s).

> The reason is that the
> scenario I'm analyzing is DIFFERENT from the scenario in the Bell's
> Paradox. In the WIKI article, it says
>
> "The distance between the spaceships does not undergo Lorentz
> contraction with respect to the distance at the start, because in S, it
> is effectively defined to remain the same, due to the equal and
> simultaneous acceleration of both spaceships in S."
>
> "S" is the initial inertial frame.
>
> So they didn't specify that the rockets had accelerometers that showed
> equal constant readings.

They didn't but those rockets accelerometers do show the same readings.
The reason is the homogeneity of space. Since their worldlines are
parallel translates of each other, any difference in accelerometer readings
would have to be attributed to the mere spatial offsetting. So the space
itself would not be homogeneous.

> And, most important, they specified that the
> separation was constant, ACCORDING TO THE INITIAL INERTIAL FRAME.

Yes, see above.

> That
> rules out the separation being constant in the trailing rocket's frame

No, it actually requires it. This is elementary Minkowski geometry.

> ... it REQUIRES that the accelerations are different, as measured by the
> accelerometers

You keep getting everything backwards.

--
Jan

On 2023-08-14 16:14:53 +0000, Mike Fontenot said:

> According to the initial inertial observers (the inertial observers who
> are stationary wrt the rockets before they are ignited), the velocity
> of the trailing rocket is
>
> v(t) = tanh(A*t).

True, and also true abaut the leading rocket.

> The distance traveled by the trailing rocket is just the integral of
> v(t), which is
>
> d(t) = (1/A) * ln( cosh[A*t] ).

True, and also true about the leading rocket.

> The equation d(t) gives the lower curve (the curve for the trailing
> rocket) on the diagram I have referred to recently, and which is given
> in https://vixra.org/abs/2307.0151 .

It also gives the correct curve for the leading rocket when the starting
point is chosen to match the initial separation.

> As required by the famous length contraction equation (the LCE),

There is no such requirement. Each rocket can move as it chooses, and they
have not chosen to respect the length contracton equation. Instead, they
have chosen to maintain the same proper acceleration A.

Mikko

On 8/14/23 11:14 AM, Mike Fontenot wrote:
> According to the initial inertial observers (the inertial observers
> who are stationary wrt the rockets before they are ignited), the
> velocity of the trailing rocket is
>
> v(t) = tanh(A*t).
>
> (ref: Taylor and Wheeler, "Spacetime Physics", p. 98.)

I did not find it there (probably different editions). t=0 is when they
started accelerating. I believe t is proper time of the rocket, not
coordinate time of that initial inertial frame.

I'll assume your equations are correct.

> The distance traveled by the trailing rocket is just the integral of
> v(t), which is
>
> d(t) = (1/A) * ln( cosh[A*t] ).

Yes, for the trailing rocket, ACCORDING TO THE INITIAL OBSERVERS IN THE
INITIAL INERTIAL FRAME. For the leading rocket this is of course:

d2(t) = (1/A) * ln( cosh[A*t] ) + D0

where D0 is the leading rocket's initial position in the initial
inertial frame.

Clearly for any t>0 the distance MEASURED SIMULTANEOUSLY IN THE INITIAL
INERTIAL FRAME remains D0.

> As required by the famous length contraction equation (the LCE), the
> initial inertial observers say that the rockets get closer together,
> by the factor
>
> gamma = 1 / sqrt(1 - v*v),

There's your mistake. "Length contraction" only applies between inertial
frames. So to apply it you must imagine the rockets cease accelerating
at the same value of t. Then in that final inertial frame, "length
contraction" applies, and THE ROCKETS ARE FURTHER APART IN THE FINAL
INERTIAL FRAME THAN THEY ARE IN THE INITIAL INERTIAL FRAME. Because that
is how "length contraction" works -- you attempted to apply it BACKWARDS.

Note carefully that "length contraction" applies to an object measured
in an inertial frame relative to which it is moving. THERE IS NO SUCH
OBJECT IN THE INITIAL INERTIAL FRAME, but the two rockets do form such
an object in the final inertial frame. So their "length contracted"
separation in the initial inertial frame is D0, hence their separation
in the final inertial frame MUST BE LARGER.

> [... repetitions of mistakes]

Tom Roberts

On August 14, Mike Fontenot wrote:
> The reason is that the scenario I'm analyzing is DIFFERENT from
> the scenario in the Bell's Paradox.

