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Which leg is longer and which is faster?

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  Which leg is longer and which is faster? "sepp623@yahoo.com" <sepp623@yahoo.com> - 2022-12-13 10:06 -0800
    Re: Which leg is longer and which is faster? patdolan <patdolan@comcast.net> - 2022-12-13 11:45 -0800
    Re: Which leg is longer and which is faster? Trevor Lange <trevorlange97@gmail.com> - 2022-12-13 12:01 -0800
      Re: Which leg is longer and which is faster? patdolan <patdolan@comcast.net> - 2022-12-13 13:23 -0800
      Re: Which leg is longer and which is faster? "sepp623@yahoo.com" <sepp623@yahoo.com> - 2022-12-17 07:05 -0800
        Re: Which leg is longer and which is faster? Trevor Lange <trevorlange97@gmail.com> - 2022-12-17 08:18 -0800
          Re: Which leg is longer and which is faster? "sepp623@yahoo.com" <sepp623@yahoo.com> - 2022-12-17 14:44 -0800
            Re: Which leg is longer and which is faster? "Dono." <eggy20011951@gmail.com> - 2022-12-17 14:58 -0800
            Re: Which leg is longer and which is faster? Trevor Lange <trevorlange97@gmail.com> - 2022-12-17 15:36 -0800
          Re: Which leg is longer and which is faster? Maciej Wozniak <maluwozniak@gmail.com> - 2022-12-17 14:52 -0800
    Re: Which leg is longer and which is faster? The Starmaker <starmaker@ix.netcom.com> - 2022-12-13 22:13 -0800
    Re: Which leg is longer and which is faster? "Paul B. Andersen" <paul.b.andersen@paulba.no> - 2022-12-18 22:30 +0100
      Re: Which leg is longer and which is faster? Maciej Wozniak <maluwozniak@gmail.com> - 2022-12-18 13:54 -0800
      Re: Which leg is longer and which is faster? "Paul B. Andersen" <paul.b.andersen@paulba.no> - 2022-12-19 12:32 +0100
        Re: Which leg is longer and which is faster? beda pietanza <bedapietanza@gmail.com> - 2022-12-19 05:17 -0800
        Re: Which leg is longer and which is faster? Richard Hachel <r.hachel@wanadou.fr> - 2022-12-19 16:04 +0000
          Re: Which leg is longer and which is faster? Python <python@invalid.org> - 2022-12-19 18:22 +0100
            Re: Which leg is longer and which is faster? Richard Hachel <r.hachel@wanadou.fr> - 2022-12-19 17:42 +0000
              Re: Which leg is longer and which is faster? Python <python@invalid.org> - 2022-12-19 22:27 +0100
                Re: Which leg is longer and which is faster? Richard Hachel <r.hachel@wanadou.fr> - 2022-12-19 21:33 +0000
                Re: Which leg is longer and which is faster? Maciej Wozniak <maluwozniak@gmail.com> - 2022-12-19 23:48 -0800
            Re: Which leg is longer and which is faster? Maciej Wozniak <maluwozniak@gmail.com> - 2022-12-19 12:08 -0800
              Re: Which leg is longer and which is faster? Python <python@invalid.org> - 2022-12-19 22:28 +0100

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#597441 — Which leg is longer and which is faster?

There is an inertial frame F0.  In that frame there is a spaceship at rest.  The spaceship has an accelerometer on board.  At time t0, the spaceship accelerates in the positive x direction at a constant rate as measured by the on board accelerometer. When the spaceship reaches a velocity V with respect to F0, the direction of the acceleration changes to the negative x direction with the accelerometer on board the spaceship showing the same magnitude as it did during the positive leg of this journey. When the spaceship has zero velocity with respect to F0, the acceleration of the spaceship stops.
    During the first leg the time to go from 0 to V is t1 as measured in F0, and the distance traveled during that leg is d1. On the leg with the negative acceleration, the time to go from V to 0 as measured in  F0 is t2 and the distance traveled during that leg is d2.
    Which is greater t1 or t2 or are they both equal?
    Which is greater d1 or d2 or are they both equal?
Thanks,
David Seppala
Bastrop TX

