I have previously described the acceleration of two separated rockets,
according to inertial observers who are stationary wrt the rockets
immediately before the rockets are fired. I described a diagram of the
motion of the two rockets, according to those inertial observers. The
diagram showed the separation of the two rockets being constant.

I recently realized that that diagram is incorrect. The length
contraction equation (LCE) of special relativity says that an inertial
observer will always conclude that a moving yardstick is shorter than
his own yardstick, by the gamma factor. So the inertial observers in
the above first paragraph must conclude that the separation of the
rockets decreases with time.

I have corrected the original diagram. There's no way I can post that
corrected diagram here, but you can see it on this link:

https://www.sciforums.com/threads/a-quandary-about-accelerated-motion-in-special-relativity.165975/

On Tuesday, June 27, 2023 at 3:31:11 PM UTC-4, Mike Fontenot wrote:
> I have previously described

I did not consult your previous scenario, and do not need to for this post.

> the acceleration of two separated rockets,
> according to inertial observers who are stationary wrt the rockets
> immediately before the rockets are fired.

This is simply classic physics, it has nothing to do with (special) relativity.
Taking the Rockets as very small (e.g.: point particles), you ascribe to them
position functions, like x(t) & X(t) resp. This is what your positing, such that these have accelerations.

> I described a diagram of the
> motion of the two rockets, according to those inertial observers.

There is no need to do that and is overkill. You have already posited that their positions
are x(t) & X(t) resp.

> The diagram showed the separation of the two rockets being constant.

Their separation is simply X(t) - x(t).
Or |x(t) - X(t)| if you prefer.
If you posited that X(t) - x(t) is constant then it is... Constant!
There is no mystery there since you have posited it in the beginning.

> I recently realized that that diagram is incorrect.

Correct or not, X(t) - x(t) is the separation.

> The length contraction equation (LCE) of special relativity says that an inertial
> observer will always conclude that a moving yardstick is shorter than
> his own yardstick, by the gamma factor.

Correct but this has nothing to do with your above scenario.

> So the inertial observers in
> the above first paragraph must conclude that the separation of the
> rockets decreases with time.

No. The inertial Observer simply applies the meaning of the word 'separation';
And concludes that their separation is X(t) - x(t).
Again, this has nothing to do with special relativity.

The Lorentz length contraction applies to the length of material objects or to the measurement of the spatial separation of two events. Your Spaceships are not events. And your spaceships have been posited to be Point particles hence have no length. And your scenario involves measuring the separation of the two ships, not their lengths. So your scenario has nothing to do with the Lorentz contraction.

On 2023-06-27 19:31:06 +0000, Mike Fontenot said:

> I have previously described the acceleration of two separated rockets,
> according to inertial observers who are stationary wrt the rockets
> immediately before the rockets are fired. I described a diagram of the
> motion of the two rockets, according to those inertial observers. The
> diagram showed the separation of the two rockets being constant.

The two rockets are able to move independently so the evolution of their
separation can be anything. If more is specified about the motion of the
rockets then something may be inferrable of their separation but above
more is not specified. If the additional specification is that the
separation of the rockets is constant in which case the separation of
the rockets is constant. More often the specification is that the
accleration
or proper acceleration as a function of time or proper time is constant.
In these cases the separation is constant, too. But more generally, it
need not be.

Mikko

On 6/28/23 2:33 AM, Mikko wrote:
>
> The two rockets are able to move independently so the evolution of their
> separation can be anything. If more is specified about the motion of the
> rockets then something may be inferrable of their separation but above
> more is not specified. If the additional specification is that the
> separation of the rockets is constant in which case the separation of
> the rockets is constant. More often the specification is that the
> accleration
> or proper acceleration as a function of time or proper time is constant.
> In these cases the separation is constant, too. But more generally, it
> need not be.
>

In the scenario that I am interested in, and which I have analyzed, the
two rockets (immediately after they are ignited) always have the same
constant acceleration, as reported by accelerometers attached to the two
rockets.

On 2023-06-28 15:58:59 +0000, Mike Fontenot said:

> In the scenario that I am interested in, and which I have analyzed, the
> two rockets (immediately after they are ignited) always have the same
> constant acceleration, as reported by accelerometers attached to the
> two rockets.