< quadruple forehead slap>

Be not dismayed, Mike, there's hope for you yet:
https://www.elitedaily.com/life/scatterbrained-more-intelligent/1238986

--
Rich

On 8/15/23 12:43 PM, Tom Roberts wrote:
>
> For the leading rocket this is of course:
>
>     d2(t) = (1/A) * ln( cosh[A*t] ) + D0
>
> where D0 is the leading rocket's initial position in the initial
> inertial frame.

No, that is incorrect. The distance between the two rockets at time "t"
(according to the initial inertial observers), is

D0 / gamma,

not just D0, as you say.

D0 would be correct for the Bell's scenario where the motion is DEFINED
that way. But a scenario defined in that way does NOT result in equal
readings on the two accelerometers, which violates the original
assumption of my scenario. In Bell's scenario, the accelerometers would
NOT show the same readings.

D0 is the initial separation of the rockets (according to everybody).
And D0 is also the separation at ALL times, according to the people on
the trailing rocket. But it is NOT the separation at all times,
according to the initial inertial observers.

The complete plot, for four choices of the initial separation (or,
alternatively, for an array of 5 rockets), is given in

https://vixra.org/pdf/2307.0151v1.pdf .

On 8/14/23 11:14 AM, Mike Fontenot wrote:
> According to the initial inertial observers (the inertial observers
> who are stationary wrt the rockets before they are ignited), the
> velocity of the trailing rocket is
>
> v(t) = tanh(A*t).
>
> (ref: Taylor and Wheeler, "Spacetime Physics", p. 98.)

I did not find it there (probably different editions). t=0 is when they
started accelerating. I believe t is proper time of the rocket, not
coordinate time of that initial inertial frame -- doesn't matter.

I'll stipulate your equations are correct.

> The distance traveled by the trailing rocket is just the integral of
> v(t), which is
>
> d(t) = (1/A) * ln( cosh[A*t] ).

Yes, for the trailing rocket, ACCORDING TO THE INITIAL OBSERVERS IN THE
INITIAL INERTIAL FRAME. For the leading rocket this is of course:

d2(t) = (1/A) * ln( cosh[A*t] ) + D0

where D0 is the leading rocket's initial position in the initial
inertial frame at t=0.

Clearly for any t>=0 the separation MEASURED SIMULTANEOUSLY IN THE
INITIAL INERTIAL FRAME remains D0.

> As required by the famous length contraction equation (the LCE), the
> initial inertial observers say that the rockets get closer together,
> by the factor
>
> gamma = 1 / sqrt(1 - v*v),

There's your mistake. "Length contraction" only applies between inertial
frames. So to apply it you must have the rockets cease accelerating at
the same value of t; they come to rest in the same final inertial frame.
In that final inertial frame, "length contraction" applies relative to
the initial inertial frame, and THE ROCKETS ARE FURTHER APART IN THE
FINAL INERTIAL FRAME THAN THEY ARE IN THE INITIAL INERTIAL FRAME.
Because that is how "length contraction" works -- you attempted to apply
it BACKWARDS.

Note carefully that "length contraction" applies to an object measured
in an inertial frame relative to which it is moving. THERE IS NO SUCH
OBJECT IN THE INITIAL INERTIAL FRAME, but the two rockets do form such
an object in the final inertial frame; as their separation is D0 in the
initial inertial frame (to which "length contraction" has been applied),
their separation must be LARGER THAN D0 in the final inertial frame
(specifically γ*D0, where γ is the usual Lorentz factor between those
two inertial frames).

> [... repetitions of mistakes]

Tom Roberts

On 8/15/23 4:23 PM, Tom Roberts wrote:

>
> There's your mistake. "Length contraction" only applies between inertial
> frames.

Not true. The length contraction equation (LCE) gives the conclusion
about the length of an object (according to the specified inertial
observer) at any instant in that inertial observer's life. The object
can be accelerating, and the LCE gives the amount of contraction of the
object at any given instant in the life of the inertial observer. The
fact that the object is accelerating only causes the amount of
contraction of the accelerating object to vary with different instants
in the inertial observer's life.

That's easy to see in the diagram I referenced. In that diagram, gamma
increases monotonically as the age "t" of the initial inertial observers
increases. At each instant "t", gamma depends on the velocity "v" of
the trailing rocket, which is just the slope of the lowest curve at that
time "t".