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#597447

On Tuesday, December 13, 2022 at 10:06:59 AM UTC-8, sep...@yahoo.com wrote:
> There is an inertial frame F0. In that frame there is a spaceship at rest. The spaceship has an accelerometer on board. At time t0, the spaceship accelerates in the positive x direction at a constant rate as measured by the on board accelerometer. When the spaceship reaches a velocity V with respect to F0, the direction of the acceleration changes to the negative x direction with the accelerometer on board the spaceship showing the same magnitude as it did during the positive leg of this journey. When the spaceship has zero velocity with respect to F0, the acceleration of the spaceship stops. 
> During the first leg the time to go from 0 to V is t1 as measured in F0, and the distance traveled during that leg is d1. On the leg with the negative acceleration, the time to go from V to 0 as measured in F0 is t2 and the distance traveled during that leg is d2. 
> Which is greater t1 or t2 or are they both equal? 
> Which is greater d1 or d2 or are they both equal? 
> Thanks, 
> David Seppala 
> Bastrop TX

I'll bite!  But first a question: is the rocket radio controlled from an observer in F0, or is the accelerometer determining when the rocket reaches V according to the equation v = at?

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#597449

On Tuesday, December 13, 2022 at 10:06:59 AM UTC-8, sep...@yahoo.com wrote:
> Which is greater t1 or t2 or are they both equal? 
Obviously equal.

> Which is greater d1 or d2 or are they both equal? 
Obviously equal.

Let me guess... now you will say "Okay, so here's my REAL question...."  Why don't you just start out saying whatever it is you have to say, or asking whatever it is you really want to ask?

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#597463

On Tuesday, December 13, 2022 at 12:01:50 PM UTC-8, Trevor Lange wrote:
> On Tuesday, December 13, 2022 at 10:06:59 AM UTC-8, sep...@yahoo.com wrote:
> > Which is greater t1 or t2 or are they both equal?
> Obviously equal.
> > Which is greater d1 or d2 or are they both equal?
> Obviously equal. 
> 
> Let me guess... now you will say "Okay, so here's my REAL question...." Why don't you just start out saying whatever it is you have to say, or asking whatever it is you really want to ask?
David is employing the Socratic method.  Just play along, old boy.

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#597713

On Tuesday, December 13, 2022 at 2:01:50 PM UTC-6, Trevor Lange wrote:
> On Tuesday, December 13, 2022 at 10:06:59 AM UTC-8, sep...@yahoo.com wrote:
> > Which is greater t1 or t2 or are they both equal?
> Obviously equal.
> > Which is greater d1 or d2 or are they both equal?
> Obviously equal. 
> 
> Let me guess... now you will say "Okay, so here's my REAL question...." Why don't you just start out saying whatever it is you have to say, or asking whatever it is you really want to ask?

If the two legs are equal that contradicts the concepts of relativity.  I was hoping someone would post the equations for this scenario.  In this scenario, if the accelerometer on board the spaceship reads a constant magnitude during the first leg versus a scenario where the acceleration is constant as measured in F0, what do the passengers on board the spaceship feel as the spaceship accelerates in the two scenarios?  With the accelerometer showing a constant value on board the spacecraft they must feel the same constant force throughout the leg of the journey.  Whereas, if the acceleration is constant as measured in F0, do they feel less and less force as the spaceship goes from 0 to V as measured in F0, and on the return leg, do the on board passengers feel more and more force as the spaceship decelerates from V to 0 if the acceleration rate is constant as measured in F0?
Thanks,
David Seppala
Bastrop TX

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#597714

On Saturday, December 17, 2022 at 7:05:52 AM UTC-8, sep...@yahoo.com wrote:
> > > Which is greater t1 or t2 or are they both equal? 
> > Obviously equal. 
> > > Which is greater d1 or d2 or are they both equal? 
> > Obviously equal. 
> > 
> > Let me guess... now you will say "Okay, so here's my REAL question...." Why don't you just start out saying whatever it is you have to say, or asking whatever it is you really want to ask?