In that case the motion of a rocket can be described with equations
t = (sinh as) / a
x = p + (cosh as - 1)/a
where s is the proper time
and p is the position of the rocket when s = 0

From these equations one can check that the acceleration is constant
and independent of p.

Answers to other interesting questions can be inferred from these equations.

Mikko

On 6/28/23 9:58 AM, Mike Fontenot wrote:
>
> In the scenario that I am interested in, and which I have analyzed, the
> two rockets (immediately after they are ignited) always have the same
> constant acceleration, as reported by accelerometers attached to the two
> rockets.
>

In that scenario, the separation of the rockets, according to the
accelerating traveler in the trailing rocket, is constant, and the
string doesn't break.

According to the initial inertial observers who are stationary wrt the
rockets immediately before they are fired, the separation of the two
rockets decreases as the acceleration proceeds, as required by the
famous length contraction equation (LCE) of special relativity. So they
also conclude that the string doesn't break.

The scenario, as given in Bell's Paradox, may be a completely different
scenario from the above scenario. As far as I know, there is no mention
of rocket accelerometer readings in that Wiki article on Bell's Paradox.
If, in Bell's Paradox, the initial inertial observers correctly
conclude that the rocket separation doesn't decrease, then the rockets
CAN'T be accelerating at the same rate (according to accelerometers
attached to the rockets) ... the leading rocket must have a greater
acceleration than the trailing rocket. If so, I have no interest in
that different scenario.

On 6/27/23 2:31 PM, Mike Fontenot wrote:
> I have previously described the acceleration of two separated
> rockets, according to inertial observers who are stationary wrt the
> rockets immediately before the rockets are fired. I described a
> diagram of the motion of the two rockets, according to those
> inertial observers. The diagram showed the separation of the two
> rockets being constant.
>
> I recently realized that that diagram is incorrect. The length
> contraction equation (LCE) of special relativity says that an
> inertial observer will always conclude that a moving yardstick is
> shorter than his own yardstick, by the gamma factor. So the
> inertial observers in the above first paragraph must conclude that
> the separation of the rockets decreases with time.

YOU are confused. In the scenario you used you are wrong and your
original diagram is correct.

Stipulate that (x,t) are the coordinates of the initial inertial frame,
and both rockets start accelerating at t=0.

First assume that the two rockets have the same acceleration measured in
their initial inertial frame (i.e. d^2x/dt^2). Then integrating the
acceleration twice (dt) shows that their positions relative to their
starting points are the same functions of t, so their separation is
constant measured in that frame. Note that no relativity is used at all,
just integral calculus. Note also that the acceleration need not be
constant, all that matters is that the acceleration as a function of t
is the same for both rockets. Note also that if their acceleration
ceases at the same value of t, the two rockets will come to rest in the
same final inertial frame, and we can apply SR to conclude that their
distance apart measured in that final frame must be LARGER than their
distance apart measured in the initial inertial frame (because the
measurement in the initial frame is "length contracted").

[Make sure you understand everything I just said, and
agree with it.]

Now switch to the scenario in which their proper accelerations are
identical functions of their proper time \tau (constant or not, does not
matter). For any value of \tau it should be obvious that the two
rockets' speeds relative to the initial inertial frame are equal. But
the transform from proper acceleration to acceleration relative to the
initial inertial frame depends only on the speed of the rocket relative
to that frame, so at any given value of \tau they have equal values of
d^2x/dt^2. So we have the same situation as the previous paragraph --
their distance apart is constant measured in their initial inertial
frame, and when they come to rest in the same final inertial frame they
are further apart in the final frame than they were in the initial frame.

Tom Roberts

On Wednesday, June 28, 2023 at 12:15:16 PM UTC-5, Mike Fontenot wrote:
> On 6/28/23 9:58 AM, Mike Fontenot wrote:
> >
> > In the scenario that I am interested in, and which I have analyzed, the
> > two rockets (immediately after they are ignited) always have the same
> > constant acceleration, as reported by accelerometers attached to the two
> > rockets.
> >
> In that scenario, the separation of the rockets, according to the
> accelerating traveler in the trailing rocket, is constant, and the
> string doesn't break.
>
> According to the initial inertial observers who are stationary wrt the
> rockets immediately before they are fired, the separation of the two
> rockets decreases as the acceleration proceeds, as required by the
> famous length contraction equation (LCE) of special relativity. So they
> also conclude that the string doesn't break.
>
> The scenario, as given in Bell's Paradox, may be a completely different
> scenario from the above scenario. As far as I know, there is no mention
> of rocket accelerometer readings in that Wiki article on Bell's Paradox.
> If, in Bell's Paradox, the initial inertial observers correctly
> conclude that the rocket separation doesn't decrease, then the rockets
> CAN'T be accelerating at the same rate (according to accelerometers
> attached to the rockets) ... the leading rocket must have a greater
> acceleration than the trailing rocket. If so, I have no interest in
> that different scenario.