On Tuesday, August 15, 2023 at 3:52:39 PM UTC-7, Mike Fontenot wrote:
> On 8/15/23 4:23 PM, Tom Roberts wrote:
>
> >
> > There's your mistake. "Length contraction" only applies between inertial
> > frames.
>
> Not true. The length contraction equation (LCE) gives the conclusion
> about the length of an object (according to the specified inertial
> observer) at any instant in that inertial observer's life.

Yes, and there is no "object" here. There are just two rockets, each with
its own engine. Again: since the accelerations are the same, the two
worldlines are two hyperbolas which are offset in the x direction.
This is easiest to see by reversing this argument: imagine two rockets
whose worldlines are offset that way. Then their accelerometers must
show the same number because *space is homogeneous*.

> The object
> can be accelerating, and the LCE gives the amount of contraction of the
> object

There is no "object" here. There are just two rockets, each with its own
propulsion system.

> at any given instant in the life of the inertial observer. The
> fact that the object is accelerating only causes the amount of
> contraction of the accelerating object to vary with different instants
> in the inertial observer's life.

There is no contraction applicable here.

> That's easy to see in the diagram I referenced. In that diagram, gamma
> increases monotonically as the age "t" of the initial inertial observers
> increases. At each instant "t", gamma depends on the velocity "v" of
> the trailing rocket, which is just the slope of the lowest curve at that
> time "t".

If this was correct, then space would not be homogeneous.

--
Jan

On 8/15/23 5:52 PM, Mike Fontenot wrote:
> On 8/15/23 4:23 PM, Tom Roberts wrote:
>> There's your mistake. "Length contraction" only applies between
>> inertial frames.
>
> Not true.

Yes, true.

> The length contraction equation (LCE) gives the conclusion about the
> length of an object (according to the specified inertial observer) at
> any instant in that inertial observer's life. The object can be
> accelerating, and the LCE gives the amount of contraction of the
> object at any given instant in the life of the inertial observer.

In the instantaneously co-moving inertial frame of the object. I repeat:
"Length contraction" only applies between an object at rest in one
inertial frame and an observer measuring it using a different inertial
frame.

For two rockets, as JanPB points out, one must be careful as they are
not really an object. Only once they are at rest in THE SAME inertial
frame can one consider them to be the endpoints of an "object".

[I have been careful to repeatedly specify that.
You have been excessively imprecise and have
confused yourself.]

> That's easy to see in the diagram I referenced. In that diagram,
> gamma increases monotonically as the age "t" of the initial inertial
> observers increases. At each instant "t", gamma depends on the
> velocity "v" of the trailing rocket, which is just the slope of the
> lowest curve at that time "t".

Yes, the gamma for the trailing rocket RELATIVE TO THE INITIAL INERTIAL
FRAME depends on t. And it is equal to the gamma for the leading rocket
relative to that frame (at equal values of t).

[As I said before, t can represent either the proper
time of the two rockets, or the coordinate time of
the initial inertial frame -- the conversion between
them is the same for the two rockets.]

But you apply "length contraction" BACKWARDS. Once the rockets cease
accelerating, one can consider them to be located at the two ends of an
"object". Since the length of that "object" is measured to be D0 in the
initial inertial frame, the length of that "object" MUST be larger in
the final inertial frame -- that's how "length contraction" works.

["Length contraction" would also apply to an observer
in the final frame looking at an object at rest in the
initial inertial frame, but there is no such object.
Note: you cannot apply "length contraction" to
coordinates, you MUST use the Lorentz Transform.]

> D0 would be correct for the Bell's scenario where the motion is
> DEFINED that way. But a scenario defined in that way does NOT
> result in equal readings on the two accelerometers, which violates
> the original assumption of my scenario. In Bell's scenario, the
> accelerometers would NOT show the same readings.

This is just plain wrong. Bell's scenario is indistinguishable from
yours -- for both scenarios the (constant) proper accelerations of the
rockets are equal, in both scenarios the separation measured
simultaneously in the initial inertial frame is constant, and in both
scenarios the separation of the rockets in the final inertial frame is
LARGER than in the initial inertial frame.

I don't know why you cannot understand this, as it is simply a basic
application of Lorentz transforms.

(This is overly repetitious. Don't expect me to continue.)

Tom Roberts