> If the accelerometer on board the spaceship reads a constant magnitude during 
> the first leg... what do the passengers on board the spaceship feel ...? 

If, as you have stipulated, the space ship is undergoing constant proper acceleration "a", then the ship and its passengers will be undergoing a constant proper acceleration "a".  This is what the passengers "feel", assuming by "feel" you mean what proper acceleration they are undergoing... which is (remember) "a".

> as the spaceship accelerates in the two scenarios.

You didn't specify two scenarios, you specified one scenario, with two legs, during one of which the spaceship has proper acceleration "a", and during the other it has proper acceleration "-a".  Now you're trying to introduce a different scenario (as I predicted you would) in which the spaceship undergoes constant coordinate acceleration, which is quite different, although equally trivial.

> With the accelerometer showing a constant value on board the spacecraft 
> they must feel the same constant force throughout the leg of the journey. 

If by "feel a force" you mean "undergo the stipulated proper acceleration", then yes.  And this acceleration would be produced by a constant force applied to the spaceship (and hence to the contents of the spaceship).

Now, as predicted, you finally get around to introducing your second scenario, involving constant coordinate (instead of proper) acceleration:

> If the acceleration is constant as measured in F0, do they feel less and less 
> force as the spaceship goes from 0 to V as measured in F0...

To the contrary, they must be subjected to more and more (not less and less) force to maintain constant coordinate acceleration.  Indeed, the required force goes to infinity as the speed increases.

> On the return leg, do the on board passengers feel more and more force 
> as the spaceship decelerates from V to 0 if the acceleration rate is constant
> as measured in F0?

Just the opposite, the spaceship would be subjected to less and less force to maintain constant coordinate deceleration as its speed is reduced.  This is the time-reversed version of the first leg.

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#597747

On Saturday, December 17, 2022 at 10:18:38 AM UTC-6, Trevor Lange wrote:
> On Saturday, December 17, 2022 at 7:05:52 AM UTC-8, sep...@yahoo.com wrote: 
> > > > Which is greater t1 or t2 or are they both equal? 
> > > Obviously equal. 
> > > > Which is greater d1 or d2 or are they both equal? 
> > > Obviously equal. 
> > > 
> > > Let me guess... now you will say "Okay, so here's my REAL question...." Why don't you just start out saying whatever it is you have to say, or asking whatever it is you really want to ask?
> > If the accelerometer on board the spaceship reads a constant magnitude during 
> > the first leg... what do the passengers on board the spaceship feel ...? 
> 
> If, as you have stipulated, the space ship is undergoing constant proper acceleration "a", then the ship and its passengers will be undergoing a constant proper acceleration "a". This is what the passengers "feel", assuming by "feel" you mean what proper acceleration they are undergoing... which is (remember) "a". 
> 
> > as the spaceship accelerates in the two scenarios. 
> 
> You didn't specify two scenarios, you specified one scenario, with two legs, during one of which the spaceship has proper acceleration "a", and during the other it has proper acceleration "-a". Now you're trying to introduce a different scenario (as I predicted you would) in which the spaceship undergoes constant coordinate acceleration, which is quite different, although equally trivial.
> > With the accelerometer showing a constant value on board the spacecraft 
> > they must feel the same constant force throughout the leg of the journey.
> If by "feel a force" you mean "undergo the stipulated proper acceleration", then yes. And this acceleration would be produced by a constant force applied to the spaceship (and hence to the contents of the spaceship). 
> 
> Now, as predicted, you finally get around to introducing your second scenario, involving constant coordinate (instead of proper) acceleration: 
> 
> > If the acceleration is constant as measured in F0, do they feel less and less 
> > force as the spaceship goes from 0 to V as measured in F0... 
> 
> To the contrary, they must be subjected to more and more (not less and less) force to maintain constant coordinate acceleration. Indeed, the required force goes to infinity as the speed increases. 
> 
> > On the return leg, do the on board passengers feel more and more force
> > as the spaceship decelerates from V to 0 if the acceleration rate is constant 
> > as measured in F0?
> Just the opposite, the spaceship would be subjected to less and less force to maintain constant coordinate deceleration as its speed is reduced. This is the time-reversed version of the first leg.
You said that in a previous reply that given the accelerometer on board the spaceship reads a constant magnitude going from 0 to V and then the same magnitude in the reverse direction of acceleration in going from V to 0 that the length and times of each leg are obviously equal (in terms of F0). Do you see why that contradicts concepts of relativity?
David Seppala
Bastrop TX

