On Thursday, June 29, 2023 at 8:16:46 PM UTC-5, Prokaryotic Capase Homolog wrote:
> On Thursday, June 29, 2023 at 6:41:01 PM UTC-5, Mike Fontenot wrote:

> > 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.

By the way. Because your ships are perpetually
accelerating, and because the distance between them
continuously increases, there is, after the start of
your version of the thought experiment, *never* any
"frame of the spaceships", so the distance between
them in YOUR presentation of the problem is *always*
somewhat ill-defined.

In my diagram Fig. 4-5, after accelerating with
constant proper acceleration (red lines), the two
ships turn off their engines after been run equal
proper times. The ships continue coasting maintaining
a constant separation in a constant reference frame
(blue lines), their separation being greater than their
original separation by the factor gamma.
https://en.wikipedia.org/wiki/Spacetime#Dewan%E2%80%93Beran%E2%80%93Bell_spaceship_paradox

On Thursday, June 29, 2023 at 8:16:46 PM UTC-5, Prokaryotic Capase Homolog wrote:
> 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.

(Deleted previous post, which may still be visible
on many systems. I replaced "constant reference
frame" in the fourth line from the bottom with
"common reference frame".)

By the way. Because your ships are perpetually
accelerating, and because the distance between them
continuously increases, there is, after the start of
your version of the thought experiment, *never* any
"frame of the spaceships", so the distance between
them in YOUR presentation of the problem is *always*
somewhat ill-defined.

In my diagram Fig. 4-5, after accelerating with
constant proper acceleration (red lines), the two
ships turn off their engines after been run equal
proper times. The ships continue coasting maintaining
a constant separation in a common reference frame
(blue lines), their separation being greater than their
original separation by the factor gamma.
https://en.wikipedia.org/wiki/Spacetime#Dewan%E2%80%93Beran%E2%80%93Bell_spaceship_paradox

On Friday, 30 June 2023 at 09:35:31 UTC+2, Prokaryotic Capase Homolog wrote:
> On Thursday, June 29, 2023 at 8:16:46 PM UTC-5, Prokaryotic Capase Homolog wrote:
> > 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.
> (Deleted previous post, which may still be visible
> on many systems. I replaced "constant reference
> frame" in the fourth line from the bottom with
> "common reference frame".)
>
> By the way. Because your ships are perpetually
> accelerating, and because the distance between them
> continuously increases, there is, after the start of
> your version of the thought experiment, *never* any
> "frame of the spaceships", so the distance between
> them in YOUR presentation of the problem is *always*
> somewhat ill-defined.
>
> In my diagram Fig. 4-5, after accelerating with
> constant proper acceleration (red lines), the two
> ships turn off their engines after been run equal
> proper times. The ships continue coasting maintaining
> a constant separation in a common reference frame
> (blue lines), their separation being greater than their
> original separation by the factor gamma.
> https://en.wikipedia.org/wiki/Spacetime#Dewan%E2%80%93Beran%E2%80%93Bell_spaceship_paradox

Fortunately, having GPS we can be sure that
this delusional mumble is just some delusional
mumble.

On 2023-06-29 21:57:37 +0000, Mike Fontenot said:

> 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.

My diagram for Bell's scenario:

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

Mikko

On 2023-06-30 01:02:56 +0000, rotchm said:

> 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.

The important difference is that a yardstick is a solid object
but a pair of rockets is not.

Mikko

On Friday, June 30, 2023 at 7:00:24 AM UTC-5, Mikko wrote:

> My diagram for Bell's scenario:
>
> https://uploads.disquscdn.com/images/7d6b7c8ae54383686bb1c5935af03c79dddfb88d1af765e126d132fca2eaed5f.png

Very nice!

1) Your diagrams are closer to Mike's scenario than
mine because you maintain continuous acceleration
throughout, whereas in my Fig 4-5, I turn off the
rocket engines to let the ships coast.
2) Mike should note that as a result, there is no
common "frame of the spaceships" in your diagram.
On the right, you present an *approximate* "frame
of the spaceships" and show how their separation
is greater than the initial proper separation of the
spaceships, represented by the orange line.
3) Could you show how the orange line shows up
in the figure on the left, or did you consider
that showing it would be distracting? I can see
why you might have chosen to omit it from the
left, but my own artistic vision would have been
to include it.
4) What software did you use to draw these graphs?

Thanks!

On 2023-06-30 13:18:32 +0000, Prokaryotic Capase Homolog said:

> On Friday, June 30, 2023 at 7:00:24 AM UTC-5, Mikko wrote:
>
>> My diagram for Bell's scenario:>>
>> https://uploads.disquscdn.com/images/7d6b7c8ae54383686bb1c5935af03c79dddfb88d1af765e126d132fca2eaed5f.png
>>
> Very nice!
>
> 1) Your diagrams are closer to Mike's scenario than
> mine because you maintain continuous acceleration
> throughout, whereas in my Fig 4-5, I turn off the
> rocket engines to let the ships coast.
> 2) Mike should note that as a result, there is no
> common "frame of the spaceships" in your diagram.
> On the right, you present an *approximate* "frame
> of the spaceships" and show how their separation
> is greater than the initial proper separation of the
> spaceships, represented by the orange line.