I consider the full Wiki article on Bell's spaceship paradox to be
somewhat overly comprehensive. Looking back at what I wrote
six years ago as a short article section on the same topic, I would
criticize my efforts then as being somewhat superficial, but I think
my short section may be somewhat useful for showing the overall
shape for the forest rather than getting you lost among the myriad
of trees that comprise the full article.
https://en.wikipedia.org/wiki/Spacetime#Dewan%E2%80%93Beran%E2%80%93Bell_spaceship_paradox

Let me know what you think.

On Thursday, 29 June 2023 at 04:39:29 UTC+2, Tom Roberts wrote:
> On 6/27/23 2:31 PM, Mike Fontenot wrote:
> > I have previously described the acceleration of two separated
> > rockets, according to inertial observers who are stationary wrt the
> > rockets immediately before the rockets are fired. I described a
> > diagram of the motion of the two rockets, according to those
> > inertial observers. The diagram showed the separation of the two
> > rockets being constant.
> >
> > I recently realized that that diagram is incorrect. The length
> > contraction equation (LCE) of special relativity says that an
> > inertial observer will always conclude that a moving yardstick is
> > shorter than his own yardstick, by the gamma factor. So the
> > inertial observers in the above first paragraph must conclude that
> > the separation of the rockets decreases with time.
> YOU are confused. In the scenario you used you are wrong and your
> original diagram is correct.
>
> Stipulate that (x,t) are the coordinates of the initial inertial frame,
> and both rockets start accelerating at t=0.
>
> First assume that

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

On 6/28/23 11:15 AM, Mike Fontenot wrote:
> On 6/28/23 9:58 AM, Mike Fontenot wrote:
>>
>> In the scenario that I am interested in, and which I have analyzed,
>> the two rockets (immediately after they are ignited) always have the
>> same constant acceleration, as reported by accelerometers attached to
>> the two rockets.
>>
>
> In that scenario, the separation of the rockets, according to the
> accelerating traveler in the trailing rocket, is constant, and the
> string doesn't break.
>
> According to the initial inertial observers who are stationary wrt the
> rockets immediately before they are fired, the separation of the two
> rockets decreases as the acceleration proceeds, as required by the
> famous length contraction equation (LCE) of special relativity.  So they
> also conclude that the string doesn't break.
>
> The scenario, as given in Bell's Paradox, may be a completely different
> scenario from the above scenario.  As far as I know, there is no mention
> of rocket accelerometer readings in that Wiki article on Bell's Paradox.
>  If, in Bell's Paradox, the initial inertial observers correctly
> conclude that the rocket separation doesn't decrease, then the rockets
> CAN'T be accelerating at the same rate (according to accelerometers
> attached to the rockets) ... the leading rocket must have a greater
> acceleration than the trailing rocket.  If so, I have no interest in
> that different scenario.
>
>

From the Wiki article:

"They start accelerating simultaneously and equally as measured in the
inertial frame S, thus having the same velocity at all times as viewed
from S. Therefore, they are all subject to the same Lorentz contraction,
so the entire assembly seems to be equally contracted in the S frame
with respect to the length at the start. At first sight, it might appear
that the thread will not break during acceleration. This argument,
however, is incorrect as shown by Dewan and Beran, and later Bell.[1][2]
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." (The bold emphasis
is mine).

I think that last sentence is very badly worded, and not even
consistent. I think they should have said "If the spaceships are forced
to maintain a constant separation according to the S inertial frame, the
leading spaceship would HAVE to accelerate faster than the trailing
spaceship, so that the combination of an increased separation due to the
higher acceleration would exactly cancel the Lorentz contraction that
must occur". IF my wording better describes their scenario, then the two
scenarios (Bell's and mine) WOULD be different, which would be a good
thing.