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#597750

On Saturday, December 17, 2022 at 2:44:29 PM UTC-8, sep...@yahoo.com wrote:
> Do you see why that contradicts concepts of relativity? 
It doesn't, you are just a hard core imbecile

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#597754

On Saturday, December 17, 2022 at 2:44:29 PM UTC-8, sep...@yahoo.com wrote:
> You said that in a previous reply that, given the accelerometer on board 
> the spaceship reads a constant magnitude going from 0 to V, and then 
>the same magnitude in the reverse direction of acceleration in going from 
> V to 0, that the length and times of each leg are obviously equal (in terms 
> of F0). 

Right.  That is self-evident.  Any other questions?

> Do you see why that contradicts concepts of relativity? 

Trolling is not considered to be respectable behavior.  Look, everything I've said is perfectly consistent with the theory of relativity, and does not entail any contradiction.  If you would like to describe what your brain thinks is contradictory, I'll be happy to explain your conceptual mistake.  But until you work up the courage to actually post what you think is contradictory, there's really not much anyone can do to help you.

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#597749

On Saturday, 17 December 2022 at 17:18:38 UTC+1, Trevor Lange wrote:
> On Saturday, December 17, 2022 at 7:05:52 AM UTC-8, sep...@yahoo.com wrote: 
> > > > Which is greater t1 or t2 or are they both equal? 
> > > Obviously equal. 
> > > > Which is greater d1 or d2 or are they both equal? 
> > > Obviously equal. 
> > > 
> > > Let me guess... now you will say "Okay, so here's my REAL question...." Why don't you just start out saying whatever it is you have to say, or asking whatever it is you really want to ask?
> > If the accelerometer on board the spaceship reads a constant magnitude during 
> > the first leg... what do the passengers on board the spaceship feel ...? 
> 
> If, as you have stipulated, the space ship is undergoing constant proper acceleration "a", then the ship and its passengers will be undergoing a constant proper acceleration "a". 


It's only as "proper" as it is "very efficient", poor halfbrain.
Like any other fanattic idiot of the world, you have
nothing to support your wild claims about a subject you
know nothing about.

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#597480

sepp623@yahoo.com wrote:
> 
> There is an inertial frame F0.  In that frame there is a spaceship at rest.  The spaceship has an accelerometer on board.  At time t0, the spaceship accelerates in the positive x direction at a constant rate as measured by the on board accelerometer. When the spaceship reaches a velocity V with respect to F0, the direction of the acceleration changes to the negative x direction with the accelerometer on board the spaceship showing the same magnitude as it did during the positive leg of this jour
>     During the first leg the time to go from 0 to V is t1 as measured in F0, and the distance traveled during that leg is d1. On the leg with the negative acceleration, the time to go from V to 0 as measured in  F0 is t2 and the distance traveled during that leg is d2.
>     Which is greater t1 or t2 or are they both equal?
>     Which is greater d1 or d2 or are they both equal?
> Thanks,
> David Seppala
> Bastrop TX


The arm bones of right side were significantly longer than those of
left. The leg bones of the left side were longer than those of the
right. Mean right-left differences of female arm bones were usually
greater than those of males.

-- 
The Starmaker -- To question the unquestionable, ask the unaskable,
 to think the unthinkable, mention the unmentionable, say the unsayable,
and challenge
 the unchallengeable.