The second diagram is in another inertial frame where the spaceships
first move backwards but later forwards.

> 3) Could you show how the orange line shows up
> in the figure on the left, or did you consider
> that showing it would be distracting? I can see
> why you might have chosen to omit it from the
> left, but my own artistic vision would have been
> to include it.

The orange line in right diagram has the same length as the distance
between the spaceships in the left diagram. It is drawn at the time
when the leading spaceship has stopped its backward movement and is
starting to move forwards.

> 4) What software did you use to draw these graphs?

I made an ad hoc program.

Mikko

On Friday, June 30, 2023 at 9:35:21 AM UTC-5, Mikko wrote:
> On 2023-06-30 13:18:32 +0000, Prokaryotic Capase Homolog said:
>
> > On Friday, June 30, 2023 at 7:00:24 AM UTC-5, Mikko wrote:
> >
> >> My diagram for Bell's scenario:>>
> >> https://uploads.disquscdn.com/images/7d6b7c8ae54383686bb1c5935af03c79dddfb88d1af765e126d132fca2eaed5f.png

> I made an ad hoc program.

All of my diagrams are ad hoc as well, written in C#, but after
struggling a bit to produce the globe in Fig 5-5 in the following
https://en.wikipedia.org/wiki/Special_relativity#Measurement_versus_visual_appearance
I've thought it would be nice to find some 3D graphics software
(commercial or freeware) that could automatically distinguish
between visible and hidden lines rather than my having to go
through the rather tedious computations. I was hoping that you
might have some experience. :-)

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.

The basic problem with your intuitive analysis is that, in applying the
gamma factor, you have completely forgotten about the relativity of
simultaneity.

The reason why Mikko's (and my) analyses using Minkowski
diagrams work is that Minkowski diagrams are a direct graphical
representation of the Lorentz transformations.

If you wanted to correct your ad hoc "intuitive" analysis, you should
add the "Rear-clock ahead" rule, which is to say, if you have a pair
of clocks separated by a constant proper distance L moving with
speed v relative to you, then at any given instant, you will observe
the rear clock reading Lv/c^2 more than the front clock.

With the three rules of "rear-clock ahead", "length contraction" and
"time dilation", you should be able to work out most simple
problems in special relativity. But ultimately, they are not a substitute
for facility with the Lorentz transformations.

On Monday, 3 July 2023 at 08:11:53 UTC+2, Prokaryotic Capase Homolog wrote:
> 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.
> The basic problem with your intuitive analysis is that, in applying the
> gamma factor, you have completely forgotten about the relativity of
> simultaneity.

Delusion of yours, nothing like that in the
world we inhabit.

Prokaryotic Capase Homolog wrote:

> On Thursday, June 29, 2023 at 6:41:01 PM UTC-5, Mike Fontenot wrote:
>> 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.
>
> The basic problem with your intuitive analysis is that, in applying the
> gamma factor, you have completely forgotten about the relativity of
> simultaneity.

there is no simultaneity in relativity. You are a fool.

*_CIA_sees_Ukraine_crisis_as_unique_‘opportunity’_*
https://%72t.com/news/579107-cia-ukraine-russia-opportunity/

Dear Ukrainians hope you have gotten the real ambition of US now. It’s
right time to kick out Zelenskyy from your county before it gets too late.

The people who should be in prison are the US officials that are robbing
us to harass the Russians.

time for Russia and China to bomb the fucking america, with nuclear heads,
out of the orbit of the planet Earth. These ugly khazar goys are not
humans. They make *_funny_money_* without work, so other countries go
hungry in bed. Reason for which they killed Jesus. Criminals from head to
tail.

take the making of fake money away from these dirty khazars, and punish
them, their vassals and supporters in the crime. It's a severe crime
*_to_make_fake_funny_money_* without work.

The US is now desperate and using all possible ways to salvage whatever
they can in what can only be described as a massive feckup. In their
dreams Russia would have already collapsed, but instead it is the US lead
criminal gang of NATO that is really falling apart.

Any opportunity to cause trouble is happy days for these subhumans called
the CIA.

Central Incompetent Agency, trying new ways to justify their salaries, a
big scam.

On 7/3/23 12:11 AM, Prokaryotic Capase Homolog wrote:

>
> If you wanted to correct your ad hoc "intuitive" analysis, you should
> add the "Rear-clock ahead" rule, which is to say, if you have a pair
> of clocks separated by a constant proper distance L moving with
> speed v relative to you, then at any given instant, you will observe
> the rear clock reading Lv/c^2 more than the front clock.
>

You've got that backwards, and your equation is wrong. When two clocks
are accelerating with the same constant acceleration "A", and are
separated by a constant distance "L", the LEADING clock runs faster, by
the factor f(L*A) > 1. Einstein (in his 1907 paper) said that

f(L*A) = exp(L*A),

but that is incorrect, as I show in

https://vixra.org/abs/2109.0076 .