On 6/28/23 9:07 PM, Prokaryotic Capase Homolog wrote:
>
> https://en.wikipedia.org/wiki/Spacetime#Dewan%E2%80%93Beran%E2%80%93Bell_spaceship_paradox
>
> Let me know what you think.

I claim that your option 2 is correct.

On 6/28/23 8:39 PM, Tom Roberts wrote:
>
> First assume that the two rockets have the same acceleration measured in
> their initial inertial frame (i.e. d^2x/dt^2).

That assumption is incorrect.

On Thursday, June 29, 2023 at 12:55:48 PM UTC-5, Mike Fontenot wrote:
> On 6/28/23 9:07 PM, Prokaryotic Capase Homolog wrote:
> >
> > https://en.wikipedia.org/wiki/Spacetime#Dewan%E2%80%93Beran%E2%80%93Bell_spaceship_paradox
> >
> > Let me know what you think.
> I claim that your option 2 is correct.

As I emphasized in my Wiki contribution, most of the
misunderstandings that people have about this puzzle will
simply vanish if people would just draw (ACCURATELY!!!) a
Minkowski diagram of their setup. My drawing (which was
ultimately based on Chris Hillman's drawing generated
using a Maple script) accurately reflects the how the Bell
spaceship paradox is usually stated, and there is NO ROOM
for any error or ambiguity in the conclusion. For you to
disagree means that *YOUR* statement of the tethered
spaceship paradox (which is no longer Bell's, therefore we
have to call it something DIFFERENT from "Bell's paradox")
differs from Bell's statement of the puzzle.

Please make an accurate Minkowski drawing of your setup
and post your drawing on a file-sharing site so that we all can
see it. The one that you describe on sciforums is obviously not
a Minkowski diagram.

I am not willing to become a member of sciforums just so
that I can see what you posted there as a drawing. I get FAR
too much junk email as it is. But just from the comments
that I see you made, I presume that your original analysis
did not take relativity of simultaneity into account.

I post step-by-step instructions on drawing Minkowski
diagrams here:
https://en.wikipedia.org/wiki/Special_relativity#Graphical_representation_of_the_Lorentz_transformation

On 6/29/23 12:58 PM, Prokaryotic Capase Homolog wrote:
>
> I am not willing to become a member of sciforums just so
> that I can see what you posted there as a drawing.

You don't have to be a member to read the posts on sciforums.

On Thursday, June 29, 2023 at 3:05:33 PM UTC-5, Mike Fontenot wrote:
> On 6/29/23 12:58 PM, Prokaryotic Capase Homolog wrote:
> >
> > I am not willing to become a member of sciforums just so
> > that I can see what you posted there as a drawing.
> You don't have to be a member to read the posts on sciforums.

How do I see your drawing? From what I understand, I have
to be a member to see attachments.

On Tuesday, June 27, 2023 at 12:31:11 PM UTC-7, Mike Fontenot wrote:
> I have previously described the acceleration of two separated rockets,
> according to inertial observers who are stationary wrt the rockets
> immediately before the rockets are fired. I described a diagram of the
> motion of the two rockets, according to those inertial observers. The
> diagram showed the separation of the two rockets being constant.
>
> I recently realized that that diagram is incorrect. The length
> contraction equation (LCE) of special relativity says that an inertial
> observer will always conclude that a moving yardstick is shorter than
> his own yardstick, by the gamma factor. So the inertial observers in
> the above first paragraph must conclude that the separation of the
> rockets decreases with time.

No. The yardstick would contract, yes, but the separation between the
spaceship is constant according to the stationary observer. That's how
SR works, think this through.

> I have corrected the original diagram. There's no way I can post that
> corrected diagram here, but you can see it on this link:
>
> https://www.sciforums.com/threads/a-quandary-about-accelerated-motion-in-special-relativity.165975/

I see no diagram there.

--
Jan

On Thursday, June 29, 2023 at 11:07:21 AM UTC-7, Mike Fontenot wrote:
> On 6/28/23 8:39 PM, Tom Roberts wrote:
> >
> > First assume that the two rockets have the same acceleration measured in
> > their initial inertial frame (i.e. d^2x/dt^2).
> That assumption is incorrect.

Read Tom's entire post. You clipped the second scenario. I'm assuming Tom
included both scenarios because in your original post you didn't say which one you
were considering.