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#597834

Den 13.12.2022 19:06, skrev sepp623@yahoo.com:
> There is an inertial frame F0.  In that frame there is a spaceship at rest.  The spaceship has an accelerometer on board.  At time t0, the spaceship accelerates in the positive x direction at a constant rate as measured by the on board accelerometer. When the spaceship reaches a velocity V with respect to F0, the direction of the acceleration changes to the negative x direction with the accelerometer on board the spaceship showing the same magnitude as it did during the positive leg of this journey. When the spaceship has zero velocity with respect to F0, the acceleration of the spaceship stops.
>      During the first leg the time to go from 0 to V is t1 as measured in F0, and the distance traveled during that leg is d1. On the leg with the negative acceleration, the time to go from V to 0 as measured in  F0 is t2 and the distance traveled during that leg is d2.
>      Which is greater t1 or t2 or are they both equal?
>      Which is greater d1 or d2 or are they both equal?
Yes and yes.

See:
https://paulba.no/pdf/TwinsByDoppler.pdf
2.4:
a = 1 ly/y

V = 0.986c  t1 = 5.915 y  d1 = 5 ly  t2 = 5.915 y d2 = 5 ly


-- 
Paul

https://paulba.no/

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#597836

On Sunday, 18 December 2022 at 22:30:24 UTC+1, Paul B. Andersen wrote:
> Den 13.12.2022 19:06, skrev sep...@yahoo.com: 
> > There is an inertial frame F0. In that frame there is a spaceship at rest. The spaceship has an accelerometer on board. At time t0, the spaceship accelerates in the positive x direction at a constant rate as measured by the on board accelerometer. When the spaceship reaches a velocity V with respect to F0, the direction of the acceleration changes to the negative x direction with the accelerometer on board the spaceship showing the same magnitude as it did during the positive leg of this journey. When the spaceship has zero velocity with respect to F0, the acceleration of the spaceship stops. 
> > During the first leg the time to go from 0 to V is t1 as measured in F0, and the distance traveled during that leg is d1. On the leg with the negative acceleration, the time to go from V to 0 as measured in F0 is t2 and the distance traveled during that leg is d2. 
> > Which is greater t1 or t2 or are they both equal? 
> > Which is greater d1 or d2 or are they both equal?
> Yes and yes. 
> 
> See: 
> https://paulba.no/pdf/TwinsByDoppler.pdf 

And in the meantime in the real world - forbidden b y
your bunch of idiots GPS and TAI keep measuring t'=t in
forbidden by your bunch of idiots old seconds.

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#597859

Den 18.12.2022 22:30, skrev Paul B. Andersen:
> Den 13.12.2022 19:06, skrev sepp623@yahoo.com:
>> There is an inertial frame F0.  In that frame there is a spaceship at 
>> rest.  The spaceship has an accelerometer on board.  At time t0, the 
>> spaceship accelerates in the positive x direction at a constant rate 
>> as measured by the on board accelerometer. When the spaceship reaches 
>> a velocity V with respect to F0, the direction of the acceleration 
>> changes to the negative x direction with the accelerometer on board 
>> the spaceship showing the same magnitude as it did during the positive 
>> leg of this journey. When the spaceship has zero velocity with respect 
>> to F0, the acceleration of the spaceship stops.
>>      During the first leg the time to go from 0 to V is t1 as measured 
>> in F0, and the distance traveled during that leg is d1. On the leg 
>> with the negative acceleration, the time to go from V to 0 as measured 
>> in  F0 is t2 and the distance traveled during that leg is d2.
>>      Which is greater t1 or t2 or are they both equal?
>>      Which is greater d1 or d2 or are they both equal?
> Yes and yes.
> 
> See:
> https://paulba.no/pdf/TwinsByDoppler.pdf

Oooops!

https://paulba.no/pdf/TwinsByMetric.pdf

> 2.4:
> a = 1 ly/y
> 
> V = 0.986c  t1 = 5.915 y  d1 = 5 ly  t2 = 5.915 y d2 = 5 ly
> 
> 