I derive the correct expression for f(L*A) in

https://vixra.org/abs/2201.0015 .

On Monday, July 3, 2023 at 12:07:43 PM UTC-5, Mike Fontenot wrote:
> On 7/3/23 12:11 AM, Prokaryotic Capase Homolog wrote:
>
> >
> > If you wanted to correct your ad hoc "intuitive" analysis, you should
> > add the "Rear-clock ahead" rule, which is to say, if you have a pair
> > of clocks separated by a constant proper distance L moving with
> > speed v relative to you, then at any given instant, you will observe
> > the rear clock reading Lv/c^2 more than the front clock.
> >
> You've got that backwards, and your equation is wrong. When two clocks
> are accelerating with the same constant acceleration "A", and are
> separated by a constant distance "L", the LEADING clock runs faster, by
> the factor f(L*A) > 1. Einstein (in his 1907 paper) said that
>
> f(L*A) = exp(L*A),
>
> but that is incorrect, as I show in
>
> https://vixra.org/abs/2109.0076 .
>
> I derive the correct expression for f(L*A) in
>
> https://vixra.org/abs/2201.0015 .

*** CAN'T YOU READ??? ***
I was NOT writing about which clock runs faster in an accelerating platform..
I was writing about which clock is AHEAD or BEHIND due to RoS.
Those are VERY different things. SHEESH!!!

On June 28, Mike Fontenot wrote:
> 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.

A thought experiment: a closed van, with high roof, travels at 100 mph.
A mirror is mounted on the roof. A floor lamp shoots a collimated light
beam straight up, at the mirror.

Does it reflect directly back to the source, in front, or behind?

--
Rich

On Wednesday, July 5, 2023 at 10:52:34 AM UTC-7, RichD wrote:
> On June 28, Mike Fontenot wrote:
> > 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.
>
> A thought experiment: a closed van, with high roof, travels at 100 mph.
> A mirror is mounted on the roof. A floor lamp shoots a collimated light
> beam straight up, at the mirror.
>
> Does it reflect directly back to the source, in front, or behind?

Can an expert juggler successfully juggle on a jetliner traveling at a steady 450 mph? I think the question about the light in the van is pretty obvious!

On Wednesday, 5 July 2023 at 22:43:56 UTC+2, Paul Alsing wrote:
> On Wednesday, July 5, 2023 at 10:52:34 AM UTC-7, RichD wrote:
> > On June 28, Mike Fontenot wrote:
> > > 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.
> >
> > A thought experiment: a closed van, with high roof, travels at 100 mph.
> > A mirror is mounted on the roof. A floor lamp shoots a collimated light
> > beam straight up, at the mirror.
> >
> > Does it reflect directly back to the source, in front, or behind?
> Can an expert juggler successfully juggle on a jetliner traveling at a steady 450 mph? I think the question about the light in the van is pretty obvious!

Of course you think. You're a brainwashed fanatic idiot.

On Wednesday, July 5, 2023 at 10:02:58 PM UTC-7, Maciej Wozniak wrote:
> On Wednesday, 5 July 2023 at 22:43:56 UTC+2, Paul Alsing wrote:
> > On Wednesday, July 5, 2023 at 10:52:34 AM UTC-7, RichD wrote:
> > > On June 28, Mike Fontenot wrote:
> > > > 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.
> > >
> > > A thought experiment: a closed van, with high roof, travels at 100 mph.
> > > A mirror is mounted on the roof. A floor lamp shoots a collimated light
> > > beam straight up, at the mirror.
> > >
> > > Does it reflect directly back to the source, in front, or behind?
> > Can an expert juggler successfully juggle on a jetliner traveling at a steady 450 mph? I think the question about the light in the van is pretty obvious!

> Of course you think. You're a brainwashed fanatic idiot.

You didn't answer the question, Woz. Can the juggler juggle on an airplane moving at a constant speed?

On Thursday, 6 July 2023 at 07:09:01 UTC+2, Paul Alsing wrote:
> On Wednesday, July 5, 2023 at 10:02:58 PM UTC-7, Maciej Wozniak wrote:
> > On Wednesday, 5 July 2023 at 22:43:56 UTC+2, Paul Alsing wrote:
> > > On Wednesday, July 5, 2023 at 10:52:34 AM UTC-7, RichD wrote:
> > > > On June 28, Mike Fontenot wrote:
> > > > > 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.
> > > >
> > > > A thought experiment: a closed van, with high roof, travels at 100 mph.
> > > > A mirror is mounted on the roof. A floor lamp shoots a collimated light
> > > > beam straight up, at the mirror.
> > > >
> > > > Does it reflect directly back to the source, in front, or behind?
> > > Can an expert juggler successfully juggle on a jetliner traveling at a steady 450 mph? I think the question about the light in the van is pretty obvious!
>
> > Of course you think. You're a brainwashed fanatic idiot.
> You didn't answer the question, Woz. Can the juggler juggle on an airplane moving at a constant speed?

Shouldn't you write "constant speed wrt [something], Al?