--
Jan

On Thursday, June 29, 2023 at 1:07:21 PM UTC-5, Mike Fontenot wrote:
> On 6/28/23 8:39 PM, Tom Roberts wrote:
> >
> > First assume that the two rockets have the same acceleration measured in
> > their initial inertial frame (i.e. d^2x/dt^2).
> That assumption is incorrect.

Then the "tethered spaceship problem" that you are
describing is NOT Bell's spaceship paradox, but
something else entirely. In Bell's problem statement,
both spaceships accelerate at the same time (as
judged from *our* frame) in the same direction
along the line between them with the same constant
proper acceleration

I therefore have NO IDEA what the scenario that you
are describing looks like. Please post your drawing
on a file-sharing site that does not require a viewer
to sign up for membership.

On 6/29/23 2:44 PM, Prokaryotic Capase Homolog wrote:

>
> How do I see your drawing? From what I understand, I have
> to be a member to see attachments.

Yeah, you're right. I just tried it, using my wife's computer without
logging in, and it showed the text but not the diagram. Sorry.

On 6/29/23 2:51 PM, JanPB wrote:
>
> The yardstick would contract, yes, but the separation between the
> spaceship is constant according to the stationary observer.

If the inertial observers say a yardstick between the two rockets is
contracted, that means they say the distance between the two rockets is
contracted, by the same percentage.

On 6/29/23 4:18 PM, Mike Fontenot wrote:
> On 6/29/23 2:51 PM, JanPB wrote:
>>
>> The yardstick would contract, yes, but the separation between the
>> spaceship is constant according to the stationary observer.
>
> If the inertial observers say a yardstick between the two rockets is
> contracted, that means they say the distance between the two rockets is
> contracted, by the same percentage.
>

I.e., if the inertial observers say a yardstick between the two rockets
is contracted by the gamma factor, they also say a long rod connecting
the two rockets is also contracted by the gamma factor, which means they
say the separation between the rockets has decreased by the gamma factor.

On 6/29/23 3:19 PM, Prokaryotic Capase Homolog wrote:

> Please post your drawing
> on a file-sharing site that does not require a viewer
> to sign up for membership.

I can just tell you how to compute it. That will be a better learning
experience anyway, and you can see why it has to be that way.

The rapidity is

theta(t) = A * t,

where "A" is the constant acceleration, and "t" is the time since the
acceleration started. As "t" grows without bound, theta(t) grows
without bound. In Newtonian physics, the rapidity is just the velocity,
which can be arbitrarily large in Newtonian physics, so there is no
problem. But in special relativity, the velocity can't equal or exceed
the speed of light.

In special relativity, given the rapidity we can get the velocity, from

v(t) = tanh( theta(t) ).

As "t" goes to infinity, theta(t) goes to infinity, but

v(t) = tanh(theta[t])

asymptotically approaches 1 lightyear/year (the speed of light) as "t"
goes to infinity.

The distance "D" traveled is just the integral of v(t), and that is
equal to

D(t) = ln( cosh( v(t) ).

If you plot D(t), you will get the lower curve (the curve for the
trailing rocket).

In the original diagram that I produced, I then plotted exactly the same
curve for the leading rocket, just shifted vertically upward by their
initial separation. So that says that the two spaceships maintain a
constant separation, according to the inertial observers who are
stationary with the rockets immediately before they are started.

But that is incorrect, because those inertial observers know that any
rod moving away from them along its long dimension will be
length-contracted by the factor

gamma(v) = 1 / sqrt{1 - v*v}.

For example, for

v = 0.866 ly/y,

gamma = 2.0 .

So they know that, for each value of "t", the distance between the
rockets must be length-contracted by the factor gamma(v[t]).

The bottom curve can't be moved upward to reduce the distance between
the two curves by the gamma factor, because as t goes to infinity, that
curve already gets arbitrarily close to 1.0, the speed of light, so we
can't make it any greater. So the upper curve, for each value of "t",
must be moved downward so that the vertical distance between the two
curves is

D2(t) = D(t) / gamma(v[t]).

That's how to compute the revised upper curve for the leading rocket.
Since gamma goes to infinity as "v" goes to 1.0 ly/y, the separation of
the rockets gets infinitesimally small as "t" goes to infinity.