-- 
Paul

https://paulba.no/

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#597860

Il giorno lunedì 19 dicembre 2022 alle 12:32:40 UTC+1 Paul B. Andersen ha scritto:
> Den 18.12.2022 22:30, skrev Paul B. Andersen: 
> > Den 13.12.2022 19:06, skrev sep...@yahoo.com: 
> >> There is an inertial frame F0.  In that frame there is a spaceship at 
> >> rest.  The spaceship has an accelerometer on board.  At time t0, the 
> >> spaceship accelerates in the positive x direction at a constant rate 
> >> as measured by the on board accelerometer. When the spaceship reaches 
> >> a velocity V with respect to F0, the direction of the acceleration 
> >> changes to the negative x direction with the accelerometer on board 
> >> the spaceship showing the same magnitude as it did during the positive 
> >> leg of this journey. When the spaceship has zero velocity with respect 
> >> to F0, the acceleration of the spaceship stops. 
> >>      During the first leg the time to go from 0 to V is t1 as measured 
> >> in F0, and the distance traveled during that leg is d1. On the leg 
> >> with the negative acceleration, the time to go from V to 0 as measured 
> >> in  F0 is t2 and the distance traveled during that leg is d2. 
> >>      Which is greater t1 or t2 or are they both equal? 
> >>      Which is greater d1 or d2 or are they both equal? 
> > Yes and yes. 
> > 
> > See: 
> > https://paulba.no/pdf/TwinsByDoppler.pdf
> Oooops! 
> 
> https://paulba.no/pdf/TwinsByMetric.pdf
> > 2.4: 
> > a = 1 ly/y 
> > 
> > V = 0.986c  t1 = 5.915 y  d1 = 5 ly  t2 = 5.915 y d2 = 5 ly 
> > 
> > 
> 
> -- 
> Paul 
> 
> https://paulba.no/
beda:
your Doppler formulae sqroot((c+-v)/(c+-v))  contains  the hidden implication A=absolute speed= 0, and that v= absolute speed of the clock B and that B clock slows as 1/sqroot(1-v^2),
during the back and forth trip of B, B will emit: 
tB= tA*1/sqroot(1-v^2)
tB=tA/sqroot(1-v^2).
the relativistic formula is identical to the classic Doppler formula, when A and B  have absolute speed versus an immobile uniform space.
sqroot((c+-v)/(c+-v))#   =    sqroot(1-va^2)/(1+-va)*(1+-vb)/sqroot(1-vb^2)##
#relativistic
##classic + the Lorentz factor; va and vb absolute
in case of absolute speeds vA = 0   and   vB(back and forth)=v,  then Doppler effect is not necessary, in any cases of vB,
is valid:
tB=tA/sqroot(1-v^2).
regards
beda pietanza
watch for mistypes

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#597865

Le 19/12/2022 à 12:32, "Paul B. Andersen" a écrit :

> Oooops!
> 
> https://paulba.no/pdf/TwinsByMetric.pdf

It is obviously very interesting.

Too bad that men take pleasure in their certainties and do not want to go 
further.

They have the intellectual capacity.

What they lack is courage.

R.H. 

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#597868

Dr Richard "Hachel" Lengrand wrote:
> Le 19/12/2022 à 12:32, "Paul B. Andersen" a écrit :
> 
>> Oooops!
>>
>> https://paulba.no/pdf/TwinsByMetric.pdf
> 
> It is obviously very interesting.
> 
> Too bad that men take pleasure in their certainties and do not want to 
> go further.
> 
> They have the intellectual capacity.
> 
> What they lack is courage.

The courage to define terms that are used? The courage to seriously
take time to study Einstein-Poicaré synchronization procedure? The
courage to not forge completely imaginary stories about Albert Einsten's
Nobel Prize? The courage to admit having goofed by stating a uniformly
accelerated rocket is inertial? The courage to seriously consider a
sound objection instead of answering "my bollocks" or "go fuck
yourself".

All the kinds of courage you definitely lack, Richard.

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#597869

Le 19/12/2022 à 18:22, Python a écrit :

>> They have the intellectual capacity.
>> 
>> What they lack is courage.
> 
> The courage to define terms that are used?

> The courage to seriously
> take time to study Einstein-Poicaré synchronization procedure? The
> courage to not forge completely imaginary stories about Albert Einsten's
> Nobel Prize? The courage to admit having goofed by stating a uniformly
> accelerated rocket is inertial? The courage to seriously consider a
> sound objection instead of answering "my bollocks" or "go fuck
> yourself".
> 
> All the kinds of courage you definitely lack, Richard.

I said that what the human spirit lacked the most was not intelligence, 
but courage.

> The courage to seriously
> take time to study Einstein-Poicaré synchronization procedure?

 The courage to admit that the synchronization procedure desired by 
physicists is biased from the outset, because it is based on the notion of 
spatial isochrony and "present time plane".

However, this concept is not physical.

If I place two watches in the same place, and set them together, they will 
mark the same time.

They will also beat simultaneously (their chronotropy being equal).

But if I discard them, they end up in different places (by definition).

They will continue to beat in the same way, between them, and in the same 
way as a central clock left there. They will have the same chronotropy.

BUT each of them will see the other watch lag behind it by a value t=x/c.

The error of the physicist will then consist in saying: "They are both in 
the same plane of present time, it is just that it takes time for the 
light to go from there to there".

The illusion is total.

> The  courage to not forge completely imaginary stories about Albert Einsten's
> Nobel Prize?

 It's an attested fact. 

>The courage to admit having goofed by stating a uniformly
> accelerated rocket is inertial?

An accelerated rocket is immobile in its accelerated reference frame, and 
inertial with all other rockets of the same type.

It is necessary, sometimes, in the face of bullshit, fatuity and human 
arrogance, to know how to send them to graze.

 R.H. 

R.H. 

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#597882

M.D. Richard "Hachel" Lengrand wrote:
> Le 19/12/2022 à 18:22, Python a écrit :
> 
>>> They have the intellectual capacity.
>>>
>>> What they lack is courage.
>>
>> The courage to define terms that are used?
> 
>> The courage to seriously
>> take time to study Einstein-Poicaré synchronization procedure? The
>> courage to not forge completely imaginary stories about Albert Einsten's
>> Nobel Prize? The courage to admit having goofed by stating a uniformly
>> accelerated rocket is inertial? The courage to seriously consider a
>> sound objection instead of answering "my bollocks" or "go fuck
>> yourself".
>>
>> All the kinds of courage you definitely lack, Richard.
> 
> I said that what the human spirit lacked the most was not intelligence, 
> but courage.

You lack both.


>> The courage to seriously
>> take time to study Einstein-Poicaré synchronization procedure?
> 
> The courage to admit that the synchronization procedure desired by 
> physicists is biased from the outset, because it is based on the notion 
> of spatial isochrony and "present time plane".

This is idiotic babbling, you've shown repeatedly you don't
understand the basic of it.

> [snip meaningless babbling]

>> The  courage to not forge completely imaginary stories about Albert 
>> Einsten's
>> Nobel Prize?
> 
> It's an attested fact.

If it was you could provide a source. You cant't.

>> The courage to admit having goofed by stating a uniformly
>> accelerated rocket is inertial?
> 
> An accelerated rocket is immobile in its accelerated reference frame, 
> and inertial with all other rockets of the same type.

Wrong. "inertial" is not a property about "another rocket of the
same type", it is a property shared with *other* inertial frames
in the Universe. Moreover you pretended that a accelerated rocket
and a rocket going at a constant speed were *both* inertial.

You contradict yourself in many ways at the same time, it's
horrendous.


> It is necessary, sometimes, in the face of bullshit, fatuity and human 
> arrogance, to know how to send them to graze.

Bullshit, fatuity and arrogance are on you side, Richard.

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#597884

Le 19/12/2022 à 22:27, Python a écrit :

> Bullshit, fatuity and arrogance are on you side, Richard.

 Sure.

 R.H. 

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