On Thursday, June 29, 2023 at 6:25:30 PM UTC-4, Mike Fontenot wrote:

> If the inertial observers say a yardstick between the
> two rockets is contracted, that means they say the distance
> between the two rockets is contracted, by the same percentage.

No. I (as others) explained this to you in my initial post, which you were unable to refute.

> I.e., if the inertial observers say a yardstick between the two rockets
> is contracted by the gamma factor, they also say a long rod connecting
> the two rockets is also contracted by the gamma factor,

Correct.

> which means they
> say the separation between the rockets has decreased by the gamma factor.

No.

Length contraction of a moving object is NOT the same concepts as the separation between two moving point particles.

On Thursday, June 29, 2023 at 6:41:01 PM UTC-5, Mike Fontenot wrote:
> On 6/29/23 3:19 PM, Prokaryotic Capase Homolog wrote:
>
> > Please post your drawing
> > on a file-sharing site that does not require a viewer
> > to sign up for membership.
> I can just tell you how to compute it. That will be a better learning
> experience anyway, and you can see why it has to be that way.
>
> The rapidity is
>
> theta(t) = A * t,
>
> where "A" is the constant acceleration, and "t" is the time since the
> acceleration started. As "t" grows without bound, theta(t) grows
> without bound. In Newtonian physics, the rapidity is just the velocity,
> which can be arbitrarily large in Newtonian physics, so there is no
> problem. But in special relativity, the velocity can't equal or exceed
> the speed of light.
>
> In special relativity, given the rapidity we can get the velocity, from
>
> v(t) = tanh( theta(t) ).
>
> As "t" goes to infinity, theta(t) goes to infinity, but
>
> v(t) = tanh(theta[t])
>
> asymptotically approaches 1 lightyear/year (the speed of light) as "t"
> goes to infinity.
>
> The distance "D" traveled is just the integral of v(t), and that is
> equal to
>
> D(t) = ln( cosh( v(t) ).
>
> If you plot D(t), you will get the lower curve (the curve for the
> trailing rocket).
>
> In the original diagram that I produced, I then plotted exactly the same
> curve for the leading rocket, just shifted vertically upward by their
> initial separation. So that says that the two spaceships maintain a
> constant separation, according to the inertial observers who are
> stationary with the rockets immediately before they are started.
>
> But that is incorrect, because those inertial observers know that any
> rod moving away from them along its long dimension will be
> length-contracted by the factor
>
> gamma(v) = 1 / sqrt{1 - v*v}.
>
> For example, for
>
> v = 0.866 ly/y,
>
> gamma = 2.0 .
>
> So they know that, for each value of "t", the distance between the
> rockets must be length-contracted by the factor gamma(v[t]).
>
> The bottom curve can't be moved upward to reduce the distance between
> the two curves by the gamma factor, because as t goes to infinity, that
> curve already gets arbitrarily close to 1.0, the speed of light, so we
> can't make it any greater. So the upper curve, for each value of "t",
> must be moved downward so that the vertical distance between the two
> curves is
>
> D2(t) = D(t) / gamma(v[t]).
>
> That's how to compute the revised upper curve for the leading rocket.
> Since gamma goes to infinity as "v" goes to 1.0 ly/y, the separation of
> the rockets gets infinitesimally small as "t" goes to infinity.

I would *normally* explain this in terms of lines of
simultaneity, but since you read my article section
and didn't grasp the arguments that I presented there
at all, I obviously need to use an alternate tactic.

A fundamental assumption that you make is that
the front and rear rockets see each other doing
exactly the same thing. *THEY DO NOT*
Since they are continuously accelerating, the rear
rocket sees the front rocket as blue-shifted. In
other words, the rear rocket sees the front rocket's
clocks as running faster than his own, which means
that the front rocket steadily pulls away, increasing
the distance between the rockets as measured in
their respective frames.

On Thursday, June 29, 2023 at 3:18:51 PM UTC-7, Mike Fontenot wrote:
> On 6/29/23 2:51 PM, JanPB wrote:
> >
> > The yardstick would contract, yes, but the separation between the
> > spaceship is constant according to the stationary observer.
> If the inertial observers say a yardstick between the two rockets is
> contracted, that means they say the distance between the two rockets is
> contracted, by the same percentage.

No. That's false. Think this through.

--
Jan